[
  {
    "id": 45001,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\frac{a^4 + 1}{b^3 + b^2 + b} + \\frac{b^4 + 1}{c^3 + c^2 + c} + \\frac{c^4 + 1}{a^3 + a^2 + a} \\ge 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\frac{a^4+1}{b^3+b^2+b} + \\frac{b^4+1}{c^3+c^2+c} + \\frac{c^4+1}{a^3+a^2+a} \\ge 3 \\cdot \\sqrt[3]{\\frac{8}{27}} = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45002,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a scalene triangle $ABC$ let $O$ be the circumcenter, $I$ be the incenter and $H$ be the orthocenter. The second intersection point of the circle which passes through $O$ and is tangent to $IH$ at $I$ and the circle which passes through $H$ and is tangent to $IO$ at $I$ is $M$. Show that $M$ lies on the circumcircle of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First observe that $\\angle MHI = \\angle MIO$ and $\\angle MIH = \\angle MOI$, hence the similarity $MIH \\sim MOI$; thus $MI/MO = IH/IO$. Now let $N$ be the midpoint of the segment $[OH]$ (so $N$ is the center of the 9-point circle), and let $S$ be the reflection of $I$ over $N$. Thus $SOIH$ is a parallelogram; therefore $\\angle IMO = \\angle HIO = \\angle IHS$ and $IM/MO = IH/IO = IH/HS$, which implies the similarity $IMO \\sim IHS$. Consequently\n$$\nMO = \\frac{HS \\cdot IO}{IS} = \\frac{IO^2}{2 \\cdot IN} = \\frac{R(R-2r)}{2(R/2-r)} = R.\n$$\nNote: The equality $IN = R/2 - r$ is Feuerbach's theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Any two pupils in a school are either friends or not, the friendship is mutual. For each integer $1 \\le l \\le 99$ there is a school pupil having exactly $l$ friends in the school. Given that there is no triple of school pupils such that any two of them are friends, find the minimal possible number of pupils in this school.",
    "options": [],
    "answer": "149",
    "solution": "**6.** First $XY \\parallel l_0$ implies that $XYBC$ is concyclic. Now let $S$ be the second intersection point of the circle ($XYZ$) with the circumcircle; then the lines $ZS, XY$ and $BC$ are concurrent since they are the pairwise radical axes of the circles ($XYBC$) and ($XYZS$) and the circumcircle (let $P$ denote the point of concurrence). Now the quadruple ($AP, AZ; AB, AC$) is harmonic, hence ($ZP, ZA; ZB, ZC$) is harmonic, thus ($S, A; B, C$) is harmonic, so $S$ is a fixed point.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two distinct circles $\\omega_1$ and $\\omega_2$ intersect at points $X$ and $Y$. Let the lines $l_1$ and $l_2$ be the common tangent lines of these circles such that $l_1$ is tangent to $\\omega_1$ at $A$ and $\\omega_2$ at $C$ and $l_2$ is tangent to $\\omega_1$ at $B$ and $\\omega_2$ at $D$. Let $Z$ be the reflection of $Y$ with respect to $l_1$. Let $BC$ and $\\omega_1$ meet at $K$ for the second time. Let $AD$ and $\\omega_2$ meet at $L$ for the second time. Prove that the line tangent to $\\omega_1$ and passing through $K$ and the line tangent to $\\omega_2$ and passing through $L$ meet on the line $XZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $\\ell_1 \\cap \\ell_2 = P$. Since $\\angle AXY = \\angle YAC$ and $\\angle CXY = \\angle YCA$ we have $\\angle AXC + \\angle AYC = 180^\\circ$. Therefore, by symmetry, we have $\\angle AZC = \\angle AYC = 180^\\circ - \\angle AXC$ hence $A, X, C, Z$ are concyclic.\n\nNow we will prove that $PX$ is tangent to this circle. Let the second intersection point of the line $PX$ with the circle $\\omega_1$ be $Q$. Since $P$ is the center of the homothety sending $\\omega_1$ to $\\omega_2$, it sends $A$ to $C$ and $Q$ to $P$. Therefore, we have $AQ \\parallel XC$ and $\\angle XCA = \\angle QAP$ and by the tangency we have $\\angle QAP = \\angle QXA$ hence $\\angle XCA = \\angle PXA$ which implies the desired tangency.\n\nSince $P$ lies on the symmetry axis of the two circles, we have $PX = PY = PZ = \\sqrt{PA \\cdot PC}$ and $PZ$ is also tangent to the circle $(AXCZ)$. Letting $XZ \\cap AC = T$, we then have that $(A, C; P, T) = -1$. Then we have $(KA, KC; KP, KT) = -1$. Let the second intersection of $PK$ and $\\omega_1$ be $S$. Then by carrying this harmonic bundle to the circle $\\omega_1$ we see that $(A, B; S, KT \\cap \\omega_1) = -1$. On the other hand, we know that $(A, B; S, K) = -1$ hence $KT$ is the tangent to $\\omega_1$ at $K$.\n\nSimilarly we can prove that $LT$ is the line passing through $L$ and tangent to $\\omega_2$, hence the proof is completed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimal value of\n$$\n\\frac{(a^2 + b^2 + 2c^2 + 3d^2)(2a^2 + 3b^2 + 6c^2 + 6d^2)}{(a+b)^2(c+d)^2}\n$$\nwhere $a, b, c, d$ are positive real numbers.",
    "options": [],
    "answer": "6",
    "solution": "**The answer is 6. The equality holds when**\n$$\n(a, b, c, d) = \\left( \\sqrt[4]{6}, \\sqrt[4]{\\frac{27}{2}}, \\frac{\\sqrt{6}}{2}, 1 \\right).\n$$\n**Solution 1.** By Cauchy-Schwarz inequality we get\n$$\n\\left[ (a^2 + 3d^2) + (b^2 + 2c^2) \\right] \\left[ \\frac{a^2}{a^2 + 3d^2} + \\frac{b^2}{b^2 + 2c^2} \\right] \\geq (a+b)^2\n$$\nand\n$$\n\\left[ (2a^2 + 6d^2) + (3b^2 + 6c^2) \\right] \\left[ \\frac{3d^2}{a^2 + 3d^2} + \\frac{2c^2}{b^2 + 2c^2} \\right] \\geq 6(c+d)^2\n$$\nThen it follows that\n$$\nS := \\frac{(a+b)^2}{a^2 + b^2 + 2c^2 + 3d^2} + \\frac{6(c+d)^2}{2a^2 + 3b^2 + 6c^2 + 6d^2} \\\\ \\le \\frac{a^2}{a^2 + 3d^2} + \\frac{b^2}{b^2 + 2c^2} + \\frac{3d^2}{a^2 + 3d^2} + \\frac{2c^2}{b^2 + 2c^2} = 2\n$$\nFinally by Arithmetic-Geometric Mean inequality, we have\n$$\n\\frac{6(a+b)^2(c+d)^2}{(a^2+b^2+2c^2+3d^2)(2a^2+3b^2+6c^2+6d^2)} \\le \\frac{S^2}{4} \\le 1\n$$\nand hence we are done.\n\n\n**Solution 2.** It follows from Cauchy-Schwarz inequality that\n$$\n(a^2 + b^2 + 2c^2 + 3d^2)(6c^2 + 6d^2 + 3b^2 + 2a^2)\n$$\n$$\n\\geq 6(ad + bc + ac + bd)^2 = 6(a+b)^2(c+d)^2\n$$\nand hence the minimal value of the expression is 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Let $D$, $E$, $F$ be points on the sides $BC$, $AC$, $AB$ respectively, such that\n$$\nDE \\parallel AB,\\ DF \\parallel AC \\text{ and } \\frac{BD}{DC} = \\frac{AB^2}{AC^2}.\n$$\nThe circumcircle of the triangle $AEF$ meets $AD$ at $R$ again and meets the line passing through $A$ and tangent to the circumcircle of the triangle $ABC$ at $S$ again. Let the line $EF$ intersect $BC$ at $L$ and $SR$ at $T$. Prove that $SR$ bisects $AB$ if and only if $BS$ bisects $TL$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFrom the condition $BD/DC = AB^2/AC^2$ we know that $AD$ is the $A$-symmedian of the triangle $ABC$. Let $a, b, c$ be the side lengths of the triangle, then we get $AF/FB = CD/DB = b^2/c^2$ hence $AF = b^2c/(b^2 + c^2)$ and similarly $AE = c^2b/(b^2 + c^2)$. Using these we get $AE \\cdot AC = AF \\cdot AB$ hence $B, E, C, F$ are concyclic. Let the circumcircle of the triangle $ABC$ be $\\Gamma$ and let the circumcircle of the triangle $AEF$ be $\\omega$. Let $AD$ meet $\\Gamma$ for the second time at $Q$ and let $AS \\cap BC = P$. Let the intersection of $\\Gamma$ and $\\omega$ other than $A$ be $M$. The radical axis of the circles $\\Gamma$, $\\omega$ and $(BCEF)$ are concurrent, let this point be $AM \\cap EF \\cap BC = L$. Let the line passing through $A$ and parallel to $BC$ intersect $\\Gamma$ for the second time at $A'$. Since $\\angle C = \\angle AFE = \\angle EAA'$ we find that $AA'$ and $\\omega$ are tangent to each other.\n\nLet us consider the spiral similarity $\\sigma$ centred at $M$ and satisfying $F \\to B, E \\to C$. It is in fact the spiral similarity that sends the circle $\\omega$ to the circle $\\Gamma$, thus $\\sigma$ also satisfies $S \\to A, A \\to A'$ and $R \\to Q$. Since $AQ$ is a symmedian, the tangent lines to $\\Gamma$ passing through $Q$ and $A$ intersect on the line $BC$. Therefore, under $\\sigma$, the tangent lines to $\\omega$ passing through $R$ and $S$ intersect on the line $EF$. In addition, simple angle chasing gives that $\\angle BDF = \\angle C$ and $\\angle AEF = \\angle B, \\angle DEC = \\angle A$ which means $\\angle FED = \\angle C$ hence, circumcircle of the triangle $DEF$ and the line $BC$ are tangent to each other. From here we find $LD^2 = LE \\cdot LF = LB \\cdot LC$, also since we know $(P,D;B,C) = -1$ we obtain that $L$ is the midpoint of the segment $PD$. We found that the lines $AA'$ and $BC$ are parallel, therefore we have $(AP;AD;AL, AA') = -1$ and projecting these lines to $\\omega$ we find that $(S,R;M,A) = -1$, which means the tangent lines to $\\omega$ passing through $S, R$ intersect on the line $AM$. In this case, this intersection point was also on the line $EF$, it must be $L$.\n\nSo, we find that the polar line of $L$ with respect to $\\omega$ is $SR$, and since $AM \\cap EF = L$, from Brokard's Theorem in quadrilateral $AMEF$, the lines $AF$ and $EM$ intersect on the polar line $SR$. Hence $AB$, $EM$, $SR$ are concurrent, and we also have $(T,L;F,E) = -1$. Furthermore, on the circle $\\omega$ we have $(S,Q;A,M) = -1$ and under $\\sigma$ its image is $(A,Q;M,A') = 1$, hence the points $A', M, P$ are collinear. Let $A'M \\cap AQ = V$, then we obtain $(P,V;M,A') = -1$. Finally, by definition we have $AA' \\parallel BC$ and under $\\sigma$ it becomes\n\nNow, we will work on the problem statement to finish the solution. We will prove that both conditions are equivalent to $\\angle B = 2\\angle C$ or $180^\\circ + \\angle B = 2\\angle C$. Solution to both cases are identical, thus we only study the case where $\\angle B > \\angle C$. First, we know that $\\angle MA' = \\angle B - \\angle C$ and under $\\sigma$ we have $\\angle SMA = \\angle B - \\angle C$. Also, we have $\\angle AME = \\angle AFE = \\angle C$. $\\angle B = 2\\angle C$ is equivalent to the following statements:\n$$\n(1)\\ \\angle B = 2\\angle C \\Leftrightarrow \\angle SMA + \\angle AMB = \\angle B - \\angle C + (180^\\circ - \\angle C) = 180^\\circ \\Leftrightarrow B, M, W, S \\text{ are collinear.}\n$$\n$$\n(2)\\ \\angle B = 2\\angle C \\Leftrightarrow \\angle MA' = \\angle B - \\angle C = \\angle C = \\angle AME \\Leftrightarrow P, M, E, A' \\text{ are collinear.}\n$$\nFrom the first condition, $BS$ bisecting $[TL]$ is equivalent to $\\angle B = 2\\angle C$. From the concurrency $AB, SR, EM$ we find that $SR$ bisecting $[AB]$ is equivalent to $EM$ bisecting $[AB]$. $EM$ bisecting $[AB]$ is equivalent to the condition $(EA, EB; EM, ED) = -1$, and reflecting these lines to the line $BC$ shows that it is in fact equivalent to $E, M, P$ being collinear. From the second condition this is equivalent to $\\angle B = 2\\angle C$ hence the required statement is proven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45007,
    "subject": "Mathematics (Multi-modal)",
    "question": "33 balls are placed on unit squares of a $10 \\times 10$ board such that no unit square contains more than one ball. For each empty unit square we calculate the total number of balls located on the same row with this unit square and the total number of balls located on the same column with this unit square and after that the sum of these two numbers we write on this unit square. What is the maximal possible sum of all numbers written on the board?",
    "options": [],
    "answer": "438",
    "solution": "7. The radical axes of the circles (ABC), (BDE), (CDE) must be concurrent at T hence T, D, E are collinear. Moreover, $TD \\cdot TE = TB \\cdot TK$ hence the power of T with respect to the circles (ABC), (ADE) are equal and it lies on their radical axis. Since DE and BC are parallel, the radical axis of (ABC), (ADE) is the common tangent at A hence TA is tangent to (ABC).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the line $l_0$ be tangent to the circumcircle of a scalene triangle $ABC$ at point $A$. A variable line $l$ parallel to $l_0$ intersects the sides $AB$ and $AC$ at points $X$ and $Y$, respectively. Let $T$ be the point of intersection of $BY$ and $CX$. Let $Z$ be the second intersection point of the line $AT$ and the circumcircle of $ABC$. Show that while $l$ changes, the circumcircles of all triangles $XYZ$ pass through some fixed point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $D$, $E$ be points on segments $AB$, $AC$ respectively, such that $DE \\parallel BC$. Let the circumcircle of $ABC$ meet the circumcircles of $BDE$ and $CDE$ again at $K$, $L$ respectively. Let $T$ be the intersection of the lines $BK$ and $CL$. Prove that $TA$ is tangent to the circumcircle of $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The radical axes of the circles $(ABC)$, $(BDE)$, $(CDE)$ must be concurrent at $T$; hence $T$, $D$, $E$ are collinear. Moreover, $TD \\cdot TE = TB \\cdot TK$; hence the power of $T$ with respect to the circles $(ABC)$, $(ADE)$ are equal and it lies on their radical axis. Since $DE$ and $BC$ are parallel, the radical axis of $(ABC)$, $(ADE)$ is the common tangent at $A$; hence $TA$ is tangent to $(ABC)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45010,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that an integer sequence $\\{a_1, a_2, \\dots\\}$ is good if for some function $f: \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ and all positive integers $i, j, n$\n$$\na_i \\equiv a_j \\pmod{n} \\Leftrightarrow i \\equiv j \\pmod{f(n)}\n$$\nFind all good sequences.",
    "options": [],
    "answer": "Exactly the following sequences are good:\n1) Two-periodic sequences: a_i equals c_0 for even indices and c_1 for odd indices (c_0 and c_1 may be equal). A valid choice is f(n) = 1 if n divides c_0 − c_1, and f(n) = 2 otherwise.\n2) Arithmetic progressions: a_i = k i + c for fixed integers k and c. A valid choice is f(n) = n / gcd(n, k).",
    "solution": "Given a good sequence, consider the following for all $n$ for fixed $i$, $j$, $k$:\n$$\nn \\mid a_{i+k} - a_i \\Leftrightarrow f(n) \\mid k \\Leftrightarrow n \\mid a_{j+k} - a_j.\n$$\nThus one has\n$$\n|a_{i+k} - a_i| = |a_{j+k} - a_j|\n$$\nfor all $i$, $j$, $k$. Now consider two cases separately:\n\na. If $a_1 = a_3$, then $a_{i+2} = a_i$ for all $i$, which means\n$$\na_i = \\begin{cases} c_0, & i \\equiv 0 \\pmod 2, \\\\ c_1, & i \\equiv 1 \\pmod 2 \\end{cases}\n$$\nfor some $c_0$, $c_1$ which are not necessarily distinct. Indeed, then the function\n$$\nf(n) = \\begin{cases} 1, & n \\mid c_0 - c_1, \\\\ 2, & n \\nmid c_0 - c_1 \\end{cases}\n$$\nsatisfies the condition.\n\nb. If $a_1 \\neq a_3$, then $a_{i+2} \\neq a_i$ for all $i$, thus $(a_{i+2} - a_{i+1}) \\neq -(a_{i+1} - a_i)$, hence $(a_{i+2} - a_{i+1}) = (a_{i+1} - a_i)$, which means\n$$\na_i = ki + c\n$$\nfor some $k$ and $c$. Indeed, then the function\n$$\nf(n) = \\frac{n}{\\gcd(n, k)}\n$$\nsatisfies the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially the equation\n$$\n\\star \\frac{1}{x-1} \\star \\frac{1}{x-2} \\star \\frac{1}{x-4} \\dots \\star \\frac{1}{x-2^{2023}} = 0\n$$\nis written on the blackboard. Asli and Zehra alternatively make moves, Asli begins. A person making move replaces one of the stars in the equation with either $+$ or $-$. What is the maximal number of real solutions of the obtained equation Asli can guarantee regardless of strategy of Zehra after all the stars have been replaced?",
    "options": [],
    "answer": "1011",
    "solution": "**Answer: 1011.**\n\nLet us describe the strategy of Aslı for guaranteeing at least 1011 solutions of the equation. Aslı makes her first move arbitrarily. After that, at each move she chooses any star with already replaced neighbour and replaces this star with the sign of this neighbour. She completes the game with 1011 such moves after her first one. With this strategy, Aslı guarantees the existence of at least 1011 intervals of the form $(2^k, 2^{k+1})$ such that the signs in front of both $\\frac{1}{x-2^k}$ and $\\frac{1}{x-2^{k+1}}$ are the same. In such an interval, by considering the values of $x$ arbitrarily close to $2^k$ and $2^{k+1}$, the equation on the board takes both arbitrarily large and arbitrarily small values. Therefore, by the Mean Value Theorem, there is a real root of this equation in this interval and hence Aslı guarantees at least 1011 solutions to this equation.\n\nNow we will describe Zehra's strategy to keep the number of real roots less than 1012. Let us enumerate the stars as 0, 1, 2, ..., 2023. Consider the pairs (0, 2023) and (1, 2), (3, 4), (5, 6), ..., (2021, 2022). Zehra's strategy is the following: For the pair (0, 2023), whenever Aslı puts a sign in one of them, Zehra puts the same sign on the other one. For all the other pairs, whenever Aslı puts a sign in one of them, Zehra puts the opposite sign. We will show that this strategy works.\n\nFirstly we consider the infinite intervals. Without loss of generality, assume that signs of the pair (0, 2023) is $+$. Consider the values $x \\in (2^{2023}, \\infty)$. We will start reading all the signs from 2023 to 0. First, we will see a $+$ sign in 2023, then for each consecutive terms we read a pair of opposite signs until the last one with a $+$ sign. At any set in the range from $k$ to 2023 for $0 \\le k \\le 2023$, we will see more positive terms than negative ones. That's because if $k$ is odd, we have $\\frac{k-1}{2}$ pairs of opposite signs with a plus sign at the end, and if $k$ is even we have $\\frac{k-2}{2}$ pairs of opposite signs with a plus sign at the end again, making sure that number of plus signs is always at least half of all signs. So, we can assign to each minus sign a plus one appearing later in the equation. Since the terms appearing later has larger absolute value, in these pairings each pair has a positive sum with two positive numbers remaining in hand, which means the function is always positive and there are no real roots. Same argument applies for the other infinite interval.\n\nNow we consider the intervals between the stars numbered $(k, k+1)$ for $0 \\le k \\le 2022$. There are two possible cases based on their signs:\n\n† If the signs of $k, k+1$ are different, there are no real solutions in the interval $(2^k, 2^{k+1})$. We will show that the equation is either always positive or always negative in this interval. We will start with the following observation.\n\n**Lemma 1:** For all $x \\in (2^k, 2^{k+1})$ we have\n$$\n-\\frac{1}{x-2^{k-2}} - \\frac{1}{x-2^{k-1}} + \\frac{1}{x-2^k} - \\frac{1}{x-2^{k+1}} + \\frac{1}{x-2^{k+2}} + \\frac{1}{x-2^{k+3}} \\ge 0\n$$\n*Proof:* Let $x = 2^{k-2}y$ then we need to show that for each $y \\in (4, 8)$ we have\n$$\n\\frac{4}{(y-4)(8-y)} \\ge \\frac{14}{(y-2)(16-y)} + \\frac{31}{(y-1)(32-y)}\n$$\nWe know that the product of two positive real numbers with a fixed sum is larger when the numbers are closer. Therefore, smallest value of LHS in this interval is 1 when $y = 3$, while the largest value of the RHS in this interval is $\\frac{14}{24} + \\frac{31}{84} < 1$ when $y = 4$, hence we are done.\n\nBack to the case, we work in the interval $x \\in (2^k, 2^{k+1})$ and W.L.O.G. assume that the sign of $k$ is $+$. Then we can swap the $-$ sign appearing in the largest element in the set of stars $\\{0, 1, \\dots, k-3\\}$ with the positive signs in the stars numbered $k-1$ and $k-2$. If there are any $+$ signs among the two. If there are less than two signs on one of the ends, we just add new terms, and because the signs are different this will only decrease the value of the function. We perform the same operation for $k+1$ (carrying the $+$ signs to the closest two places), then we see that the new function on the blackboard takes values less than the original one for each value of $x$ in this interval. Moreover, new function is always positive for the values in this interval, since the 6 consecutive terms appearing in the lemma has positive sum and for all the other terms, just like the argument in the infinite interval claim, all positive terms can be paired up with a negative term having less absolute value (and same for the other part with opposite signs), hence the function is always positive in this interval and have no roots.\n\n**Lemma 2:** For all $x \\in (2^k, 2^{k+1})$ we have\n$$\n\\frac{1}{(x - 2^k)^2} - \\frac{1}{(x - 2^{k-1})^2} - \\frac{1}{(x - 2^{k-2})^2} \\geq 0\n$$\nand\n$$\n\\frac{1}{(x - 2^{k+1})^2} - \\frac{1}{(x - 2^{k+2})^2} - \\frac{1}{(x - 2^{k+3})^2} \\geq 0\n$$\n*Proof:* Again we will use the transformation $x = 2^{k-2}y$. Then, the first inequality becomes\n$$\n\\frac{1}{(y-4)^2} \\geq \\frac{1}{(y-2)^2} + \\frac{1}{(y-1)^2}\n$$\nfor $y \\in (4, 8)$. We can even prove the stronger statement as $\\frac{1}{(y-4)^2} \\geq \\frac{2}{(y-2)^2}$ since the linear inequality $\\sqrt{2}(y-4) \\leq y-2$ is held in the interval $y \\in (4, 8)$.\nThe second inequality becomes\n$$\n\\frac{1}{(y-8)^2} \\geq \\frac{1}{(y-16)^2} + \\frac{1}{(y-32)^2}\n$$\nLHS gets its minimum value at $y = 4$, RHS gets its maximum at $x = 8$ and when we compare these two values we still get that LHS is greater than the RHS.\nNow, when we take the derivative we will see that by the lemma above, the value of the derivative is always positive for these 6 consecutive terms and by the same swapping and pairing arguments used previously, the derivative of the function on the blackboard has a value greater than 0 hence the function is strictly increasing (or decreasing) in this interval, hence it must provide exactly 1 root.\n\nFinally, Zehra's strategy shows that there can be at most 1002 real solutions to the equation at the end. That's because there are no real solutions are coming from the infinite intervals, and 1001 among the 2023 finite intervals have opposite signs at its ends thus they provide no solutions as well. This lefts at most 1002 intervals with same signs at both ends where each such interval can provide at most one real solution, bounding the number of real solutions above by 1002. However, at the end, the numerator is a polynomial of degree 2023 since the number of signs are not equal and we will have either $2x^{2023}$ or $-2x^{2023}$ as the leading coefficient after expanding. Since the degree is odd, the number of real roots must be odd, this forces the number of real solutions to be at most 1011 since 1012 real solutions is not possible and this completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45012,
    "subject": "Mathematics (Multi-modal)",
    "question": "For the points $A$, $B$, $K$, $L$, $X$ located on the circle $\\Gamma$ in the given order the lengths of arcs $\\widehat{BK}$ and $\\widehat{KL}$ are equal. The circle which passes through $A$ and is tangent to $BK$ at $B$ intersects the line segment $KX$ at points $P$ and $Q$. The second intersection point of the circle which passes through $A$ and is tangent to $BL$ at $B$ with the line segment $BX$ is $T$. Show that $\\angle PTB = \\angle XTQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The oriented angle equalities $\\angle QXA = \\angle KBA = \\angle BPA$ and $\\angle XQA = \\angle PBA$ imply the oriented similarity $AXQ \\sim APB$, and the oriented angle equalities $\\angle XTA = \\angle LBA$ and $\\angle TXA = \\angle BLA$ imply the oriented similarity $AXT \\sim ALB$. If we put a C-coordinate system on the plane such that $A$ is at the origin, these similarities would be expressed as $bx = pq = tl$. Then letting $s := tb/q = tp/x$, one gets the oriented similarities $SBP \\sim TQX$ and $STP \\sim BQL$. Consequently one gets $\\angle PST = \\angle LBQ = \\angle LBK + \\angle KBQ = \\angle KXB + \\angle BPQ = \\angle PBT$, hence $S$ is on the circle $(BPT)$, hence $\\angle XTQ = \\angle PSB = \\angle PTB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45013,
    "subject": "Mathematics (Multi-modal)",
    "question": "In scalene triangle $ABC$, the incenter is $I$ and the circumcenter is $O$. $AI$ intersects the circumcircle of $ABC$ a second time at $P$. The line passing through $I$ and perpendicular to $AI$ intersects $BC$ at $X$. The foot of the perpendicular from $X$ to $IO$ is $Y$. Show that the points $A$, $P$, $X$, $Y$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Claim 1.* $A$, $X$, $P$, $E$ are concyclic.\n\n*Proof.* $X$, $I$, $M$, $P$ are concyclic because $\\angle XIP = \\angle XMP = 90^\\circ$. It is well known that $S$ lies on the incircle and since $DI = SI$ and $DM = EM$, we get $IM \\parallel AE$. So, $\\angle EAP = \\angle PIM = \\angle PXE$, which means that $A$, $X$, $P$, $E$ are concyclic.\n\nLet $XP$ intersect the circumcircle a second time at $Z$.\n\n![](attached_image_1.png)\n\nLet $M$ be the midpoint of $BC$, let the incircle touch $BC$ at $D$ and let $E$ be the reflection of $D$ over $M$. Let $DI$ and $AE$ intersect at $S$.\n\n*Claim 2.* $IZ \\perp XP$\n\n**Proof.** $\\triangle XIB \\sim \\triangle XCI$ because $\\angle XIB = 90^\\circ - \\angle BIP = \\frac{1}{2} \\angle C = \\angle XCI$. So, $XI^2 = XB \\cdot XC = XZ \\cdot XP$. In $\\triangle XIP$, by the Euclidean theorem, $IZ \\perp XP$.\n\nLet $J$ be the reflection of $I$ over $O$. Let $T$ be the midpoint of $PZ$.\n\n*Claim 3.* $X$, $Y$, $J$, $P$, $E$ are concyclic.\n\n**Proof.** $OT \\perp PZ$ and by claim 2, $IZ \\perp PZ$. Therefore, by Thales' theorem, $JP \\perp PZ$. So, $\\angle JPX = \\angle JEX = \\angle JYX = 90^\\circ$, therefore $X$, $Y$, $J$, $P$, $E$ are concyclic.\n\nBy claims 1 and 3, it is concluded that $A$, $P$, $X$, $Y$ are concyclic, as desired.\n![](attached_image_2.png)\n\nLet $XY$ and $AP$ intersect at $K$. Let $R$ be the circumradius of $ABC$.\n\n*Claim 1.* $XI^2 = XB \\cdot XC$\n\n*Proof.* $\\triangle XIB \\sim \\triangle XCI$ because $\\angle XIB = 90^\\circ - \\angle BIP = \\frac{1}{2} \\angle C = \\angle XCI$.\n\n*Claim 2.* $KI^2 = KA \\cdot KP$\n\n*Proof.* By Claim 1 and power of a point, $XI^2 = XB \\cdot XC = XO^2 - R^2$. Also, $XO^2 - XI^2 = YO^2 - YI^2 = KO^2 - KI^2$. Hence, $KI^2 = KO^2 - R^2$. By power of point $K$, $KO^2 - R^2 = KA \\cdot KP$.\n\nBy the Euclidean Theorem, $KI^2 = KX \\cdot KY$ which is by Claim 2 also equal to $KA \\cdot KP$. Therefore, $A$, $P$, $X$, $Y$ are concyclic, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, $P$ be the midpoint of the side $BC$ and $K$ be the foot of the altitude from $A$. Let $D$ be a point on the segment $AP$ such that $\\angle BDC = 90^\\circ$. Let the second intersection point of the circumcircle of $ADK$ and line $BC$ be $E$. Let the second intersection point of the circumcircle of $ABC$ and line $AE$ be $F$. Prove that $\\angle AFD = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nSince $A$, $D$, $K$, $E$ are concyclic we have\n$$\n\\angle AKD = \\angle ADE = 90^\\circ\n$$\nand hence $\\triangle ADE$ is a right triangle. Since $D$ lies on the circle centered at $P$ and $ED \\perp PD$, we can see that $ED$ is tangent to the circumcircle of $BDC$. Using the power of the point $E$ with respect to the circles $(BDC)$ and $(ABC)$ we get\n$$\nED^2 = EB \\cdot EC = EF \\cdot EA\n$$\nand from the Euclid relations in the triangle $ADE$ we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x, y$,\n$$f(x+y)^3 = (x+2y)f(x^2) + f(f(y))(x^2+3xy+y^2).$$",
    "options": [],
    "answer": "f(x) = 0 and f(x) = x",
    "solution": "Answer: $f(x) = 0$ and $f(x) = x$.\n\n$f(x + y)^3 = (x + 2y)f(x^2) + f(f(y))(x^2 + 3xy + y^2)$.\n\nBy putting $(x, y) = (0, 0)$ to the original equation we get $f(0) = 0$.\n\nBy putting $(x, y) = (x, 0)$ to the original equation we get $f(x)^3 = x f(x^2)$.\n\nBy putting $(x, y) = (0, x)$ to the original equation we get $f(x)^3 = f(f(x)) x^2$.\n\nReplace this in the original equation to get\n$$\nf(x + y)^3 = (x^2 + 2xy)f(f(x)) + (x^2 + 3xy + y^2)f(f(y)).\n$$\nReversing the $x, y$ order, we get\n$$\nf(x + y)^3 = (x^2 + 3xy + y^2)f(f(x)) + (2xy + y^2)f(f(y)).\n$$\nSince the RHS in both expressions are equal we get\n$$\nf(f(x))(x + y)y = f(f(y))(x + y)x\n$$\nfor all real numbers $x, y$.\n\nSetting $y = 1$ and letting $f(f(1)) = k$, we have $f(f(x)) = kx$ for every $x \\in \\mathbb{R} \\setminus \\{-1\\}$, setting $y = 2$ and $x = -1$ gives that this is true for all real numbers.\n\nPutting this into $f(x)^3 = f(f(x))x^2$, we get $f(x) = cx$ for a constant $c = \\sqrt[3]{k}$, and putting this into $x f(x^2) = x^2 f(f(x))$ we get $c = 0$ or $c = 1$.\n\nTherefore, the solutions are $f(x) = 0$ and $f(x) = x$ and they satisfy the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45016,
    "subject": "Mathematics (Multi-modal)",
    "question": "A real number is written on each square of a $2024 \\times 2024$ board such that sum of all real numbers on the board is equal to $2024$. The board is also entirely covered by $1 \\times 2$ or $2 \\times 1$ dominoes each consisting $2$ unit squares of the board such that no square is covered by two different dominoes. For each domino, Asli erases the two numbers it covers, writes $0$ on one of the squares and writes the sum of the two numbers on the other square. Find the maximal possible number $k$ such that regardless of how the real numbers were written and the dominos were placed initially Asli can guarantee that after her moves there exists a column or row such that the sum of all numbers on it is at least $k$.",
    "options": [],
    "answer": "3/2",
    "solution": "Answer: $\\frac{3}{2}$.\n\nFirst, we will give an example showing that the answer is at most $\\frac{3}{2}$. Suppose that initially the number $\\frac{1}{2024}$ is written on each unit square. Let us divide the whole board to $4$ equal pieces each of sizes $1012 \\times 1012$ and cover the top-left and bottom-right pieces with horizontal dominoes and the remaining ones with vertical dominoes. Then, initially the sum of numbers on each row or column is equal to $1$ and for any given row or column Asli can increase this sum by at most $\\frac{1}{2}$.\n\nNow, let us prove that in any initial case Asli can get a row or column with sum at least $\\frac{3}{2}$. We use the method of double counting. Consider the maximal possible sum of entries for any given row or column. Let $S$ be the sum of all these $4048$ sums. Each number $t$ written on any unit square contributes $3t$ to $S$ since it contributes to either two rows and one column or two columns and one row. Hence, $S = 3 \\cdot 2024 = 6072$. Therefore, by pigeonhole principle, there exists at least one row or column that can achieve the sum $\\frac{6072}{4048} = \\frac{3}{2}$. We are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45017,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $S$ is a set consisting of $12$ positive integers, then what is the maximum number of pairs $(a, b)$ such that $a, b \\in S$ and $\\frac{b}{a}$ is a prime number?",
    "options": [],
    "answer": "20",
    "solution": "Answer: $20$.\n\nExample: $S = \\{1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 42\\}$.\n\nConstruct a graph $G$ whose vertex set is $S$ and edge set consists of noble pairs. We will show that $|E(G)| = e \\le 20$.\n\nWe first prove that $G$ is bipartite. Assume that $G$ has an odd cycle. Start from one of the vertices and move along the cycle in clockwise direction. For any prime number $p$ as we pass through an edge we either multiply the number by $p$, divide the number by $p$ or do an operation with some other prime. As we come back to the starting point, number of multiplications by $p$ must be equal to the number of divisions by $p$. Therefore, contribution of any prime $p$ to the cycle is even, and hence the cycle must be an even cycle, contradiction.\n\nLet parts of $G$ be $A$ and $B$ with $|A| = a$, $|B| = b$, $a \\le b$. Now consider a $K_{2,2}$ in $G$ with vertices $a_1, a_2 \\in A$ and $b_1, b_2 \\in B$. It is easy to see that $\\{a_1, a_2\\} = \\{n, npq\\}$ and $\\{b_1, b_2\\} = \\{np, nq\\}$ (or vice versa) for some positive integer $n$ and distinct primes $p$ and $q$. Also note that we have $a_1 a_2 = b_1 b_2$.\n\nSuppose that a $K_{2,3}$ exists in $G$ with vertices $a_1, a_2 \\in A$ and $b_1, b_2, b_3 \\in B$. Then $a_1 a_2 = b_1 b_2 = b_1 b_3$ which implies that $b_1 = b_3$, contradiction. Thus $G$ has no $K_{2,3}$. Consequently, sum of degrees of any two vertices in $A$ is at most $b+2$, and hence, we get $e \\le a(b+2)/2$. Then, when $a = 1,2,3,4$ we get $e \\le 20$.\n\nCase 1: $a = 5$ and $b = 7$. Let $m$ be the largest degree in $A$. If $m = 7$, then $e \\le 7 + 4 \\cdot 2 = 15$. If $m = 6$, then $e \\le 6 + 4 \\cdot 3 = 18$. If $m = 5$, then the other degrees are at most $4$ and hence $e \\le 5 + 4 \\cdot 4 = 21$ and $e = 21$ holds only if the remaining four vertices all have degree $4$.\n\nLet $B = \\{b_1, \\dots, b_7\\}$ and suppose that $b_1, \\dots, b_5$ are adjacent to a vertex in $A$. Then, as $G$ has no $K_{2,3}$, any vertex in $A$ of degree $4$ must be adjacent to both of $b_6$ and $b_7$. But then there occurs to be a $K_{2,3}$, contradiction. Thus, we have $e \\le 20$. Finally, if $m \\le 4$, then obviously $e \\le 20$.\n\nCase 2: $a = b = 6$. If $m = 6$, then $e \\le 6 + 5 \\cdot 2 = 16$. If $m = 5$, then $e \\le 5 + 5 \\cdot 3 = 20$. If $m \\le 3$, then clearly $e \\le 18$. Finally, let us consider the case when $m = 4$. We will show that $A$ can not have three vertices of degree $4$. Assume $a_1, a_2, a_3 \\in A$ all have degree $4$. Let the neighbors of $a_1$ be $b_1, b_2, b_3, b_4$. Since $G$ has no $K_{2,3}$, $a_2$ is adjacent to $b_5$ and $b_6$. Let the other two neighbors of $a_2$ be $b_3$ and $b_4$. Again as $G$ has no $K_{2,3}$, we easily see that neighbors of $a_3$ are $b_1, b_2, b_5, b_6$. Note that $a_1, a_2, a_3$ do not have a common neighbor and any two of them belong to some $K_{2,2}$. Suppose $a_1 < a_2 < a_3$. The ratio $a_1/a_j$ where $1 \\le j < i \\le 3$ is either product or quotient of two distinct primes. As $a_3/a_1 = (a_3/a_2) \\cdot (a_2/a_1)$, we obtain that $\\{a_1, a_2, a_3\\} = \\{np, nq, nr\\}$ or $\\{n, npq, npr\\}$ for some positive integer $n$ and distinct primes $p, q, r$. But then $a_1, a_2, a_3$ are all adjacent to $n$ or $np$, contradiction.\nLet $f(n)$ be the maximum number of such pairs in a set with $n$ positive integers. We will show that $f(12) \\le 20$. By the example in the first solution, then we can conclude that $f(12) = 20$.\n\nNote that we can assume that the greatest common divisor of all the numbers in the set is $1$. Let $a/b = p$ be a prime and $a$ and $b$ be from the set. Divide the set into two sets, $A$ consisting of the ones not divisible by $p$, and $B$ consisting of the ones divisible by $p$. Let $x/y = q$ be a prime number. Then we can not have $x \\in A$ and $y \\in B$. Also note that $x \\in B$ and $y \\in A$ implies $q = p$ and there exists a unique $x$ for given $y$ and vice versa. Therefore, we obtain the following inequality\n$$\nf(n) \\le \\max_{1 \\le k \\le n-1} \\{f(k) + f(n-k) + \\min(k, n-k)\\}.\n$$\n\nClearly $f(1) = 0$ and $f(2) = 1$. By using that inequality one can easily obtain $f(12) \\le 20$ as desired.\n\n*Remark:* By using the inequality above one can determine the exact value of $f(n)$. Let $g(k)$ be the number of ones in the binary representation of $k$. Then\n$$\nf(n) = \\sum_{k=0}^{n-1} g(k).\n$$\nAnd the example is as follows: Suppose that $n-1$ has $m$ digits in its binary representation. Let $p_1, p_2, \\dots, p_m$ be distinct prime numbers. By adding zeros to the left extend binary representations of $0, 1, \\dots, n-2$ to be with $m$ digits. Then for any $0 \\le k \\le n-1$, if its binary representation is $a_1 a_2 \\dots a_m$ pick the number $\\prod_{i=1}^m p_i^{a'_i}$ for the set. One can easily see that number of pairs whose quotient is prime is exactly $\\sum_{k=0}^{n-1} g(k)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of positive real numbers $x$, $y$, $z$ for which the triples\n$$\n(23x + 24y + 25z, 23y + 24z + 25x, 23z + 24x + 25y)\n$$\nand\n$$\n(x^5 + y^5, y^5 + z^5, z^5 + x^5)\n$$\nare permutations of each other.",
    "options": [],
    "answer": "(sqrt(6), sqrt(6), sqrt(6))",
    "solution": "$(x, y, z) = (\\sqrt{6}, \\sqrt{6}, \\sqrt{6})$.\nSince these two triples are permutations of each other, sum of the elements of the first triple must be equal to sum of the elements of the second triple. Therefore we obtain $2(x^5 + y^5 + z^5) = 72(x + y + z)$ and dividing by 2 we get $x^5 + y^5 + z^5 = 36(x + y + z)$. Combining this equation with the Power Mean Inequality we find\n$$\n12(x + y + z) = \\frac{x^5 + y^5 + z^5}{3} \\geq \\left(\\frac{x + y + z}{3}\\right)^5\n$$\nhence $x + y + z \\leq 3\\sqrt{6}$. Combining this equality with the AM-GM inequality we get\n$$\nxyz \\leq \\left(\\frac{x+y+z}{3}\\right)^3 \\leq 6\\sqrt{6} \\quad (1)\n$$\n\nNow, we recall the equation $x^5 + y^5 + z^5 = 36(x + y + z)$. If we subtract the elements of the triple $(23x + 24y + 25z, 23y + 24z + 25x, 23z + 24x + 25y)$ from $36(x + y + z)$ we get the triple $(13x + 12y + 11z, 13y + 12z + 11x, 13z + 12x + 11y)$. Also, if we subtract the elements of the triple $(x^5+y^5, y^5+z^5, z^5+x^5)$ from $x^5+y^5+z^5$ we get the triple $(x^5, y^5, z^5)$. Since the initial triples are permutations of each other and the numbers we subtracted from are equal, we get that $(x^5, y^5, z^5)$ and $(13x+12y+11z, 13y+12z+11x, 13z+12x+11y)$ are permutations of each other as well. Hence, the product of the elements of the first triple must be equal to the product of the elements of the second triple:\n$$\n(xyz)^5 = (13x + 12y + 11z)(13y + 12z + 11x)(13z + 12x + 11y) \\quad (2)\n$$\n\nFrom the AM-GM inequality we have\n$$\n\\sqrt{13x + 12y + 11z} \\ge 36 \\sqrt[36]{x^{13}y^{12}z^{11}}\n$$\nsimilarly we find\n$$\n13y + 12z + 11x \\ge 36 \\sqrt[36]{y^{13}z^{12}x^{11}}\n$$\nand\n$$\n13z + 12x + 11y \\ge 36 \\sqrt[36]{z^{13}x^{12}y^{11}}\n$$\nMultiplying all these inequalities and using it in (2) we find\n$$\n(xyz)^5 = (13x+12y+11z)(13y+12z+11x)(13z+12x+11y) \\ge 6^6 xyz \\\\\n\\text{hence } xyz \\ge 6\\sqrt{6}. \\text{ Combining this with (1), we get that } xyz = 6\\sqrt{6}. \\\\\n\\text{Therefore, all inequalities above should be equalities and hence} \\\\\nx = y = z. \\text{ We are done.}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 45019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $k$ be the number of positive divisors of $n$. If any two positive divisors of $n$ have different remainders when divided by $k + 1$, then show that either $k + 1$ is a prime number or equal to $4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We consider two cases.\n\nCase 1: $x = 0$ is one of the remainders.\n\nThen $k+1$ divides exactly one of the divisors. Thus $k+1$ divides $n$ and indeed $k+1 = n$. So $k+1$ has $k$ positive divisors. Then $k|k+1$ or $k-1|k+1$. Therefore, $k=1,2$ or $3$, that is $k+1=2,3$ or $4$.\n\nCase 2: $0$ is not one of the remainders.\n\nSuppose that $k+1$ is not prime. Then $k+1 = pm$ where $p$ is a prime number and $m > 1$ is an integer. The ones with the remainders $p, 2p, \\dots, p(m-1)$ are the ones which are divisible by $p$. Let $n = p^a r$ where $\\gcd(p, r) = 1$. Let $s$ be the number of positive divisors of $r$. Then $k = (\\alpha+1)s$ and $m-1 = \\alpha s$ is the number of positive divisors of $n$ which are divisible by $p$. Then we have\n$$\n2 < \\frac{pm-1}{m-1} = \\frac{k}{m-1} = \\frac{(\\alpha+1)s}{\\alpha s} = \\frac{\\alpha+1}{\\alpha} \\le 2,\n$$\nwhich is a contradiction. Thus, in this case $k+1$ has to be prime.\n\n*Remark:* By Dirichlet's theorem and primitive roots, for any prime number $p$ there exists a prime number $q$ such that $n = q^{p-2}$ has $p-1$ positive divisors and they all have distinct remainders when divided by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(a, b)$ such that\n$$\n\\frac{10^{a} - 3^{b} + 1}{2^{a}}\n$$\nis a perfect square.",
    "options": [],
    "answer": "(1, 1), (1, 2)",
    "solution": "Answer: $(a, b) = (1, 1)$ and $(1, 2)$.\nAssume that\n$$\n\\frac{10^{a} - 3^b + 1}{2^a} = m^2\n$$\nholds for some integer $m$. Let us rewrite the equation as\n$$\n3^b = 10^{a} - m^2 2^a + 1.\n$$\nIf $a=1$ then clearly the only solutions are $(a, b) = (1, 1)$ and $(1, 2)$. If $a \\ge 2$ then $a$ is even. Then, since $3$ divides $10^{a} - m^2 2^a + 1$, we get $m^2 2^a \\equiv 2 \\pmod{3}$ which means that $a$ is odd. If $a=3$ then by using modulo $4$ we get that $b$ is even. Since $10^6 + 1 = 3^6 + m^2 2^a = c^2 + 2d^2$, we have no integer solutions because $-2$ is not a quadratic residue modulo $101$ and $101 \\nmid 3^{b/2}$. Therefore, $a \\ge 5$. In that case, by using modulo $16$ we get that $4 \\mid b$ and hence $5 \\mid 3^b - 1$. Then $5 \\mid m$ and hence $25 \\mid 3^b - 1$. Let $b=4k$, then $25 \\mid 81^k - 1$ and from the LTE lemma,\n$$\nv_5(81^k - 1) = v_5(80) + v_5(k) \\ge 2\n$$\nand hence $5 \\mid k$, which implies $5 \\mid b$. Hence $3^b \\equiv 1 \\pmod{11}$ and $11 \\mid 10^{a} - m^2 2^a = x^2 - 2y^2$ but $2$ is not a quadratic residue modulo $11$ so we must have $11 \\mid 10^{a/2}$, which is a contradiction. Thus, there is no solution for $a \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle, $H$ be its orthocenter and $G$ be its centroid. Let $A_b$ and $A_c$ be points on $AB$ and $AC$, respectively, such that $B$, $C$, $A_b$, $A_c$ are cyclic and the points $A_b$, $A_c$, $H$ are collinear. Let $O_a$ be the circumcenter of the triangle $AA_bA_c$. Define $O_b$ and $O_c$ analogously. Prove that the centroid of the triangle $O_aO_bO_c$ lies on the line $HG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet us define $B_a$, $B_c$, $C_a$, $C_b$ similarly. Let the circle passing through $B$, $C$, $A_b$, $A_c$ be $\\omega_a$ and let us define $\\omega_b$, $\\omega_c$ analogously. We will start with a lemma that we will use repeatedly.\n\n**Lemma:** Let $ABC$ be a triangle, $O$ be its circumcenter and $AO \\cap BC = D$. Let $E$, $F$ be points on the lines $AB$, $AC$ respectively such that $DB = DE$ and $DC = DF$. Then we have $BC \\parallel EF$.\n\n*Proof:* Let the midpoints of the segments $BE$ and $CF$ be $K$ and $L$, respectively. Since $\\angle AKL = \\angle ADL = \\angle B$.\n![](attached_image_2.png)\nLet $BA_c \\cap CA_b = X$.\n\n*Claim 1:* $X$ lies on the radical axis of the circles $\\omega_b$ and $\\omega_c$.\n\n*Proof:* Let $\\omega_c \\cap CA_b = C_1$ and $\\omega_b \\cap BA_c = B_1$. We will prove that $B$, $C$, $B_1$, $C_1$ are concyclic which implies the claim. Since $\\angle C_a C_1 C = \\angle C_a C_b C = \\angle A$ we have that $A$, $B$, $C_a$, $C_1$ are concyclic. Using the lemma, we have $A_b C_b \\parallel B_a B_c$ hence $A_b$, $C_b$, $A$, $C$ are concyclic and $B$ lies on the radical axis of the circles $\\omega_c$ and $(AA_b C_1 C_a)$ therefore $B$, $C_1$, $C_a$ are collinear. Hence we have $\\angle BC_1 C = 180^\\circ - \\angle A$ and $\\angle BB_1 C = 180^\\circ - \\angle A$ can be obtained similarly. $\\square$\n\nLet $M_a$ be the midpoint of the segment $O_b O_c$ and let $O$ be the circumcenter of the triangle $ABC$.\n\n*Claim 2:* $HM_a \\parallel OO_a$.\n\n*Proof:* $A$ lies on the radical axis of the circles $\\omega_b$, $\\omega_c$ and from Claim 1 we have $AX \\perp O_b O_c$. Let $BE$, $CF$ be the altitudes of the triangle and $A_b A_c \\cap BC = T$. Let the second intersection of the line $HM_a$ with the circle $(AFHE)$ be $R$ and the foot of the perpendicular from $H$ to the line $AX$ be $S$. Since the points $O_a$, $O_b$, $O_c$ lie on the altitudes of the triangle $ABC$, we have $(O_b, O_c; M_a, \\infty) = -1$ and projecting this to the circle $(AFHE)$ from the point $H$ we obtain $(E, F; R, S) = -1$. Then projecting this from $A$ to the line $BC$ we have $(C, B; AR \\cap BC, AX \\cap BC) = -1$ hence we must have $AR \\cap BC = T$. Since $R$ lies on the circle with diameter $AH$ we obtain $HM_a \\perp AT$. Finally, since the line $AT$ is the radical axis of the circles $\\omega_a$ and $(ABC)$ we have $OO_a \\perp AT$ which implies\n\n$$HM_a \\parallel OO_a.$$ $\\Box$\n\nNow, let $G_1$ be the centroid of the triangle $O_aO_bO_c$, we will prove that $G_1$ is the midpoint of $HG$. Triangles $HM_aM_c$ and $OO_aO_c$ have sides that are parallel to each other, hence they are similar and the similarity ratio is equal to $\\frac{O_bO_c}{M_bM_c} = 2$. Therefore the intersection of the lines $O_aM_a$ and $HG$ divides both segments with the ratio $2 : 1$ and this point is in fact $G_1$ since $O_aM_a$ is a median, and it is also the midpoint of $HG$ hence we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer and $a_1, a_2, \\dots, a_n$ be real numbers. For each $1 \\le k \\le n$ the real numbers $b_1, b_2, \\dots, b_{n+1}$ are defined by\n\n$$\nb_k = \\frac{a_k + \\max\\{a_{k+1}, a_{k+2}\\}}{2}\n$$\nand $b_{n+1} = b_1$ ($a_{n+1} = a_1$ and $a_{n+2} = a_2$). Find the smallest $\\lambda$ such that the inequality\n$$\n\\lambda \\left[ \\sum_{i=1}^{n} (a_i - a_{i+1})^{2024} \\right] \\geq \\sum_{i=1}^{n} (b_i - b_{i+1})^{2024}\n$$\nis held for each $n \\ge 3$ and all real numbers $a_1, a_2, \\dots, a_n$.",
    "options": [],
    "answer": "2^{2024}",
    "solution": "Answer: $\\lambda = 2^{2024}$.\nLet $x_i = a_i - a_{i+1}$. Since for all real numbers\n$$\n\\max\\{x, y\\} = \\frac{x+y}{2} + \\left|\\frac{x-y}{2}\\right|\n$$\nwe have\n$$\nb_i - b_{i+1} = x_i + \\frac{x_{i+1}}{2} + \\frac{x_{i+2}}{2} + \\left| \\frac{x_{i+1}}{2} - \\frac{x_{i+2}}{2} \\right| \\quad (1)\n$$\n\nLet $y_i = \\frac{x_i}{2} + \\left|\\frac{x_i}{2}\\right|$ and $z_i = \\frac{x_i}{2} - \\left|\\frac{x_i}{2}\\right|$. Then, $y_i = x_i$, $z_i = 0$ for $x_i \\ge 0$ and $y_i = 0$, $z_i = x_i$ for $x_i < 0$. Inserting it to (1) we get\n$$\n(b_i - b_{i+1})^{2024} = (x_i + y_{i+1} + z_{i+2})^{2024} \\le 2^{2023} (x_i^{2024} + (y_{i+1} + z_{i+2})^{2024}) \\quad (2)\n$$\nby the Power Mean Inequality. Because $y_i$ is always a non-negative real number and $z_i$ is always a non-positive real number, we have\n$$\n|y_{i+1} + z_{i+2}| \\le |y_{i+1}| + |z_{i+2}|\n$$\nwhich in turn implies that\n$$\n(y_{i+1} + z_{i+2})^{2024} \\le y_{i+1}^{2024} + z_{i+2}^{2024}\n$$\n\nSince $y_i^{2024} + z_i^{2024} = x_i^{2024}$ for all indices $i$, by summing the inequalities (2) for all indices $i$ we get the desired inequality.\nTo prove that $\\lambda = 2^{2024}$ is the best possible value, consider $n = 2m$ and choose the sequence as $\\{a_i\\} = \\{1, 2, \\dots, n-1, n, n-1, \\dots, 1\\}$. Then, $\\lambda$ has to satisfy the condition\n$$\n\\lambda \\cdot n \\geq 2^{2024} (n-4) + 1 + 1 + 0 + 0\n$$\nfor every $n$. Taking $n$ to infinity, we see that $\\lambda$ should be at least $2^{2024}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n=0}^{\\infty}$ and $(b_n)_{n=0}^{\\infty}$ be sequences of real numbers such that $a_0 = 40$, $b_0 = 41$ and for all $n \\ge 0$ we have\n$$\na_{n+1} = a_n + \\frac{1}{b_n} \\quad \\text{and} \\quad b_{n+1} = b_n + \\frac{1}{a_n}\n$$\nFind the smallest positive integer $k$ such that $a_k > 80$.",
    "options": [],
    "answer": "2460",
    "solution": "Answer: 2460.\nSince\n$$\n\\frac{a_{n+1}}{b_{n+1}} = \\frac{a_n + \\frac{1}{b_n}}{b_n + \\frac{1}{a_n}} = \\frac{a_n}{b_n}\n$$\nwe get that $\\frac{a_n}{b_n}$ is a constant and hence equals $\\frac{40}{41}$. Therefore, $a_k > 80$ is equivalent to $b_k > 82$ and $a_k b_k > 80 \\cdot 82 = 6560$. Multiplying both sequences we get\n$$\na_{n+1} b_{n+1} = a_n b_n + 2 + \\frac{1}{a_n b_n}\n$$\nHence, for any $k$ we have $a_k b_k > 40 \\cdot 41 + 2k$ which means $a_k b_k > 6560$ for $k = 2460$. Moreover, if $k = 2459$, we have\n$$\na_k b_k = 6560 - 2 + \\sum_{i=0}^{k-1} \\frac{1}{a_i b_i} < 6560 - 2 + \\frac{2459}{1640} < 6560.\n$$\nThus, the minimal value of $k$ satisfying conditions is 2460.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45024,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $r \\ge 2$ each positive integer is coloured into one of $r$ colours. Suppose that for each integer number $n$ and each pair $(a, b)$ of two colours $a$ and $b$ the difference between the number of divisors of $n$ coloured $a$ and the number of divisors of $n$ coloured $b$ is at most $1$. Find all possible values of $r$.",
    "options": [],
    "answer": "2, 3",
    "solution": "Answer: $r = 2, 3$.\n\nSuppose that $r \\geq 4$. Since there are infinitely many primes, there are same coloured primes $p$ and $q$. Since all divisors of $n = pq$ are $1, p, q, pq$, and hence no divisor of $n = pq$ is coloured into some of $r \\geq 4$ colours. Thus, the problem conditions are not satisfied.\n\nLet $r = 3$. Each positive integer $n$ with prime factorization $n = p_1^{a_1} \\cdot p_2^{a_2} \\cdots p_m^{a_m}$ we colour to $(a_1 + a_2 + \\cdots + a_m) \\pmod{3}$. The number $n = 1$ with $t = 0$ we colour to $0$. By induction over the number of different primes in the prime factorization of $n$ we will show that this colouring satisfies the problem conditions. Let $d_n(0), d_n(1), d_n(2)$ be the number of divisors of $n = p_1^{a_1} \\cdot p_2^{a_2} \\cdots p_m^{a_m}$ coloured to $0, 1, 2$, respectively.\n\nIf $t = 0$ then the corresponding $d(0) = 1, d(1) = 0, d(2) = 0$ and conditions are held. Suppose that problem conditions are held for all integer numbers with $t = m$ and consider $a = p_1^{a_1} \\cdot p_2^{a_2} \\cdots p_m^{a_m} \\cdot p_{m+1}^{a_{m+1}}$, with $d_a(0), d_a(1), d_a(2)$. Let $d_b(0), d_b(1), d_b(2)$ be the number of divisors of $b = p_1^{a_1} \\cdot p_2^{a_2} \\cdots p_m^{a_m}$ of colours $0, 1, 2$, respectively. By induction hypothesis\n$$\n|d_b(i) - d_b(j)| \\leq 1 \\text{ for all pairs } (i, j), \\text{ where } i, j = 0, 1, 2. \\quad (\\dagger)\n$$\n\nAny divisor of $a$ has a form $d \\cdot p_{m+1}^u$, where $d$ is a divisor of $b$ and $0 \\le u \\le a_{m+1}$. Then\n$$\nd_{p_{m+1}^{3l+0}}(0) = d_b(0), \\quad d_{p_{m+1}^{3l+0}}(1) = d_b(1), \\quad d_{p_{m+1}^{3l+0}}(2) = d_b(2)\n$$\n$$\nd_{p_{m+1}^{3l+1}}(0) = d_b(2), \\quad d_{p_{m+1}^{3l+1}}(1) = d_b(0), \\quad d_{p_{m+1}^{3l+1}}(2) = d_b(1) \\quad (\\ddagger)\n$$\n$$\nd_{p_{m+1}^{3l+2}}(0) = d_b(1), \\quad d_{p_{m+1}^{3l+2}}(1) = d_b(2), \\quad d_{p_{m+1}^{3l+2}}(2) = d_b(0)\n$$\nLet $d(0) + d(1) + d(2) = D$.\n\nCase 1: $a_{m+1} = 3l$. Then by (\\ddagger)\n$$\nd_a(0) = d_b(0) + lD, \\quad d_a(1) = d_b(1) + lD, \\quad d_a(2) = d_b(2) + lD\n$$\nand hence by (\\dagger) the conditions are held.\n\nCase 2: $a_{m+1} = 3l+1$. Then by (\\ddagger)\n$$\nd_a(0) = d_b(0) + lD + d_b(2), \\quad d_a(1) = d_b(1) + lD + d_b(0), \\quad d_a(2) = d_b(2) + lD + d_b(1)\n$$\nand hence by (\\dagger) the conditions are held.\n\nCase 3: $a_{m+1} = 3l+2$. Then by (\\ddagger)\n$$\nd_a(0) = d_b(0) + lD + d_b(2) + d_b(1), \\quad d_a(1) = d_b(1) + lD + d_b(0) + d_b(2), \\quad d_a(2) = d_b(2) + lD + d_b(1) + d_b(0)\n$$\nand hence the conditions are held.\n\nLet $r = 2$. Each positive integer $n$ with prime factorization $n = p_1^{\\alpha_1} \\cdot p_2^{\\alpha_2} \\cdots p_m^{\\alpha_m}$ we colour to the colour $(\\alpha_1 + \\alpha_2 + \\cdots + \\alpha_m) \\pmod{2}$. As in the case $r = 3$, it can be readily shown that this colouring also satisfies the problem conditions.\n\nTherefore, the only possible values of $r$ are $2$ and $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and let the midpoints of $AB$, $BC$, $CD$ and $DA$ be $K$, $L$, $M$ and $N$, respectively. Let the reflections of the point $M$ with respect to the lines $AD$ and $BC$ be $P$ and $Q$, respectively. Finally the circumcenter of the triangle $KPQ$ be $R$. Prove that $RN = RL$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start by noting that $KLMN$ is a parallelogram, and by the symmetry we have $NP = NM = KL$ and $LQ = LM = KN$. Moreover, we have $\\angle KNM = \\angle KLM$.\n\nSince $ABCD$ is cyclic, $MN \\parallel AC$ and $LM \\parallel BD$ we get that $\\angle DNM = \\angle DAC = \\angle DBC = \\angle MLC$ and from the symmetry we get $\\angle MNP = 2\\angle MND = 2\\angle MLC = \\angle MLQ$ and finally we obtain $\\angle KNP = \\angle KLQ$.\n\nCombining this with the side equalities above we find $\\angle KNP \\cong \\angle QLK$ and hence $KP = KQ$. Since $R$ is the circumcenter of the triangle $KPQ$ we also have $\\angle RPK \\cong \\angle RKQ$ and hence $\\angle RKN \\cong \\angle RQL$ which implies $RN = RL$. We are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45026,
    "subject": "Mathematics (Multi-modal)",
    "question": "For an integer $n$, let $\\sigma(n)$ be the sum of all positive divisors of $n$. A sequence of positive integers $(a_i)_{i=0}^{\\infty}$ is defined as follows: $a_0 = 1$ and for each $n > 1$, $a_n$ is the smallest integer greater than 1 such that\n$$\n\\sigma(a_0 a_1 \\cdots a_{n-1}) \\mid \\sigma(a_0 a_1 \\cdots a_n).\n$$\nDetermine the number of terms of this sequence dividing the number $2024^{2024}$.",
    "options": [],
    "answer": "36",
    "solution": "Answer: 36.\n\nWe will show that this sequence is increasing and consists of all the numbers of the form $p^{2^n}$, where $p$ is a prime and $n$ is a non-negative integer. Thus, all the terms satisfying the condition are $1, 2, 11, 23, \\dots, 2^{1024}, 11^{1024}, 23^{1024}, 2^{2048}, 2^{4096}$.\n\nLet $S$ be the set of positive integers of the described form $p^{2^n}$. We will use induction to prove that $a_n$ is the $n^{\\text{th}}$ smallest number of $S$. It is easy to verify that for $n=1, 2$ the statement is correct. Assume that it is true for $a_1, a_2, \\dots, a_{n-1}$. Note that the $n^{\\text{th}}$ smallest term in the set $S$ already satisfies the problem condition. We will consider two cases.\n\nFor the first case, assume that $a_n$ has only one prime divisor, say $p$. If there are no other terms among the first $n-1$ that is not divisible by $p$, then we must have $a_n = p$ by the minimality assumption. Otherwise, assume it is equal to $p^y$ for some $y$. By the induction hypothesis, all the previous terms that are divisible by $p$ are $p, p^2, \\dots, p^{2^{x-1}}$ for some $x$. Then, since the $\\sigma$ function is multiplicative for relatively prime numbers we should only consider the powers of $p$ for the divisibility condition. Then, we must have $p^{2^x-1} \\mid p^{2^x+y}-1$ which means $2^x\\mid 2^x+y$ hence $y \\ge 2^x$. Which means, we must have $a_n = p^{2^x}$ since it is the smallest possible choice for $a_n$ and it satisfies the problem conditions. Note that in this case $a_n$ is in the set $S$, it is a number that has not appeared before, and it is the minimal possible number in $S$ satisfying these conditions as claimed.\n\nFor the second case, assume that $a_n$ has at least two distinct prime divisors. Let $\\prod_{i=0}^{n-1} a_i = c_n$. By the minimality condition, we can assume that $a_n$ has no prime divisors that is relatively prime to $c_{n-1}$, because in that case that prime divisor itself would satisfy the condition and it is smaller than $a_n$. Assume that the prime divisors of $a_n$ represented in the product $c_{n-1}$ be $p_i^{x_i}$ for $1 \\le i \\le k$ and represented in the product $c_n$ be $p_i^{y_i}$ for $1 \\le i \\le k$. By the induction hypothesis, we have $x_i = 2^{z_i} - 1$. Therefore, using the divisibility condition we have\n$$\n\\prod_{i=1}^{k} (p_i^{2z_i} - 1) \\mid \\prod_{i=1}^{k} (p_i^{y_i} - 1).\n$$\nSince there are at least two primes, at least one of them are odd and both sides are even. Comparing the $v_2$ valuation of both sides and using the LTE Lemma, we have\n$$\nv_2(p_i^{2z_i} - 1) = v_2(p - 1) + v_2(p + 1) + z_2 - 1\n$$\nand\n$$\nv_2(p_i^{y_i} - 1) \\le v_2(p - 1) + v_2(p + 1) + v_2(y_i) - 1\n$$\ntherefore at least one index $i$ must satisfy $v_2(y_i) \\ge z_i$ and since we have non-negative powers this implies $y_i > 2^{z_i}$ and $y_i \\ge 2^{z_i+1}$. Then, choosing $a_n = p_i^{2^{z_i}}$ also satisfies the condition and it contradicts the minimality condition. Therefore, $a_n$ can not have two distinct prime divisors. We are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45027,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 207 boxes on the table numbered $1, 2, \\ldots, 207$. Asli puts one red ball to each of some 100 boxes and one white ball to each of the remaining 107 boxes. After that Zehra writes $N$ pairs $(i, j)$, $1 \\le i < j \\le 207$ to the board. Finally for each pair $(i, j)$ on the board Asli informs Zehra whether balls in the boxes $i$ and $j$ are same-coloured. Find the minimal value of $N$ for which Zehra can guarantee to determine the colours of balls in all boxes.",
    "options": [],
    "answer": "205",
    "solution": "**Answer: 205.**\n\nLet us show that Zehra can guarantee to determine the colours of all balls for $N = 205$. Zehra writes the pairs $(1,2), (1,3), \\dots, (1,206)$ on the board. Let us assign a number $i$ to the ball contained in the box number $i$. Suppose that out of balls numbered $2, 3, \\dots, 206$ the number of balls same-coloured with $i$ is $a$ and the number of differently coloured balls is $b = 205 - a$. Then the colours of all balls can be easily determined: One of these numbers $a$ and $b$ is greater or equal to 103. If $a \\ge 103$ then ball 1 is white. $a+1$ is either 106 or 107. In the first case ball 207 is white, in the second case ball 207 is red. If $b \\ge 103$ then ball 1 is red. $b$ is either 106 or 107. In the first case ball 207 is white, in the second case ball 207 is red.\n\nNow we show that if $N \\le 204$ then Zehra cannot guarantee to determine colours of all balls. Suppose that $N$ pairs $(i, j)$ are already chosen by Zehra. Let us define a graph $G$ consisting of 207 vertices representing balls where vertices $i$ and $j$ are connected by an edge if and only if there is a pair $(i, j)$ on the board. Since the total number of edges is $N \\le 204$ the graph $G$ has at least 3 connected components. Asli divides these connected components into 3 non-empty groups $C_1, C_2$ and $C_3$ with number of vertices $|C_1|, |C_2|$ and $|C_3|$. Assuming $|C_1| \\ge |C_2| \\ge |C_3|$ we get that $|C_2| + |C_3| \\le 138$. Obviously at least one of groups $C_2, C_3$ and $C_2 \\cup C_3$ contains even number of vertices. Asli can divide vertices of this group into two parts $A_1$ and $A_2$, where $|A_1| = |A_2| < 100$. Since Zehra is going to guarantee to determine ball colours we can suppose that Asli knows all pairs on the board before her ball distribution. Thus, Asli can make all balls in $A_1$ red coloured and all balls in $A_2$ white coloured or alternatively all balls in $A_1$ white coloured and all balls in $A_2$ red coloured and the remaining balls will have arbitrary colours just to provide 100 red and 107 white balls in total. In both cases the answers of Asli will be the same. Therefore, Zehra cannot determine the ball colours of this group containing even number of boxes. Done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle, $I$ be its incenter and $O$ be its circumcenter. The line $IO$ intersects the lines $BC$, $CA$, $AB$ at points $D$, $E$, $F$, respectively. Let $A_1$ be the intersection of $BE$ and $CF$. The points $B_1$ and $C_1$ are defined similarly. The incircle of $ABC$ is tangent to sides $BC$, $CA$, $AB$ at points $X$, $Y$, $Z$, respectively. Let the lines $XA_1$, $YB_1$ and $ZC_1$ intersect $IO$ at points $A_2$, $B_2$, $C_2$ respectively. Prove that the circles with diameters $AA_2$, $BB_2$ and $CC_2$ have a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the Miquel point of the quadrilateral defined by the lines $AB$, $AC$, $BC$, $IO$. We will prove that $M$ lies on all three circles. Since the statement is symmetric, we will only show that $M$ lies on the circle with diameter $AA_2$.\n\nDefine $S = A_1 \\cap BC$, or equivalently, as the point satisfying $(B, C; D, S) = -1$. Let $A_3$ be the foot of the perpendicular from $S$ to $IO$. We will eventually show that $A_2 = A_3$.\n\n![](attached_image_1.png)\n\n**Claim 1:** $\\angle BA_3S = \\angle SA_3C = \\angle BAC$.\n**Proof:** Since $(B, C, D, S) = -1$, $SA_3$ is the polar of $D$ with respect to the circle $(ABC)$. Therefore we have $OA_3 \\cdot OD = R^2$, where $R$ is the circumradius of $ABC$. Then we obtain $DA_3 \\cdot DO = DO^2 - OA_3 \\cdot DO = DO^2 - R^2 = DB \\cdot DC$ hence $B$, $C$, $O$, $A_3$ are concyclic and $\\angle BA_3C = 2 \\angle BAC$. The conclusion then follows as we have $DA_3 \\perp A_3S$ and $(B, C, D, S) = -1$. $\\square$\n\n**Claim 2:** $\\angle AMA_3 = 90^\\circ$.\n**Proof:** Let $Q$, $R$ be the projections from $A_3$ to $AB$, $AC$ respectively. Then, we have\n$$\n\\frac{BF}{BQ} = \\frac{BF}{BA_3} \\cdot \\frac{BA_3}{BQ} = \\frac{\\cos BAC}{\\sin AFE} \\cdot \\frac{1}{\\sin AEF} = \\frac{CE}{CA_3} \\cdot \\frac{CR}{CR} = \\frac{CE}{CB}\n$$\n\nLet $A'$ be the reflection of $A$ with respect to the line $IO$.\n\n**Claim 3:** $A_3$, $A_1$, $A'$ are collinear.\n**Proof:** Since $AA'$, $A_3S$ are both perpendicular to $IO$, it suffices to show that $\\frac{AA'}{A_3S} = \\frac{AA_1}{A_1S}$. Let $AS \\cap IO = K$. We have $\\frac{AA'}{A_3S} = 2\\frac{AA'/2}{A_3S} = 2\\frac{AK}{KS}$. Hence we need to prove that $2 \\cdot AK \\cdot A_1S = KS \\cdot AA_1$, which is well known since these points satisfy $(A, A_1; K, S) = -1$. $\\square$\n\nLet $T$ be the circumcircle of $ABC$ and let $T$ be the second intersection of $AD$ and $R$.\n\n**Claim 4:** $T$ lies on the line $\\overline{A'A_1A_3}$.\n**Proof:** From Claim 1, we know that $Pow(D, T) = DT \\cdot DA = DA_3 \\cdot DO$ hence $T$, $A$, $A_3$, $O$ are concyclic. Then we find $\\angle A'TA_3 = 180^\\circ - \\angle AOA'/2 = \\angle A'TA'$ hence $T \\in A_3A'$. $\\square$\n\n**Claim 5:** $T$, $A'$, $X$ are collinear.\n**Proof:** Take the reflection of the triangle $ABC$ with respect to line $IO$. Then, $A'B'C$ is also a triangle with the same incircle and circumcircle as $ABC$. Let $T_0$ be the second intersection of $A'X$ and $R$. Then, from the Dual of Desargues' Involution Theorem in the degenerate complete quadrilateral $ABXC$, there is an involution swapping pairs $(A'B, A'C)$, $(A'B', A'C')$ and $(A'A, A'X)$, and involutions on circles are inversions hence $BC$, $B'C'$, $AT_0$ must concur and $T = T_0$. $\\square$\n\nTherefore, $A_3 = A_2$ and $M$ lies on the circle with diameter $AA_2$ and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve (in real numbers) the equation\n$$\n\\left|x - \\frac{\\pi}{6}\\right| + \\left|x + \\frac{\\pi}{3}\\right| = \\arcsin \\frac{x^3 - x + 2}{2}.\n$$",
    "options": [],
    "answer": "x = -1, 0",
    "solution": "Використовуючи властивості модуля числа та властивості функції $\\arcsin$, маємо:\n$$\n\\frac{\\pi}{2} \\leq \\left|x - \\frac{\\pi}{6}\\right| + \\left|x + \\frac{\\pi}{3}\\right| = \\arcsin \\frac{x^3 - x + 2}{2} \\leq \\frac{\\pi}{2}.\n$$\nЗвідси випливає, що\n$$\n\\arcsin \\frac{x^3 - x + 2}{2} = \\frac{\\pi}{2}, \\text{ тобто, } \\frac{x^3 - x + 2}{2} = 1.\n$$\nСеред трьох коренів останнього рівняння тільки два зазначені вище задовольняють умову.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real numbers $x$ and $y$ such that\n$$\n\\frac{x-2}{y} + \\frac{5}{xy} = \\frac{4-y}{x} - \\frac{|y-2x|}{xy}.\n$$",
    "options": [],
    "answer": "(1, 2)",
    "solution": "Зауважимо, що при **допустимих** значеннях $x$ і $y$ вихідна рівність рівносильна рівності\n$$\n(x-1)^2 + (y-2)^2 + |y-2x| = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha, \\beta, \\gamma \\in [0; \\frac{\\pi}{2}]$ satisfy the conditions\n$$\n\\sin \\alpha + \\sin \\beta + \\sin \\gamma = 1, \\sin \\alpha \\cos 2\\alpha + \\sin \\beta \\cos 2\\beta + \\sin \\gamma \\cos 2\\gamma = -1.\n$$\nFind all the possible values of the expression $\\sin^2 \\alpha + \\sin^2 \\beta + \\sin^2 \\gamma$.",
    "options": [],
    "answer": "1",
    "solution": "З умови задачі випливають рівності\n$$\n\\sin \\alpha - 2\\sin^3 \\alpha + \\sin \\beta - 2\\sin^3 \\beta + \\sin \\gamma - 2\\sin^3 \\gamma = -1, \\\\ \\sin^3 \\alpha + \\sin^3 \\beta + \\sin^3 \\gamma = 1.\n$$\nОскільки для $\\varphi \\in [0; \\frac{\\pi}{2}]$ справджуються нерівності $\\sin^3 \\varphi \\le \\sin^2 \\varphi \\le \\sin \\varphi$, то\n$$\n1 = \\sin^3 \\alpha + \\sin^3 \\beta + \\sin^3 \\gamma \\le \\sin^2 \\alpha + \\sin^2 \\beta + \\sin^2 \\gamma \\le \\sin \\alpha + \\sin \\beta + \\sin \\gamma = 1.\n$$\nЗалишається зауважити, що, наприклад, при $\\alpha = \\beta = 0$, $\\gamma = \\frac{\\pi}{2}$ маємо рівність\n$$\n\\sin^2 \\alpha + \\sin^2 \\beta + \\sin^2 \\gamma = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45032,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the Cartesian plane $xOy$, draw the locus of all points $M(x;y)$ such that\n$$\n(x^2 - 1)(|y| - 1) \\geq 0.\n$$",
    "options": [],
    "answer": "Locus = {(x, y): (|x| ≥ 1 and |y| ≥ 1) or (|x| ≤ 1 and |y| ≤ 1)}; equivalently, the union of the closed square |x| ≤ 1, |y| ≤ 1 and the four closed exterior corner regions |x| ≥ 1, |y| ≥ 1.",
    "solution": "$$\n(x^2 - 1)(|y| - 1) \\geq 0 \\iff \\left[ \\begin{cases} x^2 - 1 \\geq 0, \\\\ |y| - 1 \\geq 0; \\\\ x^2 - 1 \\leq 0, \\\\ |y| - 1 \\leq 0; \\end{cases} \\right] \\iff \\left[ \\begin{cases} |x| \\geq 1, \\\\ |y| \\geq 1; \\\\ |x| \\leq 1, \\\\ |y| \\leq 1. \\end{cases} \\right]\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45033,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of positive real numbers $a$, $b$, $c$ equals $1$. Prove the inequality\n$$\n\\sqrt{\\frac{1}{a} - 1}\\sqrt{\\frac{1}{b} - 1} + \\sqrt{\\frac{1}{b} - 1}\\sqrt{\\frac{1}{c} - 1} + \\sqrt{\\frac{1}{c} - 1}\\sqrt{\\frac{1}{a} - 1} \\ge 6.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{(b+c)(c+a)}{ab}} + \\sqrt{\\frac{(c+a)(a+b)}{bc}} + \\sqrt{\\frac{(a+b)(b+c)}{ca}} \\right) \\geq 2.\n$$\nЗа нерівністю Коші\n$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{(b+c)(c+a)}{ab}} + \\sqrt{\\frac{(c+a)(a+b)}{bc}} + \\sqrt{\\frac{(a+b)(b+c)}{ca}} \\right) \\geq \\sqrt[3]{\\frac{(a+b)(b+c)(c+a)}{abc}}.\n$$\nПотрібна нам нерівність є наслідком нерівності\n$$\n(a + b)(b + c)(c + a) \\geq 8abc,\n$$\nяка, у свою чергу, випливає з нерівностей\n$$\na + b \\geq 2\\sqrt{ab}, \\quad b + c \\geq 2\\sqrt{bc}, \\quad c + a \\geq 2\\sqrt{ca}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $m$ and $n$ such that\n$$\n\\sqrt{m} + \\frac{2005}{\\sqrt{n}} = 2006.\n$$",
    "options": [],
    "answer": "(m, n) ∈ {(1, 1), (2576025, 25), (4004001, 160801), (4020025, 4020025)}",
    "solution": "$$\n\\sqrt{n} = \\frac{2006^2 n + 2005^2 - m n}{2 \\cdot 2005 \\cdot 2006},\n$$\nso $\\sqrt{n} \\in \\mathbb{Q}$, and therefore, as is known, the number $n$ is the square of a natural number. It also follows that $m = k^2$, $k \\in \\mathbb{N}$, and moreover $k \\mid 2005$. By checking all natural divisors of $2005$, we obtain the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AD$ be the median of a triangle $ABC$, and let $\\angle ADB = 45^\\circ$, $\\angle ACB = 30^\\circ$. Find $\\angle BAD$ (in degrees).",
    "options": [],
    "answer": "30",
    "solution": "Нехай $BK$ — висота трикутника $ABC$. Оскільки точка $D$ є серединою гіпотенузи $BC$ прямокутного трикутника $KBC$, то трикутник $BKD$ є рівностороннім. Отже, як нескладно бачити, $\\angle ADK = \\angle DAK = 15^\\circ$, $AK = DK = BK$. А тому $\\angle BAK = 45^\\circ$, $\\angle BAD = 45^\\circ - 15^\\circ = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45036,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given $n$ ($n \\ge 3$) points in the plane not all lying on the same line. For an arbitrary point $M$ we define $f(M)$ to be the sum of the distances from these $n$ points to $A$. It is known that there exists a point $M_1$ such that for every point $M$ of the plane the inequality $f(M_1) \\le f(M)$ holds. Let $M_2$ be a point such that $f(M_1) = f(M_2)$. Prove that the points $M_1$ and $M_2$ coincide.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to prove that if a point $M$ is the midpoint of a segment $BC$, then for any point $A$ the inequality $AM \\le \\frac{1}{2}(AB + AC)$ holds. Suppose that the points $M_1$ and $M_2$ mentioned in the problem statement do not coincide. Let $M_3$ be the midpoint of the segment $M_1M_2$. Then $A_kM_3 \\le \\frac{1}{2}(A_kM_1 + A_kM_2)$, $1 \\le k \\le n$, and at least one of these inequalities is strict (since the points $A_1, A_2, \\dots, A_n$ do not all lie on the same line). Adding these $n$ inequalities, we obtain\n$$\nf(M_3) < \\frac{1}{2}(f(M_1) + f(M_2)) = f(M_1).\n$$\nThis is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45037,
    "subject": "Mathematics (Multi-modal)",
    "question": "For an acute triangle $ABC$: its circumcircle $\\omega$ with center $O$, the circumcircle $\\omega_1$ of the triangle $AOC$ and the diameter $OQ$ of $\\omega_1$ were drawn. The points $M$ and $N$ were taken on the lines $AQ$ and $AC$ respectively in such a way that the quadrilateral $AMBN$ is a parallelogram. Prove that the intersection point of the lines $MN$ and $BQ$ belongs to the circle $\\omega_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the line $BQ$ meet the circle $\\omega_1$ again at the point $T$. We will show that the points $M$, $T$, $N$ are collinear. Since $OQ$ is the diameter of the circle $\\omega_1$, $MQ$ and $CQ$ are tangents to the circle $\\omega$. Let $\\angle ABC = \\beta$. Then $\\angle CAQ = \\angle ACQ = \\beta$ (as angles between tangents and a chord), $\\angle ATQ = \\angle ACQ = \\beta$ and $\\angle CTQ = \\angle CAQ = \\beta$ (as inscribed angles in the circle $\\omega_1$), $\\angle BNA = \\angle AMB = \\angle CAQ = \\beta$ (as angles at parallel lines). Hence $\\angle AMB + \\angle ATB = \\beta + (180^\\circ - \\beta) = 180^\\circ$, so the quadrilateral $AMBT$ is cyclic and $\\angle BTM = \\angle MAB = \\angle ACB$.\n\nNext, depending on the order of the points $A$, $C$, and $N$ on the line, we consider the following cases. If the point $C$ lies between $A$ and $N$, then $\\angle BNC + \\angle BTC = \\beta + (180^\\circ - \\beta) = 180^\\circ$. Thus, the quadrilateral $BNCT$ is cyclic, and $\\angle BTN = \\angle BCN = 180^\\circ - \\angle ACB = 180^\\circ - \\angle BTM$. If the point $N$ lies between $A$ and $C$, then $\\angle BNC = 180^\\circ - \\beta = \\angle BTC$. Then the quadrilateral $BCNT$ is cyclic, and $\\angle BTN = 180^\\circ - \\angle ACB = 180^\\circ - \\angle BTM$. We see that in both cases, the equality $\\angle BTM + \\angle BTN = 180^\\circ$ holds, which proves that $T$ is the intersection point of the lines $MN$ and $BQ$.\n\n\nAnother solution, proposed by 11th grade student Roman Cheplyaka (Odessa), is based on an additional construction. Namely, construct the circle $\\omega_2$ on the segment $OB$ as a diameter. Let $T$ be the point of intersection of the circles $\\omega_1$ and $\\omega_2$ different from $O$. Since $\\angle OTB = \\angle OTQ = 90^\\circ$ as inscribed angles subtended by the diameter, the point $T$ lies on $BQ$ (so this is the same point $T$ as in the first solution). Let $R$ be the midpoint of the side $AB$ (and the intersection point of the diagonals of the parallelogram $AMBN$). Then the line $OR$ is the perpendicular bisector of $AB$. Hence, $\\angle ORB = 90^\\circ$, and the point $R$ also lies on $\\omega_2$. Since the inscribed angles $RTB$ and $BOR$ in the circle $\\omega_2$ are equal, and $\\angle BOR = \\frac{1}{2}\\angle AOB = \\angle ACB$, we have $\\angle BTR = \\angle ACB$. In addition, $\\angle AQT = \\angle ACT$ as angles inscribed in $\\omega_1$. Let $P$ be the intersection point of the lines $AC$ and $RT$. Since the angle $CAR$ is acute and the angle $ART$ is right, these lines do intersect, and one of two cases occurs: either the point $P$ lies between $A$ and $C$, or the point $C$ lies between $A$ and $R$.\n\nIn the first case, we have the equalities $\\angle BTP = 180^\\circ - \\angle BTR = 180^\\circ - \\angle BCP$, so the quadrilateral $BTPC$ is cyclic and $\\angle ACT = \\angle PCT = \\angle PBT$. In the second case, $\\angle BTP = 180^\\circ - \\angle BTR = 180^\\circ - \\angle ACB = \\angle BCP$, so the quadrilateral $BTCP$ is cyclic and $\\angle ACT = 180^\\circ - \\angle PCT = \\angle PBT$. We see that in both cases the equality $\\angle PBT = \\angle ACT = \\angle AQT$ holds, from which it follows that $BP \\parallel AQ$. Thus, the point $P$ coincides with the point $N$, and the line $PR$ coincides with the line $MN$ (which contains the diagonal of the parallelogram $AMBN$). Therefore, the point $T$ lies simultaneously on the circle $\\omega_1$ and on the lines $BQ$ and $MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45038,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square board $6 \\times 6$ is filled out with positive integer numbers (by one number in every cell). Each step, it is allowed to choose a square larger than $1 \\times 1$ and increase by 1 all numbers in its cells. Is it always possible to reach, after several steps, a situation when all numbers on the board are divisible by 3?",
    "options": [],
    "answer": "No",
    "solution": "Будемо позначати через $S$ суму чисел, які містяться в клітинках зі знаком “+”, а через $Q$ — суму чисел, які містяться в клітинках зі знаком “-”. Нескладно перевірити, що на кожному кроці різниця $S-Q$ або не змінюється, або збільшується на 3, або ж — зменшується на 3. А тому, якщо спочатку $S-Q \\neq 0 \\pmod{3}$, то ми не зможемо досягти того, щоб усі числа ділилися без остачі на 3.\n\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45039,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which real numbers $x > 1$ there exists a triangle with sides of lengths $x^4 + x^3 + 2x^2 + x + 1$, $2x^3 + x^2 + 2x + 1$ and $x^4 - 1$?",
    "options": [],
    "answer": "All real numbers greater than 1",
    "solution": "Дійсно, при всіх $x > 1$, як нескладно переконатися, мають місце нерівності\n$$\nx^4 + x^3 + 2x^2 + x + 1 > 2x^3 + x^2 + 2x + 1,\n$$\n$$\nx^4 + x^3 + 2x^2 + x + 1 > x^4 - 1,\n$$\n$$\nx^4 + x^3 + 2x^2 + x + 1 < (x^4 - 1) + (2x^3 + x^2 + 2x + 1).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two different points $P$ and $Q$ are chosen inside a parallelogram $ABCD$ in such a way that $\\angle ABP = \\angle ADP$, $\\angle CBQ = \\angle CDQ$, and those points do not lie on the diagonal $AC$. Prove that $\\angle PAQ = \\angle PCQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**10.4. Лема.** Якщо всередині паралелограма $ABCD$ відмічено таку точку $M$, що $\\angle ABM = \\angle ADM$, то $\\angle MAD = \\angle MCD$.\n\nДійсно, нехай $N$ — така точка, що відрізок $MN$ паралельний і рівний відрізку $BC$. Тоді чотирикутники $AMND$ і $MBCN$ є паралелограмами. Звідси встановлюємо, що\n$$\n\\angle ABM = \\angle DCN, \\quad \\angle ADM = \\angle NMD.\n$$\nЗ умови задачі випливає, що $\\angle DCN = \\angle DMN$. Отже, навколо чотирикутника $DMCN$ можна описати коло. Таким чином, $\\angle MCD = \\angle MND = \\angle MAD$, що й треба було довести.\n\nА тепер перейдемо до розв’язування задачі. Оскільки $\\angle QBC = \\angle QDC$, то маємо, що $\\angle ABQ = \\angle ADQ$. Далі, оскільки $\\angle ABP = \\angle ADP$ і $\\angle ABQ = \\angle ADQ$, то за допомогою попереднього твердження одержуємо, що\n$$\n\\angle PAD = \\angle PCD \\text{ і } \\angle QAD = \\angle QCD.\n$$\nЯкщо $\\angle PAD > \\angle QAD$, то $\\angle PCD > \\angle QCD$, тобто — точка $Q$ лежить усередині чотирикутника $APCD$, і $\\angle PAQ = \\angle PAD - \\angle QAD = \\angle PCD - \\angle QCD = \\angle PCQ$.\n\nЯкщо $\\angle QAD > \\angle PAD$, то $\\angle QCD > \\angle PCD$, а тому точка $P$ лежить усередині чотирикутника $AQCD$, і $\\angle PAQ = \\angle QAD - \\angle PAD = \\angle QCD - \\angle PCD = \\angle PCQ$.\n\nВипадок $\\angle QAD = \\angle PAD$ є неможливим, оскільки при цьому $\\angle QCD = \\angle PCD$, і точки $P$ і $Q$ співпадають, що суперечить умові.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mykolka the numismatist possesses 241 coins of total amount 360 tugriks (value of each coin is a positive integer quantity of tugriks). May it be claimed for sure that all those coins can be divided into three heaps of equal amount?",
    "options": [],
    "answer": "Yes",
    "solution": "Розіб’ємо коло на 360 рівних частин, і 241 з 360 точок поділу відмітимо червоним кольором так, щоб 241 дуга з червоними кінцями відповідала — за довжиною — вартостям монет. Усі 360 точок поділу на колі позначимо по порядку таким чином: $A_1, A_2, ..., A_{120}, B_1, B_2, ..., B_{120}, C_1, C_2, ..., C_{120}$. За принципом Діріхле, знайдеться такий номер $k$, що три точки $A_k, B_k, C_k$ будуть пофарбованими червоним кольором. Вони і будуть визначати розбиття монет на три однакові за вартістю купки.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-decreasing and all non-increasing functions $f : [0; +\\infty) \\to \\mathbf{R}$ such that for all $x, y \\ge 0$ the equality\n$$\nf(x+y) - f(x) - f(y) = f(xy+1) - f(xy) - f(1)\n$$\nholds, and additionally $f(3) + 3f(1) = 3f(2) + f(0)$.",
    "options": [],
    "answer": "All solutions are quadratic polynomials f(u) = A u^2 + B u + C for u ≥ 0. The functional equation holds for any real A, B, C, and the extra condition forces f(0) = C. Monotonicity constraints: nondecreasing if and only if A ≥ 0 and B ≥ 0; nonincreasing if and only if A ≤ 0 and B ≤ 0.",
    "solution": "**Лема.** Якщо функція $F:(0;+\\infty) \\rightarrow \\mathbf{R}$ задовольняє на проміжку $(0;+\\infty)$ **функціональне рівняння Коші**\n$$\nF(u + v) = F(u) + F(v)\n$$\nі є обмеженою на інтервалі $(0;1)$ (тобто, існує така стала $M$, що $|F(u)| \\le M$ для всіх $u \\in (0;1)$), то $F(u) = F(1)u$ для всіх $u \\in (0;+\\infty)$.\n\n**Доведення леми.** Розглянемо функцію $G(u) = F(u) - F(1)u$. Легко переконатися, що $G(1) = 0$ і що функція $G$ так само, як і $F$, задовольняє на $(0;+\\infty)$ функціональне рівняння Коші: для всіх $u, v \\in (0;+\\infty)$\n$$\nG(u + v) = F(u + v) - (u + v)F(1) = F(u) + F(v) - uF(1) - vF(1) = G(u) + G(v).\n$$\nКрім того, для всіх $u \\in (0; +\\infty)$\n$$\nG(u + 1) = F(u + 1) - (u + 1)F(1) = F(u) + F(1) - uF(1) - F(1) = G(u).\n$$\nФункція $G$ є обмеженою на всьому проміжку $(0;+\\infty)$. Дійсно, будь-яке дійсне число $u > 0$ можна подати у вигляді $k + \\alpha$, де $k$ є цілим невід'ємним числом, $0 < \\alpha \\le 1$, причому $|G(u)| = |G(k + \\alpha)| = |G(\\alpha)| \\le |F(\\alpha)| + \\alpha|F(1)| \\le M + |F(1)|$ для довільного дійсного $u > 0$. Але за індукцією легко довести, що $G(nu) = nG(u)$ для довільного дійсного $u > 0$ і довільного натурального числа $n$. Отже, нерівність $|G(u)| = \\frac{1}{n} |G(nu)| \\le \\frac{1}{n} (M + |F(1)|)$ виконується для довільного натурального $n$ і довільного дійсного $u > 0$. З цього випливає, що $G(u) = 0$ на $(0; +\\infty)$.\n\nПерейдемо до розв'язання задачі.\nЗ умови виводимо, що\n$$\n\\begin{align*}\nf(x+y+z) - f(x) - f(y) - f(z) &= \\\\\n&= f(x+y+z) - f(x+y) - f(z) + f(x+y) - f(x) - f(y) \\\\\n&= f(xz + yz + 1) - f(xz + yz) - f(1) + f(xy + 1) - f(xy) - f(1)\n\\end{align*}\n$$\nдля будь-яких $x, y, z > 0$.\nЗвідки\n$$\n\\begin{align*}\nf(xz + xy + 1) - f(xz + xy) - f(1) + f(yz + 1) - f(yz) - f(1) &= \\\\\n&= f(xz + yz + 1) - f(xz + yz) - f(1) + f(xy + 1) - f(xy) - f(1).\n\\end{align*}\n$$\nОтже, для довільних $a,b,c > 0$ — якщо покласти $x = \\sqrt{ca/b}$, $y = \\sqrt{bc/a}$, $z = \\sqrt{ab/c}$ —\nодержуємо рівність\n$$\nf(a+b+1)-f(a+b)-f(b+1)+f(b)=f(a+c+1)-f(a+c)-f(c+1)+f(c).\n$$\nРозглянемо першу різницю функції $f$ як функцію $\\Delta f$, що визначається на $[0;+\\infty)$ співвідношенням $\\Delta f(u) = f(u+1) - f(u)$. Тоді, як нескладно отримати, для всіх $a,b,c > 0$ виконується рівність\n$$\n\\Delta f(b+a) - \\Delta f(b) = \\Delta f(c+a) - \\Delta f(c).\n$$\nПрава частина цієї тотожності не залежить від $b$, тому й ліва її частина від $a$ не залежить. Отже, $\\Delta f(b+a) - \\Delta f(b) = \\phi(a)$ для певної дійсної функції $\\phi$, визначеної на проміжку $(0;+\\infty)$. Помінявши місцями $a$ та $b$, отримаємо, що $\\Delta f(a+b) - \\Delta f(a) = \\phi(b)$, звідки випливає, що $\\Delta f(a) - \\phi(a) = \\Delta f(b) - \\phi(b)$ для всіх $a, b > 0$. Знову ж таки, права частина останньої рівності не залежить від $a$, отже, ліва також не залежить, тому є сталою: $\\Delta f(a) - \\phi(a) = C_1$ для якогось дійсного $C_1$. Таким чином, $\\Delta f(a+b) = \\Delta f(b) + \\Delta f(a) - C_1$ для всіх $a, b > 0$. Легко переконатися, що функція $g(u) = \\Delta f(u) - C_1$ задовольняє умови леми (обмеженість $g$ на $(0;1]$ випливає з монотонності $f$), тому $g(u) = g(1)u$, і функція $\\Delta f$ є лінійною на $(0;+\\infty)$: $\\Delta f(u) = C_2u + C_1$, де $C_2 = g(1)$. Тепер з вихідної рівності для довільних $x, y > 0$ виводимо, що $f(x+y)-f(x)-f(y)=C_2xy+C_3$, де $C_3 = C_1 - f(1)$. Функція $h(u) = f(u) - \\frac{1}{2}C_2u^2 + C_3$, як нескладно довести, задовольняє умови леми. Тому $h(u) = h(1)u$, і на проміжку $(0; +\\infty)$ функція $f$ має вигляд $f(u) = Au^2 + Bu + C$, де $A = \\frac{1}{2}C_2$, $B = h(1)$, $C = -C_4$. Згідно з умовою задачі маємо також, що\n$$\nf(0) = f(3) - 3f(2) + 3f(1) = (9A + 3B + C) - 3(4A + 2B + C) + 3(A + B + C) = C.\n$$\nОтже, $f(u) = Au^2 + Bu + C$ для всіх $u \\ge 0$. Підстановка такого тричлена в вихідну рівність дає тотожність для будь-яких дійсних $A, B$ і $C$. Залишається відкинути випадок $AB < 0$, коли $f$ не є монотонною функцією на $[0; +\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Can the number $1$ be represented as a sum of $2005$ different terms, each one of the form $\\frac{1}{3n-1}$ with a positive integer $n$?",
    "options": [],
    "answer": "No",
    "solution": "Якщо доданки даної суми звести до спільного знаменника, то за mod $3$ чисельник матиме остачу $1$, а знаменник — остачу $2$. Таким чином, така сума не може дорівнювати $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ satisfying the inequality\n$$\n\\cos(\\pi\\sqrt{n^2+n}) \\ge 0.\n$$",
    "options": [],
    "answer": "all even positive integers",
    "solution": "Легко довести, що $n < \\sqrt{n^2 + n} < n + \\frac{1}{2}$ для всіх натуральних $n$. Отже, маємо, що $\\pi\\sqrt{n^2 + n} \\in (\\pi n; \\pi n + \\frac{\\pi}{2})$. Залишається тільки зауважити, що для $k \\in \\mathbb{N}$ на проміжках $(2\\pi k; \\frac{\\pi}{2} + 2\\pi k)$ косинус набуває додатні значення, а на проміжках $(2\\pi k + \\pi; \\frac{3\\pi}{2} + 2\\pi k)$ — від’ємні.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : (0; +\\infty) \\to \\mathbf{R}$ such that the equality\n$$\nf(x)f(y) = f(xy) + 2005 \\left( \\frac{1}{x} + \\frac{1}{y} + 2004 \\right)\n$$\nholds for all positive real $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 1/x + 2005 or f(x) = -(2005/2006) (1/x + 2005), for all x > 0",
    "solution": "При $y=1$ із вихідного співвідношення одержимо, що для всіх $x > 0$\n$$\nf(x) \\cdot f(1) = f(x) + 2005 \\left( \\frac{1}{x} + 2005 \\right).\n$$\nЗвідси дістаємо рівність $f(1)^2 = f(1) + 2005 \\cdot 2006$. Таким чином, $f(1) = -2005$ або $f(1) = 2006$.\nЯкщо $f(1) = -2005$, то при всіх $x > 0$\n$$\nf(x) = -\\frac{2005}{2006} \\left( \\frac{1}{x} + 2005 \\right),\n$$\nа якщо $f(1) = 2006$, то при всіх $x > 0$\n$$\nf(x) = \\frac{1}{x} + 2005.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist integers $a, b, c, d, t, x, y, z$ such that each of expressions\n$$\n|ay - bx|, |az - cx|, |at - dx|, |bz - cy|, |bt - dy|, |ct - dz|\n$$\nwould be valued either 1 or 2005?",
    "options": [],
    "answer": "No",
    "solution": "Помітимо, що серед чисел $x, y, z, t$ принаймні три є непарними. Аналогічно, непарними будуть щонайменше три з чотирьох чисел $a, b, c, d$. Отже, у принаймні двох з чотирьох пар $(a;x), (b;y), (c;z), (d;t)$ обидві компоненти є непарними числами. Але ж тоді один із заданих шести виразів має набувати парне значення. Отримали суперечність.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which minimal quantity of sides of even lengths can have a polygon on a squared paper built of $2005$ dominoes $1 \\times 2$? (Each domino covers two adjacent unit squares of the paper, and the boundary of the polygon is a connected closed polygonal line that does not touch and does not cross itself.)",
    "options": [],
    "answer": "2",
    "solution": "Розфарбуємо дошку як шахівницю, а разом з нею і многокутник. Проведемо в усіх клітинках дошки, які входять до складу многокутника, діагоналі, тим самим розбивши кожну з цих клітинок на чотири рівнобедрених прямокутних трикутнички з одиничною гіпотенузою. Оскільки многокутник утворено з додоміно, то кількість чорних клітинок в його складі дорівнює кількості білих, а тому кількість чорних трикутничків дорівнює кількості білих. Усі трикутнички, гіпотенуза яких не входить до складу межі многокутника, розбиваються на пари тих, що мають спільну гіпотенузу. Очевидно, що кожна така пара складається з чорного та білого трикутничка, отже, кількість чорних трикутничків у парах дорівнює кількості білих трикутничків у парах. Тому серед трикутничків, які не мають пари, тобто — прилягають до межі многокутника, — кількість чорних дорівнює кількості білих. Легко бачити, що уздовж кожної сторони многокутника білі й чорні трикутнички, що прилягають до неї, чергуються. Якщо така сторона має непарну довжину, то прилеглих до неї трикутничків одного кольору на $1$ більше, ніж другого. Якщо дві сторони непарної довжини є суміжними, то внаслідок шахового розфарбування крайні трикутнички, прилеглі до них, будуть одного кольору. Якщо припустити, що в многокутника є не більше від одної сторони парної довжини, то всі його $N$ сторін непарної довжини виявляться послідовно суміжними, отже, трикутничків одного кольору буде рівно на $N$ більше, аніж іншого (при цьому, зрозуміло, $N \\ge 1$). Ця суперечність доводить, що серед сторін нашого многокутника знайдуться дві такі, що мають парну довжину. Приклад прямокутника розміром $1 \\times 4010$ показує, що їх може бути рівно дві.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45048,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex quadrilateral $ABCD$ with $BC = CD$, $\\angle CBA + \\angle DAB > 180^\\circ$ is given. Let the points $W$ and $Q$ (different from the quadrilateral's vertices) belong to the sides $BC$ and $DC$ respectively, and let $AD = QD$, $WQ \\parallel AD$. It is known that the intersection point $M$ of segments $AQ$ and $BD$ is equidistant from the lines $AD$ and $BC$. Prove that $\\angle BWD = \\angle ADW$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай $F$ — точка перетину прямих $BC$ і $AD$ (точки $F$ і $S$ лежать по різні боки від прямої $AB$). Проведемо через точку $D$ пряму $\\ell \\parallel BC$. Позначимо через $S$ точку перетину прямих $WQ$ та $\\ell$. Легко бачити, що $QM$ і $DM$ є бісектрисами зовнішніх кутів трикутника $SDQ$. Звідси випливає, що $SM$ — бісектриса кута $WSD$. Із умови задачі випливає, що $FM$ — бісектриса кута $WFD$. Оскільки чотирикутник $FWSD$ є паралелограмом і бісектриси його протилежних кутів $F$ і $S$ мають спільну точку $M$, то прямі $FM$ і $SM$ збігаються. Отже, $FWSD$ — ромб. Звідси й отримуємо твердження задачі.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive integer $n$ the inequality\n$$\n\\frac{3}{1!+ 2!+ 3!} + \\frac{4}{2!+ 3!+ 4!} + \\dots + \\frac{n+2}{n!+ (n+1)!+ (n+2)!} < \\frac{1}{2}\n$$\nholds. (Here $1!=1$, $k!=1 \\cdot 2 \\cdot \\dots \\cdot (k-1) \\cdot k$, $k \\ge 2$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{aligned}\n& \\frac{3}{1!+ 2!+ 3!} + \\frac{4}{2!+ 3!+ 4!} + \\dots + \\frac{n+2}{n!+ (n+1)!+ (n+2)!} = \\\\\n&= \\left( \\frac{1}{(1+1)!} - \\frac{1}{(1+2)!} \\right) + \\left( \\frac{1}{(2+1)!} - \\frac{1}{(2+2)!} \\right) + \\dots + \\left( \\frac{1}{(n+1)!} - \\frac{1}{(n+2)!} \\right) = \\\\\n&= \\frac{1}{2} - \\frac{1}{(n+2)!} < \\frac{1}{2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45050,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $M$ is taken on the midperpendicular of the side $AC$ of an acute triangle $ABC$. The points $M$ and $B$ belong to the same half-plane with respect to the line $AC$, $\\angle BAC = \\angle MCB$ and $\\angle ABC + \\angle MBC = 180^\\circ$. Find $\\angle BAC$ (in degrees).",
    "options": [],
    "answer": "30",
    "solution": "З умови задачі випливає, що вершина **B** не лежить на серединному перпендикулярі відрізка **AC**. Тому залишилося розглянути такі два випадки.\n\n**Перший випадок.** Припустимо, що $AB < BC$. Нехай точка **M** лежить усерединні трикутника **ABC**. Тоді $\\angle MCB < \\angle ACB < \\angle BAC$, що суперечить умові. Нехай тепер точка **M** лежить зовні трикутника **ABC**. Оскільки за умовою $\\angle ABC + \\angle MBC = 180^\\circ$, то точки **A**, **B**, **M** колінеарні, і $\\angle BAC = \\angle MCA > \\angle MCB$, що суперечить умові задачі.\n\n**Другий випадок.** Припустимо, що $AB > AC$. Нехай точка **M** лежить усередині трикутника **ABC**. Тоді $\\angle MBC < \\angle ABC < 90^\\circ$, $\\angle ABC + \\angle MBC < 90^\\circ + 90^\\circ = 180^\\circ$, що суперечить умові. Нехай тепер точка **M** лежить зовні трикутника **ABC**. Розглянемо точку **N**, симетричну точці **M** відносно прямої **BC**, і нехай\n\n$\\varphi = \\angle BAC$. Тоді $\\angle MBC = \\varphi = \\angle NBC$. Оскільки $\\angle MBC = 180^\\circ - \\angle ABC$, то точки A, B, N є колінеарними. Далі, оберемо на продовженні сторони CB за точку B таку точку K, що $\\angle MKC = \\varphi$. Тоді $\\angle CKN = \\varphi$, і, позаяк $\\angle CKN = \\angle CAN$, робимо висновок щодо циклічності четвірки точок C, A, K, N.\n\nОтже, $\\angle KAN = \\angle KCN = \\varphi$. Зауважимо, що $MA = MC$, і, крім того, $MC = MK$ ($\\angle MCK = \\varphi = \\angle MKC$). Таким чином, точка M — центр описаного кола трикутника AKC. Маємо: $\\angle KMC = 2 \\cdot \\angle KAC = 2 \\cdot 2\\varphi = 4\\varphi$. Розглядаючи трикутник CMK, знаходимо, що має виконується рівність $4\\varphi + \\varphi + \\varphi = 180^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove for every integer $n \\ge 2$, that such a set $A_n$ of $n$ pairwise distinct positive integers can be collected that for any different $i, j \\in A_n$ the sum of their squares $i^2 + j^2$ is divisible by their difference $i - j$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\nA_n = \\{1 \\cdot n!; 2 \\cdot n!; \\dots; n \\cdot n!\\}\n$$\nsatisfies the condition of the problem. Indeed, take any two distinct numbers from the set $A_n$: $k \\cdot n!$ and $m \\cdot n!$, where $1 \\le k < m \\le n$. Since $m-k \\le n$, it follows that $m-k|n!$. Therefore, obviously, $(k^2 + m^2) \\cdot (n!)^2$ is divisible by their difference $(m-k) \\cdot n!$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45052,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane, $2005$ points were marked, no three of which are collinear. Straight lines were drawn through all the pairs of marked points. Prove that all the marked points can be colored into two colors in such a way that for any two points of the same color the number of the drawn lines separating them is even. (We say that a line separates two points if none of those points belongs to it, and they lie in different half-planes with respect to it.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Крім того, покладемо\n$$\n\\delta(a,P,Q,R) = \\delta(a,P,Q) + \\delta(a,Q,R) + \\delta(a,P,R).\n$$\nРозглянемо довільні три відмічені точки $P$, $Q$, $R$. Очевидно, що\n$$\nn_{PQ} + n_{QR} + n_{PR} = \\sum_{a} \\delta(a,P,Q,R),\n$$\nде сума береться по всіх проведених прямих.\nЛегко бачити, що якщо пряма $a$ проходить через дві з точок $P$, $Q$, $R$, то маємо рівність $\\delta(a,P,Q,R)=0$. Якщо пряма $a$ проходить рівно через одну з них, то $\\delta(a,P,Q,R) \\in \\{0;1\\}$. А якщо пряма $a$ не проходить через жодну, то маємо, що $\\delta(a,P,Q,R) \\in \\{0;2\\}$.\nРозглянувши всі сім усіляких випадків взаємного розташування чотирьох точок на площині, одержуємо, що для будь-якої точки $X$ з решти $2002$ відмічених точок або серед трьох прямих $XP$, $XQ$, $XR$ тільки одна розділяє дві точки з числа точок $P$, $Q$, $R$, або ж кожна з прямих $XP$, $XQ$, $XR$ має таку властивість. Отже, парність числа $n_{PQ} + n_{QR} + n_{PR}$ визначається кількістю зазначених вище точок $X$, а їх саме $C_{2002}^1 = 2002$. Таким чином, ми довели, що число $n_{PQ} + n_{QR} + n_{PR}$ є парним для будь-яких трьох відмічених точок $P$, $Q$, $R$. А тому серед чисел $n_{PQ}$, $n_{QR}$, $n_{PR}$ парними є або всі три, або ж — тільки одне. З цього випливає, що якщо для якоїсь точки $P$ пофарбувати всі ті точки $Q$, для яких $n_{PQ}$ парне, у той самий колір, що й точку $P$, а решту точок — у другий колір, то таке розфарбування буде задовольняти умову.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45053,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the space, $2005$ points were marked, none four of which lie in the same plane. Planes were drawn through all the triples of marked points. Prove that all the marked points can be colored into two colors in such a way that for any two points of the same color the number of the drawn planes separating them is odd. (We say that a plane separates two points if none of those points belongs to it, and they lie in different half-spaces with respect to it.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Доведення леми.** Розглянемо три можливі випадки взаємного розташування п’яти точок.\n1. Пряма $XY$ паралельна площині $PQR$. Тоді серед трьох прямих, що проходять через точки $P, Q, R$ паралельно $XY$, тільки одна лежить між двома іншими. Відповідна площина розділяє точки, а дві інші площини цього не роблять.\n2. Пряма $XY$ перетинає площину $PQR$ в точці $O$, що лежить зовні трикутника $PQR$. Тоді три промені $OP, OQ, OR$ лежать усередині певного розгорнутого кута з вершиною в точці $O$, отже рівно один з них проходить між двома іншими. Відповідна площина розділяє точки, а дві інші площини цього не роблять.\n3. Пряма $XY$ перетинає площину $PQR$ в точці $O$, що лежить усередині трикутника $PQR$. Тоді кожна з прямих $OP, OQ, OR$ перетинає одну зі сторін трикутника, тому кожна з трьох площин розділяє точки.\n\n**Розв’язання задачі.** Нехай $n_{PQ}$ позначає кількість проведених площин, які розділяють точки $P$ і $Q$. Покладемо $\\delta(\\alpha, P, Q) = 0$, якщо площина $\\alpha$ не розділяє точки $P$ і $Q$, і нехай $\\delta(\\alpha, P, Q) = 1$, якщо розділяє.\nНехай $\\delta(\\alpha, P, Q, R) = \\delta(\\alpha, P, Q) + \\delta(\\alpha, Q, R) + \\delta(\\alpha, P, R)$. Розглянемо довільні три відмічені точки $P, Q, R$. Очевидно, що $n_{PQ} + n_{QR} + n_{PR} = \\sum_{\\alpha} \\delta(\\alpha, P, Q, R)$, де сума береться по всіх проведених площинах. Легко бачити, що якщо площина $\\alpha$ проходить через три або дві з точок $P, Q, R$, то $\\delta(\\alpha, P, Q, R) = 0$, а якщо не проходить через жодну, то $\\delta(\\alpha, P, Q, R) \\in \\{0; 2\\}$.\n\nРозглядаючи площини, які проходять рівно через одну з точок $P, Q, R$, застосуємо лему і одержимо, що для кожної пари точок $X, Y$, відмінних від $P, Q, R$, $\\delta(XYP,P,Q,R) + \\delta(XYQ,P,Q,R) + \\delta(XYR,P,Q,R) \\in \\{1; 3\\}$. З цього робимо висновок, що парність числа $n_{PQ} + n_{QR} + n_{PR}$ визначається кількістю зазначених вище пар $X, Y$, а таких пар буде $C_{2002}^2$ (а таке число є непарним). Таким чином ми довели, що число $n_{PQ} + n_{QR} + n_{PR}$ є непарним для будь-яких трьох відмічених точок $P, Q, R$. Інакше кажучи, серед чисел $n_{PQ}$, $n_{QR}$ і $n_{PR}$ непарними є або всі три, або тільки одне. З цього випливає, що якщо для якоїсь точки $P$ пофарбувати всі ті точки $Q$, для яких $n_{PQ}$ непарне, у той самий колір, що й точку $P$, а решту точок — у другий колір, то таке розфарбування буде задовольняти умову.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45054,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a circle, two non-intersecting chords $AB$ and $CD$ are drawn. On the chord $AB$, a point $E$ (different from $A$ and $B$) is taken. Consider the arc $AB$ that does not contain the points $C$ and $D$. With a compasses and a straightedge, find a point $F$ on that arc such that $\\frac{PE}{EQ} = \\frac{1}{2}$, where $P$ and $Q$ are the points in which the chord $AB$ meets the segments $FC$ and $FD$ respectively.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Побудова і подальше доведення очевидним чином ґрунтуються на наступному аналізі.\n\nПерший випадок. Нехай точка $E$ ділитиме відрізок $PQ$ внутрішнім чином. На промені $CE$ оберемо таку точку $K$, що $EK = 2 \\cdot CE$. Тоді, як легко довести, $CF \\parallel QK$, і, крім того, з точки $Q$ хорди $AB$ відрізок $KD$ видно під кутом $180^\\circ - \\angle CFD = 180^\\circ - \\angle CAD$.\n\nДругий випадок. Нехай точка $E$ ділитиме відрізок $PQ$ зовнішнім чином. Візьмемо середину відрізка $DE$ — точку $M$. Тоді $PM \\parallel FD$, і з точки $P$ відрізок $CK$ видно під кутом $\\angle CPK = \\angle CFD = \\angle CBD$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45055,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that the equality $\\{x\\} + \\{2y\\} = \\{y\\} + \\{2x\\}$ for real numbers $x$ and $y$ implies $\\{x\\} = \\{y\\}$.\n\nb) Do there exist integers $n \\ge 3$ such that the equality $\\{x\\} + \\{ny\\} = \\{y\\} + \\{nx\\}$ for real numbers $x$ and $y$ implies $\\{x\\} = \\{y\\}$?\n\n(Here $\\{a\\} = a - [a]$, where $[a]$ stands for the greatest integer that does not exceed the real number $a$.)",
    "options": [],
    "answer": "a) {x} = {y}. b) No, such integers do not exist.",
    "solution": "a) Suppose that for some real numbers $x$ and $y$ the equality $\\{x\\} + \\{2y\\} = \\{y\\} + \\{2x\\}$ holds. Let $\\{x\\} = \\alpha$, $\\{y\\} = \\beta$, $\\alpha, \\beta \\in [0;1)$. It suffices for $\\alpha, \\beta \\in [0;1)$ to consider the equality $\\alpha + \\{2\\beta\\} = \\beta + \\{2\\alpha\\}$.\n\nIf $0 \\le \\gamma < \\frac{1}{2}$, then $\\{2\\gamma\\} = 2\\gamma$, and for $\\frac{1}{2} \\le \\gamma < 1$, $\\{2\\gamma\\} = 2\\gamma - 1$. Note that for the cases $0 \\le \\alpha < \\frac{1}{2} \\le \\beta < 1$ and $0 \\le \\beta < \\frac{1}{2} \\le \\alpha < 1$, the equality $\\alpha + \\{2\\beta\\} = \\beta + \\{2\\alpha\\}$ cannot hold. If $0 \\le \\alpha, \\beta < \\frac{1}{2}$ or $\\frac{1}{2} \\le \\alpha, \\beta < 1$, then the equality $\\alpha + \\{2\\beta\\} = \\beta + \\{2\\alpha\\}$ takes the form $\\alpha = \\beta$.\n\nb) Answer: no, such integers do not exist. Take $x=0$, $y = \\frac{1}{n-1}$. Then $\\{x\\} \\neq \\{y\\}$, but, as is easy to see, $\\{x\\} + \\{ny\\} = \\{y\\} + \\{nx\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any rational numbers $a$ and $b$ the graph of the function\n$$\nf(x) = x^3 - 6abx - 2a^3 - 4b^3, \\quad x \\in \\mathbb{R},\n$$\nhas exactly one common point with the $x$-axis.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Оскільки $f'(x) = 3x^2 - 6ab$, то при $ab \\le 0$ функція $f$ строго зростає на всій числовій прямій, набуває, очевидно, і від’ємних, і додатних значень, а тому твердження задачі справджується.\n\nЯкщо $a > 0$ і $b > 0$, то точками локального екстремуму будуть $x = \\pm\\sqrt{2ab}$, причому $f(\\pm\\sqrt{2ab}) = - (\\sqrt{2a^3} \\mp \\sqrt{4b^3})^2 \\le 0$. Рівність в останній нерівності досягатися, як легко помітити, не може через ірраціональність числа $\\sqrt{2}$. Отже, в обох точках екстремуму кубічний многочлен набуває від’ємних значень, а тому, зрозуміло, його графік має з віссю $Ox$ єдину спільну точку.\n\nАналогічно розглядається випадок, коли $a < 0$ і $b < 0$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 45057,
    "subject": "Mathematics (Multi-modal)",
    "question": "A toy factory hired three painters to paint $2006$ cubic bricks. A painter needs $5$ minutes to paint a face of a brick. Each face has to be painted by just one of the painters. It is acceptable that different painters paint different faces of the same brick; however they cannot do it at the same time. Find the minimal amount of time needed for painters to complete their task.",
    "options": [],
    "answer": "20060 seconds",
    "solution": "Відповідь: $20060$ секунд. Час, необхідний для розфарбування, не може бути меншим за $\\frac{2006 \\cdot 6 \\cdot 5}{3} = 20060$ секунд (зрозуміло, що час розфарбування буде найменшим, якщо вдасться так розподілити роботу, щоб кожні $5$ секунд фарбувалася максимальна кількість граней). Цього можна досягти, наприклад, таким чином. Спочатку художники розфарбовують повністю по $667$ кубиків кожний. На це витрачається $667 \\cdot 6 \\cdot 5 = 20010$ секунд, потім протягом $30$ секунд двоє розфарбовують ще по одному кубику кожний, а третій у цей же час розфарбовує по дві грані у трьох кубиків, що залішилися. Насамкінець, кожен художник бере собі по одному з кубиків з нерозфарбованими гранями (таких кубиків саме три, і в кожному чотири нерозфарбовані грані), і протягом $20$ секунд роботу буде завершено.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(f(x) - y^2) = (x - y)^2 \\cdot f(x + y)\n$$\nfor any real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x^2 for all real x",
    "solution": "Відповідь: $f(x) = 0$, $x \\in \\mathbb{R}$; $f(x) = x^2$, $x \\in \\mathbb{R}$.\n\nЗ вихідного співвідношення отримуємо, що\n$$\n(x - y)^2 f(x + y) = (x + y)^2 f(x - y).\n$$\nПокладемо $x = \\frac{t+1}{2}$, $y = \\frac{t-1}{2}$ і одержимо, що $1^2 \\cdot f(t) = t^2 \\cdot f(1)$ при всіх $t \\in \\mathbb{R}$. Звідси випливає з необхідністю, що функцію $f$ слід шукати у вигляді $f(x) = \\lambda x^2$, де $\\lambda \\in \\mathbb{R}$. Таким чином, для будь-яких $x, y \\in \\mathbb{R}$ має виконуватися рівність $\\lambda(\\lambda x^2 - y^2)^2 = (x - y)^2 \\cdot \\lambda(x + y)^2$. Беручи $x = 1$ і $y = 1$, знаходимо, що $\\lambda \\in \\{0; 1\\}$. Перевірка свідчить, що обидві наведені у відповіді функції задовольняють умову.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45059,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle is divided into $2006$ equal arcs by $2006$ points. Baron Munchausen claims that he can construct a closed polygonal curve with the set of vertices consisting of these $2006$ points such that amongst its $2006$ edges there cannot be found any two, which are parallel to each other. Is his claim true or false?",
    "options": [],
    "answer": "false",
    "solution": "Відповідь: ні, барон помиляється. Позначимо $2006$ точок поділу числами $0, 1, 2, \\ldots, 2005$, записаними послідовно. Ланки ламаної будемо позначати номерами її кінців. Нескладно перевірити, що якщо $i + j = k + l$ (mod $2006$), то ланки $i j$ та $k l$ будуть паралельними. Припустимо, що барон Мюнхгаузен правий. Тоді існує така перестановка $n_1, n_2, \\ldots, n_{2006}$ чисел $0, 1, 2, \\ldots, 2005$, що всі суми $n_1 + n_2, n_2 + n_3, \\ldots, n_{2005} + n_{2006}, n_{2006} + n_1$ попарно неконгруентні за mod $2006$. Тоді цей набір також є деякою перестановкою чисел $0, 1, 2, \\ldots, 2005$. Таким чином,\n$$\n(n_1 + n_2) + (n_2 + n_3) + \\dots + (n_{2006} + n_1) \\equiv 0 + 1 + \\dots + 2005 \\equiv 1003 \\pmod{2006}\n$$\nАле, з іншого боку,\n$$\n2n_1 + 2n_2 + 2n_3 + \\dots + 2n_{2006} \\equiv 2(0 + 1 + 2 + \\dots + 2005) \\equiv 0 \\pmod{2006}.\n$$\nОдержана суперечність і доводить, що барон Мюнхгаузен помиляється.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45060,
    "subject": "Mathematics (Multi-modal)",
    "question": "The segments $BC$ and $AD$ are the bases of a trapezoid $ABCD$. Through the point of intersection of its diagonals, a straight line is drawn parallel to the bases. This line intersects the sides $AB$ and $CD$ at points $M$ and $N$ respectively. The segments $DP$ and $AQ$ are the heights of the triangles $DMC$ and $ABN$ respectively. Prove that $AP = DQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай $K$ — спільна точка прямих $AD$ і $MC$. Оскільки, як легко встановити, $\\frac{KA}{BC} = \\frac{AD}{BC}$, то $KA = AD$, а тому $AP = AK = AD$. Аналогічно доводиться, що $DQ = DA$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45061,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find all real numbers $x$ such that\n$$\nx^2 + \\frac{1}{x^2 + 2x} = 1\n$$\n\nb) Find all pairs of positive real numbers $x$ and $y$ such that\n$$\nx^2 + y^2 + \\frac{1}{x^2 + 2x} + \\frac{1}{y^2 + 2y} = 2\n$$",
    "options": [],
    "answer": "a) x = (-1 ± √5)/2; b) x = y = (-1 + √5)/2",
    "solution": "a. Відповідь: $x = \\frac{-1 \\pm \\sqrt{5}}{2}$.\nДане рівняння можна записати у вигляді\n$$\nx^2 + 2x + \\frac{1}{x^2 + 2x} = 2(x + 1) - 1.\n$$\n$$\n\\frac{1}{x^2 + 2x} \\left( (x^2 + 2x) - (x + 1) \\right)^2 = 0.\n$$\n\nb. Відповідь: $x = y = \\frac{-1 + \\sqrt{5}}{2}$.\nМаємо:\n$$\nx^2 + 2x + y^2 + 2y + \\frac{1}{x^2 + 2x} + \\frac{1}{y^2 + 2y} = 2(x + 1) + 2(y + 1) - 2,\n$$\n$$\n\\frac{1}{x^2 + 2x} \\left( (x^2 + 2x) - (x + 1) \\right)^2 + \\frac{1}{y^2 + 2y} \\left( (y^2 + 2y) - (y + 1) \\right)^2 = 0.\n$$\nОтже, оскільки $x > 0$ і $y > 0$, то $x^2 + x - 1 = 0$ і $y^2 + y - 1 = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triplets of real positive numbers $x$, $y$ and $z$ such that\n$$\n\\begin{cases}\n\\sqrt{2x - \\frac{2}{y}} + \\sqrt{2y - \\frac{2}{z}} + \\sqrt{2z - \\frac{2}{x}} = \\sqrt{3(x + y + z)}, \\\\\nx^2 + y^2 + z^2 = 6.\n\\end{cases}\n$$",
    "options": [],
    "answer": "x = y = z = √2",
    "solution": "Відповідь: $x = y = z = \\sqrt{2}$. Із системи випливає, що\n$$\n\\sqrt{x - \\frac{1}{y}} + \\sqrt{y - \\frac{1}{z}} + \\sqrt{z - \\frac{1}{x}} = \\frac{1}{2} \\sqrt{x + y + z} \\cdot \\sqrt{x^2 + y^2 + z^2}.\n$$\nЗвідси за нерівністю Коші-Буняковського маємо:\n$$\n\\sqrt{x - \\frac{1}{y}} + \\sqrt{y - \\frac{1}{z}} + \\sqrt{z - \\frac{1}{x}} \\ge \\frac{1}{2}(x\\sqrt{y} + y\\sqrt{z} + z\\sqrt{x}), \\\\\n\\left(x\\sqrt{y} - 2\\sqrt{x - \\frac{1}{y}}\\right) + \\left(y\\sqrt{z} - 2\\sqrt{y - \\frac{1}{z}}\\right) + \\left(z\\sqrt{x} - 2\\sqrt{z - \\frac{1}{x}}\\right) \\le 0, \\\\\n\\frac{\\left(\\sqrt{xy} - 1\\right)^2}{\\sqrt{y}} + \\frac{\\left(\\sqrt{yz} - 1\\right)^2}{\\sqrt{z}} + \\frac{\\left(\\sqrt{zx} - 1\\right)^2}{\\sqrt{x}} \\le 0.\n$$\n\nОтже, з необхідністю $xy = yz = zx = 2$. Заливається переконатися, що трійка $x = y = z = \\sqrt{2}$ задовольняє умову задачі.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45063,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that for any rational number $\\alpha \\in (0; 1)$ there exists an infinite set of real numbers that satisfy the equation $\\{x[x\\{x\\}]\\} = \\alpha$ and any two of them have the same fractional part.\n\nb) Prove that for any rational number $\\alpha \\in (0; 1)$ there exists an infinite set of real numbers that satisfy the equation $\\{x[x\\{x\\}]\\} = \\alpha$ and any two of them have different fractional parts.\n\n(The fractional part of a real number $a$ is given by $\\{a\\} = a - \\lfloor a \\rfloor$, where $\\lfloor a \\rfloor$ is its integer part, i.e., the greatest integer that does not exceed $a$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $\\alpha = \\frac{p}{q}$, where $p, q \\in \\mathbb{N}$, $p < q$. Consider $y = pq + \\frac{1}{q}$. Then all $x = y + m$, where $m = q^n$, $n \\in \\mathbb{N}$, $n \\ge 2$, have equal fractional parts and satisfy the given equation. Indeed, then we have:\n$$\n\\{x\\} = \\{y\\}, \\quad m\\{y\\} \\in \\mathbb{N}, \\quad my\\{y\\} \\in \\mathbb{N},\n$$\n\n\\begin{align*}\nx\\{x\\} &= (y+m)\\{y\\} = y\\{y\\} + m\\{y\\}, & [x\\{x\\}] &= [y\\{y\\}] + m\\{y\\}, \\\\\nx[x\\{x\\}] &= (y+m)([y\\{y\\}] + m\\{y\\}) = y[y\\{y\\}] + m[y\\{y\\}] + my\\{y\\} + m^2\\{y\\}, & \\\\\n\\{x[x\\{x\\}]\\} &= \\{y[y\\{y\\}]\\} = \\frac{p}{q} = \\alpha.\n\\end{align*}\n$$\n\nb) For $\\alpha = \\frac{p}{q}$, $p, q \\in \\mathbb{N}$, $p < q$, consider $x = pqn^2 + \\frac{1}{qn}$, $n \\in \\mathbb{N}$.\nThen\n$$\n\\begin{align*}\n\\{x\\} &= \\frac{1}{qn}, \\quad x\\{x\\} = \\left( pqn^2 + \\frac{1}{qn} \\right) \\frac{1}{qn} = pn + \\frac{1}{q^2 n^2}, \\quad [x\\{x\\}] = pn, \\\\\nx[x\\{x\\}] &= \\left( pqn^2 + \\frac{1}{qn} \\right) pn = p^2 q n^3 + \\frac{p}{q}, \\quad \\{x[x\\{x\\}]\\} = \\frac{p}{q} = \\alpha.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mykolka and Andriyko are playing the following game. They write positive integers turn by turn thus forming a sequence $a_1, a_2, ..., a_{2006}$ obeying the following restrictions: $a_1 = 1$ (this first turn is fixed, and it is made by Mykolka), and $a_n \\le a_{n+1} \\le 3a_n$ for $1 \\le n \\le 2005$. If, after the last turn was made, the sums $a_1 + a_2 + ... + a_{2004} + a_{2005}$ and $a_1 + a_2 + ... + a_{2005} + a_{2006}$ appear to be mutually prime numbers, then the winner is Andriyko, otherwise it is Mykolka. Who of the players can secure his victory whatever way the other one plays?",
    "options": [],
    "answer": "Andriyko",
    "solution": "Андрійко може забезпечити собі перемогу. Для доведення досить показати, що Андрійко зможе записати число $a_{2006} = M - 1$, де $M = \\sum_{k=1}^{2005} a_k$. Очевидно, що $M - 1 \\ge a_{2005}$. Доведемо, що Андрійко може забезпечити й виконання нерівності $a_{2006} = M - 1 \\le 3a_{2005}$. Нехай Миколка своїм черговим ходом записує число $a_{2k-1}$, тоді Андрійко відповідає записом числа $a_{2k} = 3a_{2k-1}$, $1 \\le k \\le 1002$. Оскільки $a_{2k+1} \\ge a_{2k} = 3a_{2k-1}$, то, додавши нерівності $a_3 \\ge 3a_1$, $a_5 \\ge 3a_3$, ..., $a_{2005} \\ge 3a_{2003}$, будемо мати, що $a_{2005} \\ge 1 + 2A$, де $A = a_1 + a_3 + ... + a_{2003}$. Таким чином,\n\n$$\n\\begin{aligned}\nM &= a_1 + 3a_1 + a_3 + 3a_3 + \\dots + a_{2003} + 3a_{2003} + a_{2005} = \\\\\n&= 4A + a_{2005} \\le 2(a_{2005} - 1) + a_{2005} = 3a_{2005} - 2 < 3a_{2005}.\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\left( \\frac{a+b}{c^2} + \\frac{b+c}{a^2} + \\frac{c+a}{b^2} \\right) \\cdot \\left( \\frac{a^2}{b+c} + \\frac{b^2}{c+a} + \\frac{c^2}{a+b} \\right) \\ge 3 + \\frac{a+b}{c} + \\frac{b+c}{a} + \\frac{c+a}{b}\n$$\nfor any positive real numbers $a$, $b$ and $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to prove the inequalities\n$$\n\\frac{a^2}{b+c} \\geq a - \\frac{b+c}{4}, \\quad \\frac{b^2}{c+a} \\geq b - \\frac{c+a}{4}, \\quad \\frac{c^2}{a+b} \\geq c - \\frac{a+b}{4},\n$$\nadding which we obtain\n$$\n\\frac{a^2}{b+c} + \\frac{b^2}{c+a} + \\frac{c^2}{a+b} \\ge \\frac{a+b+c}{2}. \\quad (1)\n$$\n\nSince, as can be easily verified, $\\frac{x}{y^2} + \\frac{y}{x^2} \\ge \\frac{1}{x} + \\frac{1}{y}$ for $x>0, y>0$, then\n$$\n\\begin{align*}\n\\frac{a+b}{c^2} + \\frac{b+c}{a^2} + \\frac{c+a}{b^2} &= \\left( \\frac{a}{b^2} + \\frac{b}{a^2} \\right) + \\left( \\frac{b}{c^2} + \\frac{c}{b^2} \\right) + \\left( \\frac{c}{a^2} + \\frac{a}{c^2} \\right) \n\\\\\n&\\ge \\left( \\frac{1}{a} + \\frac{1}{b} \\right) + \\left( \\frac{1}{b} + \\frac{1}{c} \\right) + \\left( \\frac{1}{c} + \\frac{1}{a} \\right) = 2 \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n\\end{align*}\n$$\nThat is,\n$$\n\\frac{a+b}{c^2} + \\frac{b+c}{a^2} + \\frac{c+a}{b^2} \\ge 2 \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right). \\quad (2)\n$$\nIt remains to multiply inequalities (1) and (2).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a$, $100 \\le a \\le 999$, such that the decimal values of $a^2$ and $(3a - 2)^2$ have the same three-digits endings.",
    "options": [],
    "answer": "[251, 313, 501, 563, 751, 813]",
    "solution": "Відповідь: $251$, $313$, $501$, $563$, $751$, $813$.\n\nМаємо: $(3a-2)^2 - a^2 = 1000k$.\nТобто, $(2a-1)(a-1) = 250k = 2 \\cdot 5^3 \\cdot k$.\nЧисла $2a-1$ і $a-1$ взаємно прості, причому перше з них непарне. Отже, друге число парне, і, відповідно, число $a$ непарне. Звідси випливає, що $a-1=250$, або ж $2a-1=125$. Таким чином, непарне число $a$ з проміжку $100 \\le a \\le 999$ задовольняє умову задачі тоді й тільки тоді, коли $a \\in \\{251; 501; 751\\}$ або $2a-1 \\in \\{375; 625; 875; 1125; 1375; 1625; 1875\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $\\omega_1$ and $\\omega_2$ intersect each other at two distinct points $A$ and $B$. The tangent line of the circle $\\omega_1$ at the point $A$ and the tangent line of the circle $\\omega_2$ at the point $B$ meet at point $C$. The first one of these two lines intersects the circle $\\omega_2$ for the second time at point $T$ (distinct from $A$). A point $X$ (distinct from $A$ and $B$) belongs to the circle $\\omega_1$, and the straight line $XA$ intersects the circle $\\omega_2$ for the second time at point $Y$ (distinct from $A$). The straight lines $YB$ and $XC$ meet at point $Z$. Prove that $TZ$ is parallel to $XY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай точка $X$ розташована на колі $\\omega_1$ так, як показано на рисунку (інші випадки її розташування на цьому колі розглядаються аналогічно). Оскільки, як нескладно помітити, $\\angle CBT = \\angle BAT = \\angle BDA$, то $BT \\parallel AD$. Далі, $\\angle AYB = \\angle ABD = 180^\\circ - \\angle DXA$. Отже, $BY \\parallel DX$. Тепер розглянемо гомотетію з центром у точці $C$ і коефіцієнтом $k = CB/CD$.\n\nПри такій гомотетії пряма $DX$ перейде в паралельну її пряму $BZ$, а образом точки $X$ буде точка $Z$. Пряма $DA$ перейде в паралельну її пряму $BT$, а тому точка $A$ перейде в точку $T$. Таким чином, $AX \\parallel TZ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45068,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $S$ lies on the side $PM$ of a trapezoid $MPQ$ with bases $PM$ and $RQ$ ($PQ < PM$, and $S$ is distinct from the vertices). The bisectors of the angles $MSQ$ and $MPQ$ meet at point $O$. It is known that the segment $OI$, where $I$ is the incenter of the triangle $PQR$, is parallel to the bases of the trapezoid. Prove that $SR = OI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай $\\angle SPO = \\alpha$. Очевидно, що $\\angle SPO = \\angle QPO = \\angle PQI = \\angle RQI = \\angle POI$.\nЗокрема, оскільки $\\angle PQI = \\angle POI$, то чотирикутник $POQI$ циклічний, причому,\nз урахуванням паралельності прямих $PO$ і $QI$, — рівнобічна трапеція. Таким\nчином, $OI = PQ$. Якщо $\\angle QOI = \\beta$, то $\\angle PQI = \\beta$, $\\angle OQP = \\angle OIP = 180^\\circ - 2\\alpha - \\beta$.\nТочка $O$ є центром зовнівписаного кола трикутника $SPQ$, звідки\n$$\n\\angle OQS = \\beta + 2\\alpha, \\quad \\angle SQR = 180^\\circ - (\\beta + 2\\alpha) = 180^\\circ - \\angle SPR.\n$$\nОтже, чотирикутник $SQRP$ є циклічним. Враховуючи, що $SP \\parallel QR$, дістаємо рівність $PQ = SR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any real numbers $x$ and $y$ the inequality\n$$\n|\\cos x| + |\\cos y| + |\\cos(x+y)| \\ge 1\n$$\nholds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{align*}\n& |\\cos x| + |\\cos y| + |\\cos(x+y)| \\ge |\\cos x \\sin y| + |\\cos y \\sin x| + \\cos^2(x+y) \\\\ \n& \\ge |\\sin(x+y)| + \\cos^2(x+y) \\ge \\sin^2(x+y) + \\cos^2(x+y) = 1.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any real numbers $x$ and $y$, both taken on the segment $[0; \\frac{\\pi}{2}]$, the inequality\n$$\n\\cos x + \\cos y + |\\cos(x + y)| \\ge 1\n$$\nholds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Для $x, y \\in [0; \\pi/2]$ маємо:\n$$\n\\cos x + \\cos y + |\\cos(x+y)| \\geq \\cos x \\sin y + \\cos y \\sin x + \\cos^2(x+y) = \\\\\n= \\sin(x+y) + \\cos^2(x+y) \\geq \\sin^2(x+y) + \\cos^2(x+y) = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45071,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $M$ lies on the side $BC$ of an equilateral triangle $ABC$ ($M$ is distinct from the vertices). A point $N$ is chosen in such a way that the triangle $BMN$ is also equilateral, and the points $A$ and $N$ belong to the different half-planes with respect to the straight line $BC$. The points $P$, $Q$ and $R$ are the midpoints of the segments $AB$, $BN$ and $CM$ respectively. Prove that the triangle $PQR$ is equilateral as well.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай $S$ — така точка, що чотирикутник $PBQS$ є паралелограмом, $L$ — точка перетину прямих $QS$ і $BM$, $F$ — середина $BC$. Тоді нескладно довести, що $PS = QL = FR$, $PF = QS = LR$, $\\angle PSQ = \\angle QLR = \\angle RFP$, а тому трикутники $PSQ$, $QLR$, $RFP$ є рівними. Звідси й випливає твердження задачі.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that positive real numbers $a_1, a_2, \\dots, a_{2006}$ satisfy the equality\n$$\n\\frac{a_1}{(a_1+a_2)a_2} + \\frac{a_2}{(a_2+a_3)a_3} + \\dots + \\frac{a_{2005}}{(a_{2005}+a_{2006})a_{2006}} + \\frac{a_{2006}}{(a_{2006}+a_1)a_1} = 2006\n$$\nFind the value of the expression\n$$\n\\frac{a_1}{a_{2006}(a_{2006} + a_1)} + \\frac{a_2}{a_1(a_1 + a_2)} + \\dots + \\frac{a_{2005}}{a_{2004}(a_{2004} + a_{2005})} + \\frac{a_{2006}}{a_{2005}(a_{2005} + a_{2006})}\n$$",
    "options": [],
    "answer": "2006",
    "solution": "Відповідь: 2006.\n\nПомітимо, що\n$$\n\\frac{a_i}{(a_i + a_{i+1})a_{i+1}} = \\frac{1}{a_{i+1}} - \\frac{1}{a_i + a_{i+1}}, \\quad \\frac{a_{i+1}}{a_i(a_i + a_{i+1})} = \\frac{1}{a_i} - \\frac{1}{a_i + a_{i+1}}\n$$\nде $i=1,2006$, $a_{2007} = a_1$. Додаванням цих рівностей одержимо, що значення сум, про які йдеться в умові задачі, будуть однаковими.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mariyka has drawn a square grid $2006 \\times 2006$ on a blackboard. During one step, it is allowed to choose any unit segment of that grid and erase it together with all adjacent segments, which were not erased before (so, at most $7$ segments can be erased in one step). Is it possible to erase the whole drawing in no more than $1300000$ steps?",
    "options": [],
    "answer": "No",
    "solution": "Відповідь: ні, не можна. Задача полягає в тому, чи можемо ми покрити всю \"сітку\" таблиці $1300000$ фігурками, зображеними на рисунку. Якщо така фігурка не лежить повністю в таблиці, то вона покриває не більше $6$ відрізків. Якщо фігурка повністю знаходиться в таблиці, тоді інші фігурки повинні покрити відрізки $BC$ і $EF$ (див. рисунок). Фігурка, яка закриє $BC$, обов'язково закриє й хоча б один з відрізків $BG$, $CH$ (це легко перевіряється невеличким перебором). Аналогічно, фігурка, яка закриє $EF$, обов'язково закриє й принаймні один з відрізків $FG$, $EH$. Кожному відрізку фігурок, що покривають відрізок таблиці без перекриття з іншими відрізками, поставимо у відповідність коефіцієнт $1$, кожному з відрізків, що покривають з перекриттям, — коефіцієнт $1/2$. Тоді сума всіх коефіцієнтів для однієї фігурки не більша за $6$, по всіх фігурках — не перевищує $7800000$.\n\nЗ іншого боку, по всій покритій таблиці сума коефіцієнтів має бути не меншою за кількість відрізків: $2 \\cdot 2006 \\cdot 2007$. Але $2 \\cdot 2006 \\cdot 2007 > 7800000$, і тому всю \"сітку\" таблиці покрити $1300000$ фігурками не можна.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x^3 + y^3) = x^2 f(x) + y f(y^2)\n$$",
    "options": [],
    "answer": "All functions of the form f(x) = k x for arbitrary real constant k.",
    "solution": "Відповідь: $f(x) = kx$, $k \\in \\mathbb{R}$ — довільне.\n\nЗ вихідного рівняння дістаємо, що $f(y^3) = y f(y^2)$, $f(x^3) = x^2 f(x)$. А тому при всіх $x, y \\in \\mathbb{R}$\n$$\nf(x^3 + y^3) = f(x^3) + f(y^3).\n$$\nОтже, ми встановили, що функція $f$ є адитивною: при всіх $u, v \\in \\mathbb{R}$\n$$\nf(u+v) = f(u) + f(v).\n$$\nДалі, оскільки\n$$\nf(0) = 0 \\quad \\text{і} \\quad f(x^3) \\equiv x^2 f(x) \\equiv x f(x^2), \\quad \\text{то} \\quad f(x^2) \\equiv x f(x). \\quad \\text{Таким чином,}\n$$\n$$\nf(x^2 + 2x + 1) \\equiv (x+1) f(x+1) \\equiv (x+1)(f(x) + f(1)).\n$$\nЗ іншого боку,\n$$\nf(x^2 + 2x + 1) \\equiv x f(x) + 2 f(x) + f(1).\n$$\nЗ цих співвідношень дістанемо, що $f(x) = f(1)x$, $x \\in \\mathbb{R}$. Залишається перевіркою переконатися, що всі функції вигляду $f(x) = kx$, $k \\in \\mathbb{R}$, задовольняють умову.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $m$ and $n$ such that\n$$\n14^n = 13 \\cdot m^n + 1.\n$$",
    "options": [],
    "answer": "m=1, n=1",
    "solution": "Відповідь: $m = n = 1$. При $n \\geq 2$ ліва частина рівняння ділиться без остачі на $4$, а тому $m'' = 3 \\pmod{4}$, що можливо лише для непарних $m$ і $n$. При непарному $n$ маємо, що $14'' = 2 \\pmod{3}$, а тому $m = 1 \\pmod{3}$. Оскільки $m < 14$, то $m \\in \\{1; 7; 13\\}$. Але при $m = 7$ права частина рівності не ділиться без остачі на $7$, а ліва ділиться. При $m = 13$ маємо рівність $14'' = 13'' + 1 = 1$. Але при непарному $n$ число $14''$ закінчується цифрою $4$, а число $13'' + 1 = 1$ — або нулем, або двійкою. Отже, $n = 1$, звідки $m = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eleven linguists were instructed to learn eleven foreign languages (initially, none of the linguists knew any of those languages). It became necessary to invite a Foreign Consultant who is able to teach (by means of hypnosis, of course!) any two linguists any two languages during one session (so that each one of those two linguists learns each one of those two languages). What is the minimal number of sessions required to teach all the eleven linguists all the eleven languages (a linguist may attend a session even if he has learnt one of the appropriate languages already)?",
    "options": [],
    "answer": "33",
    "solution": "Кожен лінгвіст має взяти участь щонайменше в $6$ сеансах, інакше він не оволодіє всіма $11$ мовами. Оскільки в одному сеансі беруть участь два лінгвісти, то кількість сеансів не менша за $33 = \\frac{6 \\cdot 11}{2}$. Покажемо, що $33$ сеансів Консультантові насправді вистачить. Кожний сеанс будемо зображати у вигляді $(a,b|c,d)$, де $a$ і $b$ — номери лінгвістів, $c$ і $d$ — номери мов $(1 \\leq a,b,c,d \\leq 11)$. Для $1 \\leq k \\leq 5$ $(2k-1)$-го й $(2k)$-го лінгвістів запрошуємо на сеанси $(2k-1,2k|2l,2l+1)$, де $k \\leq l \\leq 5$. Маємо вже $5+4+3+2+1=15$ сеансів. Далі, для кожного з цих сеансів розглянемо \"доповняльний\" для сеансу $(a,b|c,d)$ розглянемо сеанс $(12-a,12-b|12-c,12-d)$. Після цих $30$ сеансів лінгвісти з парними номерами опанують усі $11$ мов, а лінгвісти з непарними номерами опанують усі мови, номер яких не співпадає з номером відповідного лінгвіста. Отже, потрібні ще такі сеанси $(4m-3,4m-1|4m-3,4m-1)$, $1 \\leq m \\leq 3$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45077,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex pentagon $ABCDE$ is inscribed into the circle $\\omega$. The diagonal $AD$ is a diameter of that circle. The diagonals $BE$ and $AC$ are perpendicular to each other. The diagonals $CE$ and $AD$ meet at point $P$. Prove that the area of the triangle $APE$ is equal to the sum of the areas of the triangles $ABC$ and $CDP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem will be solved if we prove that the area of triangle $ACE$ is equal to the area of quadrilateral $ABCD$.\n\nLet $K$ be the intersection point of segments $AC$ and $BE$. Since $S(ACE) = \\frac{1}{2} AC \\cdot KE$, $S(ABCD) = \\frac{1}{2} AC \\cdot BK + \\frac{1}{2} AC \\cdot CD$, it remains to show that $KE = BK + CD$.\n\nLet point $H$ on diagonal $BE$ be symmetric to point $B$ with respect to line $AC$. Then, as is known, $H$ is the orthocenter of triangle $ACE$, and therefore $CH \\perp AE$, $DE \\perp AE$, $CH \\parallel DE$, $CD \\perp AC$, $EH \\perp AC$, $EH \\parallel CD$. Thus, quadrilateral $CDEH$ is a parallelogram. Hence, $CD = EH$, and we have $KE = KH + HE = BK + CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45078,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $M$ lies inside a cube $ABCD-A_1B_1C_1D_1$. Points $A', B', C', D', A'_1, B'_1, C'_1$ and $D'_1$ belong to the half-lines $MA, MB, MC, MD, MA_1, MB_1, MC_1$ and $MD_1$ respectively. Prove that if the polyhedron $A'B'C'D'A'_1B'_1C'_1D'_1$ is a parallelepiped (i.e., all its faces are parallelograms), then in fact it is a cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Доведемо спочатку, що якщо основою чотирикутної піраміди $SABCD$ є квадрат $ABCD$, і точки $A_0, B_0, C_0, D_0$ належать променям $SA, SB, SC, SD$ відповідно, причому $A_0B_0C_0D_0$ є паралелограмом, то $A_0B_0C_0D_0$ — квадрат. Нехай $l = (SCD) \\cap (SAB)$ і $m = (SAD) \\cap (SBC)$. За відомою лемою про три паралельні, $CD \\parallel BA \\parallel l$, $BC \\parallel AD \\parallel m$, $C_0D_0 \\parallel B_0A_0 \\parallel l$, $B_0C_0, A_0D_0 \\parallel m$. Отже, $(ABCD) \\parallel (A_0B_0C_0D_0)$, а тому, як легко зрозуміти (наприклад, з міркувань перетворення гомотетії), $A_0B_0C_0D_0$ — квадрат. З цього допоміжного твердження випливає, що всі грані нашого паралелепіпеда — квадрати. Таким чином, цей паралелепіпед $A'B'C'D'A'_1B'_1C'_1D'_1$ насправді є кубом.\n\nНаведемо й один з можливих векторних способів розв'язування цієї задачі. Нехай $MABCD$ — чотиригранний кут з вершиною $M$ і неколінеарними ребрами $MA, MB, MC$ і $MD$. Досить показати, що якщо два його перерізи $A_1B_1C_1D_1$ і $A_2B_2C_2D_2$ ($A_1, A_2 \\in MA$, $B_1, B_2 \\in MB$, $C_1, C_2 \\in MC$, $D_1, D_2 \\in MD$) є паралелограмами, то ці паралелограми подібні. Розглянемо в просторі гомотетію з центром у точці $M$ і коефіцієнтом $k = MA_1/MA_2$. Нехай образом паралелограма $A_2B_2C_2D_2$ при такій гомотетії буде паралелограм $A_1B_3C_3D_3$. Доведемо, що $B_1 = B_3$, $C_1 = C_3$, $D_1 = D_3$. Оскільки $A_1B_3 = D_3C_3$, то $MA_1 = MB_3 - MC_3 + MD_3$. Нехай $u, v, w$ — напрямні орти променів $MB, MC, MD$ відповідно. Тоді для деяких дійсних чисел $x, y$ і $z$ $MB_3 = x \\cdot u$, $MC_3 = y \\cdot v$, $MD_3 = z \\cdot w$, звідки $MA_1 = x \\cdot u - y \\cdot v + z \\cdot w$. Оскільки $A_1B_1 = D_1C_1$, то $MA_1 = MB_1 - MC_1 + MD_1$, а тому існують такі $x_0, y_0$ і $z_0$, що $MB_1 = x_0 \\cdot u$, $MC_1 = y_0 \\cdot v$, $MD_1 = z_0 \\cdot w$, $MA_1 = x_0 \\cdot u - y_0 \\cdot v + z_0 \\cdot w$. Враховуючи, що трійка векторів $u, v, w$ утворює базис у просторі, дістаємо, що $x = x_0$, $y = y_0$, $z = z_0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45079,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\ge 3$ soldiers in the captain Petrenko's squad, and none two of them have the same height. The captain has drawn them all up into a single rank (not necessarily sorted by height). We call a \"wave\" any subsequence of soldiers in this rank (they are not supposed to stand next to each other) such that the first (leftmost) soldier in the wave is higher than the second soldier in it, but the second soldier in it is lower than the third one, who is in turn higher than the fourth one, and so on. (For example, if $n=9$, the soldiers are enumerated in ascending order by height, and the captain aligned them as 9-3-5-7-1-2-6-4-8, then one of the longest waves for this rank is 9-3-7-1-6-4-8. However, if the captain aligns them as 1-2-3-4-5-6-7-8-9, then each longest wave will consist of a single soldier, who can be anyone.) For every $n$, consider the number of possible ranks with the longest waves of even lengths and the number of possible ranks with the longest waves of odd lengths. Which of these numbers is bigger?",
    "options": [],
    "answer": "They are equal; the numbers are the same for every n.",
    "solution": "Спочатку покажемо, що одна з найдовших \"хвиль\" шеренги містить найвищого солдата $A$ цієї шеренги.\nДійсно, нехай солдат $A$ не входить до жодної найдовшої \"хвилі\". Якщо вся найдовша \"хвиля\" знаходиться справа від $A$, то першого солдата \"хвилі\" замінюємо на $A$. Якщо вся \"хвиля\" знаходиться зліва від $A$, то \"хвиля\" має закінчуватися на підйом (інакше, додавши $A$, ми зробили б \"хвилю\" довшою), і ми останнього солдата \"хвилі\" замінюємо на $A$. Розглянемо тепер двох солдатів $B$ і $C$ найдовшої \"хвилі\", які найближче стоять до $A$ по різні боки від цього. Тоді вищого в парі $B$ і $C$ ми можемо замінити на $A$, і зрозуміло, що при цьому довжина \"хвилі\" не зміниться.\n\nДалі розглядаються тільки такі найдовші “хвилі”, котрі містять солдата $A$. Серед усіх солдатів шеренги, що стоять справа від $A$ (якщо такі існують), оберемо найнижчого. Легко довести, що існує найдовша “хвиля” нашої шеренги, яка містить цього солдата. Серед тих, хто стоїть в шерензі після нього, ми обираємо найвищого. Аналогічно, можемо вважати, що одна з найдовших “хвиль” шеренги містить і цього солдата. Продовжуючи описаний процес, дістанемося такою “хвилею” найбільшої довжини до, принаймні, одного з двох останніх солдатів шеренги. Нехай, наприклад, він був обраний як найвищий у черговому “хвості” шеренги (випадок, коли його обирали як найнижчого в такому “хвості”, розглядається аналогічно). Позначимо цього солдата через $S$.\nЯкщо $S$ стоїть на передостанньому місці шеренги, то “хвиля” дістанеться до кінця шеренги, причому буде закінчуватися на “спад”, а тому — міститиме парну кількість солдатів. Якщо ж $S$ стоїть на останньому місці шеренги, то найдовша “хвиля” шеренги ним закінчується, причому — на “підйом”, тобто, містить непарну кількість солдатів.\nЗвідси випливає, що коли ми поміняємо місцями останніх двох солдатів шеренги, парність найдовшої “хвилі” шеренги зміниться.\nДля кожного $n$ ($n \\ge 3$) існує рівно $n!$ різних способів вишикувати $n$ солдатів у шеренгу. Тому всі $n!$ перестановок солдатів можна розбити на такі пари, що шеренги однієї пари відрізняються розташуванням лише останніх двох солдатів. За доведеним, найдовша “хвиля” однієї шеренги пари буде складатися з парної кількості солдатів, а найдовша “хвиля” другої шеренги пари — з непарної. Це й завершує розв’язання задачі.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45080,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $D$ is chosen on the side $BC$ of an acute triangle $ABC$ (and is distinct from the vertices). Points $P$ and $Q$ are defined as the centers of the circumcircles of the triangles $ABD$ and $ACD$ respectively. Prove that, as soon as the triangle $ABC$ is fixed, there exists a point in the plane, which is different from $A$ and belongs to the circumcircles of all the possible triangles $APQ$ (generated by the different positions of the point $D$).",
    "options": [],
    "answer": "The fixed point is the circumcenter of triangle ABC.",
    "solution": "We claim that this point is the circumcenter $O$ of triangle $ABC$.\n\nIf $AD$ is an altitude, the statement is obvious. We may assume that triangle $ADC$ is acute-angled and triangle $ADB$ is obtuse-angled. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$ respectively. Since $\\angle ADC > \\angle ABC$, as can be easily shown, $\\angle NAQ < \\angle NAO$, and the point $O$ lies inside the angle $QAM$. Then $\\angle AQN = \\angle ADC$, $\\angle APM = 180^\\circ - \\angle ADB = \\angle ADC$. It follows that the quadruple of points $A$, $P$, $O$, $Q$ is concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45081,
    "subject": "Mathematics (Multi-modal)",
    "question": "We know that $|x+y|+|x-y|=1$. Find the least and the greatest value of expression $x^2 - 6x + y^2 - 6y$.",
    "options": [],
    "answer": "Least value is -11/2; greatest value is 13/2.",
    "solution": "Expression $f = x^2 - 6x + y^2 - 6y = (x-3)^2 + (y-3)^2 - 18 = R^2 - 18$ (fig.6) is at maximum (minimum) if expression $R^2 = (x-3)^2 + (y-3)^2$ being an equation of the circle of radius $R$ with the center at point $(3,3)$ is at maximum (minimum). The graph of equation $|x+y|+|x-y|=1$ is a square formed by lines $x = \\pm \\frac{1}{2}$, $y = \\pm \\frac{1}{2}$, see fig.6.\n\n![](attached_image_1.png)\n\nAmong all the circles intersecting the square, the circle passing through the point $B(-\\frac{1}{2}, -\\frac{1}{2})$ has the maximal radius, and the circle passing through the point $A(\\frac{1}{2}, \\frac{1}{2})$ has the minimal radius. Thus we find $R^2 = (\\frac{7}{2})^2 + (\\frac{7}{2})^2 = \\frac{49}{2}$ and $f_{max} = \\frac{49}{2} - 18 = \\frac{13}{2}$ for maximum, and $R^2 = (\\frac{5}{2})^2 + (\\frac{5}{2})^2 = \\frac{25}{2}$, $f_{min} = \\frac{25}{2} - 18 = -\\frac{11}{2}$ for minimum.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Side $BC$ of the parallelogram $ABCD$ is extended beyond the point $C$ and point $K$ is placed on the extension so that $\\triangle CDK$ is isosceles triangle with base $CK$. Side $DC$ is extended beyond the point $C$ and point $L$ is placed on the extension so that $CBL$ is isosceles triangle with base $CL$. Bisectors of angles $\\angle LBC$ and $\\angle CDK$ intersect at point $Q$. Find radius of a circle circumscribed about triangle $ALK$, if $\\angle BQD = \\alpha$ and $KL = a$.\n\n**Answer:** $\\frac{a}{2\\sin 2\\alpha}$.",
    "options": [],
    "answer": "a/(2 sin 2 alpha)",
    "solution": "Since trapezium $ABKD$ is equilateral, points $A, B, K, D$ are on one circle. Thereafter, points $B, L, D, A$ reside on one circle too. Thus all the five points $B, L, D, A, K$ are on the same circle $w$ circumscribed about the triangle $\\triangle ALK$. Since trapeziums $ABKD$ and $BLDA$ share one of the diagonals, their diagonals have equal length. Therefore, $AL = AK$ and $\\triangle ALK$ is isosceles.\n\nAs $BQ$ is a bisector of the isosceles triangle $\\triangle LBC$ (fig.7), $BQ \\perp LC$ and $LC \\parallel AB$. Therefore, $BQ \\perp AB$ and $\\angle QBA = 90^\\circ$, similarly $\\angle QDA = 90^\\circ$. This implies that points $Q, B, D, A$ are on one circle, and this circle is $w$. As angles $\\angle BQD$ and $\\angle ALK$ rest upon chords of equal length: $BD = AK$, $\\angle ALK = \\alpha$ and $\\angle LAK = 180^\\circ - 2\\alpha$. According to the law of sines, $R = \\frac{LK}{2\\sin \\angle LAK} = \\frac{a}{2\\sin 2\\alpha}$ for $\\triangle ALK$.\n\n![](attached_image_1.png)\nFig.7",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's examine all the possible sequences $a_1, a_2, ..., a_{2008}$ of non-negative integers where $a_1 \\le a_2 \\le ... \\le a_{2008}$ and $a_k \\le (k-1)$ for all values of $k$ from $1$ to $2008$. Prove that the number of such sequences exceeds\n\na) $2^{2007}$;\nb) $2^{2008}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nTo begin, we form $2^{2007}$ different sequences satisfying the given conditions. For all the sequences let $a_1 = 0$. Let each successive member of the sequence be \"by 0 greater\" or \"by 1 greater\" than the preceding one. Thus we obtain $2^{2007}$ different sequences. It's easy to show that $a_k \\le k-1$ for each of these sequences. To prove that the number of sequences is greater than $2^{2007}$, one sequence will be enough, which is not in the list of $2^{2007}$ sequences, but satisfies the given conditions. E.g., it can be a sequence: $(0,0,2,2,...,2)$.\n\nb.\nLet's name some random non-decreasing sequence of non-negative integers $a_1 \\le a_2 \\le ... \\le a_n$ obeying $a_k \\le (k-1)$, $k=1, n$ as fine. Note that a random fine sequence with $(n+1)$ members can be obtained from fine sequence with $n$ members if you add a member $a_{n+1}$ for which $a_n \\le a_{n+1} \\le n$ is true. Similarly, if we remove the last member of some sequence with $(n+1)$ members, we obtain fine sequence with $n$ members.\n\nLet the number of fine sequences with $n$ members be $x_n$. We can obviously build at least two different fine sequences with $(n+1)$ members if we assign the two allowable possibilities for $a_{n+1}$: they will be $a_n$ or $(a_n + 1)$. Similarly, proceeding analyzing, we find $x_{n+m} \\ge 2^m x_n$.\nNow you only have to find out the number of sequences for a small value of $n$. $x_1 = 1$ as there is only one such a sequence: $(0)$. Again you'll easily find that $x_2 = 2$ because only sequences $(0,0)$ and $(0,1)$ can be regarded as fine. Thus we have an assessment $x_{2008} \\ge 2^{2006} x_2 = 2^{2007}$, but it is not enough. To obtain a sequence with 3 members, we may add one of the numbers $0,1,2$ to the sequence $(0,0)$, and one of the numbers $0,1$ to the sequence $(0,1)$. I.e., $x_3 = 5$ and $x_{2008} \\ge 2^{2005} x_3 = 5 \\cdot 2^{2005} > 2^{2007}$.\nAfter that we continue calculating the number of fine sequences for small values of $n$: $(0,0,0) \\rightarrow 0,1,2,3$, $(0,0,1), (0,1,1) \\rightarrow 1,2,3$, $(0,0,2), (0,1,2) \\rightarrow 2,3$. I.e. in all we obtain 14 different fine sequences and therefore $x_4 = 14$ and $x_{2008} \\ge 2^{2004} x_4 = 14 \\cdot 2^{2004} > 5 \\cdot 2^{2005}$.\nAs one of these fine sequences ends in $0$, we can add $0,1,2,3,4$. As three of the sequences end in $1$, we can add $1,2,3,4$. As five of the sequences end in $2$, we can add $2,3,4$: as another five end in $3$, we can add $3,4$. So in all we have $5 \\cdot 1 + 3 \\cdot 4 + 5 \\cdot 3 + 5 \\cdot 2 = 42$, which implies the first proven assessment: $x_{2008} \\ge 2^{2003} x_5 = 42 \\cdot 2^{2003} > 32 \\cdot 2^{2003} = 2^{2008}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's consider all the increasing geometric progressions and select only those that have the maximal number $M$ of elements in the set $A = \\{1,2,3,...,2008\\}$. Find $M$.",
    "options": [],
    "answer": "11",
    "solution": "Let one of the progressions $\\{a_n\\}$ which has the required maximal number $M$ of common members have the first member $a_0$ and denominator $q_0 > 1$. The progression has some common elements with set $A$. Let $n$ be the least number in $A$, which can also be found in the progression. Then you can define a new progression with the first member $b_0 = n$ and the same denominator $q_0$. It has as many members in the set $A$ as progression $\\{a_n\\}$ and therefore we can take it as an initial one. Let $m$ be the next common member of the set $A$ that also belongs to the progression, i.e. in a new progression $b_0 = n$, $b_k = b_0 q_0^k = n q_0^k = m \\Rightarrow q_0^k = \\frac{m}{n} = \\frac{u}{v} \\in \\mathbb{Q}$ where $\\frac{u}{v}$ is an irreducible fraction and $n \\le \\frac{u}{v} = m$.\n\nLet's show that $q_0^i \\notin \\mathbb{Q}$ if $1 \\le i < k$. By contradiction, let $q_0^i = \\frac{s}{t}$, which is also an irreducible fraction. In this case $\\left(\\frac{u}{v}\\right) = q_0^{ki} = \\left(\\frac{s}{t}\\right)^k$. Then $u^i t^k = s^k v^i$. This implies that $u^i = s^k$ and $v^i = t^k$, since $(u,v)=1 \\Rightarrow t^k \\nmid v^i$ and vice versa $v^i \\nmid t^k$. As $i < k$, $v > t$, and $b_0 = n \\nmid v \\Rightarrow b_0 = n \\nmid t$. Therefore $b_i = b_0 q_0^i = n \\frac{s}{t} \\in \\mathbb{N}$, which contradicts the following condition: the first natural number after $n$ will be $b_k = b_0 q_0^k = n \\frac{u}{v} = m$.\n\nIt follows that the rational members of the chosen geometric progression may only be members of such a set as $b_0, b_0 q_0^k, b_0 q_0^{2k}, \\dots$. Therefore we can choose $q = q_0^k \\in \\mathbb{Q}$ and consider the progression $(b_0, q)$ with the first integer member $b_0 = n$ and rational denominator $q$ where $q = \\frac{p}{r}$ is an irreducible fraction.\n\nThe progression $(1,2)$ obviously has the maximum members in $A$ among all the progressions with integer denominators. They share 11 common elements. All these elements are powers of two: $1, 2, 4, \\dots, 1024$. Let there be a progression with a rational exponent, which has more elements in $A$. Let it be progression $(n, \\frac{p}{r})$. Thus, if $c_k = n\\left(\\frac{p}{r}\\right)^k$ is an integer, i.e. $n : r^k$, all the preceding values of $c_i$ are integers too, $i = 0, k$. Therefore, if the progression has at least 12 elements in $A$, then $c_{12} = n\\left(\\frac{p}{r}\\right)^{11}$ should be an integer, which implies that $n : r^{11}$. As $\\frac{p}{r} \\notin \\mathbb{N}$, $r \\ge 2$. Therefore, $r^{11} \\ge 2048$ and $n \\ge 2048$. We've finished proving the given condition (An increasing geometric progression cannot have more than 11 members in the set $A$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Take the chequered board $10 \\times 10$ and consider all the possible patterns of painting 10 sectors so that there is only one painted sector in every row and every column. For each pattern we find rectangle of maximal area the sides of which are on the lines of the grid. Besides, this rectangle must have no painted sectors. What can be the maximal area of such a rectangle?\n\n**Answer:** 25.",
    "options": [],
    "answer": "25",
    "solution": "Let's consider rectangles with their sides along the grid lines.\nOn the board let there be a rectangle with dimensions $A \\times B$, which has no painted sectors ($A$ is its width, $B$ is its height). According to the problem statement $A$ columns which the rectangle occupies must have $A$ painted sectors. On the other hand, these $A$ sectors can not be in those $B$ rows that the rectangle occupies. The rest is $(10-B)$ rows. The obligatory condition is $10-B \\ge A$ (otherwise we are not able to place $A$ painted sectors in $(10-B)$ rows, one at a row). $A+B \\le 10$ (you'll obtain the same result if you consider $B$ rows). Thus rectangle $5 \\times 5$ has the maximal area (this area $A(10-A) \\le (\\frac{A+10-A}{2})^2 = 25$ is represented in the Cauchy inequality or as a parabola with its branches down). We can easily draw an example of the board with such a rectangle (fig.3).\n\n![](attached_image_1.png)\nFig.3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove inequality $\\frac{a^2}{bc(a^2+b^2)} + \\frac{b^2}{ca(b^2+c^2)} + \\frac{c^2}{ab(c^2+a^2)} \\ge \\frac{9}{2}$ for random positive numbers $a, b, c$ obeying equality $ab + bc + ca = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's consider the secondary inequality $\\frac{a^3}{a^2+b^2} \\ge a - \\frac{b}{2}$ which is proved by means of rearrangement: $2a^3 \\ge (a^2+b^2)(2a-b) = 2a^3 + 2ab^2 - a^2b - b^3 \\Leftrightarrow b(a-b)^2 \\ge 0$.\n\n$$\n\\frac{a^2}{bc(a^2+b^2)} + \\frac{b^2}{ca(b^2+c^2)} + \\frac{c^2}{ab(c^2+a^2)} = \\frac{1}{abc} \\left( \\frac{a^3}{a^2+b^2} + \\frac{b^3}{b^2+c^2} + \\frac{c^3}{c^2+a^2} \\right) \\ge \\frac{1}{abc} \\left( a - \\frac{b}{2} + b - \\frac{c}{2} + c - \\frac{a}{2} \\right) = \\frac{a+b+c}{2abc} = \\frac{(a+b+c)(ab+bc+ca)}{2abc} \\ge \\frac{3\\sqrt{abc} \\cdot \\sqrt[3]{a^2b^2c^2}}{2abc} \\ge \\frac{9}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be a point which altitudes of the acute-angled triangle $ABC$ intersect at. Points $A_1, B_1, C_1$ are midpoints of the sides $BC, CA, \\text{and } AB$ respectively. Let $A_2$\n\nand $C_2$ be such points for which $A_2A \\perp AC$ and $A_2C_1 \\perp AB$, $C_2C \\perp AC$ and $C_2A_1 \\perp BC$. Prove the following:\n\na) midpoint of the sector BH is on line $A_2C_2$;\n\nb) let line $BB_1$ intersect a circle circumscribed about triangle $A_1B_1C_1$ at points $B_1$ and $B_3$, point $B_3$ is then on line $A_2C_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H_2$ be a midpoint of the segment $BH$. Let $w_A$, $w_C$ be circles of radius $A_2A$ and $C_2C$ with centers at points $A_2$, $C_2$ respectively. Then they are tangent to line $AC$ at endpoints of the segment $AC$ and pass through point $B$, since points $A_2$, $C_2$ are on the respective perpendicular bisectors. Furthermore, power of point $B_1$ about these circles is equal and therefore the radical axis of the circles is the line $BB_1$. Let the second point which the circles intersect at be $B_4$. Then according to the theorem about the tangent and chord we find $\\angle B_4AC = \\angle B_4BA$, $\\angle B_4CA = \\angle B_4BC \\Rightarrow \\angle CB_4A = \\pi - \\angle B_4BA - \\angle B_4BC = \\pi - \\angle ABC = \\angle AHC$. Then points $A$, $B_4$, $C$, $H$ are on one circle. Dilation with center $B$ and ratio $\\frac{1}{2}$ (fig.4) converts the circle circumscribed about $\\triangle AHC$ into a circle circumscribed about $\\triangle A_1B_2C_1$ and respectively segment $B_4H$ into segment $B_3B_2$. In this case $B_1B_2$ is a diameter of the last circle which implies $B_3B_2 \\perp BB_1$. Therefore, $B_2$ is on the perpendicular bisector to $BB_4$ which has points $A_2$, $C_2$ on it as they are centers of the circles passing through $B$, $B_4$.\n\n![](attached_image_1.png)\nFig.4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45088,
    "subject": "Mathematics (Multi-modal)",
    "question": "What should be the value of the parameter $\\alpha$ in order that equation $|x - \\frac{1}{2}| + |x - \\sin \\alpha| = \\cos 3\\alpha$ has a single solution? Find this solution.",
    "options": [],
    "answer": "All such parameters are α = (-1)^n·π/6 + πn for n ∈ ℤ, and the unique solution is x = 1/2.",
    "solution": "![](attached_image_1.png)\nFig.8\n\nThe graph of functions $y = |x - a| + |x - b|$ for $b > a$ is shown in **fig.8**. Thus it is clear that equation $|x - a| + |x - b| = c$ may have a single solution only if $a = b$ and $c = 0$ (**fig.9**).\n\nTherefore for our equation the following condition should be satisfied:\n$$\n\\begin{cases}\n\\sin \\alpha = \\frac{1}{2} \\\\\n\\cos 3\\alpha = 0\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\n\\alpha = (-1)^n \\frac{\\pi}{6} + \\pi n, & n, k \\in \\mathbb{Z} \\\\\n\\alpha = \\frac{\\pi}{6} + \\frac{1}{3} \\pi k\n\\end{cases}\n$$\nThis system has the following common solution: $\\alpha = (-1)^n \\frac{\\pi}{6} + \\pi n, n \\in \\mathbb{Z}$ and $x = \\frac{1}{2}$.\n\n![](attached_image_2.png)\nFig.9",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the functions $f: \\mathbb{R} \\to \\mathbb{R}$ so that\n$$ f(f(y) + 2 + x) + f(f(y) - x) = y f(y)(x + 1) $$\nis true for any real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 0 for all real x",
    "solution": "Let's make substitution $x = -2 - t$ where $t$ is an arbitrary real number. We find that $f(f(y) - t) + f(f(y) + 2 + t) = -y f(y)(t + 1)$. We can see that the left side of the equation has not changed while a minus sign appeared on its right side. Thus for all $y$ and $t$ the following equation should be true: $y f(y)(t + 1) = 0$, which implies that $f(y) = 0$ for all $y \\neq 0$.\nLet's make another substitution $x = -2$, $y = 1$. We find: $f(0) + f(2) = 0$. As $f(2) = 0$, $f(0) = 0$. Evidently, the identically zero function fulfills the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find out how many solutions may equation\n$$\n[a^2, b^2] + [b^2, c^2] + [c^2, a^2] = (a^2, b^2)(b^2, c^2)(c^2, a^2)\n$$\nhave in natural numbers, if $[m, n]$ and $(m, n)$ stands respectively for the LCM and GCD of the natural numbers $m$ and $n$.",
    "options": [],
    "answer": "Infinitely many solutions",
    "solution": "Let $a = px$, $b = py$, $c = pz$ where $p$, $x$, $y$, $z$ are pairwise coprime numbers. Then the equation will be as follows: $(pxy)^2 + (pxz)^2 + (pyz)^2 = (p^2)^3 \\Leftrightarrow x^2y^2 + x^2z^2 + y^2z^2 = p^4$\n$$ \\Leftrightarrow x^2(y^2 + z^2) = (p^2 - yz)(p^2 + yz). $$\nIf $p^2 = y^2 + yz + z^2$, $x^2(y^2 + z^2) = (y^2 + 2yz + z^2)(y^2 + z^2) \\Rightarrow x = y + z$. Under such conditions the solution will be the following set of three $(p(y+z), py, pz)$.\nWe just have to show that there are infinitely many sets of three natural numbers $(p, y, z)$ for which $p^2 = y^2 + yz + z^2$ is true. Let's denote $u = \\frac{y}{p}$, $v = \\frac{z}{p}$, i.e. we have to show that equation $u^2 + uv + v^2 = 1$ has infinitely many solutions in rational coordinates. One point is $(1, 0)$. Let's choose rational number $k$, then besides $(1, 0)$ line $v = k(u-1)$ intersects curve (ellipse) $u^2 + uv + v^2 = 1$ at one more point. According to the Vieta theorem this point is rational.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequence $\\{f_n\\}$ is defined as follows: $f_1 = 1$, $f_2 = 2$, $f_{n+2} = f_{n+1} + f_n$ for random natural $n$. What may be the greatest number of members of the sequence $\\{f_n\\}$ among the consecutive members of the increasing arithmetic progression?",
    "options": [],
    "answer": "3",
    "solution": "Let the first consecutive members be $f_1 = a$, $f_m = a + d$. Since $a > 0$, $d > 0$, $f_{m+1} > f_m$, $f_{m+2} = f_{m+1} + f_m > 2a + 2d > a + 2d$, the third member may only be $f_{m+1}$.\n\nSo $f_{m+1} = a + 2d$. It's clear that the next member of progression does not satisfy the condition because $f_{m+2} = 2a + 3d > a + 3d$. But we can easily find the three consecutive members 1, 2, 3, and 2, 5, 8.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45092,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ cities in a country. A monopoly company intends to establish air traffic between some of these cities. The government demands that the air traffic satisfies the following conditions. People should be able to get from each city to any other of these cities by one or several flights. Besides there must be an equal number of flights from each city. Note that two flights — from $A$ to $B$ and from $B$ to $A$ — are considered as one flight between cities $A$ and $B$. Moreover, there may be at most one flight between two cities. But the company wants to cancel as little flights as possible so that you can not get from each city to any other city. What minimal number of flights has the company to cancel if:\na) $n = 2008$;\nb) $n = 2007$?",
    "options": [],
    "answer": "a) 1; b) 2",
    "solution": "a) Let's divide all the cities into two groups consisting of $1003$ and $1005$ cities respectively. Let's connect all the cities in each group in cycle. Thus, we obtain the degree $2$ of each vertex. Next lets select one city in each component and connect the two cities with a flight \"007\". Let's connect the rest of the cities with additional diagonals inside the group so that the degree of each vertex is $3$ (fig.5). It's clear that you just have to cancel flight \"007\" and the problem is solved. Thus, we obtain answer $1$.\n\n![](attached_image_1.png)\n\nFig.5\n\nBy contrary let us assume that $\\Delta = 1$. If you exclude this flight, the system falls into two groups of connected cities. In one of these groups there are $2m$ cities, i.e. even number of cities. If the degree of each vertex was $k$ before we excluded the flight, in all there would be $\\frac{1}{2}((2m-1)k + (k-1)) = \\frac{1}{2}(2mk - 1)$ flights in this group. But this is not an integer as it should be.\n\nb) Let this minimum be $\\Delta$. It's obvious that $\\Delta \\le 2$. To prove this you just have to connect the cities in any cycle. It means that there are only two flights from each city. Thus, if you exclude any two flights, the system falls into two groups of cities, which are not connected between each other. Let's show that in this case $\\Delta = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that inequality\n$$ \\frac{x}{\\sqrt{x^2 + y + z}} + \\frac{y}{\\sqrt{x + y^2 + z}} + \\frac{z}{\\sqrt{x + y + z^2}} \\le \\sqrt{3} $$\nis true for any random nonnegative real numbers $x, y, z$ which obey equality $x^2 + y^2 + z^2 = 3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Cauchy-Bunyakowsky inequality implies that you just have to prove the following inequality:\n$$\n\\left( \\frac{x}{x^2 + y + z} + \\frac{y}{y^2 + z + x} + \\frac{z}{z^2 + x + y} \\right) (x + y + z) \\le 3. \\text{ And since} \\\\ (x + y + z)^2 \\le 3(x^2 + y^2 + z^2) = 9, \\text{ it's enough to prove that}\n$$\n$$\n\\frac{x}{x^2 + y + z} + \\frac{y}{y^2 + z + x} + \\frac{z}{z^2 + x + y} \\le 1.\n$$\n\nWe obtain $(a^2 + b + c)(1 + b + c) \\geq (a + b + c)^2$ for non-negative $a, b, c$. Therefore, $\\frac{a}{a^2+b+c} \\leq \\frac{a(1+b+c)}{(a+b+c)^2}$. If we add all these inequalities, we obtain that it's enough to prove that: $\\frac{x(1+y+z)+y(1+z+x)+z(1+x+y)}{(x+y+z)^2} \\leq 1$. This inequality is equivalent to $x+y+z \\leq x^2+y^2+z^2=3$, which has already been proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A_0, B_0, C_0$ are feet of altitudes in an acute-angled triangle $ABC$. Points $A_1, B_1, C_1$ are placed inside the triangle so that $\\angle A_1BC = \\angle A_1AB$, $\\angle A_1CB = \\angle A_1AC$, $\\angle B_1CA = \\angle B_1BC$, $\\angle B_1AC = \\angle B_1BA$, $\\angle C_1BA = \\angle C_1CB$, $\\angle C_1AB = \\angle C_1CA$. Points $A_2, B_2$, and $C_2$ are midpoints of the segments $AA_1, BB_1$, and $CC_1$ respectively. Prove that lines $A_0A_2, B_0B_2$, and $C_0C_2$ intersect at one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the orthocenter of $\\triangle ABC$, lines $AA_1$ and $BC$ intersect at point $A_3$. Then $\\triangle A_3BA_1 \\sim \\triangle ABA_3$, $\\triangle A_1A_3C \\sim \\triangle ACA_3 \\Rightarrow CA_3^2 = A_3A_1 \\cdot AA_3 = BA_3^2 \\Rightarrow CA_3 = BA_3$ (fig. 6).\n\nSince $\\angle BA_1C = 180^\\circ - \\angle BAC$, by means of the triangles $BB_0A$ and $CC_0A$ we can easily find that $\\angle BHC = 180^\\circ - \\angle BAC$, i.e. $\\angle BA_1C = \\angle BHC$. Therefore,\n\n![](attached_image_1.png)\n\nFig. 6\n\n$B, H, A_1, C$ are circular points and $\\angle HA_1C = 180^\\circ - \\angle HBC = 90^\\circ + \\gamma$. Since $\\angle A_3A_1C = \\angle A_1AC + \\angle A_1CA = \\gamma \\Rightarrow \\angle HA_1A_3 = \\angle HA_1C - \\angle A_3A_1C = 90^\\circ - \\gamma + \\gamma = 90^\\circ = \\angle HA_1A$. Hence $A, C_0, H, A_1$ are circular points. Therefore, $\\angle AA_1C_0 = \\angle AHC_0 = 180^\\circ - \\angle AHC = 180^\\circ - A_0HC_0 = \\beta$ since points $B, C_0, H, A_1$ are also circular. Therefore, $\\triangle AA_1C_0 \\sim \\triangle ABA_3$.\n\nIf we find point $C_3 = CC_1 \\cap AB$ similarly to the previous constructions, it will be a midpoint of the side $AB$. Therefore, $C_3$ and $A_2$ are midpoints of the respective sides of the similar triangles. Thus, $\\angle C_0A_2A_1 = \\angle A_3C_3B = \\alpha$. But $\\angle C_0A_0A_3 = 90^\\circ + \\angle AA_0C_0 = 90^\\circ + \\angle C_0BH = 180^\\circ - \\alpha$ and therefore $A_0, C_0, A_2, A_3$ are circular points. This implies that point $A_2$ belongs to the circle of nine points of $\\triangle ABC$. Similarly points $B_2$ and $C_2$ also belong to this circle.\n\nThen we obtain that $\\angle C_0A_1A_2 = \\beta$, $\\angle C_0A_2A_1 = \\alpha$, therefore $\\triangle A_1A_2C_0 \\sim \\triangle ABC \\Rightarrow \\frac{C_0A_2}{AC} = \\frac{A_2A_1}{AB}$. Similarly $\\frac{B_0A_2}{AB} = \\frac{A_2A_1}{AC} \\Rightarrow \\frac{C_0A_2}{B_0A_2} = \\frac{AC^2}{AB^2} = \\frac{\\sin \\angle C_0A_0A_2}{\\sin \\angle B_0A_0A_2}$ since all the points $A_0, B_0, C_0,$ and $A_2$ are on one circle. Similarly $\\frac{B_0C_2}{A_0C_2} = \\frac{BC^2}{AC^2} = \\frac{\\sin \\angle B_0C_0C_2}{\\sin \\angle A_0C_0C_2}$ and $\\frac{A_0B_2}{C_0B_2} = \\frac{AB^2}{CB^2} = \\frac{\\sin \\angle A_0B_0B_2}{\\sin \\angle C_0B_0B_2}$.\n\nLet's multiply the last equalities $\\frac{\\sin \\angle C_0A_0A_2}{\\sin \\angle B_0A_0A_2} \\cdot \\frac{\\sin \\angle B_0C_0C_2}{\\sin \\angle A_0C_0C_2} \\cdot \\frac{\\sin \\angle A_0B_0B_2}{\\sin \\angle C_0B_0B_2} = 1$. The Ceva theorem implies that lines $A_0A_2$, $B_0B_2$, and $C_0C_2$ intersect at one point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45095,
    "subject": "Mathematics (Multi-modal)",
    "question": "(Lishunov Vitalii) $\\frac{1}{x-y} + \\frac{1}{y-z} + \\frac{1}{z-x} = \\frac{3}{2}$ is true for some real numbers $x$, $y$, $z$. What value may expression $\\frac{1}{(x-y)^2} + \\frac{1}{(y-z)^2} + \\frac{1}{(z-x)^2}$ have?\n**Answer:** $\\frac{9}{4}$.",
    "options": [],
    "answer": "9/4",
    "solution": "Let $B = \\frac{1}{x-y} + \\frac{1}{y-z} + \\frac{1}{z-x} = \\frac{3}{2}$ and $A = \\frac{1}{(x-y)^2} + \\frac{1}{(y-z)^2} + \\frac{1}{(z-x)^2}$ then\n$$B^2 = \\frac{9}{4} = A + 2\\left(\\frac{1}{(x-y)(y-z)} + \\frac{1}{(x-y)(z-x)} + \\frac{1}{(y-z)(z-x)}\\right).$$\nLet's rearrange the second summand of the last expression:\n$$\\frac{1}{(x-y)(y-z)} + \\frac{1}{(x-y)(z-x)} + \\frac{1}{(y-z)(z-x)} = \\frac{(z-x)+(y-z)+(x-y)}{(x-y)(y-z)(z-x)} = 0.$$ \nTherefore, $A = \\frac{9}{4}$.\n\nThough it is not required in the problem, we should check that there are numbers $x$, $y$, $z$ which satisfy the given condition. Let $x=2$, $y=1$, $z=0$ then $(x - y) = 1$, $(y - z) = 1$, $(z - x) = -2$, $B = 1 + 1 - \\frac{1}{2} = \\frac{3}{2}$, and $A = 1 + 1 + \\frac{1}{4} = \\frac{9}{4}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45096,
    "subject": "Mathematics (Multi-modal)",
    "question": "(Malitskii Yrii) Find out what five-digit numbers are there more: (1) those in which all digits follow from left to right in the order of increasing, or (2) those in which each digit is not greater than $5$ and the order those digits follow from left to right is not descending (e.g. $12459$ satisfies the first condition, while $22589$ and $01234$ don't; $11145$ satisfies the second condition, while $21224$, $12346$, and $01234$ don't).",
    "options": [],
    "answer": "They are equal in number.",
    "solution": "In order to prove the answer let's associate each number in one set of numbers to a single number in the other set. It's enough to associate the number $abcde$ from the first set to the number $a(b-1)(c-2)(d-3)(e-4)$. This implies that both sets have equal power.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $M$ is placed on the side $BC$ of the triangle $ABC$ so that $BM = AC$. $H$ is the foot of the perpendicular dropped on $AM$ from point $B$. We know that $BH = CM$, and $\\angle MAC = 30^\\circ$. Find degree measure of the $\\angle ACB$.\n\n**Answer:** $15^\\circ$ or $105^\\circ$.",
    "options": [],
    "answer": "15° or 105°",
    "solution": "Let $AC = BM = b$, $BH = MC = h$. We can easily show that point $H$ can not be on the segment $AM$. Let's consider two cases.\n\n1) Point $H$ is on the ray $MA$ (fig.1).\nLet's draw perpendicular $LC \\perp AM$, point $L$ is on the line $AM$. As one of the angles of the right-angled triangle $ALC$ is $30^\\circ$, $LC = \\frac{1}{2}b$. Therefore, $\\angle MCL = \\angle MBH \\Rightarrow$\n\n![](attached_image_1.png)\nFig.1\n\n$\\cos \\angle MCL = \\frac{LC}{MC} = \\frac{b}{2h} = \\cos \\angle MBH = \\frac{BH}{BM} = \\frac{h}{b} \\Rightarrow \\frac{b}{2h} = \\frac{h}{b}$ and $b^2 = 2h^2$, which means that $\\cos \\angle MCL = \\frac{b}{2h} = \\frac{1}{\\sqrt{2}}$ and $\\angle MCL = 45^\\circ$. Since $\\angle LCA = 60^\\circ$, $\\angle ACB = 15^\\circ$.\n\n2) The point $H$ is on the ray $AM$ (fig.2). Let's draw perpendicular $LC \\perp AM$, the point $L$ is on the line $AM$. Further we solve the problem as in the first case, and find that $\\angle MCL = 45^\\circ$. Since $\\angle LCA = 60^\\circ$, $\\angle ACB = 15^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45098,
    "subject": "Mathematics (Multi-modal)",
    "question": "We know that at some natural $n$ the number $n^2 + 2008n$ written in decimal notation ends with 4. Find what digit is in the ten's place of the number.",
    "options": [],
    "answer": "8",
    "solution": "It's clear that the number $2000n$ does not influence the answer, which implies that the sought digits will be the same for numbers $A = n^2 + 2008n$ and $B = n^2 + 8n^2$. As number $(B+16)$ equals $(n+4)^2$ (being the square of the natural number) and ends in $0$, this number should end in $00$. Thus $B = \\overline{X00}-16 = \\overline{Y84}$ where $X, Y$ are some natural numbers. Therefore the last two digits of the number are $84$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45099,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are five piles made of $42$, $70$, $105$, $462$, and $2009$ stones respectively. You are allowed to do the following: instead of two piles consisting of $a$ and $b$ stones you can make\n* two piles of $a$ and $(a+b)$ stones;\n* two piles of $a$ and $|a-b|$ stones.\nYou can also choose $k$ different piles, in which the total number of stones equals $nk$ where $n$ is some natural number, and create\n* $k$ piles of $n$ stones in each one instead.\nCan we make five piles each of which will consist of\na) $2008$ stones;\nb) $2009$ stones\nin finite number of steps?",
    "options": [],
    "answer": "a) No; b) Yes",
    "solution": "a) We can easily find that the GCD of all the numbers $42$, $70$, $105$, $462$, and $2009$ is $7$. For every operation you are allowed to perform the number of stones in piles stays divisible by $7$. Therefore the answer is \"No\" as the number $2008$ is not divisible by $7$.\n\nb) In the second case the answer is \"Yes\". One of the ways to obtain the necessary set of piles is as follows: $(42,70,105,462,2009) \\to (42,35,105,462,2009) \\to (7,35,105,462,2009)$. Now we have a pile of $7$ stones and can increase the other pile by $7$ stones $1061$ times: $\\to (7,35+7 \\times 1061,105,462,2009)$. After that we divide the five derived piles into equal parts, in which the total number of stones equals $10045$, and arrive at the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Diagonals $AC$ and $BD$ of the quadrangle $ABCD$ intersect at point $O$. We know that diagonal $BD$ is perpendicular to the side $AD$, $\\angle BAD = \\angle BCD = 60^\\circ$, $\\angle ADC = 135^\\circ$. Find the ratio $DO:OB$.\n![](attached_image_1.png)\n\nFig. 1\n**Answer:** 1:2.",
    "options": [],
    "answer": "1:2",
    "solution": "Under the problem statement we can easily find the following angles (fig.1): $\\angle ABD = 30^\\circ$, $\\angle BDC = 45^\\circ$, $\\angle DBC = 75^\\circ$. Let's draw rays $ADE$ and $ABF$. Then $\\angle EDC = 45^\\circ$, $\\angle FBC = 75^\\circ$. Therefore $BC$ is a bisector of $\\angle DBF$ and $DC$ is a bisector of $\\angle BDE$, which implies that $AC$ is a bisector of $\\angle BAD$. The last statement is easily proved by the locus of the bisector. Thus $\\angle BAO = 30^\\circ$ and $\\angle AOB$ is an isosceles triangle. Therefore $AO = BO$ and $DO = \\frac{1}{2}AO$ as $\\triangle ADO$ is right-angled triangle with an angle of $30^\\circ$. From this we find that $\\frac{DO}{OB} = \\frac{1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Peter has several equal squares with dimensions: $4 \\times 4$. Each square is divided into sectors $(1 \\times 1)$. He paints each of these sectors red or blue so that there are no similar patterns in all columns and all rows of all the squares. Rotating the squares is forbidden. How many squares can Peter paint in that way?\n\n**Answer:** 4.",
    "options": [],
    "answer": "4",
    "solution": "Altogether there are $2^4 = 16$ different patterns of painted columns. Since each square has 4 columns, in all there can not be more than $\\frac{2^4}{4} = 4$ differently painted squares.\n\n| 0001 | 1110 | 0110 | 1001 |\n|------|------|------|------|\n| 0010 | 1101 | 0011 | 1100 |\n| 0100 | 1011 | 0101 | 1010 |\n| 1000 | 0111 | 0000 | 1111 |\n\nLet's prove that we can paint this number of squares. An example is shown in the table. Checking the columns is enough, if their patterns are different, the symmetry about the diagonal proves the difference of the row patterns.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\frac{a+b}{2b+c} + \\frac{b+c}{2c+a} + \\frac{c+a}{2a+b} \\geq 2\n$$\nis true for any positive real numbers $a, b, c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Applying the Cauchy-Bunyakowsky inequality to sets $\\frac{a_1}{\\sqrt{b_1}}, \\dots, \\frac{a_n}{\\sqrt{b_n}}$ and $\\sqrt{b_1}, \\dots, \\sqrt{b_n}$ in which numbers $a_1, \\dots, a_n, b_1, \\dots, b_n$ are positive we find that $\\frac{a_1^2}{b_1} + \\dots + \\frac{a_n^2}{b_n} \\ge \\frac{(a_1+\\dots+a_n)^2}{b_1+\\dots+b_n}$. Next we rearrange the inequality as follows:\n$$\n\\frac{a+b}{2b+c} + \\frac{b+c}{2c+a} + \\frac{c+a}{2a+b} = \\frac{(a+b)^2}{(a+b)(2b+c)} + \\frac{(b+c)^2}{(b+c)(2c+a)} + \\frac{(c+a)^2}{(c+a)(2a+b)}\n$$\n$\\ge$ (we use the mentioned above inequality here)\n$$\n\\ge \\frac{((a+b)+(b+c)+(c+a))^2}{(a+b)(2b+c)+(b+c)(2c+a)+(c+a)(2a+b)} \\ge 2.\n$$\nIn order to verify the last rearrangement, simply expand the brackets and summarize similar summands.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45103,
    "subject": "Mathematics (Multi-modal)",
    "question": "We know that at some natural $n$ the number $n^3 + 2009n^2 + 27n$ written in decimal notation ends with digit 3. Find what digits are in the hundred's and ten's place of the number.",
    "options": [],
    "answer": "hundreds digit = 9; tens digit = 7",
    "solution": "It's clear that the number $2000n^2$ does not influence the answer therefore the sought digits are the same for numbers $A = n^3 + 2009n^2 + 27n$ and $B = n^3 + 9n^2 + 27n$.\n\nAs the number $(B + 27)$ equals $(n + 3)^3$ (being the cube of the natural number) and ends in 0, this number should end in 000. Thus $B = \\overline{X000} - 27 = \\overline{Y073}$ where $X, Y$ are some natural numbers. Therefore the last three digits of the number are 073.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45104,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ natural numbers are written on the board. You can add only natural numbers in the form $\\frac{a+b}{a-b}$ where $a$ and $b$ are the numbers already written on the board. It appears that by doing so you can make any natural number appear on the board. Calculate the least value of $n$ and find the numbers initially written (consider all the cases).",
    "options": [],
    "answer": "Minimum n is 2; the valid initial sets are {1,2} and {1,3}.",
    "solution": "As $(a + b) > (a - b)$, you can not obtain $1$ performing the operations allowed. Therefore it should be written on the board, but one number is not enough. Let's show that two numbers will be enough. Let another number of the two be $x$. $\\frac{x+1}{x-1}$ is the only number which can be obtained in the first step. Since it is a natural number, $\\frac{x+1}{x-1} \\ge 2 \\Rightarrow (x+1) \\ge 2x-2$ or $x \\le 3$. Thus the second number should be $2$ or $3$. We obtain the two possible sets: $\\{1,2\\}$ and $\\{1,3\\}$.\n\nLet's prove that they both satisfy the condition. As $\\frac{2+1}{2-1} = 3$ and $\\frac{3+1}{3-1} = 2$, in the first step we obtain the set $\\{1,2,3\\}$ in both cases. Now we have to prove that any natural number greater than $3$ can be obtained from these three numbers.\n\nLet's assume that we've already obtained the set $\\{1,2,3,...,(2k+1)\\}$. Let's show how we can obtain the next two numbers. We obtain number $\\frac{(k+2)+(k+1)}{(k+2)-(k+1)} = 2k+3$ from numbers $(k+1), (k+2)$. Next we obtain $\\frac{(2k+3)+(2k+1)}{(2k+3)-(2k+1)} = 2k+2$ from numbers $(2k+3), (2k+1)$. This implies the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45105,
    "subject": "Mathematics (Multi-modal)",
    "question": "$BB_1$ is an altitude of the acute-angled scalene triangle $ABC$. Point $D$ is placed on the side $BC$ so that $\\angle BAD = \\angle CBB_1$. Segments $AD$ and $BB_1$ intersect at point $F$. Line $l$ is drawn through point $B$ at right angle to the side $AB$. This line intersects line $CF$ at point $K$. Prove that line $DK$ intersects segment $BF$ at a midpoint.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let line $AD$ intersect a circle circumscribed about triangle $\\triangle ABC$ at point $N$. Then $\\angle BCN = \\angle BAN = \\angle CBB_1$. This implies that $BB_1 \\parallel CN$ (fig.2), and therefore $\\angle ACN = 90^\\circ \\Rightarrow AN$ is a diameter. Thus $\\angle NBA = 90^\\circ$, which implies that $N,B,K$ are on the same line. As diagonals of trapezium $CFBN$ intersect at point $D$, its sides $CF$ and $BN$ extended intersect at point $K$. The well-known properties of trapezium imply the rest of the proof.\n\n![](attached_image_1.png)\n\n**Fig.2**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45106,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are discounts in the supermarket. An active customer will be given a discount of $4\\%$, if he buys goods for the sum of over $300$; and if the same customer buys goods for the sum of over $600$, he will be given a discount of $10\\%$. What is the biggest amount of money that he can spend for goods if he has\n\na) $594$;\nb) $534$?",
    "options": [],
    "answer": "a) 660; b) 556.25",
    "solution": "a) If a customer buys goods for the sum of $x \\ge 600$, he'll pay $10\\%$ less the cost of these goods. Consequently, he can buy goods for the sum $x$ if $0.90x \\le 594 \\Rightarrow x \\le 660$ at the most.\n\nb) Analogously to the item a) we obtain that a customer will be given a discount of $4\\%$, to buy goods for the amount $x$ if $0.96x \\le 534$, that is $x \\le 556.25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the biggest 3-digit number which satisfies the following three conditions:\n1) the number is prime;\n2) the 3-digit number obtained by writing the digits of our number in the reverse order is a prime too;\n3) the product of the digits is a prime too.",
    "options": [],
    "answer": "311",
    "solution": "As the product of the digits this number has is prime, then these digits are $1, 1, p$ (not necessarily in this order) and the digit $p$ is a prime number. Obviously, $p \\neq 7$ because otherwise the sum of the digits is divisible by $3$ and the number isn't prime. The digits $2$ and $5$ can stay only between both digits $1$, then we obtain the numbers $121$ and $151$. If $p=3$, then we have numbers $113$, $131$ and $311$. The biggest from these numbers is $311$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45108,
    "subject": "Mathematics (Multi-modal)",
    "question": "A line $l_1$ intersects the parabola $y = ax^2 + bx + c$ ($a \\neq 0$) at the two points $A$ and $B$. A line $l_2$ is parallel to the line $l_1$ and tangent to this parabola at the point $C$. Prove that the arithmetic mean of abscissas of points $A$ and $B$ equals the abscissa of point $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the equation of the line $l_1$ be $y = kx + d$. Then abscissas of points $A$ and $B$ are defined using the equality $ax^2 + bx + c = kx + d$. Therefore these abscissas $x_1, x_2$ satisfy that quadratic equation, then according to Vieta's formula we have equality $x_1 + x_2 = \\frac{-b + k}{a}$.\n\nThe abscissa of point $C$ is defined using the same equality\n\n$ax^2 + bx + c = kx + d$\n\non condition that the line $l_2$ is tangent to this parabola. Then the abscissa of point $C$ is $\\frac{k-b}{2a} = \\frac{x_1 + x_2}{2}$, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system of equations:\n$$\n\\begin{cases}\n\\frac{9}{2(x+y)} = \\frac{1}{x} + \\frac{1}{y}, \\\\\n\\sqrt{x^2 - 2} = \\sqrt{3 - y^2}.\n\\end{cases}\n$$",
    "options": [],
    "answer": "[(2, 1), (-2, -1)]",
    "solution": "Consider the first equations of the system. It is easy to see that $xy \\neq 0$, and we\n$$\n\\text{get } \\frac{9}{2(x+y)} = \\frac{x+y}{xy} \\Leftrightarrow 9xy = 2x^2 + 4xy + 2y^2 \\Leftrightarrow 2y^2 - 5xy + 2x^2 = 0\n$$\n\n$y = \\frac{5x \\pm \\sqrt{25x^2 - 16x^2}}{4} = \\frac{5x \\pm 3x}{4}$. It follows that all couples $(x, 2x)$ and $(x, \\frac{1}{2}x)$ without $(0,0)$ are solutions of the first equation. We are putting these solutions in the second equation. From this equation we have $x^2 + y^2 = 5$, $|x| \\ge \\sqrt{2}$ and $y \\le \\sqrt{3}$. Further, putting $(x, 2x)$ the first condition becomes $5x^2 = 5 \\Leftrightarrow x = \\pm 1$. And we have solutions $(1,2)$ and $(-1,-2)$ but these couples don't satisfy the condition $|x| \\ge \\sqrt{2}$. Analogously putting $(x, \\frac{1}{2}x)$ the first condition becomes $x^2 + \\frac{1}{4}x^2 = 5 \\Leftrightarrow x^2 = 4$. Here we have solutions $(2,1)$ and $(-2,-1)$ which satisfy our conditions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Compare two numbers:\n$$\n\\sqrt{2008 + \\sqrt{2009}} + \\sqrt{2009 + \\sqrt{2008}} \\text{ and } \\sqrt{2008 + \\sqrt{2008}} + \\sqrt{2009 + \\sqrt{2009}}\n$$",
    "options": [],
    "answer": "sqrt(2008 + sqrt(2009)) + sqrt(2009 + sqrt(2008)) > sqrt(2008 + sqrt(2008)) + sqrt(2009 + sqrt(2009))",
    "solution": "Reformulate this task in the general case: for the positive distinct real numbers $a, b$ compare two numbers: $A = \\sqrt{a+\\sqrt{b}} + \\sqrt{b+\\sqrt{a}}$ and $B = \\sqrt{a+\\sqrt{a}} + \\sqrt{b+\\sqrt{b}}$. Prove that $A > B$.\n\nIs equivalent to $A^2 > B^2 \\Leftrightarrow \\sqrt{a+\\sqrt{b}} \\cdot \\sqrt{b+\\sqrt{a}} > \\sqrt{a+\\sqrt{a}} \\cdot \\sqrt{b+\\sqrt{b}}$.\n\nAfter some simple calculations we have: $a\\sqrt{a} + b\\sqrt{b} > a\\sqrt{b} + b\\sqrt{a} \\Leftrightarrow (a-b)(\\sqrt{a}-\\sqrt{b}) > 0$ if $a \\ne b$.\n\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45111,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $m$ denote by $\\sigma(m)$ the sum of all its divisors, and by $\\phi(m)$ the number of positive integers less than $m$ and coprime with $m$. Prove that there exists infinitely many such positive integer numbers $n$ for which $\\phi(\\sigma(n)) > n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider number $n = 2^{p-1}$, where $p$ is a prime number. Then $\\sigma(n) = 2^p - 1$. Let's prove that $\\phi(2^p - 1) > 2^{p-1}$ for infinitely many primes $p$.\n\nLet $m = 2^p - 1 = p_1^{\\alpha_1} \\dots p_k^{\\alpha_k}$. Then\n$$\n\\phi(m) = (2^p - 1) \\frac{p_1 - 1}{p_1} \\dots \\frac{p_k - 1}{p_k}.\n$$\nNotice that every $p_i > p$, $i = 1, \\dots, k$. Really, as it follows from the little Fermat theorem, $2^{p_i-1} \\equiv 1 \\pmod{p_i}$, and $2^p - 1 \\equiv 0 \\pmod{p_i}$ as well, therefore $(p_i - 1) \\nmid p$, because $p$ is a prime number.\n\nHence $m = 2^{p-1} = p_1^{\\alpha_1} \\dots p_k^{\\alpha_k} > p^k$ and so $k \\le p \\log_p 2$. I.e. $k \\le \\frac{p}{2}$ for sufficiently large $p$.\n\nThen\n$$\n\\frac{\\phi(2^p-1)}{2^{p-1}} \\ge \\frac{2^{p-1}}{2^p} \\left(1 - \\frac{1}{p}\\right)^p.\n$$\nBut $\\left(1 - \\frac{1}{p}\\right)^p \\to \\frac{1}{e}$ as $p \\to \\infty$, therefore the number $n = 2^{p-1}$ satisfies the condition for sufficiently large $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45112,
    "subject": "Mathematics (Multi-modal)",
    "question": "The knight stands in the left lower corner of a $2009 \\times 2009$ chessboard and it is known that the first column and the last row are colored in red. The knight moves on the chessboard according to the chess rules but it can't get on the cell which is already colored in red. After every move of the knight we color in red the column and the row that contain the cell on which the knight has just got. Is it possible to color the whole chessboard in red if we act as described above?\n\n![](attached_image_1.png)\nFig.15",
    "options": [],
    "answer": "Yes",
    "solution": "As it is shown on the fig.15, we can color in red the first four columns and the first four rows of the chessboard. Now consider our chessboard without these four columns and rows. As we can see on the same figure, the fifth move of the knight brings it to the left lower corner of the new reduced $2005 \\times 2005$ chessboard. If we continue reducing our chessboard in the same way, then after 501 iterations we will receive a $5 \\times 5$ chessboard, which can be colored in red as shown on the fig.16.\n\n![](attached_image_2.png)\nFig.16",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, such that for all real $x, y$ the following equality holds:\n$$\nf(x + x y + f(y)) = (f(x) + \\frac{1}{2})(f(y) + \\frac{1}{2}).\n$$",
    "options": [],
    "answer": "f(x) = x + 1/2",
    "solution": "If we take $y = -1$ we'll get that: $f(f(-1)) = (f(x) + \\frac{1}{2})(f(-1) + \\frac{1}{2})$. So if $f(-1) \\neq -\\frac{1}{2}$, then $f$ is constant. If we substitute $f = c$ in our equality, then we'll get that $c = (c + \\frac{1}{2})^2$ which is impossible. Therefore $f(-1) = -\\frac{1}{2}$.\n$$\nx = 0 \\Rightarrow f(f(y)) = (f(0) + \\frac{1}{2})(f(y) + \\frac{1}{2}). \\quad (1)\n$$\nSubstituting $y = -1$ in (1) we see that $f(-\\frac{1}{2}) = 0$. Now suppose that for some $y_0 \\neq -1$ we have $f(y_0) = -\\frac{1}{2}$. Then substituting $y = y_0$ in the given equality we will obtain: $f(x(1 + y_0) - \\frac{1}{2}) = 0$, which means that $f$ is constant which was proved to be impossible. Thus $f(y) = -\\frac{1}{2} \\Leftrightarrow y = -1$. Now take some $y \\neq -1$, and substitute in the given equality\n$$\nx = \\frac{-\\frac{1}{2} - f(y)}{y + 1} \\text{. Then } 0 = f(-\\frac{1}{2}) = (f(y) + \\frac{1}{2}) \\left( f\\left(\\frac{-\\frac{1}{2} - f(y)}{y+1}\\right) + \\frac{1}{2} \\right) \\text{. Since } y \\neq -1, \\\\ f(y) + \\frac{1}{2} \\neq 0, \\text{ and so } f\\left(\\frac{-\\frac{1}{2} - f(y)}{y+1}\\right) = -\\frac{1}{2} \\Rightarrow \\frac{-\\frac{1}{2} - f(y)}{y+1} = -1 \\Rightarrow f(y) = y + \\frac{1}{2} \\text{. An easy}\n$$\ncheck shows that this function meets all the requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Distinct real numbers $a, b, c$ satisfy the condition $a + \\frac{1}{b} = b + \\frac{1}{c} = c + \\frac{1}{a}$. Find all possible values of the product $abc$:\n1) for all real $a, b, c$;\n2) for positive real $a, b, c$?",
    "options": [],
    "answer": "For real a, b, c: the product can be −1 or 1. For positive a, b, c: no such triples exist, so there are no possible values.",
    "solution": "a) See problem 9.5\n\nb) We will show that there are no such positive numbers that satisfy the condition of the problem. Without loss of generality we suppose that $a > b$, then from the equality $(a - b) = \\frac{b - a}{b - c}$ we conclude that $b > c$. From the condition $b - c = \\frac{c - a}{a - c}$, we can get a contradiction: $c > a > b > c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45115,
    "subject": "Mathematics (Multi-modal)",
    "question": "A parallelogram $ABCD$ with $AC > BD$ is given. Let $k_1$ be the circle with diameter $AC$ and $k_2$ the circle with diameter $DC$. $k_1$ meets line $AB$ at the point $E$, $k_2$ meets the line $AC$ at the points $C$ and $O$, and line $AD$ at the point $F$. Suppose that $AO = a$, $FO = b$ and $\\angle BAC = 45^\\circ$. Find the ratio of the areas of the triangles $AOE$ and $COF$.\n\n**Answer:** $\\left(\\frac{a}{b}\\right)^2$.",
    "options": [],
    "answer": "(a/b)^2",
    "solution": "Finally, $EF$ and $AC$ are chords in the circle $k_1$, hence $EO \\cdot OF = AO \\cdot OC$, which implies:\n$$\n\\frac{S(AOE)}{S(COF)} = \\frac{AO \\cdot GE}{FO \\cdot CO} = \\left(\\frac{AO}{FO}\\right)^2 = \\left(\\frac{a}{b}\\right)^2\n$$\n\n![](attached_image_1.png)\nFig.25",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Angular bisectors $BD$ and $CE$ are drawn in the triangle $ABC$ with angle $\\angle BAC \\ge 60^\\circ$. Prove that $AD + AE \\le BC$.\n![](attached_image_1.png)\n\nFig.30",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the sides of triangle by $a$, $b$, $c$. By the property of angular bisectors,\n\n$AD = \\frac{ch}{a+c}$, $AE = \\frac{ch}{a+b}$ (pic.30), thus it is enough to prove: $\\frac{ch}{a+c} + \\frac{ch}{a+b} \\le a$.\n\nAfter simple transformations, we have: $(a^2 - bc)(a+b+c) \\ge 0$, and so it is enough to prove: $a^2 - bc \\ge 0$.\n\nBy the cosine theorem, $a^2 = b^2 + c^2 - 2bc \\cos A \\ge b^2 + c^2 - 2bc \\cdot \\frac{1}{2} \\ge 2bc - bc = bc$, which required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45117,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which values of parameter $a$ the system of equations\n$$\n\\begin{cases} x + y + z = 0 \\\\ (xy + yz) + a xz = 0 \\end{cases} \\text{ has a unique solution?}\n$$",
    "options": [],
    "answer": "(0,4)",
    "solution": "It is evident that for every value of parameter $a$ there exists solution $x = y = z = 0$, and so all we need is to find out when this solution is unique.\n\nFrom the first equation we have that $y = -x - z$. Substitute $(-x - z)$ instead of $y$ into the second equation:\n\n$-x^2 - 2xz - z^2 + a xz = 0$\n\nor\n\n$x^2 + xz(2 - a) + z^2 = 0$.\n\nWe have obtained a quadratic equation with respect to $x$. Its discriminant $D = z^2(2 - a)^2 - 4z^2 = z^2(4a - a^2)$.\n\nIf $z = 0$ then the system has zero solution, thus for all $z \\neq 0$ our quadratic equation shouldn't have any solutions, that is $D = z^2(4a - a^2) < 0$.\n\nAnd the last inequality holds for $a \\in (0,4)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the set $M = \\{1,2,3,4,6,8,12,16,24,48\\}$ and all of its four-element subsets. Denote by $n$ the number of these subsets such that the product of their elements is greater than $2009$ and by $m$ the number of these subsets such that the product of their elements is smaller than $2009$. Which number is bigger - $n$ or $m$?",
    "options": [],
    "answer": "n > m",
    "solution": "Let $M_1 = \\{a, b, c, d\\}$ be a four-element subset of $M$ such that $abcd < 2009$. Then the subset $M_2 = \\{\\frac{48}{a}, \\frac{48}{b}, \\frac{48}{c}, \\frac{48}{d}\\}$ is also a four-element subset of $M$ because all of its elements are distinct and belong to $M$ and $\\frac{48}{a} \\cdot \\frac{48}{b} \\cdot \\frac{48}{c} \\cdot \\frac{48}{d} = \\frac{48^4}{abcd} > \\frac{48^4}{2009} > 2009$. It follows that $n \\ge m$. But also we have the subset $\\{1, 2, 24, 48\\}$ which has product of elements $1 \\cdot 2 \\cdot 24 \\cdot 48 = 48^2 > 2009$ and doesn't have any corresponding pair with product of elements less than $2009$. Thus $n > m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45119,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ points $M$ and $N$ are the midpoints of the sides $BC$ and $AC$ respectively. Inside $\\triangle ABC$ a point $P$ is taken such that $\\angle BAP = \\angle PBC = \\angle PCA$. It is known that $\\angle PNA = \\angle AMB$. Prove that $ABC$ is an isosceles triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us draw the line $l \\parallel BC$ through the point $A$ and denote $W = BP \\cap l$. Then\n\n$\\angle BPC = 180^\\circ - (\\angle PBC + \\angle PCB) =$\n$180^\\circ - (\\angle PCA + \\angle PCB) = 180^\\circ - \\angle BCA \\Rightarrow$\n$\\angle CPW = \\angle CAW$. And so points $A, P, C, W$ are cyclic. Thus $\\angle AWC = 180^\\circ - \\angle APC$,\n$\\angle APC = 180^\\circ - (\\angle PAC + \\angle PCA) =$\n$180^\\circ - (\\angle PAC + \\angle PAB) = 180^\\circ - \\angle BAC$, and\nit follows that (fig.17) $\\angle AWC =$\n$180^\\circ - \\angle APC = \\angle BAC$. Now $AW \\parallel CB$ implies\nthat $\\angle WAC = \\angle BCA$. And so $\\triangle ABC \\sim \\triangle ACW$.\nSince $M, N$ are the midpoints of the corresponding sides of similar triangles we have that\n$\\angle WNA = \\angle AMC \\Rightarrow$\n$\\angle WNA + \\angle ANP = \\angle AMC + \\angle AMB = 180^\\circ$. And so\n$B, P, N, W$ lie on the same line.\nTherefore $\\angle BNA = \\angle BMA$, which implies that\n$A, B, M, N$ are cyclic. Since $MN \\parallel AB$ as the centerline, $ABMN$ is an isosceles trapezoid, whence\n$AN = BM \\Rightarrow AC = BC$, what was to be proved.\n\n![](attached_image_1.png)\nFig.17\nLet $Q$ be a point of the median $AM$ such that $\\angle QCB = \\angle PCA$. Since $\\angle QMB = \\angle PNA$ we have that $\\angle QMC = \\angle PNC$ and so $\\triangle PNC \\sim \\triangle QMC$. From this similarity $\\frac{PC}{CQ} = \\frac{CN}{CM} = \\frac{BC}{AC}$ and thus $\\triangle QCB \\sim \\triangle PCA$. Then we can obtain that (fig.18) $\\angle BQC = \\angle APC =$\n![](attached_image_2.png)\nFig.18\n\n$\\angle PAC - \\angle PCA = 180^\\circ - \\angle PAC + \\angle PAB - \\angle PCA = 180^\\circ - \\angle BAC$. Now let $Q'$ be symmetric to $Q$ with respect to $M$. Then $BQCQ'$ is a parallelogram and $\\angle BQ'C = \\angle BQC = 180^\\circ - \\angle BAC$. From this it also follows that the quadrilateral $ABQ'C$ is cyclic and so $\\angle MAC = \\angle Q'AC = \\angle Q'BC = \\angle OBC$. Denote $T = NP \\cap BC$. Then the triangles $NTC$ and $MAC$ are similar and thus $\\angle PTC = \\angle NTC = \\angle MAC = \\angle PBC$. Points $T$ and $B$ lie on the circumcircle of the triangle $PBC$, and also on the line $BC$. Therefore $T$ coincides with one of the points $B$ or $C$. It is evident that $T$ cannot coincide with $C$, so $T=B$, which means that the points $N, P$ and $B$ lie on a line. Since $\\angle ANB = \\angle ANP = \\angle AMB$, points $A, N, M$ and $B$ are cyclic, and it follows that $CN \\cdot NA = CM \\cdot CB$, which implies that $CA = CB$, i.e. that $ABC$ is an isosceles triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ with real coefficients, such that for every distinct natural $x, y, z, t$ satisfying $x^2 + y^2 + z^2 = 2t^2$ and $\\text{GCD}\\{x, y, z, t\\} = 1$, the following equality holds:\n$$\n2P^2(t) + 2P(xy + yz + zx) = P^2(x + y + z).\n$$",
    "options": [],
    "answer": "P(x) = x; P(x) = 0; P(x) = -2",
    "solution": "Take $t = 5k$ and $z = 7k$, then we want the equality $x^2 + y^2 = k^2$ to hold. Consider two series of solutions: $x = n^2 - 1$, $y = 2n$, $k = n^2 + 1$ and $x = 3n^2 + 4n + 1$, $y = 2n(2n+1)$, $k = 5n^2 + 4n + 1$. It is evident that in both cases $(x, y, z, t) = 1$ (here and on $(x, y, z, t)$ denotes $\\text{GCD}\\{x, y, z, t\\}$) for even $n$. Consider the first series and denote $Q = 2P^2(t) + 2P(xy + yz + zx) - P^2(x + y + z)$ and $P(x) = a_m x^m + a_{m-1} x^{m-1} + \\dots + a_1 x + a_0$, $a_m \\neq 0$. Then polynomial $Q_1(n) = 2P^2(5(n^2+1)) + 2P(7n^4 + 16n^3 - 12n - 7) - P^2(8n^2 + 2n + 6)$ equals to zero for every $n = 2s$ with $s \\in \\mathbb{N}$, whence $Q_1(n) = 0$. Consider the coefficient at $x^{4m}$ in $Q_1(x)$: on the one hand it is equal to $2 \\cdot 5^{2m} a_m^2 + 2 \\cdot 7^m a_m - 8^{2m} a_m^2$ and on the other $0$, therefore $2 \\cdot 5^{2m} a_m^2 + 2 \\cdot 7^m a_m - 8^{2m} a_m^2 = 0$. Analogously we can obtain that $Q_2(n) = 2P^2(5(5n^2+4n+1)) + 2P(257n^4 + T_2(n)) - P^2(42n^2 + 34n + 8)$, with $\\deg(T_2) \\le 3$, is equal to zero and thus the coefficient at $x^{4m}$ in this polynomial also equals zero: $2 \\cdot 25^{2m} a_m^2 + 2 \\cdot 257^m a_m - 42^{2m} a_m^2 = 0$. So, $a_m = \\frac{2 \\cdot 7^m}{64^m - 2 \\cdot 25^m} = \\frac{2 \\cdot 257^m}{42^{2m} - 2 \\cdot 25^{2m}}$, whence $2 \\cdot 7^m \\cdot (42^{2m} - 2 \\cdot 25^{2m}) = (64^m - 2 \\cdot 25^m) \\cdot 2 \\cdot 257^m$. For $m \\ge 2$ we have: $(64^m - 2 \\cdot 25^m) \\cdot 2 \\cdot 257^m > (64^m - 2 \\cdot 25^m) \\cdot 2 \\cdot 252^m$, and so $42^{2m} - 2 \\cdot 25^{2m} > 36^m (64^m - 2 \\cdot 25^{2m})$. $7^{2m} > 7^{2m} - \\frac{2 \\cdot 25^{2m}}{36^m} > 64^m - 2 \\cdot 25^m = (7^2 + 15)^m - 2 \\cdot 25^m > 7^{2m} + m \\cdot 7^{2m-2} \\cdot 15 - 2 \\cdot 25^m \\Rightarrow 2 \\cdot 25^m > m \\cdot 49^{m-1} \\cdot 15 \\ge 2 \\cdot 49^{m-1} \\cdot 15 \\Rightarrow \\frac{49^{m-1}}{15} \\ge \\frac{49}{25}$, which is evidently false. Thus, $m \\le 1$. For $m = 1$, we have $P(x) = a_1 x + a_0$ and $a_1 = 1$, i.e. $P(x) = x + a_0$. Substitute $P(x)$ into the initial equality and take $(x, y, z, t)$ from the first series: $(n^2 - 1, 2n, 7(n^2 + 1), 5(n^2 + 1))$. After some easy transformations we will have: $4a_0(5n^2 + 5) + 2a_0^2 + 2a_0 - 2a_0(8n^2 + 2n + 6) - a_0^2 = 0$, which implies that $a_0 = 0$ or that $a_0$ depends on $n$, which is impossible. So for $m = 1$ we have solution $P(x) = x$. Now if $n = 0$ then $a_0^2 + 2a_0 = 0$, and we obtain two more answers: $P(x) = 0$ and $P(x) = -2$. An easy check shows that these solutions meet all the requirements.\nConsider the equation $x^2 + y^2 + z^2 = 2t^2$ and make use of the method of Diophantine secants: $\\left(\\frac{x}{t}\\right)^2 + \\left(\\frac{y}{t}\\right)^2 + \\left(\\frac{z}{t}\\right)^2 = 2$, i.e. $\\alpha^2 + \\beta^2 + \\gamma^2 = 2$, where $\\alpha, \\beta, \\gamma \\in \\mathbb{Q}^+$. Evidently, $\\alpha = 0$, $\\beta = \\gamma = 1$ is a solution of the last equation. Now take $\\beta = -k_1\\alpha + 1$, $\\gamma = -k_2\\alpha + 1$, with $k_1, k_2 \\in \\mathbb{Q}$. Then $\\alpha^2 + (1-k_1\\alpha)^2 + (1-k_2\\alpha)^2 = 2 \\Leftrightarrow \\alpha(k_1^2\\alpha + k_2^2\\alpha - 2k_1 - 2k_2 + \\alpha) = 0$. If $\\alpha \\neq 0$, we have $\\alpha = \\frac{2(k_1+k_2)}{1+k_1^2+k_2^2} \\Rightarrow \\beta = \\frac{1+k_2^2-k_1^2-2k_1k_2}{1+k_1^2+k_2^2}$, $\\gamma = \\frac{1+k_1^2-k_2^2-2k_1k_2}{1+k_1^2+k_2^2}$.\nTake $k_1 = \\frac{n}{p}$, $k_2 = \\frac{m}{p}$, $n, m, p \\in \\mathbb{N}$. Then $\\alpha = \\frac{2(m+n)p}{p^2+n^2+m^2}$, $\\beta = \\frac{p^2+m^2-n^2-2mn}{p^2+n^2+m^2}$, $\\gamma = \\frac{p^2+n^2-m^2-2mn}{p^2+n^2+m^2}$. Thus, we see that $x = 2(m+n)p$, $y = p^2+m^2-n^2-2mn$, $z = p^2+n^2-m^2-2mn$, $t = p^2+m^2+n^2$, with $n, m, p \\in \\mathbb{N}$ are the solutions of the equation $x^2 + y^2 + z^2 = 2t^2$.\nFirst consider $m = 2$, $n = 1$. Then $x_1 = 6p$, $y_1 = p^2-1$, $z_1 = p^2-7$, $t_1 = p^2+5$. It is evident that for $p \\neq 6$ we have $(x_1, y_1, z_1, t_1) = 1$. Then substituting these $(x_1, y_1, z_1, t_1)$ into the problem statement we obtain that $Q = 2P^2(t) + 2P(xy + yz + zx) - P^2(x + y + z) = 0$, which means that $Q_1(p) = 2P^2(p^2+5) + 2P(p^4+12p^3-8p^2-48p+7) - P^2(2p^2+6p-8) = 0$. From the problem conditions it follows that all numbers $p$ of the form $p=6k$ should be the roots of $Q_1$, thus $Q_1 \\equiv 0$.\nLet $P(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0$, with $a_n \\neq 0$. Then $Q_1(p) = 2(a_n(p^2+5)^n + \\dots + a_0)^2 + 2(a_n(p^4+12p^3-8p^2-48p+7)^n + \\dots + a_0)^2 - (a_n(2p^2+6p-8)^n + \\dots + a_0) = (2a_n^2 + 2a_n - 2^{2n}a_n^2)p^{4n} + R_1(p)$, where $\\deg R_1 \\le 4n-1$. Since $Q_1(p) \\equiv 0$ and $a_n \\neq 0$ we have that\n$$\n2a_n + 2 - 2^{2n}a_n = 0. \\quad (1)\n$$\nConsider $p = 2m$, $n = 1$. Then $x_2 = 4m(m+1)$, $y_2 = 5m^2 - 2m - 1$, $z_2 = 3m^2 - 2m + 1$, $t_2 = 5m^2 + 1$. For $m$ large enough we have $t_2 > y_2 > x_2 > z_2$, and so these numbers are distinct. If $m$ is even, $(x_2, z_2) = (4m(m+1), 3m^2 - 2m + 1) = (4(m+1), 3m^2 - 2m + 1) = (m+1, 3m^2 - 2m + 1) = (m+1, 5m-1) = (m+1, 6)$, thus for $m \\neq 6$ $(x_2, y_2, z_2, t_2) = 1$. If we follow the same lines as for the first series $(x, y, z, t)$, we will obtain that $Q_2(m) = 2P^2(5m^2+1) + 2P(47m^4 + T(m)) - P^2(12m^2) = 0$, where $\\deg T \\le 3$. Let us compute the coefficient at $m^{4n}$ in $Q_2(m)$ - it's equal to $2a_n^2 \\cdot 5^{2n} + 2a_n \\cdot 47^n - 12^{2n}a_n^2 = 0$. Since $a_n \\ne 0$ we have that:\n$$\n2a_n \\cdot 5^{2n} + 2 \\cdot 47^n - 12^{2n} a_n = 0. \\qquad (2)\n$$\nNow from (1) and (2) we obtain that $a_n = \\frac{1}{2^{2n-1}}$ and $(2^{2n-1}-1) \\cdot 2 \\cdot 47^n = 12^{2n} - 2 \\cdot 5^{2n}$, whence $2 \\cdot 5^{2n} + (4 \\cdot 47)^n = 144^n + 2 \\cdot 47^n$. It's easy to see that the last equality is impossible for arbitrary natural $n \\ge 2$, since $2 \\cdot 5^{2n} + (4 \\cdot 47)^n > 188^n = (144 + 44)^n \\ge 144^n + 44 \\cdot 144^{n-1} > 144^n + 2 \\cdot 47^n$.\nSo $\\deg P \\le 1$. First assume that $\\deg P = 1$. Then the equality $a_n = \\frac{1}{2^{2n-1}}$ implies that $a_1 = 1$, i.e. that $P(x) = x + a_0$. Substitute this polynomial into the initial equality:\n$$\n2P^2(t) + 2P(xy + yz + zx) = P^2(x + y + z) \\Rightarrow \\\\\n2(t^2 + 2a_0t + a_0^2) + 2(xy + yz + zx) + 2a_0 = (x + y + z)^2 + 2(x + y + z)a_0 + a_0^2 \\Leftrightarrow \\\\\n4a_0t + a_0^2 + 2a_0 = 2(x + y + z)a_0.\n$$\nIf $a_0 = 0$ then $P(x) = x$, otherwise $4t + a_0 + 2 = 2(x + y + z)$. Substitute here the known series $x_1 = 6p$, $y_1 = p^2 - 1$, $z_1 = p^2 - 7$, $t_1 = p^2 + 5$ with $p = 6$. Then $4(p^2 + 5) + a_0 + 2 = 2(2p^2 + 6p - 8)$, and we get that the constant $a_0$ depends on $p$, which is impossible.\nIf $\\deg P = 0$ then $P = a_0 = \\text{const}$. From the problem conditions we obtain that $2a_0^2 + 2a_0 = a_0^2 \\Rightarrow a_0 = -2$ also $a_0 = 0$, which means that $P(x) = 0$ or $P(x) = -2$. An easy check shows that all three polynomials obtained meet all the requirements.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system of equations over real numbers:\n$$\n\\begin{cases}\nx^3 = 2y^3 + z - 2, \\\\\ny^3 = 2z^3 + x - 2, \\\\\nz^3 = 2x^3 + y - 2.\n\\end{cases}\n$$",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "From the first two equations of our system we obtain: $x^3 - y^3 = 2y^3 - 2z^3 + y - z = (y-z)(2y^2 + 2yz + 2z^2 + 1)$, analogously, we can easily get equalities: $y^3 - z^3 = (z-x)(2z^2 + 2zx + 2x^2 + 1)$, $z^3 - x^3 = (x-y)(2x^2 + 2xy + 2y^2 + 1)$. It is easy to see that $(2y^2 + 2yz + 2z^2 + 1) = y^2 + z^2 + (y+z)^2 + 1 > 0$. If we suppose that $y > z$, then from the first equality we have that $x > y$, then applying third equality we conclude: $z > x > y > z$ - contradiction. The case when $y < z$ is completely analogous. Hence we obtain $x = y = z$. and solving the equation we get the answer $(1,1,1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f$, mapping the set of integer numbers into itself, such that for all integers $m$ and $n$ the following equality holds: $f(n|m|) + f(n(|m|+2)) = 2f(n(|m|+1))$.",
    "options": [],
    "answer": "All solutions are piecewise linear with a common intercept: f(n) = a n + b for n > 0, f(0) = b, and f(n) = c n + b for n < 0, where a, b, c are integers.",
    "solution": "Put $n=1$: $f(|m|) + f(|m|+2) = 2f(|m|+1)$. If we denote $l=|m| \\ge 0$, then we can easily get: $f(l+2) - f(l+1) = f(l+1) - f(l)$, so our sequence is an arithmetic progression, hence $f(l) = al + b$, where $a,b \\in \\mathbb{Z}$, $l \\in \\mathbb{Z}^+$. \n\nIn the same way, substituting $n=-1$ we obtain: $f(-|m|) + f(-(|m|+2)) = 2f(-(|m|+1))$. If we denote $k=-|x| \\le 0$, then $\\forall k \\in \\mathbb{Z}^-$, $f(k) + f(k+2) = 2f(k+1)$, and we get: $f(k+2) - f(k+1) = f(k+1) - f(k)$, which implies $f(k) = ck+d$, where $c,d \\in \\mathbb{Z}$, $k \\in \\mathbb{Z}^-$. \n\nSubstituting $|m|=0$ we get on the one hand that $f(0) = 2f(1) - f(2)$, and on the other hand that $f(0) = 2f(-1) - f(-2)$, and so $l=k=0$ $al+b=ck+d$ or $b=d$. \n\nThus, we get the following answers: $f(n) = \\begin{cases} an+b, & n \\in \\mathbb{Z}^+, \\\\ cn+b, & n \\in \\mathbb{Z}^- \\end{cases}$, $a,b,c \\in \\mathbb{Z}$. An easy check shows that all such functions satisfy the condition of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45123,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\frac{AO^2}{BC} + \\frac{BO^2}{CA} + \\frac{CO^2}{AB} \\ge \\frac{AO + BO + CO}{\\sqrt{3}}\n$$\n\nwhere $O$ is the point inside triangle $ABC$ such that $\\angle AOB = \\angle BOC = \\angle COA = 120^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the length of $AO$ by $x$, $BO$ by $y$ and $CO$ by $z$. From the cosine law we have: (fig.26)\n![](attached_image_1.png)\n\nFig.26\n\n$$\nAB = \\sqrt{x^2 + xy + y^2}, \\quad BC = \\sqrt{y^2 + yz + z^2},\n$$\n$CA = \\sqrt{z^2 + zx + x^2}$ and so we can rewrite our inequality:\n$$\n\\frac{x^2}{\\sqrt{y^2 + yz + z^2}} + \\frac{y^2}{\\sqrt{z^2 + zx + x^2}} + \\frac{z^2}{\\sqrt{x^2 + xy + y^2}} \\ge \\frac{x+y+z}{\\sqrt{3}} \\quad (1)\n$$\n\nWithout loss of generality we can suppose that $x \\ge y \\ge z$. Then it is easy to see that\n$$\n\\frac{1}{\\sqrt{y^2 + yz + z^2}} \\ge \\frac{1}{\\sqrt{x^2 + xz + z^2}} \\ge \\frac{1}{\\sqrt{x^2 + xy + y^2}}. \\text{ And also it is obvious that } x^2 \\ge y^2 \\ge z^2. \\text{ Thus}\n$$\nwe can use an obvious inequality:\n$$\n\\frac{x^2}{\\sqrt{y^2 + yz + z^2}} + \\frac{y^2}{\\sqrt{x^2 + xz + z^2}} + \\frac{z^2}{\\sqrt{x^2 + xy + y^2}} \\ge \\frac{y^2}{\\sqrt{y^2 + yz + z^2}} + \\frac{z^2}{\\sqrt{x^2 + xz + z^2}} + \\frac{z^2}{\\sqrt{x^2 + xy + y^2}}, \\\\\n\\frac{x^2}{\\sqrt{y^2 + yz + z^2}} + \\frac{y^2}{\\sqrt{x^2 + xz + z^2}} + \\frac{z^2}{\\sqrt{x^2 + xy + y^2}} \\ge \\frac{z^2}{\\sqrt{y^2 + yz + z^2}} + \\frac{x^2}{\\sqrt{x^2 + xz + z^2}} + \\frac{y^2}{\\sqrt{x^2 + xy + y^2}}\n$$\nTherefore the left hand side of the inequality (1) is no less than\n$$ \\frac{1}{2} \\left( \\frac{y^2+z^2}{\\sqrt{y^2+yz+z^2}} + \\frac{x^2+z^2}{\\sqrt{x^2+xz+z^2}} + \\frac{z^2}{\\sqrt{x^2+xy+y^2}} \\right). $$\nWe will now show that $\\frac{y^2+z^2}{\\sqrt{y^2+yz+z^2}} \\ge \\frac{y+z}{\\sqrt{3}}$. To prove this let us use the power mean inequality:\n$$\n\\frac{y^2+z^2}{\\sqrt{y^2+yz+z^2}} \\ge \\sqrt{\\frac{2}{3}(y^2+z^2)} = \\frac{2}{\\sqrt{3}} \\sqrt{\\frac{y^2+z^2}{2}} \\ge \\frac{2}{\\sqrt{3}} \\cdot \\frac{y+z}{2} = \\frac{y+z}{\\sqrt{3}}\n$$\nIn the same way we can get analogous inequalities for pairs $(x, y)$ and $(x, z)$. And so we finally get:\n$$\n\\frac{y^2+z^2}{\\sqrt{y^2+z^2}} + \\frac{x^2+z^2}{\\sqrt{x^2+xz+z^2}} + \\frac{x^2+y^2}{\\sqrt{x^2+xy+y^2}} \\ge \\frac{2(x+y+z)}{\\sqrt{3}},\n$$\nwhich was to be proved.\nHere we will give another proof of the inequality (1). Using obvious inequalities $yz \\le \\frac{y^2+z^2}{2}$, $xz \\le \\frac{x^2+z^2}{2}$ and $xy \\le \\frac{x^2+y^2}{2}$ we can obtain:\n$$\n\\frac{x^2}{\\sqrt{y^2 + yz + z^2}} + \\frac{y^2}{\\sqrt{z^2 + zx + x^2}} + \\frac{z^2}{\\sqrt{x^2 + xy + y^2}} \\ge \\sqrt{\\frac{2}{3}} \\left( \\frac{x^2}{\\sqrt{y^2 + z^2}} + \\frac{y^2}{\\sqrt{z^2 + y^2}} + \\frac{z^2}{\\sqrt{x^2 + y^2}} \\right).\n$$\nNow denote $S = x^2 + y^2 + z^2$ and consider the function $f(t) = \\frac{1}{\\sqrt{S-t}}$. Since\n$$\nf''(t) = \\frac{1}{\\sqrt{(S-t)^3}} + \\frac{3}{4} \\frac{1}{\\sqrt{(S-t)^5}} = \\frac{4(S-t)+3t}{4\\sqrt{(S-t)^5}} = \\frac{4S-t}{4\\sqrt{(S-t)^5}} \\ge 0\n$$\nfor $0 \\le t < S$, function $f$ is convex on $[0, S)$. And so applying the Jensen's inequality for $(0 \\le x^2, y^2, z^2 < S$ we can get:\n$$\n\\frac{1}{3} \\left( \\frac{x^2}{\\sqrt{y^2+z^2}} + \\frac{y^2}{\\sqrt{z^2+x^2}} + \\frac{z^2}{\\sqrt{x^2+y^2}} \\right) = \\frac{1}{3} \\left( \\frac{x^2}{\\sqrt{S-x^2}} + \\frac{y^2}{\\sqrt{S-y^2}} + \\frac{z^2}{\\sqrt{S-z^2}} \\right) = \\\\\n= \\frac{1}{3} \\left( f(x^2) + f(y^2) + f(z^2) \\right) \\ge f\\left(\\frac{x^2+y^2+z^2}{3}\\right) = \\frac{8}{3\\sqrt{S-\\frac{8}{3}}} = \\sqrt{\\frac{(x^2+y^2+z^2)^2}{6}} \\ge \\frac{x+y+z}{3\\sqrt{2}}\n$$\nAnd finally\n$$\n\\frac{x^2}{\\sqrt{y^2+yz+z^2}} + \\frac{y^2}{\\sqrt{z^2+zx+x^2}} + \\frac{z^2}{\\sqrt{x^2+xy+y^2}} \\ge \\sqrt{\\frac{2}{3}} \\cdot \\frac{1}{\\sqrt{2}} (x+y+z) = \\frac{x+y+z}{\\sqrt{3}},\n$$\nAnd we're done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45124,
    "subject": "Mathematics (Multi-modal)",
    "question": "A country has a road network with the following properties: along each road we can go in both directions; between two different towns may exist at most one road; there does not exist a road which connects some town with itself; the road network is a connected graph; there are at least $m$ roads out of any town $(m = \\text{const} \\ge 3)$. The king of this country wants to go round all the towns, but it turned out that it is impossible. What is the minimum number of towns the country can have?",
    "options": [],
    "answer": "2m+2",
    "solution": "Consider a road network of this country as a graph. Denote the chain of maximum length by $L$ in this graph. Denote $L$ by sequence of connected vertices $A_1 \\to A_2 \\to \\dots \\to A_n$.\n\n![](attached_image_1.png)\nFig.27\n\nThat is, all other chains have the same or less length as $L$. Obviously, there is at least one outside vertex, for instance, some $A_0$, that is outside chain $L$. Besides, as the graph is connected, we can obtain a chain which connects $A_0$ with one of the vertices of chain $L$. Define the properties of the chain obtained. Call the vertex of chain $L$ an inside one, if there are no edges outside chain $L$.\n\n1) Vertices $A_1$ and $A_n$ are not connected with each other, otherwise there will be a chain longer than chain $L$. Really, if there is an edge $A_1A_n$ in the graph, and the vertex $A_0$ is connected with some vertex $A_i$, $i \\in \\{1, \\dots, n\\}$, then there exists the following chain $A_0 \\to A_i \\to A_{i+1} \\to \\dots \\to A_n \\to A_1 \\to \\dots \\to A_{i-1}$ which is longer than chain $L$, a contradiction.\n\n2) Vertices $A_1$ and $A_n$ are inside ones, that is, vertex $A_1$ is connected with $A_2$ and with $(m-1)$ other vertices $A_{k_1}, \\dots, A_{k_m}$ where $2 < k_1 < \\dots < k_m < n$. Then $n \\ge m+2$.\n\n3) Vertices with numbers $(k_1-1), \\dots, (k_m-1)$ are also inside ones. Really, let there be an edge $A_1A_i$, if $A_{i-1}$ isn't an inside vertex, then there is an edge $A_{i-1}A_0'$ in the graph and vertex $A_0'$ isn't inside. Then it is possible to build a chain which is longer than chain $L$: $A_0 \\to A_{i-1} \\to A_{i-2} \\to \\dots \\to A_1 \\to A_i \\to A_{i+1} \\to \\dots \\to A_n$. Therefore, in this chain $L$ there are at least $(m+1)$ inside vertices.\n\n4) As there is an outside vertex which is also connected with $m$ vertices, that are not included in inside vertices mentioned above, totally there must be at least $(2m+2)$ vertices.\n\nShow that there is a corresponding graph with such number of vertices. Consider two groups of vertices $A_1, \\dots, A_m$ and $B_1, \\dots, B_{m+2}$. The graph is built in such a way that each vertex $B_i$, $i=1, \\dots, m+2$ is connected with each vertex $A_j$, $j=1, \\dots, m$, that is, the degree of each vertex $B_i$ equals $m$, and the degree of each vertex $A_j$ equals $(m+2)$. If there is a chain which includes all the vertices, then there must be a vertex $B_i$ after each vertex $A_j$ and vice versa, because vertices with the same letter are not connected with each other. But there are $m+2$ vertices $B_i$, so we must have at least $m+1$ vertices $A_j$, and there are less. The example given concludes our proof.\nShow another proof of the existence of the chain which includes all towns if $n \\le 2m+1$. Let $A_1, A_2, \\dots, A_n$ be the vertices of our graph. We add one more vertex $B$ to the graph and connect vertex $B$ with all vertices $A_1, A_2, \\dots, A_n$. The new graph satisfies the condition that it has vertices less than $n+1 \\le 2m+2$ and the degree of each vertex more than $m+1$. According to the theorem about a Hamiltonian circuit, in the new graph there exists a Hamiltonian circuit. Therefore, we can represent its circuit as the cycle $B - A_1 - A_2 - \\dots - A_n - B$. Among vertices $A_i$, $k=1, \\dots, n$, we have not added vertex $B$. Then the chain $A_1 - A_2 - \\dots - A_n$ is a circuit which goes round all the towns.\n\nThe example we can take from the first solution.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 45125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Decimal representation of a number $a$ is written one or several times on the blackboard. As a result, the binary representation of the same number $a$ is obtained. Find all possible values of $a$.",
    "options": [],
    "answer": "1 and 10",
    "solution": "Let decimal representation of $a$ consist of exactly $k$ digits, and binary representation consists of exactly $l$ times more digits, i.e. of $kl$ digits. Since all digits of $a$ equal to either $0$ or $1$, then the number $a$ should belong to, from one side, the interval $[10^{k-1}, \\frac{1}{6}(10^k - 1)]$, and from the other side, to the interval $[2^{k/l-1}, 2^{k/l} - 1]$. Hence we come to two inequalities: $10^{k-1} \\le 2^{k/l} - 1$ and $\\frac{1}{6}(10^k - 1) \\ge 2^{k/l-1}$. After transformation, we get $\\frac{2}{3} < (\\frac{10}{27})^k < 10$. This inequality, obviously, is wrong for $l \\ge 4$.\n\nFor $l=1$ we have the only possible value $k=1$ and the first answer - number $1$.\n\nIf $l=2$, then $k=2$. From all two-digit numbers, having only $0,1$ as digits, only $10$ suits - the second answer.\n\nFor the last case, $l=3$, possible values are $k \\in \\{7,8,9,10\\}$. Let's examine them.\n\nFor $k=10$ we have $10^9 > 2^{29} + 2^{28}$. That is, if $a$ is a ten-digit number such that written thrice it represents binary form of itself, then it has to have $1$ in the second-highest decimal position, i.e. $a > 11 \\cdot 10^8 > 2^{30}$ - a contradiction (binary representation of $a$ should be a $30$-digit number).\n\nFor all other values of $k$ we see that $a = (2^{2k} + 2^k + 1)$. For $k=9$, $2^{2k} + 2^k + 1 = 262657$. Numbers $10^0, 10^1, ..., 10^8$ give remainders $1, 10, ..., 100000, 212029, 19034$ and $190340$ under the division by $262657$, respectively. Since $a$ is a sum of several powers of $10$ with the highest $10^8$, we have to choose from $1, 10, ..., 100000, 212029, 19034, 190340$ such that they sum up either to $262657-190340$, or to $2 \\cdot 262657-190340$. By examination of options we see that it is impossible to get such numbers. Similarly, we can check all other cases. Let's see (for example) the case $k=8$. $2^{10} + 2^8 + 1 = 65793$. The remainders of $10^0, 10^1, ..., 10^7$ under the division by $65793$ equal $1, 10, ..., 10000, 234207, 13105$ and $-536$. We have to select a combination of powers, summing up to $536$ - others impossible. By checking two last digits ($00, 01, 10, 07, 05$) we see that it is impossible to get the required two last digits $36$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eight teams are playing single round-robin tournament, that is each team plays every other team once. And at that the schedule is arranged by rounds, in every round all teams play. What is the least possible number of rounds should finish to determine the best and the worst teams (that is, for any results of the games in uncompleted round, this teams will have strictly highest and strictly lowest numbers of points), if the tournament is:\n\na) volleyball (that is, winner receives 1 point, loser - 0 and there are no ties);\nb) handball (that is, winner receives 2 points, loser - 0, and for a tie each team receives 1 point)?",
    "options": [],
    "answer": "a) 6 rounds; b) 5 rounds",
    "solution": "a) Lets show that at least 6 rounds are required. It is easy to construct an example for 6 rounds. We need to show that 5 rounds are not enough. Indeed, even if the best team won all games, to ensure first place, it should have at least 3 point advantage. Similarly, the worst team should have at least 3 point disadvantage. But such situation is impossible because the highest possible difference between the best and the worst team after 5 rounds is 5 points.\n\nb) Lets show that at least 5 rounds required. It is easy to construct an example (one team gets 10 points, one - zero points and all other 6 teams get 5 points each. In the following 2 rounds each team can get at most 4 points). Absolutely similarly to the part a) it is possible to show that 4 rounds aren't enough.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of pairs of positive integer numbers $(n, m)$, satisfying $(2^k)! = 2^n m$.",
    "options": [],
    "answer": "2^k - 1",
    "solution": "Note that if $(2^k)! : 2^l$, then the pair $(l, (2^k)!)$ satisfies the equation and vice versa: if a pair $(n, m)$ is the solution of equation, then $(2^k)! : 2^n$. That is, the number of solutions equals to the number of divisors of the kind $2^l$ of $(2^k)!$. By the Legendre theorem the latter equals: $[2^k/2] + [2^k/2^2] + [2^k/2^3] + \\ldots = 2^{k-1} + 2^{k-2} + \\ldots + 2^1 + 2^0 = 2^k - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove for any positive integer $n$:\n$$\n\\sum_{k=0}^{n-1} \\frac{\\left(\\binom{n-1}{k}\\right)^2}{k+1} = \\frac{\\binom{2n-1}{n}}{2n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\frac{n \\binom{n-1}{k}}{k+1} = \\frac{n(n-1)!}{(k+1)k!(n-k-1)!} = \\frac{n!}{k!(n-k-1)!} = \\binom{n}{k+1}$, the left-hand-side sum takes the form\n$$\n\\sum_{k=0}^{n-1} \\binom{n-1}{k} \\binom{n}{k+1} = \\sum_{k=0}^{n-1} \\binom{n-1}{k} \\binom{n}{n-k-1} = \\binom{2n-1}{n}.\n$$\nThe latter equality follows from the properties of binomial coefficients. On the other hand,\n$$\n\\binom{2n}{n} = \\binom{2n-1}{n-1} + \\binom{2n-1}{n} = 2\\binom{2n-1}{n-1},\n$$\nwhich finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(x, y)$ satisfying the equation:\n$$x^y + y^x = 2008.$$",
    "options": [],
    "answer": "(1, 2007) and (2007, 1)",
    "solution": "Let $x \\le y$. $x=1$ gives us an obvious solution $y=2007$.\n\nIn what follows, suppose that $2 \\le x \\le y$. We can check for small values of $x$:\n\n$2^{10} = 1024 < 2008 < 2^{11} = 2048$,   $3^6 = 729 < 2008 < 3^7 = 2187$, \n$4^5 = 1024 < 2008 < 4^6 = 4096$,   $5^4 = 625 < 2008 < 5^5 = 3125$,   $6^4 = 1296 < 2008 < 6^5 = 7776$,   $7^3 = 343 < 2008 < 7^4 = 2401$,   $8^3 = 512 < 2008 < 8^4 = 4096$,   $9^3 = 729 < 2008 < 9^4 = 6561$,   $10^3 = 1000 < 2008 < 10^4 = 10000$.\n\nIt is clear, that there is no need to continue. That is, it follows from the relations $2^{10} + 10^2 = 1124 < 2008$, $3^6 + 6^3 = 945 < 2008$, $4^5 + 5^4 = 1649 < 2008$ that there are no more new solutions (for further values of $x$, the condition $2 \\le x \\le y$ breaks).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45130,
    "subject": "Mathematics (Multi-modal)",
    "question": "The integers $a, b, c$ satisfy the following condition: $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 0$. Prove that the number $a^2 + b^2 + c^2$ is a perfect square of some integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We note the following facts: $abc \\neq 0$ and $ab + bc + ca = 0$, then\n$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = a^2 + b^2 + c^2$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimum positive integer number which has a product of its digits equals $5120$.",
    "options": [],
    "answer": "25888",
    "solution": "Since $5120 = 2^{10} \\cdot 5$, then the main question is – how many digits have its number? Obviously among these digits it hasn't $0$ and has $5$ because only $5$ from these digits divides by $5$. From the ten factors $2$ we can produce at the least four digits – $2$, $8$, $8$, $8$ or $4$, $4$, $8$, $8$. Therefore from two sets of digits $5$, $2$, $8$, $8$, $8$ and $5$, $4$, $4$, $8$, $8$ we must produce a minimum of numbers. The minimum of first figure is $2$ so that the required minimum integer number is $25888$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45132,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Four positive integer numbers $a$, $b$, $c$, $d$ satisfy the condition: every number $ab$, $bc$, $cd$, $da$ is a perfect cube. Are all the numbers $a$, $b$, $c$, $d$ perfect cubes?\n\nb) Five positive integer numbers $a$, $b$, $c$, $d$, $e$ satisfy the condition: every number $ab$, $bc$, $cd$, $de$, $ea$ is a perfect cube. Are all the numbers $a$, $b$, $c$, $d$, $e$ perfect cubes?",
    "options": [],
    "answer": "a) No; for example a=c=2 and b=d=4. b) Yes; all five numbers must be perfect cubes.",
    "solution": "a) The example that it isn't necessary: $a = c = 2$, $b = d = 4$.\n\nb) It shows that if a positive integer number $n$ has its square $n^2$ as a perfect cube of an integer number, then $n$ itself is a perfect cube of some integer number. Really, consider a factorization of $n$ into prime numbers $n = p_1^{m_1} \\ldots p_k^{m_k}$, its square is $n^2 = p_1^{2m_1} \\ldots p_k^{2m_k}$, if $p_1^{2m_1} \\ldots p_k^{2m_k}$ is a perfect cube then for each $i = 1, k$ we have to hold the condition $2m_i \\equiv 0 \\pmod{3}$ which is equal to the condition $m_i \\equiv 0 \\pmod{3}$. Hence, the number $n$ is a perfect cube.\n\nAs $\\frac{ab \\cdot cd \\cdot ea}{bc \\cdot de} = a^2 = a_1^3$, then each of the numbers $a$, $b$, $c$, $d$, $e$ is a perfect cube.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integer numbers $n$ such that the number $3\\pi$ is a period of the function $f(x) = \\cos nx \\cdot \\sin \\frac{2009x}{n^2}$.",
    "options": [],
    "answer": "1, 7",
    "solution": "$3\\pi$ is a period of the function $f(x)$ if and only if $f(x+3\\pi) = f(x)$ for arbitrary real number $x$. Putting $x=0$ we obtain that $f(0)=0$, since $f(3\\pi) = \\cos 3\\pi n \\cdot \\sin \\frac{3 \\cdot 2009\\pi}{n^2}$ we have $0 = \\cos 3\\pi n \\cdot \\sin \\frac{3 \\cdot 2009\\pi}{n^2} \\Leftrightarrow 0 = \\sin \\frac{3 \\cdot 2009\\pi}{n^2} \\Leftrightarrow \\frac{6027\\pi}{n^2} = \\pi k,\\ k \\in \\mathbb{Z} \\Leftrightarrow 6027 : n^2$.\n\nAs $6027 = 3 \\cdot 7^2 \\cdot 41$ thus the possible values for $n$ can only be numbers $1$ or $7$. It is directly verified that the numbers $1$ and $7$ satisfy the same initial conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the sequence of polynomials $P_0(x), P_1(x), \\dots, P_n(x)$, $n \\ge 2$. It is known that for every integer $i$ ($0 \\le i \\le n$): $\\deg(P_i(x)) = n-i$, and $P_n(x) \\ne 0$. It is also known that for every integer $i$ ($2 \\le i \\le n$) there exists polynomial $Q_i(x)$ such that $P_i(x) = P_{i-2}(x) + P_{i-1}(x)Q_i(x)$. If polynomials $R(x)$ and $S(x)$ satisfy $P_0(x)R(x) + P_1(x)S(x) = 1$ for all real $x$, prove that $\\deg(R(x)) \\ge n-2$ and $\\deg(S(x)) \\ge n-1$. (Where by $\\deg P(x)$ we denote the degree of polynomial $P(x)$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is possible that the degree of the sum of two polynomials is less than the degree of one summand only if their degrees equal. For shortening we pull down $(x)$ in the notation. $\\deg P_i < \\deg P_{i-2}$ and $P_i = P_{i-2} + P_{i-1}Q_i$, therefore $\\deg(P_{i-1}Q_i) = n-i+2$, hence $\\deg(Q_i) = 1$ for all integer $i = 2, n$.\n\nLet's prove by induction by $i=2, n$, that there exist polynomials $R_i$ and $S_i$ of degrees $i-2$ and $i-1$ respectively, satisfying $P_i = P_0 R_i + P_1 S_i$.\n\nBase. For $i=2$ we can get $R_i = 1$ and $S_i = Q_2$.\n\nInduction step. Let for all $i$, $2 \\le i \\le k$ ($k < n$) there exist required $R_k$ and $S_k$. We will construct required polynomials for $i = k+1$. We have from the statement: $P_{k+1} = P_{k-1} + P_k Q_{k+1}$.\n\nIn the case $k > 2$ we get from the induction assumption:\n\n$P_{k+1} = (P_0 R_{k-1} + P_1 S_{k-1}) + (P_0 R_k + P_1 S_k)Q_{k+1} = (R_{k-1} + R_k Q_{k+1})P_0 + (S_{k-1} + S_k Q_{k+1})P_1$.\n\nAnd we can set $R_{k+1} = R_{k-1} + R_k Q_{k+1}$ and $S_{k+1} = S_{k-1} + S_k Q_{k+1}$.\n\nFor $k=2$: $P_{k+1} = P_1 + (P_0 R_2 + P_1 S_2)Q_3 = (R_2 Q_3)P_0 + (1 + S_2 Q_3)P_1$. And we can set $R_{k+1} = R_2 Q_3$ and $S_{k+1} = 1 + S_2 Q_3$. In both cases degrees of $R_{k+1}$ and $S_{k+1}$ are as required. The statement proved.\n\nIf we set $i = n$, then $P_i = P_0 R_i + P_1 S_i$, or, taking into account $P_i \\neq 0$ and $\\deg(P_i) = 0$, for $R_{n0} = \\frac{R_n}{P_n}$ and $S_{n0} = \\frac{S_n}{P_n}$ it holds $P_0 R_{n0} + P_1 S_{n0} = 1$ (and at that $\\deg(R_{n0}) = n - 2$ and $\\deg(S_{n0}) = n - 1$).\n\nLet for some other $R$ and $S$ it holds $P_0 R + P_1 S = 1$. Then $P_0(R - R_{n0}) + P_1(S - S_{n0}) = 0$, or $P_0 \\tilde{R} + P_1 \\tilde{S} = 0$ for $\\tilde{R} = R - R_{n0}$, $\\tilde{S} = S - S_{n0}$.\n\nLet's prove that $\\tilde{R}$ divides by $P_1$, i.e. there exists polynomial $C$, for which $\\tilde{R} = P_1 C$. By multiplying both sides by $R_{n0}$, we get $P_0 R_{n0} \\tilde{R} + P_1 R_{n0} \\tilde{S} = 0$. Taking into account preceding formulas: $P_0 R_{n0} = 1 - P_1 S_{n0}$. Let's substitute this equality into previous one: $(1 - P_1 S_{n0}) \\tilde{R} + P_1 R_{n0} \\tilde{S} = 0$, whence $\\tilde{R} - P_1 S_{n0} \\tilde{R} + P_1 R_{n0} \\tilde{S} = 0$, or $\\tilde{R} = P_1 (S_{n0} \\tilde{R} - R_{n0} \\tilde{S})$, which is required.\n\nAnd so either $\\tilde{R} = 0$ or $\\deg \\tilde{R} \\ge \\deg P_1 = n - 1$. If $\\tilde{R} = 0$ then $\\deg R = \\deg R_{n0} = n - 2$. If $\\deg \\tilde{R} \\ge n - 1$ then $\\deg R = \\deg \\tilde{R} \\ge n - 1$. Therefore $\\deg R \\ge n - 2$. It can be proved in the similar manner that $\\tilde{S}$ divides by $P_0$, and from this that $\\deg S \\ge n - 1$, which is to be proven.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all such positive integer $n$ that for every polynomial $g(x)$ with real coefficients there exist such polynomials $p_1(x), p_2(x),..., p_n(x)$ and $q_1(x), q_2(x),..., q_n(x)$ that $g(x) = \\sum_{i=1}^{n} (p_i^2(x) + q_i^3(x))$.",
    "options": [],
    "answer": "All positive integers n greater than or equal to 2",
    "solution": "For $n \\ge 2$ let $q_1(x) = q_2(x) = ... = q_n(x) = 0$ and $p_2(x) = p_3(x) = ... = p_n(x) = 0$, $q_2(x) = -\\frac{x+1}{3}$, $q_1(x) = \\frac{x-2}{3}$, $p_1(x) = \\frac{1}{\\sqrt{3}}(x+1)$. Then\n\n$$\n\\sum_{i=1}^{n} (p_i^2(x) + q_i^3(x)) = x, \\quad (1)\n$$\nand changing $x$ by $g(x)$ for all polynomials $p_i(x)$ and $q_i(x)$, any polynomial $g(x)$ can be obtained:\n\n$$\nq_2(x) = -\\frac{g(x)+1}{3}, \\quad q_1(x) = \\frac{g(x)-2}{3}, \\quad p_1(x) = \\frac{1}{\\sqrt{3}}(g(x)+1),\n$$\nfrom (1) it is easy to see that we get what is required.\n\nLets show that $n = 1$ does not suit. Suppose that there exist such polynomials $p(x)$ and $q(x)$ that\n$$\np^2(x) + q^3(x) = x. \\quad (2)\n$$\nChanging $x$ by $x^2$ in (2), we get: $(x - p(x^2))(x + p(x^2)) = q^3(x^2)$. Suppose that both polynomials $x - p(x^2)$ and $x + p(x^2)$ have common (possibly complex) root $\\alpha$, then $\\alpha = p(\\alpha^2) = -p(\\alpha^2)$, hence $\\alpha = 0$ and $p(0) = p(\\alpha^2) = \\alpha = 0$, $q(0) = 0$. But in such case the left-hand side of (2) is divisible by $x^2$, and the right-hand side is not, which is\n\nimpossible. So $x - p(x^2)$ and $x + p(x^2)$ have no common roots. Then each of them has to be a cube of some polynomial (with, possibly, complex coefficients). I.e. $x - p(x^2) = q_1^3(x)$ and $x + p(x^2) = q_2^3(x)$. Adding up these equalities we get\n$$\n2x = q_1^3(x) + q_2^3(x) = (q_1(x) + q_2(x))(q_1^2(x) - q_1(x)q_2(x) + q_2^2(x)),\n$$\nand 2 cases are possible:\n1) $q_1(x) + q_2(x) = c$;\n2) $q_1(x) + q_2(x) = 2cx$.\nFor the first case, $\\frac{2x}{c} = q_1^2(x) - q_1(x)q_2(x) + q_2^2(x) = c^2 - 3q_1(x)(c - q_1(x))$, which is impossible because the degree of the left-hand side is odd and of the right-hand side is even.\nFor the second case $\\frac{1}{c} = q_1^2(x) - q_1(x)q_2(x) + q_2^2(x) = 4c^2x^2 - 3q_1(x)(2cx - q_1(x))$, thus $\\deg q_1(x) = 1$, i.e. $q_1(x) = ax + b$ ($a \\neq 0$). Then the equality $p(x^2) = x - q_1^3(x)$ is impossible, because the right-hand side is the polynomial of the third degree, and the left-hand side is the polynomial of even degree.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's call the filling of the square $2009 \\times 2009$ which is divided into unit squares \"correct\" if it is filled with numbers $1,2,\\ldots,2009$ in such a way that each of these numbers appears in every row and every column. Let's consider the distance from the central square to the nearest square with number «1» (by the distance we mean the least number of moves of the king to get to the square). What is the maximum value the distance can have?",
    "options": [],
    "answer": "502",
    "solution": "The square with number «1» which is the nearest to the center is located on the longest diagonal. The distance to the border is $1004$, therefore the distance to the nearest square is $502$. Let us show that this distance is the shortest possible. By condition the distance between two squares is determined by the minimum difference between horizontal and vertical lines of squares' location. Let us assess the distance from the nearest number «1» to the center. Note that in upper and lower $502$ horizontals there are $2 \\cdot 502 = 1004$ numbers «1», because each row on the square has exactly one number «1». Analogously in $502$ left and right verticals there are no more than $1004$ numbers «1». Therefore there is at least one number «1» in the central square $1005 \\times 1005$. And the distance to each square of this \"big\" square to the center is not more than $502$. Our solution is complete.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45137,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Consider the square of the numbers $1$, $2$, $\\ldots$, $2009$ written one after another in some random order as one number. Is it possible to obtain a perfect square of an integer in this way?\n\nb) Consider the numbers $1$, $2$, $\\ldots$, $2009$ written one after another in some random order as one number. Is it possible to obtain a perfect square of an integer in this way?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Consider remainders of these squares divided by $3$. In the big number which we have generated each square can be represented as $k^2 \\cdot 10^n$, and the big number is the sum of such terms. If $k^2 \\equiv l \\pmod{3}$ then $k^2 \\cdot 10^n \\equiv l \\pmod{3}$. It is well-known that $k^2 \\equiv 0 \\pmod{3}$ is equivalent to $3|k$, and $k^2 \\equiv 1 \\pmod{3}$ is equivalent to $3 \\nmid k$. There are $669$ numbers from $1$ to $2009$ that are divisible by $3$, and $1340$ there are not. As $1340 \\equiv 2 \\pmod{3}$ the remainder of the big number modulo $3$ is $2$, which is impossible for a perfect square, which completes the proof.\n\nb) Consider remainders of these numbers divided by $9$. In the big number which we have generated each number can be represented as $k \\cdot 10^n$, and the big number is the sum of such terms. $k \\equiv l \\pmod{9}$ if and only if $k \\cdot 10^n \\equiv l \\pmod{9}$. Thus we need to find the sum of the remainders which equals to the sum of all digits of the big number. If we break the number into the groups of nine $1$, $2$, $\\ldots$, $9$; $10$, $11$, $\\ldots$, $18$; $\\ldots$, $1999$, $2000$, $\\ldots$, $2007$, then the sum of the numbers in each group is divisible by $9$. The sum of the remaining numbers $2008$ and $2009$ is divisible by $3$ but is not divisible by $9$, which is impossible for a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all solutions in positive integer $m, n$ to the equation\n$$\nm n^2 = 2009(n + 1).\n$$",
    "options": [],
    "answer": "(m, n) = (4018, 1) and (328, 7)",
    "solution": "Obviously, $n$ and $(n+1)$ are coprime numbers. Consequently, for this equation we have that $m$ is divisible by $(n+1)$ and $2009$ is divisible by $n^2$. Since $2009 = 7^2 \\cdot 41$, this can be achieved in two ways: $n=1$ or $n=7$.\n\nIf $n=1$, then $m = 2009 \\cdot 2 = 4018$.\n\nIf $n=7$, then $49m = 2009 \\cdot 8$, therefore $m = 41 \\cdot 8 = 328$.\n\nSo this equation has two solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45139,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane let be given a circle with a chord $AB$ and a point $P$ on this chord such that $AP = 2PB$. The chord $DE$ is normal to the chord $AB$ and passing through the point $P$. Prove that the midpoint of the segment $AP$ is the orthocenter of the triangle $AED$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be a midpoint of the segment $AP$. Let us the line $EM$ intersects the segment $AD$ at a point $Q$. $MPE$ and $BPE$ are the congruent triangles because both of them are right and both of them have the equal legs. Then $\\angle QAM = \\angle DAB = \\angle DEB = \\angle PEB = \\angle PEM$ (Fig.9).\n\nSince $\\angle QMA = \\angle EMP$ then triangles $QMA$ and $EMP$ have in twos equal angles. Consequently both of them have equal third angles, then $\\angle AQM = \\angle MPE = 90^\\circ$, then $EQ$ and $AP$ are the altitudes of the triangle $ADE$ which proves that $M$ is the orthocenter of that triangle. That completes the proof.\n\n![](attached_image_1.png)\nFig.9",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a polynomial $f(x) = x^3 + a x^2 + b x + c$ satisfying simultaneously the following conditions: $|c| \\leq 2009$, $f$ has 3 integer roots and $|f(34)|$ is a prime number.",
    "options": [],
    "answer": "No",
    "solution": "Let $f(x) = (x - \\alpha)(x - \\beta)(x - \\gamma)$ be a polynomial which satisfies the necessary conditions. Then $\\alpha, \\beta, \\gamma$ are integer numbers and $|f(34)| = |(34 - \\alpha)(34 - \\beta)(34 - \\gamma)|$ is a prime number. Without loss of generality we have $|34 - \\alpha| = |34 - \\beta| = 1$ and $|34 - \\gamma|$ is a prime, it follows $\\alpha, \\beta \\geq 33$. The nearest prime numbers for $34$ are $31$ and $37$ then $|\\gamma| \\geq 3$.\nTherefore $|c| = |\\alpha \\beta \\gamma| \\geq 33 \\times 33 \\times 3 = 3267 > 2009$. A contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45141,
    "subject": "Mathematics (Multi-modal)",
    "question": "A board $2009 \\times 2009$ is divided into unit squares. Two players play the following game. They take turns to paint in yellow colour an unpainted unit segment which is a side of a unit square. If after a turn a unit square with all four sides painted in yellow is obtained, then the player who made this turn wins. Who has a winning strategy?",
    "options": [],
    "answer": "The second player has a winning strategy.",
    "solution": "The second player can follow the following rules for his steps.\n\n1) If he can, he paints the side which is the fourth yellow side in some square. He does it and he wins in this game.\n\n2) If the first player paints some side, the second player paints the side which is symmetric to the one just painted about the centre of the field $2009 \\times 2009$. On the Fig. 4 we can see the corresponding steps.\n\n![](attached_image_1.png)\n\nFig. 4\n\nThus we have that 1) in this game certainly the first or the second player wins; 2) the second player can always make the next step which satisfies his rules. Suppose the first player can win. Then his last step is painting the side which is the fourth yellow side in some square. Let it be the segment $AB$ (Fig. 5). Before that, the second player had to paint one of the segments $BC$, $CD$ or $DA$. Otherwise, as soon as these three segments were painted by the first player, the second player would have painted the segment $AB$ following the first rule and the game would be finished. So he had to paint one of the segments $BC$, $CD$ or $DA$. It means that the square $A'B'C'D'$ which is symmetric to the square $ABCD$ about the centre of the board had painted sides $B'C'$, $C'D'$ and $D'A'$ before a move of the second player. And so the second player would have painted the segment $A'B'$ to win this game. Obviously, this argument also works for the central square.\n\n![](attached_image_2.png)\n\nFig. 5",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b \\in [-1,1]$. Prove that $a\\sqrt{1-b^2} + b\\sqrt{1-a^2} \\le 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By substitution: $a = \\sin \\alpha$, $b = \\sin \\beta$ the given inequality becomes: $\\sin \\alpha \\cos \\beta + \\sin \\beta \\cos \\alpha \\le 1$. This inequality holds for an arbitrary $\\alpha, \\beta$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist infinitely many such prime numbers $p$ that amongst $\\{0,1,2,\\dots, p-1\\}$ can be found at least 2008 numbers $x$ such that $x^a \\equiv 1 \\pmod{p}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p$ be some prime divisor of the number $2^{2^n} + 1$. Consider numbers $x$ of the form $2^k$. By the statement of the problem we need $x^x - 1 = 2^{2^s} - 1 \\equiv 0 \\pmod{p}$. If $s \\ge n+1$ then\n$$\n2^{2^{2^k}} - 1 = (2^{2^{n+1}})^{2^{2^k-(n+1)}} - 1 : 2^{2^{n+1}} - 1 = (2^{2^k} - 1)(2^{2^k} + 1),\n$$\nwhere $a: p$ denotes \"a is divisible by $p\", i.e. $p|a$. Thus $2^{2^k} - 1$ is divisible by $p$. So if all the numbers $2^{2^{n+1}}, 2^{2^{n+2}}, \\dots, 2^{2^{2008}}$ are less than $p$, the problem is solved. The solution of the problem follows from the lemma, stated below, for $A = 2^{2008}$.\n**Lemma 1.** For any positive integer number $A$ there exists such $n_0$ that all numbers of a kind $2^{2^n} + 1$, where $n \\ge n_0$, have prime divisor greater than $A \\cdot 2^{2^n}$.\nTo prove lemma 1, we need an additional lemma.\n**Lemma 2.** If the prime number $p$ is a divisor of the number $2^{2^n} + 1$, then $2^{n+1} \\le p-1$.\n*Proof* (of lemma 2). Let $\\Delta$ be the least positive integer such that $p|2^\\Delta - 1$. We show that $\\Delta | p-1$. It follows from Fermat's (little) theorem that $2^{p-1} \\equiv 1 \\pmod{p}$. Let $\\Delta_1$ be the remainder in division of $p-1$ by $\\Delta$. Then from $2^\\Delta \\equiv 1 \\pmod{p}$ and $2^{p-1} \\equiv 1 \\pmod{p}$ it follows that $2^{\\Delta_1} \\equiv 1 \\pmod{p}$. From the supposition of minimality of $\\Delta$ it follows that $\\Delta_1 = 0$, i.e. $\\Delta | p-1$.\n\n$2^{2^{n+1}} - 1 = (2^{2^n} - 1)(2^{2^n} + 1)$, so $p \\mid 2^{2^{n+1}}$. Reasoning by analogy with the above, one gets $\\Delta \\mid 2^{n+1}$, that is $\\Delta = 2^k$, $k \\le n+1$. If $k \\le n$ then\n$2^{2^n} + 1 = (2^{2^k})^{2^n-2^k} + 1 \\equiv 1^{2^n-2^k} + 1 \\equiv 2 \\pmod p$, which is impossible due to $p \\mid 2^{2^{n+1}}$. Hence $\\Delta = 2^{n+1}$, which (with $\\Delta \\mid p-1$ proven above) finishes the proof of the lemma.\n*Proof* (of lemma 1). It follows from lemma 2 that all prime divisors of the number $2^{2^n} + 1$ can be expressed as $2^{n+1} \\beta_j + 1$. Let\n$$\n2^{2^n} + 1 = p_1^{\\alpha_1} \\cdots p_s^{\\alpha_s}, \\text{ where } p_j = 2^{n+1} \\beta_j + 1.\n$$\nThen\n$$\np_j^{\\alpha_j} = (2^{n+1} \\beta_j + 1)^{\\alpha_j} = 1 + \\binom{\\alpha_j}{1} 2^{n+1} \\beta_j + \\binom{\\alpha_j}{2} 2^{2(n+1)} \\beta_j^2 + \\dots \\equiv 1 + \\alpha_j 2^{n+1} \\beta_j \\pmod{2^{2(n+1)}}.\n$$\nThus,\n$$\np_1^{\\alpha_1} \\cdots p_s^{\\alpha_s} \\equiv \\prod_{j=1}^s (1 + \\alpha_j 2^{n+1} \\beta_j) \\equiv 1 + 2^{n+1} \\sum_{j=1}^s \\alpha_j \\beta_j \\pmod{2^{2(n+1)}}. \\quad (1)\n$$\nIf $2^{2n+2} < 2^{2^n}$ (which is true for $n > 2$), then $2^{2^n} + 1 \\equiv 1 \\pmod{2^{2(n+1)}}$, hence we can get from (1) $2^{n+1} \\sum_{j=1}^s \\alpha_j \\beta_j \\equiv 0 \\pmod{2^{2(n+1)}}$, therefore $\\sum_{j=1}^s \\alpha_j \\beta_j \\equiv 0 \\pmod{2^{n+1}}$. Note that $\\alpha_j > 0$ and $\\beta_j > 0$, so $\\sum_{j=1}^s \\alpha_j \\beta_j \\ge 2^{n+1}$.\nSuppose that all $\\beta_j < A$ (otherwise, when one of the $\\beta_j \\ge A$, let it be $\\beta_{j_0}$, the number $2^{2^n} + 1$ has a prime divisor $2^{n+1} \\beta_{j_0} + 1 > 2^{n+1} A$ and the lemma is proved). Then\n$$\n2^{n+1} \\le \\sum_{j=1}^s \\alpha_j \\beta_j < A \\sum_{j=1}^s \\alpha_j \\le A s \\alpha_k, \\text{ where } \\alpha_k = \\max_{1 \\le j \\le s} \\alpha_j.\n$$\nAll $p_j$ are distinct numbers, thus all $\\beta_j$ are also distinct numbers, and the number of $\\beta_j$ is $s$. By supposition all $\\beta_j < A$, whence we get $s < A$. So $2^{n+1} < A^2 \\alpha_k$, that is $\\alpha_k > \\frac{2^{n+1}}{A^2}$, but then $p_k^{\\alpha_k} = (2^{n+1} \\beta_k + 1)^{\\alpha_k} > (2^{n+1})^{\\alpha_k} > 2^{(n+1)2^{n+1}/A^2}$. Since $p_k^{\\alpha_k} \\le 2^{2^n} + 1$, then $2^{(n+1)2^{n+1}/A^2} < 2^{2^n} + 1$. (2)\nThere exists such $n_0$ that for $n \\ge n_0$ we have $\\frac{n+1}{A^2} > 1$. Then it follows from (2) that\n$$\n2^{2^n} + 1 \\ge 2^{(n+1)2^{n-1}/A^2} > 2^{2^{n+1}} > 2^{2^n} + 1.\n$$\na contradiction, which proves the lemma.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let be given positive numbers $a$, $b$, $c$. Prove that\n$$\n\\frac{a}{3a^2 + b^2 + 2ac} + \\frac{b}{3b^2 + c^2 + 2ab} + \\frac{c}{3c^2 + a^2 + 2bc} \\le \\frac{3}{2(a+b+c)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using for the first denominator such inequality $3a^2 + b^2 + 2ac = 2a^2 + a^2 + b^2 + 2ac \\ge 2a^2 + 2ab + 2ac = 2a(a+b+c)$ and analogously for the rest two denominators. Then the left-hand side becomes in the form of:\n$$\n\\begin{aligned}\n\\frac{a}{3a^2 + b^2 + 2ac} + \\frac{b}{3b^2 + c^2 + 2ab} + \\frac{c}{3c^2 + a^2 + 2bc} &\\le \\frac{3}{2a(a+b+c)} + \\frac{3}{2b(a+b+c)} + \\frac{3}{2c(a+b+c)} = \\frac{3}{2(a+b+c)} \\quad \\text{that completes the proof.}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a given triangle $A_1A_2A_3$ and a point $X$ which doesn't lie on each of lines $A_1A_2$, $A_1A_3$ and $A_2A_3$ be given. Denote by $l_i$ line which passes through the point $A_i$ and is normal to the chord line $XA_i$, $i=1,2,3$. Let are the points $B_1 = l_2 \\cap l_3$, $B_2 = l_3 \\cap l_1$, $B_3 = l_1 \\cap l_2$, and $H_1, H_2, H_3$ be the orthocenters the triangles $B_1A_2A_3$, $B_2A_3A_1$, $B_3A_1A_2$, respectively. Prove that the triangle $H_1H_2H_3$ is congruent to the triangle $A_1A_2A_3$.\n\n![](attached_image_1.png)\nFig.10",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $A_3H_2 \\parallel A_1X$ as the lines which are normal to the $l_i$, analogously $A_1H_2 \\parallel A_3X$ then the quadrangle $A_1XA_3H_2$ is a parallelogram. Then we have that $A_1XA_2H_3$ and $A_3XA_2H_1$ are the parallelogram too. Therefore the $A_3H_2$ and $A_2H_3$ are the equal and parallel segments. Then quadrangle $A_2A_3H_2H_3$ is a parallelogram, therefore $A_2A_3 = H_2H_3$. Analogously for the rest pairs of sides of triangles $A_1A_2A_3$ and $H_1H_2H_3$, hence they are the congruent triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let be given positive numbers $a, b, c$. Prove that:\n$$\n\\frac{a}{2a^2 + b^2 + c^2} + \\frac{b}{2b^2 + c^2 + a^2} + \\frac{c}{2c^2 + a^2 + b^2} \\le \\frac{9}{4(a+b+c)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the inequality $a^2 + b^2 + c^2 \\ge ab + bc + ca$ we have\n$$\n\\frac{a}{(a+b)(a+c)} + \\frac{b}{(b+c)(b+a)} + \\frac{c}{(c+a)(c+b)} \\le \\frac{9}{4(a+b+c)}\n$$\n, is equivalent to\n$$ 8(a+b+c)(ab+bc+ca) \\le 9(a+b)(b+c)(c+a) $$\nand this equivalent to inequality\n$$ 6abc \\le a^2b + a^2c + b^2a + b^2c + c^2a + c^2b $$\nwhich follows immediately by the AM-GM inequality applied to the six numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist infinitely many natural numbers $n$ with the following properties: we can represent number $n$ as a sum $n = a^2 + b^2$, and a sum $n = c^3 + d^3$, and can't represent it as a sum $n = x^6 + y^6$, where $a,b,c,d,x,y$ natural numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the equality: $n = 8(s^6 + t^6)$. We have a representation of number $n$ as a sum of cubes. For the representation of number $n$ as a sum of squares we rewrite last equality as $n = 8(s^6 + t^6) = 2(2s^3)^2 + 2(2t^3)^2 = (2s^3 - 2t^3)^2 + (2s^3 + 2t^3)^2$.\n\nLet us show that it is impossible to represent the number as a sum of sixth powers.\nAssume that $n = 8(s^6 + t^6) = x^6 + y^6$, among all similar representations we choose that which has the minimal value of the sum $s + t + x + y$. We now prove that the numbers $s,t,x,y$ are even. Implying that $x$ and $y$ are odd, i.e. $x,y \\equiv \\pm 1 \\pmod 4$ then $x^6 + y^6 \\equiv 2 \\pmod 4$ not divisible by 4, a contradiction. Therefore $x = 2x_1, y = 2y_1$ and we have the equality $8(s^6 + t^6) = 64(x_1^2 + y_1^2)$. Similarly we prove that $s,t$ are even too, then $s = 2s_1, t = 2t_1$, and for the numbers $s_1,t_1,x_1,y_1$ holds the equality $8(s_1^6 + t_1^6) = x_1^6 + y_1^6$ but for all that $x_1 + y_1 + s_1 + t_1 = \\frac{x+y+s+t}{2}$, which contradicts to the choice of numbers $s,t,x,y$ which have the minimal value of the sum $s + t + x + y$. It shows that such representations is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45148,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the infinite checked paper $2006$ sides of $1 \\times 1$ squares are painted black, all others are painted white. It is allowed to choose a square and to paint all its sides into opposite color. It is known that it is possible to make all square sides white in such manner. Whether is it enough\na) $260000$ repainting of square sides;\nb) $250000$ repainting of square sides\nfor such repainting?",
    "options": [],
    "answer": "a) yes; b) no",
    "solution": "a) We look at the projections of painted black squares onto coordinate axis. Denote the lengths of these projections by $a$ and $b$. In such case area of all squares with marked sides does not exceed $ab$, and the number of black sides of squares is at least $2(a+b) \\le 2006$ (it is easy to see that there should be either zero or at least two painted black horizontal lines in any column and the same for any row). Whence the required number of repainting does not exceed $ab \\le \\frac{(a+b)^2}{4} \\le \\frac{1003^2}{4} < 260000$.\n\nb) Let's assume that all squares that should be repainted odd number of times are marked with blue color. Then in the initial arrangement only those sides of squares are painted black that neighbor one unmarked and one marked blue square. We get the least possible number of repainting if each square, marked blue, will be repainted only once, and unmarked squares won't be repainted at all. Let's estimate the number of squares, marked blue for the following initial painting: the perimeter of the $501 \\times 502$ rectangle. There are at least as many marked squares as the area of rectangle is. And this area is equal to $501 \\times 502 > 250000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45149,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides $AB$, $BC$, $CA$ of triangle $ABC$ such points $C_1$, $A_1$, $B_1$ are chosen that the lines $AA_1$, $BB_1$ and $CC_1$ meet at point $K$. The perpendiculars are dropped from the point $K$ onto the sides of triangle. Lines $l_1$, $l_2$, $l_3$ are drawn through the feet of these perpendiculars, parallel to lines, symmetrical to $AA_1$, $BB_1$ and $CC_1$ with respect to angular bisectors of $∠A$, $∠B$ and $∠C$ respectively. Prove that the lines $l_1$, $l_2$, $l_3$ meet at a point.\n![](attached_image_1.png)\n\nFig.28",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The points $A$, $M$, $K$, $N$ are cyclic, therefore\n$$\n\\begin{aligned}\n\\angle KAN &= \\angle NMK = \\\\\n&= 90^\\circ - \\angle AMN = \\angle MAP, \\text{ and so } AP \\perp MN. \\text{ Due to } l_1 \\parallel AP = u_1 \\text{ we have } l_1 \\perp MN, \\\\\n\\text{therefore the lines } l_1, l_2, l_3 \\text{ are altitudes of } \\triangle MNK \\text{ (Fig.28).}\n\\end{aligned}",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45150,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $(u_n)$ is defined by conditions: $u_0 = 0$, $u_1 = \\frac{1}{3}$ and $\\frac{1}{2} u_n = \\frac{1}{2}(u_{n+1} + u_{n-1})$. Prove that for all positive integers $n$ it holds $|u_n| \\le 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let an angle $\\varphi \\in (0, \\frac{\\pi}{2})$ such that $\\sin \\varphi = \\frac{\\sqrt{5}}{3}$. Then $\\cos \\varphi = \\frac{2}{3}$ and from the recurrent relation we have: $u_{n+1} = 2u_n \\cos \\varphi - u_{n-1}$. Moreover, $u_1 = \\sin \\varphi$, $u_2 = \\frac{2}{\\sqrt{5}} \\sin \\varphi \\cos \\varphi = \\frac{1}{\\sqrt{5}} \\sin 2\\varphi$.\n\nSuppose that for every $n \\in \\mathbb{N}$ it holds: $u_n = \\frac{1}{\\sqrt{5}} \\sin n\\varphi$. Then by mathematical induction we can get:\n\n$u_{n+1} = \\frac{2}{\\sqrt{5}} \\sin n\\varphi \\cos \\varphi - \\frac{1}{\\sqrt{5}} \\sin(n-1)\\varphi = \\frac{1}{\\sqrt{5}} \\sin(n+1)\\varphi$, which is required.\n\nThe proof follows from the properties of sine function.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of natural numbers $n$, $k$, for which the following equality holds:\n$$\nn^3 - 2 = k!\n$$",
    "options": [],
    "answer": "n=2, k=3",
    "solution": "If $k \\ge 2$ then the sum $(k!+2)$ is divisible by $2$, and so $n=2m$. Then for $k \\ge 4$ in the equality $(2m)^3 - k! = 2$ the left part is divisible by $4$, but the right one is not. Thus, all that we need is to consider the cases $k=1,2,3$. An easy check shows that there is an only solution $n=2, k=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45152,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex $2009$-gon is given.\na) Find the maximal number of vertices of this polygon which can be marked in such a way that there are no two marked vertices connected by the side of the polygon.\nb) Find the maximal number of vertices of this polygon which can be marked in such a way that among any three of marked vertices there exists at least one which is not connected by the side of the polygon with any of the two others.",
    "options": [],
    "answer": "a) 1004; b) 1339",
    "solution": "a) Let us enumerate all the vertices from $1$ to $2009$. Now mark those which have even numbers: $2, 4, 6, \\ldots, 2008$. This marking obviously satisfies the above condition. We'll now show that more than $1004$ vertices cannot be marked. Indeed, if we, for example, mark the first vertex, then we cannot mark vertices with numbers $2$ and $2009$. And then remaining vertices can be divided into pairs $3$-$4$, $5$-$6$, $\\ldots$, $2007$-$2008$ in each of which we can mark at most one vertex. Thus, the maximal number of vertices that we can mark is $1003+1=1004$.\n\nb) Analogously to the first item, we need to mark vertices in such a way that among them there will be no three consecutive ones. Let us mark them in the following way: $1, 2, 4, 5, 7, 8, \\ldots, 2005, 2006, 2008$. i.e. we don't mark vertices whose numbers are divisible by $3$ and the vertex with number $2009$. So we have $1339$ marked vertices. We'll show that we cannot mark more than $1339$ vertices. It is clear that we should mark at least two consecutive vertices, let them be $1$ and $2$. Then we clearly cannot mark vertices with numbers $3$ and $2009$. Now from each of the following triples of vertices - $4$-$6$, $7$-$9$, $2005$-$2007$ we can take at most two and also we can mark the vertex with number $2008$. So, the maximal number of vertices that we can mark is $2+2 \\cdot 668+1=1339$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45153,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the New year's party each boy gave each girl one candy and each girl gave each boy one cake. After that each boy ate two of given cakes and each girl ate three of given candies. It turned out that kids had eaten quarter of all given things. What is the maximal number of kids that could participate in this party?",
    "options": [],
    "answer": "35",
    "solution": "Let us denote the number of boys at the party by $x$ and the number of girls by $y$. Then from the problem statement we can get the following equality: $2xy = 4(3y + 2x)$, which is equivalent to the equation $(x-6)(y-4) = 24$. Now it is enough to consider all divisors of $24$ and to get all possible $x$ and $y$, among which the greatest value of $(x+y)$ is achieved in the two cases:\n$$\n\\begin{cases} x-6=1 \\\\ y-4=24 \\end{cases} \\text{ or } \\begin{cases} x-6=24 \\\\ y-4=1 \\end{cases}\n$$\nor this greatest value is:\n$$\nx+y=7+28=30+5=35.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45154,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ $\\angle ABC = 120^\\circ$. The bisector of this angle intersects side $AC$ at $M$, and the bisector of angle adjacent to $\\angle BCA$, intersects line $AB$ at $P$. Segment $MP$ intersects side $BC$ in $K$. Prove that $\\angle AKM = \\angle KPC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us prove that $AK$ is the bisector of $\\angle BAM$. Point $P$ is equidistant from the lines $BC$ and $AC$, and also from the lines $BM$ and $BC$, since $\\angle ABM = \\angle MBC = \\angle CBP = 60^\\circ$ (fig. 13). Thus $P$ is equidistant from the lines $BM$ and $MC$, which means that $P$ belongs to the bisector of $\\angle BMC$. Therefore $K$ is equidistant from the lines $AC$ and $BM$, and also from $BM$ and $AP$, which proves that $AK$ is the bisector of $\\angle BAC$.\nDenote $\\angle BAK = \\angle KAC = \\alpha$. Then step by step we can calculate the following:\n\n$$\n\\begin{align*} \n\\angle KMC &= \\frac{1}{2} \\angle BMC = \\frac{1}{2} (60^\\circ + 2\\alpha) = 30^\\circ + \\alpha. \\\\ \n\\angle BCA &= 60^\\circ - 2\\alpha. \\\\ \n\\angle BCP &= \\frac{1}{2} (180^\\circ - 60^\\circ + 2\\alpha) = 60^\\circ + \\alpha, \\\\ \n\\angle ACP &= 60^\\circ - 2\\alpha + 60^\\circ + \\alpha = 120^\\circ - \\alpha. \\\\ \n\\angle MPC &= 180^\\circ - (120^\\circ - \\alpha + 30^\\circ + \\alpha) = 30^\\circ; \\\\ \n\\angle AKM &= 180^\\circ - \\angle AMK - \\alpha = \\\\ \n&= \\angle KMC - \\alpha = 30^\\circ + \\alpha - \\alpha = 30^\\circ = \\angle MPC, \\text{ what was to be proved.} \n\\end{align*}\n$$\n\n![](attached_image_1.png)\nFig. 13",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers $a, b, c$ satisfy the condition $ab + bc + ca = 1$. Prove that the number $(1 + a^2)(1 + b^2)(1 + c^2)$ is a perfect square of some natural number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us add $a^2$ to both sides of our equality:\n$$a^2 + ab + bc + ca = 1 + a^2 \\quad \\text{or} \\quad (a+b)(a+c) = 1 + a^2.$$ \nAnalogously,\n$$(a+b)(b+c) = 1 + b^2 \\quad \\text{and} \\quad (c+b)(a+c) = 1 + c^2.$$ \nIf we now multiply together all these equalities, we will obtain that $(1 + a^2)(1 + b^2)(1 + c^2) = ((a+b)(b+c)(a+c))^2$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45156,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the side $AB$ of acute-angled triangle $ABC$ there is a point $K$, $M$ is the midpoint of $BC$, segments $AM$ and $CK$ intersect at a point $F$. It is known that $KF = AK$. Prove that $CF = AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On the ray $AM$ take a point $Q$, such that $MQ = AM$ (fig.19).\n\nThen $ACQB$ is a parallelogram and $\\angle KAF = \\angle FQC = \\angle CFQ \\Rightarrow CF = CQ = AB$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45157,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that for every natural number $n$ there exist natural $m$, $k$, that satisfy the equation\n$$\nk + m^k + n^{m^k} = 2009^n.\n$$\n\nb) Prove that there exist infinitely many natural $n$, for which such pair $m, k$ is unique.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nIf we take $m=1$, then we will get that $k+1+n^1=2009^n$ or that $k=2009^n-1-n$. It is clear that this will be a solution: $(n, m, k) = (n, 1, 2009^n - n - 1)$.\n\nb.\nIf $m=1$, then our equation comes to $k+1+n=2009^n$, which holds for the only value of $k$. Thus, new solutions can exist only when $m \\ge 2$.\n\nConsider $n=2009^s$, where $s$ is natural. Then $(2009^{s m^k}) < k + m^k + n^{m^k} = 2009^{2009^s}$, whence $s \\cdot m^k < 2009^s$, and so $m^k < 2009^s$. Let us make use of the following lemma:\n\n**Lemma.** For every natural $n$: $2^n \\ge n$.\n\nThis lemma implies that for $m \\ge 2$ we have $m^k \\ge 2^k \\ge k$, i.e. $k + m^k < 2009^s + 2009^s = 2 \\cdot 2009^s$. But $k + m^k = 2009^{2009^s} - n m^k = 2009^{2009^s} - 2009^{s - m^k}$. Now since $2009^s > 2 \\cdot 1000^s > 2 \\cdot 2^s \\ge 2s$ and $m \\ge 2$, we get that the right side is divisible by $2009^{2s}$. But at the same time $1 < k + m^k < 2 \\cdot 2009^s$ and thus the left side cannot be divisible by $2009^{2s}$. The obtained contradiction shows that for $n=2009^s$ the solution is unique. And it is evident that there are infinitely many such $n$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given acute-angled triangle $ABC$. $O$ is the circumcenter, $H$ is the orthocenter and $AH_A$, $BH_B$, $CH_C$ are the altitudes of $\\triangle ABC$. Denote by $A_1$, $B_1$, $C_1$ the circumcenters of triangles $BOC$, $COA$ and $AOB$ respectively. Prove that the lines $A_1H_A$, $B_1H_B$, $C_1H_C$ meet at a point, which lies on the Euler line of $\\triangle ABC$. (Euler line of $\\triangle ABC$ is the line passing through the circumcenter and centroid of a triangle).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to see that point $O$ is the incenter of $\\triangle A_1B_1C_1$, since $OC$ is perpendicular to the tangent of the circumcircle of $\\triangle ABC$ at the point $C$ and is parallel to $H_AH_B$.\n\nTherefore $OC \\perp H_A H_B \\Rightarrow A_1 B_1 \\parallel H_A H_B$. And so $\\Delta A_1 B_1 C_1 \\sim \\Delta H_A H_B H_C$ (Fig.29).\n\nHence $A_1 H_A \\cap B_1 H_B \\cap C_1 H_C = X$, where point $X$ is the homothety center of these triangles. Since $O$ is the incenter of $\\Delta A_1 B_1 C_1$, and $H$ is the incenter of $\\Delta H_A H_B H_C$, then $X \\in OH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45159,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the coordinate space $XOY$ draw the set of points, whose coordinates satisfy the following equality:\n$$\n\\sqrt{1-x^2} + \\sqrt{1-y^2} = 2 - x^2 - y^2.\n$$",
    "options": [],
    "answer": "All points with x and y each in {−1, 0, 1}, i.e., the nine points formed by all pairs of −1, 0, and 1.",
    "solution": "Since $0 \\leq (1 - x^2) \\leq 1$, $0 \\leq (1 - y^2) \\leq 1$, we get that\n$$\n\\sqrt{1-x^2} + \\sqrt{1-y^2} \\geq (1-x^2) + (1-y^2), \\text{ and so the equality holds if } \n\\begin{cases}\n\\sqrt{1-x^2} = 1-x^2 \\\\\n\\sqrt{1-y^2} = 1-y^2\n\\end{cases} \\Rightarrow \n\\begin{cases}\nx^2 \\in \\{0;1\\} \\\\\ny^2 \\in \\{0;1\\}\n\\end{cases}.\n\\text{ Thus we get the points from the answer.}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45160,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ points $M$ and $N$ are the midpoints of $BC$ and $AC$ respectively. Inside this triangle point $P$ is chosen in such a way that $\\angle BAP = \\angle PCA = \\angle MAC$. Prove that $\\angle PNA = \\angle AMB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to see that $\\angle APC = 180^\\circ - (\\angle PCA + \\angle PAC) = 180^\\circ - \\angle BAC = \\angle ANM$, since $MN \\parallel AB$ as the centerline of the triangle (fig.14).\n\nWe know that $\\angle MAC = \\angle PCA$, and so $\\triangle MNA \\sim \\triangle APC$. $N$ and $K$ are the midpoints of the respective sides in similar triangles, thus $\\angle NKM = \\angle PNA$.\n\nNow since $KN \\parallel BC$, we have $\\angle NKM = \\angle BMA = \\angle PNA$, and we are done.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Distinct real numbers $a$, $b$, $c$ satisfy the condition $a + \\frac{1}{b} = b + \\frac{1}{c} = c + \\frac{1}{a}$. Find which values the product $abc$ can attain.",
    "options": [],
    "answer": "{-1, 1}",
    "solution": "From the first equality we can get that $a-b = \\frac{1}{c} - \\frac{1}{b}$ or that $(a-b) = \\frac{b-c}{bc}$.\nAnalogously $b-c = \\frac{c-a}{ac}$ and $c-a = \\frac{a-b}{ab}$. Thus we obtain: $a-b = \\frac{a-b}{(abc)^2}$. Since $a$, $b$ are distinct, $(abc)^2 = 1$, which means that $abc = \\pm 1$. Let us show that there exist numbers $a$, $b$, $c$ for which both of the values can be achieved.\n\nTake $a = 1$. Then we have two equations on $b$, $c$: $1 + \\frac{1}{b} = c + 1$ and $1 + \\frac{1}{b} = b + \\frac{1}{c}$.\nFrom the first one we get that $c = \\frac{1}{b} \\Rightarrow 2b = 1 + \\frac{1}{b}$ or $2b^2 - b - 1 = 0$. This equation has two roots $-b_1 = -\\frac{1}{2}$ and $b_2 = 1$. The second root coincides with $a$, thus we take $b = -\\frac{1}{2}$ and $c = -2$. For these $a$, $b$, $c$ we have $abc = 1$ and it is easy to check that they satisfy the problem conditions.\n\nNow take $a = -1 \\Rightarrow -1 + \\frac{1}{b} = c - 1$ and $-1 + \\frac{1}{b} = b + \\frac{1}{c} \\Rightarrow c = \\frac{1}{b}$ and $2b = \\frac{1}{b} - 1$. So $2b^2 + b - 1 = 0$. The roots of this equation are $b_1 = \\frac{1}{2}$ and $b_2 = -1$. Again we choose only one value $b = \\frac{1}{2} \\Rightarrow c = 2$ and $abc = -1$. A direct check shows that these $a$, $b$, $c$ fulfill the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime $p$ and natural $m$, that satisfy the equation:\n$$\n2p^2 + p + 9 = m^2.\n$$",
    "options": [],
    "answer": "p=5, m=8",
    "solution": "Let us rewrite our equation in the following way: $p(2p+1) = (m-3)(m+3)$. Since $p$ is prime, we have that $(m-3) \\nmid p$ or $(m+3) \\nmid p$.\n\n$$1) \\quad (m-3) \\nmid p \\quad \\Rightarrow \\quad m-3 = kp \\quad \\Rightarrow \\quad (m+3) > kp \\quad \\text{and}$$\n$$3p^2 > p(2p+1) = (m-3)(m+3) > k^2p^2, \\quad \\text{therefore} \\quad 3p^2 > k^2p^2 \\Rightarrow k=1, \\quad \\text{and thus}$$\n$$\\begin{cases} m-3 = p \\\\ m+3 = 2p+1 \\end{cases} \\Rightarrow \\begin{cases} p=5 \\\\ m=8 \\end{cases} \\quad \\text{this is the first solution.}$$\n\n2) $(m+3)p$, so $m+3=kp$. If $p>5$, then $m-3=kp-6 > kp-k = p(k-1)$.\nAnalogously, $3p^2 > p(2p+1) = (m-3)(m+3) > (k-1)kp^2$, which implies that $k=1$ or\n$k=2$. Number $p(2p+1) = (m-3)(m+3)$ is odd, thus $k \\neq 2$. For $k=1$ we have that\n$$ \\begin{cases} m+3=p \\\\ m-3=2p+1 \\end{cases} $$\nwhich is impossible.\n\nIt remains to consider the cases $p=2,3,5$. It can be easily seen that in these cases there're no other solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45163,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the rectangular trapezoid $ABCD$ the side $CD$ is perpendicular to the bases. Circle with diameter $AB$ intersects $AD$ at $A$ and $P$, the tangent to this circle at the point $P$ intersects line $CD$ at $M$. From the point $M$ the line $l$ is drawn, which is tangent to our circle at the point $Q$. Prove that the line $BQ$ halves the segment $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the point of intersection of $BQ$ and $CD$, $L$ be the point of intersection of $BQ$ and $AD$ and $O$ be the midpoint of $AB$. Denote the radius of our circle by $r$. We have that (fig.22)\n$$\\angle NBC = \\angle NLD = \\frac{1}{2}(\\angle AQ - \\angle BP) = \\frac{1}{2}(\\angle AOQ - \\angle BOP) = \\frac{1}{2}(180^\\circ - \\angle QOB - \\angle BOP) = 90^\\circ - \\frac{1}{2}(\\angle QOB + \\angle BOP) = 90^\\circ - \\frac{1}{2}\\angle QOP =$$\n![](attached_image_1.png)\nFig.22\n$$90^\\circ - \\angle QOM = \\angle QMO \\Rightarrow \\angle NBC \\sim \\angle QOM = \\angle POM$$, by two angles, thus $$\\frac{CN}{BC} = \\frac{OQ}{OM} \\Rightarrow CN = \\frac{BC}{OM}r = \\frac{PD}{PM}r$$. Since $\\angle BAP = \\angle BPM$, as they share on the same arc, we deduce that $\\angle ABP = \\angle MPD \\Rightarrow \\angle ABP \\sim \\angle MPD$, and so $$CN = \\frac{PD}{PM}r = \\frac{r}{AB}BP = \\frac{1}{2}BP = \\frac{1}{2}CD$$, what was to be proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45164,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a plane are placed two triangles $ABC$ and $BKL$ such that segment $AK$ is divided into three equal parts by the intersection point of the medians of triangle $ABC$ and the intersection point of the bisectors of triangle $BKL$ ($AK$ is a median of $ABC$, $KA$ is a bisector of $ABKL$), and quadrilateral $KALC$ is a trapezium. Find all angles of triangle $BKL$.",
    "options": [],
    "answer": "All angles of triangle BKL are 60°",
    "solution": "Let $I$ be the center of $ABKL$, contextual from the characteristic of the centroid of the triangle and from the condition of the task: $AI = x$, $KI = 2x$. Let $BA = x$, then from the characteristic of the bisector $BK = 2x$ and $KC = 2x$. Then $KALC$ is a trapezium. That's why $KA \\parallel LC$ (see Fig. 20). From Thales' theorem $AL = x$. Therefore, in $ABKL$, $KA$ is a bisector by condition and a median because $BA = AL$. Then $ABKL$ is an equilateral triangle, and so if $BK = 2x = BL$, therefore it is an equilateral triangle and all its angles are $60^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45165,
    "subject": "Mathematics (Multi-modal)",
    "question": "For what real values $a$ and $b$, maximum among $3a^2 + 2b$ and $3b^2 + 2a$ takes minimum value?",
    "options": [],
    "answer": "a = b = -1/3",
    "solution": "Let $M(a, b) = \\max\\{3a^2 + 2b, 3b^2 + 2a\\}$. Then $M(a, b) \\ge 3a^2 + 2b$ and $M(a, b) \\ge 3b^2 + 2a$. From the last two inequalities we get $2M(a, b) \\ge 3a^2 + 2b + 3b^2 + 2a$.\n\nWe now have\n$$\n\\frac{2}{3}M(a,b) + \\frac{2}{9} \\geq \\left(a + \\frac{1}{3}\\right)^2 + \\left(b + \\frac{1}{3}\\right)^2 \\geq 0,\n$$\nor $M(a,b) \\ge -\\frac{1}{3}$ and $M(a,b) = -\\frac{1}{3}$, if $a=b=-\\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers from $1$ to $2007$ are arbitrarily written down in strip cells. Two players by turns mark cells of this strip. That player after whose move there are such two natural numbers $m < n$ loses that the sum of numbers of all noted cells from $m$ to $n$ divides by $2008$. Prove that there is at least one thousand initial orders of numbers in cells, such that for any non-negative integer $i \\le j$: the sum of numbers of cells with numbers $i, i+1, \\dots, j$ does not divide by $2008$ at which the first player has a winning strategy.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let in a cell with number $n$ number $a_n$ is written down. We will consider such sequence $(a_n)$: $a_{2k-1} = 2k-1$, $k = \\overline{1,1004}$, $a_{2k} = 2008 - 2k$, $k = \\overline{1,1003}$. We will prove that such placing approaches us. Let $S_n = a_1 + a_2 + \\dots + a_n$. Then $S_{2k} = 2007k$, $k = \\overline{1,1003}$, $S_{2k-1} = S_{2k-2} + a_{2k-1} = 2009k - 2008$, $k = \\overline{1,1004}$. And as $S_{2k-1} \\equiv -2007k \\pmod{2008}$, from equality $S_n \\equiv S_M \\pmod{2008}$ follows that $m = n$. Therefore the sequence $(a_n)$ satisfies the condition that for any non-negative integer numbers $i$ and $j$ such that $1 \\le i < j \\le 2007$, the sum of numbers of cells $i, i+1, \\dots, j$ does not divide by $2008$.\n\nNow we will describe a winning strategy for the first player. The first move he marks a cell with number $1004$. Further, if the second player moves in the cell with number $i$, the first player moves into the cell with number $2008-i$. As the sum of numbers with numbers $i$ and $2008-i$ is equal to $2008$ for all $i = \\overline{1,2007}$, it is easy to see that the described strategy for the first player is winning.\n\nWe will show how from this sequence to receive $999$ more other sequences which also satisfy the statement of the problem. Let $(l, 2008) = 1$. We put $b_n = l \\cdot a_n \\pmod{2008}$, $n = \\overline{1,2007}$. Then the sequence $(b_n)$ obviously satisfies the condition that $\\forall i, j \\in \\mathbb{N}: 1 \\le i < j \\le 2007$, the sum of numbers of cells $i, i+1, \\dots, j$ does not divide by $2008$. On the other hand, the strategy of the first player does not change. It is obvious that for different $l = \\overline{1,2007}$, we receive different sequences $(b_n)$. Therefore, we receive $\\varphi(2008) = 1000$ sequences.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an arbitrary arrangement of the brackets. Numbers $2010$ and $2009$ always have signs '+' and '-', respectively, that's why the sum reaches the maximum value when the other terms will have '+' sign.\n\nIn the expression\n$$\n\\frac{2010 - 2009 - 2010 - 2009 - 2010 - 2009 - \\dots - 2010 - 2009}{2010 \\text{ numbers}}\n$$\nthere are somehow placed brackets and the value is calculated. Find the maximum value that can be reached. Justify your answer.\n\n*Notice.* Left bracket can be placed only before a number and right - only after. For example, expressions $-2010(-2009 - 2010)$ and $-(2010 - 2009-)2010$ are incorrect.",
    "options": [],
    "answer": "4035077",
    "solution": "$$\n\\frac{2010 - (2009 - 2010 - 2009 - 2010 - 2009 - \\dots - 2010 - 2009)}{2010 \\text{ numbers}} = 4035077.\n$$\n\n**Solution.** Consider an arbitrary arrangement of the brackets. The first numbers $2010$ and $2009$ always have signs '+' and '-', respectively, that's why the sum reaches the maximum value when other terms will have '+' signs. We can achieve this as it was made in the answer. Calculating maximum value:\n$$\n\\begin{align*}\n& \\underbrace{2010 - (2009 - 2010 - 2009 - 2010 - 2009 - \\dots - 2010 - 2009)}_{2010 \\text{ numbers}} = \\\\\n&= 2010 - 2009 + \\underbrace{2010 + 2009 + 2010 + 2009 + \\dots + 2010 + 2009}_{2008 \\text{ numbers}} = \\\\\n&= 1 + (2010 + 2009) \\cdot 1004 = 4035077.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$, such that $n^2 - 10n + 23$, $n^2 - 9n + 31$ and $n^2 - 12n + 46$ are primes.",
    "options": [],
    "answer": "n = 3 or n = 7",
    "solution": "Let $n$ be a natural number that satisfies the condition of the problem. We first calculate the sum of all given numbers:\n$$\n3n^2 - 31n + 100 = 2n^2 - 30n + n(n-1) + 100\n$$\nwhich is even. This implies that at least one of them is even, in other words, it equals to $2$. Solving the following three equations\n$$\nn^2 - 10n + 23 = 2, \\quad n^2 - 9n + 31 = 2, \\quad n^2 - 12n + 46 = 2,\n$$\nwe get $n = 3$, or $n = 7$. Simple check shows that these values of $n$ meet all the requirements. Indeed,\n\nfor $n = 3$ we have: $n^2 - 10n + 23 = 2$, $n^2 - 9n + 31 = 13$ and $n^2 - 12n + 46 = 19$.\n\nfor $n = 7$ we have: $n^2 - 10n + 23 = 2$, $n^2 - 9n + 31 = 17$ and $n^2 - 12n + 46 = 11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that:\n$$\n-2^0 + 2^1 - 2^2 + 2^3 - 2^4 + \\dots - (-2)^n = 4^0 + 4^1 + 4^2 + \\dots + 4^{2010}.\n$$",
    "options": [],
    "answer": "n = 4021",
    "solution": "**Answer:** $n = 4021$.\n\nUsing the formula for sum of a geometric progression, we have:\n$$\n-2^0 + 2^1 - 2^2 + 2^3 - 2^4 + \\dots + (-2)^n = \\frac{-1 \\cdot ((-2)^{n+1}-1)}{(-2)^{n+1}}\n$$\nThe right side is:\n$$\n4^0 + 4^1 + 4^2 + \\dots + 4^{2010} = \\frac{4^{2011}-1}{4-1} = \\frac{4^{2011}-1}{3}\n$$\nSo, equating both sides:\n$$\n\\frac{-1 \\cdot ((-2)^{n+1}-1)}{(-2)^{n+1}} = \\frac{4^{2011}-1}{3}\n$$\nMultiply both sides by $(-2)^{n+1}$:\n$$\n-(( -2 )^{n+1} - 1) = \\frac{4^{2011}-1}{3} \\cdot ( -2 )^{n+1}\n$$\nBut from the context, the solution proceeds:\n\n$(-2)^{n+1} = 4^{2011}$\n\nIt follows that $n$ must be odd. Then $n = 2m + 1 \\Rightarrow (-2)^{2m+2} = 4^{2011} \\Rightarrow 4^{m+1} = 4^{2011}$ implying $m = 2010$ and $n = 4021$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Parabola $y = ax^2 + bx + c$ passes through the points $A(-2, 1)$ and $B(2, 9)$, and does not intersect $x$-axis. Find all possible values of $x$ coordinate of the vertex of the parabola.",
    "options": [],
    "answer": "(-4, -1)",
    "solution": "We first write analytically the conditions that our parabola passes through the given points:\n$$\n\\begin{cases} 4a - 2b + c = 1, \\\\ 4a + 2b + c = 9, \\end{cases}\n$$\nWe can now find $b = 2$ and $4a + c = 5$. From the condition, that our parabola does not have real zeros we get $D = b^2 - 4ac = 4 - 4a(5 - 4a) < 0$, thus, the following inequalities hold: $\\frac{1}{4} < a < 1$. The only thing that is left now is to solve the inequality for $x$ coordinate of the vertex of the parabola. Due to the fact that $x_v = -\\frac{b}{2a} = -\\frac{1}{a} \\Rightarrow x_v \\in (-4, -1)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Draw the graphic of the equation: $$\\frac{x}{|x|} + \\frac{|y|}{y} = 2y.$$",
    "options": [],
    "answer": "The graph is the union of two rays: all points with ordinate equal to one and abscissa positive, and all points with ordinate equal to negative one and abscissa negative.",
    "solution": "Obviously $x \\neq 0$ and $y \\neq 0$. Now we have to consider three cases:\n\n1st case. $x > 0$, $y > 0$. Hence $2y = 2$, and we have the first ray of the answer.\n\n2nd case. $x < 0$, $y < 0$, analogous to the first case. It gives us the second ray of the answer.\n\n3rd case. $xy < 0$. It follows that the equality $2y = 0$ must be held, but this is impossible.\n\n![](attached_image_1.png)\n\nFig.05",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 45172,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the expression\n$$\n12 - 11 - 10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1\n$$\nthere are somehow placed brackets and the value is calculated. Find the maximum value that can be reached. Justify your answer.\n\n*Notice.* Left bracket can be placed only before a number and right - only after. For example, expressions $-4(-3-2)$ and $-(4-3-2)$ are incorrect.",
    "options": [],
    "answer": "56",
    "solution": "$$\n12 - (11 - 10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1) = 46\n$$\n\nConsider an arbitrary arrangement of the brackets. Numbers $12$ and $11$ always have signs '+' and '-', respectively, that's why the sum reaches the maximum value when the other terms will have '+' sign.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the inequality:\n$$\n2008^x + \\log_{2008}(x+1) > \\log_{x+2} 2008 + \\sin^{2007} x + \\cos^{2008} x.\n$$",
    "options": [],
    "answer": "x > 0",
    "solution": "**Answer:** $x > 0$.\n\nEasy to understand, that range of a function in our expression is set $x > 0$. We will show that this is the solution. Really when $x > 0$ $2008^x > 1$, $\\log_{2008}(x+1) > 0$, because under such conditions, the logarithm base satisfies the conditions $0 < \\frac{x}{x+2} < 1$, $\\sin^{2007}x + \\cos^{2008}x \\le \\sin^2x + \\cos^2x = 1$. Thus the left part is greater than 1, and right - lower. Therefore the solution of inequality is every point $x > 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45174,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive numbers $a$, $b$, $c$, such that $abc = 1$, prove the inequality:\n$$\n\\frac{a^2 + b^2}{c^2 + a + b} + \\frac{b^2 + c^2}{a^2 + b + c} + \\frac{c^2 + a^2}{b^2 + c + a} \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$abc = 1 \\Rightarrow \\frac{1}{3}(a+b+c)$. Then:\n$$\n\\begin{aligned}\nc^2 + a + b &\\le c^2 + (a+b)\\frac{1}{3}(a+b+c) \\\\\n&= \\frac{1}{3}(3c^2 + a^2 + b^2 + 2ab + ac + bc) \\\\\n&\\le \\frac{1}{6}(8c^2 + 5a^2 + 5b^2).\n\\end{aligned}\n$$\nSo\n$$\n\\frac{a^2 + b^2}{c^2 + a + b} + \\dots \\ge \\frac{a^2 + b^2}{\\frac{1}{6}(8c^2 + 5a^2 + 5b^2)} + \\dots + \\frac{6(a^2 + b^2)}{8c^2 + 5a^2 + 5b^2} + \\dots \\ge 2,\n$$\nthe last inequality must be proved. As three points meant a cyclic permutation of variables. For simplification denote $x = a^2$, $y = b^2$, $z = c^2$. Then we need to prove such inequality:\n$$\nA = \\frac{x+y}{8z+5x+5y} + \\dots \\ge \\frac{1}{3}.\n$$\n$$\nA = \\frac{(x+y)^2}{(x+y)(8z+5x+5y)} + \\dots \\ge \\frac{((x+y)^2 + \\dots)^2}{(x+y)(8z+5x+5y)} = \\frac{4p+8q}{10p+26q} \\ge \\frac{1}{3},\n$$\nthis inequality supervenes from Cauchy-Schwarz inequality. There are used next denote: $p = x^2 + y^2 + z^2$, $q = xy + yz + zx$. The last inequality equivalent $p \\ge q$, which are well known.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all solutions of the equation $\\sqrt[3]{x} + \\sqrt[3]{y} = \\sqrt[3]{z}$, where $x$, $y$, $z$ are integer numbers.",
    "options": [],
    "answer": "All integer solutions are x = d a^3, y = d b^3, z = d (a + b)^3 for integers a, b, d.",
    "solution": "Any group of three $(da^3, db^3, dc^3)$ if $a + b = c$ satisfies the condition of the problem.\n\nLet's find solutions of the equation $\\sqrt[3]{x} + \\sqrt[3]{y} + \\sqrt[3]{z} = 0$. From the known equation\n$$\na^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\n$$\nit follows that $x + y + z = 3\\sqrt[3]{xyz}$, thus $(x + y + z)^3 = 27xyz$. We can assume that any two from $x$, $y$, $z$ are coprime. Really, if the prime $p$ divides $x$, $y$, then it can divide the equation $(x + y + z)^3 = 27xyz$ and therefore $z \\equiv 0 \\pmod{p}$.\n\nTherefore, we can think that $x = d x_1$, $y = d y_1$, $z = d z_1$, where $x_1$, $y_1$, $z_1$ are coprime. Then $x_1$, $y_1$, $z_1$ are the third powers of integers $a$, $b$, $c$. On the other side, any group of three $(da^3, db^3, dc^3)$, if $a + b = c$, satisfy the condition of the problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45176,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ on the side $AC$ the points $F$ and $L$ with $AF = LC < \\frac{1}{2}AC$ are chosen. Find an angle $FBL$, if $AB^2 + BC^2 = AL^2 + LC^2$.\n\n**Answer:** $\\angle FBL = 90^\\circ$.",
    "options": [],
    "answer": "90°",
    "solution": "Let $AB = c$, $BC = a$, $FM = ML = c$ ($M$ is a middle of $AC$), $AF = LC = x$. According to known formula find length of median (Fig.25)\n\n![](attached_image_1.png)\n\nFig.25\n\n$$\n\\begin{aligned}\nBM^2 = m^2 &= \\frac{1}{4}(2c^2 + 2a^2 - (2b + 2x)^2) = \\frac{1}{4}(2(2b + x)^2 + 2x^2 - (2b + 2x)^2) = \\\\\n&= \\frac{1}{4}(8b^2 + 8bx + 4x^2 - 4b^2 - 8bx - 4x^2) = b^2.\n\\end{aligned}\n$$\n\nTherefore in the $\\triangle FBL$ median is half of base, in which it has marked, that's why this triangle is right and $\\angle FBL = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove, that the equation $a^3 + b^3 + c^3 = a^2 + b^2 + c^2$ has infinite number of solutions in the integer numbers, if the biggest common divisor of numbers $a, b, c$ is 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lets choose such values: $b = 1 + x$, $c = 1 - x$ and put it in desired equation:\n$$\na^3 + 2(1 + 3x^2) = 2(1 + x^2) + a^2 \\Leftrightarrow a^2 - a^3 = 4x^2.\n$$\nWe put $1 - a = 4p^2$, then\n$$\na^2(1 - a) = (4p^2 - 1)^2 4p^2 = (2x)^2.\n$$\nWe put in a number $x = p(4p^2 - 1)$. Then three integer numbers $(a, b, c) = (1 - 4p^2, 1 + p(4p^2 - 1), 1 - p(4p^2 - 1))$ satisfy equation, i.e. there are infinite number of solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45178,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a table $n \\times n$ two players fill the lines one by one with numbers \"+1\" and \"1\". At first the first player fills the first line. Then second player -- second line, then first player fills third line. Then second -- forth line etc. In the end of filling lines, first player gets 1 point for every line or column, in which product of numbers is positive, in another way, this point gets another player. Each of them try to collect points as much as possible. In a melting way of game, how much points each of them can collect?",
    "options": [],
    "answer": "Let n be the board size. The optimal scores are:\n- If n ≡ 0 (mod 4): first player n/2, second player 3n/2.\n- If n ≡ 1 (mod 4): first player (3n + 1)/2, second player (n − 1)/2.\n- If n ≡ 2 (mod 4): first player n/2 + 1, second player 3n/2 − 1.\n- If n ≡ 3 (mod 4): first player (3n − 1)/2, second player (n + 1)/2.\nEquivalently, writing n = 2k:\n- If n = 2k with k even: first k, second 3k; if k odd: first k + 1, second 3k − 1.\nAnd for n = 2k + 1:\n- If k even: first 3k + 2, second k; if k odd: first 3k + 1, second k + 1.",
    "solution": "**Answer:** By even $k$ the first player collects $(3k + 2)$, the second $k$; by the odd $k$ the first $(3k + 1)$, second $(k + 1)$.\n\nLet $n = 2k$.\n\nIn the beginning we can look into such a strategy for every player. The first player fills numbers in arbitrary way. In this way he collects $k$ points. The second -- in this way to last line collects $(k-1)$ points, but in the last line he fills numbers to win in every column. In this way he collects $2k + (k-1) = 3k-1$ points. Only we have to find out who with this strategy will win in last line. As product of all numbers in table is \"+1\" because product in all columns is $1$ and there is even quantity. Odd lines have product \"+1\"; in this way the product of all \"even\" lines is \"+1\". There are $k$ pieces.\n\nIn this way with even $k$ the product of last line will be $1$ too. Definitely first player collects at least $k$ points, and second -- $3k$. In attempt of changing strategy, each of them can only reduce his result, because the opponent can use this for improvement of his result. In the way of odd $k$ the product of last line must be \"+1\", i.e. the first player will win. The final result in this way will be: first -- $(k+1)$, second -- $(3k-1)$. Whether the second can collect more? In way of another strategy, first player will collect his $k$ points for the lines. If the second loses only one column or line, he would collect no more than $(3k-1)$. But he can't win all of them, in consequence of valuation mentioned here. The first can't collect more than $(k+1)$, because the second can collect $(3k-1)$ at least.\n\nLet $n = 2k + 1$.\n\nThe strategy remains here, but only in result of odd size of the table, the last move does the first player. In this way he collects his points for all $(2k+1)$ columns and $k$ lines, without last. The second wins his $k$ columns. We only have to check who will win in the last line with these strategies. The product of columns of all numbers is \"+1\" -- all of them positive. The product of all lines, except last is $(-1)^k$. That's why with even $k$ the first wins the last line too, but with odd -- the second wins. Thereby answer is: in the way of paired $k$ the first collects $(3k+2)$, the second -- $k$; with odd $k$, the first -- $(3k+1)$, the second -- $(k+1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45179,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive numbers $a$, $b$ with $a + b = ab$ prove inequality:\n$$\n\\frac{a}{b^2 + 4} + \\frac{b}{a^2 + 4} \\ge \\frac{1}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$ab = a + b \\ge 2\\sqrt{ab} \\Rightarrow ab \\ge 4$, then\n$$\n\\begin{aligned}\n\\frac{a}{b^2+4} + \\frac{b}{a^2+4} &\\ge \\frac{a}{b^2+ab} + \\frac{b}{a^2+ab} = \\\\\n&= \\frac{a}{b(a+b)} + \\frac{b}{a(a+b)} = \\frac{a^2+b^2}{(a+b)^2} \\ge \\frac{1}{2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45180,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let's look at a table of 3 rows and 2008 columns. In the first row, a random integer is written in each cell of the table, in a non-decreasing order. The numbers of the second row are obtained in the following way: under every number $A$ from the first row is written a number $B$, which equals the number of numbers in the first row that are less than $A$ and located to the left of $A$. Similarly, the numbers of the third row are obtained with respect to the second row: under every number $C$ of the second row is written a number $D$, which equals the number of numbers in the second row that are less than $C$ and located to the left of $C$. Prove that the second and third row of this table are filled equally.\n\nb) We have tables that consist of 2 rows and 2008 columns and are built in such a way that in every table, in the first row, random integers are written (in every cell we have exactly one number) in a non-decreasing order. The numbers of the other row are obtained in the way that under every number $A$ in the first row is written a number $B$, which equals the number of integers in the first row that are less than $A$ and located to the left of $A$. We know that every table has pairwise different second rows. Find the maximum possible number of tables.\n\n**Answer:** $2^{2007}$.",
    "options": [],
    "answer": "2^{2007}",
    "solution": "a.\n1) If $A_{i+1} = A_i$, then the quantity of numbers from the first row smaller than $A_i$ and situated to the left of $A_i$ is the same as the quantity of numbers from the first row smaller than $A_{i+1}$ and situated to the left of $A_{i+1}$. As number $A_i$ can't increase this quantity, that's why $B_{i+1} = B_i$.\n\n2) If $A_{i+1} > A_i = A_{i-1} = \\dots = A_{i-k+1}$, then the quantity of numbers from the first row smaller than $A_i$ and situated to the left of $A_i$ is $k$ less than the quantity of numbers from the first row smaller than $A_{i+1}$ and situated to the left of $A_{i+1}$, as this quantity is increased by $k$ numbers $A_{i-k+1}, \\dots, A_i$. That's why $B_{i+1} = B_i + k$ and $B_{i+1} > B_i = B_{i-1} = \\dots = B_{i-k+1}$.\n\n3) If $A_{i+1} > A_i > A_{i-1}$, then the quantity of numbers from the first row smaller than $A_i$ and situated to the left of $A_i$ is one less than the quantity of numbers from the first row smaller than $A_{i+1}$ and situated to the left of $A_{i+1}$, as this quantity is increased by the number $A_i$. That's why $B_{i+1} = B_i + 1$ and $B_{i+1} = B_i > B_{i-1}$.\n\nAs we can see, numbers $C_i$ are built the same way. Before finishing, we should note that $B_1 = C_1 = 0$.\n\nb.\nLet's denote the first and second row of the table as $A_1, \\dots, A_{2008}$ and $B_1, \\dots, B_{2008}$. As we know $B_1, \\dots, B_i$, $B_{i+1}$ depends on $A_i$ and $A_{i+1}$. Let's look at some variants.\n\n1) If $A_{i+1} = A_i$, then the quantity of numbers from the first row smaller than $A_i$ and situated to the left of $A_i$ is the same as the quantity of numbers from the first row smaller than $A_{i+1}$ and situated to the left of $A_{i+1}$. As number $A_i$ can't increase this quantity, that's why $B_{i+1} = B_i$.\n\n2) If $A_{i+1} > A_i = A_{i-1} = \\dots = A_{i-k+1}$, then the quantity of numbers from the first row smaller than $A_i$ and situated to the left of $A_i$ is $k$ less than the quantity of numbers from the first row smaller than $A_{i+1}$ and situated to the left of $A_{i+1}$, as this quantity is increased by $k$ numbers $A_{i-k+1}, \\dots, A_i$. That's why $B_{i+1} = B_i + k$ and $B_{i+1} > B_i = B_{i-1} = \\dots = B_{i-k+1}$.\n\n| $A_i$ | $A_{i+1}$ |\n|-------|-----------|\n| $B_i$ | $B_{i+1}$ |\n| $C_i$ | $C_{i+1}$ |\n\nFig. 19\n\nThat's why at each step there exist two different possibilities for the next element, which means that there are exactly $2^{2007}$ such tables.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45181,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n \\ge 5$ real numbers are written in a row. It turns out, that sum of any three consecutive numbers is positive and sum of any five consecutive numbers is negative. Find the largest $n$ for which it is possible?\n\n**Answer:** $n = 6$.",
    "options": [],
    "answer": "6",
    "solution": "We first construct the example for $n = 6$: $3, -5, 3, 3, -5, 3$.\n\nSuppose that there exist $n \\ge 7$ real numbers, that satisfy the conditions of the problem and choose any 5 consecutive numbers $a, b, c, d, e$. Using the conditions we get: $a+b+c > 0$, $c+d+e > 0 \\Rightarrow (a+b+c+d+e) + c > 0$ and $a+b+c+d+e < 0$. This implies that $c > 0$, therefore for any 5 consecutive numbers one, which is in the middle, is always positive. From the last observation, it follows that all numbers except last two from both ends are positive.\n\nLet us now consider 6 consecutive numbers: $a, b, c, d, e, f$. Using again our given conditions we get: $a+b+c+d+e < 0$ and $(a+b+c)+(d+e+f) > 0$. Thus, $f > 0$. By analogy, we have $a > 0$. This implies that all $n$ numbers must be positive and we get a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45182,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 16 consecutive positive integers written on the board. Andrew calculates their product and Olesya – their sum. Can it happen that in both numbers there coincide\n\na) three last digits,\nb) four last digits?",
    "options": [],
    "answer": "a) yes; b) no",
    "solution": "**Answer:** a) yes; b) no.\n\nIt's obvious that a number received by Andrew is divisible by $16$ and by $125$, because from $16$ consecutive numbers more than four are divisible by $2$ and at least $3$ are divisible by $5$. It also implies that three last digits in Andrew's number are $0$.\n\na) Let the numbers $a, a+1, a+2, \\dots, a+15$ be written on the board. Then Olesya obtained the number $8(2a+15)$. Putting $a=55$ we have that three last digits of this number are $0$, so the answer to the case a) is 'yes'.\n\nb) Since the number obtained by Andrew is divisible by $16$, then it's true for the number obtained by four last digits of this number. If we assume that the answer to b) is 'yes' then the number obtained by four last digits of the number $8(2a+15)$ is divisible by $16$. It follows that $8(2a+15) \\div 16$. This contradiction gives the answer 'no'.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least natural number $n > 100$ for which the next statement is true: maximal sum of digits amongst the numbers $n - 100, n - 99, \\ldots, n, \\ldots, n + 99, n + 100$ has $n$. Justify your answer.",
    "options": [],
    "answer": "999",
    "solution": "Consider an arbitrary number. If the two last digits differ from $99$ then replacing them by $99$ we obtain a number with bigger sum of digits. Numbers $199, 299, \\ldots, 899$ don't satisfy this condition, because for each $n \\in 199, 299, \\ldots, 899$ a number $n + 100$ has the sum of digits bigger by $1$. And now, we can see that the desired number is $999$. This number has the sum of digits bigger than any other three-digit number (because all digits are $9$) and it has the sum of digits bigger than the first $100$ four-digit numbers (because the maximum sum of digits among such numbers has $1099$ and it's only $20$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Cube with side length $11$ is partitioned into unit cubes. The external slice of cubes is colored with yellow, the next slice (all cubes that share a vertex with the external slice) is colored with blue. Slice of cubes that touches the previous slice is colored with yellow and so on. Find the number of yellow and blue cubes.",
    "options": [],
    "answer": "yellow: 846, blue: 485",
    "solution": "Obviously, there are $6$ slices of cubes. It is easy to see that the external yellow slice has height $11$, the next blue - $9$, the next yellow - $7$, the next blue - $5$, the next yellow - $3$ and the last one (6th slice) blue has height $1$, e.g. it's a unit cube. Further, we can calculate the number of cubes in each slice in such a way:\n\n5th slice, yellow, with height $3$, has $3^3 - 1^3 = 26$ unit cubes.\n\n4th slice, blue, with height $5$, has $5^3 - 3^3 = 98$ unit cubes.\n\n3rd slice, yellow, with height $7$, has $7^3 - 5^3 = 218$ unit cubes.\n\n2nd slice, blue, with height $9$, has $9^3 - 7^3 = 386$ unit cubes.\n\n1st slice, yellow, with height $11$, has $11^3 - 9^3 = 602$ unit cubes.\n\nFinally, the number of blue cubes is $1 + 98 + 386 = 485$, and the number of yellow cubes is $602 + 218 + 26 = 846$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Olesya and Andrew have thrown a dice once. Find the probability that the number of points that was thrown out by Olesya is strictly greater than the number of points that was thrown out by Andrew. Justify the answer.",
    "options": [],
    "answer": "5/12",
    "solution": "There are $6^2 = 36$ different combinations. Let us calculate the number of cases that satisfy the statement.\n\nOlesya was thrown out $1$, that is impossible.\n\nOlesya was thrown out $2$, the only possible case is $1$.\n\nOlesya was thrown out $3$, possible cases are $1$ and $2$.\n\nOlesya was thrown out $4$, possible cases are $1$, $2$ and $3$.\n\nOlesya was thrown out $5$, possible cases are $1$, $2$, $3$ and $4$.\n\nOlesya was thrown out $6$, possible cases are $1$, $2$, $3$, $4$ and $5$.\n\nAltogether we have $15$ variants. And by definition, the desired probability is $\\frac{15}{36} = \\frac{5}{12}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$, $Q(x)$ and $R(x)$ be polynomials, such that $Q(x)$ and $R(x)$ take nonnegative values only. It is known, that the equation\n$$\nP(x) + \\sqrt{Q(x)} + \\sqrt{Q(x) + \\sqrt{R(x)}} = 0\n$$\nhas infinitely many solutions. Is it true, that every real number is a root of this equation?",
    "options": [],
    "answer": "No",
    "solution": "Not necessary.\n\nWe construct the following example: $P(x) = x$, $Q(x) = \\frac{1}{4}x^2$, $\\sqrt{Q(x)} = \\frac{1}{2}|x|$, $R(x) = 0$. Then, $\\sqrt{Q(x) + \\sqrt{R(x)}} = \\frac{1}{2}|x|$ and we have: \n$$\nP(x) + \\sqrt{Q(x)} + \\sqrt{Q(x) + \\sqrt{R(x)}} = x + |x| = \\begin{cases} 2x, & x \\ge 0, \\\\ 0, & x < 0. \\end{cases}\n$$\nHence, our left hand side does not equal to $0$ for all $x$, but the equation has infinitely many solutions.\n\nNote, that this example is not unique, one can also take $Q(x) = \\frac{1}{8}x^2$, $R(x) = \\left(\\frac{1}{8}x^2\\right)^2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all nonnegative integers $k$, $n$ that satisfy the following equality:\n$$\n2^{2k+1} + 9 \\cdot 2^k + 5 = n^2.\n$$",
    "options": [],
    "answer": "k = 0, n = 4",
    "solution": "Consider our equation modulo $8$. For $k \\geq 3$ the left hand side is divisible by $8$, but the right hand side is not $8$. So we have to consider $k \\in \\{0, 1, 2\\}$.\n\nIf $k = 0$ we have $n^2 = 16$ or $n = 4$.\n\nFor $k = 1$ we have $n^2 = 31$, that gives no solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation:\n$$\n(x + 1)^5 + (x + 1)^4(x - 1) + (x + 1)^3(x - 1)^2 + \\\\\n+ (x + 1)^2(x - 1)^3 + (x + 1)(x - 1)^4 + (x - 1)^5 = 0.\n$$",
    "options": [],
    "answer": "x = 0",
    "solution": "**Answer:** $x = 0$.\n\nMultiplying both sides by $2 = ((x + 1) - (x - 1))$ yields $(x+1)^6 - (x-1)^6 = 0$ or equivalently $(x + 1)^2 = (x - 1)^2$. Solving this equation, we obtain the unique solution $x = 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy has cards which are the same from one side, and numbers $1, 3, 5, \\ldots, 2009$ are written on the other side. Lesya has the same cards but with numbers $2, 4, 6, \\ldots, 2010$. Lesya places cards with numbers down (so we can not see them at the beginning), after that Andriy places his cards on the Lesya's (so, each Lesya's card is covered by exactly one Andriy's card). So we get $1005$ pairs of cards. After that, we compare numbers in pairs and add $1$ point to the person, who has greater number. Andriy knows, that Lesya places cards in increasing order starting from some place and after she gets $2010$, Lesya starts placing cards in increasing order again, starting from the least number which has not appeared before.\nFor example: $20, 22, 24, \\ldots, 2010, 2, 4, \\ldots, 18$. Andriy wants to get as many points as he can. What is the maximum number of points can he guarantee to get?\n**Answer:** $502$ points.",
    "options": [],
    "answer": "502",
    "solution": "In order to get at least $502$ points Andriy can place his cards as follows, starting from the left: $2009, 2007, 2005, \\ldots, 3, 1$. We prove by induction, that he gets $502$ points in this case. Lesya has $1005$ options to place her cards and we present all of them in the following table:\n\n| 2    | 4    | 6    | 2008 | 2010 |\n|------|------|------|------|------|\n| 4    | 6    | 8    | 2010 | 2    |\n| 6    | 8    | 10   | 2    | 4    |\n| 2008 | 2010 | 2    | 2004 | 2006 |\n| 2010 | 2    | 4    | 2006 | 2008 |\n\nIf we compare Andriy's variant with the first row, we obtain: $2, 4, \\ldots, 2010$ and $2009, 2007, 2005, \\ldots, 3, 1$. Hence, Andriy gets exactly $502$ in the first columns. This is our base. Let us suppose, that in $k$-th row Andriy gets $502$ points.\n\nLet us prove that in $(k+1)$-th row he gets the same number of points. We have that the number of points decreases by $1$ at the beginning of the row and increases by $1$ at the end, comparable to $k$-th row.\n\nWe now show that Andriy can not get better result. Lesya knows the position of Andriy's cards, hence, she can choose the row which gives maximum points to her. So we now compute how many points can Andriy get over all $1005$ Lesya's options. Let us place all Andriy's cards under our $1005 \\times 1005$ table.\n\nWe compute total number of points: when we have the number $2009$, Andriy will get $1004$ points, Andriy's number $2007$ will get $1003$ points, and so on, number $1$ will get $0$ points. So our total number is $1004 + 1003 + \\ldots + 1 + 0 = \\frac{1005 \\cdot 1004}{2} = 502 \\cdot 1005$ points.\n\nHence, Andriy's average number of points over all $1005$ Lesya's options is $502$, which proves our result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45190,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangle with the lengths of sides $2010$ and $11$ is partitioned into unit squares. The external slice of squares is colored with yellow, the next slice (all squares that share a vertex with external slice) are colored with blue. A slice of squares that touches the previous slice is colored with yellow and so on. Find the number of yellow and blue squares.",
    "options": [],
    "answer": "Yellow: 12066; Blue: 10044",
    "solution": "Do allocate external slices of squares of each color step by step:\n$2010 \\times 11$ - yellow; $2008 \\times 9$ - blue; $2006 \\times 7$ - yellow; $2004 \\times 5$ - blue; $2002 \\times 3$ - yellow;\n$2000 \\times 1$ - blue.\n\nNow calculate the number of squares in each slice as difference between numbers of squares of external and internal rectangles. Bottom-up:\n\nBlue: $2000 \\cdot 1 = 2000$.\nYellow: $2002 \\cdot 3 - 2000 \\cdot 1 = 4006$.\nBlue: $2004 \\cdot 5 - 2002 \\cdot 3 = 4014$.\nYellow: $2006 \\cdot 7 - 2004 \\cdot 5 = 4022$.\nBlue: $2008 \\cdot 9 - 2006 \\cdot 7 = 4030$.\nYellow: $2010 \\cdot 11 - 2008 \\cdot 9 = 4038$.\n\nAnd simultaneously: yellow $4038 + 4022 + 4006 = 12066$, blue $4030 + 4014 + 2000 = 10044$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45191,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of a cyclic quadrilateral $ABCD$ are perpendicular. Points $K, L, M, Q$ are the orthocenters of the triangles $ABD$, $ACD$, $BCD$, $ABC$ respectively. Prove that quadrilaterals $KLMQ$ and $ABCD$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the diagonals of the quadrilateral $ABCD$ intersect at $O$. Altitudes of the triangles $BCD$ and $ACD$ lie on $AC$, so their orthocenters $K$ and $M$ too. Analogously, points $L$, $Q$ lie on $BC$ (Fig.06).\n\nNote that $BK \\parallel CL$, because $BK \\perp AD$ and $CL \\perp AD$. It follows that $\\angle BKC = \\angle KCL = \\angle ACL = 90^\\circ - \\angle CAD = 90^\\circ - \\angle DBC = 90^\\circ - \\angle OBC = \\angle BCK$.\n\n![](attached_image_1.png)\nFig.06\n\nNow it's easy to see that $\\triangle LOC = \\triangle BOC = \\triangle BOK$, which implies $OC = OK$ and $BO = OL$. Since $CK \\perp CL$ it follows that the quadrilateral $LCBD$ is a rhombus. Analogously, the quadrilateral $ADMQ$ is a rhombus too. Finally, we have that the quadrilateral $KLMQ$ is an image of $ABCD$ according to the central symmetry relative to the point $O$, which implies that the quadrilaterals $KLMQ$ and $ABCD$ are equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each girl: Oksana, Olesya, Olya and Olexandra has a rectangle with sides $2010$ and $10$. They were given a task: to cut this rectangle into two pieces from which one can make a triangle without overlaps. All of them succeeded in this task. Can they make pairwise distinct triangles?",
    "options": [],
    "answer": "Yes",
    "solution": "**Answer:** Yes. The required example on the fig.01.\n\n![](attached_image_1.png)\nFig.01",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's designate through $P(n)$ the product of digits of the integer non-negative number $n$. Prove that sets $A$ and $B$ are unbounded, where:\n\na. $A = \\left\\{ \\frac{P(n)}{P(n^2)} \\right\\}$, where $n$ belongs to the set of such whole non-negative numbers that the number $n^2$ does not contain zero in the decimal record;\n\nb. $B = \\left\\{ \\frac{P(n^2)}{P(n)} \\right\\}$, where $n$ belongs to the set of such whole non-negative numbers that the number $n$ does not contain zero in the decimal record.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Both points are proved with the help of corresponding examples which are in turn proved by a method of mathematical induction.\n\na.\nLet's consider the equality:\n$$\n(2\\underbrace{66\\dots68}_{n-1})^2 = 7\\underbrace{11\\dots18}_{n-1}\\underbrace{22\\dots24}_{n-1},\n$$\n\nFurther, it is enough to calculate the corresponding ratio as $n \\to \\infty$:\n$$\n\\frac{P(n)}{P(n^2)} = \\frac{16 \\cdot 6^{n-1}}{7 \\cdot 8 \\cdot 4 \\cdot 2^{n-1}} \\to +\\infty.\n$$\n\nb.\nLet's consider the equality:\n$$\n(\\underbrace{66\\dots67}_{n-1})^2 = \\underbrace{44\\dots48}_{n-1}\\underbrace{\\dots89}_{n-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nonnegative real numbers $a$, $b$, $c$ satisfy the equation $a + b + c = 1$. Prove the inequality\n$$\n(1 - a)^2 + (1 - b)^2 + (1 - c)^2 \\geq 6\\sqrt{abc}.\n$$\n\nCan the case of equality occur?",
    "options": [],
    "answer": "No",
    "solution": "Removing the brackets we will have $(1 - a)^2 + (1 - b)^2 + (1 - c)^2 = 3 + a^2 + b^2 + c^2 - 2(a + b + c) + 3 = a^2 + b^2 + c^2 + a + b + c \\geq 6\\sqrt{abc}$, equality is achieved when $a = b = c = a^2 = b^2 = c^2$, which is impossible, because $a + b + c = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45195,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $P$ lies inside triangle $ABC$. Denote by $O_A, O_B, O_C$ the circumcenters of triangles $PBC, PAC, PAB$ respectively. Let $O_P$ be the circumcenter of triangle $O_A O_B O_C$. Prove that the point $P$ satisfies the condition $O_P = P$ if $P$ is the orthocenter of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A' = PA \\cap O_B O_C$, $B' = PB \\cap O_A O_C$, $C' = PC \\cap O_B O_A$. $O_A O_C$ is a perpendicular bisector of $BP$, thus $B'$ is a midpoint of $BP$. By analogy, $A', C'$ are midpoints of $PA, PC$ (Fig.07). If $P$ is a circumcenter of $\\triangle O_A O_B O_C$, then the perpendicular from $P$ to $O_A O_C$ passes through the midpoint of $O_A O_C$, hence $B'$ is also a midpoint of $O_A O_C$. Following the same lines, we get that $A', C'$ are midpoints of $O_B O_C, O_A O_B$.\n\nWe also have $PB \\perp O_A O_C \\parallel A'C' \\parallel AC$, because $A'C'$ is a midline of triangles $O_A O_B O_C$ and $APC$. By analogy, $PA \\perp BC, PC \\perp AB$, thus $P$ is an orthocenter of $\\triangle ABC$. Obviously, if $P$ is an orthocenter then $O_P = P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ and $Q$ be two points inside parallelogram $ABCD$, that are symmetric with respect to the point of intersection of the diagonals. Prove that the circumcircles of triangles $ABP$, $CDP$, $BCQ$ and $ADQ$ have a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $X$ the second point of intersection of the circumcircles of $\\triangle ADQ$ and $BCQ$, $QE$ the ray which has the same direction as ray $CB$ (Fig.12).\n\nThen $\\angle EQB = \\angle QBC$, $\\angle EQA = \\angle QAD$\n$\\Rightarrow \\angle AQB = \\angle QAD + \\angle QBC = 180^\\circ - \\angle QXD + 180^\\circ - \\angle QXC = \\angle CXD$.\nMoreover, $\\angle AQB = \\angle CPD$, because triangles $AQB$ and $CPD$ are symmetric with respect to the center of our parallelogram.\nHence, points $D$, $P$, $X$, $C$ are cyclic. By analogy, $A$, $P$, $X$, $B$ also lie on the same circle, so all 4 circles pass through the common point $X$.\n\nIt is worth noting that this proof is valid for the specific configuration of given points only. For a complete solution, one has to consider at least three more configurations. However, it is possible to handle all these cases simultaneously using oriented angles.\nIndeed:\n$\\angle(QB; QA) = \\angle(QB; BC) + \\angle(BC; QA) = \\angle(QB; BC) + \\angle(AD; QA) = \\angle(QX; XC) + \\angle(DX; XQ) = \\angle(DX; XC)$.\nThe rest coincides with what we have done before, because $\\angle(QB; QA) = \\angle(PD; PC)$ ($AQB$ and $CPD$ are symmetric with respect to the center of parallelogram). Hence, $D$, $P$, $X$, $C$ are cyclic as well as $A$, $P$, $X$, $B$. Therefore, all circles pass through $X$, and we are done.\n\n![](attached_image_1.png)\nFig.12",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ for which among the numbers $n, n+1, n+2, \\dots, n^2$ there exist 4 pairwise distinct numbers $a, b, c, d$ such that equality $ab = cd$ holds. Justify the answer.",
    "options": [],
    "answer": "n ≥ 3",
    "solution": "Answer: $n \\ge 3$.\n\nIf $6n \\le n^2$ or equivalently $n \\ge 6$, then after putting $a = n$, $c = 2n$, $d = 3n$, $b = 6n$ we will have $ab = cd$. So every $n \\ge 6$ satisfies the statement.\n\nIn the case $n = 5$ we have numbers $5, 6, 7, \\dots, 25$. Considering $a = 6$, $b = 20$, $c = 8$, $d = 15$ we have the desired result.\n\nIn the case $n = 4$ we have numbers $4, 5, 6, \\dots, 16$. Considering $a = 4$, $b = 15$, $c = 6$, $d = 10$ we have the desired result.\n\nIn the case $n = 3$ we can put $a = 3$, $b = 8$, $c = 4$, $d = 6$.\n\nIn the case $n = 1, 2$ in the set $n, n+1, n+2, \\dots, n^2$ at most three numbers, so we can't find four pairwise distinct numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Acute-angled triangle $ABC$ is given. On the perpendicular bisectors to sides $AB$ and $BC$ respectively, points $P$ and $Q$ are chosen. Let $M$ and $N$ be the projections of $P$ and $Q$ onto $AC$ (Fig.09). It turns out, that $2MN = AC$. Prove, that circumcircle of triangle $PBQ$ passes through the circumcenter of triangle $ABC$.\n\n![](attached_image_1.png)\nFig.09",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcenter of $ABC$. Let our perpendicular bisectors meet sides $AB$ and $BC$ at points $K$ and $L$ respectively. Then $KL$ is a midline of triangle $ABC$. Therefore, $KL \\parallel AC$ and $2KL = AC$. Thus, $MKLN$ is parallelogram, because $KL \\parallel MN$ and $KL = MN$. Due to the fact, that $PM \\perp AC$ and $QN \\perp AC$, we have $PN \\parallel QN$. Moreover, $PM \\parallel QN$ and $KM \\parallel LN$, thus, $\\angle PMK = \\angle QNL$ (Fig.10).\n\n$\\angle AKP = \\angle AMP = 90^\\circ$, thus $P, K, M$ and $A$ lie on the same circle with diameter $PA$. By analogy, we obtain that $Q, N, C$ and $L$ lie on the same circle with diameter $CQ$, points $O, K, B$ and $L$ belong to the circle with diameter $BO$. From these observations, it follows $\\angle PAK = \\angle PMK = \\angle QNL = \\angle QCL$.\n\n$P$ and $Q$ belong to perpendicular bisectors of the segments $AB$ and $BC$ respectively, thus $APB$ and $BQC$ are isosceles triangles. From this, we obtain $\\angle PBA = \\angle PAB = \\angle QCB = \\angle QBC$. Hence, $\\angle PBQ = \\angle KBL = 180^\\circ - \\angle KOL = 180^\\circ - \\angle POQ$, in other words, points $P, B, Q, O$ are cyclic, which provides desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ lies inside triangle $ABC$. Let $I_A$, $I_B$, $I_C$ be incenters of triangles $PBC$, $PAC$, $PAB$ respectively. Let $I_P$ denote the incenter of triangle $I_A I_B I_C$. Prove that for point $P$ which satisfies the condition $I_P = P$, the following equalities hold:\n$$\nAP - BP = AC - BC, \\quad BP - CP = BA - CA, \\quad CP - AP = CB - AB.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote the points $A' = AP \\cap I_B I_C$, $B' = BP \\cap I_A I_C$, $C' = CP \\cap I_B I_A$ (see Fig. 11).\n\n$I_P = P$ implies that $I_A P$ is a bisector of $\\angle BPC$ and $\\angle I_B I_A I_C$, therefore two couples of lines $I_A I_B$, $I_A I_C$, and also $BP$, $CP$ are symmetric with respect to $I_A P$. Thus, $\\angle (I_A I_B, CP) = \\angle (BP, I_A I_C)$ (oriented angles).\n\nBy analogy,\n$\\angle (I_B I_C, AP) = \\angle (CP, I_B I_A)$,\n$\\angle (I_C I_A, BP) = \\angle (AP, I_C I_B)$,\nhence,\n$\\angle (I_A I_B, CP) = \\angle (BP, I_A I_C) = -\\angle (AP, I_C I_B) = \\angle (CP, I_B I_A) \\Rightarrow CP \\perp I_B I_A$.\n\nFeet of perpendicular from incenter to the side of triangle coincide with the point where incircle touches this side, hence, incircles of $\\triangle PAC$ and $PBC$ touch $CP$ at $C'$. Therefore, they touch each other at $C'$. Analogously, incircles of $\\triangle PAB$, $\\triangle PBC$ touch $PB$ at $B'$, and $A'$ - incircles of $\\triangle PAC$, $\\triangle PAB$.\n\nLet $A''$ be the point where incircle of $\\triangle PBC$ touches $BC$. We define $B''$, $C''$ in the same way. Then\n\n![](attached_image_1.png)\nFig. 11\n\n$$\nAP + BC = PA' + AA' + BA'' + A''C = PB' + AB'' + BB' + B''C = BP + AC.\n$$\n\nFollowing the same lines, we get $AP + BC = CP + AB$. Hence, $P$ is such a point that $AP - BP = AC - BC$, $BP - CP = BA - CA$, $CP - AP = CB - AB$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vitalii, Michael and Olexandr were given $n$ dollars by their mother each one. Vitalii had to spend all his money for buying books, Michael for buying notebooks, Olexandr for pens. Vitalii had bought 1 book, Michael 2 notebooks and Olexandr 5 pens. After that, they together had $n$ dollars. Prove that one of the boys can buy one more item.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assuming the contrary. Denote by $q_1, q_2$ and $q_3$ the costs of one book, notebook and pen, respectively, and by $r_1, r_2, r_3$ the change our boys were given. Obviously $r_1 < q_1, r_2 < q_2, r_3 < q_3$. From the statement we have\n$$\nn = q_1 + r_1, \\quad n = 2q_2 + r_2, \\quad n = 5q_3 + r_3\n$$\nand we can write the next inequalities: $n = q_1 + r_1 > 2r_1$, $n = 2q_2 + r_2 > 3r_2$, $n = 5q_3 + r_3 > 6r_3 \\Rightarrow r_1 < \\frac{1}{2}n, r_2 < \\frac{1}{3}n, r_3 < \\frac{1}{6}n$. Adding them we will have\n$$\nn = r_1 + r_2 + r_3 < \\frac{n}{2} + \\frac{n}{3} + \\frac{n}{6} = n\n$$\nwhich is impossible. This contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the value of parameter $a$, at which the equation $x^2 - 3x[x] + 2x = a$ has two positive roots.",
    "options": [],
    "answer": "(0, 2) ∪ (-49/4, -12)",
    "solution": "Let's build the graph of the equation, namely the function $y = x^2 - 3x[x] + 2x = a$. It is easy to understand that we are interested only when $x > 0$.\n$$\n\\begin{align*}\nx \\in [0, 1) &\\Rightarrow [x] = 0 \\Rightarrow y = x^2 + 2x \\\\\nx \\in [1, 2) &\\Rightarrow [x] = 1 \\Rightarrow y = x^2 - x \\\\\nx \\in [2, 3) &\\Rightarrow [x] = 2 \\Rightarrow y = x^2 - 4x \\\\\nx \\in [3, 4) &\\Rightarrow [x] = 3 \\Rightarrow y = x^2 - 7x\n\\end{align*}\n$$\nWhile $x \\ge 4$, the apex of the parabola $y = x^2 + x(2 - 3[x])$ is situated at the point $x_b = \\frac{3|x| - 2}{2} \\ge [x] + 1$, which is equivalent to $[x] \\ge 4$. From this, it follows that the function on every interval like $[n, n + 1)$ under the condition $n \\ge 4$ is downward. In addition, it is downward when $x \\ge 4$, which is shown by the following transformations:\n$$\nx \\in [n - 1, n) \\Rightarrow [x] = n - 1 \\Rightarrow y_{n-1} = x^2 + x(5 - 3n) \\\\\n\\text{and } y_{n-1}^{\\min} > y_{n-1}(n) = n^2 + n(5 - 3n);\n$$\n$$\nx \\in [n, n + 1) \\Rightarrow [x] = n \\Rightarrow y_n = x^2 + x(2 - 3n) \\text{ and } y_n^{\\max} = y_n(n) = n^2 + n(2 - 3n) < y_{n-1}^{\\min}.\n$$\nThus, exactly two positive solutions of the function can exist only when $x \\le 4$, and here it is easy to portray the study graph function. For easy perception, the scale is changed. So we can see exactly two positive solutions under the condition $0 < a < 2$, also on the interval $-\\frac{49}{4} < a < -12$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real values of $x$ for which the value of the function $y = (\\sqrt{x})^{2009} + (\\sqrt{1-x})^{2010}$ is integer?",
    "options": [],
    "answer": "x = 0 or x = 1",
    "solution": "Only $x = 0$ and $x = 1$.\n\nObviously $x \\in [0, 1]$ and $0 \\le (\\sqrt{x})^{2009} + (\\sqrt{1-x})^{2010} < 1$. On the other hand we have $x \\le 1$ and $1-x \\le 1$ which implies $(\\sqrt{x})^{2009} + (\\sqrt{1-x})^{2010} \\le x < (1-x) = 1$. The case $(\\sqrt{x})^{2009} + (\\sqrt{1-x})^{2010} = 0$ is impossible, so we have to consider only the case $(\\sqrt{x})^{2009} + (\\sqrt{1-x})^{2010} = 1$. Equality in the previous inequality obtains iff $(\\sqrt{x})^{2009} = x$ and $(\\sqrt{1-x})^{2010} = 1-x$. This is possible only when $x = 0$ or $x = 1$. It's easy to see that both values satisfy the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all positive integers $n$ equation $a^n + 2010b^n = c^{n+1}$ has infinitely many natural solutions $a, b, c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $a = b = 2011k^{n+1}$, where $k$ is a positive integer, then\n$$\na^n + 2010b^n = 2011^n k^{n(n+1)} + 2010 \\cdot 2011^n k^{n(n+1)} = 2011^{n+1} k^{n(n+1)}.\n$$\nSince $2011 \\nmid 2011^n k^{n(n+1)}$ and $2011 \\nmid 2010$, we have\n$$\n2011^{n+1} k^{n(n+1)} = (2011k^n)^{n+1} = c^{n+1} \\Rightarrow c = 2011k^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Natural numbers $a, b$ are chosen in such a way, that the number $m = a + b + 2\\sqrt{ab+1}$ is natural. Prove that $m$ is composite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us suppose that we can find such numbers $a, b$, that $p = a + b + 2\\sqrt{ab+1}$ is prime. Then $ab+1$ is complete square, $a$ and $b$ are distinct. By Cauchy-Schwartz inequality $a + b \\ge 2\\sqrt{ab}$. Besides this, we have $2\\sqrt{ab} + 1 > 2\\sqrt{ab+1} \\Rightarrow a + b \\ge 2\\sqrt{ab} + 1$. $p$ is an odd number, because it is greater than $2$, thus $a + b - 2\\sqrt{ab} + 1$ is natural. From the equality $$(a + b + 2\\sqrt{ab} + 1)(a + b - 2\\sqrt{ab} + 1) = (a - b + 2)(a - b - 2) \\ne 0$$ it follows, that either $a - b - 2$ or $a - b + 2$ is divisible by $p$, but absolute value of each of these two numbers does not exceed $a + b + 2$, which is less than $p$. This contradiction finishes the proof.\nFrom the condition of the problem, it follows that $2\\sqrt{ab+1}$ is a natural number, which immediately gives us that $z = \\sqrt{ab+1}$ is also a natural number. Therefore, $b = \\frac{z^2 - 1}{a} = \\frac{(z - 1)(z + 1)}{a}$. This implies that $a$ can be factored $a = a_1 a_2$ in such a way that $z - 1$ is divisible by $a_1$, and $z + 1$ is divisible by $a_2$. \n\n$m = a + \\frac{(z - 1)(z + 1)}{a} + 2z = \\frac{(a + z)^2 - 1}{a} = \\frac{(a + z - 1)(a + z + 1)}{a} = \\frac{a + z - 1}{a_1} + \\frac{a + z + 1}{a_2}$. Both factors $\\frac{a + z - 1}{a_1}$ and $\\frac{a + z + 1}{a_2}$ are natural and greater than one, thus, $m$ is composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45205,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given irrational number $\\alpha$ for which there exist real $x, y$, such that $x + y = \\alpha$ and $x^k + y^k$ is rational for all $k$ from $2$ to $n$. Find maximal $n$ for which it is possible?\n**Answer:** $n = 3$.",
    "options": [],
    "answer": "3",
    "solution": "We show that for $n = 4$ it cannot hold.\n\nSuppose that $xy = 0$, or $y = 0$. For $n = 2$ it is possible, as an example we can take $x = \\sqrt{2} \\in \\mathbb{R} \\setminus \\mathbb{Q}$, $x^2 = 2 \\in \\mathbb{Q}$. But, if $x^2$ and $x^3$ are rational, then $\\frac{x^3}{x^2} = x$ is also rational.\n\nLet us consider the case $xy \\neq 0$. For $n = 4$ we have $x^2 + y^2$, $x^3 + y^3$, and $x^4 + y^4$ are rational. From the equality $(x^2 + y^2)^2 = (x^4 + y^4) + 2x^2y^2$ it follows that $(xy)^2 \\in \\mathbb{Q}$. But then, $(x^6 + y^6) = (x^2 + y^2)((x^4 + y^4) - (xy)^2) \\in \\mathbb{Q}$, and the equality $(x^3 + y^3)^2 = (x^6 + y^6) + 2(xy)^3$ implies that $(xy)^3 \\in \\mathbb{Q}$, hence, $\\left(\\frac{(xy)^3}{(xy)^2}\\right) = xy \\in \\mathbb{Q}$, and finally we have $x + y = \\frac{x^3 + y^3}{(x^2 + y^2) - xy} = \\alpha \\in \\mathbb{Q}$ which provides a contradiction.\n\nWe now show that for $n = 3$ such numbers do exist. Denote $a = x + y$, $b = xy$. $x^2 + y^2 = (x + y)^2 - 2xy = (a^2 - 2b) \\in \\mathbb{Q}$, $x^3 + y^3 = (x + y)((x + y)^2 - 3xy) = a(a^2 - 3b) \\in \\mathbb{Q}$. Now we find irrational $a, b$, that satisfy the last equalities. Let $(a^2 - 2b) = p$, $a(a^2 - 3b) = q$, then $b = \\frac{1}{2}(a^2 - p) \\Rightarrow a(3p - a^2) = 2q$, in this way, we have the following cubic equation\n\nfor $a$: $a^3 - 3ap + 2q = 0$. Take $a = 2 - \\sqrt{2}$ and find $p$ and $q$: $p = \\frac{14}{3}$, $q = 4$. Then, we get $b = \\frac{2}{3} - 2\\sqrt{2}$. To find $x, y$, we have to solve the following system of equations:\n$$\n\\begin{cases}\nx + y = 2 - \\sqrt{2}, \\\\\nxy = \\frac{2}{3} - 2\\sqrt{2},\n\\end{cases}\n$$\nhence, they are roots of quadratic equation: $t^2 - at + b = 0$. We have: $D^2 = a^2 - 4b = \\frac{10}{3} + 4\\sqrt{2} > 0$ which gives us that our roots are, indeed, real.\nFinally, we have $x^2 + y^2 = (a^2 - 2b) = \\frac{14}{3} = p$ and $x^3 + y^3 = a(a^2 - 3b) = 4 = q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45206,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ are given. Positive numbers $x_0, x_1, \\dots, x_n$ such that $x_0 x_1 \\dots x_n = 1$. Find all positive $\\gamma$ such that inequality\n$$\nx_0^\\gamma + x_1^\\gamma + \\dots + x_n^\\gamma \\ge \\frac{1}{x_0} + \\frac{1}{x_1} + \\dots + \\frac{1}{x_n}\n$$\nholds for any set of numbers $x_0, x_1, \\dots, x_n$.",
    "options": [],
    "answer": "γ ≥ n",
    "solution": "**Answer:** $\\gamma \\ge n$.\n\nAt first we will show that for $0 < \\gamma < n$ there exists a set $x_0, x_1, \\dots, x_n$, for which the inequality from the statement of the problem is not held.\n\nLet $x_0 = x^{-n}$, $x_1 = x_2 = \\dots = x_n = x$ for some $x > 0$. Then we have\n$$\n\\begin{gathered}\nx_0^\\gamma + x_1^\\gamma + \\dots + x_n^\\gamma = x^{-n\\gamma} + n x^\\gamma, \\\\\n\\frac{1}{x_0} + \\frac{1}{x_1} + \\dots + \\frac{1}{x_n} = x^n + n x^{-1}.\n\\end{gathered}\n$$\nFor $\\gamma < n$ we want to find $x > 0$ for which $x^n + n x^{-1} > x^{-n\\gamma} + n x^\\gamma$. For this purpose it is enough to choose $x > 1$: $x^n + n x^{-1} > x^n > (n + 1)x^\\gamma > n x^\\gamma + x^{-n\\gamma}$, i.e. $x^{-n\\gamma} > n + 1$ also $x > \\sqrt[n]{n+1}$. From here we find a corresponding example.\n\nLet now $\\gamma = n$ then by the Cauchy's inequality we receive\n$$\n\\begin{gathered}\n\\frac{x_0^n + x_1^n + \\dots + x_{n-1}^n}{n} \\ge x_0 x_1 \\dots x_{n-1} = \\frac{1}{x_n}, \\\\\n\\frac{x_0^n + x_1^n + \\dots + x_{n-2}^n + x_n^n}{n} \\ge x_0 x_1 \\dots x_{n-2} x_n = \\frac{1}{x_{n-1}}, \\dots,\n\\end{gathered}\n$$\n$$\n\\frac{x_1^n + x_2^n + \\dots + x_n^n}{n} \\geq x_1 x_2 \\dots x_n = \\frac{1}{x_0}.\n$$\nIf all these inequalities are added then we will receive the required inequality.\n\nNow we consider the case $\\gamma > n$, if we prove that for any set of positive numbers $x_0, x_1, \\dots, x_n$ such that $x_0 x_1 \\dots x_n = 1$ the inequality $x_0^\\gamma + x_1^\\gamma + \\dots + x_n^\\gamma \\ge x_0^n + x_1^n + \\dots + x_n^n$ holds, then we will receive a necessary inequality for $\\gamma$ if we use the already proved inequality for $n$:\n$$\nx_0^\\gamma + \\dots + x_n^\\gamma \\ge x_0^n + \\dots + x_n^n \\Leftrightarrow \\\\\nx_0^\\gamma + \\dots + x_n^\\gamma \\ge x_0^{n+\\frac{\\gamma-n}{n+1}} x_1^{\\frac{\\gamma-n}{n+1}} \\dots x_n^{\\frac{\\gamma-n}{n+1}} + \\dots + x_0^{\\frac{\\gamma-n}{n+1}} \\dots x_{n-1}^{\\frac{\\gamma-n}{n+1}} x_n^{n+\\frac{\\gamma-n}{n+1}}.\n$$\nTo prove the last inequality we will take advantage of a weighted Cauchy's inequality\n$$\nx_0^{n+\\frac{\\gamma-n}{n+1}} x_1^{\\frac{\\gamma-n}{n+1}} \\dots x_n^{\\frac{\\gamma-n}{n+1}} \\le \\frac{n+\\frac{\\gamma-n}{n+1}}{\\gamma} x_0^{\\gamma} + \\frac{\\gamma-n}{\\gamma(n+1)} x_1^{\\gamma} + \\dots + \\frac{\\gamma-n}{\\gamma(n+1)} x_n^{\\gamma}.\n$$\nWe will write down similar inequalities for every item, we will add them and we will receive the requisite evidence.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45207,
    "subject": "Mathematics (Multi-modal)",
    "question": "For what minimum integer number $N$ one can change \"/\" to \"+\" and \"-\" in the following expression: $1 \\ast 2 \\ast 3 \\ast \\dots \\ast N$ in order to get:\na) $2010$; b) $2011$?",
    "options": [],
    "answer": "a) 63; b) 65",
    "solution": "a) If we have all $+$, then: $1 + 2 + 3 + \\dots + 62 = 1953 < 2010$, $1 + 2 + 3 + \\dots + 63 = 2016 > 2011$. Hence, $N \\ge 63$. From the other hand, $N = 63$ will do: $1 + 2 - 3 + \\dots + 62 + 63 = 2016 - 6 = 2010$.\n\nb) $N = 63, 64$ are not the solutions, because change of sign before any number does not change the parity of the expression itself and $1+2+3+\\dots+63 = 2016$ and $1+2+3+\\dots+64$ are even. Hence, $N \\ge 65$. Now it's easy to locate signs. Since $1+2+3+\\dots+65 = 2145$, then $-2 \\cdot (2+65)+1+2+3+\\dots+65 = 2145-134=2011$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45208,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are three runners in different vertices of an equilateral triangle with side $1$: First, Second and Third. They start moving simultaneously in the same direction (Second in First's direction, Third in Second's direction, First in Third's direction). Is it necessary that they all meet in one point at the same time, if:\n\na) First, Second and Third have velocity $2008$, $2009$ and $2010$ respectively?\n\nb) They are moving with distinct natural velocities?",
    "options": [],
    "answer": "a) necessary; b) not necessary.",
    "solution": "Answer: a) necessary; b) not necessary.\n\na) We first write the condition, which implies that they eventually meet at one point: $2008t = 2010t + 1 - 3m = 2009t + 2 - 3n$, where $m, n \\in \\mathbb{Z}, t \\in \\mathbb{R}$. We have: $t = 3n - 2$ or $2t = 6n - 4$. Moreover, $2t = 3m - 1 \\Rightarrow 3m - 1 = 6n - 4 \\Leftrightarrow m = 2n - 1$, from that, we can easily find solutions, for instance, if $m = n = 1$ then $t = 1$. If $t = 1$ then, indeed, First will run $2008$, Second $2009$, Third $2010$, hence, they will meet at one point.\n\nb) Let us suppose that First, Second and Third have velocities $1$, $2$ and $4$ respectively. Then, the condition that they meet at one point can be rewritten as follows: $t = 4t + 1 - 3m = 2t + 2 - 3n$, $m, n \\in \\mathbb{Z}, t \\in \\mathbb{R}$. Then, $t = 3n - 2$ and $3t = 3m - 1$. So we obtain the following equation: $9n - 6 = 3m - 1$ for integer $m, n$. This equation has no solutions, therefore, our runners will not meet at one point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given natural numbers $a$, $b$, $c$, $d$ for which $ab^2 + ad^2 + cb^2 = ba^2 + bd^2 + ca^2$ and the number $a^2 + b^2 + c^2 + d^2$ is prime. Prove that $a = b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume the contrary. Let $a \\ne b$. The condition $ab^2 + ad^2 + cb^2 = ba^2 + bd^2 + ca^2$ we can rewrite in the form: $(a-b)(d^2 - ab - ac - bc) = 0$. Since $a \\ne b$ it implies that $d^2 = ab + ac + bc$. Then $a^2 + b^2 + c^2 + d^2 = a^2 + b^2 + c^2 + ab + ac + bc = (a+b+c)^2 - d^2 = (a+b+c+d)(a+b+c-d)$. The number $(a+b+c+d)(a+b+c-d)$ is prime, $a$, $b$, $c$, $d$ are natural numbers, so $a+b+c-d=1$ hence $ab + ac + bc = d^2 = (a+b+c-1)^2$. Removing the brackets in the equality $(a+b+c-1)^2 = ab + ac + bc$, we have: $a^2 + b^2 + c^2 + 1 + 2ab + 2ac + 2bc - 2a - 2b - 2c = ab + ac + bc$ or\n$$\na(a+b-2) + b(b+c-2) + c(c+a-2) + 1 = 0.\n$$\nBut $a$, $b$ and $c$ are natural numbers, so the left side of the last equality is not less than 1. It follows that our assumption was wrong and consequently $a = b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45210,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum number of vertexes can convex polygon have if it is known, that all its angles have integer degree measure?",
    "options": [],
    "answer": "360",
    "solution": "Sum of all angles of convex $n$-gon is $180^{\\circ}(n-2)$, maximal angle, which has integer degree measure is $179^{\\circ}$. So we can write the following inequality $180^{\\circ}(n-2) \\le 179^{\\circ}n \\Rightarrow n \\le 360$. Hence, $n = 360$ maximal possible value. Regular 360-gon provides an example.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist pairwise distinct natural numbers $a_1, a_2, \\dots, a_k$, greater than $1$, for which\n$$\na_1 + a_2 + \\dots + a_k = 2010 \\cdot \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} \\right)\n$$\n\na) if $k=2$;\nb) if $k=12$.",
    "options": [],
    "answer": "a) Yes: any two distinct divisors of 2010 whose product is 2010, for example 2 and 1005. b) Yes: for example 2, 3, 5, 6, 10, 15, 134, 201, 335, 402, 670, 1005.",
    "solution": "**Answer:** a) any pair of divisors of the number $2010$ with product $2010$, for example $a_1 = 2$, $a_2 = 1005$;\nb) $2$, $3$, $5$, $6$, $10$, $15$, $134$, $201$, $339$, $402$, $670$, $1005$.\n\nThe main idea is: if $n$ is not a square of an integer and $d$ is its divisor, $1 < d < n$, then $d' = \\frac{n}{d}$ is also a divisor of $n$, for which $1 < d' < n$ and $d \\neq d'$.\n\na) In such a way, for $a_1$ we can take any divisor of $2010$, that differs from $1$ and $2010$, and then put $a_2 = \\frac{2010}{a_1}$.\n\nb) Analogously, we can write out pairwise distinct pairs of divisors of $2010$, that satisfy the condition above. The answer was given before.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Cut a square into triangles in such a way, that each triangle has exactly 3 neighbor triangles (two triangles are neighbors if they share a line segment; if they share only one point they aren't).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFig.04",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $\\sin \\frac{\\pi\\sqrt{x}}{4} + \\cos \\frac{\\pi\\sqrt{2-x}}{4} = \\sqrt{2}$.",
    "options": [],
    "answer": "x = 1",
    "solution": "We have $x \\in [0, 2]$, for such $x$ both functions $\\sin \\frac{\\pi\\sqrt{x}}{4}$ and $\\cos \\frac{\\pi\\sqrt{2-x}}{4}$ are increasing, thus, their sum is also increasing function. Hence, our equation has at most one real root. From the other hand, one can easily check that $x = 1$ is, indeed, the solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be real numbers such that $x \\in (0, 1]$, $y \\in (0, 1]$, $z \\in (0, 1]$. Prove the inequality:\n$$\n\\frac{x}{2+xy+yz} + \\frac{y}{2+yz+zx} + \\frac{z}{2+zx+xy} \\le \\frac{x+y+z}{x+y+z+xyz}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the obvious inequality:\n$$\n\\begin{aligned}\n0 \\le (1-x)(1-y)(1-z) &= 1 + xy + yz + zx - x - y - z - xyz \\\\\n&\\Rightarrow 1 + xy + yz + zx \\ge x + y + z + xyz \\\\\n&\\Rightarrow 2 + xy + yz \\ge x + y + z + xyz \\\\\n&\\Rightarrow \\frac{x}{2+xy+yz} \\le \\frac{x}{x+y+z+xyz}.\n\\end{aligned}\n$$\nAnalogously:\n$$\n\\begin{aligned}\n\\frac{y}{2+yz+zx} &\\le \\frac{y}{x+y+z+xyz}, \\\\\n\\frac{z}{2+zx+xy} &\\le \\frac{z}{x+y+z+xyz}.\n\\end{aligned}\n$$\nSumming up yields the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least possible natural number $n$, such that each number from the set $1, 2, \\dots, 10$ can be expressed as a digit or as a sum of consecutive digits of $n$.",
    "options": [],
    "answer": "11134",
    "solution": "It is obvious, that we can not find number with given property and three digits. Suppose, that it has 4 digits: $a b c d$, then we can construct 10 different sums of consecutive digits, more precisely: $a$, $b$, $c$, $d$, $a+b$, $b+c$, $c+d$, $a+b+c$, $b+c+d$, $a+b+c+d$. To satisfy the conditions of the problem these 10 sums have to be distinct. Hence, all digits have to be distinct either, and their sum equals 10. We have only one option: 1, 2, 3, 4. But, in order to get 9 as a sum, 1 should be the last or the first digit of our number. In the same way, to represent 8 as a sum of consecutive digits, 2 should be the first or the last digit. Therefore, we have 4 options: 1342, 1432, 2341, 2431. However, for first and fourth numbers we can not get 5 as a sum of consecutive digits, for second and third - 6.\n\nHence, the least possible number of digits is at least 5. Our number can not start with 1111, because, in order to get 10 as a sum, we have the following options: 11116, 11117, 11118, 11119, but we can not express 5 then. Following the same lines, if our number starts with 1112, then we have the following options - 11125, 11126, 11127, 11128, 11129. But we can not express 5 (except second number) and 6 for second number. Therefore, our minimal number starts with 1113 at least. Numbers 11131, 11132, 11133 do not give us 10 as a sum, and 11134 meets all the requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be a circumscribed circle of $ABC$. Chord $AD$ is a bisector of triangle $ABC$ and meets $BC$ at $L$, chord $DK$ is perpendicular to $AC$ and meets it at $M$. Find the ratio $\\frac{AM}{MC}$, if $\\frac{BL}{LC} = \\frac{1}{2}$.\n\n![](attached_image_1.png)\nFig.15",
    "options": [],
    "answer": "3",
    "solution": "Using the property of our bisector, we get $\\frac{BL}{AB} = \\frac{LC}{AC} \\Rightarrow \\frac{AC}{AB} = \\frac{LC}{BL}$, or $\\frac{AC}{AB} = 2$, hence, $AC = 2AB$ (Fig.15).\n\n$$\n\\angle BAD = \\angle DAC \\ (AD \\text{ is a bisector of } \\angle BAC), \\text{ thus } BD = DC. \\text{ Consider } S \\text{ - midpoint of } AC. \\text{ Then } AS = SC = AB. \\text{ Thus, } \\triangle BAD = \\triangle SAD \\text{ (they have two equal sides and angle between them). This implies, that } BD = DS \\text{ and } DC = BD = DS.\n$$\n\nHence, triangle $\\triangle SDC$ is isosceles with base $SC$. Then, its altitude $DM$ is also median. Thus, $MC = \\frac{1}{2}SC = \\frac{1}{2} \\cdot \\frac{1}{2}AC = \\frac{1}{4}AC$. We have $AM = AC - MC = AC - \\frac{1}{4}AC = \\frac{3}{4}AC$. Hence, $\\frac{AM}{MC} = \\frac{\\frac{3}{4}AC}{\\frac{1}{4}AC} = 3$.\n\nNote: In acute-angled triangle $ABC$, $AC = 2AB$, thus $\\angle ACB$ does not exceed $30^\\circ$ (Fig.16). Last observation implies that perpendicular from point $D$ to the line $AC$ meets segment $AC$.\n\n![](attached_image_2.png)\nFig.16",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45217,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$ angles $\\angle ABC$ and $\\angle BCD$ are not less than $120^\\circ$. Prove, that $AC + BD > AB + BC + CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AB = a$, $BC = b$, $CD = c$ (Fig.18).\n\nThen, $AC^2 = a^2 + b^2 - 2ab \\cos \\angle B \\ge a^2 + ab + b^2$.\n\nBy analogy, $BD^2 \\ge b^2 + bc + c^2 \\Rightarrow AC + BD \\ge \\sqrt{a^2 + ab + b^2} + \\sqrt{b^2 + bc + c^2}$.\n\nSince $\\sqrt{a^2 + ab + b^2} > a + \\frac{1}{2}b$ and $\\sqrt{b^2 + bc + c^2} > c + \\frac{1}{2}b$, then $AC + BD > a + b + c = AB + BC + CD$.\n\n![](attached_image_1.png)\nFig.18",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $ABCD$ be a quadrilateral inscribed in a circle with center $O$, and let $P = AC \\cap BD$, $BC \\parallel AD$. Rays $AB$ and $DC$ intersect at the point $E$. A circle with the center $I$ is inscribed in the triangle $EBC$ and is tangent to the line $BC$ at point $T_1$. The excircle of the triangle $EAD$ with center $J$ is tangent to the side $AD$ at point $T_2$. The lines $IT_1$ and $JT_2$ intersect at the point $Q$. Prove that points $O$, $P$ and $Q$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P_1$, $P_2$ be respectively the feet of perpendiculars from $P$ to lines $BC$ and $AD$, $M_1$, $M_2$ respectively are midpoints of sides $BC$ and $AD$ respectively, the excircle of the triangle $EBC$ touches the side $BC$ at the point $T'$. $T'C = BT_1 \\Rightarrow \\frac{BT_1}{T_1C} = \\frac{CT'}{T'B}$. $ABCD$ is a cyclic quadrilateral then $\\triangle EBC \\sim \\triangle EAD \\Rightarrow \\frac{BT_1}{T_1C} = \\frac{CT'}{T'B} = \\frac{AT_2}{T_2D}$ (Fig.22).\n\nMoreover $\\triangle PBC \\sim \\triangle PAD \\Rightarrow \\frac{BP}{P_1C} = \\frac{AP_2}{P_2D}$. Since $M_1$, $M_2$ are the midpoints of segments $BC$ and $AD$ respectively then $\\frac{T_1M_1}{M_1P_1} = \\frac{T_2M_2}{M_2P_2}$.\n\nLet the point $O_1$ be located on the segment $PQ$ such that $\\frac{QO_1}{O_1P} = \\frac{T_1M_1}{M_1P_1}$. Then the projections of this point on the segments $BC$ and $AD$ respectively coincide with points $M_1$ and $M_2 \\Rightarrow O$ coincides with $O_1$. Then points $O$, $P$ and $Q$ are collinear.\n\n![](attached_image_1.png)\nFig.22",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least possible value $k$ for which there exist $2010$ distinct natural numbers that satisfy the following condition: the product of any $k$ numbers from the chosen set is divisible by the product of the rest $2010 - k$ numbers.",
    "options": [],
    "answer": "1006",
    "solution": "From one hand, $k$ cannot be less than $1006$ (otherwise, the product of the $k$ smallest numbers from our set is less than the product of the $2010 - k$ numbers which are left). We construct an example for $k = 1006$ that will do.\n\nLet $p_1, p_2, \\dots, p_{2010}$ be $2010$ distinct prime numbers and $a_i = p_1 p_2 \\dots p_{i-1} p_{i+1} \\dots p_{2010} = \\frac{p_1 p_2 \\dots p_{2010}}{p_i}$. Then, the product of $1006$ numbers $a_{n_1}, a_{n_2}, \\dots, a_{n_{1006}}$ equals\n$$\n\\frac{(p_1 p_2 \\dots p_{2010})^{1006}}{p_{n_1} p_{n_2} \\dots p_{n_{1006}}} = p_1^{a_1} p_2^{a_2} \\dots p_{2010}^{a_{2010}},\n$$\nwhere the degree $a_i$ of each prime number $p_i$, $1 \\le i \\le 2010$, equals $1005$ or $1006$, which is greater than $1005$.\n\nBy analogy, the product of the rest $1004$ numbers can be expressed as $p_1^{\\beta_1} p_2^{\\beta_2} \\dots p_{2010}^{\\beta_{2010}}$, where each degree $\\beta_i$, $1 \\le i \\le 2010$, equals $1003$ or $1004$, which is less than $1004$.\n\nTo finish the proof, note that if $i \\ne j$, then $a_i = \\frac{p_1 p_2 \\dots p_{2010}}{p_j} \\ne p_1^{\\beta_1} p_2^{\\beta_2} \\dots p_{2010}^{\\beta_{2010}} = a_j$, therefore, all $\\{a_i\\}$ are distinct.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $K$ and $M$ are chosen on the sides $AB$ and $BC$ of triangle $ABC$ in such a way, that $AK = KM = MC$. Let $N$ be the point of intersection of $AM$ and $CK$, $P$ – the feet of perpendicular from point $N$ to the line $KM$, and $Q$ is such point inside segment $KM$, that $MQ = KP$. Prove, that incircle of $KMB$ touches $KM$ at point $Q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be incenter of $KMB$. Then, $KI$ and $MI$ are bisectors of $\\angle BKM$ and $\\angle BMK$ respectively. $AKM$ and $CMK$ are isosceles triangles, thus, $\\angle KAM = \\angle KMA = \\angle MKI = \\angle BKI$ and $\\angle MKC = \\angle MCK = \\angle KMI = \\angle BMI$, moreover, these angles are acute (Fig.17).\n\n![](attached_image_1.png)\n\nFig.17\n\nThis implies that $P$ belongs to the segment $KM$. $\\angle IKM = \\angle KMA$ and $\\angle IMK = \\angle MKC$, thus $KI \\parallel AM$ and $MI \\parallel CK$, hence $KIMN$ is parallelogram. So $KN = IM$. $KP = MQ$ (by the condition of the problem), therefore, triangles $PKN$ and $QMI$ are equal (they have two equal corresponding sides and angle between them). This implies that $\\angle IQM = \\angle NPK = 90^\\circ$ or $Q$ is a touching point, what we wanted to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions such that $f: \\mathbb{Z} \\to \\mathbb{Z}$ that satisfy the following two conditions:\n1) $f(x + f(x + 2y)) = f(2x) + f(2y)$ for all integers $x, y$;\n2) $f(0) = 2$.",
    "options": [],
    "answer": "f(k) = k + 2",
    "solution": "**Answer:** $f(k) = k + 2$.\n\nSubstitute $x = 0$ and $y = 0 \\Rightarrow$\n$$\nf(f(2y)) = f(2y) + 2, \\qquad (1)\n$$\n$$\nf(x + f(x)) = f(2x) + 2, \\qquad (2)\n$$\nLet us prove by induction, that $f(2n) = 2n + 2$. Base $n = 0$ is already proved. Let us suppose that we have $f(2n) = 2n + 2$ and prove that $f(2n + 2) = 2n + 4$.\nFor $n = 2m$ we have to show that $f(4m + 2) = 4m + 4$. Substitute $x = 2m$ in (2): we'll get\n$$\nf(2m + f(2m)) = f(2m + 2m + 2) = f(4m + 2) = f(4m) + 2 = 4m + 4.\n$$\nFor $n = 2m - 1$ $f(4m) = 4m + 2$. Substitute $y = 2m - 1$ in (1):\n$$\nf(f(4m - 2)) = f(4m) = f(4m - 2) + 2 = 4m + 2.\n$$\nFollowing the same lines, we can prove that $f(2n) = 2n + 2$ for negative $n$. Let $k$ be an odd number, then substitute $x = 2z + k$, $y = -z$, $z \\in \\mathbb{Z}$, we get\n$$\nf(2z + k + f(k)) = f(4z + 2k) + f(-2) = 4z + 2k + 2 - 2z + 2 = 2z + 2k + 4, \\quad (3)\n$$\nWe consider the following two cases: $f(k)$ is even. Substitute to (3): $z = -\\frac{f(k)}{2}$, then $f(k) = -f(k) + 2k + 4$, therefore $f(k) = k + 2$ is odd and we get a contradiction. $f(k)$ is odd. Then $2z + k + f(k)$ is even and $f(2z + k + f(k)) = 2z + k + f(k) + 2$. Recalling (3), we have $f(2z + k + f(k)) = 2z + k + f(k) + 2 = 2z + 2k + 4$, and, finally, $f(k) = k + 2$, $k \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ is chosen inside isosceles triangle $ABC$ with base $BC$ and $\\angle BAC < 90^\\circ$ in such a way, that $\\angle BPC = 2\\angle BAC$. Let $K$ be the feet of perpendicular from $A$ to the line which contains the bisector of adjacent angle of $\\angle BPC$. Prove, that $BP + PC = 2AK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that point $K$ is always inside triangle $\\triangle ABC$. Consider the case, when $P = P_1$ lies on the side $AB$ and compute some angles. Let $\\angle BAC = \\alpha$ (Fig.13), then $\\angle BP_1C = 2\\alpha$, $\\angle ABC = \\frac{\\pi}{2} - \\frac{\\alpha}{2}$, $\\angle BCP_1 = \\frac{\\pi}{2} - \\frac{3\\alpha}{2} \\Rightarrow \\angle P_1CA = \\alpha$, hence, $\\triangle ACP_1$ is an isosceles triangle, thus the altitude $P_1K_1$ is also a bisector, therefore $K = K_1$ belongs to the side $AC$. Therefore, as long as $P$ is inside $\\triangle ABC$ the bisector $PK$ of adjacent angle of $\\angle BPC$ forms the least angle with line $BC$. Thus, perpendicular from $A$ to this line lies between sides $AB$ and $AC$ of triangle $ABC$. Let us extend $AK$ to the intersection with $CP$ (Fig.14) (without loss of generality, we can assume that $P$ is closer to $B$ than to $C$).\n\n![](attached_image_1.png)\nFig. 14\n\nIn the same way, $BP$ meets $AK$ at point $X$. Then, in $\\triangle PXY$ segment $PK$ is a bisector and an altitude, thus, it is isosceles or $\\angle PXY = \\angle PYX$, from which it follows that $\\angle BXA = \\angle AYC$. If we now denote $x = \\angle XBA$, $y = \\angle YCA$, $x_1 = \\angle CAY$, $y_1 = \\angle BAX$, then $x_1 + y_1 = \\alpha$, $\\pi - \\alpha = x + y + \\angle PBC + \\angle PCB = x + y + \\pi - 2\\alpha \\Rightarrow x + y = \\alpha = x_1 + y_1$, moreover, $x + y_1 = y + x_1$. From last two identities, we get $x = x_1$, $y = y_1$. Hence, $\\triangle ABX = \\triangle CAY$ (they have two equal sides and angle between them).\n\nWe have the following equalities: $2AK = AX + AY = CY + BX = CY + BP + PX = CY + YP + BP = CP + BP$, which is exactly what we wanted to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $a_1, a_2, \\dots, a_n$ satisfy the following conditions $a_1 + a_2 + \\dots + a_n = n$ and $a_1 \\ge a_2 \\ge \\dots \\ge a_n \\ge 0$. Prove the inequality:\n$$\nna_1 \\ge a_1^2 + a_2^2 + \\dots + a_n^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Proof immediately follows from next transformations: $a_1^2 + a_2^2 + \\dots + a_n^2 \\le a_1^2 + a_2^2 + \\dots + a_n^2 + a_1(a_1 - a_1) + a_2(a_2 - a_2) + a_3(a_3 - a_3) + \\dots + a_n(a_n - a_n) = a_1(a_1 + a_2 + \\dots + a_n) = na_1$ and we are done.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45224,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the least possible value of\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + \\dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2,\n$$\nif $x_1, x_2, \\dots, x_n$ are distinct integer numbers.",
    "options": [],
    "answer": "4n - 6",
    "solution": "**Answer:** $4n - 6$.\n\nWe prove by induction that\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + \\dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \\geq 4n - 6,\n$$\nif $x_1, x_2, \\dots, x_n$ are distinct integer numbers.\n\nThe base is trivial. Indeed, $S_2 = (x_1 - x_2)^2 + (x_2 - x_1)^2 \\geq 2$, because all numbers are integer and distinct.\n\nLet us now suppose that our assumption holds for $n$, in other words,\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + \\dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \\geq 4n - 6.\n$$\nLet $x_1, x_2, \\dots, x_n, x_{n+1}$ be distinct integer numbers. WLOG, we can assume that $x_{n+1}$ is a maximum among our numbers. We now have\n$$\n(x_n - x_{n+1})^2 + (x_{n+1} - x_n)^2 - (x_n - x_1)^2 = (x_{n+1} - x_n)(x_{n+1} - x_1) \\geq 4.\n$$\nSumming all such inequalities and using our induction hypothesis we get the desired result for $n + 1$ numbers.\n\nWe now construct example, that proves sharpness of our bound.\nFor $n = 2k - 1$ one can take $x_j = 2j - 2$ for $j \\leq k$ and $x_j = -2j + 4k - 1$ for $j \\geq k + 1$.\nFor $n = 2k$ we take $x_j = 2j - 2$ for $j \\leq k$ and $x_j = -2j + 4k + 1$ for $j \\geq k + 1$.\n\n\nAlternative solution:\nLet us assume, that $x_1$ is a maximum and $x_k$ is a minimum among our numbers. Then, we have $x_1 - x_k \\geq n - 1$. By AM-GM:\n$$\n\\begin{aligned} & \\sum_{j=1}^{n} (x_j - x_{j+1})^2 \\geq \\frac{1}{n} \\left( \\sum_{j=1}^{n} |x_j - x_{j+1}| \\right)^2 \\\\ & \\geq \\frac{((x_1 - x_2) + \\dots + (x_{k-1} - x_k) + (-x_k + x_{k+1}) + (-x_{k+1} + x_{k+1}) + \\dots + (-x_n + x_1))^2}{n} \\\\ & = \\frac{(2(x_1 - x_k))^2}{n} \\geq \\frac{(2(n-1))^2}{n} = 4n - 8 + \\frac{4}{n}. \\end{aligned}\n$$\n$\\sum_{j=1}^{n} (x_j - x_{j+1})^2$ is natural, thus $\\sum_{j=1}^{n} (x_j - x_{j+1})^2 \\geq 4n - 7$. Moreover, squaring does not change a parity of an expression and $\\sum_{j=1}^{n} (x_j - x_{j+1}) = 0$ is even, then $\\sum_{j=1}^{n} (x_j - x_{k+1})^2$ is also even. Hence, it is not less than $4n - 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $h$ be the altitude of triangle $ABC$ passing through the vertex $A$ and $\\alpha = \\angle BAC$. Prove that the following inequality holds:\n$$\nAB + AC \\ge BC \\cdot \\cos \\alpha + 2h \\sin \\alpha.\n$$\nIn what triangles does equality hold?",
    "options": [],
    "answer": "Equilateral triangles",
    "solution": "If $\\alpha \\ge 90^\\circ$, then $AB + AC > 2h \\ge 2h \\sin \\alpha \\ge BC \\cdot \\cos \\alpha + 2h \\sin \\alpha$, because $\\cos \\alpha \\le 0$, that is, the inequality is strict.\n\nLet now $\\alpha < 90^\\circ$, denote by $H_1, H_2, H_3$ the feet of altitudes from vertices $A, B, C$ respectively, $K_2$ be the point symmetric to $H_2$ with respect to line $AB$ and analogously $K_3$ be the point symmetric to $H_3$ with respect to line $AC$. Denote by $\\omega_1$ and $\\omega_2$ the circumscribed circles of the triangles $ABH_1$ and $ACH_1$ respectively (Fig.23). Then $AB$ and $AC$ are diameters of these circles, as $\\angle AK_2B = \\angle AK_3C = 90^\\circ$, so $K_2 \\in \\omega_1$ and $K_3 \\in \\omega_2$. Then $AB \\ge K_2H_1$ and $AC \\ge K_3H_1$.\n\n![](attached_image_1.png)\nFig.23\n\nNext, by Ptolemy's theorem $AB \\cdot K_2H_1 = AK_2 \\cdot BH_1 + BK_2 \\cdot AH_1$. This implies\n$$\nK_2H_1 = \\frac{AK_2 \\cdot BH_1}{AB} + \\frac{BK_2 \\cdot AH_1}{AB} = \\frac{AH_2}{AB} \\cdot BH_1 + \\frac{BH_2}{AB} \\cdot AH_1 = BH_1 \\cos \\alpha + h \\sin \\alpha.\n$$\nAnalogously, $K_3H_1 = CH_1 \\cos \\alpha + h \\sin \\alpha$. Then\n$$\nAB + AC \\geq K_2H_1 + K_3H_1 = (BH_1 \\cos \\alpha + h \\sin \\alpha) + (CH_1 \\cos \\alpha + h \\sin \\alpha) = BC \\cos \\alpha + 2h \\sin \\alpha,\n$$\nthe proof is finished.\n\nEquality holds if and only if the segments $K_2H_1$ and $K_3H_1$ are the diameters of circles $\\omega_1$ and $\\omega_2$ respectively. In this case\n$$\n\\angle ABC = 90^\\circ - \\angle BAH_1 = \\angle BAK_2 = \\angle BAC = \\angle CAK_3 = 90^\\circ - \\angle CAH_1 = \\angle ACB,\n$$\nthen the triangle $ABC$ is equilateral. It is easily checked that the equality holds.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $1$, $2$, ..., $n$ are placed in a row in a certain order. We are allowed to do the following operation: take any two pairs of consecutive elements, that have no common elements, and exchange their positions. Is it possible to get a monotone sequence of elements after a finite number of steps if:\n\na) $n = 2009$;\nb) $n = 2010$?",
    "options": [],
    "answer": "a) not always; b) always possible.",
    "solution": "**Answer:** a) not always; b) always possible.\n\nIf we consider $5$ consecutive elements, then from combination $12345$ we can get the following:\n$$\n12345 \\rightarrow 14523 \\rightarrow 23514 \\rightarrow 51234 \\rightarrow 53412 \\rightarrow 12453.\n$$\nLast three permutations were cyclically rearranged, so, similarly, we can get three numbers $12534$. If we make symmetric permutations, then we can also permute first three elements of our five numbers:\n$$\n12345 \\rightarrow 34125 \\rightarrow 25134 \\rightarrow 23451 \\rightarrow 45231 \\rightarrow 31245.\n$$\nConsider some four positions with numbers $1001$-$1004$, for example. We first locate $1$, $2$, $3$, ..., $1000$. The following shows how this can be done on the example of $500$. Let us suppose that numbers $1$-$499$ are already on their positions. Further we will not touch their permutations. If $500$ is next right from $499$ (on its position), everything is done. If $500$ is on the last position or $n$-labeled, then we swap positions of pairs that occupy positions with numbers $(n-3, n-2)$ and $(n-1, n)$ and $500$ is already not on the last position. If $500$ is on one position after $499$ (labeled as $501$), then we change positions of pairs with following labels at first: $(500,501)$ and $(502,503)$. Let us suppose that $500$ is already on some position $k \\in \\{502, 503, ..., n-1\\}$, then we are swapping positions of pairs labeled as: $(500,501)$ and $(k, k+1)$ and number $500$ is taking its position. So we will place on their seats all the numbers $1$-$1000$. After this, we similarly symmetrical interchange places of numbers $1005$-$n$. That means that we always can get next permutation:\n\n$1$, $2$, ..., $1000$, $a$, $b$, $c$, $d$, $1005$, $1006$, ..., $n$,\n\nwhere the set of numbers $(a, b, c, d)$ is some permutation of numbers $(1001, 1002, 1003, 1004)$. Let's see what permutations of $(a, b, c, d)$ we can get. There are just $24$ of them. If we will take in attention next $5$ numbers $(1000, a, b, c, d)$, then using the scheme above we will gain also: $(1000, a, c, d, b)$ and $(1000, a, d, b, c)$. If we will add right-side $1005$ to gained sets of four numbers $(a, b, c, d)$, $(a, c, d, b)$ and $(a, d, b, c)$ and use the scheme above, so we will obtain next permutations: $(b, c, a, d)$, $(c, a, b, d)$, $(c, d, a, b)$, $(d, a, c, b)$, $(d, b, a, c)$, $(b, a, d, c)$. Apply again the scheme to some of obtained permutations and we will get another $3$ variants: $b, d, c, a$, $c, b, d, a$ and $d, c, b, a$. So, we could get $12$ out of $24$. We now have to show that to arrange remaining $12$ numbers in ascending order is impossible.\n\nFor any permutation of numbers $1$, $2$, ..., $n$ call as inversion the case when a larger number is left of a smaller, for example the position of $5$ numbers $23514$ has an inversion: $(2,1)$, $(3,1)$, $(5,1)$ and $(5,4)$. Call the quantity of such inversions as \"position invariant\". Let's show that after performing any authorized operation, the invariant of positions varies by a number which is divisible by $2$. Let us suppose that we have changed numbers $(a, b)$ and $(c, d)$, that occupied positions $(k, k+1)$ and $(l, l+1)$, $k < l$. Then the quantity of inversions will not change after permutation. We have only mentioned inversions associated with numbers located on positions $1$, $2$, ..., $(k-1)$ and $(l+2)$, $(l+3)$, ..., $n$. That means that the quantity of inversions is constant. The quantity of inversions that use the numbers on positions $(k+2)$, $(k+3)$, ..., $(l-1)$ is not changing relative to each other. But if we consider some number $e$, which occupies one of mentioned positions, then with attention to each of those pairs of numbers $(a, b)$ and $(c, d)$, that are transposed, it will change. And this is changing necessarily from $1$ to $0$ or vice versa from $0$ to $1$. Hence the total quantity of changes will be even, so the parity of position invariant will not change. And about the numbers $a$, $b$, $c$, $d$, behind $a$ and $b$, and $c$ and $d$ nothing is changing and behind all other pairs ($4$ at all) will be the same situation. So, the parity of inversions will stay the same.\n\nSo, it's easy to understand that we would not move on from position $1$, $2$, $3$, $4$, ..., $n$ (invariant is equal to $0$) to position $2$, $1$, $3$, $4$, $5$, ..., $n$ (invariant is $1$). So we would not obtain other $12$ permutations inside of $a$, $b$, $c$, $d$ permutation.\n\nBut all this applies only to try to arrange the numbers in ascending order. Completely the same way you can try to arrange them in descending order. As we see, all the positions were broken into two large groups (with even and odd invariant), that cannot be translated.\n\nLet's count the invariant of such a position: $n$, $n - 1$, $n - 2$, ..., $3$, $2$, $1$. Every pair here is an inversion, so their total amount is $C_n^2 = \\frac{n(n-1)}{2}$. If it's an even number, then both of monotone positions locate in the same half and we can transform them between each other. But we cannot get all the positions in general. If this number is odd, vice versa, then any start position with its invariant would be transformed into increasing or decreasing one. So, we just have to find the parity of number $C_n^2 = \\frac{n(n-1)}{2}$. So if $n$ and $n - 1$ have different parity, then that number will be even if one of them is divisible by $4$. That means that $n = 4m$ or $n = 4m + 1$, else it's odd. Hence for $n = 2009 \\equiv 1 \\pmod{4}$ not all the positions could be transformed one to another. And for $n = 2010 \\equiv 2 \\pmod{4}$ from any position with the same scheme we can obtain increasing or decreasing set of numbers instead.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the infinite sequences of positive integer numbers, in which each positive integer number to meet among elements of this sequence is equal once. Let $\\{a_n\\}$, $n \\ge 1$ be such a sequence. Name it sequence \"consecutive\" if for each positive integer number $k$ and for any positive integer numbers $n$ and $m$ such that $a_n < a_m$, the following inequality holds: $a_{kn} < a_{km}$. For example, the sequence $a_n = n$ is \"consecutive\".\n\na) Prove that there exists a \"consecutive\" sequence which is different from $a_n = n$.\n\nb) Does there exist a \"consecutive\" sequence for which the inequality $a_n \\neq n$, $n \\geq 2$ holds?\n\nc) Does there exist a \"consecutive\" sequence for which the inequality $a_n \\neq n$, $n \\geq 1$ holds?",
    "options": [],
    "answer": "a) Yes, a nontrivial consecutive sequence exists (e.g., swap the exponents of two and three in prime factorizations). b) Yes, such a sequence exists with a_n ≠ n for all n ≥ 2. c) No, such a sequence does not exist for all n ≥ 1.",
    "solution": "Answer: b) such sequence exists; c) such sequence doesn't exist.\n\n**a.** Show the example of such a \"consecutive\" sequence which is different from $a_n = n$. Let $n = 2^{\\alpha_1}3^{\\alpha_2}5^{\\alpha_3}\\dots p_r^{\\alpha_r}$ be the unique representation of $n$ as a product of primes, then the sequence $\\{a_n\\}$, $a_n = 2^{\\alpha_2}3^{\\alpha_1}5^{\\alpha_3}\\dots p_r^{\\alpha_r}$ satisfies the statement of the problem.\n\nReally, if $n = 2^{\\alpha_1}3^{\\alpha_2}5^{\\alpha_3}\\cdots p_r^{\\alpha_r}p_{r+1}^0p_{r+2}^0\\cdots$, $m = 2^{\\beta_1}3^{\\beta_2}5^{\\beta_3}\\cdots p_s^{\\beta_s}p_{s+1}^0p_{s+2}^0\\cdots$, $k = 2^{\\gamma_1}3^{\\gamma_2}5^{\\gamma_3}\\cdots p_t^{\\gamma_t}p_{t+1}^0p_{t+2}^0\\cdots$ and $a_n < a_m$, then $a_{kn} < a_{km}$, as\n$$\na_{kn} = 2^{\\alpha_2+\\gamma_2} 3^{\\alpha_1+\\gamma_1} 5^{\\alpha_3+\\gamma_3} \\cdots p_{\\max(r,t)}^{\\alpha_{\\max(r,t)}+\\gamma_{\\max(r,t)}} = a_k a_n,\n$$\n$$\na_{km} = 2^{\\beta_2 + \\gamma_2} 3^{\\beta_1 + \\gamma_1} 5^{\\beta_3 + \\gamma_3} \\cdots p_{\\max(s,t)}^{\\beta_{\\max(s,t)} + \\gamma_{\\max(s,t)}} = a_k a_m\n$$\nand $a_{kn} = a_k a_n < a_k a_m = a_{km}$. The constructed sequence is non-trivial, because $a_2 = 3$, and is constructed using a permutation of the exponents of $n$ in its prime factorization by means of shifting the exponents of two and three.\n\n**b.** Next, let\n$$\nN(k) = \\begin{cases} k-2, & \\text{if } k = 4, 6, 8, \\dots, \\\\ 1, & \\text{if } k = 2, \\\\ k+2, & \\text{if } k = 1, 3, 5, \\dots \\end{cases}\n$$\nThen $N(k_1) \\neq N(k_2)$ if $k_1 \\neq k_2$. If now we construct a sequence in which for the number $n$ represented as a product of primes, we shift the exponents of $p_k$ to $p_{N(k)}$, then the constructed sequence satisfies the statement of the problem. Really, let for $n = 2^{\\alpha_1}3^{\\alpha_2}5^{\\alpha_3}7^{\\alpha_4}\\dots p_{r-1}^{\\alpha_{r-1}}p_r^{\\alpha_r}$\n$$\na_n = 5^{\\alpha_1} 2^{\\alpha_2} 11^{\\alpha_3} 3^{\\alpha_4} \\cdots p_{N(r-1)}^{\\alpha_{r-1}} p_{N(r)}^{\\alpha_r}.\n$$\nThen the sequence $\\{a_n\\}$ is \"consecutive\", because $a_{kn} = a_k a_n$ and $a_{kn} = a_k a_n < a_k a_m = a_{km}$.\n\nMoreover, if $n \\ge 2$ then $a_n \\ne n$, because if $a_n = n$, then the following equalities hold: $\\alpha_{N(k)} = \\alpha_k$, $\\alpha_{N(N(k))} = \\alpha_{N(k)} = \\alpha_k$, $\\dots$, $\\alpha_{N(...N(N(k)))} = \\alpha_k$. But $N(...N(N(k)))$ is unrestrictedly large beginning with some iteration, so if there exists nonzero $\\alpha_k$ in the prime factorization of $n$, then in its representation must exist nonzero powers of arbitrarily large prime numbers, which is a contradiction.\n\n**c.** If for some sequence holds $a_1 \\neq 1$, then there exists $l$ such that $a_l = 1$. Then $a_l < a_1$ and $a_{l2} < a_l = 1$ — contradiction, because $a_{l2} \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Olesya wrote down a natural number $N$. After this, Andriy wrote down one sixth, one fifth, one fourth, one third and a half of $N$. It turns out that the sum of all numbers that are written is integer. What is the least possible number that Olesya could write?",
    "options": [],
    "answer": "20",
    "solution": "The sum is given by: $\\left(\\frac{1}{6} + \\frac{1}{5} + \\frac{1}{4} + \\frac{1}{3} + \\frac{1}{2}\\right)N = \\frac{29N}{20}$, and therefore $N$ is divisible by $20$ and the result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Olesya was given a homework to add two canonical fractions $\\frac{a}{b}$ and $\\frac{c}{d}$. Her classmate Andriy who missed the class asked her by phone about the homework, and due to bad connection he heard that they need to add $\\frac{b}{a}$ and $\\frac{d}{c}$. After he added them, Andriy asked Olesya for the answer. It turns out that the answers are the same. Was Andriy right in his calculations, if Olesya got an excellent mark, and all 4 fractions, they have been working with are distinct?",
    "options": [],
    "answer": "No",
    "solution": "Suppose his calculations are correct, then the following equality holds: $\\frac{a}{b} + \\frac{c}{d} = \\frac{b}{a} + \\frac{d}{c}$ or $\\frac{ad+bc}{bd} = \\frac{bc+ad}{ac}$. Therefore, we have $bd = ac$ or $\\frac{b}{a} = \\frac{c}{d}$, which contradicts to the assumption that all fractions are distinct.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $[x^2] - 2x + 1 = 0$, where $[x^2]$ denotes the greatest integer that does not exceed $x^2$.",
    "options": [],
    "answer": "1/2, 1, 3/2",
    "solution": "We have $x = \\frac{[x^2] + 1}{2}$. Hence, $x = t$, or $x = t + \\frac{1}{2}$, where $t$ is an integer.\n\nIf $x = t$, we have\n$$\n[x^2] - 2x + 1 = [t^2] - 2t + 1 = t^2 - 2t + 1 = (t - 1)^2 = 0 \\Leftrightarrow t = 1 \\Leftrightarrow x = 1.\n$$\n\nIf $x = t + \\frac{1}{2}$, then\n$$\n[x^2] - 2x + 1 = [(t + \\frac{1}{2})^2] - 2(t + \\frac{1}{2}) + 1 = [t^2 + t + \\frac{1}{4}] - 2t - 1 + 1.\n$$\nSince $t^2 + t + \\frac{1}{4}$ is not an integer, $[t^2 + t + \\frac{1}{4}] = t^2 + t$.\nSo,\n$$\n[x^2] - 2x + 1 = t^2 + t - 2t - 1 + 1 = t^2 - t.\n$$\nThus,\n$$\nt^2 - t = 0 \\Leftrightarrow t = 0 \\text{ or } t = 1 \\Leftrightarrow x = \\frac{1}{2} \\text{ or } x = \\frac{3}{2}.\n$$\n\nTherefore, the solutions are $x = 1$, $x = \\frac{1}{2}$, and $x = \\frac{3}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Olesya writes down numbers $1, 2, 3, 4, 5, 6$ at the vertices of a prism. After this, at each edge Andriy writes down the sum of numbers that are written at the vertices that form this edge. Can Olesya write numbers in such a way that all Andriy's numbers are different?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes. See fig. 10.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find minimal number $n$ such that $n^3 + n^2 + 330n + 330$ is divisible by $2011$?",
    "options": [],
    "answer": "41",
    "solution": "$$\nn^3 + n^2 + 330n + 330 = (n+1)(n^2 + 330),\n$$\nSo this expression is divisible by $2011$ if at least one bracket is divisible by $2011$. If first bracket is divisible by $2011$ then minimal $n=2010$, for $n^2 + 330$, since $n^2$ is increasing, we find that for $n=41$, $n^2 + 330 = 2011$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of parameter $b$, such that for all $x$ at least one function $f_1(x) = x^2 + 2011x + b$ or $f_2(x) = x^2 - 2011x + b$ is positive.",
    "options": [],
    "answer": "b > 0",
    "solution": "For $x = 0$ we have $f_1(0) = f_2(0) = b$, thus all $b \\le 0$ does not satisfy the condition of the problem.\n\nLet $b > 0$. We add two values up and get $f_1(x) + f_2(x) = 2x^2 + 2b > 0$, thus at least one function is positive.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find maximal natural number, with all distinct digits such that the difference between any two consecutive digits is at least $2$.",
    "options": [],
    "answer": "9758642031",
    "solution": "It is clear that our number has to have $10$ digits.\nWe start with $975$ (it is clear that $8$ cannot be neither second nor third digit).\n$975864$ - are first six digits. We can't have $3$ after, so we have $2$, and finally the answer is: $9758642031$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45235,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of a few consecutive integers (that are greater than $1$) is $2011$. Find all such numbers.",
    "options": [],
    "answer": "1005 + 1006",
    "solution": "Let $a_1$ be the first number and $a_n$ is the last one. We arrive at:\n$$\n\\frac{a_1 + a_n}{2} n = 2011 \\Leftrightarrow (a_1 + a_n)n = 2 \\cdot 2011.\n$$\nSince $2$ and $2011$ are prime, we have two options: $a_1 + a_n = 2$ and $n = 2011$ or $a_1 + a_n = 2011$ and $n = 2$. First case is clearly impossible, while the second case gives the answer: $1005 + 1006$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find maximal natural even number, with all distinct digits such that the difference between any two consecutive digits is at least 2.",
    "options": [],
    "answer": "9758641302",
    "solution": "We start in the same way as in 7-1. We have the first piece $975864$.\nIf the next digit is $2$, then the next one is $0$ and all even digits are used. So the next digit should be $1$. And we get the answer: $9758641302$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $x, y$ satisfy the inequality:\n$$\nx^2 + 3xy + 4y^2 \\leq \\frac{7}{2}.\n$$\n\nProve that $x + y \\leq 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $t = x + y$, and put $x = t - y$ into the given inequality. Then we get:\n$$(t - y)^2 + 3(t - y)y + 4y^2 - \\frac{7}{2} \\le 0$$\nor, equivalently,\n$$\n2y^2 + ty + t^2 - \\frac{7}{2} \\le 0.\n$$\nThe left-hand side of the last inequality can be considered as a quadratic polynomial in $y$. This polynomial has a positive leading coefficient and at least one non-positive value, so its determinant is non-negative:\n$$\nD = 28 - 7t^2 \\geq 0 \\quad \\Rightarrow \\quad t^2 \\leq 4.\n$$\nThis proves that $t = x + y \\le 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45238,
    "subject": "Mathematics (Multi-modal)",
    "question": "An equilateral triangle with the side length $7$ is divided into $49$ small equilateral triangles with the side length $1$, as it is shown on fig. 28. Parallelograms with side lengths $1$ and $2$ are cut from the triangle along the grid lines. What is the greatest number of parallelograms that can be cut in this way?\n\n![](attached_image_1.png)\nFig. 28",
    "options": [],
    "answer": "10",
    "solution": "Paint the small triangles in black and white as it is shown on fig. 29. Since every parallelogram with the side lengths $1$ and $2$ contains two white triangles, and the total number of white triangles is $21$, the number of parallelograms does not exceed $10$.\n\nOn the fig. 30 it is shown how one can cut $10$ parallelograms.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45239,
    "subject": "Mathematics (Multi-modal)",
    "question": "For the natural number $N = p_1^{a_1} p_2^{a_2} \\dots p_n^{a_n}$, written in the canonical form ($p_i$ are distinct primes and $a_i$ are naturals, $1 \\le i \\le n$), we denote $T(N) = a_1 + a_2 + \\dots + a_n$. For some distinct natural $a, b, c, d$ the number $ab + cd$ is divisible by $ac + bd$. Prove that $T(ab + cd) \\ge 3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To the contrary, assume that $T(ab + cd) \\le 2$.\n\nProblem 8-3 implies that $T(ac + bd) \\ge 2$. Now, since $(ac + bd) \\mid (ab + cd)$, we have that $T(ab + cd) \\ge T(ac + bd) \\ge 2$, which by the assumption means that $T(ab + cd) = T(ac + bd) = 2$. But then $(ac + bd) = (ab + cd) \\Rightarrow (a - d)(b - c) = 0$, which is a contradiction to the fact that our numbers are distinct.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Through a point $F$ outside a circle $k$, a tangent $FA$ and a secant $\\ell$ are drawn. $\\ell$ intersects $k$ at the points $B$ and $C$ ($C$ is between $F$ and $B$). The tangent to the circle $k$ at the point $C$ intersects the line segment $FA$ at the point $E$. $FX$ is a bisector of the triangle $AFC$. If the points $E,X$ and $B$ are collinear, prove that the product of two side lengths of the triangle $ABC$ equals to the square of the third side length.\n![](attached_image_1.png)\nFig. 31",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that $BX$ is a \"simedian\" of $\\triangle ABC$ (a cevian symmetric to the median with respect to the bisector, all drawn from the vertex $B$). By a known property of the \"simedian\" (which we will also prove) $\\frac{CX}{AX} = \\frac{CB^2}{AB^2}$. By the angle bisector theorem for $FX : \\frac{CX}{AX} = \\frac{FC}{FA}$ (fig. 31).\n\n$\\triangle AFC$ is similar to $\\triangle BFA$ since they have a common angle $\\angle CFA$, and $\\angle CAF = \\angle FBA$.\nSo, $\\frac{FC}{AF} = \\frac{CX}{AX} = \\frac{CB^2}{AB^2}$.\n$$\n\\frac{AC}{AB} = \\frac{FC}{AF} = \\frac{CX}{AX} = \\frac{CB^2}{AB^2},\n$$\nwhich provides the required\n$$\nAB \\cdot AC = CB^2.\n$$\n\nNow we will prove the necessary properties of the “simedian”.\n1) Let $AS = s$ be the “simedian”, as on fig. 32., $AL = l$ be the bisector, $AM = m$ — the median of $\\triangle ABC$, then $\\angle MAL = \\angle LAS$.\nConsider the quotient of the areas of the triangles $ASC$ and $AMB$ :\n$$\n\\frac{S_{AMB}}{S_{ASC}} = \\frac{\\frac{1}{2}mc \\sin \\angle BAM}{\\frac{1}{2}sb \\sin \\angle SAC} = \\frac{cm}{bs},\n$$\nbecause the angles $\\angle BAM$ and $\\angle SAC$ are equal (fig. 32).\nSimilarly, consider the areas of the triangles $ASB$ and $AMC$ to obtain that\n$$\n\\frac{S_{AMC}}{S_{ABS}} = \\frac{bm}{cs}.\n$$\nSince $AM$ is a median, we have that $S_{AMB} = S_{AMC}$, and so dividing the equalities we get:\n$$\n\\frac{S_{ABS}}{S_{ASC}} = \\frac{c^2}{b^2} = \\frac{BS}{SC},\n$$\nas required.\n\nLet $O$ be the center of the circle. Then in the right triangle $FOC$ we have $OM \\cdot OF = OC^2 = R^2 = OB^2$, and so\n$$\n\\frac{OM}{OB} = \\frac{OB}{OF}.\n$$\nNow consider the triangles $BOM$ and $FOB$. They have a common angle and proportional adjacent sides, hence, they are similar. Let\n$$\n k = \\frac{OM}{OB} = \\frac{OB}{OF} = \\frac{BM}{BF},\n$$\n![](attached_image_2.png)\nFig. 32\n![](attached_image_3.png)\nFig. 33\n\nthen $OB = kOF$, $OM = kOB = k^2OF$. This implies (see fig. 33)\n$$\n\\frac{DM}{DF} = \\frac{BO-OM}{OF-OB} = \\frac{kOF-k^2OF}{OF-kOF} = k = \\frac{BM}{BF},\n$$\nand by the angle bisector theorem, $BD$ is the bisector of $\\angle FBM$. This means that the angle between $BF$ and the bisector $BD$ equals to the angle between the bisector $BD$ and the median $BM$, and so $BF$ is indeed a \"simedian\" of the triangle $ABC$.\n**Alternative solution.** Consider the triangle $ACF$. The points $E$, $X$ and $B$ are collinear, so, by the Menelaus' theorem, we have\n$$\n\\frac{AE}{EF} \\cdot \\frac{FB}{BC} \\cdot \\frac{CX}{XA} = 1. \\quad (1)\n$$\nSince $FX$ is the bisector of the angle $AFC$, we get $\\frac{CX}{XA} = \\frac{EF}{FB}$. $EA = EC$, as they are tangents to the same circle from the same point, and $BF = \\frac{AF^2}{FC}$ from the secant-tangent theorem. So we can rewrite (1) in the form\n$$\n\\frac{EC}{EF} \\cdot \\frac{AF}{BC} = 1. \\quad (2)\n$$\nSince $\\angle ECA = \\angle CBA$, we have $\\angle FCE = \\pi - \\angle ECA - \\angle ACB = \\pi - \\angle CBA - \\angle ACB = \\angle CAB$. So by the sine theorem for the triangle $CEF$ we get\n$$\n\\frac{EC}{\\sin \\angle AFB} = \\frac{EF}{\\sin \\angle FCE} = \\frac{EF}{\\sin \\angle CAB}, \\quad (3)\n$$\nand, on the other hand, by the sine theorem for the triangle $AFB$\n$$\n\\frac{AF}{\\sin \\angle FBA} = \\frac{AB}{\\sin \\angle AFB}. \\quad (4)\n$$\nUsing (2), (3), and (4) together with the sine theorem for the triangle $ABC$, we obtain\n$$\n\\frac{AB}{BC} = \\frac{EF \\sin \\angle AFB}{EC \\sin \\angle FBA} = \\frac{\\sin \\angle CAB}{\\sin \\angle CBA} = \\frac{BC}{AC},\n$$\nwhich implies the necessary identity.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45241,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ with $AC > BC > AB$. Points $D$ and $K$ are chosen on the sides $BC$ and $AC$ respectively so that $CD = AB$, $AK = BC$. $F$ and $L$ are the midpoints of the segments $BD$ and $KC$ respectively. $R$ and $S$ are the midpoints of the sides $AC$ and $AB$ respectively. The line segments $SL$ and $FR$ intersect at the point $O$, and it is known that $\\angle SOF = 55^\\circ$. Find $\\angle BAC$.",
    "options": [],
    "answer": "70°",
    "solution": "![](attached_image_1.png)\nFig. 41\n\n$\\angle FOS = \\angle BJT = 55^\\circ \\Rightarrow \\angle BJC = 125^\\circ$, and because $J$ is the incenter $\\angle BJC = 90^\\circ + \\frac{1}{2}\\angle BAC \\Rightarrow \\angle BAC = 70^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45242,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive real numbers $x, y, z$ with $xyz=1$ prove the following inequality:\n$$\n(-x+y+z)(x-y+z) + (x-y+z)(x+y-z) + (x+y-z)(-x+y+z) \\le 3.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, assume that $z = \\min\\{x, y, z\\}$. The inequality can be rewritten as\n$$\n4xy \\le 3 + (x+y-z)^2.\n$$\nSince $x+y-z \\ge 2\\sqrt{xy-z} > 0$, we have that $(x+y-z)^2 \\ge (2\\sqrt{xy-z})^2$. So, it is sufficient to prove that $3 + (2\\sqrt{xy-z})^2 \\ge 4xy$. From the problem condition $xy = \\frac{1}{z}$, hence, we need to prove the following inequality\n$$\n3 + \\left( \\frac{2}{\\sqrt{z}} - z \\right)^2 \\ge \\frac{4}{z}.\n$$\nExpanding the brackets and multiplying by $z$, we obtain the inequality: $z^2 + 3 \\ge 4\\sqrt{z}$, which follows from the AM-GM inequality:\n$$\nz^2 + 1 + 1 + 1 \\ge 4\\sqrt{z^2 \\cdot 1 \\cdot 1 \\cdot 1} = 4\\sqrt{z}.\n$$\nWe can rewrite our inequality in the form\n$$\n2xy + 2yz + 2zx \\le 3 + x^2 + y^2 + z^2.\n$$\nWe will use the Schur's inequality $(u, v, w \\ge 0)$:\n$$\nu^3 + v^3 + w^3 + 3uvw \\ge u^2v + v^2u + w^2u + w^2v + w^2u + w^2v + w^2u,\n$$\nand the AM-GM inequality for two numbers. With $x = a^3$, $y = b^3$, $z = c^3$, we have\n$$\n\\begin{aligned}\nx^2 + y^2 + z^2 + 3 &= a^6 + b^6 + c^6 + 3a^2b^2c^2 \\\n&\\ge (a^4b^2 + a^2b^4) + (b^4c^2 + b^2c^4) + (c^4a^2 + c^2a^4) \\\\\n&\\ge 2a^3b^3 + 2b^3c^3 + 2c^3a^3 = 2xy + 2yz + 2xz.\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation:\n$$\n\\cos \\pi x = \\left[ \\frac{x}{2} - \\left[ \\frac{x}{2} \\right] - \\frac{1}{2} \\right].\n$$\n\nHere $[a]$ stands for the greatest integer number that does not exceed $a$.",
    "options": [],
    "answer": "x = 3/2 + 2n, where n is any integer",
    "solution": "**Answer:** $x = \\frac{3}{2} + 2n, n \\in \\mathbb{Z}$.\n\nSince $a - [a] = \\{a\\}$, where $\\{a\\}$ is the fractional part of $a$, we can rewrite our equation in the following way:\n$$\n\\cos \\pi x = \\left[ \\frac{\\{x\\}}{2} - \\frac{1}{2} \\right].\n$$\nObviously, $0 \\leq \\frac{\\{x\\}}{2} < 1$ for every real $x$. Consider two cases:\n\n1) Let $0 \\leq \\frac{\\{x\\}}{2} < \\frac{1}{2}$. Then $-\\frac{1}{2} \\leq \\frac{\\{x\\}}{2} - \\frac{1}{2} < 0$, and thus $[\\frac{\\{x\\}}{2} - \\frac{1}{2}] = -1$. So in this case we get the equation $\\cos \\pi x = -1$. The solutions of this equation are $x = 1 + 2k, k \\in \\mathbb{Z}$. But for such $x$ we have that $\\{\\frac{x}{2}\\} = \\{\\frac{1}{2} + k\\} = \\frac{1}{2}$, which contradicts our assumption. So, we obtain that there are no solutions in this case.\n\n2) Let $\\frac{1}{2} \\leq \\frac{\\{x\\}}{2} < 1$. Then $0 \\leq \\frac{\\{x\\}}{2} - \\frac{1}{2} < \\frac{1}{2}$, which implies that $[\\frac{\\{x\\}}{2} - \\frac{1}{2}] = 0$. So, in this case our equation reduces to the equation $\\cos \\pi x = 0$. The solutions for this equation are $x = \\frac{1}{2} + k$, $k \\in \\mathbb{Z}$. For such $x$ we have:\n$$\n\\begin{cases} \\frac{x}{2} \\\\ \\frac{1}{2} \\end{cases} = \\begin{cases} \\frac{1}{4} + k \\\\ \\frac{1}{2} \\end{cases} = \\begin{cases} \\frac{1}{4}, & k = 2n, \\\\ \\frac{3}{4}, & k = 2n+1. \\end{cases}\n$$\nSo, for $\\frac{1}{2} \\leq \\frac{\\{x\\}}{2} < 1$ we should take $k = 1 + 2n$, $n \\in \\mathbb{Z}$. Therefore, $x = \\frac{3}{2} + 2n$, $n \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Natural numbers $a, b, c, d$ satisfy\n$$\n0 < |ad - bc| < \\min\\{c, d\\}.\n$$\nProve that for any coprime natural numbers $x, y > 1$ the number $x^a + y^b$ is not divisible by $x^c + y^d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove this by contradiction. Denote the sum $x^c + y^d$ by $s$. Then, obviously, $x^c = -y^d \\pmod{s}$ and also if $(x^a + y^b):s$, then $x^a = -y^b \\pmod{s}$. This implies that $x^{ad} = (-1)^d y^{bd} \\pmod{s}$ and $x^{bc} = (-1)^b y^{bd} \\pmod{s}$, hence $(-1)^d x^{ad} = y^{bd} = (-1)^b x^{bc} \\pmod{s} \\Rightarrow x^{ad} = (-1)^{b-d} x^{bc} \\pmod{s}$. It is clear that $x$ and $s$ are coprime, so we can divide the last congruence by $x$ raised to the power $\\min\\{ad, bc\\}$, and obtain $x^{\\max\\{ad, bc\\}} = (-1)^{b-d} \\pmod{s}$. Similarly, we get $y^{\\max\\{ad, bc\\}} = (-1)^{a-c} \\pmod{s}$. Therefore, $y^{\\max\\{ad-bc\\}} - x^{\\max\\{ad-bc\\}}:s$ or $y^{\\max\\{ad-bc\\}} + x^{\\max\\{ad-bc\\}}:s$. By the statement of the problem, we have $0 < |ad-bc| < \\min\\{c, d\\}$. But then\n$$\n|y^{\\max\\{ad-bc\\}} - x^{\\max\\{ad-bc\\}}| < y^{\\max\\{ad-bc\\}} + x^{\\max\\{ad-bc\\}} < y^d + x^c = s.\n$$\nSo, the expression $|y^{\\max\\{ad-bc\\}} \\pm x^{\\max\\{ad-bc\\}}|$ can be divisible by $s$ only if it is equal to zero. But this contradicts the fact that $x$ and $y$ are coprime, and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45245,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a trapezoid $ABCD$ with the bases $AD$ and $BC$, a point $F$ is chosen on the side $CD$. Let $E$ be the point of intersection of the lines $AF$ and $BD$. A point $G$ is chosen on the side $AB$ so that $EG \\perp AD$. Let $H$ be the point of intersection of the lines $CG$ and $BD$, and let $I$ be the point of intersection of the lines $FH$ and $AB$. Prove that the lines $CI$, $FG$ and $AD$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, assume that $BC < AD$. Let $S$ be the point of any intersection of the lines $AB$ and $CD$, and $T$ the point of intersection of the lines $AF$ and $DG$ (fig. 34).\n\nFirst we prove that the points $S$, $H$, and $T$ are collinear. To this end, we use Menelaus' theorem for the triangle $ABE$ and three points $S$, $H$, $T$, that lie on the lines that contain its sides: the points $S$, $H$, $T$ will be collinear if and only if\n$$\n\\frac{AT}{TE} \\cdot \\frac{EH}{HB} \\cdot \\frac{BS}{SA} = 1.\n$$\nBecause $EG \\parallel AD$, $GE \\parallel BC$ and $AD \\parallel BC$, we have that $\\triangle ATD \\sim ETG$, $\\triangle GHE \\sim CHB$ and $\\triangle ASD \\sim BSC$. This implies that\n$$\n\\frac{AT}{TE} = \\frac{AD}{GE}, \\quad \\frac{EH}{BH} = \\frac{GE}{BC}, \\quad \\frac{BS}{SA} = \\frac{BC}{AD}.\n$$\nSo,\n$$\n\\frac{AT}{TE} \\cdot \\frac{EH}{HB} \\cdot \\frac{BS}{SA} = \\frac{AD}{GE} \\cdot \\frac{GE}{BC} \\cdot \\frac{BC}{AD} = 1,\n$$\nwhich proves that the points $S$, $H$, $T$ are collinear.\n\nNext, consider the triangles $AFI$ and $DGC$. Because $IF \\cap CG = H$, $FA \\cap GD = T$, $AI \\cap CD = S$, and these points are collinear, the Desargues' theorem implies that the lines $CI$, $FG$ and $AD$ are concurrent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45246,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle touches the side $BC$ of a triangle $ABC$ at the vertex $B$ and intersects the side $AC$ at the vertex $A$ and a point $E$. Another circle touches the side $BC$ at the vertex $C$ and intersects the side $AB$ at the vertex $A$ and a point $D$. $F$ is the point of intersection of the line segments $BE$ and $CD$. Prove that the triangle $BCF$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $G$ the second point of intersection of the circles and join $G$ with all the vertices of the triangle $ABC$ (fig. 42). Then we have: $\\angle GBC = \\angle BAG = \\alpha$ (the angle subtended by $BG$ of the first circle). On the other hand, $\\angle GCD = \\angle BAG = \\alpha$ (the angle subtended by $DG$ of the second circle). Similarly, $\\angle GBF = \\angle CAG = \\angle GCB = \\beta$. So, $\\angle FBC = \\angle FCB = \\alpha + \\beta$, which implies that the triangle $BCF$ is isosceles ($BF = CF$).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any collection $a_1, a_2, \\dots, a_{2011}$ of real numbers with $a_{2011} \\neq 0$ there exists a function $f: \\mathbb{R} \\to \\mathbb{R}$, such that for any real $x$ we have:\n$$\na_1 f(x) + a_2 f(f(x)) + \\dots + a_{2011} \\underbrace{f(f(f\\dots f(x)\\dots))}_{2011} = x.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will search for the function $f$ in the form $f(x) = kx$ with $k \\neq 0$. Then\n$$\n\\underbrace{f(f(f\\dots f(x)\\dots))}_{n} = k^n x.\n$$\nSo the equality from the problem statement becomes:\n$$\na_1 kx + a_2 k^2 x + \\dots + a_{2011} k^{2011} x = x.\n$$\nWe cancel $x$ and obtain the following equation\n$$\na_{2011}k^{2011} + a_{2010}k^{2010} + \\dots + a_2k^2 + a_1k - 1 = 0,\n$$\nwhich has a non-zero solution, because the left-hand side is a polynomial of an odd degree with non-zero leading and free coefficients. So, for this $k$ the function $f(x) = kx$ will solve the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive real numbers $a$, $b$, $c < 1$ satisfying\n$$\n2(a+b+c) + 4abc = 3(ab+bc+ca) + 1,\n$$\nthe following inequality holds: $a+b+c \\le \\frac{3}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We can rewrite the given equality as follows:\n$$4abc - 4(ab + bc + ca) + 4(a+b+c) - 4 = -(ab + bc + ca) + 2(a+b+c) - 3,$$\n$$4(a-1)(b-1)(c-1) = -(a-1)(b-1) - (b-1)(c-1) - (c-1)(a-1),$$\n$$4 = \\frac{1}{1-a} + \\frac{1}{1-b} + \\frac{1}{1-c}.$$\nThe function $f(x) = \\frac{1}{1-x}$ is convex on the interval $x < 1$ because $f''(x) = \\frac{2}{(1-x)^3} \\ge 0$. So, by the Jensen's inequality,\n$$\n4 = f(a) + f(b) + f(c) \\ge 3f\\left(\\frac{a+b+c}{3}\\right) = \\frac{9}{3-(a+b+c)}.\n$$\nThe last inequality implies the required one.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45249,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a number of toy kittens in a box. Each kitten has its head and tail painted with one of 2011 colors, not necessarily the same. One can take some kittens out of the box to form a collection. A collection is called \"right\", if it consists of exactly 2011 kittens so that their heads are painted with different colors, and their tails are also painted with different colors. We know that one can choose a \"right\" collection of kittens from the box in more than one way. Prove that some kittens (maybe none) can be removed from the box so that there are exactly two ways to choose a \"right\" collection of kittens from the remaining ones.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Enumerate the colors with the natural numbers from $1$ to $2011$. Construct a graph with the vertices $A_1, A_2, \\dots, A_{2011}, B_1, B_2, \\dots, B_{2011}$. For each kitten we draw an edge in the graph in the following way. If its head is painted with the color $i$, and the tail – with the color $j$, then the edge will join the vertices $A_i$ and $B_j$. This way, for each kitten there is a corresponding edge in the graph. A \"right\" collection of kittens corresponds to a subset of $2011$ edges such that no two edges are adjacent to the same vertex. We call this a \"right\" subset of edges. What we need is to remove several (or none) edges from the graph in such a way that there will be exactly two \"right\" subsets of edges in the remaining graph.\n\nBy the problem condition, there exist two \"right\" subsets in the graph. Remove all the edges that do not belong to any of these two subsets. For every \"right\" subset and every vertex of the graph, we have exactly one edge in the subset that is adjacent to this vertex. So, in the remaining graph, for each vertex we either have an adjacent edge that belongs to both \"right\" subsets, or we have two adjacent edges, one from each of the \"right\" subsets. Now, if we have exactly one edge adjacent to a vertex $V$, then the other vertex adjacent to this edge does not have any other adjacent edges. So, the graph is composed of the cycles and isolated edges that do not have common vertices with other edges. Note that every such cycle will have an even length because the edges from the first and the second \"right\" subsets have to alternate.\n\nEvery \"right\" subset has the following form: it contains all isolated edges, and a half of the edges from each cycle so that these edges do not have common vertices. There are exactly two ways to choose edges from a cycle for a \"right\" subset. The graph contains at least one cycle, because otherwise there are no two different \"right\" subsets. If there is more than one cycle, we remove half of the edges (so that the remaining edges have no common vertices) from all of the cycles except one. Clearly, the remaining graph will have exactly one cycle, and hence, exactly two \"right\" subsets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer $n$, that satisfy the following equality:\n$$\n(n-1)(n-3)(n-5)\\dots(n-2011) = n(n+2)(n+4)\\dots(n+2010).\n$$",
    "options": [],
    "answer": "no integer solutions",
    "solution": "If $n$ is even, then LHS is odd, and RHS is even, and vice versa (for odd $n$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find distinct numbers $a$, $b$, $c$, $d$ such that:\n$$\n\\frac{1}{2011} = \\frac{a}{a+1} + \\frac{b}{b+1} - \\frac{c}{c+1} - \\frac{d}{d+1}\n$$",
    "options": [],
    "answer": "One valid choice is: a = 2010, b = 2011*2012 - 1, c = 2010*2011, d = 2010*2011*(2010*2011 + 1) - 1.",
    "solution": "We use the following formulas:\n$$\n\\frac{1}{n} = \\frac{n-1}{n(n-1)} = \\frac{1}{n-1} - \\frac{1}{n(n-1)}, \\quad \\frac{1}{n} = \\frac{n+1}{n(n+1)} = \\frac{1}{n+1} + \\frac{1}{n(n+1)}\n$$\nUsing them we arrive at:\n$$\n\\frac{1}{2011} = \\frac{1}{2010} - \\frac{1}{2010 \\cdot 2011} = \\frac{1}{2011} + \\frac{1}{2011 \\cdot 2012} - \\frac{1}{2010 \\cdot 2011 + 1} \\\\\n- \\frac{1}{2010 \\cdot 2011(2010 \\cdot 2011+1)} = \\frac{1}{a+1} + \\frac{1}{b+1} - \\frac{1}{c+1} - \\frac{1}{d+1}\n$$\nWe add 1 to the first two fractions and subtract one from the last two, we get the following representation:\n$$\n\\frac{1}{2011} = \\frac{2010 \\cdot 2011 \\cdot 2012 - 1}{2011 \\cdot 2011 \\cdot 2012} + \\frac{2010 \\cdot 2011}{2010 \\cdot 2011 + 1} + \\\\\n+ \\frac{2010 \\cdot 2011(2010 \\cdot 2011 + 1) - 1}{2010 \\cdot 2011(2010 \\cdot 2011 + 1)} = \\frac{a}{a+1} + \\frac{b}{b+1} - \\frac{c}{c+1} - \\frac{d}{d+1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of natural numbers $a$ and $b$, such that the difference is $2011$ and the product is a complete square.",
    "options": [],
    "answer": "(a, b) = (1006^2, 1005^2)",
    "solution": "Since $a - b = 2011$ and $2011$ is prime, then $(a, b)$ is $1$ or $2011$.\n\n1 Case. If $(a, b) = 1$ then $a = c^2$, $b = d^2$, and we have $a - b = c^2 - d^2 = (c - d)(c + d) = 2011 = 1 \\cdot 2011$ or\n$$\n\\begin{cases}\nc - d = 1, \\\\\nc + d = 2011.\n\\end{cases}\n$$\nSolving this system we get the answer.\n\n2 Case. If $(a, b) = 2011$, then $a = 2011c$, $b = 2011d$, where $(c, d) = 1$. We have $ab = 2011^2 cd$, hence $c = e^2, d = f^2$. We have $a - b = 2011(c - d) = 2011$, or $c - d = (e - f)(e + f) = 1$, which is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45253,
    "subject": "Mathematics (Multi-modal)",
    "question": "For nonnegative real numbers $a, b, c$, prove that\n$$\n(a+1)^2 + (b+1)^2 + (c+1)^2 \\ge 3(1 + a\\sqrt{b} + b\\sqrt{c} + c\\sqrt{a}).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By AM-GM:\n$$\n\\begin{cases} a + a^2 + b \\ge 3a\\sqrt{b}, \\\\ b + b^2 + c \\ge 3b\\sqrt{c}, \\\\ c + c^2 + a \\ge 3c\\sqrt{a}. \\end{cases}\n$$\nAdding them up we get what is required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Using the numbers $1, 2, \\ldots, 20$ (each number once) as denominators and numerators, construct $10$ fractions with integer sum.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\frac{17}{3} + \\frac{13}{2} + \\frac{11}{6} + \\frac{19}{1} + \\frac{14}{7} + \\frac{18}{9} + \\frac{20}{10} + \\frac{16}{8} + \\frac{15}{5} + \\frac{12}{4} = 14 + 19 + 2 + 2 + 2 + 2 + 3 + 3 = 47.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45255,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known, that $x_1, x_2, x_3$ are distinct real numbers.\n\na) $x_2, x_3$ are zeros of the function $f_1(x) = x^2 + p_1x + q_1$; $x_3, x_1$ are zeros of the function $f_2(x) = x^2 + p_2x + q_2$; $x_1, x_2$ are zeros of the function $f_3(x) = x^2 + p_3x + q_3$. Does the function $f(x) = f_1(x) + f_2(x) + f_3(x)$ always have zeros?\n\nb) $x_2, x_3$ are zeros of the function $f_1(x) = a_1x^2 + b_1x + c_1$; $x_3, x_1$ are zeros of the function $f_2(x) = a_2x^2 + b_2x + c_2$; $x_1, x_2$ are zeros of the function $f_3(x) = a_3x^2 + b_3x + c_3$. Does the function $f(x) = f_1(x) + f_2(x) + f_3(x)$ always have zeros?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We write our function in the form\n$$\nf(x) = (x - x_2)(x - x_3) + (x - x_1)(x - x_3) + (x - x_1)(x - x_2),\n$$\nWLOG, $x_1 < x_2 < x_3$. Then $f(x_2) = (x_2 - x_1)(x_2 - x_3) < 0$, which is equivalent to the existence of the roots of the function.\n\n\nb) Consider the following three functions:\n$$\nf_1(x) = x^2 + x, \\quad f_2(x) = x^2 - x, \\quad f_3(x) = 1 - x^2,\n$$\nThey have the roots $0, -1$, $0, 1$ and $-1, 1$ respectively. Their sum\n$$\nf(x) = x^2 + x + x^2 - x + 1 - x^2 = x^2 + 1,\n$$\ndoes not have zeros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Five years ago the total age of all sons in the family was two years more than the total age of all daughters. Since that time, one more child was born and now the total age of all daughters is two years more than the total age of all sons. What was the difference between the total age of sons and daughters two years ago?",
    "options": [],
    "answer": "0 or 1",
    "solution": "Suppose that five years ago family had $n$ sons and $m$ daughters, total age of sons was $N$, total age of daughters was $M$.\nIf $k$ years ago one daughter was born, then the total age of sons now is $N+5n$, for daughters it is $M+5m+k$. Since $N=M+2$, we get that\n$$\n(N+5n)+2 = M+5m+k \\Rightarrow M+5n+4 = M+5m+k \\Rightarrow k = 5(n-m)+4\n$$\nand $0 \\le k \\le 5$, then $k=4$ and $n-m=0$. So two years ago the total age of sons was $N+3n$ and daughters was $M+3m+2$ were equal:\n$$\nN+3n-(M+3m+2)=(N-M)+3(n-m)-2=2-2=0.\n$$\nIf $k$ years ago one son was born. The total age of sons now is $N+5n+k$, daughters - $M+5m$. We have\n$$\n(N+5n+k)+2=M+5m \\Rightarrow M+5n+k+4=M+5m \\Rightarrow k=5(m-n)-4.\n$$\nSince $0 \\le k \\le 5$, then $k=1$ and $m-n=1$. Two years ago the difference was:\n$$\nM+3m-(N+3n)=(M-N)+3(m-n)=-2+3=1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given natural number $N$ we write two different nonzero digits from the right. It turns out, that new number is divisible by $N$. What is maximal value of $N$ can be?",
    "options": [],
    "answer": "98",
    "solution": "Assume that two digits form the number $\\overline{ab}$. Then we have $\\overline{Nab}:N$, or $100N+\\overline{ab}:N$, which implies $\\overline{ab}:N$. Since $\\overline{ab}$ is form by two distinct digits then $a \\le 98$, hence $N \\le 98$. The number $N=98$ satisfies all the requirements.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\tau(n)$ be the number of divisors of a natural number $n$. Prove that there exists infinitely many natural numbers $N$ such that $(\\tau(N)+\\tau(N+1)+1) \\equiv 3 \\pmod{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "So, finally, we have that every number $N = 3n^3$ with $n = 24m+1$ satisfies the problem condition, which evidently means, that there are infinitely many such numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists infinitely many squares that can be represented as $2^n + 2^m$, where $n, m$ are distinct natural numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = m + 3$, then $N = 2^n + 2^m = 2^m(2^3 + 1) = 2^m \\cdot 9$, and we can take $m = 2s$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequence $(a_n)$ is defined as follows:\n$$\na_1 = 1,\\ a_2 = 2,\\ a_{n+2} = (n+1)(a_n + a_{n+1})\n$$\nfor each natural $n$. How many zeros does $a_{2011}$ end with?",
    "options": [],
    "answer": "501",
    "solution": "We first prove that $a_n = n!$ by PMI:\n$$\na_{n+2} = (n+1)(n! + (n+1)!) = n!(n+1)(n+2) = (n+2)!\n$$\nOne can now find the number of zeros using the well-known formula:\n$$\n\\left[ \\frac{2011}{5} \\right] + \\left[ \\frac{2011}{5^2} \\right] + \\left[ \\frac{2011}{5^3} \\right] + \\dots = 402 + 80 + 16 + 3 = 501.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a quadruple with non parallel sides, that can be partitioned into 2011 isosceles triangles?",
    "options": [],
    "answer": "Yes",
    "solution": "**Answer:** See fig. 13.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45262,
    "subject": "Mathematics (Multi-modal)",
    "question": "$M$ and $N$ are chosen on the sides $AD$ and $BC$ of the square $ABCD$, such that $AM = BN$. Point $X$ is a feet of perpendicular from the point $D$ onto $AN$. Prove that angle $MXC$ is right.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the diagonals of $MNCD$, $O$ is the point of its intersection, which is a center of the circle with diameter $DN$ (fig. 17). We have the following equalities:\n$$\n\\angle NXC = \\angle NDC = \\angle MCD = \\angle MXD. \\text{ Hence,}\n$$\n$$\n\\angle MXC = \\angle MXD + \\angle DXC = \\angle CXN + \\angle DXC = \\angle DXN = 90^\\circ,\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $SH$ be an altitude in a tetrahedron $SABC$ and $H$ be inside the base $ABC$. A point $O$ on $SH$ is chosen so that $\\angle AOS + \\alpha = \\angle BOS + \\beta = \\angle COS + \\gamma = 180^\\circ$, where $\\alpha, \\beta, \\gamma$ are the dihedral angles corresponding to the edges $BC, AC, AB$ respectively. Let $A_1, B_1, C_1$ be the points of intersection of the following lines and planes: $A_1 = AO \\cap SBC$, $B_1 = BO \\cap SAC$, $C_1 = CO \\cap SBA$. If the planes $ABC$ and $A_1B_1C_1$ are parallel to each other, prove that $SA = SB = SC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_2$ be a projection of the point $H$ onto the edge $BC$. Then $\\angle AOS = 180^\\circ - \\angle HA_2S = 180^\\circ - \\alpha$, and so $\\angle AOH = \\alpha$. This implies that $\\triangle AOH \\sim \\triangle SA_2H \\Rightarrow \\frac{AH}{SH} = \\frac{OH}{HA_2} \\Rightarrow AH \\cdot HA_2 = OH \\cdot SH$. (fig. 43).\n\nAnalogously, $BH \\cdot HB_2 = CH \\cdot HC_2 = OH \\cdot SH$. We will next prove that $H$ is the orthocenter of $\\triangle ABC$. Since $\\frac{AH}{HC_2} = \\frac{CH}{HA_2}$ we have that $\\sin \\angle HAC_2 = \\sin \\angle HCA_2$. In view of the fact that $\\angle HAC_2 + \\angle HCA_2 < 180^\\circ$ this implies that $\\angle HAB = \\angle HCB$, from which $\\angle AHC_2 = \\angle CHA_2$. Similarly we obtain that $\\angle AHB_2 = \\angle BHA_2$ and $\\angle BHC_2 = \\angle CHB_2$. From this it easily follows that $\\angle HAB = \\angle HCB = 90^\\circ - \\angle B$, which implies that $H$ belongs to the altitude of $\\triangle ABC$ starting at $A$. Analogously, $H$ belongs to the other two altitudes of $\\triangle ABC$. So, we see that the projection $H$ of the point $S$ onto the base of the tetrahedron is the orthocenter of the base, and the line segments $AA_2, BB_2$ and $CC_2$ are the altitudes of the base (fig. 44).\n\nConsider the $\\triangle ASA_2$ (see fig. 45). Since $\\triangle AOH \\sim \\triangle SA_2H$, we have that $AA_1 \\perp SA_2$.\n\n![](attached_image_1.png)\nFig. 44\n\nSo the lines $SA_2$ and $AA_2$ are both perpendicular to the line $BC$, which means that the whole plane $ASA_2$ is perpendicular to $BC$. This implies that $AO \\perp BC$. Similarly, in the plane $SBC$ we have two non-parallel lines $BC$ and $SA_2$ that are perpendicular to $AO$, and so $AO \\perp SBC$. Analogously, $BO \\perp SAC, CO \\perp SBA$.\n\nNow, since $ABC \\parallel A_1B_1C_1$, we have that $AB \\parallel A_1B_1$, and so the triangles $AOB$ and $A_1OB_1$ are similar. It follows that $\\frac{OA_1}{AO} = \\frac{OB_1}{BO} = t$. Also we have that $OA_1 = SO \\sin \\angle OSA_1$ and $OH = AO \\sin \\angle OAH = AO \\sin \\angle OSA_1$. This implies that $OA_1 \\cdot OA = SO \\cdot OH$. Analogously we can obtain that $SO \\cdot OH = OA \\cdot OA_1 = OB \\cdot OB_1 = OC \\cdot OC_1 \\Rightarrow SO \\cdot OH = t \\cdot OA^2 = t \\cdot OB^2 = t \\cdot OC^2$.\n\nIt follows that $OA = OB = OC \\Rightarrow AH = BH = CH$ and finally that $SA = SB = SC$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $a_1, a_2, a_3, a_4, a_5$ and $b_1, b_2, b_3, b_4, b_5$ are permutations of $1, 2, 3, 4, 5$. Prove that among five numbers $a_1b_1, a_2b_2, a_3b_3, a_4b_4, a_5b_5$ at least two have the same remainders modulo $5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "WLOG, $a_5 = 5$. If $b_5 \\neq 5$, then there will be two numbers that are divisible by $5$. Let $b_5 \\neq 5$ and suppose that the statement of the problem is false. Then the numbers $a_1b_1, a_2b_2, a_3b_3, a_4b_4$ have remainders $1, 2, 3$ and $4$ modulo $5$ (in some order). Then from one hand the product has remainder $4$.\n\nFrom the other, since the numbers $a_1, a_2, a_3, a_4$ and $b_1, b_2, b_3, b_4$ are permutations of $1, 2, 3$ and $4$, the product is $a_1b_1 \\cdot a_2b_2 \\cdot a_3b_3 \\cdot a_4b_4 = 24 \\cdot 24 = 576$ has remainder $1$ modulo $5$. This contradiction finishes the proof.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 45265,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are candies on the table. At each step Petro can take away some of them. At the first step he takes away one candy and on each next step he can take away either the same amount or twice the amount that he has taken at the previous step. What is the minimal number of steps does Petro need in order to take from the table away exactly $2011$ candies?",
    "options": [],
    "answer": "17",
    "solution": "**Answer:** $17$.\n\nNote, that on each step the number of candies that Petro takes from the table is a power of $2$. Moreover, if at some step Petro takes away $2^n$ candies, then there were the moves where he was taking away $2, 4, \\ldots, 2^{n-1}$. Thus, the maximum number that Petro could take can not exceed $512$, since $1+2+4+\\ldots+1024 = 2047 > 2011$.\n\nWe assume that Petro keeps optimal strategy. Note, that the number of steps when Petro takes away exactly $2^n$ candies can not exceed $2$, otherwise, he could reduce the number of moves by taking $2^{n+1}$ candies instead. This implies that he could not take less than $512$ candies each time because $2 \\cdot (1+2+4+\\ldots+256) = 1022 < 2011$.\n\nTherefore Petro was taking $1, 2, \\ldots, 512$ candies. We write this in the following form: $a_0 + 2a_1 + 4a_2 + \\ldots + 512a_9 = 2011$, where $a_i \\in \\{1, 2\\}$ is the number of times he was taking $2^i$ candies. This is equivalent to: $a'_0 + 2a'_1 + 4a'_2 + \\ldots + 512a'_9 = 2011 - (1+2+4+\\ldots+512) = 988$, where $a'_i = a_i - 1, a'_i \\in \\{0, 1\\}$.\n\nSince binary representation is unique we have $a'_i, i = 0, \\ldots, 9: 0, 0, 1, 1, 1, 0, 1, 1, 1, 1$, since\n$$\n988 = 512 + 256 + 128 + 64 + 16 + 8 + 4.\n$$\nHence, $a_i, i = 0, \\ldots, 9$ have values $1, 1, 2, 2, 2, 1, 2, 2, 2, 2$ and optimal number of steps is $3 \\cdot 1 + 7 \\cdot 2 = 17$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45266,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2012$ piles of stones. The first pile contains $2^0$ stones, the second pile contains $2^1$ stones, the third pile contains $2^2$ stones, and so on. The $2012$-th pile contains $2^{2011}$ stones. At one step, one can pick three piles and add $2$ stones to the first pile, $3$ stones to the second pile, and $4$ stones to the third pile, or pick any three piles and take $2$ stones out of the first pile, $3$ stones out of the second pile, and $4$ stones out of the third pile, provided that each pile has enough stones. Is it possible after a finite number of such operations to get exactly $3^{1005}$ stones in each pile?",
    "options": [],
    "answer": "No",
    "solution": "After each operation the total number of stones changes by a number which is divisible by $9$. At the end, the total number is $2012 \\cdot 3^{1005}$, which is divisible by $9$. However, at the starting moment the total number is $2^0 + 2^1 + 2^2 + \\ldots + 2^{2011} = 2^{2012} - 1$, which is not divisible by $9$. Indeed, $2012 = 335 \\cdot 6 + 2$, $2^6 = 64 = 1 \\pmod{9}$, therefore\n$$\n2^{2012} - 1 = (2^6)^{335} \\cdot 2^2 - 1 = 4 - 1 = 3 \\pmod{9}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45267,
    "subject": "Mathematics (Multi-modal)",
    "question": "The trapezoid $ABCD$ with parallel sides $BC = a$ and $AD = 2a$ is drawn on the plane. Using only ruler construct triangle with the area equals to the area of trapezoid.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first construct the point $O$ of the intersection of diagonals, and the point $T$ of intersection of sides $AB$ and $CD$. It is well known that $OT$ contains midpoints $P$ and $E$ of parallel sides (fig. 11). We have $AE = ED = BC = a$. Hence $ABCE$ and $BCDE$ are parallelograms, $M$ and $N$ are the points of intersection of their diagonals respectively. We can draw the line $MN$ which is a midline of trapezoid and intersects two sides at $K$ and $L$ respectively. We then draw line $BL$, that intersects $AD$ at point $F$. Thus $\\square ABF$ is desired, since $\\square ABF = \\square BCL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45268,
    "subject": "Mathematics (Multi-modal)",
    "question": "In volleyball tournament there are 8 teams, that play one-round tournament (each team plays exactly one game with another). Each win worth 1 point, each lose worth 0 points, there are no draws in volleyball. After tournament is finished, if the difference between the first and the second place, does not exceed 1 point, then they play one extra game. The same applies for the teams that scored 3-d and 4-th, 5-th and 6-th, and 7-th and 8-th respectively. What is the least number of extra games can occur?\n\n*Note.* After tournament is finished each place takes only one team, even though two teams can have the same number of points.",
    "options": [],
    "answer": "1",
    "solution": "If we assume, that there were no extra games then the difference between 1-st and 2-nd, 3-d and 4-th, 5-th and 6-th, and 7-th and 8-th is at least 2 points and therefore the difference between 1-st and 8-th places is at least 8 points, while the first place can not have more than 7 points. The following example shows that 1 extra game is indeed possible:\n\n| Team      | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Points |\n|-----------|---|---|---|---|---|---|---|---|--------|\n| 1 place   | XX| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 7      |\n| 2 place   | 0 | XX| 0 | 1 | 1 | 1 | 1 | 1 | 5      |\n| 3 place   | 0 | 1 | XX| 0 | 0 | 1 | 1 | 1 | 4      |\n| 4 place   | 0 | 0 | 1 | XX| 1 | 0 | 1 | 1 | 4      |\n| 5 place   | 0 | 0 | 1 | 0 | XX| 1 | 1 | 1 | 4      |\n| 6 place   | 0 | 0 | 0 | 1 | 0 | XX| 0 | 1 | 2      |\n| 7 place   | 0 | 0 | 0 | 0 | 0 | 1 | XX| 1 | 2      |\n| 8 place   | 0 | 0 | 0 | 0 | 0 | 0 | 0 | XX| 0      |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45269,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Rectangle $ABCD$ is partitioned into squares, each of which has integer perimeter. Is it true that $ABCD$ has integer perimeter?\n\nb) Square $ABCD$ is partitioned into squares, each of which has integer perimeter. Is it true that $ABCD$ has integer perimeter?",
    "options": [],
    "answer": "a) No. b) Yes.",
    "solution": "a)\nWe construct a counterexample. Consider two squares $ABMN$ and $NMCD$ with side length $\\frac{1}{4}$. Then, perimeter of the rectangle $ABCD$ is $2 \\cdot (\\frac{1}{2} + \\frac{1}{4}) = \\frac{3}{2}$ — non-integer, while both squares have integer perimeter.\n\nb)\nConsider the side of the external square, and all squares that have one side which belongs to the side of the external square (see figure 1).\n\n![](attached_image_1.png)\n\nLet $a$ be the side of the external square, and the sides of small squares are $a_1, a_2, ..., a_n$. Then $a = a_1 + a_2 + ... + a_n$, and each side $a_i$, after multiplying by $4$, is integer. Therefore, perimeter $P$ of our square $ABCD$ is $P = 4a = 4a_1 + 4a_2 + ... + 4a_n$ — integer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45270,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number $n = 11^{2011} \\cdot 2011^{11}$ is given. How many divisors that are less than $n$ and are not divisors of $n$, does $n^2$ have?",
    "options": [],
    "answer": "22121",
    "solution": "The number of divisors of $n^2 = 11^{4022} \\cdot 2011^{22}$ is:\n$$\nN = (4022 + 1)(22 + 1) = 92529.\n$$\nWe can pair the divisors of $n^2$ in a way $d$ and $\\frac{n^2}{d}$ where one divisor is greater than $n$, another one is less. The number $n$ does have a pair, therefore the number of divisors that are less than $n$ is $\\frac{92529-1}{2} = 46264$.\n\nThe number $N_1$ of divisors of $n$ that are less than $n$ is:\n$$\nN_1 = (2011+1)(11+1)-1 = 24143.\n$$\nSo the number we are looking for is $46264 - 24143 = 22121$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45271,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$ consider points $X$ and $Y$ such that $ABXY$ is a parallelogram. Prove that the circumradii of $BXD$ and $CYA$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ABCD$ is cyclic, then $\\angle ABD = \\angle ACD$. Moreover, $\\angle ABD = \\angle ABY = \\angle ABY$. Hence, $CXYD$ is cyclic because $\\angle XCD = \\angle XYB$ (fig. 12).\n\n![](attached_image_1.png)\n\nThis implies that $\\angle XCY = \\angle XDY$, from what it follows that $\\sin \\angle ACY = \\sin \\angle BDX$. Moreover, $BX = AY$, and applying sine Law for triangles $BXD$ and $CAY$ we get:\n$$\nR_{BDX} = \\frac{BX}{2\\sin \\angle BDX} = \\frac{AY}{2\\sin \\angle ACY} = R_{CYA}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$, $q$, $r$, such that:\n$$\np(p+1) + q(q+1) = r(r+1).\n$$",
    "options": [],
    "answer": "p = 2, q = 2, r = 3",
    "solution": "WLOG, $p \\le q$, clearly $r > p$. We can write:\n$$\np(p+1) = (r-q)(r+q+1). \\quad (1)\n$$\nSince $p$ is prime, then either $r-q$ or $r+q+1$ is divisible by $p$. If $p$ divides $r-q$, then $r-q \\ge p$,\n$$\np(p+1) \\le (r-q)(r-q+1) < (r-q)(r+q+1),\n$$\nwhich contradicts to (1).\n\nLet suppose that $r+q+1$ is divisible by $p$. If $p=2$, then $r+q+1$ has to be even, which implies that $r$ is odd, hence $q$ is even, thus $q=2$ and $r=3$.\n\nIf $p > 2$, let $r+q+1 = kp$. Then $r$ and $q$ are odd and $k$ is odd, $k > 1$. Then $p+1 = k(r-q)$ and we get $k^2(r-q) = kp+k$ or $r+q+1+k = k^2r-k^2q$ which is equivalent to\n$$\n(k^2+1)q = (k^2-1)r - (k+1).\n$$\nRHS is divisible by $(k+1)$, thus LHS is divisible by the same number, since $k$ is odd, then $\\frac{k^2+1}{2}$ is divisible by $\\frac{k+1}{2}$. Observe, that\n$$\n\\left(\\frac{k^2+1}{2}, \\frac{k+1}{2}\\right) = \\left(\\frac{k^2+1}{2} - \\frac{(k-1)(k+1)}{2}, \\frac{k+1}{2}\\right) = \\left(1, \\frac{k+1}{2}\\right) = 1.\n$$\n$\\frac{k^2+1}{2}$ and $\\frac{k+1}{2}$ are coprime, so $\\frac{k+1}{2}$ is divisible by $q$ and since $k > 1$, then $\\frac{k+1}{2} > 1$ and $q = \\frac{k+1}{2}$.\n\nWe get $r = kp - q - 1$ and $q = \\frac{k+1}{2}$. Plugging in this into $p+1 = k(r-q)$ we get $p+1 = (kp - k - 2)k$. But the last equality is impossible because $k \\ge 3$, $p \\ge 3$ implies\n$$\n(kp - k - 2)k > kp - k - 2 = k(p-1) - 2 \\ge 3(p-1) - 2 = p + (2p-5) \\ge p+1.\n$$\nThis completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three circles touch externally, $A_1A_2$, $B_1B_2$, $C_1C_2$ are their diameters that have the same directions. Prove that $A_1B_2$, $B_1C_2$, $C_1A_2$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let circles with diameters $B_1B_2$, $C_1C_2$ touch at point $A$. By analogy we define $B$, $C$ (fig. 22).\n\nSince touching point is a center of homothety that transforms one circle into another, then $A = B_1C_2 \\cap B_2C_1$ and so on. Let $D = A_2C_1 \\cap A_1B_2$. Points $A$, $B$, $C$, $D$ are cyclic, because\n$$\n\\angle CDB = \\angle B_2DC_1 = \\pi - \\angle DB_2C_1 - \\angle DC_1B_2 = \\pi - \\angle CAE - \\angle BAE = \\pi - \\angle CAB,\n$$\n\n$$\n\\angle CFA = \\angle FA_1C_2 + \\angle FC_2A_1 = \\angle CA_1B + \\angle FC_2B = \\angle CA_1B + \\angle BC_1A = \\\\\n= \\angle CBE_1 + \\angle E_1BA = \\angle CBA.\n$$\n\nBut except point $C$, the line $A_1B_2$ intersects the circle at point $D$, hence $A_1B_2$, $B_1C_2$, $C_1A_2$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles touch each other externally at point $C$. Consider two diameters $A_1A_2$, $B_1B_2$ of the same direction. Circle with the center on the common internal tangent passes through the point of intersection of $A_1B_2$, $A_2B_1$, and meets these lines at points $M$, $N$. Prove that $MN$ is perpendicular to $A_1A_2$, $B_1B_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since point $C$ is a center of homothety that transforms one circle into another, then $C = A_1B_2 \\cap A_2B_1$, $A_1B_2 \\perp A_2B_1$ (fig. 21).\n\nLet $D = MN \\cap B_1B_2$. Then, $\\angle DB_2C = \\angle B_2A_1A_2 = \\angle A_2CO = \\angle CND$. Therefore $DB_2NC$ is cyclic and $\\angle B_2DN = \\angle B_2CA_2 = \\frac{\\pi}{2}$, which implies that $MN \\perp B_1B_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadruple. Let us denote the midpoints of $AB$, $BC$, $CD$ and $DA$ by $M$, $L$, $N$ and $K$ respectively. It is known, that $\\angle BMN = \\angle MNC$. Prove that:\n\na) $\\angle DKL = \\angle CLK$;\n\nb) $ABCD$ has a pair of parallel sides.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Using the properties of inscribed angles we get $KM \\perp BD$, $KN \\perp AC$, $\\angle ABD = \\angle ACD \\Rightarrow \\angle AMK = \\angle ABD = \\angle ACD = \\angle KND$. Thus $\\angle KMN = \\pi - \\angle AMK - \\angle BMN = \\pi - \\angle KND - \\angle MNC = \\angle KNM$, hence $\\square KMN$ is isosceles and $KM = KN$, which implies that $KMLN$ is rhombus and $\\angle NKL = \\angle NLK$ (fig. 8). Since by analogy we have that $\\angle AMK = \\angle KND$, $\\angle DKN = \\angle NLC$, then $\\angle KLD = \\angle DKN + \\angle NKL = \\angle NLC + \\angle LNK = \\angle CLK$, which proves part a).\n\nb) Since $KM = KN$, then $DB = 2KM = 2KN = AC$ and $\\angle ABC + \\angle DAB = \\pi$. We have $DA \\perp BC$, or $\\angle ABC = \\angle DAB$. But since $\\angle DAC = \\angle DBC$, then $\\angle CAB = \\angle DAB - \\angle DAC = \\angle ABC - \\angle DBC = \\angle ABD = \\angle ACD$, and $AB \\perp BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45276,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ $AL$, $BM$, $CN$ are medians. Prove, that $\\angle ANC = \\angle ALB$ if and only if $\\angle ABM = \\angle LAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lines $LN$ and $AC$ are parallel, hence $\\angle NLA = \\angle LAC$. We have to show the following implication (fig. 7):\n$$\n\\angle ANC = \\angle ALB \\Leftrightarrow \\angle ABM = \\angle NLA.\n$$\nLet $G$ be a centroid, then we have the following equivalences. $\\angle ANC = \\angle ALB \\Leftrightarrow \\angle BNC + \\angle ALB = \\pi \\Leftrightarrow BNLG$ - cyclic $\\Leftrightarrow \\angle ABM = \\angle NLA$, since they share the common segment. The statement is proved\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a polynomial with integer coefficients. Given that for some integer $a$ there exists $n$ such that $\\underbrace{P(P(...P(a)...))}_{n} = a$, prove that $P(P(a)) = a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define a sequence $(a_i)$ in the following way: $a_0 = a$, $a_i = P(a_{i-1})$ for all natural $i$. Consider minimal number $m$, for which there exists $j < m$, such that $a_j = a_m$. From the condition of the problem it follows that this number does exist. Observe, that $a_0, a_1, ..., a_{m-1}$ are distinct.\nWe show that $j=0$. If this is not the case, then our sequence has the following form:\n$$\na_0, a_1, ..., a_{j-1}, a_j, a_{j+1}, ..., a_{m-1}, a_m = a_j, a_{j+1}, ..., a_{m-1}, a_j, a_{j+1}, ..., a_{m-1}, a_j, a_{j+1}, ...\n$$\nand $a_0$ is not going to show up in this sequence, which contradicts to our assumption. Thus, $a_0 = a_m$, and our sequence is $m$-periodic, in other words $a_p = a_q$ if $p-q:m$. Let $a_l$ be a maximum, $a_k$ be a minimum among $a_0, a_1, ..., a_{m-1}$. If $a_l = a_k$, then $P(a_l) = a$ and $P(P(a_l)) = a$. Suppose that $a_l > a_k$. Note, that $P(a_l) - P(a_k)$ is divisible by $a_l - a_k$ and $P(a_l), P(a_k)$ are members of our sequence, thus $|P(a_l) - P(a_k)| \\le |a_l - a_k|$, and the difference $P(a_l) - P(a_k)$ is 0, or $\\pm(a_l - a_k)$.\n\nIf $P(a_l) - P(a_k) = a_l - a_k$, then $P(a_k) - a_l$ and $m=1$ with $a_l = a_k$, and the result follows from previous observation.\nIf $P(a_l) - P(a_k) = 0$, namely $a_{l+1} = P(a_l) = P(a_k) = a_{k+1}$, and we must have that $(l+1)-(k+1)=l-k$ is divisible by $m$, which implies that $l=k$ and $m=1$.\nIf $P(a_l) - P(a_k) = -(a_l - a_k)$, then $P(a_l) = a_k$, $P(a_k) = a_l$ and $m=2$, which implies that $P(P(a)) = a$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inscribed circle $\\omega$ of a triangle $ABC$ touches its sides $AB$, $BC$, $CA$ at the points $K$, $L$, $M$ respectively. On the arc $KL$ of the circle $\\omega$ that does not contain the point $M$ a point $S$ is chosen. Let $P$, $Q$, $R$, $T$ be the points of intersection of the lines $AS$ and $KM$, $ML$ and $SC$, $LP$ and $KQ$, $AQ$ and $PC$ respectively. If the points $R$, $S$ and $M$ are collinear, prove that $T$ also belongs to the line $SM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the triangle $SLM$. $LC$ and $MC$ are the tangent lines to the circumscribed circle of this triangle drawn at the points $L$ and $M$, therefore $SC$ is a simedian, and so $\\frac{MQ}{CL} = \\frac{MS^2}{SL^2}$.\n\nSimilarly, $\\frac{KP}{PM} = \\frac{KS^2}{SM^2}$. Let $E$ be the point of intersection of the lines $KL$ and $MR$. Since the lines $ME$, $KQ$ and $LP$ are concurrent, by the Ceva's theorem we have that $\\frac{KP}{PM} \\cdot \\frac{MQ}{QL} \\cdot \\frac{LE}{EK} = 1$, hence applying the above equalities we obtain that $\\frac{LE}{EK} = \\frac{SL^2}{KS^2}$. This implies that the line $SE$ is a simedian of the triangle $SKL$, therefore it passes through the point of intersection of the tangents to the circumscribed circle of the triangle $KSL$ that are drawn at the points $L$ and $K$, that is through the point $B$ (Fig. 49). So, we have proved that the points $B$, $S$, $R$ and $M$ are collinear.\n\nNext we will prove that the point $T$ also belongs to this line. By the Ceva's theorem, it is sufficient to prove that $\\frac{AP}{\\sin \\angle AMP} = \\frac{AM}{\\sin \\angle APM}$, $\\frac{PS}{\\sin \\angle KMB} = \\frac{MS}{\\sin \\angle SPM}$. Dividing the last two inequalities we obtain that $\\frac{AP}{PS} = \\frac{AM \\sin \\angle AMK}{MS \\sin \\angle KMB}$. Similarly, $\\frac{CQ}{QS} = \\frac{MC \\sin \\angle CML}{MS \\sin \\angle LMB}$. Substitute the last two relations into the equality that we need to prove:\n$$\n\\frac{AP}{PS} \\cdot \\frac{SQ}{QC} \\cdot \\frac{CM}{MA} = \\frac{AP}{PS} = \\frac{AM \\sin \\angle AMK}{MS \\sin \\angle KMB} \\cdot \\frac{MS \\sin \\angle LMB}{MC \\sin \\angle CML} \\cdot \\frac{CM}{MA} = \\frac{\\sin \\angle AMK \\cdot \\sin \\angle LMB}{\\sin \\angle KMB \\cdot \\sin \\angle CML}\n$$\nBy the sine theorem for the triangles $BLM$ and $BMK$, we obtain:\n$$\n\\frac{BL}{BM} = \\frac{\\sin \\angle LMB}{\\sin \\angle BLM} = \\frac{\\sin \\angle LMB}{\\sin \\angle MLC} = \\frac{\\sin \\angle LMB}{\\sin \\angle CML}, \\text{ since the triangle MLC is isosceles. Similarly, } \\frac{BK}{BM} = \\frac{\\sin \\angle KMB}{\\sin \\angle AMK}\n$$\nTherefore,\n$$\n\\frac{AP}{PS} \\cdot \\frac{SQ}{QC} \\cdot \\frac{CM}{MA} = \\frac{\\sin \\angle AMK \\cdot \\sin \\angle LMB}{\\sin \\angle KMB \\cdot \\sin \\angle CML} = \\frac{BM}{BK} \\cdot \\frac{BL}{BM} = 1,\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that for two natural numbers $m, n$ the following equality holds\n$$\nm+n = [m,n] + (m,n),\n$$\nWhere $[m,n]$ and $(m,n)$ are the least common multiple and the greatest common divisor of $m,n$ respectively. Prove that one number is divisible by another.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d = (m,n)$, then $m = ad$, $n = bd$, and using the formula $mn = [m,n] \\cdot (m,n)$ we get, that $[m,n] = abd$. This implies that $abd + d = ad + bd \\Leftrightarrow d(a-1)(b-1) = 0$, which is possible if either $a=1$ or $b=1$ and the result follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of prime numbers $(a, b)$, such that $a^b = b^a + 1$ is prime.",
    "options": [],
    "answer": "(2,3), (3,2), (2,2)",
    "solution": "**Answer:** $(2,3)$, $(3,2)$, $(2,2)$.\n\nClearly, either $a$ or $b$ is even. WLOG, $a = 2$. It is easy to see that $b = 2$ and $b = 3$ satisfy the condition. Suppose that $b > 3$. We have: $2^b = b^2 + 1$, $b > 3$. It is easy to see that the last expression is divisible by $3$ and cannot be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45281,
    "subject": "Mathematics (Multi-modal)",
    "question": "For nonnegative real numbers $a, b, c$, with the sum that does not exceed $2$, prove\n$$\nab(a^2 + b^2) + bc(b^2 + c^2) + ca(c^2 + a^2) \\leq 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is enough to prove the inequality for $a + b + c = 2$.\nLet us denote $x = ab + bc + ac$, then\n$$\na^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ac) = 4 - 2(ab + bc + ac) = 2(2 - x).\n$$\nFrom the obvious inequality $x(2 - x) \\leq 1$, we get\n$$\n\\begin{aligned}\n2 + 2abc &\\geq 2 \\geq 2x(2 - x) = (ab + bc + ac)(a^2 + b^2 + c^2) = \\\\\n&= ab(a^2 + b^2) + cb(c^2 + b^2) + ac(a^2 + c^2) + abc(a + b + c),\n\\end{aligned}\n$$\nUsing the equality $2abc = abc(a + b + c)$, we arrive at the conclusion.\n\nEquality occurs, for example, $a = b = 1$, $c = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45282,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many integer solutions does the equation have\n$$\nx^2 + y^2 + z^2 - xy - yz - zx = x^3 + y^3 + z^3 + s,\n$$\nif\n$$\na) s = 0;\n$$\n$$\nb) s = 1?\n$$",
    "options": [],
    "answer": "a) Infinitely many integer solutions (e.g., z = 3t^2 + 1, x = t(3t^2 + 1), y = -t(3t^2 + 1) for any integer t). b) No integer solutions.",
    "solution": "a) Take $x = -y$, we get\n$$\n3x^2 + z^2 = z^3.\n$$\nTake $x = tz$, we get: $3z^2t^2 + z^2 = z^3$, or $3t^2 + 1 = z$, therefore\n$$\nz = 3t^2 + 1, x = t(3t^2 + 1), y = -t(3t^2 + 1)\n$$\n\nb) Note that $t^3 \\equiv t \\pmod{3}$. Suppose that there exists solution and denote and consider equation modulo $3$. We have\n$$\n\\begin{aligned}\na+1 &\\equiv x+y+z+1 \\equiv x^3+y^3+z^3+1 \\equiv x^2+y^2+z^2-xy-yz-zx \\equiv \\\\\n&\\equiv x^2+y^2+z^2+2(xy+yz+zx) \\equiv (x+y+z)^2 \\equiv a^2,\n\\end{aligned}\n$$\nso $a^2 - a - 1 = 3$ which is impossible.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $N$, that have two digits and are equal to the sum of their digits added to the cube of this sum.",
    "options": [],
    "answer": "30",
    "solution": "Let $n$ be the sum of the digits, then: $N = n + n^3$. The cube has to be less than $99$, that gives us four options: $1$, $2$, $3$, $4$. An easy check shows that the only possible answer is $30$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an increasing sequence of integers $0 = a_0 < a_1 < a_2 < \\dots$, such that the following two conditions are satisfied:\n\n1) every natural number can be written in the form $a_i + a_j$ for some (maybe equal) indices $i, j \\ge 0$;\n\n2) $a_n > \\frac{n^2}{16}$ for all natural $n$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $(a_n)$ be the sequence of all natural numbers $k$ that in binary representation have $1$s only on even places or only on odd places. For instance, this sequence contains the numbers that have the following binary representations: $10000$, $10100$, $101$, $1000$. Evidently, the first condition is satisfied for this sequence. We will show that the estimate from the second condition is also valid.\n\nConsider all non-negative numbers that are less than $2^{2r}$, that is the numbers that have no more than $2r$ digits in their binary representations. We count the number of elements of our sequence among these numbers: there are $2^r$ elements that have $0$s on all even places and $2^r$ elements that have zeros on all odd places, and zero is the only number that was counted twice. Hence, there are $2^{r+1}-1$ elements of our sequence that are less than $2^{2r}$ and so $a_{2^{r+1}-1} = 2^{2r}$.\n\nFor any natural number $n$ we can find an integer $r$, such that: $2^{r+1}-1 \\le n < 2^{r+2}-1$. Then we have: $2^r > \\frac{n}{4} \\Rightarrow a_n \\ge a_{2^{r+1}-1} = 2^{2r} > \\frac{n^2}{16}$, and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45285,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 10 piles of stones, with $3$, $4$, $5$, $\\ldots$, $12$ stones respectively. At one step one can pick three piles and add $1$ stone to the first pile, $2$ stones to the second pile, $3$ stones to the third pile, or pick any three piles and take $1$ stone out of first pile, $2$ stones out of second pile, $3$ stones out of third pile, provided that each pile has enough stones. Is it possible after finite number of such operations to get exactly $2011$ stones in each pile?",
    "options": [],
    "answer": "No",
    "solution": "After each operation the total number of stones changes by a number which is divisible by $3$. At the end the total number is $2011 \\cdot 10$ which is not divisible by $3$. However, at the starting moment the total number $75$ is divisible by $3$, which provides a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$, such that:\n\n1) For all real $x, y$ the following equality holds\n$$\nf(2x) = f(x+y)f(y-x) + f(x-y)f(-x-y).\n$$\n\n2) $f(x) \\ge 0$ for all $x$.",
    "options": [],
    "answer": "f(x) = 0 for all x, or f(x) = 1/2 for all x",
    "solution": "Take $x = y$, we get:\n$$\nf(2x) = f(0)f(2x) + f(0)f(-2x). \\quad (1)\n$$\nTake in (1) $x = 0$ and we get $f(0) = 2f^2(0)$. So $f(0) = 0$ or $f(0) = \\frac{1}{2}$.\nIf $f(0) = 0$, (1) implies that $f(2x) = 0$, hence $f = 0$.\n\nIf $f(0) = \\frac{1}{2}$, (1) implies that $f(2x) = \\frac{1}{2}f(2x) + \\frac{1}{2}f(-2x)$ and $f(2x) = f(-2x)$.\nTaking $x = 0$ and using the fact that our function is even, we arrive at\n$$\n\\frac{1}{2} = f(0) = (f(y))^2 + (f(-y))^2 = 2(f(y))^2.\n$$\nThis gives us that $f = \\frac{1}{2}$.\n\nIt is easy to check that both functions satisfy all the requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given natural number $k > 1$. Find all integer pairs $(x, y)$, that satisfy the following equation:\n\n$$\ny^k = x^2 + x.\n$$",
    "options": [],
    "answer": "All solutions are (x, y) = (0, 0) and (−1, 0) for any integer exponent greater than one.",
    "solution": "We have $y^k = x(x+1)$. If $x = 0$ and $x = -1$ then $y = 0$.\n\nSuppose that $x \\neq 0, -1$.\n\n$x$ and $x+1$ are coprime, therefore $x = a^k$ and $x+1 = b^k$ for some integer nonzero $a, b$, that are of the same sign. Then\n$$\n(x+1) - x = 1 = b^k - a^k = (b - a)(b^{k-1} + b^{k-2}a + \\dots + a^{k-1})\n$$\nWhich is impossible, since the absolute value of the second bracket is greater than $1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle **ABC** is inscribed into the circle. Tangents at points **A** and **B** meet at point **T**. The line, that passes through the point **T** and is parallel to **AC**, intersects **BC** at point **D**. Prove that $AD = CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since triangle $ATB$ is isosceles, $\\angle TBA$ and $\\angle BCA$ share common arc, then:\n\n$$\n\\angle TBA = \\angle TAB = \\angle BCA = \\angle BDT.\n$$\n\nThis implies, that $BTAD$ is cyclic (fig. 9). Hence $\\angle BTA = \\angle ADC$ and triangles $ABT$ and $ACD$ are similar. Triangle $ATB$ is isosceles, thus so is $\\square ACD$. This implies that $AD = CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside an inscribed quadrilateral $ABCD$ a point $P$ is chosen so that $\\angle PBC = \\angle PDA$, $\\angle PCB = \\angle PAD$. Prove that there exists a circle that touches the lines $AB$, $CD$ and also touches the circumscribed circles of the triangles $ABP$, $CDP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$, $F$ be the points of intersection of the lines $AB$, $CD$ and $AD$, $BC$ respectively, and $O$ be the center of the circumscribed circle of the quadrilateral $ABCD$ (Fig. 50). On the half-line $OE$ we choose the point $P'$, such that $OE \\cdot OP' = OA^2$. Let $\\angle AOB = 2\\alpha$, $\\angle BOC = 2\\beta$, $\\angle COD = 2\\gamma$, $\\angle OEA = x$. Then the triangles $OP'B$, $OEB$ are similar by the angle and two adjacent sides $\\Rightarrow \\angle OEB = \\angle OBP'$. The triangles $OAE$, $OAP'$ are also similar $\\Rightarrow \\angle OEA = \\angle OAP'$. Hence the points $O$, $A$, $B$, $P'$ lie on the same circle. Analogously, the points $D$, $C$, $P'$ lie on the same circle since $\\angle OED = \\angle ODP' = \\angle OCP'$. Therefore, we have:\n$$\n\\angle P'BC = \\angle OBC - \\angle OBP' = 90^\\circ - \\beta - x; \\quad \\angle OED = \\angle AED - \\angle AEO = 180^\\circ - 2\\beta - \\gamma - \\alpha - x;\n$$\n$$\n\\angle EOC = \\angle OCD - \\angle OED = 90^\\circ - \\gamma - (180^\\circ - 2\\beta - \\gamma - \\alpha - x) = 2\\beta + \\alpha + x - 90^\\circ;\n$$\n$$\n\\angle P'DA = \\angle EDA - \\angle EDP' = \\angle EDA - \\angle EOC = \\alpha + \\beta - (2\\beta + \\alpha + x - 90^\\circ) = 90^\\circ - \\beta - x.\n$$\n\nSimilarly, $\\angle P'CB = \\angle P'AD$. So, the points $A$, $P'$, $C$, $F$ lie on the same circle, and the points $B$, $P'$, $D$, $F$ lie on the same circle. The same is true for the point $P' \\Rightarrow P = P'$.\n\nLet $w$ be the circle that touches the lines $AB$, $CD$ at points $M$, $N$ respectively and touches the circumscribed circle of the triangle $OAB$ at a point $M_1$. By a known lemma the points $M$, $M_1$, $O$ are collinear and $OM \\cdot OM_1 = OA^2$.\n\nApply the inversion with the center $O$ and radius $OA$ (Fig. 51). Then the circumscribed circles of the triangles $OAB$, $OCD$ will be mapped to the lines $AB$, $CD$ respectively, the point $M$ will be mapped to the point $M_1$, and so the circle $w$ will be mapped to $w$ itself.\n\nSo $w$ touches the image of the line $CD$, that is the circumscribed circle of the triangle $OCD$. Thus $w$ is the circle that we were looking for.\n\n![](attached_image_1.png)\nFig. 50\n\n![](attached_image_2.png)\nFig. 51",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the point of intersection of the altitudes $AP$ and $CQ$ of an acute-angled triangle $ABC$. On the median $BM$ points $E$ and $F$ are chosen so that $\\angle APE = \\angle BAC$, $\\angle CQF = \\angle BCA$, where the point $E$ lies inside the triangle $APB$, and the point $F$ lies inside the triangle $CQB$. Prove that the lines $AE$, $CF$ and $BH$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that the lines $AE$ and $CF$ divide the line segment $BH$ in the same ratio (it is easy to see that the points of intersection of the lines $AE$ and $CF$ with the line $BH$ will belong to this segment). Let $T$ be the point of intersection of the lines $AE$ and $BH$ (Fig. 52). Then\n\n$$\n\\frac{BT}{TH} = \\frac{S_{\\triangle BAT}}{S_{\\triangle TAH}} = \\frac{BA \\cdot \\sin \\angle BAT}{AH \\cdot \\sin \\angle TAH}.\n$$\nThe lines $AE$, $BE$ and $CE$ are concurrent and so by the trigonometric version of the Ceva's theorem we have:\n$$\n\\frac{\\sin \\angle BAT}{\\sin \\angle TAH} \\cdot \\frac{\\sin \\angle APE}{\\sin \\angle EPB} \\cdot \\frac{\\sin \\angle PBM}{\\sin \\angle MBA} = 1,\n$$\nand so\n\nSince $S_{\\square ABM} = S_{\\square CBM}$, we have that\n$AB \\sin \\angle ABM = BC \\sin \\angle CBM$, hence\n$$\n(*) = \\frac{BA}{AH} \\cdot \\frac{\\sin(90^\\circ - \\angle BAC)}{\\sin \\angle BAC} \\cdot \\frac{BC}{AB} = \\frac{BC}{AH} \\cdot \\frac{\\sin \\angle ABH}{\\sin \\angle BAC} = (**).\n$$\nThen using the sine theorem for the triangle $ABH$ and\ntaking into account that $\\angle AHB = 180^\\circ - \\angle ACB$, we obtain:\n$$\n\\frac{AH}{\\sin \\angle ABH} = \\frac{AB}{\\sin \\angle AHB} = \\frac{AB}{\\sin \\angle ACB} = \\frac{BC}{\\sin \\angle BAC},\n$$\nthat is $\\frac{BT}{TH} = (**)=1$. So, we have proved that the line $AE$ passes through the midpoint of the line segment $BH$. One can similarly prove that the line $CF$ also passes through this point, which proves that the lines $AE$, $CF$ and $BH$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $(x, y)$ that satisfy the following equality:\n$$\n|x+|x+|x|| \\cdot ||-y|-y|-y|=2011.\n$$",
    "options": [],
    "answer": "(-1, 2011) and (-2011, 1)",
    "solution": "Since $2011$ is a prime number, every multiple of the left-hand side must be equal to either $1$ or $2011$.\nIf $x \\ge 0$, the first multiple equals $3x$, and there are no solutions. Similarly, there are no solutions if $y \\le 0$ (the second multiple is $-3y$).\nSuppose now that $x < 0$ and $y > 0$. Then\n$$\n|x+|x+|x|| \\cdot ||-y|-y|-y|=|x+|x-x|| \\cdot ||y-y|-y|=|x| \\cdot |-y| = -xy = -2011.\n$$\nRecalling that $2011$ is prime, we get the above answers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45292,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ the angle $A$ is twice as big as the angle $B$, and $CD$ is the bisector of the angle $C$. Prove that $BC = AC + AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$ be a point on $BC$ such that $AE$ is perpendicular to $CD$. Then $\\triangle ACE$ is isosceles, since the bisector of the angle $C$ is also an altitude of the triangle (fig. 23). Hence, $AC = CE$. $\\triangle ADE$ is isosceles, since the line $CD$ is perpendicular to $AE$ and divides $AE$ in half (altitude is a median). So, we have:\n$$\nAD = DE. \\quad (1)\n$$\nLet $\\angle B = \\alpha$, $\\angle A = 2\\alpha$, $\\angle C = 180^\\circ - 3\\alpha$. Then $\\angle ADC = \\alpha + 90^\\circ - \\frac{3\\alpha}{2} = 90^\\circ - \\frac{\\alpha}{2}$, so $\\angle ADE = 180^\\circ - \\alpha$, $\\angle EDB = \\alpha = \\angle B$, hence, $\\triangle DEB$ is isosceles, i.e.,\n$$\nBE = DE. \\quad (2)\n$$\nFrom (1) and (2), we have $AD = BE$. Finally, $BC = CE + BE = AC + AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d$ be distinct natural numbers such that $ab + cd$ is divisible by $ac + bd$. Prove that $ac + bd$ is a composite number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To the contrary, assume that $ac + bd$ is prime. Then $ac + bd \\mid ab + cd \\Rightarrow ac + bd \\mid ac + bd + ab + cd = (a + d)(b + c)$, and so $ac + bd \\mid a + d$ or $ac + bd \\mid b + c$. But this is impossible because for distinct $a, b, c, d$ we have that $ac + bd > a + d$ and $ac + bd > b + c$. This contradiction completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45294,
    "subject": "Mathematics (Multi-modal)",
    "question": "One corner cell is removed from a $2011 \\times 2011$ square. Is it possible to cut the obtained figure along the lines of the grid into less than $121$ squares?",
    "options": [],
    "answer": "Yes",
    "solution": "We will write $a \\to x$, if an $a \\times a$ square without one corner cell can be cut along the lines of the grid into $x$ squares.\n\nEvidently, $2 \\to 3$. Also note that $7 \\to 8$ and $9 \\to 9$ (fig. 24 and 25). We will show that if $a \\to x$ and $b \\to y$, then $ab \\to x+y$. Indeed, a square with the side length $ab$ without a corner cell can be cut into two parts: a square with the side length $b$ without a corner cell, and a square with the side length $ab$, with a $b \\times b$ square removed from its corner. The first part can be cut into $y$ squares, and the second part can be cut into $x$ squares, since it is a square with the side length $a$ without a corner, magnified by a factor of $b$. So,\n$$\na \\to x \\Rightarrow 2a \\to x+3. \\quad (*)\n$$\nWe will also show that\n$$\na \\to x \\Rightarrow 2a-1 \\to 2x+2. \\quad (**)\n$$\nIndeed, an \"incomplete\" square can be partitioned into 2 squares with the side length $a-1$ and 2 \"incomplete\" squares with the side length $a$ (fig. 26).\nSo: $7 \\to 8$, $9 \\to 9$, $63 \\to 17$, $126 \\to 20$, $252 \\to 23$, $503 \\to 48$, $1006 \\to 51$, $2011 \\to 104$, i.e., we have shown that our original figure can be cut into $104 < 121$ squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45295,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\max\\{|x|-1, x^2-1\\} \\leq \\min\\{x^2-1, 1-x\\},\n$$\nwhere $\\max\\{a,b\\} = \\begin{cases} a, & \\text{if } a \\ge b \\\\ b, & \\text{if } a < b \\end{cases}$, and $\\min\\{a,b\\} = \\begin{cases} b, & \\text{if } a \\ge b \\\\ a, & \\text{if } a < b \\end{cases}$.",
    "options": [],
    "answer": "x ∈ {-1, 0, 1}",
    "solution": "It is clear that\n$$\n\\max\\{|x|-1, x^2-1\\} \\geq x^2-1 \\geq \\min\\{x^2-1, 1-x\\},\n$$\nso the inequality from the problem condition can be satisfied only if\n$$\n\\max\\{|x|-1, x^2-1\\} = x^2-1 = \\min\\{x^2-1, 1-x\\},\n$$\nand this, in turn, implies that\n$$\n|x|-1 \\leq x^2-1 \\leq 1-|x|.\n$$\nTherefore, $1-|x| \\geq |x|-1$, which means that $|x| \\leq 1$. Also we should have that $|x| \\leq |x|^2 = |x| \\cdot |x|$, and since we already know that $|x| \\leq 1$, this can happen only if $|x|=1$ or $|x|=0$, so $x=1, x=0$ or $x=-1$.\n\nIt is also easy to solve this problem using graphs (fig. 35). Here the graph of the left-hand side is drawn with dash-and-dash line, and the graph of the right-hand side – with dash-and-dot line. They intersect exactly at the specified values of $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45296,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a natural number a *twin* if it has two natural divisors whose difference is equal to $2$. Determine whether there are more twin numbers or the numbers that are not twin among the first $20112012$ natural numbers.",
    "options": [],
    "answer": "There are more twin numbers than non-twin numbers among the first 20112012 natural numbers.",
    "solution": "Let $N$ be the number of twin numbers that do not exceed $M$. Then $N \\ge N_3 + N_4 - N_{12}$, since all the numbers from $M_3$ and $M_4$ are twin, as they have divisors $1, 3$ and $2, 4$ respectively, and $M_2$ consists of all the numbers that belong to both $M_3$ and $M_4$. But there are also twin numbers that do not exceed $M$ and do not belong neither to $M_3$ nor to $M_4$, for example $35 = 5 \\cdot 7$, and so we even have that $N > N_3 + N_4 - N_{12}$. Since $M$ is divisible by $12$, we obtain:\n$$\nN > \\frac{M}{3} + \\frac{M}{4} - \\frac{M}{12} = \\frac{M}{2},\n$$\nwhich precisely means that there are more twin numbers than the numbers that are not twin among the first $M = 20112012$ natural numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45297,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\ge 3$ burrows on a straight line. Mouse Jerry is hiding in one of these burrows. Cat Tom has a possibility to put his paw into one of the burrows and to catch Jerry if he is hiding in this burrow. After every Tom's attempt Jerry necessarily runs to the neighboring (left or right) burrow. Can Tom always catch Jerry?",
    "options": [],
    "answer": "Yes, Tom can always catch Jerry.",
    "solution": "Enumerate the burrows from left to right with numbers from $1$ to $n$ and define the \"distance\" between the burrows $i$ and $j$ by $i - j$ (it can be negative).\n\nFirst Tom checks all the burrows from $1$ to $n$ one after another. If at the moment when Tom checks the first burrow Jerry is in a burrow with an odd number, then Tom will necessarily catch Jerry during this check.\n\nIndeed, if at the moment when Tom checks the first burrow there is no Jerry, then at this moment the \"distance\" between them is a positive even number. If he still did not catch Jerry when he checks the last burrow, this would mean that the \"distance\" between them at this moment is a negative even number. Note, that after every Jerry's run and Tom's move the \"distance\" between them either does not change (if they move in the same direction) or changes by $2$ (if they move in different directions). So if we start with a positive even number and end up with a negative even number, there should be a moment when the \"distance\" between them is zero, which means that Tom catches Jerry.\n\nIf at the beginning Jerry is in a burrow with an even number, then at the moment when Tom checks the $n$-th burrow, Jerry will be in a burrow with a number of the different parity than $n$. After this Jerry moves to the burrow with a number of the same parity as $n$. So, if now Tom checks all the burrows from $n$ to $1$ one after another, he will necessarily catch Jerry by the above arguments because the \"distance\" between them is now even.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45298,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a parallelogram $ABCD$ $\\angle ABC = 105^\\circ$. It is known that inside this parallelogram there is a point $M$, such that the triangle $BMC$ is equilateral and $\\angle CMD = 135^\\circ$. Let $K$ be a midpoint of the side $AB$. Find $\\angle BKC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "45°",
    "solution": "Drop the perpendicular $CL$ to the line $DM$. Then in the triangle $MCL$ we know: $\\angle MLC = 90^\\circ$, $\\angle LMC = 45^\\circ$. Therefore, $\\angle LCM = 45^\\circ$ and $CL = \\frac{1}{\\sqrt{2}} CM$ (fig. 36).\n\nSince $\\angle LCD = 60^\\circ$, from the right triangle $LCD$ we find that $CD = 2CL = \\sqrt{2} CM$. So, $AB = CD = \\sqrt{2} CM = \\sqrt{2} BM$, and since $\\angle ABM = 45^\\circ$, we obtain that $\\angle BMA = 90^\\circ$.\n\nThen $MK$ is a median from the vertex of the right angle in the right triangle $AMB$, and so $KM = BK = AK$. Therefore, the quadrilateral $KBCM$ is a deltoid, its diagonals are perpendicular and intersect at the point $O$.\n\nThus in the triangle $BOK$ we have that $\\angle BOK = 90^\\circ$ and $\\angle OBK = \\angle ABC - \\angle MBC = 105^\\circ - 60^\\circ = 45^\\circ$. This implies that $\\angle BKO = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$, $Q(x)$ be polynomials with real coefficients such that $P(0) > 0$ and all coefficients of the polynomial $S(x) = P(x)Q(x)$ are non-negative. Prove that for any positive $x$ the following inequality holds:\n$$\nS(x^2) - S^2(x) \\le \\frac{1}{4}(P^2(x^3) + Q(x^3)).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $S = 0$, then $Q = 0$, and the inequality is evident. Suppose now that $S$ is not identically zero. Then $\\forall x > 0$ $S(x) > 0$. If for some $y > 0$ $P(y) < 0$, then the polynomial $P$, and so the polynomial $S$, have roots on the interval $(0, y)$, which is impossible. So, $P$ and $Q$ are positive for $x > 0$. Rewrite our inequality in the following way:\n$$\n4(2P(x^2)2Q(x^2)) - (2P(x)2Q(x))^2 \\le (2P(x^3))^2 + 2(2Q(x^3)).\n$$\nDenote $\\alpha = 2P$, $\\beta = 2Q$, $\\gamma = \\alpha\\beta = 4PQ$. Then the last inequality becomes:\n$$\n4\\gamma(x^2) \\le \\gamma^2(x) + \\alpha^2(x^3) + 2\\beta(x^3).\n$$\nEstimate both sides of this inequality:\n$$\n\\begin{aligned}\n\\gamma^2(x) + \\alpha^2(x^3) + 2\\beta(x^3) &= \\gamma^2(x) + \\beta(x^3) + \\alpha^2(x^3) + \\beta(x^3) \\ge \\\\\n&\\ge 4\\sqrt[4]{\\gamma^2(x)\\alpha^2(x^3)\\beta^2(x^3)} = 4\\sqrt[4]{\\gamma^2(x^3)\\gamma^2(x)} = 4\\sqrt{\\gamma(x)\\gamma(x^3)}.\n\\end{aligned}\n$$\nIf $\\gamma(x) = a_0 + a_1x + \\dots + a_nx^n$, then\n$$\n(a_0 + a_1x + \\dots + a_nx^n)(a_0 + a_1x^3 + \\dots + a_nx^{3n}) \\ge (\\sqrt{a_0a_1} + \\sqrt{a_1a_2} + \\dots + \\sqrt{a_n a_{n+1}})^2\n$$\n(the Cauchy-Schwartz inequality), which implies the required inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the following equation:\n$$\n|[x]| = |[x]|,\n$$\nwhere $[a]$ stands for the greatest integer that does not exceed $a$.",
    "options": [],
    "answer": "All non-negative real numbers and negative integers",
    "solution": "**Answer:** All non-negative reals and negative integers.\n\nConsider 3 cases.\n\n1) $x \\ge 0$. Then $|x| = x$, $[x] \\ge 0$, and so $|[x]| = [x] = [|x|]$, which means that any non-negative $x$ is a solution of our equation.\n\n2) $x$ is a negative integer. Then $[x] = x$, $[-x] = -x$, $|x| = -x$, and so $|[x]| = [-x] = (-x = |x| = [x]|)$, proving that these values of $x$ are also solutions.\n\n3) $x$ is negative, but not an integer. In this case $|x| = -x$, $|[x]| < -x$. On the other hand, $[x] < x < 0$, $|[x]| > -x$, and the equality is not satisfied.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45301,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex 100-gon all the vertices, as well as some other points inside the polygon, are selected. No three of them are collinear. The selected points are joined (by straight line segments) in such a way that the 100-gon is partitioned into 2011 convex polygons. Prove that at least one of these polygons has an even number of sides.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Compute a number $N$, which is the sum of numbers $e_j$ of sides of all the polygons of the partition. It is even because it is the sum of 100 (boundary segments) and the doubled number of all segments that are inside the 100-gon. But there is an odd number of summands $e_j$ (exactly 2011), and so it cannot happen that all of them are odd as they have an even sum. So, at least one of $e_j$ is even, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45302,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\sqrt{1-a} + \\sqrt{1-b} + \\sqrt{1-c} \\le \\sqrt{2}(\\sqrt{ab+bc+ca} + 2\\sqrt{a^2+b^2+c^2}).\n$$\nfor non-negative $a, b, c$ with $a + b + c = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the Cauchy-Schwartz inequality for the collections $(\\sqrt{\\alpha}, \\sqrt{\\beta}, \\sqrt{\\gamma})$ and $(\\sqrt{\\alpha x}, \\sqrt{\\beta y}, \\sqrt{\\gamma z})$ we obtain:\n$$\n(\\alpha\\sqrt{x} + \\beta\\sqrt{y} + \\gamma\\sqrt{z}) \\le \\sqrt{(\\alpha + \\beta + \\gamma)(\\alpha x + \\beta y + \\gamma z)},\n$$\nand so\n$$\n\\begin{aligned}\n& a\\sqrt{a+b} + b\\sqrt{b+c} + c\\sqrt{c+a} \\le \\sqrt{a\\sqrt{a+b} + b\\sqrt{b+c} + c\\sqrt{c+a}} \\\\\n& \\le \\sqrt{(a+b+c)(a^2+ab+b^2+bc+c^2+ca)} \\le \\sqrt{2(a^2+b^2+c^2)}.\n\\end{aligned}\n$$\nsimilarly,\n$$\n\\begin{aligned}\n& b\\sqrt{a+b} + c\\sqrt{b+c} + a\\sqrt{c+a} \\le \\sqrt{2(a^2+b^2+c^2)}, \\\\\n& c\\sqrt{a+b} + a\\sqrt{b+c} + b\\sqrt{c+a} \\le \\sqrt{2(ab+bc+ca)}.\n\\end{aligned}\n$$\n\nAdding all three inequalities, we get the required one.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $C$, $E$ and $A$ be points on the sides $XY$, $YZ$ and $ZX$ of the triangle $XYZ$ respectively. Points $B$, $D$ and $F$ are chosen on the segments $AX$, $CY$ and $EZ$ respectively, so that $BC \\perp AD$, $DE \\perp CF$, $AF \\perp BE$. Can it happen that the lines $XF$, $YB$ and $ZD$ are concurrent?",
    "options": [],
    "answer": "No",
    "solution": "**Answer:** No, it cannot.\n\nSuppose, it is possible (fig. 27). Let $O$ be the point of intersection of the lines $XF$, $YB$ and $ZD$. Then by the Ceva's theorem, we have\n$$\n\\frac{YD}{DX} \\cdot \\frac{XB}{BZ} \\cdot \\frac{ZF}{FY} = 1.\n$$\n\n\\begin{align*}\n\\frac{YD}{DX} &= \\frac{YD}{CD} \\cdot \\frac{CD}{DX} = \\frac{YE}{EF} \\cdot \\frac{CD}{DX}, \\\\\n\\frac{XB}{BZ} &= \\frac{XB}{AB} \\cdot \\frac{AB}{BZ} = \\frac{XC}{CD} \\cdot \\frac{AB}{BZ}, \\\\\n\\frac{ZF}{FY} &= \\frac{ZF}{FE} \\cdot \\frac{FE}{FY} = \\frac{ZA}{AB} \\cdot \\frac{FE}{FY}.\n\\end{align*}\n$$\nTherefore,\n$$\n1 = \\frac{YD}{DX} \\cdot \\frac{XB}{BZ} \\cdot \\frac{ZF}{FY} = \\left( \\frac{YE}{EF} \\cdot \\frac{CD}{DX} \\right) \\cdot \\left( \\frac{XC}{CD} \\cdot \\frac{AB}{BZ} \\right) \\cdot \\left( \\frac{ZA}{AB} \\cdot \\frac{FE}{FY} \\right) = \\frac{YE}{YF} \\cdot \\frac{XC}{XD} \\cdot \\frac{ZA}{ZB} < 1,\n$$\nsince each of the fractions $\\frac{YF}{XC}$ and $\\frac{ZA}{XB}$ is less than 1. This contradiction proves that the lines $XF$, $YB$ and $ZD$ cannot be concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45304,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\max\\{x^2+3x+3, x^{2011}+x^4+x^2+x+1\\} \\le \\min\\{1-x-x^2, x^{2011}+x^4+x^2+x+1\\},\n$$\nwhere $\\max\\{a,b\\} = \\begin{cases} a, & \\text{if } a \\ge b \\\\ b, & \\text{if } a < b \\end{cases}$, and $\\min\\{a,b\\} = \\begin{cases} b, & \\text{if } a \\ge b \\\\ a, & \\text{if } a < b \\end{cases}$.",
    "options": [],
    "answer": "-1",
    "solution": "For our convenience we denote $P(x) = x^{2011} + x^4 + x^2 + x + 1$. It is clear that\n$$\n\\max\\{x^2+3x+3, P(x)\\} \\ge P(x) \\ge \\min\\{1-x-x^2, P(x)\\},\n$$\nso the inequality from the problem condition can be satisfied only if\n$$\n\\max\\{x^2+3x+3, P(x)\\} = P(x) = \\min\\{1-x-x^2, P(x)\\},\n$$\nand this, in turn, implies that\n$$\nx^2+3x+3 \\le P(x) \\le 1-x-x^2.\n$$\nSo, $x^2+3x+3 \\le 1-x-x^2$, which yields that $2x^2+4x+2 \\le 0 \\Leftrightarrow (x+1)^2 \\le 0 \\Leftrightarrow x=-1$. Hence, the only possible solution is $x=-1$. A simple verification shows that this value of $x$ satisfies the original problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45305,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ $M$ is the midpoint of the side $BC$, and on the side $AB$ a point $N$ is chosen so that $NB = 2AN$. If $\\angle CAB = \\angle CMN$, find $\\frac{AC}{BC}$?",
    "options": [],
    "answer": "1/2",
    "solution": "Let $D$ be a point on the half-line $CA$ such that $CA = AD$ (fig. 37).\n\n![](attached_image_1.png)\n\nThen $DM$ and $BA$ are medians of the triangle $BCD$, and so the point of their intersection divides both of them in the ratio $2:1$ from the vertex. This means that they intersect at the point $N$. Hence $\\angle DAB = \\pi - \\angle CAB = \\pi - \\angle CMN = \\angle DMB$, and so the quadrilateral $BMAD$ is cyclic. Since $AM$ is parallel to $BD$ as a midline, we have that $\\angle CDB = \\angle CAM = \\angle CBD$. This implies that\n$$\nCD = CB, \\text{ therefore } \\frac{AC}{BC} = \\frac{AC}{CD} = \\frac{1}{2}.\n$$\nIt is easy to see that the problem condition is satisfied for every triangle with this ratio of the sides.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be arbitrary integers. Prove that at least one of the numbers $a^5b - ab^5$, $b^5c - bc^5$, $c^5a - ca^5$ is divisible by $8$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Серед трьох чисел $a, b$ і $c$ існують два числа, які мають однакову парність. Будемо вважати, що це числа $a$ і $b$. Тоді числа $a-b$, $a+b$, $a^2+b^2$ є парними, і $a^5b-ab^5 = ab(a-b)(a+b)(a^2+b^2):8$.\n\nЗауваження Неважко довести, що в розглядуваному випадку $a^5b-ab^5:240$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45307,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the coordinate plane $xOy$, depict the set of all points such that $|y - [x]| = 2 - y - [x]$, where $[x]$ is the integer part of $x$, i.e. the largest integer not exceeding $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Вихідне співвідношення рівносильне системі\n$$\n\\begin{cases}\ny = 1, \\\\\n[x] = 1; \\\\\ny \\le 2 - [x].\n\\end{cases}\n$$\nСлід зобразити множину точок $\\{(x; y) : 1 \\le x < 2, y < 1\\} \\cup \\{(x; 1) : x < 2\\}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45308,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the decimal form of the number\n$$\n34! = 295232799039*041408476186096435*0000000\n$$\ntwo digits are replaced by stars. Find these digits.",
    "options": [],
    "answer": "first digit = 6, second digit = 2",
    "solution": "Нехай перша пропущена цифра — $a$, друга — $b$. Число $34!$ ділиться без остачі на $9$ та на $11$. Оскільки сума записаних цифр від ділення на $9$ дає остачу $1$, то сума $a$ і $b$ від ділення на $9$ має давати остачу $8$, звідки $(a, b)$ — одна з таких пар цифр\n$$(0,8), (1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1), (8,0), (8,9), (9,8).$$\nЗа ознакою подільності на $11$, сума цифр на парних позиціях дає таку саму остачу від ділення на $11$, що й сума цифр, записаних на непарних позиціях. Звідси $a - b$ від ділення на $11$ дає остачу $4$. Нескладним перебором одержуємо, що $(a, b) = (6, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45309,
    "subject": "Mathematics (Multi-modal)",
    "question": "1. In the plane 5 circles are given such that no three of them have a common point. Can it happen that they have exactly:\n\na) 12;\n\nb) 24 different intersection points?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "a) Yes, it is possible. It is enough to take 4 circles, each pair of which intersects in two points, and a fifth circle such that it does not intersect the others.\n\nb) No, it is not possible. The first circle can intersect the others in at most 8 points, the second adds at most 6 intersection points, and so on. In total, we have at most $8 + 6 + 4 + 2 = 20$ different intersection points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45310,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is written on a blackboard. Each minute Andrew looks at his clock and adds the number of minutes the clock is showing (an integer between $0$ and $59$) to the number written on the blackboard.\n\na) Prove that at some moment a composite number will be written on the blackboard.\n\nb) Will necessarily a number divisible by $5$ be written?",
    "options": [],
    "answer": "No",
    "solution": "a) У першому, третьому, п'ятому і т.д. записаних числах буде чергуватись парність (адже сума двох послідовних кількостей хвилин непарна). Отже, серед них будуть парні числа. При цьому лише перше з них може дорівнювати $2$ (отримане як $2 + 0$ або $1 + 1$), наступні парні числа будуть більшими за $2$, а тому складеними.\n\nб) Не обов'язково. Якщо на дошці написано число $1$, а перша побачена Андрієм кількість хвилин дорівнює $0$, то остачі записаних чисел від ділення на $5$ будуть періодично повторюватись: $1$, $2$, $4$, $2$, $1$, $1$, $2$, $4$, $2$, $1$, .... Серед цих остач не буде остачі $0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $x, y \\in (0, \\pi)$ satisfy the equality\n$$\n\\cos 2x \\cos y - \\cos 2y \\cos x = \\cos y - \\cos x.\n$$\nShow that $x = y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Запишемо дану рівність у вигляді $\\cos^2 x \\cos y - \\cos^2 y \\cos x = \\cos y - \\cos x$, $(\\cos x \\cos y + 1)(\\cos x - \\cos y) = 0$. Оскільки для $x \\in (0; \\pi)$ і $y \\in (0; \\pi)$ $\\cos x \\cos y > -1$, то $\\cos x = \\cos y$, і тому, враховуючи спадання функції $f(t) = \\cos t$ на проміжку $(0; \\pi)$, маємо, що $x = y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45312,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the coordinate plane $xOy$, depict the set of all points such that $|y - x| = 2 - y - x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Вихідне співвідношення рівносильне системі\n$$\n\\begin{cases}\n  y = 1, \\\\\n  x = 1; \\\\\n  y \\le 2 - x.\n\\end{cases}\n$$\nСлід зобразити множину точок $\\{(x; 1) : x \\le 1\\} \\cup \\{(1; y) : y \\le 1\\}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45313,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane six lines are given such that no three of them are coincident. Can it happen that they have exactly:\n\na) $12$;\n\nb) $16$ different intersection points?",
    "options": [],
    "answer": "a) Yes; b) No",
    "solution": "a) It is possible. It is enough to take $3$ lines that have $3$ intersection points among themselves, and also $3$ parallel lines, each of which intersects each of the $3$ initial lines.\n\nb) It is not possible. The first line intersects the others in at most $5$ points, the second adds at most $4$ intersection points, and so on. In total, we have at most $5 + 4 + 3 + 2 + 1 = 15$ different intersection points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to write numbers $1, 2, 3, 4, 5, 6, 7, 8$ at the vertices of an octagon (one number at each vertex) so that the sum of numbers in each three consecutive vertices is greater than:\n\na) $13$;\nb) $11$;\nc) $12$?",
    "options": [],
    "answer": "a) No; b) Yes; c) No",
    "solution": "Нехай біля вершин правильного восьмикутника послідовно записані числа $a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8$.\n\na) Припустимо, що числа можна розташувати потрібним чином. Тоді\n$$\n\\begin{cases} a_1 + a_2 + a_3 \\geq 14, \\\\ a_2 + a_3 + a_4 \\geq 14, \\\\ \\vdots \\\\ a_8 + a_1 + a_2 \\geq 14. \\end{cases}\n$$\nДодаючи всі ці нерівності, одержимо: $3(a_1 + a_2 + \\ldots + a_8) \\geq 8 \\cdot 14$. Але $a_1 + a_2 + \\ldots + a_8 = 1 + 2 + \\ldots + 8 = 36$, тому дістаємо нерівність $3 \\cdot 36 \\geq 8 \\cdot 14$, тобто $108 \\geq 112$. Суперечність.\n\nб) Можна покласти, наприклад, $a_1 = 1$, $a_2 = 5$, $a_3 = 6$, $a_4 = 2$, $a_5 = 4$, $a_6 = 7$, $a_7 = 3$, $a_8 = 8$.\n\nв) Припустимо, що потрібне розміщення чисел існує. Числа $2$ і $3$ не можна записати в сусідніх вершинах, інакше і зліва і справа від них мало б стояти число $8$, яке не може зустрітись двічі. Далі, не порушуючи загальності, можна вважати, що $a_1 = 1$, тоді жодне з чисел $a_2, a_3, a_7, a_8$ не дорівнює ані $2$ ані $3$, а тому або $a_4 = 2$ і $a_6 = 3$, або $a_4 = 3$ і $a_6 = 2$. У обох цих випадках маємо $a_5 = 8$. Отже, $a_2, a_3, a_7$ та $a_8$ — це записані в деякому порядку числа $4, 5, 6, 7$, і тому $(a_1 + a_2 + a_3) + (a_7 + a_8 + a_1) = 24$ та сума принаймні в одній з дужок є меншою за $13$. Суперечність.\n\nВідповідь: а) Не можна. б) Можна. в) Не можна.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = x^2 - 8x + 20$. Solve the equation $f(f(f(x))) = 5$.",
    "options": [],
    "answer": "3, 5",
    "solution": "Неважко помітити, що $f(x) = (x - 4)^2 + 4$. Тоді рівняння $f(f(f(x))) = 5$ рівносильне рівнянню $(x - 4)^8 + 4 = 5$, звідки $x - 4 = \\pm 1$, тобто $x = 3$ або $x = 5$.\n\nВідповідь: $x = 3, x = 5$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = \\frac{x^3}{3x^2 - 3x + 1}$. Compute\n$$\nf\\left(\\frac{1}{2012}\\right) + f\\left(\\frac{2}{2012}\\right) + \\dots + f\\left(\\frac{2012}{2012}\\right).\n$$",
    "options": [],
    "answer": "2013/2",
    "solution": "Зауважимо, що $f\\left(\\frac{i}{2012}\\right) = \\frac{i^3}{i^3 + (2012 - i)^3}$, $1 \\le i \\le 2012$. Тоді\n$$\nf\\left(\\frac{1006}{2012}\\right) = \\frac{1}{2}, \\quad f\\left(\\frac{2012}{2012}\\right) = 1, \\quad f\\left(\\frac{i}{2012}\\right) + f\\left(\\frac{2012 - i}{2012}\\right) = 1, \\quad 1 \\le i \\le 1005.\n$$\nВідповідь: $\\frac{2013}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist a point $A$ on the graph of $f(x) = x^4$ and a point $B$ on the graph of $g(x) = x^4 + x^2 + x + 1$ such that the distance between $A$ and $B$ is less than $\\frac{1}{100}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Розглянемо точку $B(100, 100010101)$, яка лежить на графіку функції $g$, і точку $A(\\sqrt[4]{100010101}, 100010101)$, яка лежить на графіку функції $f$. Відстань між цими точками дорівнює\n$$\n\\sqrt[4]{100010101} - 100 = \\frac{100010101 - 100^4}{(\\sqrt[4]{100010101} + 100)(\\sqrt{100010101} + 100^2)} < \\frac{10101}{200 \\cdot 20000} < \\frac{1}{100}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 45318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y, z \\in \\mathbb{R}$\n$$\nf(xy) + f(xz) \\geq f(x)f(yz) + 1.\n$$",
    "options": [],
    "answer": "f(x) = 1 for all real x",
    "solution": "Покладемо у вихідній нерівності $x = y = z = 0$. Тоді $(f(0) - 1)^2 \\le 0$, тобто $f(0) = 1$.\n\nВізьмемо $y = z = 0$ і одержимо, що $f(0) + f(0) \\ge f(x)f(0) + 1$. Звідси $f(x) \\le 1$ для всіх $x \\in \\mathbb{R}$.\n\nЯкщо $x = y = z = 1$, то $(f(1) - 1)^2 \\le 0$, $f(1) = 1$.\n\nДля $y = z = 1$ маємо: $f(x) + f(x) \\ge f(x)f(1) + 1$, $f(x) \\ge 1$, $x \\in \\mathbb{R}$.\n\nМи встановили, що для всіх $x \\in \\mathbb{R}$ $f(x) = 1$.\n\nПеревірка показує, що така функція задовольняє умову задачі.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three boys picked nuts. After discovering that $420$ nuts are picked in total, the boys decided to share them evenly. First, one of the boys gave each of two others one fourth of the nuts he picked and a nut. Then another boy gave each of two others one fourth of the nuts he collected (those he picked and those he obtained from the first boy) and a nut. After the last boy had made this operation, it turned out that, indeed, the nuts were distributed evenly. Determine, how many nuts each boy picked.",
    "options": [],
    "answer": "first boy: 68, second boy: 122, third boy: 230",
    "solution": "З рівностей\n$$\nx_3 - \\frac{1}{4}x_3 - \\frac{1}{4}x_3 - 2 = 140,\n$$\n$$\nx_2 + \\frac{1}{4}x_3 + 1 = 140.\n$$\n$$\nx_1 + \\frac{1}{4}x_3 + 1 = 140\n$$\nзнайдемо кількість горіхів у кожного з хлопчиків на передостанньому етапі. Аналогічно відновлюємо весь «ланцюжок»:\n$$\n(140, 140, 140) \\leftarrow (68, 68, 284) \\leftarrow (32, 140, 248) \\leftarrow (68, 122, 230).\n$$\n**Відповідь:** Перший хлопчик зібрав $68$ горіхів, другий хлопчик — $122$ горіхи, третій хлопчик — $230$ горіхів.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45320,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $y = f(x)$ is such that its graph is symmetric to the graph of $y = x^2$ with respect to the point $(1; 1)$. Solve the equation $f(f(x)) = f(x)$.",
    "options": [],
    "answer": "1, 2, 3",
    "solution": "Let us find the explicit form of the function $y = f(x)$. Let $P(p; p^2)$ be an arbitrary point of the parabola $y = x^2$, and $Q(q; f(q))$ be the point on the graph $y = f(x)$, symmetric to $P$ with respect to the point $(1; 1)$. Then:\n$$\n\\frac{p + q}{2} = 1, \\quad \\frac{p^2 + f(q)}{2} = 1.\n$$\nEliminating $p$ from these equations, we get $f(q) = -q^2 + 4q - 2$.\n\nNext, we need to solve the equation $f(f(x)) = f(x)$. We have:\n$$\n-(f(x))^2 + 4f(x) - 2 = f(x), \\quad (f(x))^2 - 3f(x) + 2 = 0, \\quad (f(x) - 1)(f(x) - 2) = 0,\n$$\n$$\n\\begin{bmatrix}\nf(x) = 1, & -x^2 + 4x - 2 = 1, \\\\\nf(x) = 2, & -x^2 + 4x - 2 = 2.\n\\end{bmatrix}\n$$\nIt remains to solve the obtained quadratic equations.\n\nAnswer: $1$, $2$, $3$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 45321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S(a)$ denote the sum of decimal digits of a positive integer $a$. Positive integer $n$ is such that $S(n) = 406$, $S(2012n) = 2012$. Find all possible values of $S(2011n)$.",
    "options": [],
    "answer": "1609, 1615, 1624",
    "solution": "Зауважимо, $2012n$ є сумою двох чисел $1000n$, одного числа $10n$ та двох чисел $n$. Якщо не відбувається перенесень у наступний розряд під час додавання цих чисел в стовпчик, то сума цифр утвореного числа буде в 5 разів більшою за суму цифр числа $n$. Під час перенесення однієї одиниці в старший розряд сума цифр зменшується на 9. Оскільки $2012 = 5 \\cdot 406 - 2 \\cdot 9$, то в старший розряд (або в старші розряди) було перенесено двійку (або, відповідно, дві одиниці).\n\nАналогічно, під час додавання двох чисел $1000n$, одного числа $10n$ та одного числа $n$ (тобто під час множення $n$ на $2011$) в старші розряди або було перенесено дві одиниці чи двійку, або одну одиницю, або нічого. Відповідно, сума цифр числа $2011n$ становитиме $1606 = 4 \\cdot 406 - 2 \\cdot 9$, $1615 = 4 \\cdot 406 - 9$ або $1624 = 4 \\cdot 406$. Покажемо, що всі ці значення можуть досягатися.\n\nВізьмемо $A_k = \\frac{40004000...4000}{2011n}$ (у записі $k$ разів повторюються цифри $4000$). Тоді сума цифр кожного з чисел $n_0 = A_{100} + 420$, $n_1 = 10A_{100} + 1050$, $n_2 = 10A_{99} + 2323$ становить $406$, сума цифр кожного з чисел $2012n_0$, $2012n_1$, $2012n_2$ становить $2012$, а суми цифр чисел $2011n_0$, $2011n_1$ та $2011n_2$ дорівнюють, відповідно, $1624$, $1615$ та $1609$.\n\nВідповідь: $1609$, $1615$, $1624$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45322,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f : \\mathbb{R} \\to \\mathbb{R}$ is such that for all $x, y \\in \\mathbb{R}$\n$$\nf(x + 2xy) = f(x) + 2f(xy).\n$$\nGiven that $f(2011) = 2012$, find $f(2012)$.",
    "options": [],
    "answer": "2012^2/2011",
    "solution": "Підставимо $x = 0$ і одержимо, що $f(0) = 0$. Підставляючи до нашого функціонального рівняння $y = -1$ та $y = -\\frac{1}{2}$, одержимо, що $f(-x) = -f(x)$ та $f(x) = 2f(\\frac{x}{2})$ для будь-яких дійсних $x$. Далі, підставимо $y = \\frac{z}{2x}$, де $x \\neq 0$, і одержимо, що $f(x+z) = f(x) + f(z)$. Оскільки $f(0+z) = 0 + f(z) = f(0) + f(z)$, то рівність $f(x+z) = f(x) + f(z)$ має місце для всіх дійсних $x$ і $z$ (позаяк кожен розв'язок функціонального рівняння адитивності, очевидно, задовольняє умову $f(x+2xy) = f(x)+2f(xy)$, то ми встановили, що множини розв'язків цих рівнянь співпадають). Легко довести, що $f(kx) = kf(x)$ для довільних $x \\in \\mathbb{R}$ і $k \\in \\mathbb{Z}$. Оскільки $f(2011) = f(2011 \\cdot 1) = 2011f(1)$, і за умовою $f(2011) = 2012$, то $2012 = 2011f(1)$, $f(1) = \\frac{2012}{2011}$, $f(2012) = 2012f(1) = \\frac{2012^2}{2011}$.\nВідповідь: $f(2012) = \\frac{2012^2}{2011}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45323,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square $11 \\times 11$ is divided into parts of sizes $4 \\times 4$, $1 \\times 3$, $3 \\times 1$ (not necessarily all these sizes must be present). Prove that there is a row of the initial square intersecting an odd number of these parts.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Оскільки рядок початкового квадрата містить непарну кількість клітинок, він буде перетинати непарну кількість частин $1 \\times 3$ та $3 \\times 1$. Якщо твердження задачі неправильне, то кожен рядок має перетинати непарну кількість частин $4 \\times 4$, тобто одну. Відтак, один квадрат $4 \\times 4$ буде лежати в перших чотирьох рядках, один — у чотирьох наступних, і ми не зможемо «вмістити» потрібний квадрат у три нижні рядки.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integers $A$ and $B$ have the following decimal forms: $A = \\overline{abcabc}$ and $B = \\overline{d00d}$, where $a, b, c, d$ are decimal digits, $a \\neq 0, d \\neq 0$. Find all possible values of $a, b, c, d$ such that $A+B$ is a full square.",
    "options": [],
    "answer": "a = 9, b = 9, d ∈ {2,3,4,5,6,7,8,9}, c = 11 − d",
    "solution": "Оскільки\n$$\nA + B = \\overline{abcabc} + \\overline{d00d} = 1001 (\\overline{abc} + d) \\le 1001 (999 + 9) = 1001 \\cdot 1008,\n$$\nто для виконання умови задачі необхідно й достатньо, щоб $\\overline{abc} + d = 1001$. Звідси $a = 9, b = 9$ і $c + d = 11$.\n\n*Відповідь:* $a = 9, b = 9, c = 11 - d, a, d \\in \\{2, 3, 4, 5, 6, 7, 8, 9\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Can a product of four consecutive odd positive integers be a cube of an integer?",
    "options": [],
    "answer": "No",
    "solution": "Нехай для деякого парного числа $k$ і натурального числа $l$ виконується рівність $(k-3)(k-1)(k+1)(k+3) = l^3$, тобто $(k^2-9)(k^2-1) = l^3$. Якщо числа $k^2-9$ і $k^2-1$ мають спільний простий дільник, то він є дільником їхньої різниці, а тому дорівнює 2. Але числа $k^2-9$ і $k^2-1$ непарні, отже вони взаємно прості і кожне з них є точним кубом. Відтак, $k^2-9 = n^3$ та $k^2-1 = m^3$, де $m, n \\in \\mathbb{N}$. Звідси $m^3-n^3=8$, тобто $(m-n)(m^2+mn+n^2) = 8$. Оскільки $m-n < m^2+mn+n^2$, то можливі лише два випадки:\n$$\n\\begin{cases} m-n=1, \\\\ m^2+mn+n^2=8; \\end{cases} \\text{або} \\begin{cases} m-n=2, \\\\ m^2+mn+n^2=4. \\end{cases}\n$$\nПерша система не має розв'язків у натуральних числах, бо коли $m-n=1$, то $m$ і $n$ мають різну парність, а тоді $m^2+mn+n^2$ — непарне число.\nДруга система також не має розв'язків у натуральних числах, бо коли $m-n=2$, то $m \\geq 3$, а тоді $m^2+mn+n^2 > m^2 \\geq 9$. Отже, вказаний добуток не може бути точним кубом.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45326,
    "subject": "Mathematics (Multi-modal)",
    "question": "2012 people stand in a line. Each of them is either a knight (and always tells truth) or a knave (and always lies). Each of the people announced: «There are more knaves on my left than knights on my right». How many knaves are in the line?",
    "options": [],
    "answer": "1006",
    "solution": "Нехай у шерензі стоять $b$ брехунів та $r$ лицарів. Оскільки найлівіший з лицарів сказав правду, маємо $b > r - 1$. Оскільки найправіший з брехунів збрехав, то маємо $b - 1 \\le r$. Отже, $r = b$ або $r = b - 1$. Другий випадок неможливий, оскільки загальна кількість людей парна. Тому $r = b = 1006$.\n\n*Відповідь:* 1006 брехунів.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram. The incircle of triangle $ABC$ touches its sides $AB$, $BC$ and $AC$ at points $M$, $N$ and $K$ respectively. The lines $MK$ and $CD$ intersect at $P$, the lines $NK$ and $AD$ intersect at $Q$. Prove that the points $B$, $E$ and $F$ are collinear, where $E$ and $F$ are the midpoints of $MP$ and $NQ$ respectively.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Оскільки $AM = AK$, $CN = CK$ як дотичні, то трикутники $MAK$ та $NCK$ рівнобедрені. Звідси $\\angle AMK = \\angle AKM$, і $\\angle CNK = \\angle CKN$.\n\nДалі, $\\angle AMK = \\angle CPK$ і $\\angle AQK = \\angle CNK$ як різносторонні при паралельних прямих. Крім того, $\\angle AKM = \\angle CKP$ і $\\angle AKQ = \\angle CKN$ як вертикальні. Отже,\n$\\angle AKQ = \\angle AQK$ і $\\angle CKP = \\angle CPK$, тобто трикутники $KAQ$ і $KCP$ рівноберені. Звідси $AM = AK = AQ$, $CN = CK = CP$, тобто трикутники $MAQ$ і $NCP$ рівноберені.\n\nНехай $I$ — центр кола, вписаного в трикутник $ABC$, тоді $BI$ — серединний перпендикуляр відрізка $MN$ і бісектриса кута $ABC$. З рівнобереного трикутника $MAQ$ знаходимо $\\angle AMQ = \\frac{180^\\circ - \\angle MAQ}{2} = \\frac{\\angle ABC}{2} = \\angle ABI$, тобто $MQ \\parallel BI$. Аналогічно $NP \\parallel BI$.\n\nВідтак, $MQ \\parallel NP \\parallel BI$ і $BI$ проходить через середину $MN$. За теоремою Фалеса, $BI$ проходить через середини відрізків $MP$ і $NQ$, що й завершує доведення.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ such that $p^5 - p = 2n!$, where $n$ is a positive integer.",
    "options": [],
    "answer": "p = 3",
    "solution": "Число $p = 2$ не задовольняє умову, оскільки $p^5 - p = 30 = 2 \\cdot 15$ та $3! < 15 < 4!$ Число $p = 3$ задовольняє умову, оскільки $p^5 - p = 240 = 2 \\cdot 5!$\n\nНехай $p \\ge 5$ та $p^5 - p = 2n!$ для деякого натурального $n$, тоді $n! = (p-1)p(p+1) \\frac{p^2+1}{2}$. Оскільки $p$ просте та $n!$ ділиться без остачі на $p$, то $n \\ge p$. Отже,\n$$\n1 \\cdot 2 \\cdots (p-2) \\cdot (p-1) \\cdot p \\cdots n = (p-1)p(p+1) \\frac{p^2+1}{2}.\n$$\nЗвідси $(p+1)(p^2 + 1)$ ділиться без остачі на $p-2$. Але $(p+1)(p^2 + 1) = p^3 + p^2 + p + 1 = (p-2)(p^2 + 3p + 7) + 15$, тому 15 ділиться без остачі на $p-2$.\n\nОскільки $p-2 \\ge 3$, то $p-2 = 3$, або $p-2 = 5$, або $p-2 = 15$ тобто $p = 5$, або $p = 7$, або $p = 17$. Перевірка показує, що ці числа не задовольняють умову.\n\nВідповідь: $p = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to paint each cell of an $8 \\times 8$ table with one of 16 colors so that for each two colors there are two cells painted with these colors and having a common side?",
    "options": [],
    "answer": "It is not possible.",
    "solution": "In the table, there are 112 unit segments that separate its neighboring cells. But from 16 given colors, one can form 120 pairs of colors, and for each of these pairs, there must be two neighboring cells, which is impossible.\n\n*Answer:* It is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vandal Peter cut a rectangular head teacher's portrait along a straight line. After this he cut one of the pieces along a straight line, then he cut one of the new pieces etc. After he had made 100 cuts, the head teacher arrived and forced Peter to pay 2 kopecks for each triangular piece and 1 kopeck for each quadrangular piece. Prove that Peter paid more than 1 hryvnya (1 hryvnya = 100 kopecks).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Очевидно, що отримані під час розрізання шматки є опуклими многокутнимами. Одним розрізанням загальна кількість вершин збільшується щонайбільше на 4. Тому після 100 розрізань многокутники матимуть не більше за 404 вершини. З іншого боку, після 100 розрізань утворився 101 многокутник. Нехай серед них $n$ трикутників та $m$ чотирикутників. Тоді утворені многокутники мають не менше за $3n + 4m + 5(101 - m - n) = 505 - 2n - m$ вершин. Отже, $505 - 2n - m \\le 404$, звідки $2n + m \\ge 101 > 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45331,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a triangle *ABC*, $I_A$ is the center of excircle opposite vertex $A$. Prove that points $B$, $C$ and the circumcenters of triangles $ABI_A$ and $ACI_A$ are cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай точка $P$ — центр описаного кола трикутника $ABI_A$. Кут $ABI_A$ тупий, а тому точки $P$ і $B$ лежать по різні боки від прямої $AI_A$. Нехай $M$ і $N$ — точки дотику даного кола зі стороною $BC$ і продовженням сторони $AC$ відповідно, а $K$ — точка перетину цього продовження з описаним колом трикутника $ABI_A$ (дивись рисунок). Оскільки $I_A M = I_A N$ і $BI_A = KI_A$, то прямокутні трикутники $BMІ_A$ та $KNI_A$ рівні, а тому $BM = KN$. Оскільки $CM = CN$, то $BC = CK$, тобто трикутник $BCK$ рівнобедрений. Нехай $\\angle BCA = \\gamma$, тоді $\\angle BKC = \\gamma/2$, $\\angle BPA = \\gamma$. Отже, $\\angle BCA = \\angle BPA$, а це й означає, що точка $P$ належить описаному колу трикутника $ABC$. Аналогічно доводиться, що центр описаного кола трикутника $ACI_A$ також лежить на цьому колі. Це й завершує доведення.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45332,
    "subject": "Mathematics (Multi-modal)",
    "question": "For arbitrary positive numbers $a$, $b$, $c$ prove the inequality\n$$\n\\frac{ab}{a+3b+2c} + \\frac{bc}{b+3c+2a} + \\frac{ca}{c+3a+2b} \\le \\frac{1}{6}(a+b+c).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Використаємо відому нерівність $(x + y + z) \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) \\geq 9$, $x, y, z > 0$, яку неважко довести за допомогою нерівності Коші. Перепишемо її у вигляді $\\frac{1}{x+y+z} \\leq \\frac{1}{9}\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right)$. Оскільки $a + 3b + 2c = 2b + (b+c) + (c+a)$, то\n$$\n\\frac{ab}{a+3b+2c} \\leq \\frac{ab}{9} \\left( \\frac{1}{2b} + \\frac{1}{b+c} + \\frac{1}{c+a} \\right) = \\frac{a}{18} + \\frac{1}{9} \\left( \\frac{ab}{b+c} + \\frac{ab}{c+a} \\right).\n$$\nДодавши цю та ще дві аналогічні нерівності, дістанемо\n$$\n\\begin{aligned}\n\\frac{ab}{a+3b+2c} + \\frac{bc}{b+3c+2a} + \\frac{ca}{c+3a+2b} &\\le \\frac{1}{18}(a+b+c) + \\frac{1}{9}\\left(\\frac{ab}{b+c} + \\frac{ab}{c+a} + \\frac{bc}{c+a} + \\frac{bc}{a+b} + \\frac{ca}{a+b} + \\frac{ca}{b+c}\\right) \\\\\n&\\le \\frac{1}{18}(a+b+c) + \\frac{1}{9}(a+b+c) = \\frac{1}{6}(a+b+c).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integer $n$ the number $5^n (5^n + 1) - 6^n (3^n + 2^n)$ is divisible by $91$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Позначимо значення даного виразу через $a_n$. Маємо:\n$$\n\\begin{aligned}\na_n &= 25^n + 5^n - 18^n - 12^n = (25^n - 18^n) - (12^n - 5^n) = \\\\\n&= (25 - 18)(25^{n-1} + 25^{n-2} \\cdot 18 + \\dots + 25 \\cdot 18^{n-2} + 18^{n-1}) - \\\\\n&\\quad -(12 - 5)(12^{n-1} + 12^{n-2} \\cdot 5 + \\dots + 12 \\cdot 5^{n-2} + 5^{n-1}).\n\\end{aligned}\n$$\nОскільки $25 - 18 = 12 - 5 = 7$, то $a_n$ ділиться без остачі на $7$. Аналогічно,\n$$\n\\begin{aligned}\na_n &= 25^n + 5^n - 18^n - 12^n = (25^n - 12^n) - (18^n - 5^n) = \\\\\n&= (25 - 12)(25^{n-1} + 25^{n-2} \\cdot 12 + \\dots + 25 \\cdot 12^{n-2} + 12^{n-1}) - \\\\\n&\\quad -(18 - 5)(18^{n-1} + 18^{n-2} \\cdot 5 + \\dots + 18 \\cdot 5^{n-2} + 5^{n-1}),\n\\end{aligned}\n$$\nтому $a_n$ ділиться без остачі на $13$. Оскільки $7$ і $13$ взаємно прості, то $a_n$ ділиться без остачі на $7 \\cdot 13 = 91$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play the following game. They start with a pile of $2012$ pebbles and take some amounts of pebbles by turns. The player starts can take $1$ or $4$ pebbles each turn at his discretion, another player can take $1$ or $3$ each turn. The player unable to make his turn loses. Which player has a winning strategy?",
    "options": [],
    "answer": "First player",
    "solution": "Перший гравець може забезпечити собі перемогу, узявши першим ходом чотири камінці, а всіма наступними — по одному камінцю. Тоді кількість узятих загалом камінців буде парною лише після ходів першого гравця, а перемагає той гравець, після ходу якого ця величина стане рівною $2012$.\n\n*Відповідь:* Перший гравець.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45335,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of an acute non-isosceles triangle $ABC$. Lines $BO$ and $CO$ meet sides $AC$ and $AB$ respectively at points $K$ and $N$. Points $P$ and $T$, different from $K$ and $N$, are chosen respectively on $AC$ and $AB$ so that $OK = OP$ and $ON = OT$. A line through $P$ parallel to $BK$ and a line through $T$ parallel to $CN$ meet at point $M$. Prove that the circumradii of triangles $AMB$, $BMC$, $CMA$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай $H$ — ортоцентр трикутника $ABC$, точка $D$ симетрична $H$ відносно прямої $AC$. Як відомо, точка $D$ лежить на описаному кілі трикутника $ABC$. Позначимо через $Q$ точку перетину відрізків $OD$ і $AC$. Тоді маємо: $\\angle QDH = \\angle QHD = \\angle OBD$. Звідси випливає, що $HQ \\parallel BK$, і $\\angle BKC = \\angle HQC$. Отже, $\\angle HQC = \\angle DQC = \\angle OQK = \\angle BKC$, а тому точки $P$ і $Q$ співпадають. Ми довели, що пряма, проведена через точку $P$ паралельно $BK$, проходить через точку $H$. Аналогічно доводиться, що точка $H$ лежить і на прямій, що проходить через точку $T$ паралельно $CN$. Відтак, точка $M$ з умови задачі є ортоцентром трикутника $ABC$. Рівність радіусів описаних кіл трикутників $AMB, BMC$ і $CMA$ є наслідком властивостей кола дев'яти точок (названі трикутники та трикутник $ABC$ мають спільне коло дев'яти точок, а радіус кола дев'яти точок будь-якого трикутника вдвічі менший за радіус його описаного кола). До того ж, рівність радіусів описаних кіл трикутників $ABC, AMB, BMC$ і $CMA$ легко встановлюється за допомогою узагальненої теореми синусів.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S(a)$ denote the sum of decimal digits of a positive integer $a$. A positive integer $n$ is such that $S(n) = 503$, $S(121n) = 2012$. Find all possible values of $S(11n)$.",
    "options": [],
    "answer": "1006",
    "solution": "Зауважимо, що $121n$ — це сума одного числа $100n$, двох чисел $10n$ і одного числа $n$. Якщо додавати ці числа в стовпчик, то в кожному розряді суми, починаючи з другого і закінчуючи передостаннім, буде сума двох цифр попереднього розряду, однієї цифри цього розряду і однієї цифри наступного розряду числа $n$. Якщо при цьому немає перенесень у наступні розряди, то кожна цифра числа $n$ додається 4 рази, і сума цифр числа $121n$ дорівнює $4S(n) = 4 \\cdot 503 = 2012$. Якщо при цьому є хоча б одне перенесення, то сума цифр числа $121n$, очевидно, зменшиться, тобто в цьому випадку $S(121n) < 4S(n) = 2012$. Оскільки за умовою задачі $S(121n) = 2012$, то перенесень немає, і всі цифри числа $n$ не перевищують 4. Отже, додавання чисел $10n$ та $n$ у стовпчик теж відбудеться без перенесень, а тому $S(11n) = s(10n + n) = 2S(n) = 1006$.\n\nВідповідь: $S(11n) = 1006$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to fill in a table $7 \\times 7$ with $24$ ones and $25$ zeros so that:\n1) for each cell containing one, sum of numbers in neighbour cells equals one,\n2) for each cell containing zero, sum of numbers in neighbour cells does not equal one?\n(Two cells are neighbours if they have a common side.)",
    "options": [],
    "answer": "Yes",
    "solution": "**Answer:** Yes, it is possible (see the figure).\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of $1005$ citizens of the Emerald city wears glasses (with two lenses). The lenses in the glasses are colored rose, blue and green evenly: in total there are $670$ lenses of each color. All citizens composed a circle so that no two neighbors in this circle have lenses of a common color. What is the maximum possible number of citizens wearing glasses with lenses of different color?",
    "options": [],
    "answer": "501",
    "solution": "Будемо позначати кольори лінз літерами Р, Б, З. Зрозуміло, що в колі не можуть стояти поруч мешканці з різнокольоровими окулярами, тому таких мешканців не більше за $502$. Якщо їх $502$, то рівно двоє людей з однокольоровими окулярами стоять поруч (нехай це мешканці з лінзами РР та ББ), а далі по колу «різнокольорові» та «однокольорові» мешканці чергуються. Ліворуч та праворуч від зазначеної пари отримуємо кольори\n\n..., РР, БЗ, РР, БЗ, РР, ББ, РЗ, ББ, РЗ, ББ, ...,\n\nі при замиканні кола з'явиться пара сусідів або з лінзами однакового кольору, або ж — з однокольоровими окулярами в обох. Тому кількість $502$ є неможливою. Приклад з $501$ «різнокольоровим» мешканцем і вказаною кількістю лінз можна побудувати за схемою:\n\nРР, БЗ, РР, БЗ, ..., БЗ, РР, ББ, РЗ, ББ, РЗ, ..., РЗ, ББ, ЗЗ, РБ, ЗЗ, РБ, ..., РБ, ЗЗ,\n\nде останній, $1005$-ий, мешканець стає поруч із першим, а між виділеними мешканцями стоять по $333$ особи.\n\nВідповідь: $501$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 1$ be integer. Sequence $a_1, a_2, \\dots, a_{2n}$ is called lucky if\n\n1) $a_1, \\dots, a_n$ is a permutation of $\\{1, 2, \\dots, n\\}$;\n\n2) $a_k = a_{n+k}$ for all $k = 1, 2, \\dots, n$;\n\n3) there exist indices $1 \\le i_1 < i_2 < \\dots < i_n \\le 2n$ such that $a_{i_k} = k$ for all $k = 1, 2, \\dots, n$.\n\nFor each $n \\geq 1$ find the number of lucky sequences.",
    "options": [],
    "answer": "2^n - n",
    "solution": "Через $m$, $0 \\le m \\le n$, позначимо кількість чисел множини $\\{1, 2, \\dots, n\\}$, які знаходяться серед індексів $i_1 < i_2 < \\dots < i_n$. Неважко довести, що такими $m$ індексами визначаються $n - m$ чисел із множини $\\{n+1, n+2, \\dots, 2n\\}$, які увійдуть до даного набору індексів $i_1 < i_2 < \\dots < i_n$. Цим повністю описуються всілякі вдалі послідовності $a_1, a_2, \\dots, a_{2n}$. Для $m = 0$ маємо вдалу послідовність $1, 2, \\dots, n$, $1, 2, \\dots, n$. Але ж ця сама вдала послідовність утворюється ще в $n$ випадках, коли для набору індексів $i_1 < i_2 < \\dots < i_n$ маємо, що\n$$\n\\{i_1, i_2, \\dots, i_n\\} \\cap \\{1, 2, \\dots, n\\} = \\{1, 2, \\dots, m\\}, \\quad 1 \\le m \\le n.\n$$\nУсіляких підмножин $n$-елементної множини $\\{1, 2, \\dots, n\\}$, відмінних від підмножин $\\{1, 2, ..., m\\}$, $1 \\le m \\le n$, існує $2^n-n$. Усім таким підмножинам відповідають різні вдалі послідовності (порожня підмножина відповідає випадку $m=0$ і також враховується серед $2^n-n$ підмножин).\nВідповідь: $2^n-n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 45340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a point $A$ lay outside a given circle $\\omega$. Through $A$ two lines are drawn, the first intersect $\\omega$ at $B$ and $C$, the second, at $D$ and $E$ ($D$ is between $A$ and $E$). The line going through $D$ and parallel to $BC$ intersects $\\omega$ at $F \\neq D$, and the line $AF$ intersects $\\omega$ at $T \\neq F$. Let $BC$ and $ET$ intersect at $M$, $N$ be symmetric to $A$ with respect to $M$, and $K$ be the midpoint of $BC$. Prove that the points $D$, $E$, $K$, $N$ are cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Оскільки $DF \\parallel BC$, то $\\angle DFA = \\angle CAF$. До того ж, $\\angle DFA = \\angle DFT = \\angle DET$ як вписані, що спираються на одну дугу. Отже, $\\triangle AMT \\sim \\triangle EMA$ за двома кутами, звідки $\\frac{AM}{MT} = \\frac{EM}{AM}$, тобто $AM^2 = EM \\cdot MT$. За властивістю січних, $ME \\cdot MT = MB \\cdot MC$. Тому\n$$ AM^2 = MB \\cdot MC = (AB - AM) \\cdot (AC - AM) = AB \\cdot AC - AM \\cdot (AB + AC) + AM^2. $$\nЗвідси $AB \\cdot AC = AM \\cdot (AB + AC)$. Оскільки $K$ — середина $BC$, то $AB + AC = 2AK$. За властивістю січних маємо $AB \\cdot AC = AD \\cdot AE$. Отже, $AD \\cdot AE = AM \\cdot 2AK$. Оскільки $M$ — середина $AN$, то останню рівність можна записати у вигляді $AD \\cdot AE = AN \\cdot AK$. Тому точки $D, E, K$ і $N$ лежать на одному колі.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inscribed circle of triangle *ABC* touches its sides *AB*, *BC*, *CA* at the points *K*, *N*, *M* respectively. Given that $\\angle MKC = \\angle MNA$, prove that triangle *ABC* is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай прямі $AN$ і $CK$ вдруге перетинають вписане коло трикутника $ABC$ в точках $P$ і $Q$ відповідно. За теоремами про вписаний кут та кут між дотичною й хордою одержуємо:\n$$\n\\angle PNM = \\angle PQM = \\angle PMA,\n$$\n$$\n\\angle QKM = \\angle QPM = \\angle QMC.\n$$\nЗа умовою задачі, $\\angle PNM = \\angle QKM$, а тому $PQ \\parallel AC$. Звідси випливає, що $\\angle CAN = \\angle QPN = \\angle QKN = \\angle CKN$. Тобто $\\angle CAN = \\angle CKN$. Це означає, що навколо чотирикутника $AKNC$ можна описати коло, і $\\angle CAK = \\angle BNK$, $\\angle ACN = \\angle BKN$. Трикутник $KBN$ є рівнобедреним, тобто $\\angle BNK = \\angle BKN$. Отже, $\\angle CAK = \\angle ACN$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in [0, 1]$. Prove that\n$$\n\\frac{1}{1+a+b} + \\frac{1}{1+b+c} + \\frac{1}{1+c+a} + a+b+c \\le 3 + \\frac{1}{3}(ab+bc+ca).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Доведемо допоміжну нерівність\n$$\n\\frac{1}{1+a+b} \\le 1 - \\frac{a+b}{2} + \\frac{ab}{3}.\n$$\nЦя нерівність рівносильна таким нерівностям:\n$$\n6 \\le (1+a+b)(6-3(a+b)+2ab),\n$$\n$$\n6 \\le 6 + 6(a+b) - 3(a+b) - 3(a+b)^2 + 2ab + 2ab(a+b),\n$$\n$$\n2a^2b + 2ab^2 - 3a^2 - 3b^2 - 4ab + 3a + 3b \\ge 0,\n$$\n$$\n2a(a-1)(b-1) + 2b(a-1)(b-1) + a(1-a) + b(1-b) \\ge 0.\n$$\nОстання нерівність є правильною, бо $a, b \\in [0, 1]$.\n\nДодаючи цю та ще дві аналогічні нерівності, дістанемо, що\n$$\n\\frac{1}{1+a+b} + \\frac{1}{1+b+c} + \\frac{1}{1+c+a} \\le 3 - (a+b+c) + \\frac{1}{3}(ab+bc+ca),\n$$\nзвідки й випливає потрібна нерівність.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45343,
    "subject": "Mathematics (Multi-modal)",
    "question": "A real number sequence $a_1, a_2, \\dots, a_n, \\dots$ is such that $a_1 = 1$, $a_2 = 9$ and $a_{n+2} = 14a_{n+1} - a_n - 4$ for all positive integers $n$. Prove that for each positive integer $n$ the number $a_n$ is a square of integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "З умови задачі випливає, що всі члени даної послідовності є цілими числами, причому $a_3 = 14 \\cdot 9 - 1 - 4 = 121$. Доведемо індукцією за $n \\ge 2$, що $a_{n-1}a_{n+1} = (a_n + 2)^2$. Справді, для $n = 2$ маємо: $a_1a_3 = 1 \\cdot 121 = (9+2)^2 = (a_n + 2)^2$. Припустимо тепер, що $a_{n-1}a_{n+1} = (a_n + 2)^2$. Тоді\n$$\n\\begin{aligned}\na_n a_{n+2} - (a_{n+1} + 2)^2 &= a_n (14a_{n+1} - a_n - 4) - (a_{n+1} + 2)^2 \\\\\n&= 14a_n a_{n+1} - a_n^2 - 4a_n - a_{n+1}^2 - 4a_{n+1} - 4 \\\\\n&= a_{n+1} (14a_n - a_{n+1} - 4) - (a_n + 2)^2 \\\\\n&= a_{n-1} a_{n+1} - (a_n + 2)^2 = 0.\n\\end{aligned}\n$$\nЗа принципом математичної індукції, $a_{n-1}a_{n+1} = (a_n + 2)^2$ для всіх $n \\ge 2$, тобто для всіх $n \\ge 2$ добуток $a_{n-1}a_{n+1}$ є квадратом цілого числа. Якщо $a_{n-1}$ — квадрат цілого числа і $a_{n-1}a_{n+1}$ — квадрат цілого числа, то і $a_{n+1}$ — квадрат цілого числа. Оскільки $a_1 = 1$ і $a_2 = 9$ — квадрати цілих чисел, за індукцією одержуємо, що $a_n$ — квадрат цілого числа для кожного натурального $n$, що й треба було довести.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to paint each cell of an $8 \\times 8$ table with one of $16$ colors so that for each two colors there are two cells painted with these colors and having a common side?",
    "options": [],
    "answer": "Not possible",
    "solution": "У таблиці є $112$ одиничних відрізків, що розділяють її сусідні клітинки. Але з $16$ даних кольорів можна утворити $120$ пар кольорів, і для кожної з цих пар повинні знайтись дві сусідні клітинки, що неможливо.\n\n*Відповідь:* Не можна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45345,
    "subject": "Mathematics (Multi-modal)",
    "question": "In Nottingham city, 100 riches reside, and each of them owns at least one million of gold coins. Robin Hood develops plans to raid them. Each plan consists in stealing same number of gold coins (not exceeding one million) from each of the riches. Prove that, no matter what wealths of the riches are, Robin Hood can develop at least 100 raiding plans such that for two different plans $P'$ and $P''$ there no two riches $B'$ and $B''$ such that $B'$ would have the same number of gold coins after the raiding plan $P'$ as $B''$ after the raiding plan $P''$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Без обмеження загальності можна вважати, що всі багатії мають різну кількість монет. Тоді задачу можна переформулювати так: нехай $A$ — довільна 100-елементна множина з натуральних чисел, кожне з яких не менше від мільйона. Тоді знайдуться такі різні числа $t_1, t_2, \\dots, t_{100}$, які не перевищують мільйона, що множини $A_i = A - t_i = \\{x - t_i \\mid x \\in A\\}$ попарно не перетинаються.\n\nНехай $D = \\{x - y \\mid x, y \\in A\\}$. Очевидно, що умова $A_i \\cap A_j = \\emptyset$ рівносильна тому, що $t_i - t_j \\notin D$. Помітимо, що в множині $D$ не більше за $100^2 = 10^4$ елементів. Будемо вибирати числа $t_i$ послідовно. Число $t_1$ виберемо довільно. Якщо числа $t_1, t_2, \\dots, t_k$ вибрано, то число $t_{k+1}$ потрібно вибрати так, щоб воно не лежало в жодній з множин $t_1 + D, t_2 + D, \\dots, t_k + D$. Для $k < 100$ такий вибір можливий, оскільки загалом у цих множинах не більше за $99 \\cdot 10^4 < 10^6$ елементів.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play the following game. They start with the polynomial $2013x^2 + 2012x + 2011$ and take moves by turns. During each move a player subtracts from the polynomial one of the following polynomials: $x^2$, $x$, $x^2 - x + 1$ or $x^2 + x - 1$, at his choice. If after a player's move the polynomial has an integer root, he loses. Which of the players has a winning strategy?",
    "options": [],
    "answer": "first player",
    "solution": "Доведемо, що виграє перший гравець. Він може ходити так, щоб після його ходу вільний член одержаного многочлена був непарним, а коефіцієнти при $x^2$ та $x$ мали однакову парність (для цього він може спочатку відняти від многочлена $x^2$, а далі повторювати ходи другого гравця). Тоді після ходів першого гравця будуть з'являтися квадратні тричлени, які в цілих точках приймають лише непарні значення, а тому не мають цілих коренів. Отже, перший гравець не програє. Зазначимо, що з кожним ходом сума коефіцієнтів одержаного многочлена зменшується на 1. Через декілька ходів ця сума стане рівною 0, а тоді одержаний многочлен матиме корінь $x = 1$. Відтак, гра гарантовано закінчиться і виграє перший гравець.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $SABC$ be a tetrahedron with $MA > 1$, $MB > 1$, $MC > 1$, where $M$ is the centroid of $ABC$. Prove that\n$$\nSA + SB + SC > 3.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Розглянемо вектори $\\overrightarrow{MA} = \\vec{a}$, $\\overrightarrow{MB} = \\vec{b}$, $\\overrightarrow{MC} = \\vec{c}$, $\\overrightarrow{MS} = \\vec{s}$, $\\overrightarrow{SA} = \\vec{a} - \\vec{s}$, $\\overrightarrow{SB} = \\vec{b} - \\vec{s}$ і $\\overrightarrow{SC} = \\vec{c} - \\vec{s}$. За умовою задачі, $|\\vec{a}| > 1$, $|\\vec{b}| > 1$ і $|\\vec{c}| > 1$. Нам потрібно довести нерівність\n$$\n|\\vec{a} - \\vec{s}| + |\\vec{b} - \\vec{s}| + |\\vec{c} - \\vec{s}| > 3.\n$$\nНехай $|\\vec{s}| \\ge 1$. Тоді, використовуючи відому рівність $\\vec{a} + \\vec{b} + \\vec{c} = \\vec{0}$ та властивості скалярного добутку, маємо:\n$$\n\\begin{aligned}\n|\\vec{a} - \\vec{s}| + |\\vec{b} - \\vec{s}| + |\\vec{c} - \\vec{s}| &\\geq -\\frac{(\\vec{a} - \\vec{s})\\vec{s}}{|\\vec{s}|} - \\frac{(\\vec{b} - \\vec{s})\\vec{s}}{|\\vec{s}|} - \\frac{(\\vec{a} - \\vec{s})\\vec{s}}{|\\vec{s}|} = \\\\\n&= 3|\\vec{s}| - \\frac{\\vec{s}}{|\\vec{s}|} (\\vec{a} + \\vec{b} + \\vec{c}) = 3|\\vec{s}| \\geq 3.\n\\end{aligned}\n$$\nРівність у цій нерівності досягається тоді й тільки тоді, коли $|\\vec{s}| = 1$ і вектори $\\overrightarrow{SA}$, $\\overrightarrow{SB}$ і $\\overrightarrow{SC}$ протилежно напрямлені до вектора $\\overrightarrow{SM}$, що, зрозуміло, неможливо.\n\nРозглянемо тепер випадок $|\\vec{s}| < 1$. Тоді\n$$\n\\begin{aligned}\n& |\\vec{a} - \\vec{s}| + |\\vec{b} - \\vec{s}| + |\\vec{c} - \\vec{s}| \\ge \\frac{(\\vec{a} - \\vec{s}) \\cdot \\vec{a}}{|\\vec{a}|} + \\frac{(\\vec{b} - \\vec{s}) \\cdot \\vec{b}}{|\\vec{b}|} + \\frac{(\\vec{c} - \\vec{s}) \\cdot \\vec{c}}{|\\vec{c}|} = \\\\\n& = |\\vec{a}| + |\\vec{b}| + |\\vec{c}| - \\vec{s} \\left( \\frac{\\vec{a}}{|\\vec{a}|} + \\frac{\\vec{b}}{|\\vec{b}|} + \\frac{\\vec{c}}{|\\vec{c}|} \\right) = \\\\\n& = 3 + \\left( |\\vec{a}| - 1 \\right) \\left( 1 - \\frac{\\vec{s} \\cdot \\vec{a}}{|\\vec{a}|} \\right) + \\left( |\\vec{b}| - 1 \\right) \\left( 1 - \\frac{\\vec{s} \\cdot \\vec{b}}{|\\vec{b}|} \\right) + \\left( |\\vec{c}| - 1 \\right) \\left( 1 - \\frac{\\vec{s} \\cdot \\vec{c}}{|\\vec{c}|} \\right) > 3.\n\\end{aligned}\n$$\nДля останньої оцінки ми врахували, що $1 > |\\vec{s}| \\ge \\frac{\\vec{s} \\cdot \\vec{a}}{|\\vec{a}|}$, $1 > |\\vec{s}| \\ge \\frac{\\vec{s} \\cdot \\vec{b}}{|\\vec{b}|}$, $1 > |\\vec{s}| \\ge \\frac{\\vec{s} \\cdot \\vec{c}}{|\\vec{c}|}$.\nЯкщо застосувати ортогональне проектування точки $S$ на площину $ABC$, то стає очевидним, що твердження задачі достатньо довести для точки $S$ цієї площини, для якої сума відстаней $SA + SB + SC$ є мінімальною. Добре відомо, що коли всі кути трикутника $ABC$ менші за $120^\\circ$, то потрібною точкою є точка Торічеллі (така точка $T$ всередині трикутника $ABC$, що $\\angle ATB = \\angle ATC = \\angle BTC = 120^\\circ$), а якщо, наприклад, $\\angle BAC \\ge 120^\\circ$, то — вершина $A$. У першому випадку позначимо $x = TA$, $y = TB$, $z = TC$ (без обмеження загальності вважаємо, що $x \\le y \\le z$), і з використанням формули для довжини медіани та теореми косинусів матимемо:\n$$\n\\begin{aligned}\nMA^2 &= \\frac{1}{9}(2AC^2 + 2AB^2 - BC^2) = \\\\\n&= \\frac{1}{9}(2x^2 + 2z^2 + 2xz + 2x^2 + 2y^2 + 2xy - y^2 - z^2 - yz) = \\\\\n&= \\frac{1}{9}(4x^2 + y^2 - z^2 + 2xz + 2xy - yz) = \\\\\n&= \\frac{1}{9}((x + y + z)^2 + 3x^2 - 2z^2 - 3yz) \\le \\frac{1}{9}(x + y + z)^2.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45348,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides $AB$ and $AC$ of an acute triangle $ABC$ triangles $ABC_1$ and $AB_1C$ are constructed outside with $\\angle BAC_1 = \\angle B_1AC = 30^\\circ$, $\\angle AC_1B = \\angle CB_1A = 60^\\circ$. A point $A_1$ inside $ABC$ is such that $\\angle CBA_1 = \\angle BCA_1 = 30^\\circ$. Prove that the points $A_1$, $B_1$ and $C_1$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Побудуємо на стороні $BC$ даного трикутника зовні нього рівносторонній трикутник $BCA_2$. Унаслідок умови задачі, $\\angle ABC_1 = \\angle ACB_1 = 90^\\circ$ і $\\angle A_1BA_2 = \\angle A_1CA_2 = 90^\\circ$, а внаслідок рівності трикутників $A_1BA_2$ і $A_1CA_2$ (за двома катетами) маємо: $\\angle A_1A_2B = \\angle A_1A_2C = 30^\\circ$. З подібності трикутників $ABC_1$ і $A_2BA_1$ (за двома кутами) випливає, що $\\frac{BC_1}{BA_1} = \\frac{BA}{BA_2}$, звідки $\\frac{BA}{BC_1} = \\frac{BA_2}{BA_1}$, і $\\angle ABA_2 = \\angle C_1BA_1$. Тому трикутники $C_1BA_1$ і $ABA_2$ подібні (за двома сторонами і кутом між ними). Звідси $\\angle C_1A_1B = \\angle AA_2B$. Аналогічно, $\\angle B_1A_1C = \\angle AA_2C$, тому\n![](attached_image_1.png)\n$$\n\\angle C_1A_1B + \\angle CA_1B_1 = \\angle AA_2B + \\angle AA_2C = \\angle BA_2C = 60^\\circ.\n$$\nВідтак,\n$$\n\\angle C_1A_1B_1 = \\angle C_1A_1B + \\angle BA_1C + \\angle CA_1B_1 = 120^\\circ + 60^\\circ = 180^\\circ.\n$$\nОтже, точки $A_1$, $B_1$ і $C_1$ лежать на одній прямій.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive numbers $x$ and $y$ satisfy equation $x^2 + y^2 + \\frac{8xy}{x+y} = 16$. Prove that $x + y = 4$.",
    "options": [],
    "answer": "4",
    "solution": "$$(x^2 + y^2)(x + y) + 8xy = 16(x + y),$$\n$$( (x + y)^2 - 2xy )(x + y) - 16(x + y) + 8xy = 0, $$\n$$(x + y)^3 - 16(x + y) - 2xy(x + y) + 8xy = 0,$$\n$$(x + y)((x + y)^2 - 16) - 2xy(x + y - 4) = 0,$$\n$$(x + y)(x + y - 4)(x + y + 4) - 2xy(x + y - 4) = 0,$$\n$$(x + y - 4)((x + y)(x + y + 4) - 2xy) = 0.$$\n\nFor $x > 0$ and $y > 0$\n$$(x + y)(x + y + 4) - 2xy = x^2 + y^2 + 4(x + y) > 0.$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45350,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides of an acute triangle $ABC$ three triangles $A'BC$, $AB'C$, $ABC'$ are constructed outside with $\\angle ABC' = \\angle A'BC = \\angle B'AC = 30^\\circ$ and $\\angle BAC' = \\angle AB'C = \\angle A'CB = 90^\\circ$. Prove that $A'C' \\perp B'M$, where $M$ is the midpoint of $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Лема.** На сторонах гострокутного трикутника як на основах побудовано рівнобедрені трикутники $АХВ$, $BYC$, $CZA$, для яких $\\angle АХВ = \\angle ВҮС = 120^\\circ$, $\\angle СZA = 60^\\circ$. Тоді $XY \\perp BZ$.\n\n**Доведення.** Див. задачу 4 для 8 класу. $\\Box$\n\nПерейдемо до розв'язування задачі. Нехай $X$ і $Y$ — середини сторін $BC'$ і $BA'$ відповідно, а $Z$ — точка, симетрична точці $C$ відносно $B'$. Тоді $A'C' \\parallel XY$, $B'M \\parallel BZ$.\n\nМедіана прямокутного трикутника дорівнює половині гіпотенузи, тому $XA = XB$ і $YB = YC$. До того ж, $\\angle АХВ = \\angle ВҮС = 180^\\circ - (30^\\circ + 30^\\circ) = 120^\\circ$, а трикутник $AZC$ правильний. Тому за лемою, $BZ \\perp XY$, звідки $B'M \\perp A'C'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integers are written on $2012$ cards, one number on each side. On the first card, numbers $1$ and $2$ are written, on the second, $3$ and $4$, ..., on the $2012$th, $4023$ and $4024$. All cards are arranged in line along a table. Two players play the following game. During each move a player chooses $5$ cards and swaps them. If after his move the set of numbers face up is the same as after one of previous turns, he loses. Which of the players has a winning strategy?",
    "options": [],
    "answer": "First player",
    "solution": "Доведемо, що виграє перший гравець. Для цього йому достатньо розбити всі можливі способи розташування чисел на пари, узявши до однієї пари два розташування, які відрізняються одне від одного лише зміною перших п'яти чисел розташувань. Тоді виграшна стратегія першого гравця полягає в тому, щоб завжди перевертати лише перші п'ять карток. Після першого ходу він створює розташування чисел, яке утворює пару з початковим розташуванням чисел. Далі, кожний наступний хід другого гравця буде створювати розташування чисел, яке не зустрічалося раніше (інакше він програє) і входить до нової пари розташувань, а наступним ходом перший гравець створює розташування чисел, яке також не зустрічалося раніше і є другою компонентою цієї пари. Оскільки усіх можливих розташувань чисел скінченна кількість (а саме, $2^{2012}$), то в деякий момент другий гравець уже не зможе створити розташування цих чисел, яке раніше не зустрічалося.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45352,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the coordinate lines points with coordinates $1, 2, \\ldots, 2n$ are marked, where $n > 3$ is a given integer. A flee starts jumping from the point with coordinate $1$ and after $2n$ jumps returns there having visited all marked points. It is known that the total length of all jumps except the last one is $n(2n - 1)$. Find the length of the last flee's jump.",
    "options": [],
    "answer": "n",
    "solution": "Нехай блоха послідовно побувала в точках:\n$$\na_1 = 1, \\ a_2, \\ \\dots, \\ a_{2n}.\n$$\nСума довжин усіх стрибків становить:\n$$\nS = |a_1 - a_2| + |a_2 - a_3| + \\dots + |a_{2n-1} - a_{2n}| + |a_{2n} - a_1|.\n$$\n\nЗрозуміло, що\n$$S \\le 2(2n + 2n - 1 + \\dots + n + 2 + n + 1) - 2(n + n - 1 + \\dots + 2 + 1) = 2n^2,$$\nа оскільки за умовою $|a_1 - a_2| + |a_2 - a_3| + \\dots + |a_{2n-1} - a_{2n}| = n(2n - 1)$, то довжина останнього стрибка $|a_{2n} - a_1|$ не перевищує $n$.\nОскільки $a_1 = 1$, то $a_{2n} \\le n + 1$. Доведемо, що $a_{2n} = n + 1$, тобто довжина останнього стрибка дорівнює $n$. Припустимо, що $a_{2n} \\le n$. Тоді в послідовності (1) знайдуться два сусідні члени $a_i$ та $a_{i+1}$, кожний з яких більший за $n$. Справді, якщо це не так, то в послідовності (2.1) між $n$ елементами, більшими за $n$, містяться принаймні $n-1$ менших за $n$ елементів, а разом з $a_1 = 1$ та $a_{2n} \\le n$ у (2.1) знайдуться $n+1$ елементів, менших за $n$, що неможливо. Отже, у послідовності (2.1) знайдуться два сусідні члени $a_i$ та $a_{i+1}$, кожний із яких більший за $n$. Перебудуємо послідовність стрибків блохи:\n$a_1, a_2, \\dots, a_i, a_{2n}, a_{2n-1}, \\dots, a_{i+1}.$\nДля нової послідовності сума стрибків дорівнює\n$$\n\\begin{aligned}\nS' &= |a_1 - a_2| + \\dots + |a_i - a_{2n}| + |a_{2n} - a_{2n-1}| + \\dots + |a_{i+1} - a_1| = \\\\\n&= n(2n-1) - |a_i - a_{i+1}| + |a_i - a_{2n}| + |a_{i+1} - 1| \\le 2n^2.\n\\end{aligned}\n$$\nПроте $1 \\le a_{2n} \\le n < a_i < a_{i+1}$ або $1 \\le a_{2n} \\le n < a_{i+1} < a_i$, і тому неважко перевірити, що $|a_i - a_{2n}| + |a_{i+1} - 1| - |a_i - a_{i+1}| > n$, звідки $S' > 2n^2$. Маємо суперечність.\nВідповідь: Довжина останнього стрибка дорівнює $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive rational numbers $x$ and $y$ such that both $x + y^{-1}$ and $y + x^{-1}$ are integers.",
    "options": [],
    "answer": "(x, y) ∈ {(1, 1), (1/2, 2), (2, 1/2)}",
    "solution": "Нехай $x = \\frac{m}{n}$, $y = \\frac{p}{q}$, де $m, n, p, q$ — натуральні числа, причому $(m; n) = (p; q) = 1$. Оскільки $x + y^{-1} = \\frac{m}{n} + \\frac{q}{p} = \\frac{mp + nq}{np}$ має бути цілим, то $np \\mid mp + nq$. Аналогічно, $y + x^{-1} = \\frac{p}{q} + \\frac{n}{m} = \\frac{pm + nq}{qm}$ має бути цілим, тобто $qm \\mid pm + nq$.\n\nОскільки $np \\mid mp + nq$ і $qm \\mid pm + nq$, а тому $p \\mid n$ і $n \\mid p$. Маємо, що $n = p$. Аналогічно доводиться, що $m = q$. Звідси тепер випливає, що $2m \\mid n$ і $2n \\mid m$. Отже, $m, n, p, q \\in \\{1; 2\\}$. Тепер уже неважко одержати відповідь.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45354,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\begin{cases} (x+y)(1+xy) + (x-y)^2 = 2, \\\\ x^3 + y^3 = 1. \\end{cases}\n$$",
    "options": [],
    "answer": "(1, 0), (0, 1)",
    "solution": "Перше рівняння системи запишемо, з урахуванням рівності $x^3 + y^3 = 1$, у вигляді $(x + y)(1 + xy) + (x - y)^2 - (x + y)(x^2 - xy + y^2) - 1 = 0$, тобто $(x + y - 1)(1 - (x - y)^2) = 0$. Подальші міркування є очевидними.\n\nВідповідь: $(1; 0)$, $(0; 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $x + \\sqrt{1-x} + 1 = \\sqrt{x} + 3\\sqrt{x - x^2}$.",
    "options": [],
    "answer": "1/2, 16/25",
    "solution": "Областю допустимих значень рівняння є відрізок $[0; 1]$. Запишемо рівняння у вигляді\n$$\n(\\sqrt{x})^2 + \\sqrt{1-x} + (\\sqrt{x})^2 + (\\sqrt{1-x})^2 - \\sqrt{x} - 3\\sqrt{x}\\sqrt{1-x} = 0.\n$$\nРозкладаючи ліву частину рівняння на множники, одержуємо:\n$$\n(\\sqrt{x} - \\sqrt{1-x})(2\\sqrt{x} - \\sqrt{1-x} - 1) = 0.\n$$\nЗалишається розв'язати стандартними методами рівняння\n$$\n\\sqrt{x} - \\sqrt{1-x} = 0, \\quad 2\\sqrt{x} = \\sqrt{1-x} + 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive numbers $a$ and $b$ satisfy the equality $a + b + a^{-1} + b^{-1} = 5$. Prove that $3\\sqrt{a + b} \\ge a + b + 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $u = a + b$ and $v = \\frac{1}{a} + \\frac{1}{b}$. Then $u + v = 5$. As is known, $uv \\ge 4$. Therefore, $5 = u + v \\ge u + \\frac{4}{u}$. Hence, $u^2 - 5u + 4 \\le 0$. Thus, $1 \\le \\sqrt{u} \\le 2$, that is, $(\\sqrt{u} - 1)(\\sqrt{u} - 2) \\le 0$, and $3\\sqrt{u} \\ge u + 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocenter of a triangle $ABC$, $M$ be the midpoint of $AB$, $N$ be the midpoint of $AC$. The rays $MH$ and $NH$ intersect the circumcircle of $ABC$ at points $P$ and $Q$ respectively. Prove that the lines $BQ$, $CP$ and $AH$ are either concurrent or parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AA_1$ be the altitude of triangle $ABC$. Consider the case when the lines $BQ$ and $CP$ intersect at some point $T$. We need to prove that the point $T$ lies on the line $AA_1$. Suppose first that points $A$ and $P$ lie on the same side of line $BC$, and points $A$ and $Q$ lie on different sides of line $BC$ (see figure).\n\nAs is known, the points symmetric to the orthocenter of a triangle with respect to the midpoints of its sides lie on the circumcircle of this triangle: indeed, if in triangle $ABC$ the point $F$ is symmetric to the orthocenter $H$ with respect to the midpoint $N$ of side $AC$, then $\\angle AFC = \\angle AHC = 180^\\circ - \\angle ABC$. Further, since $FC \\parallel AH$, we have $\\angle FCB = 90^\\circ$, and points $B$ and $F$ are diametrically opposite. Hence, $HQ \\perp BT$. Similarly, $HP \\perp CT$.\n\nThe quadrilaterals $HQT P$ and $BQ A_1 H$ are cyclic. Therefore, we have: $\\angle QBC = \\angle QH A_1$, $\\angle QHT = \\angle QPC$. Since $\\angle QPC = \\angle QBC$, it follows that $\\angle QHT = \\angle QH A_1$, which completes the proof.\n\nIf the chord $PQ$ and point $A$ lie on different sides of line $BC$, then $\\angle QBC = \\angle QH A_1$, $\\angle QHT = \\angle QPT$, $\\angle QPT = \\frac{\\angle QP}{2} + \\frac{\\angle PC}{2} = \\angle QBC$. Other cases of the arrangement of points $P$ and $Q$ on the circumcircle of triangle $ABC$ are considered similarly.\n\nIf the lines $BQ$ and $CP$ are parallel, it is not difficult to prove that the point $H$ lies on the segment $MN$, points $P$ and $F$ coincide, and therefore $AH \\parallel PC$.\nLet $\\omega$ be the circumcircle of triangle $ABC$, and $\\omega_B$ and $\\omega_C$ be the circles constructed on diameters $HB$ and $HC$ respectively. Since $\\angle HPC = 90^\\circ$, we have $P \\in \\omega_C$, and the line $CP$ is the radical axis of circles $\\omega$ and $\\omega_C$. Similarly, $BQ$ is the radical axis of circles $\\omega$ and $\\omega_B$. Further, the line $AA_1$ is the radical axis of circles $\\omega_B$ and $\\omega_C$. Therefore, the statement of the problem follows from the theorem on the radical axes of three circles.\n(This solution was proposed by olympiad participant Anna Mitrushchenkova.)\n\nThe case when points $P$ and $F$ coincide, i.e., $AH \\parallel PC$, was considered in Solution 1. Let $Y$ be such a point that $H$ is the midpoint of segment $AY$. Then $BY \\parallel PH$, and $CP \\perp BY$. Similarly, $BQ \\perp CY$. Moreover, $AH \\perp BC$. The required result follows from the concurrency of the altitudes of triangle $BCY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45358,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$, $\\angle C = 90^\\circ$, $AC < BC$. Point $K$ inside $BC$ is such that $CK = CA$, and point $D$ inside $CK$ is such that $\\angle DAK = \\angle BAK$. $F$ and $P$ are, respectively, the feet of perpendiculars from $D$ to $AB$ and from $A$ to $FK$. Prove that $CP = \\frac{1}{2}(AF + FD + DA)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Нехай пряма $DF$ перетинає прямі $AC$ і $AP$ в точках $N$ і $M$ відповідно (нескладно довести, що такі точки перетину існують). Нехай $\\angle BAC = \\alpha$, $\\alpha > 45^\\circ$, $\\angle BAK = \\angle KAD = \\delta$. Тоді $\\angle FDB = \\alpha$, $\\angle NDC = \\alpha$. Далі, $\\alpha - \\delta = 45^\\circ$, $90^\\circ - \\alpha + 2\\delta = \\alpha$, а тому $\\angle ADC = \\alpha$. Звідси випливає, що точка $C$ — середина відрізка $AN$.\n\nОскільки промінь $AK$ — бісектриса внутрішнього кута $A$ трикутника $FDA$, а промінь $DK$ — бісектриса його зовнішнього кута при вершині $D$, то промінь $FK$ є бісектрисою його зовнішнього кута при вершині $F$. Тому $FP$ — медіана трикутника $MFA$. Отже, $CP$ є середньою лінією трикутника $AMN$, і $CP = \\frac{1}{2}NM$. Залишається врахувати, що $ND = AD$ і $AF = FM$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $[x]$ denote the integer part of $x$ (i.e. the largest integer not exceeding $x$), $\\{x\\} = x - [x]$. Solve the equation $\\{x\\}^2 + 2\\{x\\} = 3x^2$.",
    "options": [],
    "answer": "{0, 1 - sqrt(10)/2}",
    "solution": "Оскільки $0 \\le \\{x\\} < 1$, то $x^2 < 1$, тобто $-1 < x < 1$.\n\nЯкщо $x \\in [0;1)$, то $x = \\{x\\}$, і знаходимо $x = 0$.\n\nДля $x \\in (-1;0)$ позначимо $u = \\{x\\}$, $0 < u < 1$, $[x] = -1$, $x = -1 + u$. Тоді з рівняння $2u^2 - 8u + 3 = 0$, з урахуванням нерівності $0 < u < 1$, одержуємо $u = 2 - \\frac{\\sqrt{10}}{2}$, $x = 1 - \\frac{\\sqrt{10}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-intersecting circles $\\omega_1$ and $\\omega_2$ of radii are inscribed into angle $BAC$ with\n\n$B \\in \\omega_1$, $C \\in \\omega_2$, and the radius of $\\omega_1$ is smaller than that of $\\omega_2$. Let $K \\neq B$ and $N \\neq C$ be the points of intersection of $BC$ with $\\omega_1$ and $\\omega_2$ respectively. Let also $P \\neq K$ and $M \\neq N$ be the points of intersection of $AK$ with $\\omega_1$ and of $AN$ with $\\omega_2$ respectively. Prove that $A$ and the circumcenters of triangles $ACM$ and $ABP$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Проведемо дотичну $AF$ до описаного кола трикутника $APB$. Тоді $\\angle PAF = \\angle PBA = \\angle PKB$, а тому $AF \\parallel BC$. Отже, $\\angle FAC = \\angle ACN = \\angle CMA$. З цього випливає, що пряма $AF$ дотикається до описаного кола трикутника $ACM$. Оскільки описані кола трикутників $АСМ$ і $АВР$ в точці $А$ мають спільну дотичну, то точка $А$ лежить на лінії центрів цих кіл.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers such that\n$$\n\\frac{a^4-1}{b+1} + \\frac{b^4-1}{a+1}\n$$\nis integer. Prove that $a^{2010}b^{2012} - 1$ is divisible by $a+1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Якщо виконується умова задачі, і принаймні одне з чисел $a, b$ дорівнює 1, то твердження є очевидним. Нехай $a > 1$ і $b > 1$. Позначимо $\\frac{a^4-1}{b+1} = \\frac{x}{y}$, $\\frac{b^4-1}{a+1} = \\frac{z}{t}$, де $x, y, z, t$ — натуральні числа, причому $(x;y) = (z;t) = 1$. За умовою задачі, сума $\\frac{x}{y} + \\frac{z}{t} = m$ є цілим числом, тобто $xt + yz = myt$. З цієї рівності випливає, що $y:t$ і $t:y$. Отже, $y = t$. Оскільки добуток\n$$\n\\frac{x}{y} \\cdot \\frac{z}{t} = \\frac{a^4 - 1}{b+1} \\cdot \\frac{b^4 - 1}{a+1} = (a-1)(b-1)(a^2+1)(b^2+1) = k\n$$\nтакож є цілим числом, то $y = t = 1$, бо $(x;y) = (z;t) = 1$. Звідси випливає, що $b^4 - 1$ ділиться без остачі на $a+1$. Звідси одержуємо, що $b^{2012} - 1 \\equiv a + 1 \\pmod{2012}$, а числа $b^{2012} - 1 \\equiv a^{2010} - 1 \\pmod{2010}$, і діляться без остачі на $a+1$ (ми враховуємо, що $a^{2010} - 1 = (a^2)^{1005} - 1$ ділиться без остачі на $a^2 - 1$), то $a^{2010}b^{2012} - 1 \\equiv a + 1$, що й треба було довести (можна також було скористатися й властивостями конгруенцій).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of triangle $ABC$. Points $M$ and $N$ are chosen on sides $AB$ and $AC$ respectively such that $M \\neq B$, $N \\neq C$ and the points $A$, $I$, $M$, $N$ are cyclic. Prove that $BM + CN = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Візьмемо на стороні $BC$ таку точку $K$, що $BM = BK$. Легко бачити, що $\\Delta BMI = \\Delta BKI$. Оскільки навколо чотирикутника $ANIM$ можна описати коло, і $\\angle MAI = \\angle NAI$, то $MI = NI = KI$. Зауважимо, що $\\angle BKI = \\angle BMI = 180^\\circ - \\angle AMI = \\angle ANI$, $\\angle CNI = \\angle CKI$. З того, що $\\angle NCI = \\angle KCI$, випливає рівність кутів $CIN$ і $CIK$. Отже, $\\Delta CNI = \\Delta CKI$, $CN = CK$, $BM + CN = BK + KC = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45363,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 2012 cities in the land of Olympia. Some of the cities are connected by direct two-way routes (each route connects exactly two cities and no two routes connect the same pair of cities). It is known that no city is directly connected to more than 8 other cities. Prove that it is possible to close at most 2012 routes in such a way that among any four cities there will be two with no direct route between them.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be the graph of the problem. Partition the set $V_G$ of its 2012 vertices into three subsets $A$, $B$, and $C$ so that the sum $S$ of the numbers of edges in the subgraphs $G(A)$, $G(B)$, and $G(C)$ is minimal (clearly, this is possible). Suppose that in one of the subsets (let it be $A$), there is a vertex $X$ such that in the subgraph $G(A)$ its degree $\\rho_{G(A)}(X) \\ge 3$. Since $\\rho_G(X) \\le 8$, the vertex $X$ cannot be connected to each of the subgraphs $G(B)$ and $G(C)$ by more than two edges. Suppose the vertex $X$ is connected to the subgraph $G(B)$ by at most two edges. Then, for the subgraphs $G(A\\setminus\\{X\\})$, $G(B\\cup\\{X\\})$, and $G(C)$, the sum $S$ decreases, which is impossible. Therefore, in the subgraphs $G(A)$, $G(B)$, and $G(C)$, the degree of all their vertices does not exceed 2. As is known, in any graph, the sum of the degrees of all vertices equals twice the number of edges, so in each of the subgraphs $G(A)$, $G(B)$, and $G(C)$, the number of edges does not exceed the number of vertices, that is, $S \\le 2012$. Remove all these $S$ edges (make the graphs with vertex sets $A$, $B$, and $C$ empty). For any four vertices, at least two lie in one of these three vertex sets, and after removing the edges, such two vertices are not connected by an edge.\n\n\nSolution 2:\n\nLet us prove the following auxiliary fact.\n\n**Lemma.** If in a graph $G$ with $n$ vertices the degree of each vertex does not exceed $p$, then all its vertices can be partitioned into $q \\le p+1$ groups so that in each group no two vertices are connected by an edge.\n\n**Proof.** We use induction on the order $n$ of the graph. The base case is obvious. Suppose the statement holds for $n = k$. Consider a graph $G$ with $k+1$ vertices that satisfies the lemma's condition. Let $X$ be any of its vertices. For the subgraph $G(V_G\\setminus\\{X\\})$, we can apply the induction hypothesis and partition the set $V_G\\setminus\\{X\\}$ of all its vertices as required into $q \\le p+1$ groups. Since $\\rho_G(X) \\le p$, the vertex $X$ is not connected by any edge to at least one of the formed groups. It remains to add $X$ to this group, and the lemma is proved. $\\square$\n\n![](attached_image_1.png)\n\nPartition — according to the lemma — the set of vertices of the graph $G$ of the problem into 9 groups (some may be empty). Call a bundle of edges all the edges that connect vertices of two different groups. In total, $\\frac{9 \\cdot 8}{2} = 36$ bundles of edges are formed (some bundles may contain no edges). Divide them into four groups of 9 bundles each as shown in the figures. In each group, the bundles of edges form three \"triangles\" that \"cover\" all 9 groups of vertices. In the graph $G$ there are at most $\\frac{2012 \\cdot 8}{2} = 8048$ edges, so in at least one of the four groups of bundles, the number of edges does not exceed $\\frac{8048}{4} = 2012$. Remove all the edges of this group. Whatever four vertices we now choose, by the pigeonhole principle at least two of them will fall into one of the formed \"triangles\". Clearly, such two vertices are not connected by an edge.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45364,
    "subject": "Mathematics (Multi-modal)",
    "question": "1. Compare $A$ to $0$, where:\n$$\na)\\ A = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \\dots + 2012 + 2013 - 2014 - 2015 + 2016;\n$$\n$$\nb)\\ A = \\frac{1}{1} - \\frac{1}{2} - \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} - \\frac{1}{7} + \\frac{1}{8} + \\frac{1}{9} - \\frac{1}{10} - \\frac{1}{11} + \\dots + \\frac{1}{2012} + \\frac{1}{2013} - \\frac{1}{2014} - \\frac{1}{2015} + \\frac{1}{2016}.\n$$\nIn each question, the signs go as follows: \"+\" before the first term, then two \"-\" and two \"+\" signs in turn, and, finally, a \"+\" sign before the last term.",
    "options": [],
    "answer": "a) A = 0; b) A > 0",
    "solution": "a.\nIf we split all numbers into $504$ groups of $4$, from left to right, each group will have numbers of the type:\n$$\n(4k+1)-(4k+2)-(4k+3)+(4k+4)=0.\n$$\nAs we can see, the sum of numbers in each group is $0$, therefore, $A = 0$.\n\nb.\nLike in the previous question, split all numbers into groups of $4$: $\\left(\\frac{1}{4k+1} - \\frac{1}{4k+2} - \\frac{1}{4k+3} + \\frac{1}{4k+4}\\right)$. Then the sum of numbers in each group is positive:\n$$\n\\frac{1}{4k+1} - \\frac{1}{4k+2} - \\frac{1}{4k+3} + \\frac{1}{4k+4} > 0 \\Leftrightarrow \\frac{1}{4k+1} - \\frac{1}{4k+2} > \\frac{1}{4k+3} - \\frac{1}{4k+4} \\Leftrightarrow \\frac{1}{(4k+1)(4k+2)} > \\frac{1}{(4k+3)(4k+4)}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation for arbitrary distinct reals $a$, $b$, $c$:\n$$\nx^3 a - x a^3 + a^3 b - a b^3 + b^3 x - b x^3 = (x-a)(x-b)(x-c)(a-b).\n$$",
    "options": [],
    "answer": "x = a or x = b",
    "solution": "Obviously the equation can be solved by trivial transformations and reduction to a quadratic one. We suggest a different approach. Denote the left-hand and right-hand sides by $f(x)$ and $g(x)$ respectively. It's easy to verify that $f(a) = g(a) = 0$ and $f(b) = g(b) = 0$. But this implies that distinct numbers $a$, $b$ are roots of this equation. By comparing the coefficients at $x^3$ for both sides, we conclude that our equation is quadratic. Hence, $x=a$ and $x=b$ are its only roots.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45366,
    "subject": "Mathematics (Multi-modal)",
    "question": "**Can real numbers $x$, $y$, $z$ satisfy**\n$$\n\\frac{1}{(x-y)(x+y)} + \\frac{1}{(y-z)(y+z)} + \\frac{1}{(z-x)(z+x)} = 0?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Denote $a = x^2 - y^2$, $b = y^2 - z^2$, then $-a - b = z^2 - x^2$ and the equation can be rewritten as:\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{a+b} \\Leftrightarrow (a+b)^2 = ab \\Leftrightarrow a^2 - ab + b^2 = 0 \\Leftrightarrow \\left(a - \\frac{b}{2}\\right)^2 + \\frac{3b^2}{4} = 0.\n$$\nThe last equality holds only with $a = b = 0$, which is impossible. Therefore, the equation has no solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three cyclists start off at the same time and ride along the sides of a triangle $ABC$ along the route $AB \\to BC \\to CA$. Their speeds on each of the segments $AB$, $BC$, $CA$ are known: the first cyclist has speeds $12$, $10$ and $20$ mph respectively on the three sides, the second one rides $15$, $15$ and $10$ mph, the third one rides $10$, $20$ and $12$ mph respectively. What can be the angle measure of $\\angle ABC$, if all three cyclists arrived back at the point $A$ simultaneously?",
    "options": [],
    "answer": "60 degrees",
    "solution": "Denote the sides of the triangle by $AB = x$, $BC = y$, $CA = z$. Then the following equality must hold:\n$$\n\\frac{x}{12} + \\frac{y}{10} + \\frac{z}{20} = \\frac{x}{15} + \\frac{y}{15} + \\frac{z}{10} = \\frac{x}{10} + \\frac{y}{20} + \\frac{z}{12} \\text{ or } 5x + 6y + 3z = 4x + 4y + 6z = 6x + 3y + 5z.\n$$\nHence, $x + 2y - 3z = 0$ and $2x - y - z = 0$, which implies $x = y$ and $z = y$. Therefore, $\\triangle ABC$ is equilateral and all its angles are equal to $60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the exact value of the expression below:\n$$\n\\frac{(6!+5!)(5!+4!)(4!+3!)(3!+2!)(2!+1!)}{(6!-5!)(5!-4!)(4!-3!)(3!-2!)(2!-1)!}\n$$\nif $n!$ denotes the product $1 \\cdot 2 \\cdot 3 \\cdots n$ for every natural number $n$.",
    "options": [],
    "answer": "21",
    "solution": "Using the obvious equality $(n+1)! = (n+1) \\cdot n!$ we can make the following transformations:\n$$\n\\frac{(6!+5!)(5!+4!)(4!+3!)(3!+2!)(2!+1!)}{(6!-5!)(5!-4!)(4!-3!)(3!-2!)(2!-1)!} = \\frac{5!(6+1) \\cdot 4! (5+1) \\cdot 3! (4+1) \\cdot 2! (3+1) \\cdot 1! (2+1)!}{5!(6-1) \\cdot 4! (5-1) \\cdot 3! (4-1) \\cdot 2! (3-1) \\cdot 1! (2-1)!} = \\frac{7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3}{5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1} = \\frac{7 \\cdot 6}{2 \\cdot 1} = 21.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the parameter $a$ such that the parabolas $y = x^2 + 2013x + a$ and $y = -x^2 + ax + 2013$ are tangent to each other.",
    "options": [],
    "answer": "2013 or 2021",
    "solution": "Two parabolas with different leading coefficients are tangent if and only if they have exactly one common point, so the equation $x^2 + 2013x + a = -x^2 + ax + 2013$, that is equivalent to the equation $2x^2 + (2013-a)x + (a-2013) = 0$, must have exactly one root. Therefore,\n$$\nD = (a - 2013)^2 - 8(a - 2013) = (a - 2013)(a - 2021) = 0. \\text{ Hence, } a = 2013, \\text{ or } a = 2021.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy and Olesia play such game. Firstly, Andriy chooses an arbitrary chessman (a king, a queen, a rook, a bishop or a knight) and places it on the chessboard. Then they move in turn by the rules of the chosen chessman. However, it is not allowed to put the chessman on the field that Andriy began from or that has already been used. The loser is the one who cannot move. Who wins if both are trying to win?\n\nRemind that when a knight moves, it can move to a square that is two squares horizontally and one square vertically, or two squares vertically and one square horizontally. The bishop has no restrictions in distance for each move, but is limited to diagonal movement. The rook moves horizontally or vertically, through any number of unoccupied squares. A king can move one square in any direction (horizontally, vertically, or diagonally).\n\n![](attached_image_1.png)\n\nFig. 19",
    "options": [],
    "answer": "Olesia",
    "solution": "For every chessman the chessboard is divided into couples of squares that are connected by the move of the chosen chessman. Then the winning strategy of Olesia is as follows: Andriy moves the chessman to the square of some couple (this also concerns the first Andriy's choice of placing the chessman) and Olesia moves the chessman to the other square of this couple. She always can move, because after her move for all chosen couples either both squares are used or none are used.\n\nFor the king, the queen, the rook it is enough to make couples of the neighboring squares horizontally in 1st and 2nd, 3rd and 4th, 5th and 6th, 7th and 8th columns (see Fig. 19).\n\nFor the knight it is enough to make couples of the neighboring squares in 1st and 3rd, 2nd and 4th, 5th and 7th, 6th and 8th columns as is shown in Fig. 20.\n\nFor the bishop the couples are made in such a way. Look at all diagonals of one color (black or white) in the direction where they have an even amount of squares. This direction is parallel to the biggest appropriate diagonal that contains 8 squares. Then make couples of the neighboring squares.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ that have more than $\\frac{n}{2}$ divisors.",
    "options": [],
    "answer": "1, 2, 3, 4, 6",
    "solution": "**Answer:** $n \\in \\{1; 2; 3; 4; 6\\}$.\n\nClearly a number cannot have divisors greater than $\\frac{n}{2}$, besides $n$ itself. Therefore, in order to have more than $\\frac{n}{2}$ divisors it must be divisible by all numbers from $1$ to $\\frac{n}{2}$ and by $n$. Denote by $m$ the integer number which equals either $\\frac{n}{2}$ or $\\frac{n-1}{2}$. Then if $n \\ge 10$, either $m$ or $m-1$ is coprime with $3$. Hence, $n \\ge 3m$, because it must be divisible by both $3$ and $m$. However, in this case $n \\ge 3 \\cdot (\\frac{n-1}{2} - 1)$, or $2n \\ge 3n - 9$, which contradicts the assumption that $n \\ge 10$. All numbers $n < 10$ can be checked by hand.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to compose two integer numbers using each of the ten digits $0, 1, \\ldots, 9$ exactly once, such that one of them is the square of the other?\n\n*0 cannot be the first digit in either number.*",
    "options": [],
    "answer": "No",
    "solution": "If a number has $3$ digits, its square can contain no more than $6$ digits, which gives $9$ in total. If our number has $4$ digits or more, its square has at least $7$, therefore, we must use at least $11$ digits. Therefore, no such number exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45373,
    "subject": "Mathematics (Multi-modal)",
    "question": "A straight line that contains the center of rectilinear triangle $ABC$ intersects straight lines $AB$, $BC$ and $CA$ in points $C_1$, $A_1$ and $B_1$ respectively. Let $A_2$ be the symmetric point to $A_1$ about the middle of $BC$; points $B_2$ and $C_2$ are defined in a similar manner. Prove that points $A_2$, $B_2$ and $C_2$ are on a line that touches the circle inscribed in the triangle $ABC$.\n\n(Serdiuk Nazar)\n\n![](attached_image_1.png)\n\nFig. 25",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let points $B_1$ and $C_1$ be on the sides, and $A_1$ is on the extension of the side $BC$ in direction of point $B$. By the Menelaus' theorem\n$$\n\\frac{CB_2 \\ AC_2 \\ BA_2}{B_2A \\ C_2B \\ A_2C} = \\frac{AB_1 \\ BC_1 \\ CA_1}{B_1C \\ C_1A \\ A_1B} = 1,\n$$\nso points $A_2$, $B_2$ and $C_2$ are collinear (fig. 25).\n\nLet $M$ be the middle of $AB$, $N$ be the middle of $AC$, $K$ be the base of the perpendicular dropped from a point $I$, the center of the inscribed circle, to a straight line $B_2C_2$. Triangles $C_2IM$ and $C_1IM$ are equal, so $C_2I = IC_1$, $\\angle C_2IC_1 = 180^\\circ - 2\\angle B_1C_1A$. Similarly, $B_2I = B_1I$ and $\\angle B_2IB_1 = 180^\\circ - 2\\angle C_1B_1A$. Then,\n$$\n\\angle B_2IC_2 = 2\\angle B_1C_1A + 2\\angle C_1B_1A - 180^\\circ = 2(180^\\circ - \\angle A) - 180^\\circ = 60^\\circ.\n$$\nSince $\\angle B_2IC_2 = \\angle B_1AC_1$ and $\\frac{B_2I}{C_2I} = \\frac{B_1I}{C_1I} = \\frac{B_1A}{C_1A}$ ($I$ is a base of the bisectrix $AI$ of the triangle $B_1AC_1$), triangles $B_2IC_2$ and $B_1AC_1$ are homothetic, hence $\\angle C_2B_2I = \\angle AB_1C_1$ and $\\angle B_2C_2I = \\angle AC_1B_1$. The segment $IB_2$ is common for the right-angled triangles $B_2MI$ and $B_2KI$, and also $\\angle NB_2I = \\angle NB_1I = \\angle KB_2I$, so $NB_2 = KB_2$. In the same way, $KC_2 = C_2M$. Since\n$$\nB_2C + C_2B = B_2N + NC + BM + MC_2 = B_2K + KC_2 + BC = B_2C_2 + BC,\n$$\nthe quadrangle $CB_2C_2B$ is circumscribed, therefore the straight line $B_2C_2$ touches the inscribed circle of the triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45374,
    "subject": "Mathematics (Multi-modal)",
    "question": "In January, Petro used to buy from one to three toy cars every day. On February 1, he tried to make a rectangle of all his cars. When he arranged them into rows of 7 cars, one car remained. When he arranged the cars into rows of 10, there were 2 excessive cars. Can Petro arrange them into rows of 4 cars?\n\n**Answer:** yes.",
    "options": [],
    "answer": "yes",
    "solution": "For some positive integers $n$, $k$ we have:\n$$\n7k+1=10n+2 \\text{ or } 7k=10n+1.\n$$\nTherefore, we need to find a number between 29 and 92 that ends with 1 and is divisible by 7. The minimal such number is 21, which is less than 30. The next one is 91. It's easy to check that no other number with these properties exists in the given interval. Therefore, Petro has\n$$\n7k+1=91+1=92 \\text{ cars.}\n$$\nSince $92 = 4 \\cdot 23$, he can arrange them into 4 columns.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Using each of ten digits exactly once build two numbers (none of them can start with $0$) so that the absolute value of their difference is smallest possible.\n\n**Answer:** $50123$ and $49876$.",
    "options": [],
    "answer": "50123 and 49876",
    "solution": "It is almost obvious that both numbers should be five-digit, since otherwise their difference will have at least five digits. Indeed, the smallest six-digit number that we can build is $102345$, the biggest four-digit number is $9876$, and their difference is $92469$. For other cases of integers with different number of digits the difference will be even bigger than that.\n\nIn order to minimize the difference of two five-digit numbers we should take their first digits to be consequent, because otherwise their difference will also have five digits.\nNow we need to consider all possibilities when a number with greater first digit is complemented by the minimal digits possible, and the other number is complemented by the maximal digits possible.\n$$\n90123-87654=2469, \\quad 80123-79654=469, \\quad 70123-69854=269, \\quad 60123-59874=249, \\\\\n50123-49876=247, \\quad 40125-39876=249, \\quad 30145-29876=269, \\quad 20345-19876=469.\n$$\nWe see that the minimal possible difference is attained for the numbers $50123$ and $49876$, and is equal to $247$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB$ be a diameter of a circle $\\omega$, and points $M$ and $C$ on $\\omega$ be in different half-planes with respect to the line $AB$. Perpendiculars $MN$ and $MK$ are dropped from the point $M$ to the lines $AB$ and $AC$ respectively. Prove that the line $KN$ bisects the line segment $CM$.\n(Igor Nagel)",
    "options": [],
    "answer": "null",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $m, n$ satisfying the equation $2^m = 7n + 4$.",
    "options": [],
    "answer": "no positive integer solutions",
    "solution": "The powers of $2$ give remainders $2$, $4$ and $1$ in division by $7$ with period $3$, in particular, $2^k \\equiv 4 \\pmod{7}$ if and only if $k \\equiv 2 \\pmod{3}$. But this means that $m^2 \\equiv 2 \\pmod{3}$, which is impossible. This contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d$ be positive integers satisfying $ab = cd$. Is it possible that $a+b+c+d$ is a prime number?",
    "options": [],
    "answer": "No",
    "solution": "It follows from the problem statement that $\\frac{ab}{c}$ is a positive integer. Then there should exist positive integers $m, n, x, y$ such that $c = mn$, $a = mx$, $b = ny$. This implies that $d = \\frac{ab}{c} = xy$, and so $a+b+c+d = mx+ny+mn+xy = (n+x)(m+y)$, which is, obviously, not prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45379,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polynomial\n$$\nP(x) = x^{2016} + 2016x^{2015} + a_{2014}x^{2014} + a_{2013}x^{2013} + \\dots + a_1x + 1\n$$\ncan be expressed as $P(x) = (x - x_1)(x - x_2)\\dots(x - x_{2016})$, where among the numbers $x_1, x_2, \\dots, x_{2016}$ at least 2015 are negative (not necessarily distinct). Find all coefficients of $P(x)$.",
    "options": [],
    "answer": "P(x) = (x + 1)^{2016}",
    "solution": "Without loss of generality assume $x_1, x_2, \\dots, x_{2015} < 0$. Vieta's theorem implies $x_1x_2x_3\\dots x_{2015}x_{2016} = 1$, that is, $x_{2016}$ is also a real negative number. Again, by Vieta's theorem:\n$$\n\\left\\{ \\begin{array}{l} x_1 + x_2 + \\dots + x_{2016} = -2016, \\\\ x_1x_2\\dots x_{2016} = 1, \\end{array} \\right. \\Rightarrow \\left\\{ \\begin{array}{l} |x_1| + |x_2| + \\dots + |x_{2016}| = 2016, \\\\ |x_1| \\cdot |x_2| \\cdot \\dots \\cdot |x_{2016}| = 1. \\end{array} \\right.\n$$\nFrom the inequality between arithmetic and geometric means:\n$$\n1 = \\frac{|x_1| + |x_2| + \\dots + |x_{2016}|}{2016} \\ge 2016 \\sqrt[2016]{|x_1| \\cdot |x_2| \\cdot \\dots \\cdot |x_{2016}|} = 1.\n$$\nThe equality implies\n$$\n|x_1| = |x_2| = \\dots = |x_{2016}| = 1, \\text{ or } x_1 = x_2 = \\dots = x_{2016} = -1,\n$$\nthus, $P(x) = (x+1)^{2016}$, and $a_k = C_{2016}^k$, $k = 0, \\ldots, 2016$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy and Olesia play such game. Firstly, Andriy chooses a chessman and place it on the chessboard. Then they moves in turn by the rules of the chosen chessman. However, it is not allowed to put the chessman on the field that Andriy began from or was already been used. Looser is the one who can not move. Who wins if both are trying to win and Andriy choose:\na) a knight; b) a bishop?\n\nRemind that when a knight moves, it can move to a square that is two squares horizontally and one square vertically, or two squares vertically and one square horizontally. The bishop has no restrictions in distance for each move, but is limited to diagonal movement.",
    "options": [],
    "answer": "Olesia wins in both cases (knight and bishop).",
    "solution": "Olesia always wins.\n\nFor every chessman the chessboard is divided into couples of squares that are connected by the move of chosen chessman. Then the win strategy of Olesia is as follows: Andriy moves the chessman to the square of some couple (it also concerns to the first Andriy's choice of placing the chessman) and Olesia moves the chessman to the other square of this couple. She always can move, because after her move for all chosen couples either both squares are used or none are used.\n\n![](attached_image_1.png)\nFig. 19\n\nFor the king, the queen, the castle it is enough to make couples of the neighboring squares horizontally in 1st and 2nd, 3rd and 4th, 5th and 6th, 7th and 8th columns (fig. 19).\nFor the knight it is enough to make couples of the neighboring squares in 1st and 3rd, 2nd and 4th, 5th and 7th, 6th and 8th columns as is shown in fig. 20.\nFor the bishop the couples are made in such a way. Look at all diagonals of one color (black or white) in direction where they have even amount of squares. This direction is parallel to the biggest appropriate diagonal that contains 8 squares. Then make couples of the neighboring squares.\n\n![](attached_image_2.png)\nFig. 20",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45381,
    "subject": "Mathematics (Multi-modal)",
    "question": "A right trapezoid is given with the following property: a square can be inscribed into it such that all its vertices lie on different edges of the trapezoid and none of them coincide with any vertex of the trapezoid. Construct this square with a ruler and a compass.\n(Mariya Rozhkova)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Analysis.** Let $ABCD$ be our trapezoid with right angles $A$ and $B$. Let $EFGH$ be the required square centered at $O$, and suppose $E \\in AB$, $F \\in BC$. In the quadrilateral $EBFO$ two opposite angles are right, hence, it's cyclic. This implies that $\\angle EFO = \\angle EBO = 45^\\circ$ as they intercept the same arc $EO$ (Fig. 4). Likewise, $\\angle EAO = 45^\\circ$. Thus $O$ is the intersection point for bisectors of angles $A$ and $B$ in the trapezoid. Also, $E$ and $G$ are symmetric with respect to $O$.\n\n**Construction.** Construct $O$ as the intersection point of two bisectors drawn from $A$ and $B$. Then we can construct a straight line $l$ symmetric to $AB$ with respect to $O$. The line $l$ intersects the segment $CD$ at a unique point $G$, which is a vertex of our square. Having the center and one vertex of the square, we can reconstruct it in a unique way.\n\n![](attached_image_1.png)\n\nFig. 4",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AD$ be a bisector in an isosceles triangle $ABC$ ($AB = BC$), and let $DE$ be another bisector in the triangle $ABD$. Find out the measures of all angles in $ABC$ if the bisectors of $ABD$ and $AED$ intersect on the straight line $AD$.",
    "options": [],
    "answer": "∠BAC = 80°, ∠BCA = 80°, ∠ABC = 20°",
    "solution": "Let $K$ be the intersection point for the bisectors of the angles $ABD$ and $AED$ (Fig. 3). Then this point lies on the segment $AD$ and is equidistant from rays $BA$ and $BC$, as well as from $EA$ and $ED$. Hence, it's equidistant from rays $DE$ and $DC$. Then $DA$ is the bisector of $\\angle CED$ (in other words, $K$ is an excenter of the triangle $EBD$). With the initial conditions, this implies $\\angle ADC = 60^\\circ$. Since $\\angle DCA = 2\\angle DAC$, we have that $\\angle DCA = 80^\\circ$. Finally, we can write $\\angle BAC = \\angle BCA = 80^\\circ$, $\\angle ABC = 20^\\circ$.\n\n![](attached_image_1.png)\n\nFig. 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Winnie-the-Pooh and Piglet play the following game. There is a 15-inch-long stick. By his first move, Piglet breaks it into two pieces, then the players in turn break one of the existing pieces into two. The rules are that the resulting pieces must have integer length (in inches) and can't be 1-inch-long. The player who can't make a move loses. Who has a winning strategy?",
    "options": [],
    "answer": "Piglet",
    "solution": "$$\n15 = 3+3+3+3+3 = 3+3+3+2+2+2 = 3+2+2+2+2+2+2.\n$$\nThis implies that in order to win Piglet needs to ensure the existence of two 3-inch and one 2-inch pieces (this will make the first and the third outcomes impossible). So, by his first move he must break the stick into 5 and 10 inches. The first piece will be eventually divided into 2 and 3 inches. If Winnie breaks the 10-inch stick by his next move, Piglet must break 3 inches off the longer part, otherwise he should just break it into 3 inches and 7 inches. In this case, regardless of Winnie's moves, in the end there will be at least two 3-inch pieces and one 2-inch piece, which is enough for Piglet to win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45384,
    "subject": "Mathematics (Multi-modal)",
    "question": "The road between $A$ and $B$ is $15$ km long, firstly the road goes up, then it is flat, and lastly it goes down. It is known that every part is no less than $1$ km. The path made by a pedestrian takes exactly $3$ hours. What are the minimum and the maximum amount of time that is taken by the path in opposite direction, if it is known that the speed of pedestrian while going up is $4$ km per hour, while going straight is $5$ per hour and is $6$ per hour while going down?\n\n(Rubliov Bogdan)",
    "options": [],
    "answer": "Minimum time = 73/24 hours, Maximum time = 97/30 hours",
    "solution": "Mark the up, flat and down parts on the way from $A$ to $B$ as $x$, $y$, $z$ respectively. Then:\n$$\nx + y + z = 15, \\frac{x}{4} + \\frac{y}{5} + \\frac{z}{6} = 3,\\ 1 \\le x, y, z \\le 13.\n$$\nFrom the first equation: $y = 15 - x - z$, substitute it into the second equation:\n$$\n\\frac{x}{4} + \\frac{15-z-x}{5} + \\frac{z}{6} = 3 \\Leftrightarrow \\frac{x}{4} - \\frac{x}{5} = \\frac{z}{5} - \\frac{z}{6} \\Leftrightarrow \\frac{x}{20} = \\frac{z}{30} \\Leftrightarrow z = \\frac{3}{2}x.\n$$\n$$\n\\text{Then } y = 15 - x - z = 15 - x - \\frac{3}{2}x = 15 - \\frac{5}{2}x.\n$$\nSo the required time is:\n$$\nt = \\frac{x}{6} + \\frac{y}{5} + \\frac{z}{4} = \\frac{x}{6} + 3 - \\frac{x}{2} + \\frac{3}{8}x = 3 + \\frac{x}{24}.\n$$\n\nThe maximum (the minimum) $t$ can be in case of $x$ is maximum (minimum).\nPut down the limitation for $x$, which follow from the condition of the problem:\n$$\n1 \\le x \\le 13,\\ 1 \\le z = \\frac{3}{2}x \\le 13 \\Leftrightarrow \\frac{2}{3} \\le x \\le \\frac{26}{3},\\ 1 \\le y = 15 - \\frac{5}{2}x \\le 13 \\Leftrightarrow \\frac{4}{5} \\le x \\le \\frac{28}{5}.\n$$\nSince all conditions have to be fulfilled simultaneously, we have such limitation for $x$:\n$$\n1 \\le x \\le \\frac{28}{5}.\n$$\nIf $x=1$, then $z = \\frac{3}{2}$ and $y = \\frac{25}{2}$. If $x = \\frac{28}{5}$, then $z = \\frac{42}{5}$ and $y = 1$.\n\n$$\nt = \\frac{x}{6} + \\frac{y}{5} + \\frac{z}{4} = \\frac{x}{6} + 3 - \\frac{x}{2} + \\frac{3}{8}x = 3 + \\frac{x}{24}.\n$$\n\nThe maximum (the minimum) $t$ can be in case of $x$ is maximum (minimum).\nPut down the limitation for $x$, which follow from the condition of the problem:\n\n$$\nt_{\\max} = 3 + \\frac{1}{24} \\cdot \\frac{28}{5} = 3 + \\frac{7}{30} = \\frac{97}{30},\\ t_{\\min} = 3 + \\frac{1}{24} \\cdot 1 = \\frac{73}{24}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $x_1, x_2, \\ldots, x_{2015}$ fulfill both equalities simultaneously:\n\n$x_1^{2014} + x_2^{2014} + \\ldots + x_{2015}^{2014} = 1$ and $x_1^{2015} + x_2^{2015} + \\ldots + x_{2015}^{2015} = -1$.",
    "options": [],
    "answer": "Exactly one number equals negative one and all the others equal zero.",
    "solution": "From the first equation $-1 \\le x_i \\le 1$, $i = 1, \\ldots, 2015$, so for all $i$: $0 \\le 1 + x_i \\le 2$. Add both equations and get\n$$\nx_1^{2014}(1+x_1) + x_2^{2014}(1+x_2) + \\ldots + x_{2015}^{2014}(1+x_{2015}) = 0.\n$$\nSince every item $x_i^{2014}(1+x_i) \\ge 0$, then their sum equals zero if and only if every item equals zero. So all $x_i \\in \\{-1, 0\\}$. From the first equation follows that exactly one variable is not null, from the second equation follows that this variable equals $-1$. Thus, the equation is satisfied only by such set of numbers: $x_i = -1$, $i = 1, \\ldots, 2015$, $x_j = 0$, $j \\ne i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let us call a year *colored* if the decimal representation of its number has no repeating digits. For example, all years from $2013$ to $2019$ are colored, unlike $2020$.\n\na) Find the nearest chain of seven consecutive colored years in the future.\n\nb) Can a chain of more than seven consecutive years happen in the future?",
    "options": [],
    "answer": "a) 2103, 2104, 2105, 2106, 2107, 2108, 2109; b) No, a chain longer than seven consecutive colored years cannot occur.",
    "solution": "a) Let us show that the nearest sequence of $7$ colored years is $2103, \\ldots, 2109$. First, we prove that in this century no sequence of more than six colored years can happen any longer. We see that digits $0$ and $2$ cannot represent units or tens. Therefore, a chain is broken at each year ending in $0$ or $2$, which gives us the only way to form a chain of $7$ years:\n$$\n20*3, 20*4, \\dots, 20*9.\n$$\nHere, $*$ can only be equal to $1$, but this is the current chain. In the next century, after $2100$, the first chain is easy to find: $2103, \\ldots, 2109$.\n\nb) Years we deal with are written with at least four digits. A colored chain cannot contain numbers ending in $99$, hence the number of hundreds doesn't change throughout the chain. This implies that there are only $8$ possible last digits.\nAssume some colored chain has $8$ numbers. We have two cases.\n1. All years have the same number of tens. Then there are only $7$ options for the last digit. Contradiction.\n2. The number of tens changes within the chain. In this case the tens digit cannot be equal to $9$. Assume this digit changes from $x$ to $x+1$. The chain has $8$ numbers, thus, it should have two numbers of the type\n$$\n\\overline{ab}x+x+1, \\overline{ab}x+1x.\n$$\nFor example, they might be $2145$ and $2154$. But then the chain has at least $10$ years (for example, $2145$ to $2154$). Again, we have a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $a$, $b$, $c$, $d$ satisfy: $ab + cd > 0$, $ac + bd > 0$, and $a^2 + d^2 = c^2 + b^2$. Prove that $ad + bc > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$(ab + cd)(ac + bd) = a^2bc + ac^2d + ab^2d + bcd^2 = bc(a^2 + d^2) + ad(c^2 + b^2) = bc(a^2 + d^2) + ad(a^2 + d^2) = (a^2 + d^2)(bc + ad) > 0,$$\nwhich implies the required inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45388,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that $\\frac{a}{b+c+d} + \\frac{b}{c+d+a} + \\frac{c}{d+a+b} + \\frac{d}{a+b+c} = 1$. Find the value of the expression\n$$\n\\frac{a^2}{b+c+d} + \\frac{b^2}{c+d+a} + \\frac{c^2}{d+a+b} + \\frac{d^2}{a+b+c}.\n$$",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of prime numbers $(p, q)$ with $p > q$, for which both numbers $p+q$ and $p-q$ are also prime.",
    "options": [],
    "answer": "(5, 2)",
    "solution": "For the number $p+q$ to be prime the numbers $p$ and $q$ must be of different parity, which automatically means that $q=2$, since $p > q$. By the problem statement we then have that the numbers $p-2, p$, and $p+2$ should be prime. Since they obviously have different remainders in division by $3$, one of them must be equal to $3$. So we have three possibilities:\n* If $p-2=3$, then $p=5$, $p+2=7$, and the pair $(5; 2)$ satisfies the problem statement;\n\n* if $p=3$, then $p-2=1$, and is not prime;\n* if $p+2=3$, then $p=1$, and is not prime.\nSo after considering all possible cases we end up with a single solution $p=5, q=2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $a$, $b$ fulfill both equalities simultaneously:\n$$\na^2 + b^2 = 1 \\text{ and } a^3 + b^3 = -1.\n$$\n\nWhat is the possible value of the expression $a^3 + b^2$?",
    "options": [],
    "answer": "{-1, 1}",
    "solution": "From the first equation $-1 \\leq a \\leq 1$ and $-1 \\leq b \\leq 1$, therefore $0 \\leq 1+a \\leq 2$ and $0 \\leq 1+b \\leq 2$. Add both equations and get\n$$\na^2(1+a) + b^2(1+b) = 0.\n$$\nAs both items are non-negative, their sum equals zero if and only if every item equals $0$. So $a, b \\in \\{-1, 0\\}$. From the first equation it follows that one variable is not zero, from the second equation it follows that this variable equals $-1$. Thus, the equations are satisfied by such pairs of numbers $(a, b): (-1, 0)$ and $(0, -1)$, hence $a^3 + b^2 = -1$ or $a^3 + b^2 = 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy and Olesia play such game. Firstly, Andriy chooses a chess king and places it on the chessboard. Then they move in turn by the rules of the king. However, it is not allowed to put the king on the field that Andriy began from or was already used. The loser is the one who cannot move. Who wins if both are trying to win?",
    "options": [],
    "answer": "Olesia always wins.",
    "solution": "Olesia always wins.\n\nFor every chessman the chessboard is divided into couples of squares that are connected by the move of the chosen chessman. Then the win strategy of Olesia is as follows: Andriy moves the chessman to the square of some couple (it also concerns the first Andriy's choice of placing the chessman) and Olesia moves the chessman to the other square of this couple. She always can move, because after her move for all chosen couples either both squares are used or none are used.\n\n![](attached_image_1.png)\nFig. 19\n\nFor the king, the queen, the rook it is enough to make couples of the neighboring squares horizontally in 1st and 2nd, 3rd and 4th, 5th and 6th, 7th and 8th columns (fig. 19).\n\nFor the knight it is enough to make couples of the neighboring squares in 1st and 3rd, 2nd and 4th, 5th and 7th, 6th and 8th columns as is shown in fig. 20.\n\nFor the bishop the couples are made in such a way. Look at all diagonals of one color (black or white) in direction where they have even amount of squares. This direction is parallel to the biggest appropriate diagonal that contains 8 squares. Then make couples of the neighboring squares.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45392,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $9 \\times 9$ square is divided into $81$ small $1 \\times 1$ squares, $8$ of which are painted black, the rest being white. We cut a fully white rectangle (possibly, a square) out of the big $9 \\times 9$ square. What is the maximal area of the rectangle that we can attain regardless of the positions of the black squares? It is only allowed to cut the rectangle along the grid lines.",
    "options": [],
    "answer": "9",
    "solution": "Cut the square into $9$ smaller $3 \\times 3$ blocks. Since there are only eight black squares, at least one of the blocks doesn't contain any of them. Therefore a white square of area $9$ can always be found.\n\nNext, we show that sometimes it is impossible to find a larger rectangle. Fig. 2 is an example. Here, one can cut out either a $3 \\times 3$ square or a $1 \\times 9$ rectangle, but not any rectangle of larger area.\n\n![](attached_image_1.png)\n\nFig. 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $w_1$ and $w_2$ with centers $O_1$ and $O_2$ respectively intersect in points $A$ and $B$. A straight line $O_1O_2$ intersects $w_1$ in a point $Q$, that is not inside $w_2$, and $w_2$ in a point $X$, that is inside $w_1$. Around the triangle $O_1AX$ a circle $w_3$ is circumscribed and intersects $w_1$ for the second time in a point $T$. A straight line $QT$ intersects $w_3$ in a point $K$, and a straight line $QB$ intersects $w_2$ a second time in a point $H$. Prove that\n\n![](attached_image_1.png)\n\nFig. 24\n\na) points $T$, $X$, $B$ are collinear;\nb) points $K$, $X$, $H$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $\\angle AO_1X = \\alpha$, then $\\angle ATX = \\alpha$, because they are subtended by the same arc of the circle $w_3$ (Fig. 24), moreover $\\angle AO_1X = \\angle ATB$, so $\\angle ATX = \\angle ATB$, therefore $T$, $X$, $B$ are collinear.\n\nb) Let $K_1 = XH \\cap TQ$, $\\angle O_2XH = \\alpha$, $\\angle HXB = \\beta$, $\\angle XHB = \\varphi$, then:\n\n$$\n\\angle XQB = \\alpha - \\varphi, \\quad \\angle QTB = 90^\\circ - \\alpha + \\varphi,\n$$\n$$\n\\angle TQX = 180^\\circ - (\\beta + \\alpha) - (90^\\circ - \\alpha + \\varphi) = 90^\\circ - \\beta - \\varphi = \\alpha,\n$$\nin other words $\\angle TQX = \\angle QTO_1$, since $O_1T = QO_1$, then $\\angle K_1XQ = \\alpha$, hence $\\angle K_1XQ = \\angle QTO_1$, that is $K_1 \\in w_3$, so $K_1 = K$ and points $K$, $X$, $H$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality for positive $a$, $b$, $c$, $d$:\n$$\n\\left(\\frac{a+b}{2c}\\right)^2 + \\left(\\frac{b+c}{2d}\\right)^2 + \\left(\\frac{c+d}{2a}\\right)^2 + \\left(\\frac{d+a}{2b}\\right)^2 \\ge 4.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consecutively use the inequalities of means:\n$$\n\\begin{aligned}\n& \\left(\\frac{a+b}{2c}\\right)^2 + \\left(\\frac{b+c}{2d}\\right)^2 + \\left(\\frac{c+d}{2a}\\right)^2 + \\left(\\frac{d+a}{2b}\\right)^2 \n\\ge \\frac{ab}{c^2} + \\frac{bc}{d^2} + \\frac{cd}{a^2} + \\frac{da}{b^2} \n\\ge \\\\\n& \\ge 2\\sqrt{ab} \\sqrt{bc} \\frac{\\sqrt{ab}}{c} \\frac{\\sqrt{bc}}{d} + 2\\sqrt{cd} \\sqrt{da} \\frac{\\sqrt{ab}}{a} \\frac{\\sqrt{bc}}{b} \n\\ge 4\\sqrt{\\frac{ab}{c} \\frac{\\sqrt{bc}}{d} \\frac{\\sqrt{cd}}{a} \\frac{\\sqrt{da}}{b}} = 4.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ that for all real $x, y$ fulfill the equality:\n$$\nf(f(x) - y^2) = f(x^2) + y^2 f(y) - 2f(xy).\n$$",
    "options": [],
    "answer": "f(x) = 0 for all x, and f(x) = x^2 for all x",
    "solution": "Substitute $y = 1$ and $y = -1$:\n$$\nf(f(x)-1) = f(x^2) + f(1) - 2f(x) \\text{ and } f(f(x)-1) = f(x^2) + f(-1) - 2f(-x). \\ (*)\n$$\nCombining both equalities, get $f(1) - 2f(x) = f(-1) - 2f(-x)$. Hence, if $x = 1$: $f(1) = f(-1)$, so $f(x) = f(-x)$, the function $f$ is even.\n\nIf $x = y = 1$, $f(f(1) - 1) = 0$, in other words there exists such number $b$ that $f(b) = 0$. If $x = b$ in $(*)$ then\n$$\nf(f(b) - 1) = f(b^2) + f(1), \\quad f(-1) = f(b^2) + f(1), \\quad \\text{so } f(b^2) = 0.\n$$\nLet substitute $x = b$ and $y = 0$ in the initial equality: $f(f(b)) = f(b^2) - 2f(0)$, so $3f(0) = f(b^2) = 0$. If $x = 0$ then\n$$\nf(y^2) = y^2 f(y). \\quad (**)\n$$\nThere are two cases.\nCase 1: there exists $b \\in \\mathbb{R}$, $b \\neq 0$ such that $f(b) = 0$. As is shown above, $f(b^2) = 0$. Let substitute $x = b$ in the initial equality:",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45396,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute-angled triangle $ABC$ has the side $BC > AB$, and the bisectrix $BL = AB$. On the segment $BL$ there exists point $M$, for which $\\angle AML = \\angle BCA$. Prove that $AM = LC$.\n\n![](attached_image_1.png)\n\nFig. 16",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On the segment $BC$ put a point such as $BD = BL$ (fig. 16). Then $\\Delta ABL = \\Delta BLD$, mark $\\angle LAB = \\angle BLA = \\angle BLD = \\angle BDL$. On the segment $BD$ choose a point $K$, such that $\\angle ALM = \\angle DLK$. Then $\\Delta ALM = \\Delta DLK$ because of $LD = AL$, but then $\\angle AML = \\angle LKD = \\angle BCA$, that's why $KL = AM = LC$, and it is exactly what we have to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that number $m^4 + 1$ has no divisors in interval $[m^2 - 2m, m^2 + 2m]$ for every natural $m > 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose the opposite. Let\n$$\nm^2 + a \\in [m^2 - 2m, m^2 + 2m] \\text{ i } m^4 + 1 \\equiv m^2 + a.\n$$\nSince\n$$\n(m^4 + 1, m^2 + a) = (-a m^2 + 1, m^2 + a) = (a^2 + 1, m^2 + a), \\text{ then } a^2 + 1 \\equiv m^2 + a, \\\\\n a^2 + 1 = (m^2 + a) r.\n$$\nLet $a \\in [-2m+3, 0]$, then $\\frac{m^4+1}{m^2+a} \\le \\frac{m^4+1}{m^2-2m+3} < m^2+2m+1$, so $m^4+1$ has a divisor that is no less than $m^2$ and fulfills the assumption, therefore instead of $m^2+a$ we can examine a divisor $m^2+b$, where $b \\ge 0$.\nIf $a = -2m + 2$, then $4m^2 - 8m + 5 \\equiv m^2 - 2m + 2$, so $3 \\equiv m^2 - 2m + 2$, which is impossible in case of $m > 1$.\nIf $a = -2m + 1$, then $4m^2 - 4m + 2 \\equiv (m-1)^2 - m - 1$, thus $2 \\equiv m - 1$, $m \\in \\{2,3\\}$. In case of $m = 2$ and $m = 3$: $m^4 + 1 = 17$ and $m^4 + 1 = 2 \\cdot 41$ - none of these numbers fulfills the condition.\nIf $a = -2m$, then $1 \\equiv m$ is a contradiction.\nThus, if number $m^4 + 1$ has a divisor $m^2 + a$, where $a \\in [-2m, 2m]$, then the number $m^4 + 1$ has a divisor $m^2 + b$, where $b \\in [0, 2m]$. It is also clear that $b \\ne 0$. In such case, $4m^2 + 1 \\ge b^2 + 1 = (m^2 + b) r \\ge m^2 r$, so $r \\le 3$.\nCase $r = 3$: $b^2 + 1 = 3(m^2 + b)$ - it is impossible, because $b^2 + 1$ is not divisible by 3.\nCase $r = 2$: $b^2 + 1 = 2(m^2 + b)$, $(b-1)^2 = 2m^2$ - impossible.\nCase $r=1$: $b^2 - b + 1 = m^2$, but $b^2 > b^2 - b + 1 = m^2 > (b-1)^2$, so this case is also impossible. Thus, we have the contradiction with the supposition, so the statement is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BM$ be a median in an acute triangle $ABC$, whose sides $AB$ and $BC$ have different length. The extension of $BM$ intersects the circumcircle of $ABC$ at a point $N$. Let $D$ be a point on the circumcircle such that $\\angle BDH = 90^\\circ$, where $H$ is the orthocenter of $ABC$. Let $K$ be a point chosen so that $ANCK$ is a parallelogram. Prove that the lines $AC$, $KH$ and $BD$ are concurrent.\n\n(Ihor Nahel)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $BC_1HA_1$ and $BANC$ are cyclic quadrilaterals, it follows that $\\angle A_1HC_1 = 180^\\circ - \\angle B$ and $\\angle ANC = 180^\\circ - \\angle B$. This implies $\\angle A_1HC_1 = \\angle ANC$. But $\\angle ANC = \\angle A_1HC_1$ (as vertical angles) and $\\angle AKC = \\angle ANC$ (as opposite angles in a parallelogram). Hence we obtain\n$\\angle AHC = \\angle AKC$, which means $AKHC$ is cyclic. This implies $\\angle KHC_1 = \\angle KAC$. Then $\\angle KAC = \\angle ACN$, because $AK \\parallel NC$ and $\\angle ACN = \\angle ABN$, as these angles are inscribed in the circumcircle of $ABC$. Therefore, $\\angle KHC_1 = \\angle KBC_1$, which implies that the point $K$ also lies on the circle with diameter $BH$.\n\nDenote by $\\omega_1$ the circumcircle of $ABC$, by $\\omega_2$ the circle with diameter $BH$, and by $\\omega_3$ the circumcircle of $AKHC$. Then $AC$, $KH$ and $BD$ are radical axes of circles $\\omega_1$ and $\\omega_3$, $\\omega_2$ and $\\omega_3$, $\\omega_1$ and $\\omega_2$. As it is known, the radical axes of three circles either intersect at the same point, which is their radical center, or are parallel.\n\nSince $AB > BC$, lines $AC$, $KH$ and $BD$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45399,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that $\\frac{a}{b+c+d} + \\frac{b}{c+d+a} + \\frac{c}{d+a+b} + \\frac{d}{a+b+c} = 1$. Find the value of the expression\n$$\n\\frac{a^2}{b+c+d} + \\frac{b^2}{c+d+a} + \\frac{c^2}{d+a+b} + \\frac{d^2}{a+b+c}.\n$$",
    "options": [],
    "answer": "0",
    "solution": "Multiplying the given equality by $(a+b+c+d)$, we obtain\n$$\n\\begin{aligned}\n& \\frac{a(a+b+c+d)}{b+c+d} + \\frac{b(a+b+c+d)}{c+d+a} + \\frac{c(a+b+c+d)}{d+a+b} + \\frac{d(a+b+c+d)}{a+b+c} = \\\\\n& = \\frac{a^2+a(b+c+d)}{b+c+d} + \\frac{b^2+b(c+d+a)}{c+d+a} + \\frac{c^2+c(d+a+b)}{d+a+b} + \\frac{d^2+d(a+b+c)}{a+b+c} = \\\\\n& = \\frac{a^2}{b+c+d} + a + \\frac{b^2}{c+d+a} + b + \\frac{c^2}{d+a+b} + c + \\frac{d^2}{a+b+c} + d = a+b+c+d.\n\\end{aligned}\n$$\nIt then follows that\n$$\n\\frac{a^2}{b+c+d} + \\frac{b^2}{c+d+a} + \\frac{c^2}{d+a+b} + \\frac{d^2}{a+b+c} = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45400,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number $2013$ is written on the board. Two players are playing the following game. A move consists of replacing the number on the board with the difference of this number and one of its divisors. The player who writes $0$ loses. Who of the two players can guarantee the win?\n\n(Oleksiy Piskun)",
    "options": [],
    "answer": "Second player",
    "solution": "It is easy to observe that odd numbers have only odd divisors. So, if the player moves from an odd number, he has to write an even number. This provides a strategy for the second player: subtract $1$ from the current number at any move. Then the first player will always deal with an odd number, so will write an even number. Since the number written on the board always decreases, eventually the first player will have to write $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several people gathered at a party. Each two of them who don't know each other have exactly two common acquaintances (if a person $A$ knows a person $B$, then necessarily the person $B$ knows the person $A$). Mykhailyk and Vitalyk know each other, but have no common acquaintances.\n\na) Prove that Mykhailyk and Vitalyk have the same number of acquaintances on the party.\n\nb) Can such a situation happen on a party with exactly 6 people?",
    "options": [],
    "answer": "a) Mykhailyk and Vitalyk have the same number of acquaintances. b) Yes, such a situation can occur with exactly 6 people.",
    "solution": "a) Denote Mykhailyk and Vitalyk by $A$ and $B$ respectively. Let $M(A)$ and $M(B)$ be the sets of acquaintances of $A$ and $B$ respectively (see Fig. 3).\n\nThere is no person in $M(A)$ that knows $B$, since in this case $A$ and $B$ would have a common acquaintance. So every person $X$ from $M(A)$ has exactly two common acquaintances with $B$. One of them is $A$, and the other person, call him $Y$, belongs to the set $M(B)$. In this way for every person $X$ from the set $M(A)$ we assign exactly one person $Y$ from $M(B)$. It is easy to see that this correspondence is one-to-one, and so the numbers of people in $M(A)$ and $M(B)$ are the same.\n\nb) Example given on Fig. 4 shows that the situation described in the problem statement can happen when there are exactly 6 people on the party.\n\n![](attached_image_1.png)\nFig. 3\n![](attached_image_2.png)\nFig. 4",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45402,
    "subject": "Mathematics (Multi-modal)",
    "question": "A committee has 4 subcommittees, each controlled by 3 leaders from the committee. For effective coordination, each two subcommittees must have exactly one leader in common. What is the least possible number of people in the committee?",
    "options": [],
    "answer": "6",
    "solution": "If we consider two subcommittees, they have exactly one leader in common, therefore, together they have exactly 5 members. Hence there are at least 5 people in the committee. Denote them by $A = \\{1, 2, 3, 4, 5\\}$. However it's impossible to choose leaders for another subcommittee out of them. Therefore, the committee must have at least 6 members. Here's an example of four subcommittees and their leaders:\n$$\nA = \\{1, 2, 3, 4, 5, 6\\} \\rightarrow \\{1, 2, 3\\}; \\{3, 4, 5\\}; \\{1, 5, 6\\}; \\{2, 4, 6\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality for positive $a$, $b$, $c$, $d$\n$$\n\\frac{(a+b)^2}{cd} + \\frac{(c+d)^2}{ab} \\ge 8.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Use two inequalities of means:\n$$\n\\frac{(a+b)^2}{cd} + \\frac{(c+d)^2}{ab} \\ge \\frac{4ab}{cd} + \\frac{4cd}{ab} \\ge 2 \\cdot 4 \\sqrt{\\frac{ab}{cd} \\cdot \\frac{cd}{ab}} = 8.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system of equations:\n$$\n\\begin{cases} x^2 + 3xy = 3y + x, \\\\ y^2 - yx = 3x + y. \\end{cases}\n$$",
    "options": [],
    "answer": "[(0, 0), (1, -1), (1, 3), (6, -2)]",
    "solution": "Add up the equations and get\n$$\nx^2 + 2xy + y^2 = 4y + 4x \\text{ or } (x+y)^2 = 4(x+y).\n$$\n\nIf $x = -y$, the second equation implies\n$$\n2y^2 = -2y, \\text{ hence } y = 0 \\text{ or } y = -1.\n$$\nWe get two solutions, (0; 0) and (1; -1).\n\nIf $x = 4 - y$, the first equation implies\n$$\ny^2 - y(4 - y) = 3(4 - y) + y, \\text{ hence, } y^2 - y - 6 = 0.\n$$\nThe final equation has two roots, $y = 3$ and $y = -2$. They give us solutions (1; 3) and (6; -2). It can be easily verified that they satisfy both equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45405,
    "subject": "Mathematics (Multi-modal)",
    "question": "The equal segments $AB$ and $CD$ intersect in the point $O$ and are divided into the ratio $AO:OB=CO:OD=1:2$. The lines $AD$ and $BC$ intersect in the point $M$. Prove that $DM = MB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the length of the segments be $AB=CD=3a$, then $AO=CO=a$ and $OB=OD=2a$. Since $\\angle AOD=\\angle COB$ as vertical (fig/ 17), then $\\triangle AOD=\\triangle COB$. Hence $\\angle ADO=\\angle CBO$. As $\\triangle BOD$ is isosceles, then $\\angle BDO=\\angle DBO$, consequently $\\angle MDB=\\angle MBD$ as the sums of the equal angles. In other words $\\triangle MDB$ is isosceles, from where the equality we need.\n\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45406,
    "subject": "Mathematics (Multi-modal)",
    "question": "2015 candies are placed along a circle and numbered $1$ to $2015$ clockwise. Andriy and Olesia play the following game. In each turn, a player can take either $2$ or $3$ candies with consecutive numbers ($1$ and $2015$ are also considered \"consecutive\"). The player who can't make a move loses. Who has a winning strategy if Andriy plays first?",
    "options": [],
    "answer": "Olesia",
    "solution": "Andriy takes $2$ (or $3$) consecutive candies. Then Olesia takes $3$ (or $2$) diametrically opposite candies, so that there are $1005$ candies on both sides between the groups of taken candies. Then she just copies Andriy's moves on the other part. If Andriy can make a move, she can as well, so she will not lose anyway. Since the game is finite, in the end she will win.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $w_1$ and $w_2$ centered at points $O_1$ and $O_2$ respectively intersect at points $A$ and $B$. Let $w$ be the circumscribed circle of $O_1O_2B$ centered at $O$, which intersects $w_1$ and $w_2$ again at points $K$ and $L$ respectively. The straight line $OA$ intersects $w_1$ and $w_2$ at points $M$ and $N$ respectively. Denote by $P$ the intersection point of lines $MK$ and $NL$. Prove that $P$ lies on $w$ and $PM = PN$.\n(Vadym Mytrofanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the following lemma by Archimedes:\n\n**Lemma (Archimedes).** Circles $w_1$ and $w_2$ intersect at points $A$ and $B$, with the center of $w_2$ lying on $w_1$. A chord $AC$ of $w_2$ intersects $w_1$ again at a point $M$. Then $CM = CB$.\n\nPut $\\alpha = \\angle KBA = \\angle KMA$, since they intercept the same arc in $w_1$. Similarly, $\\beta = \\angle ABL = \\angle ANL$, because they intercept the same arc in $w_2$. Hence $\\angle MPN = 180^\\circ - \\alpha - \\beta$, which implies that $P \\in w$ (Fig. 8).\n\nBy Archimedes' lemma, for circles $w_1$ and $w_2$ we can write $KJ = JA = LJ$, but then $\\alpha = \\beta$, and from the circle $w$ we have that $PM = PN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to construct a triangle with sides $x$, $y$, $z$ satisfying the condition:\n$$\n3x^2y^2 + 3y^2z^2 + 3z^2x^2 = x^4 + y^4 + z^4?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Rewrite the equation as:\n$$\n2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4 = -(x^2y^2 + y^2z^2 + z^2x^2).\n$$\nThe left-hand side can be decomposed as:\n$$(x+y+z)(x+y-z)(y+z-x)(z+x-y) = -(x^2y^2 + y^2z^2 + z^2x^2).$$\nHence, the left-hand side is negative, therefore, at least one of the multipliers is negative too. The first one is always positive, then, without loss of generality, we can assume the second one is negative, i.e. $x+y-z<0$, which contradicts the triangle inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $w_1, w_2$ are externally tangent at a point $Q$. A common external tangent line to these circles (that doesn't pass through $Q$) is tangent to $w_1$ at a point $B$, and $BA$ is a diameter of this circle. The point $A$ belongs to the line tangent to the circle $w_2$ at a point $C$ such that $B$ and $C$ are in the same half-plane with respect to the line $AQ$. Prove that the circle $w_1$ bisects the segment $BC$.\n\n![](attached_image_1.png)\nFig. 6.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the point where the common tangent touches the circle $w_2$ (Fig. 6). Consider the common tangent line to the two circles that passes through the point $Q$. Suppose it intersects the line $BK$ at a point $P$. By the properties of lines tangent to circles,\n$$\nPB = PQ = PK.\n$$\nTherefore, $\\angle BQK = 90^\\circ$. Since $AB$ is a diameter of $w_1$, we have that $\\angle AQB = 90^\\circ$, and so the points $A, Q, K$ are collinear. It is given in the problem statement that $AB \\perp BK$, which implies that\n$$\nAB^2 = AQ \\cdot AK \\text{ and } AC^2 = AQ \\cdot AK\n$$\n(from the properties of the right triangle $\\triangle ABK$ and the properties of secant and tangent lines to the circle $w_2$). It then follows that $AB = AC$. If $w_1$ intersects $BC$ at a point $M$, then $\\angle BMA = 90^\\circ$, and so $AM$ is the altitude of the isosceles triangle $\\triangle ABC$. This proves that $BM = MC$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45410,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 9 weights with labels $1$ g, $2$ g, $\\ldots$, $9$ g respectively. It's known that one of the weights is lighter than the label says, while the other eight labels are correct. Is it possible to detect the counterfeit weight using scales with no additional weights no more than twice?",
    "options": [],
    "answer": "Yes",
    "solution": "At the first move, put $1+4+9$ on the left side and $2+5+7$ on the right side. If the total weights are equal, the counterfeit weight is among the other three. Then put on scales the combinations $3+4$ and $1+6$.\n\n* If $3+4=1+6$, they are genuine, hence, $8$ is counterfeit.\n\n* If $3+4<1+6$, then $3$ is counterfeit, because it is lighter and $4$ is known to be genuine.\n* Similarly, $3+4>1+6$ would imply that $6$ is counterfeit.\n\nIf the total weights at the first weighing are not equal, the counterfeit weight is among the three lighter ones. The second move involves putting one potentially counterfeit weight on each side and balancing them with genuine ones. For example, $1+5$ and $4+2$. Then we use the same arguments: the counterfeit one is where the total weight is less. If the scales are balanced, the unused weight is counterfeit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $a, b$ for which there exist integers $x, y$ such that the following equation holds:\n$$\n8x^4 + 8y^4 = a^4 + 6a^2b^2 + b^4.\n$$",
    "options": [],
    "answer": "All integer pairs a, b with the same parity.",
    "solution": "If $a, b$ have the same parity, define $x, y$ as:\n$$\nx = \\frac{a+b}{2}, \\quad y = \\frac{a-b}{2}.\n$$\nThen they are obviously integers, and by substituting them we verify that the equation does hold.\n\nIf $a, b$ don't have the same parity, the right-hand side of the equation is an odd integer. For example, if $a$ is even and $b$ is odd, then $a^4 + 6a^2b^2$ is even and $b^4$ is odd, therefore, the equality cannot hold.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45412,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\operatorname{ctg}[x] \\cdot \\operatorname{ctg}\\{x\\} = 1.\n$$\nWhere $[a]$ is the integer part of $a$, and $\\{a\\} = a - [a]$.",
    "options": [],
    "answer": "x = pi/2 + pi*k, where k is any integer",
    "solution": "Transform the equation to the form:\n$$\n\\cos[x] \\cdot \\cos\\{x\\} = \\sin[x] \\cdot \\sin\\{x\\},\n$$\nif $\\sin[x] \\cdot \\sin\\{x\\} \\ne 0$. Then we have\n$$\n\\cos[x] \\cdot \\cos\\{x\\} - \\sin[x] \\cdot \\sin\\{x\\} = \\cos([x] + \\{x\\}) = \\cos x = 0.\n$$\nHence $x = \\frac{1}{2}\\pi + \\pi k, k \\in \\mathbb{Z}$.\nRemains that all these values satisfy the initial condition. Indeed, none of them are integer, hence $\\{x\\} \\ne 0 \\Rightarrow \\{x\\} \\in (0; 1) \\Rightarrow \\sin\\{x\\} \\ne 0$. Also since $\\sin y = 0$ implies $y = \\pi n, n \\in \\mathbb{Z}$, hence the expression is equal to zero for integer $y$ only when $y = 0$. Notice that for $k \\ge 0$ $x = \\frac{1}{2}\\pi + \\pi k > 1$ and $[x] > 0$, and for $k < 0$ $x = \\frac{1}{2}\\pi + \\pi k < -1$ and $[x] < 0$. Hence all $x$ that were found satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside an equilateral triangle $ABC$ point $M$ is chosen. Let points $M_1$, $M_2$, $M_3$ be symmetric corresponding to sides $BC$, $AC$, $AB$ of the triangle. Prove that $\\overrightarrow{MM_1} + \\overrightarrow{MM_2} + \\overrightarrow{MM_3}$ is equal to $\\overrightarrow{MA} + \\overrightarrow{MB} + \\overrightarrow{MC}$.\n(Tereshin Dmitro)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us draw through point $M$ lines that are parallel to sides $ABC$. Let them intersect $AB$, $BC$, $AC$ at $C_1$, $C_2$; $A_1$, $A_2$; $B_1$, $B_2$ (Fig. 30).\n\n![](attached_image_1.png)\n\nHence, $C_1A_2 \\parallel AC$, $A_1B_2 \\parallel BA$, $B_1C_2 \\parallel BC$, so $C_1A_2$, $A_1B_2$ and $B_1C_2$ intersect at $M$. Consider $\\triangle A_1MA_2$. Obviously, this triangle is equilateral. Line $M_1M$ contains its altitude, because it is perpendicular to $BC$. So $M_1M$ is doubled median,\n\nhence $\\overrightarrow{MA_1} + \\overrightarrow{MA_2} = \\overrightarrow{MM_1}$, similarly $\\overrightarrow{MB_1} + \\overrightarrow{MB_2} = \\overrightarrow{MM_2}$ and $\\overrightarrow{MC_1} + \\overrightarrow{MC_2} = \\overrightarrow{MM_3}$. On the other hand $MC_1AB_2$ is a parallelogram, so $\\overrightarrow{MA} = \\overrightarrow{MC_1} + \\overrightarrow{MB_2}$. Similarly $\\overrightarrow{MB} = \\overrightarrow{MC_2} + \\overrightarrow{MA_1}$ and $\\overrightarrow{MC} = \\overrightarrow{MA_2} + \\overrightarrow{MB_1}$. Hence:\n\n$$\n\\overrightarrow{MM_1} + \\overrightarrow{MM_2} + \\overrightarrow{MM_3} = \\overrightarrow{MA_1} + \\overrightarrow{MA_2} + \\overrightarrow{MB_1} + \\overrightarrow{MB_2} + \\overrightarrow{MC_1} + \\overrightarrow{MC_2} = \\overrightarrow{MA} + \\overrightarrow{MB} + \\overrightarrow{MC}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integers $a$, $p$ satisfy: $p = 2^a - 1$. Find all $a$ such that $\\frac{1}{2}(p^2+1)$ is a square of an integer.",
    "options": [],
    "answer": "a = 1 and a = 3",
    "solution": "For $a=1$ we have $p=1$ and $\\frac{1}{2}(p^2+1)=1$ satisfies the problem.\n\nFor $a=2$ we have $p=3$ and $\\frac{1}{2}(p^2+1)=5$ doesn't satisfy the problem.\n\nAssume now $a \\ge 3$. Let $\\frac{1}{2}(p^2+1) = p_1^2$, then $p^2 - 2p_1^2 = -1$. Hence: $2^{2a} - 2^{a+1} + 1 - 2p_1^2 = -1$ or $2^{2a-1} - 2^a = p_1^2 - 1$. So $2^a(2^{a-1}-1) = (p_1-1)(p_1+1)$. LHS is even, so as RHS. So $\\gcd(p_1-1, p_1+1) = 2$. So only the following cases are possible.\n\n1) $p_1+1=2l$, and $kl = 2^{a-1}-1$.\n\nIf $k \\ge 2$ $p_1 \\ge 2^a+1$ and $l \\ge 2^{a-1}+1$, a contradiction with $kl = 2^{a-1}-1$.\n\nIf $k=1$ $p_1 = 2^{a-1}+1$ and $l=2^{a-2}+1$ and $kl = 2^{a-2}+1 = 2^{a-1}-1$. Hence $a=3$.\n\n2) $p_1-1=2k$, $p_1+1=2^{a-1}l$, and $kl = 2^{a-1}-1$.\n\nIf $l \\ge 2$ $p_1 \\ge 2^a-1$ and $k \\ge 2^{a-1}-1$, so $2^{a-1}-1 = kl \\ge 2^a-2$, hence $a=1$.\n\nIf $l=1$ $p_1 = 2^{a-1}-1$ and $k = 2^{a-2}-1$ and $kl = 2^{a-2}-1 = 2^{a-1}-1$ - contradiction.\nSubstitute $p = 2^a - 1$ into equality $\\frac{1}{2}(p^2+1) = n^2$:\n$$\n(2^a - 1)^2 + 1 = 2n^2. \\quad (*)\n$$\nIf $a=1$ we have $n=1$ satisfies the problem.\n\nIf $a>1$ we have that equality (*) becomes $2^{2a-1}-2^a+1=n^2$, so $n$ is odd. The last equality can be transformed to: $(2^{a-1})^2 + (2^{a-1}-1)^2 = n^2$ or\n$$\n2^{2a-2} = n^2 - (2^{a-1}-1)^2 = (n-2^{a-1}+1)(n+2^{a-1}-1) \\Leftrightarrow 2^{2a-2} = xy.\n$$\nAlso $x = n-2^{a-1}+1 < y = n+2^{a-1}-1$. If $x, y$ are even, then they are powers of two, and also $y-x = 2^a-2 \\equiv 2 \\pmod 4$. Then $x=2$, and $y = 2^a = 2^{2a-3} \\Rightarrow$ the second solution is $a=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the maximum positive integer that is divisible by $7$, and all digits of which are odd, and their sum equals $2015$.",
    "options": [],
    "answer": "A = (10^2013 − 1)/9 + 2·10^2008, i.e., the 2013-digit number with four ones, then a three, followed by 2008 ones",
    "solution": "It is obvious that the greater the number is, the more digits it has. Since every digit equals at least $1$, the maximum amount of digits is $2015$. Such number is unique: $\\underline{11\\dots1}$, since every digit is $1$ and their sum is $2015$.\n\nHowever, $111\\dots1$ (with $2015$ digits) is not divisible by $7$. For example, $11111 = 15873 \\cdot 7$, but $11111\\underline{\\hspace{2cm}}7$ and $2015 = 335 \\cdot 6 + 5$, so this number is not divisible by $7$.\n\nIt is not possible that the number has $2014$ digits, since the sum of $2014$ odd numbers cannot be odd. That means that the maximum amount of digits in the number is $2013$. Obviously, it consists of $2012$ digits $1$ and one digit $3$, since no other cases are possible. We will find the maximum number with such structure that is divisible by $7$. That means we have to find such number with the digit $3$ in the maximum possible position. Thus the number will look like $A = \\underline{11\\dots1} + 2 \\cdot 10^k$, where $k \\le 2012$, and $A$ is divisible by $7$ and $k$ is the maximum possible number.\n\nSince $A = \\overline{11\\dots1} + 2 \\cdot 10^k$, and $2010 = 335 \\cdot 6$ and $11111:7$, then $B = 111 + 2 \\cdot 10^k$. Since $111 \\equiv 6 \\pmod{7}$, then $2 \\cdot 10^k \\equiv 1 \\pmod{7}$. Consider periodicity of remainders of such numbers modulo $7$ (remembering that $10 \\equiv 3 \\pmod{7}$):\n$$\n2 \\cdot 10^0 = 2 \\pmod{7},\\quad 2 \\cdot 10^1 = 6 \\pmod{7},\\quad 2 \\cdot 10^2 = 4 \\pmod{7},\\quad 2 \\cdot 10^3 = 5 \\pmod{7},\\\\\n2 \\cdot 10^4 = 1 \\pmod{7},\\quad 2 \\cdot 10^5 = 3 \\pmod{7},\\quad 2 \\cdot 10^6 = 2 \\pmod{7}, \\dots\n$$\nAnd so on. Thus values for $k$ will look like $k = 6l + 4$. Since $2010 = 335 \\cdot 6$, then the maximum possible value for $k = 334 \\cdot 6 + 4 = 2008$, therefore, the maximum number is $11113\\overline{11\\dots1}$ (with $2008$ digits $1$ after the $3$).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45416,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Andrii and Olesia both received the set of cards, on which all integer numbers from $1$ till $2015$ are written. After that Olesia leaves herself some amount of cards (but not all) from her set, and the rest she puts aside. Andrii does the same. There are $2015^2$ points on the coordinate plane, and coordinates are integer and bounded from $1$ till $2015$. Olesia paints in blue color the points whose first coordinate equals the number on one of her cards and the second coordinate is on one of Andrii's cards, and Andrii paints in blue color the points whose first coordinate equals the number on one of his cards and the second coordinate is on one of Olesia's cards (some points can be painted in both yellow and blue colors). Prove that no matter which cards Andrii and Olesia will choose they can't make all the $2015^2$ points painted at least in one color.\n\nb) Oxana joins Andrii and Olesia and takes all the cards that Olesia and Andrii put aside. Then they paint the coordinate plane with the following rule: Olesia paints in blue color the points, the first coordinate of which equals the number on one of her cards and the second coordinate is on one of Andrii's cards, Andrii paints in yellow color the points the first coordinate of which equals the number on one of his cards, and the second coordinate is on one of Oxana's cards, and Oxana paints in green color the points, first coordinate of which equals the number on one of her cards and the second coordinate is on one of Olesia's cards (again some points can be painted in more than one color). How do Olesia and Andrii have to choose the cards in order to paint every point from $2015^2$ at least in one color?",
    "options": [],
    "answer": "a) It is impossible to color all points; for any number not chosen by one player the point with both coordinates equal to that number remains uncolored. b) Olesia and Andrii must ensure that no number is omitted by both of them, i.e., their discarded sets are disjoint so that every number is chosen by at least one of them; under this condition all points are colored.",
    "solution": "a) Without loss of generality, we will consider that Olesia didn't choose the card with the number $a$. Then the point $(a, a)$ cannot be painted. Olesia doesn't paint it because the first coordinate is $a$, Andrii doesn't paint it because the second coordinate is $a$.\n\nb) If there is one number, for instance $1$, that was not chosen by either Olesia or Andrii, then the point $(1, 1)$ cannot be painted by Olesia, nor by Andrii (since the first coordinate is $1$), nor by Oxana, because the second coordinate is $1$.\n\nLet all the cards in the sets that were not chosen by Olesia and Andrii be pairwise distinct. Thus, it is obvious that every number from $1$ till $2015$ will be chosen at least once. We will show that under such a condition all points will be painted.\n\nConsider the point $(a, b)$ and possible cases.\n\nIf $b$ is not chosen by Andrii, then Olesia chose it and it goes to Oxana. If $a$ is the one Andrii also chose, then $(a, b)$ is painted by Andrii. If not, then $a$ goes to Oxana and Olesia, since Andrii didn't choose it, thus $(a, b)$ is painted by Oxana.\n\nIf $a$ was not chosen by Olesia, then Andrii and Oxana both have it. If $b$ is the one Olesia has, then $(a, b)$ is painted by Oxana; if Olesia doesn't have it, then Oxana has it and the point $(a, b)$ is painted by Andrii. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle and altitudes $AA_1$ and $BB_1$ intersect at $H$. Consider circles $w_1$ and $w_2$ with centers $H$ and $B$ and with radii $HB_1$ and $BB_1$ respectively. Let $CN$ and $CK$ be the tangent lines from $C$ to circles $w_1$ and $w_2$ respectively $(N \\neq B_1, K \\neq B_1)$. Prove that $A_1, N$ and $K$ are collinear.\n(Igor Nagel)\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $CC_1$ be the altitude (Fig. 42). Since quadrilateral $AB_1HC_1$ is cyclic, we have $\\angle A = 180^\\circ - \\angle C_1HB_1 = \\angle B_1HC_1$. Since $\\angle B_1HC_1 = \\angle NHC_1$, we obtain $\\angle A = \\angle CHN_1$.\n\nTaking into account $\\angle HA_1C = 90^\\circ$, $\\angle HNC = 90^\\circ$, $\\angle AC_1C = 90^\\circ$, we get that quadrilaterals $HA_1NC$ and $AC_1A_1C$ are cyclic. Hence $180^\\circ - \\angle C_1A_1C = \\angle BAC = \\angle CHN = \\angle CA_1N$ and therefore points $C_1, A_1$ and $N$ are collinear.\n\nSince $BA_1HC_1$ is cyclic, we have $\\angle HC_1A_1 = \\angle HBA_1$. Note that $\\angle A_1BK = \\angle HBA_1$. Since $C, C_1, B, K$ are cyclic, it follows that $\\angle CBK = \\angle KC_1C$. This means that $C_1, A_1$ and $K$ are collinear. Notice that points $C_1$ and $A_1$ are different. This proves that $C_1, A_1, N$ and $K$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\n4f(x+f(y)) = f(x)+f(y)+f(xy)+1\n$$\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "f(x) ≡ 1",
    "solution": "**Answer:** $f(x) \\equiv 1$.\n\nFirst suppose that there exists $a \\neq 0$ such that $f(a) = f(0) = b$. By (0) we denote the main equation.\nSubstituting $y=0$ in (0), we get:\n$$\n4f(x+b) = f(x)+b+b+1.\n$$\n(1)\n\nSubstituting $y=a$ in (0), we get:\n$$\n4f(x+b) = f(x)+b+f(xa)+1.\n$$\n(2)\n\nCombining (1) and (2), we obtain $f(ax) = b$, and we obviously have that $f$ is constant. Supposing that $f(x)=b$ and using (0) we get $4b=3b+1$. The result is $f(x) \\equiv 1$.\n\nNow, it follows that $f(a) = f(0) \\Rightarrow a = 0$.\nIf we replace $x$ by $y$ in (0) and combine the result and (0), we obtain:\n\n$f(x + f(y)) = f(y + f(x))$.\n\nSubstituting $x = -f(y)$ in the last equation, we get\n$$\n\\begin{aligned}\nf(0) &= f(-f(y)+f(y)) = f(y+f(-f(y))) \\Rightarrow y+f(-f(y)) = 0 \\\\\nf(-f(y)) &= -y.\n\\end{aligned}\n\\quad (3)\n$$\n\nSubstituting $x = 1$ and $y = -f(t)$ in (3), we obtain:\n$$\n4f(1+f(-f(t))) = f(1)+f(-f(t))+f(-f(t))+1.\n$$\nUsing (3), we get\n$$\n4f(1-t) = f(1)-t-t+1. \\quad (4)\n$$\nReplace $t = 1-z$ in (4), we have\n$$\n4f(z) = f(1) - 2(1-z) + 1 = 2z + c_1. \\text{ That is } f(z) = \\frac{1}{2}z + c.\n$$\nIt can be shown in the usual way that such function satisfies the condition of the question.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real $(x; y)$, such that:\n$$\n\\{x\\} + \\{y\\} = [x+y].\n$$",
    "options": [],
    "answer": "All pairs are: (n, −n) for any integer n, and (x, 1 − x) where x is a non-integer real.",
    "solution": "From the statement of the problem we conclude that $\\{x\\} + \\{y\\}$ is an integer, but $0 \\le \\{x\\} < 1$, so: $\\{x\\} + \\{y\\} = 0$ or $\\{x\\} + \\{y\\} = 1$.\n\n**Case 1:** $\\{x\\} + \\{y\\} = 0$. So $\\{x\\} = \\{y\\} = 0$, hence $x, y \\in \\mathbb{Z}$. But then $[x+y] = x+y=0$, so the solutions are $(n; -n)$, $n \\in \\mathbb{Z}$.\n\n**Case 2:** $\\{x\\} + \\{y\\} = 1$. Obviously $x, y \\notin \\mathbb{Z}$. Assume $x \\notin \\mathbb{Z}$, then\n$$\nx + y = [x] + \\{x\\} + [y] + \\{y\\} = [x] + [y] + 1,\n$$\nso $x+y$ is an integer. Hence $x+y = [x+y] = 1$. In this case the solutions are $(x; 1-x)$, $x \\notin \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45420,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a white square $8 \\times 8$. In one move Dmitry can choose a totally white square $2 \\times 2$ and paint in black color any two cells of this square, located on the diagonal. What is the maximum number of cells according to the following rules Dmitry will paint?\n\n![](attached_image_1.png)\n\nFig. 31",
    "options": [],
    "answer": "42",
    "solution": "**Answer:** 42.\n\nFirst show how to achieve the required number of colorings. In each box $4 \\times 4$ will painting in such sequence. First select all four squares $2 \\times 2$, on which the $4 \\times 4$ square is split and paint diagonals, as shown in Fig. 31 (black squares). After this there is a completely white square $2 \\times 2$ inside. We paint any of the diagonals (gray squares). Upon completion of painting all four squares $4 \\times 4$ in the center of $8 \\times 8$ square there is completely white $2 \\times 2$ square, where we can paint two more cells. In total: $10 \\cdot 4 + 2 = 42$ of squares $1 \\times 1$.\n\nNow show that the greater amount of paint cannot be achieved. Along with a given square $8 \\times 8$ (call him «initial») consider $7 \\times 7$ square, that formed from centers of the squares of the initial square (we call it «central»). Coloring the diagonal of the square $2 \\times 2$ in the initial square corresponds to drawing a diagonal in the central square (Fig. 32).\n\n![](attached_image_2.png)\nFig. 32\n\nNow it is easy to see that in two adjacent squares $1 \\times 1$ of the central square the diagonals can't be drawn. Let us consider rectangle $3 \\times 4$ of the central square. We will show that it is impossible to draw diagonals in exactly half of all squares. Assume the contrary. Then the diagonals are drawn in six squares $1 \\times 1$. Let us choose four squares A, B, C, D, that contains diagonals and there is no corners among them (Fig. 33). Let us show that it is impossible to draw diagonals in all of them. WLOG the first diagonal is drawn in the square A, in the direction that is shown at Fig. 8. Then in the square C it is impossible to draw a diagonal. So in the rectangle $3 \\times 4$ it is possible to draw not more than 5 diagonals. Then divide $7 \\times 7$ square into four rectangles $3 \\times 4$ and square $1 \\times 1$, as it shown at Fig. 34. Then the maximum number is $5 \\cdot 4 + 1 = 21$ in total, that is exactly 42 colored squares.\n\n![](attached_image_3.png)\nFig. 33\n\n![](attached_image_4.png)\nFig. 34",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45421,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the convex quadrilateral $ABCD$ with angles $ABC$ and $BCD$ equal $120^\\circ$, $O$ is the intersection of diagonals, $M$ is the midpoint of $BC$, $K$ is the point of intersection of $MO$ and $AD$. It happens that $\\angle BKC = 60^\\circ$. Prove that $\\angle BKA = \\angle CKD = 60^\\circ$.\n\n(Serduk Nazar)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let lines $AB$ and $CD$ intersect at $E$. We have that $\\angle BKC = 60^\\circ$. Let us prove that $BK > \\frac{1}{2}BC$. Indeed, one of the angles $\\angle KBC$ or $\\angle KCB$ is at least $60^\\circ$. WLOG, this angle is $\\angle KBC$ (Fig. 35). Then in $\\triangle BMK$ $\\angle KBM \\ge 60^\\circ = \\angle BKC > \\angle BKM$. So $MK > BM$, hence on the segment $BK$ there is point $O'$, such that $MK \\cdot MO' = BM^2 = MC^2$. Then circumcircles of $\\triangle BO'K$ and $\\triangle CO'K$ are tangent to $BC$. Hence, $\\angle BKM = \\angle O'BC$, $\\angle CKM = \\angle O'CB$. So $\\angle BO'C = 180^\\circ - \\angle O'BC - \\angle O'CB = 180^\\circ - \\angle BKM - \\angle MKC = 180^\\circ - 60^\\circ = 120^\\circ$\n\n![](attached_image_1.png)\n\nFig. 35\n\nThen point $O'$ is on circumcircle of $\\triangle EBC$. Let line $BO'$ intersect $CE$ at $D'$, $CO'$ intersect $BE$ at $A'$. From the fact that $BECO'$ is cyclic we conclude that (Fig. 36)\n$$\n\\angle BO'E = \\angle BCE = 60^\\circ = \\angle EBC = \\angle EO'C.\n$$\nConsider $\\triangle O'EB$ and $\\triangle EBD'$. Two pairs of angles are equal. Hence $\\angle BEO' = \\angle ED'B$. Then $\\angle BEO' = \\angle BCO' = \\angle CKO' = \\angle ED'B$. So point $D'$ is on circumcircle of $\\triangle KO'C$. Similarly, point $A'$ is on the circle $\\triangle BKO'$. Then $\\angle A'KB = \\angle A'O'B = 60^\\circ = \\angle CO'D' = \\angle CKD'$. It follows that points $A'$, $K$ and $D'$ are on the same line.\n\nPoint $O$ lies on segment $MK$. If this point is inside $MO'$, then $A$ lies inside $A'B$, $D'$ inside $CD'$. Then $K$ lies strictly outside $\\triangle EAD$, but by the problem $K$ is inside $AD$, a contradiction. Similarly, if point $O$ is outside $MO'$, then $A$, $D$ are outside $EA'$ and $ED'$, so $K$ is strictly inside $\\triangle EAD$, again a contradiction. Hence $O = O'$. Then $A = A'$ and $D = D'$. But this is already proved that $\\angle A'KB = \\angle CKD' = 60^\\circ$, which is needed, because $A = A'$ and $D = D'$.\nLet us draw external bisector of $\\angle BKC$. Let it intersect lines $BA$ and $CD$ at $A'$ and $D'$ (Fig. 36). We draw on sides $\\triangle BKC$ outside equilateral triangles $BPC$, $BRK$ and $CQK$. Obviously $R$ and $Q$ are on $A'D'$, and $P$ is an intersection of $AB$ and $CD$.\n\nConsider $\\triangle BKC$. We saw, that\n![](attached_image_2.png)\n\n$$\n\\angle CBR = \\angle CBK + 60^\\circ = 180^\\circ - (120^\\circ - \\angle CBK) = \\angle KBA'\n$$\n\nIt implies that $BR$ and $BA'$ are isogonal, also $KR$ is external bisector. So $A'$ and $R$ are isogonal, hence $CA'$ and $CR$ are isogonal. Similarly we have that segments $BD'$ and $BQ$ are isogonal. It is easy to see that $P$ - the point of intersection of tangents to circumcircle of $\\triangle BKC$, so $\\angle PBC = \\angle BKC = \\angle PCB = 60^\\circ$. For this $KP$ is symmedian of $\\triangle BKC$, so $KP$ and $KM$ are isogonal. As it is known $KP$, $BQ$ and $CR$ intersect in the same point, namely at Fermat point, so as lines $KM$, $BD'$ and $CA'$ intersect in the same point. Let this be point $O'$. Assume that $MO > MO'$ (Fig. 37). Then points $A'$ and $D'$ are on sides $BA$ and $CD$, but in this case $AD$ and $A'D'$ don't intersect each other. Similarly if $MO < MO'$. Hence $MO = MO' \\Rightarrow A = A', D = D' \\Rightarrow \\angle AKB = \\angle DKC$.\n\n![](attached_image_3.png)\n\nFig. 37",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45422,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Determine whether there exist positive integer numbers $a_1, a_2, ..., a_{2015}$ such that any two of them are co-prime and $a_1a_2...a_{2015}-1$ is a product of two consequent odd numbers?\n\nb) Determine whether there exist positive integer numbers $a_1, a_2, ..., a_{2015}$ such that: any two of them are co-prime and $a_1a_2...a_{2015}-1$ is a product of two consequent even numbers?",
    "options": [],
    "answer": "a) Yes. For example, take the squares of the first 2015 primes including two; then the product minus one equals the product of two consecutive odd numbers. b) Yes. For example, take the squares of the first 2015 odd primes; then the product minus one equals the product of two consecutive even numbers.",
    "solution": "a) Let $a_1 = p_1^2, a_2 = p_2^2, ..., a_{2015} = p_{2015}^2$, where $p_1 = 2, p_2, ..., p_{2015}$ are the first 2015 prime numbers. It is clear that every two of these numbers are co-prime and $a_1a_2...a_{2015}-1 = (p_1p_2...p_{2015}-1)(p_1p_2...p_{2015}+1)$ is a product of two consequent odd numbers.\n\nb) Let $a_1 = p_1^2, a_2 = p_2^2, ..., a_{2015} = p_{2015}^2$, where $p_1 = 3, p_2 = 5, ..., p_{2015}$ are the first 2015 odd prime numbers. It is clear that every two of these numbers are co-prime and $a_1a_2...a_{2015}-1 = (p_1p_2...p_{2015}-1)(p_1p_2...p_{2015}+1)$ is a product of two consequent even numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(a, b)$ for which the following condition holds: there exists index $n_0$ such that the sequence $(a_n + a_{n+1}), n \\ge n_0$ is non-decreasing.\n\nConsider a sequence $(a_n)$: $a_1 = a, a_2 = b$, where $a$ and $b$ are positive integers, and for all $n \\ge 2$ $a_{n+1}$ equals the number of indexes $i, 1 \\le i \\le n$, such that $a_i = a_n$.",
    "options": [],
    "answer": "(1, 1)",
    "solution": "If $a = b = 1$ then $(a_n)$ has terms $(1; 1; 2; 1; 3; 1; 4; 1; 5; 1; ...)$, and $(a_n + a_{n+1})$ has terms $(2; 3; 3; 4; 4; 5; 5; 6; 6; 7; ...)$.\nTherefore $a = b = 1$ satisfies the condition of the question.\n\nLet $a_1 = a$ and $a_2 = b$. Suppose that $a_1 \\ne 1$ or $a_2 \\ne 1$. Without loss of generality it can be assumed that $a_1 \\ne 1$ (if not then we just swap $a_1, a_2$). It is clear that $(a_n)$ is unbounded. Indeed, if $\\max\\{a_i : i \\in \\mathbb{N}\\} = s$ then there must be at least $s+1$ equal numbers among $\\{a_1, a_2, ..., a_{s+1}\\}$. Thus the next number is greater or equal to $s+1$. Contradiction.\nNow, assume that for $n \\ge n_0$ sequence $(a_n + a_{n+1})$ is non-decreasing. Consider a number $k \\ge n_0 + 1$ such that $a_k = t \\ge \\max\\{a, b\\} + 1$ and this is the first number $t$ in the sequence. Obviously, all numbers before $a_k$ are lower than $t$. Then $a_{k+1} = 1$ and since $(a_n + a_{n+1})$ is non-decreasing, it follows that $a_{k-1} = 1$. Thus there exist exactly $t$ indexes $\\{1 < i_1 < i_2 < ... < i_t = k-1\\}$ such that $a_{i_1} = a_{i_2} = ... = a_{i_t} = 1$. Hence numbers $\\{a_{i_1-1}; a_{i_2-1}; ..., a_{i_t-1}\\}$ are lower or equal to $t-1$ and there must be at least two equal numbers among $\\{a_{i_1-1}; a_{i_2-1}; ..., a_{i_t-1}\\}$. Therefore, there must be at least two equal pairs $\\{..., l, 1, ...\\}$ in our sequence. However, after the second number $l$ it must be at least 2. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45424,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every real numbers $a \\ne b$ solve the system:\n$$\n\\begin{cases} \n3x+z = 2y+(a+b), \\\\\n3x^2 + 3xz = y^2 + 2(a+b)y + ab, \\\\\nx^3 + 3x^2z = y^2(a+b) + 2yab.\n\\end{cases}\n$$\n(Rublyov Bogdan)",
    "options": [],
    "answer": "x = y = a, z = b; and x = y = b, z = a",
    "solution": "**Answer.** $x = y = a, z = b$ and $x = y = b, z = a$.\n\nLet $(x, y, z)$ be a solution of the system. Consider polynomials\n$$\n\\begin{aligned}\nP(t) &= (t-x)^3(t-z) = t^4 + p_1t^3 + q_1t^2 + r_1t + s_1 \\text{ and} \\\\\nQ(t) &= (t-y)^2(t-a)(t-b) = t^4 + p_2t^3 + q_2t^2 + r_2t + s_2,\n\\end{aligned}\n$$\nand let:\n$$\nf(t) = P(t) - Q(t) = (p_1 - p_2)t^3 + (q_1 - q_2)t^2 + (r_1 - r_2)t + (s_1 - s_2).\n$$\nNumbers $x, x, x, z$ are roots of $P(t)$ and $y, y, a, b$ are roots of $Q(t)$. Using Vieta's formulas, we get:\n$$\n\\begin{aligned}\np_1 &= -(3x+z) = -(2y+(a+b)) = p_2 \\\\\nq_1 &= 3x^2 + 3xz = y^2 + 2(a+b)y + ab = q_2 \\\\\nr_1 &= -(x^3 + 3x^2z) = -(y^2(a+b) + 2yab) = r_2\n\\end{aligned}\n$$\n\n$$\nP'(t) = (t-x)^2(4t-3z-x) = Q'(t) = (t-y)(4t^2-(3a+3b+2y)t+2ab+y(a+b)).\n$$\nConsider $R(t) = 4t^2 - (3a+3b+2y)t + 2ab + y(a+b)$ and take $t_0 = \\frac{1}{2}(a+b)$. We obtain\n$$\nR(t_0) = (a+b)^2 - 3(a+b)\\frac{a+b}{2} - 2y\\frac{a+b}{2} + 2ab + y(a+b) = -\\frac{(a+b)^2}{2} + 2ab = -\\frac{(a-b)^2}{2} < 0.\n$$\nThus $R(t)$ has two different roots and thus the discriminant of $R(t)$ is positive. But we have $P'(t) \\equiv Q'(t)$ and $P'(t) \\ne (t-x)^2$. Then $Q'(t) \\ne (t-x)^2$. Since $R(t)$ has two different roots, we get that exactly one of them equals to $x$ and $y = x$. Moreover, $R(y) = 0$, that is\n$$\nR(y) = 4y^2 - (3a + 3b + 2y)y + 2ab + y(a + b) = 2(y^2 - (a + b)y + ab) = 0.\n$$\n\nBut this is possible only if $y = a$ or $y = b$. Finally, we obtain\n$$\nf(t) = P(t) - Q(t) = (t - x)^3 (z - c) = C,\n$$\nwhere $c = a$ or $c = b$. But this is possible only if $z = c$. Therefore $P$ equals to $Q$ and we have two solutions\n$$\nx = y = a, z = b, \\text{ or } x = y = b, z = a.\n$$\nThe second part also could be done with application of Rolle's theorem. Consider polynomial $Q(t) = (t - y)^2(t - a)(t - b)$. Suppose that $y, a, b$ are pairwise distinct. Then derivative $Q'(t)$ has three distinct roots. One of them $t = y$ and two other belong to the intervals between $y, a, b$. But $P'(t)$ equals to $Q'(t)$ and thus $P'(t)$ also has three distinct roots. But $P'(t): (t - x)^2$. Contradiction. And thus some of the numbers $y, a, b$ are equal. Since $a \\neq b$, it follows that $y = a$ or $y = b$. Consider the case $y = a$. Then $Q(t) = (t - y)^3(t - b)$ and $Q'(t): (t - x)^2$. Using that $Q'(t) = P'(t):(t - x)^2$, we get $x = y$, and thus:\n$$\nC = (t - x)^3 (t - z) - (t - x)^3 (t - b) = (t - x)^3 (z - b),\n$$\nthat is only possible if $z = b$. And we obtain the solution $x = y = a, z = b$. Similarly, in the second case we get the solution: $x = y = b, z = a$.\n**Alternative solution.** From the first and second equations we get:\n$$\n\\begin{cases} a+b=3x+z-2y \\\\ ab=3x^2+3y^2-6xy+3xz-2yz. \\end{cases} \\qquad (1)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer $n$, such that:\n$$\n(n-2013)(n-2014)(n-2016)(n-2017)=4.\n$$",
    "options": [],
    "answer": "2015",
    "solution": "If integer number $n$ satisfies the given condition, then $4$ can be presented as product of four pairwise distinct integers. Since the integer divisors of this number are only $\\pm1$, $\\pm2$ and $\\pm4$, we have that sought-for divisors are $\\pm1$ and $\\pm2$. Indeed, if the absolute value of one of divisors is equal to $4$ then others are not less than $1$ by absolute value, a contradiction. Since $n-2013$ is the biggest factor, it should be equal to $2$. Also we see that $n=2015$ satisfies the condition of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nastia has 5 yellow coins, among which all are authentic. She has also 5 blue coins, among which three are authentic and two are fake. All 8 authentic coins have the same weight, one of the fake coins is heavier than authentic by 1 gram, and the other is lighter by 1 gram. Can Nastia, using scales without weights and 3 weighings, determine both fake coins and indicate which one is heavier and which lighter?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote blue coins by $B_1, \\ldots, B_5$. Firstly, weigh 3 yellow and 3 blue coins $B_1, B_2, B_3$:\n$$\n1) \\text{Ж}_1 + \\text{Ж}_2 + \\text{Ж}_3 \\ ? \\ B_1 + B_2 + B_3.\n$$\n\n$$\n1a) \\text{Ж}_1 + \\text{Ж}_2 + \\text{Ж}_3 = B_1 + B_2 + B_3.\n$$\nFake coins are among $B_1, B_2, B_3$, or among $B_4$ and $B_5$. In a group where fakes are, all coins have different weight.\n\n2) $B_1 \\ ? \\ B_2$.\n$$\n\n$$\n2a) B_1 = B_2.\n$$\nIt means that these coins are authentic, so $B_4$ and $B_5$ are fake. We only have to compare their weight.\n\n$$\n2b) B_1 > B_2.\n$$\nConsider three possible cases. $B_1$ is heavier, $B_2$ lighter, or $B_1$ is heavier, $B_2$ authentic, or $B_1$ authentic, $B_2$ lighter. Compare total weight of coins $B_1, B_2$ and $\\text{Ж}_1, \\text{Ж}_2$ (both authentic). In the first case they are equal, in the second $B_1, B_2$ heavier, in the third - lighter than $\\text{Ж}_1, \\text{Ж}_2$, so we get answers to all questions after the third weighing:\n$$\n3) \\text{Ж}_1 + \\text{Ж}_2 \\ ? \\ B_1 + B_2.\n$$\n\n$$\n2c) B_1 < B_2.\n$$\n\n$$\n16) \\text{Ж}_1 + \\text{Ж}_2 + \\text{Ж}_3 > B_1 + B_2 + B_3.\n$$\nAmong $B_1, B_2, B_3$ there is a lighter fake coin, and among $B_4, B_5$ - a heavier one. Then we use the following weighings:\n$$\n2) B_1 \\ ? \\ B_2 \\text{ and } 3) B_4 \\ ? \\ B_5.\n$$\nThe end in this case is clear.\n\n$$\n1c) \\text{Ж}_1 + \\text{Ж}_2 + \\text{Ж}_3 < B_1 + B_2 + B_3.\n$$\nThis case is similar to 1b), but now among $B_1, B_2, B_3$ there is a heavier fake coin.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation:\n$$\n\\frac{1}{\\sqrt{2014}-\\sqrt{x}} + \\frac{1}{\\sqrt{2015}-\\sqrt{x+1}} + \\frac{1}{\\sqrt{2016}-\\sqrt{x+2}} = \n\\frac{1}{\\sqrt{2015-x}-\\sqrt{1}} + \\frac{1}{\\sqrt{2016-x}-\\sqrt{2}} + \\frac{1}{\\sqrt{2017-x}-\\sqrt{3}}\n$$",
    "options": [],
    "answer": "1",
    "solution": "Multiply both numerator and denominator by the conjugate:\n$$\n\\frac{\\sqrt{2014+\\sqrt{x}}}{2014-x} + \\frac{\\sqrt{2015+\\sqrt{x+1}}}{2015-(x+1)} + \\frac{\\sqrt{2016+\\sqrt{x+2}}}{2016-(x+2)} = \\frac{\\sqrt{2015-x+\\sqrt{1}}}{(2015-x)-1} + \\frac{\\sqrt{2016-x+\\sqrt{2}}}{(2016-x)-2} + \\frac{\\sqrt{2017-x+\\sqrt{3}}}{(2017-x)-3}\n$$\nAfter multiplying by the common denominator we have:\n$$\n\\sqrt{2014+\\sqrt{x}} + \\sqrt{2015+\\sqrt{x+1}} + \\sqrt{2016+\\sqrt{x+2}} = \n\\sqrt{2015-x} + \\sqrt{1} + \\sqrt{2016-x} + \\sqrt{2} + \\sqrt{2017-x} + \\sqrt{3},\n$$\n\n1) $x=1$ is the solution of the equation:\n$$\n\\sqrt{2014} + \\sqrt{1} + \\sqrt{2015} + \\sqrt{2} + \\sqrt{2016} + \\sqrt{3} = \\sqrt{2014} + \\sqrt{1} + \\sqrt{2015} + \\sqrt{2} + \\sqrt{2016} + \\sqrt{3}.\n$$\n\n2) Denote $S = \\sqrt{2014} + \\sqrt{1} + \\sqrt{2015} + \\sqrt{2} + \\sqrt{2016} + \\sqrt{3}$. For $x > 1$ we have\n$$\n\\sqrt{2014} + \\sqrt{x} + \\sqrt{2015} + \\sqrt{x+1} + \\sqrt{2016} + \\sqrt{x+2} > \n\\sqrt{2014} + \\sqrt{1} + \\sqrt{2015} + \\sqrt{2} + \\sqrt{2016} + \\sqrt{3} = S.\n$$\nSo the left-hand side is greater than $S$, and in the same way we analyze the right-hand side:\n$$\n\\sqrt{2015-x} + \\sqrt{1} + \\sqrt{2016-x} + \\sqrt{2} + \\sqrt{2017-x} + \\sqrt{3} < \n\\sqrt{2014} + \\sqrt{1} + \\sqrt{2015} + \\sqrt{2} + \\sqrt{2016} + \\sqrt{3} = S.\n$$\nSo in this interval there are no roots. The same holds for $x < 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Trapezoid $ABCD$ with $BC \\parallel AD$ is given. On diagonals $AC$ and $BD$ denote points $P$ and $Q$ respectively, so that $AC$ bisects $\\angle BPD$, $BD$ bisects $\\angle AQC$. Prove that $\\angle BPD = \\angle AQC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M_1$ and $M_2$ be the middles of the diagonals (Fig. 26). Then $M_1M_2 \\parallel BC$. Consider circumcircles of $\\angle AQC$ and $\\angle BPD$. Let them intersect $BD$ and $AC$ in points $X$ and $Y$. Since $BD$ contains the bisector of $\\angle AQC$, then point $X$ is the middle of the bigger arc $AC$ of the circumcircle of $\\angle AQC$. Then $X$ is on the perpendicular bisector of $AC$. In the same way, $Y$ is on the perpendicular bisector of $BD$. So $XM_1M_2Y$ is inscribed, as $\\angle XM_1Y = \\angle XM_2Y = 90^\\circ$. Then $\\angle M_1XM_2 = \\angle M_1YM_2$. Also $\\angle BXY = \\angle CM_1M_2 = \\angle M_1CB$, so $XBCY$ is inscribed. Hence, $\\angle BXC = \\angle BYC$.\n\n$$\n\\angle AQC = 180^\\circ - 2\\angle M_1XC = 180^\\circ - 2(\\angle M_1XM_2 + \\angle M_2XC) = \\\\\n= 180^\\circ - 2(\\angle M_1YM_2 + \\angle M_1YB) = 180^\\circ - 2\\angle M_2YB = \\angle BPD.\n$$\n\n\nAlternative solution.\n\nDenote on $BD$ point $Q'$ such that $\\angle BQ'C = \\angle BPC$. Then $BCQ'P$ is inscribed (Fig. 27). Hence $\\angle Q'PC = \\angle Q'BC = \\angle Q'DA$, so $APQ'D$ is also inscribed. Then\n$\\angle CPD = \\angle CPQ' + \\angle Q'PD = \\angle Q'DA + \\angle Q'AD = \\angle BQ'A.$\nSo $\\angle BPD = \\angle CQ'A$. Also $BD$ bisects $\\angle CQ'A$. Let us show that $Q'=Q$. Assume these points are distinct. Because of the fact that\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45429,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that the arithmetic average of the numbers *a*, *b* is equal to the number *c*, so $c = \\frac{1}{2}(a+b)$, and that the harmonious average number of *a*, *c* is equal to the number *b*, so $b = \\frac{2}{\\frac{1}{a}+\\frac{1}{c}}$. Is it necessary that numbers *a*, *b*, *c* are equal?\n\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's rewrite the condition of harmonious average: $b = \\frac{2ac}{a+c}$, and now use the fact $c = \\frac{a+b}{2}$:\n$$\n2a \\cdot \\frac{a+b}{2} = b\\left(a + \\frac{a+b}{2}\\right) \\Leftrightarrow a^2 + ab = ba + \\frac{ab+b^2}{2} \\Leftrightarrow 2a^2 = ba + b^2 \\Leftrightarrow (a-b)(2a+b) = 0.\n$$\nLet's denote, for example, $a=2$, which means $b=-4$ and $c=-1$, hence we receive three different numbers satisfying the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45430,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the trapezoid $ABCD$ with perpendicular diagonals, points $P$, $N$, $Q$, $M$ are the middles of sides $AB$, $BC$, $CD$, $DA$ respectively. On the base $CD$ there is a point $L$ (different from the point $Q$) for which the angle $MLN$ is straight. Find the angle $LPA$.\n\n(Bogdan Rublyov)",
    "options": [],
    "answer": "90°",
    "solution": "By Varignon's Theorem quadrangle $MPNQ$ is a parallelogram and its sides are parallel to the diagonals of trapezoid $ABCD$, so it is a rectangle (Fig. 38). Denote $O$ the intersection point of $AC$ and $BD$. Then from the properties of rectangular $\\triangle MNL$ we get that: $OL = OM = ON \\Rightarrow OL = OP = OQ$, so $\\triangle PQL$ is rectangular. That means $PL \\perp LQ$, and since $AB \\parallel CD$ then $PL \\perp AB$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all three prime numbers $p$, $q$, $r$ that satisfy\n$$\n\\frac{q}{p-1} + \\frac{r}{p+1} = \\frac{q+r+1}{p}.\n$$",
    "options": [],
    "answer": "(p, q, r) = (2, 2, 3) and (3, 3, 2)",
    "solution": "Let's rewrite the given equality as follows:\n$$\n\\frac{q}{p-1} - \\frac{q}{p} = \\frac{r}{p} - \\frac{r}{p+1} + \\frac{1}{p} \\Leftrightarrow \\frac{q}{p(p-1)} = \\frac{r}{p(p+1)} + \\frac{1}{p} \\Leftrightarrow \\frac{q}{p-1} = \\frac{r}{p+1} + 1 \\Leftrightarrow q = \\frac{(p-1)r}{p+1} + p-1 \\Leftrightarrow q = p + r - 1 - \\frac{2r}{p+1} \\quad (1)\n$$\n\nSince $p$, $q$, $r$ are prime, then the number $\\frac{2r}{p+1}$ is a positive integer. The number $2r$ has only four divisors: $1$, $2$, $r$ and $2r$. Since $p+1 \\ge 3$, then two cases are possible: $p+1 = r$ or $p+1 = 2r$.\n\n1) Suppose that $p+1 = r$ which means $p = r-1$. $p+1 \\ge 3$, so the only pair of consecutive prime numbers is $2$ and $3$, and thus, $p = 2$ and $r = 3$. Then from (1) we find that $q = 2$. After checking we make certain that $p = q = 2$, $r = 3$ is an answer.\n\n2) Suppose that $p+1 = 2r$ which means $p = 2r-1$. Then from (1) we find that $q = 3r-3 \\div 3$, and since $q$ is prime then $q = 3$. Further, consistently find that $r = 2$, $p = 3$. Checking shows that $p = q = 3$, $r = 2$ is an answer as well.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45432,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $N$ points on the circle. Andriy and Olesya play the game as follows. Andriy goes first. They take turns and connect two of the given points with a chord, if it does not cross any of the prior chords. The one after whose move there is a triangle formed by the drawn chords wins. Who will win the fair game if:\n\na) $N=14$;\nb) $N=15$?",
    "options": [],
    "answer": "a) Andriy wins; b) Olesya wins.",
    "solution": "**Answer:** a) Andriy wins, b) Olesya wins.\n\nClearly, the one who is forced to draw a chord from the point from which there is another chord drawn already, will lose (Fig. 39). For example, one player draws $AC$ and chord $AB$ has already been drawn before (we assume that there are no other chords drawn from point $A$ except these two). Then the other player just draws a chord $BC$ and wins. It is obvious that this chord can be drawn, because if else, it would have crossed some other chord - $MN$, for example, and then one of the points, let's say $M$, would belong to the arc $\\cup BC$, and another one wouldn't.\nWhich means the point $N$ would belong to the arc $\\cup AB$, and then $MN$ would cross $AB$, or it would belong to $\\cup AC$ and then $MN$ would cross $AC$.\n\nSo the game will continue until one of the players will be forced to draw a second chord from one point and, as the result, to lose. It means that in the beginning both players will draw only chords that connect the \"free\" points (those points that have no chords drawn from them yet). At any moment all the free points can be divided into \"groups\" within each of which we can connect any two points. Moreover, the chords within the group are independent of other groups and do not affect other groups. So the position after each move can be written as a list of groups, by the number of free points in it. If the next chord is drawn in a group containing $m$ points, then instead of the old group there are two groups created containing $m_1$ and $m_2$ points, where $m_1 + m_2 = m-2$ (note that one or both groups may be empty). Thus, the starting point for $N=15$ can be written as (15), and after the first chord may form, for example, a position (10; 3), (8; 5), or (13; 0), etc. After the next move there will be 3 groups, and note that groups with (0) or (1) can be ignored, because we can't draw a chord there that would not lead to defeat.\n\nNow let us show some simple properties of groups of points that will shorten the process of analyzing the current situation.\n\n**Property 1.** In the group with 3 points one can always draw one chord, in the group with 5 points one can always draw 2 chords.\n\n**Property 2.** In the group with 4 points the one who draws a chord within this group may choose any of the two options for partitioning, (1; 1) or (2; 0), so there can be drawn either one or two chords, depending on this player's move.\n\n**Property 3.** In the group with 6 points the one who draws a chord within this group may choose one of the these options for partitioning: (2; 2) or (3; 1), so that there can be drawn 3 or 2 chords.\n\n**Property 4.** If there is a symmetrical situation (i.e. groups of points can be divided into pairs containing equal number of points), the winner is the one whose move resulted in such symmetrical situation.\n\n![](attached_image_1.png)\n\nProving these properties is almost obvious and can be easily conducted by going through each of those variants. Let us explain the property 4. If after the move of one of the players we get the situation $(k; k)$, he wins by using moves symmetric to his/her opponent's moves. There are other symmetric situations, but they are just this situation's analogues.\n\nLet's agree to not further specify the groups (1) and (0), because they do not affect the further course of the game.\n\na) Andriy wins due to move (6; 6), as his victory follows from property 4 through further symmetrical strategy.\n\nb) Olesya wins. Let's look at the possible first moves of Andriy.\n\nOption 1. (13): Olesya wins due to move (8; 3). Let's look at the possible answers of Andriy.\n\nVersion 1.1. (8): (Actually it was a move (8; 1; 0) but we do not specify the last two groups). Then Olesya goes (3; 3) and wins due to the symmetrical strategy.\n\nVersion 1.2. (3; 6): Since there is a group (3), where one move is possible, and the group (6) where Olesya can choose how many moves the game will last, then she breaks group (6) accordingly - making a move into position (3; 6) $\\to$ (3; 3).\n\nVersion 1.3. (3; 5): The game lasts exactly three moves, Olesya's move is the last and she wins.\n\nVersion 1.4. (3; 4; 2): due to group (4) Olesya wins. She should leave the groups (2,3).\n\nVersion 1.5. (3; 3; 3): Obviously, Olesya wins.\n\nOption 2. (12) $\\to$ (5; 5) and Olesya wins.\n\nOption 3. (11; 2) $\\to$ (8; 2) Andriy has such options:\n\n(6; 2), Olesya goes (3; 2) and wins.\n\n(5; 2), exactly three moves left. Thus, Olesya wins.\n\n(4; 2; 2), Olesya makes a move in a group with 4 points into (2; 2) and wins.\n\n(3; 2; 2), exactly three moves left. Which means Olesya wins.\n\nOption 4. (10; 3) $\\to$ (5; 3; 3) and the game will continue for another four moves, meaning Olesya wins.\n\nOption 5. (9; 4) $\\to$ (9). Then there are several sub-variants possible, which easily lead to Olesya's victory: (9) $\\to$ (7) $\\to$ (5) or (9) $\\to$ (6) $\\to$ (2; 2) or (9) $\\to$ (4; 3) $\\to$ (2; 3).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45433,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the tennis tournament in one circle attended the 8 girls (i.e. every tennis player has played with each other exactly once, draws in tennis does not happen). Oksana took the second place recruited points and there is no other participant with the same number of points. What is the maximum number of games could lose Olesya who won in this tournament?",
    "options": [],
    "answer": "1",
    "solution": "Suppose that Olesya lost 2 games. Then she scored 5 points. Oksana could not score 4 points, because then all together all teams had supplied the maximum $5+4+3 \\cdot 6 = 27$ wins. But in total we have $\\frac{1}{2}(8 \\cdot 7) = 28$ games, a contradiction. Similarly Oksana could not score 3 or less points, or Olesya lose 3 more games. Thus, only Olesya could lose up to 1 game. This option is possible, as evidenced by the table (Fig. 28) example.\n\n| M | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | p |\n|---|---|---|---|---|---|---|---|---|---|\n| 1 | X | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 6 |\n| 2 | 1 | X | 1 | 0 | 1 | 0 | 1 | 0 | 4 |\n| 3 | 0 | 0 | X | 1 | 0 | 1 | 0 | 1 | 3 |\n| 4 | 0 | 1 | 0 | X | 1 | 0 | 1 | 0 | 3 |\n| 5 | 0 | 0 | 1 | 0 | X | 1 | 0 | 1 | 3 |\n| 6 | 0 | 1 | 0 | 1 | 0 | X | 1 | 0 | 3 |\n| 7 | 0 | 0 | 1 | 0 | 1 | 0 | X | 1 | 3 |\n| 8 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | X | 3 |\n\nFig. 28",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45434,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the side $BC$ of acute triangle $ABC$ one chooses an arbitrary point $D$. Let $O$ be the circumcenter of $\\triangle ABC$, $Z$ the point on this circle which is diametrically opposite to $A$. Let $X$, $Y$ be points on segments $BO$, $CO$ such that:\n$$\n\\angle BXD + \\angle ABC = 180^{\\circ} = \\angle CYD + \\angle ACB.\n$$\n\nProve that the measure of angle $\\angle XZY$ doesn't depend on point $D$.\n(Khilko Danylo)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Here is the solution for the location of points showed in the picture. For other cases, the location of the solution will be similar. Prove that for any selected point $D$ quadrilateral $XOYZ$ is inscribed. Then $\\angle XZY = 180^{\\circ} - 2\\angle A$, so doesn't depend on $D$.\n\nFrom the statement of the problem we have that circles $\\triangle BXD$ and $\\triangle DYC$ are tangent to lines $AB$ and $AC$. Denote these circles as $w_1$ and $w_2$, and their centers are $O_1$ and $O_2$. Let $K$ be the second point of intersection of these circles other than $D$ (see figure below).\n\n![](attached_image_1.png)\n\nThen\n$$\n\\angle BKC = \\angle BKD + \\angle DKC = \\angle B + \\angle C = 180^{\\circ} - \\angle A,\n$$\nso $K$ is on the circumcircle $\\triangle ABC$. Also we have\n$$\n\\angle XKY = \\angle XKD + \\angle DKY = \\angle XBD + \\angle DCY = 180^{\\circ} - \\angle XOY,\n$$\nso quadrilateral $XOYK$ is inscribed. Because $AZ$ is the diameter of the big circle, we have that $AB \\perp BZ$, so $O_1$ is on line $BZ$. Similarly $O_2$ is on the line $CZ$. We prove that $O_1$, $O_2$ are on the circle $OXKY$. Indeed,\n$$\n\\angle XO_1K = 2\\angle XBK = 2(90^{\\circ} - \\frac{1}{2}\\angle BOK) = 180^{\\circ} - \\angle XOK,\n$$\nSimilarly for point $O_2$. Now we prove that point $Z$ is on circle $OXKY$, which will finish the solution.\n\nFor this purpose we prove that $\\angle O_1ZC + \\angle O_1KO_2 = 180^{\\circ}$. Indeed,\n$$\n\\angle O_1KO_2 = \\angle O_1KD + \\angle DKO_2 = 90^{\\circ} - \\angle DBK + 90^{\\circ} - \\angle DCK = 180^{\\circ} - \\angle A = 90^{\\circ} - \\angle BOC\n$$\n$$\n\\begin{align*}\n&= \\angle BZC = 180^{\\circ} - \\angle O_1ZC \\\\\n&\\text{10th form} \\\\\n&\\text{10th form}\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45435,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that the arithmetic average of the numbers *a*, *b* is equal to the number *c*, so $c = \\frac{1}{2}(a+b)$, and that the geometric average number of *a*, *c* is equal to the number *b*, so $b = \\sqrt{ac}$. Is it necessary that numbers *a*, *b*, *c* are equal?",
    "options": [],
    "answer": "No; for example, a = -4, b = 2, c = -1.",
    "solution": "Let's rewrite the equality $b^2 = ac$ using $c = \\frac{a+b}{2}$:\n\n$$\nb^2 = a \\cdot \\frac{a+b}{2} \\Leftrightarrow 2b^2 = ba + a^2 \\Leftrightarrow (b-a)(2b+a) = 0.\n$$\nNow let's denote, for example, $b=2$, which means $a=-4$ and $c=-1$, hence we receive three different numbers satisfying the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1$ and $C_1$ be the points on the sides $BC$ and $AB$ of the triangle $ABC$ respectively, so that $AA_1 = CC_1$. Segments $AA_1$ and $CC_1$ meet at the point $F$. Prove that if $\\angle CFA_1 = 2\\angle ABC$ then $AA_1 = AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider translation by vector $\\vec{A_1A}$ (Fig. 40). Thus $AA_1CC_3$ is a parallelogram, $CC_1C_2C_3$ is a rhombus. Then $\\angle C_1CC_3 = \\angle C_1FA = \\angle CFA_1 = 2\\alpha$, therefore\n$$\n\\angle C_1AC_3 = 180^\\circ - \\angle ABC = 180^\\circ - \\frac{1}{2}\\angle CFA_1 = 180^\\circ - \\alpha.\n$$\nIf we build a circle where the point $C$ is its centre, then the central angle $\\angle C_1CC_3 = 2\\alpha$, thus the inscribed angle, that subtends the arc $C_1C_3$, equals $\\alpha$, therefore the point $A$ belongs to the circle. Therefore, $AC = CC_3 = AA_1$, this completes the proof.\n\n\nAlternative solution. Let $l$ be the bisector of the vertical angles $A_1FC$ and $C_1FA$. Let the line intersect the segments $AC_1$ and $CA_1$ at the points $P$ and $Q$ respectively (Fig. 41).\n![](attached_image_1.png)\nFig. 40\nSince from the conditions of the problem $\\angle CFA_1 = 2\\angle ABC$, then $\\angle QFA_1 = \\angle PBA_1$. That means that quadrilateral $PBA_1F$ is inscribed. Similarly the quadrilateral $QBC_1F$ is also inscribed. $AA_1 = AC$ if and only if $\\angle AA_1C = \\angle ACA_1$. Since quadrilateral $PBA_1F$ is inscribed, then $\\angle AA_1C = \\angle QPB$. It is enough to prove that $\\angle ACB = \\angle BPQ$. The equality of angles is equivalent to the congruence of the triangles $ABC$ and $QBP$, that is equivalent to $\\frac{AB}{BC} = \\frac{QB}{BP}$. We will use the fact that the quadrilateral $PBA_1F$ is inscribed. From secant-secant theorem $AP \\cdot AB = AF \\cdot AA_1$. Similarly, $CQ \\cdot CB = CF \\cdot CC_1$. Thus\n$$\n\\frac{AF}{CF} = \\frac{AP}{CQ} \\cdot \\frac{AB}{BC} \\text{ (using } AA_1 = CC_1).\n$$\nNow it is enough to prove that $\\frac{AF}{CF} = \\frac{AP}{CQ} \\cdot \\frac{QB}{BP}$.\n![](attached_image_2.png)\n\n$\\frac{QB}{BP} = \\frac{AF}{PA} \\cdot \\frac{CQ}{CF}$. Let $\\angle ABC = \\angle QFC = \\angle PFA = \\beta$, $\\angle FQB = \\angle FC_1A = \\varphi$, $\\angle FPB = \\angle FA_1C = \\psi$. From the sine rule for the triangle $QBP$ $\\frac{QB}{BP} = \\frac{\\sin \\psi}{\\sin \\varphi}$. From the sine rule for the triangle $FQC$ $\\frac{CQ}{CF} = \\frac{\\sin \\beta}{\\sin(180^\\circ - \\varphi)} = \\frac{\\sin \\beta}{\\sin \\varphi}$. From the sine rule for the triangle $FPA$ $\\frac{AF}{PA} = \\frac{\\sin(180^\\circ - \\psi)}{\\sin \\beta} = \\frac{\\sin \\psi}{\\sin \\beta}$. Therefore, $\\frac{QB}{BP} = \\frac{AF}{PA} \\cdot \\frac{CQ}{CF} \\Leftrightarrow \\frac{\\sin \\psi}{\\sin \\varphi} = \\frac{\\sin \\psi}{\\sin \\beta} \\cdot \\frac{\\sin \\beta}{\\sin \\varphi}$, and the last equality, obviously, is correct.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all prime numbers $p < q < r$ so that $A = (r-p)(r-q)(q-p)+1$ and $B = 3p+5q$ equal the same prime number.",
    "options": [],
    "answer": "p=2, q=5, r=7",
    "solution": "Let $p, q, r$ be the prime numbers that satisfy the conditions of the problem. If $p > 2$ then all $p, q, r$ are odd thus the number $A = 3p + 5q$ is even and the number $B = (r-p)(r-q)(q-p)+1$ is odd, which contradicts the conditions of the problem. Therefore, $p=2$, thus:\n\n$$\n(r-2)(r-q)(q-2)+1=6+5q.\n$$\n\n$p < q < r$, that means $r-2 > q-2$ and $r-q \\ge 2$. Therefore, $6+5q > 2(q-2)^2+1$. After solving the last inequality, we will have that $q < 7$. Prime numbers that satisfy this inequality are $q=3$ and $q=5$.\n\nIf $q=3$, then $B = 3p+5q = 21$ is not prime.\n\nIf $q=5$, then $B = 3p+5q = 31$ is prime, that implies $A = (r-2)(r-5) \\cdot 3+1 = 31$, and $r^2-7r=0$, that means $r=7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45438,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many three-digit numbers are there with non-zero digits which have the following property: after any permutation of its digits one obtains three-digit number which is divisible by $4$?",
    "options": [],
    "answer": "8",
    "solution": "Obviously, all digits of such number are even because only even digit can be the last one. Also we cannot use $2$ or $6$ because if a number is divisible by $4$ then the last two digits of it are $12$, $32$, $...$, $92$ or $16$, $36$, $...$, $96$. So such number consists of the digits $4$ and $8$. So:\n\nThree digits $4$ (or $8$) – two numbers.\n\nTwo $4$ and $8$ (or vice versa) – six numbers.\n\nOverall – $8$ numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriiko has unlimited amount of chips painted in 6 colours. He wants to put some of the chips in one row so that for any two different colours in that row there are two adjacent chips of those colours. What is the minimum amount of chips he can put in a row?\n**Answer:** 18.",
    "options": [],
    "answer": "18",
    "solution": "The chip of each colour has to be adjacent at least 5 times, to each of other colour. But any chip has maximum 2 neighbours, thus there is at least 3 chips of each colour in a row. Here is an example that 18 is a correct answer: 123456246325164135.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist real $x$, such that both $x + \\sqrt{2}$ and $x^4 + \\sqrt{2}$ are rational?",
    "options": [],
    "answer": "No",
    "solution": "Let us suggest that there exist rational $a, b$, such that: $a = x + \\sqrt{2}$ and $b = x^4 + \\sqrt{2}$. Thus $x = a - \\sqrt{2}$. After substitution in another equality we obtain:\n$$\nb = a^4 - 4a^3\\sqrt{2} + 12a^2 - 8a\\sqrt{2} + 4 + \\sqrt{2}.\n$$\nRight part of equality has to be rational, thus the sum of components that contain $\\sqrt{2}$ has to equal 0. It is enough to answer the question, whether there exist rational solution of $4a^3 + 8a - 1 = 0$. It is not difficult to check that there is no solutions. We have a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45441,
    "subject": "Mathematics (Multi-modal)",
    "question": "In every cell of the table $5 \\times 9$ either $0$ or $1$ is written. Then the sums of the numbers for every row and column are calculated. What is the maximum amount of different values that can be among these $14$ numbers?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "9",
    "solution": "The sums vary from $0$ to $9$, thus there exist $10$ different values for the sum. Let us show that all $10$ values cannot be present simultaneously. Let us suppose that there is a row or a column with the sum $0$, thus all the cells of that row or column are zeroes. If it is a column, then the maximum value for the row is $8$. If it is a row, then in columns can be only values from $0$ to $4$. But then there can be $4$ more different values in rows. Together we have not more than $9$ different values. Example is shown on the picture (Fig. 20).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be a circumcenter of a right triangle. The circle with smaller radius and center at point $O$ is tangent to the greater cathetus and the height of the triangle from the right angle.\nFind the acute angles of the right triangle and the relation between the radii of the circumcircle and the other circle.\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Acute angles: 30° and 60°; radii ratio: 2:1",
    "solution": "The acute angles are $30^\\circ$ and $60^\\circ$; the ratio of the radii is $2:1$.\n\nLet the right triangle be $CMD$ (see fig. 16).\nLet $ON$ be the radius of the smaller circle that is tangent to the cathetus.\nThus $\\triangle ODN = \\triangle OMN$, and $\\triangle AMO = \\triangle OMN$, therefore, $2AM = MN + ND = MD$, hence $\\angle ADM = 30^\\circ$.\nTherefore, the acute angles equal $30^\\circ$ and $60^\\circ$.\nFrom $\\triangle OND$ it is clear that $OD = R$ (hypotenuse), $ON = r$ (cathetus), that opposes angle $30^\\circ$, thus $\\frac{R}{r} = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vertices of a regular $(6n+3)$-gon, which are renumbered clockwise by numbers $1; 2; \\ldots; 6n+3$, are the game field. Vertices with numbers $2n+1; 4n+2; 6n+3$ are called holes. At the start of the game there are $3$ chips on the field. $2$ players in turn choose any one of the $3$ chips and move it clockwise to a neighbouring vertex, if this one isn't occupied by another chip. The first player wins, if at least $2$ chips get in holes after anyone's turn. Can the first player always win, if at the start of the game they are placed in the form of a regular triangle?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's call a number of steps that each chip needs to reach the nearest hole, moving clockwise, if there are no other chips in the way, the distance to the hole. For example, for a chip that is placed on the vertex $4n$ the distance to the hole is $2$, if positions $4n+1$ and $4n+2$ are free from other chips; if it's placed on a vertex with number $3n$, then its distance is $4n+2-3n=n+2$, if on vertices with numbers $3n+1; \\ldots; 4n+2$ there are no chips.\n\nLet's prove with MMI that the position when $2$ chips have the same distance to the holes is the winning position for the first player. The induction will be done on this same distance to the holes. Let's say there are $2$ chips that are placed at a distance $k$ from the hole. Let's call these chips **hot**, let's call the **first** - the hot chip on the way from which to other hot chip (let's call it **second**) there is a not hot chip. Let's call the last chip **usual** and **third**.\n\nFor $k=0$ we have a situation when the first player wins.\n\nFor $k=1$ let's consider the cases of the next step in this situation.\nIf in this situation it's the first player's turn, then, if there is such an opportunity, he makes a turn with a usual chip (if it's not situated near to the hot chip). And we come to the same situation with the second player's turn. Similarly in the next variation, the situation when the usual chip is situated near to the hot chip is analysed.\nThus in the described situation when it's the turn of the second player and he moves one of the hot chips and then gets it to the hole, the first player then moves another hot chip to the hole and wins.\nSo in order not to lose, the second player has to move the usual chip. First then also move the usual chip. It holds until the usual chip approaches the nearest hot chip and won't be able to move. We will not analyse situations, when in the process of the move of the usual chip the first player could have already won (for example, when the usual chip gets into the hole after the turn of the second player). Thus, without generality limitation, we have the following situation: the first chip is placed on the vertex $6n+2$, the third - on the vertex $6n+1$. For the second there are location options on vertices $2n$ or $4n+1$. Let's call this position **zugzwang**. In any case, after the second player's turn he will have lost because it's impossible to move the third chip. Thus, it remains to consider the case of the first player's turn. He moves the first chip. It gets to the hole. The second, in order not to immediately lose, moves the same first chip. Then the first player moves the third chip and we have the following situation. Again two hot chips are in positions $6n+2$ or $4n+1$. The third chip is in position $1$. Thus, now the chip in the position $2n$ or $4n+1$ become the first, and the chip in the vertex $6n+2$ - the second. Both players have to move only the third chip, the second player is forced, and the first - due to his strategy. To situation zugzwang the third chip must reach the position $2n-1$ or $4n$. In the first case they together with the third chip will make $2n-1-1=2n-2$ moves. When the second player starts, he must make a turn in a position zugzwang and he loses. In the second case both together with the third chip will make $4n-1=4n-1$ moves. Here the first player has to make a turn in a zugzwang position. We have a following situation of chips location: $6n+2, 4n+1$ - hot and $4n$ - third chip. Then the first player moves the second chip to the hole $4n+2$, the second player has to move the same chip, it gets in $4n+3$. Then the first player moves the third chip in $4n+1$, and we have a situation of the first case of zugzwang. The second loses. Base of induction is proved.\n\nLet's say everything is proved for some $k$. Let's consider the situation where two chips have $k+1$ moves to the hole. Let's conditionally move the holes to the next position from the chips. Then we have the situation, which was considered in the case $k=1$. Thus, we will always get to the situation when the hot chips will be on $k$ distance from the holes.\n\nIt remains to make clear that the initial situation satisfies the conditions of MMI, so the statement is proved.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 45444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Petro has to plant $8$ trees in a row: apple trees or oak trees. There is one restriction: there has to be no apple trees between any two oak trees. For example, such planting $AAOOAOAA$ or $OAOAAAAA$ are not allowed, and $AAOOAAAA$ is allowed. How many different plantings are possible?",
    "options": [],
    "answer": "37",
    "solution": "Obviously, all oaks have to be planted together as one group no matter where this group will be situated. Let us count how many oaks can possibly be.\n\nIf there are no oak trees, then the option of planting is only one.\n\nIf there are $k$ oaks, $1 \\leq k \\leq 8$, there are $9 - k$ options:\n\n$OOOAAAAA$, $AOOOAAAA$, $\\ldots$, $AAAAAOOO$.\n\nTogether we have $1 + 1 + 2 + 3 + \\ldots + 8 = 37$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45445,
    "subject": "Mathematics (Multi-modal)",
    "question": "The twins Pete and Ostap had had an argue and started to go to school different ways. Pete goes $210$ meters to the South, then $70$ meters to the East and reaches the school. Ostap goes to the North for a while and then he head to the school directly. How many meters Ostap has to go to the North if both twins have equal speeds and come to the school simultaneously?",
    "options": [],
    "answer": "30",
    "solution": "Let us denote the points: home - $D$, school - $S$, the point where Pete turns to the East - $B$, the point where Ostap begins to walk straight to the school - $N$ (fig. 12). Then $DB = 210$, $BS = 70$, $DN = x$, $NS = y$. Also we have $x + y = 280$.\n\n![](attached_image_1.png)\n\nBy Pythagorean theorem: $BN^2 + BS^2 = NS^2$ or\n$$ (210 + x)^2 + 70^2 = y^2 \\Rightarrow (210 + x)^2 + 70^2 = (280 - x)^2. $$\n$$ 44100 + 420x + x^2 + 4900 = 78400 - 560x + x^2, \\text{ hence } x = 30. $$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Son, his Dad and his Grandfather has run from their home to a shop and back. Son's velocity was constant. Grandfather's velocity was two times greater than Son's while he was running to the shop and three times less when he was running back. Dad's velocity was two times less than Son's on the way to shop and 3 times greater when he was running back. Who was the first and who was the last to come home?",
    "options": [],
    "answer": "Son first, Grandfather last",
    "solution": "Denote the velocity of Son by $x$ and the distance by $S$. Then we can compute the time which was spent on running:\n$$\nt_1 = \\frac{S}{x} + \\frac{S}{x}, \\quad t_2 = \\frac{S}{\\frac{1}{2}x} + \\frac{S}{3x}, \\quad t_3 = \\frac{S}{2x} + \\frac{S}{\\frac{1}{3}x}\n$$\nHence, we need to compare the following numbers:\n$a_1 = 2$, $a_2 = 2 + \\frac{1}{3}$, $a_3 = \\frac{1}{2} + 3$\nAnd get the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Compare the following numbers: $A = 11$, $B = \\log_2 3 \\cdot \\log_3 4 \\cdot \\log_4 5 \\cdots \\log_{2015} 2016$ and $C = \\log_3 2 \\cdot \\log_4 3 \\cdot \\log_5 4 \\cdots \\log_{2016} 2015$.",
    "options": [],
    "answer": "C < B < A",
    "solution": "Obviously, for any integer $n > 0$ $\\log_{n+1} n < 1$. Hence, $C < 1$. Also we have that $B \\cdot C = 1$, so $B > 1$, moreover\n$$\nB = \\frac{\\lg 3}{\\lg 2} \\cdot \\frac{\\lg 4}{\\lg 3} \\cdot \\frac{\\lg 5}{\\lg 4} \\cdots \\frac{\\lg 2016}{\\lg 2015} = \\frac{\\lg 2016}{\\lg 2} < 11 = A \\Leftrightarrow \\lg 2016 < 11 \\lg 2 = \\lg 2^{11} \\Leftrightarrow 2016 < 2^{11} = 2048.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all natural $n \\ge 2$ the following number is composite:\n$$\n\\frac{n^{1003} + n^{1002} + n^{1001} + 1}{n+1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since\n$$\n\\begin{aligned}\nn^{1003} + n^{1002} + n^{1001} + 1 &= n^{1001}(n^2 + 1) + n^{1002} + 1 \\\\\n&= n^{1001}(n^2 + 1) + (n^2 + 1)(n^{1000} - n^{998} + n^{996} - \\dots - n^2 + 1)\n\\end{aligned}\n$$\nhence, the numerator is divisible by $n^2 + 1$. We can see that the numerator is also divisible by $n+1$, as follows from the equation:\n$$\n\\begin{aligned}\nn^{1003} + n^{1002} + n^{1001} + 1 &= n^{1002}(n+1) + n^{1001} + 1 \\\\\n&= n^{1002}(n+1) + (n+1)(n^{1000} - n^{998} + n^{996} - \\dots - n + 1).\n\\end{aligned}\n$$\n\n$$\n\\frac{n^{1003} + n^{1002} + n^{1001} + 1}{n+1} = \\frac{(n+1)(n^2+1)P(n)}{n+1} = (n^2+1)P(n).\n$$\nIt is easy to see that $P(n) > 1$. The statement has been proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Number $2016$ is written on the board. Olesya and Andriy are playing game: they take turns (Olesya is the first to start) to reduce the number on the board to an integer, not exceeding the number of the move (on the first move Olesya must reduce the number to $1$, Andriy, in his turn, to $1$ or $2$, then Olesya to $1$, $2$ or $3$ etc.). The winner is the first one to be able to write number $0$ on the board. Who wins with the right game of both opponents?\n\n(Oleksiy Piskun)\n\n**Answer:** the second player wins.",
    "options": [],
    "answer": "the second player wins",
    "solution": "Let us define the position for this game with a pair of numbers $(m, n)$, where $m$ – the number written on the board, and $n$ – the number of a move. We assume a position being winning, if after having got into it, a player still has a winning strategy. A losing one – when having got into it, an opponent will have a winning strategy. Each player tries to reach the position $(0, n)$ after his turn (that is to write down $0$), that is why these positions are winning. Let us show that the map of winning-losing positions looks like in fig. 25.\n![](attached_image_1.png)\nFig. 25\nLet us note, that from the cell $(m, n)$ one can get into the cell $(m-1, n+1)$, $(m-2, n+1)$, ..., $(m-n, n+1)$ that is, one can move into the next right column, or one, two, ..., $n$ cells down.\nHaving considered the values $n=1, n=2, n=3$, one can suggest that periods of winning and losing positions in the $n$-th column are as follows:\n$$\n\\textit{winning:} [k^2 + nk; k^2 + nk + k], \\textit{losing:} [k^2 + nk + k + 1; k^2 + nk + 2k + n], k \\in \\mathbb{Z}^{+}.\n$$\nWe can prove this. That is, show that from each defined as a winning position one can get only into defined as a losing one, and from each position, defined as a losing, there exists a way into a position, defined as a winning.\nLet on the $n$-th step, number $m$ is written on the board, and there exists $k \\in \\mathbb{N}$:\n$$\nk^2 + nk \\le m \\le k^2 + nk + k \\text{ (that is, the position is winning).}\n$$\nThen we can subtract number $l$, which is $1 \\le l \\le n$. The next move after that will be the $(n+1)$th. We have got:\n$$\nm-l \\ge m-n \\ge k^2 + nk - n = (k-1)^2 + (n+1)(k-1) + (k-1) + 1,\n$$\n$$\nm-l \\le m-1 \\ge k^2 + nk + k-1 = (k-1)^2 + (n+1)(k-1) + 2(k-1) + (n+1).\n$$\nAs these are the final points of one losing range, then any move out of the winning position leads into the losing range for $(k-1)$.\nLet on the $n$-th step, number $m$ is written on the board and there exists $k \\in \\mathbb{N}$:\n$$\nk^2 + nk + k + 1 \\le m \\le k^2 + nk + 2k + n \\text{ (that is, the position is losing).}\n$$\nLet us discover what happens if we subtract $1$ or $n$. The move after that will be the $(n+1)$th. We have got:\n$$k^2 + (n+1)k + 2k + (n+1) \\ge k^2 + nk + n-1 \\ge m-1 \\ge k^2 + nk + k = k^2 + (n+1)k,$$ that is, we will get into a losing range for $k$, or into a winning position.\n$$k^2 + (n+1)k + k = k^2 + nk + 2k \\ge m-n \\ge k^2 + nk + k + 1 - n \\ge$$ \n$$\\ge (k-1)^2 + (n+1)(k-1) + (k-1) + 1,$$ \nthat is, we will get into a losing range for $(k-1)$, or into a winning position.\nWe can also get from number $m$ to any number from $(m-n)$ to $(m-1)$. As these numbers belong to various losing ranges, there exists a winning range between them.\nThus, the chosen ranges are really winning or losing positions.\nLet us see, into which range the position $(2016; 1)$ will get. Just as\n$$[44^2 + 44 = 1980 < 2016 < 44^2 + 44 + 44 = 2024,$$ \nso this position is winning, hence the second player wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there a function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ exist, that for any real numbers $x$, $y$ the following inequality is fulfilled:\n$$\nf(x - f(y)) \\le x - y f(x)?\n$$",
    "options": [],
    "answer": "it doesn't exist",
    "solution": "**Answer:** it doesn't exist.\n\nLet us assume such function exists. Let us substitute $y = 0$ in the given inequality:\n$$\nf(x - f(0)) \\le x,\n$$\nThen, substitute $x = x + f(0)$ there:\n$$\nf(x) \\le x + f(0). \\quad (1)\n$$\nLet us substitute $x = f(y)$ in the initial inequality. We will get:\n$$\n\\begin{cases}\nf(0) \\le f(y) - y f(f(y)) \\le y + f(0) - y f(f(y)) \\text{ or} \\\\\ny f(f(y)) \\le y.\n\\end{cases}\n$$\nFrom the last inequality while $y < 0$ we will have:\n$$\n1 \\le f(f(y)) \\le f(y) + f(0) \\le y + 2f(0). \\quad (1)\n$$\nThe last inequality has to be satisfied for any $y < 0$, that is surely impossible.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $x[x] = 2016$.\nHere $[x]$ is the integer part of the number $x$, i.e. the largest integer number not greater than $x$.",
    "options": [],
    "answer": "-224/5",
    "solution": "**Answer:** $x = -\\frac{224}{5}$.\nSuppose that $x \\ge 0$. Let us denote $t = [x] \\ge 0$, and then we can write the following estimations:\n$$\nt \\le x < t+1 \\Rightarrow t^2 \\le x[x] < t^2 + t.\n$$\n$44^2 + 44 = 1980 < 2016 < 45^2$, therefore there are no solutions of the equation among positive numbers.\nSuppose that $x < 0$. Then denote $t = [x] < 0$, i.e. $-[x] = -t > 0$. Hence,\n$$\nt \\le x < t+1 \\Rightarrow -t-1 \\le -x < -t \\Rightarrow t^2 + t \\le x[x] \\le t^2.\n$$\n$44^2 = 1936 < 2016$ and $46^2 = 1936 < 2016 < 46^2 - 46$, therefore only possible value is $t = -45$. Indeed, $45^2 - 45 = 1980 < 2016 < 45^2 = 2025$, i.e. the solution is possible. Let us denote $x = -45 + y$, $0 \\le y < 1$. Then it should hold that:\n$$\nx[x] = -45 \\cdot (-45 + y) = 2025 - 45y = 2016 \\Rightarrow 45y = 9 \\Rightarrow y = \\frac{1}{5}.\n$$\nThus, the desired solution is $x = -45 + \\frac{1}{5} = -\\frac{224}{5}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Before 1995, in Football Championship of Ukraine in all leagues the team received 2 points for winning, 1 point for a draw and 0 points for a defeat. Since 1995, winning would give 3 points to the team, 1 point for a draw and 0 for a defeat. Football Federation has decided to transpose all the championships that were held in all leagues during all years to the new grading system, believing that the distribution of places will not change drastically. However, it turned out that in the championship in 1927 in one of the regional leagues $n$ teams held the championship in one round, i.e. each team played with each another once. There team \"A\" gained the largest amount of points, and the team \"B\" gained the least number of points. But after the points calculation according to the new rules team \"B\" gained the most points, and team \"A\" gained the least amount of points. For which the least $n$ this could happen?\nHere the fact that the team has the lowest or the highest score, means that no other team gained this number of points.",
    "options": [],
    "answer": "12",
    "solution": "Team \"A\" is the winner when 2 points are calculated for a victory, and takes the last place when 3 points are calculated for a victory, while for the team \"B\" it is all the way around. Let the team \"A\" win $x$ matches and have $y$ draws. Team \"B\" will have $a$ wins and $b$ draws, respectively. Since there could not be only two teams, the rest of the teams gained the numbers of points that between the results of teams \"A\" and \"B\" in both cases of calculation. So we have the following inequalities:\n$$\n2x + y \\ge 2a + b + 2 \\text{ and } 3x + y + 2 \\le 3a + b \\Rightarrow 2(a - x) + 2 \\le y - b \\le 3(a - x) - 2.\n$$\nConsider the possible values of $(y-b)$ for different values of $(a-x)$.\nFor $a-x=0$ we have that $2 \\le y-b \\le -2$ which is impossible.\nFor $a-x=1$ we have that $4 \\le y-b \\le 1$ which is impossible.\nFor $a-x=2$ we have that $6 \\le y-b \\le 4$ which is impossible.\nFor $a-x=3$ we have that $8 \\le y-b \\le 7$ which is impossible.\nFor $a-x=4$ we have that $10 \\le y-b \\le 10$.\n\nThus, the minimum possible value for $y-b=10$. So either $y-b \\ge 10$ or $y \\ge 10+b$. Thus the smallest value for $y=10$. Since it is clear that $x \\ge 1$, then $x+y \\ge 11$, i.e. the should be at least 12 teams.\n\nWhat is left to show that for 12 teams there exists such distribution of points. In the table (fig. 41) victory is denoted by 1, defeat by 0 and empty cells correspond to draws between the teams.\nThen in the championship when calculating by the rule \"2 points for a win\" team \"A\" (number 2 in the table) gains 12 points, \"B\" (number 1 in the table) gains 10 points, and the rest of the teams gain 11 points each.\nIn the championship when calculating by the rule \"3 points for a win\" team \"A\" gains 13 points, the team \"B\" gains 15 points, and the rest of the teams gain 14 points each.\n\n![](attached_image_1.png)\nFig. 41",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let us denote by $deg(n) = \\alpha_1 + \\alpha_2 + \\ldots + \\alpha_k$ the degree of the number $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\ldots p_k^{\\alpha_k}$, where $p_i$ are pairwise different prime numbers, and $\\alpha_1, \\alpha_2, \\ldots, \\alpha_k$ are positive integers. Prove that there exist $2016$ consecutive positive integers among which there are precisely $1000$ numbers with the degree less than $11$.\n\n(Bogdan Kivva)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First let us prove the following lemma.\n\n**Lemma.** There are $l$ consecutive positive integers, among which there are no numbers with the degree less than $t$ ($t \\ge 2$).\n\n*Proof* is conducted by Mathematical Induction with respect to $t$.\n\nBase for $t=2$. The statement is equivalent to the fact that there are $l$ consecutive positive integers, none of which is prime. So it is sufficient to take $(l+1)!+2, (l+1)!+3, \\ldots, (l+1)!(l+1)$.\n\nAssume that we have proved the statement for $2, 3, \\ldots, (t-1)$. Let us prove it for $t$. Suppose for $(t-1)$ $x+1, \\ldots, x+l$ are the numbers with the degree not less than $(t-1)$. Now consider the numbers $(x+l)!+x+1, \\ldots, (x+l)!+x+l$. Obviously, each of them has the degree not less than $t$ and there are $l$ of them. Thus, the lemma is proved.\n\nNote that $2^{11} = 2048 > 2016$. So among the first $2016$ positive integers all numbers have the degree less than $11$. Let $w(x)$ be the amount of numbers with the degree less than $11$ among the numbers $x+1, \\ldots, x+2016$. Hence, $w(0) = 2016$. At the same time we have shown that there are $2016$ consecutive positive integers which do not include any of the numbers with the degree less than $11$. So there exists a positive integer $y$ such that $w(y)=0$. It is also clear that $w(z)-1 \\le w(z+1) \\le w(z)+1$. Thus, as $w(z)$ takes only positive integer values or zero, in $2$ consecutive points differs by not less than $1$ and takes the values $0$ and $2016$, it has to also take all the intermediate values, therefore $1000$ as well.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45454,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the plane there is a triangle $APQ$ and a rectangle $ABCD$ such that the midpoint of the segment $PQ$ belongs to the diagonal $BD$ of the rectangle, and one of the rays $AB$ and $AD$ is a bisector of the angle $PAQ$. Prove that one of the rays $CB$ and $CD$ is a bisector of the angle $PCQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFig. 42\nLet us consider the case where the ray $AB$ is a bisector of the angle $PAQ$, and $\\angle BAC < \\angle BAQ$ (fig. 42). We will prove that in this case the ray $CD$ is the bisector of the angle $PCQ$. In the case where $AD$ is the bisector of the angle $PAQ$ all the proof is analogous.\n\nLet $M$ be the midpoint of $PQ$, and $O$ the intersection point of the diagonals of the rectangle. Let us select point $S$ symmetrically to $Q$ with respect to the point $O$. In this case $AQCS$ is a parallelogram.\n\nLet us first prove that $\\angle APC = \\angle AQC$. For this we should prove that the points $A$, $S$, $P$, $C$ are cyclic. Indeed, let $\\angle BAC = \\alpha$, and $\\angle BAQ = \\beta$. By the assumption, $\\beta > \\alpha$. Then $\\angle CAQ = \\beta - \\alpha$.\n\nAlso $CS \\parallel AQ$, then $\\angle SCA = \\angle CAQ = \\beta - \\alpha$. $AB$ is the bisector of the angle $PAQ$, therefore $\\angle PAB = \\beta$. Moreover, $ABCD$ is a rectangle, which leads to $\\angle ABD = \\angle BAC = \\alpha$. $M$ is the midpoint of $PQ$, and $O$ is the midpoint of $QS$, hence $MO$ is the mid-segment of $PQS$. This brings the fact that $PS \\parallel MO$, i.e. $PS \\parallel BD$. Which leads to $\\angle PTB = \\angle TBD = \\alpha$, where $T = (PS) \\cap (AB)$.\n\nAccording to the Triangle exterior angle theorem, we obtain that\n$$\n\\angle APS = \\angle APT = \\angle PAB - \\angle PTA = \\beta - \\alpha.\n$$\nThus, we proved that $\\angle APS = \\angle ACS$, i.e. the points $A$, $S$, $P$, $C$ are cyclic. From which it follows that $\\angle APC = \\angle ASC = \\angle AQC$ (because inscribed angles that have the same arc are of the same length, and also the opposite angles of a parallelogram are the same).\n\n![](attached_image_2.png)\nFig. 43\nWhat is left is to prove that when in the convex quadrangle $APCQ$ the opposite angles $APC$ and $AQC$ are equal, then the bisectors of the angles $PAQ$ and $PCQ$ are parallel.\n\nLet the bisector of the angle $PAQ$ intersect lines $PC$ and $CQ$ in the points $N$ and $F$ respectively (fig. 43), and let the bisector of the angle $PCQ$ intersect $AQ$ in the point $L$. Let $K$ be the point symmetrical to $P$ with respect to $AF$, then the point $K$ lies on $AQ$ (as $AF$ is the bisector of the angle $PAQ$). This means that $\\angle AKN = \\angle APN$. $\\angle APC = \\angle AQC$, therefore $\\angle AKN = \\angle AQC$. Thus, $NK \\parallel CQ$. Which means $\\angle CFN = \\angle KNA = \\angle PNA = \\angle CNF$, i.e. $\\angle CFN = \\angle CNF$ and triangle $NCF$ is isosceles. The bisector of the exterior angle adjacent to the vertex of isosceles triangle is parallel to its base, therefore $AF \\parallel CL$, Q.F.D.\n\nIn the case when $\\alpha > \\beta$, the points $P$ and $Q$ will lie in one direction from the line $AC$ and it will hold that $\\angle APC + \\angle AQC = 180^\\circ$. The ray $CB$ will be the bisector of the angle $PCQ$ (all proofs are analogous).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45455,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle, inscribed into a triangle $ABC$, touches its sides $AB$, $BC$ and $CA$ in the points $N$, $P$, $K$ respectively. A segment $BK$ intersects the inscribed circle the second time in a point $L$.\nLet us define points $T = AL \\cap NK$, $Q = CL \\cap KP$. Prove that straight lines $BK$, $NQ$ and $PT$ intersect in one single point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote the angles as shown in fig. 26. For triangles $AKT$ and $ATN$ let us apply Law of sines and get:\n$$\n\\frac{NT}{TK} = \\frac{AN \\cdot \\sin \\alpha_1}{AK \\cdot \\sin \\alpha_2} = \\frac{\\sin \\alpha_1}{\\sin \\alpha_2}\n$$\nSimilarly we can get:\n$$\n\\frac{NF}{FP} = \\frac{NB \\cdot \\sin \\beta_1}{PB \\cdot \\sin \\beta_2} = \\frac{\\sin \\beta_1}{\\sin \\beta_2}, \\quad \\frac{PQ}{QK} = \\frac{PC \\cdot \\sin \\gamma_1}{KC \\cdot \\sin \\gamma_2} = \\frac{\\sin \\gamma_1}{\\sin \\gamma_2}\n$$\nIf we multiply all these expressions, we will get:\n$$\n\\frac{NT \\cdot KQ \\cdot PF}{TK \\cdot QP \\cdot FN} = \\frac{\\sin \\alpha_1 \\cdot \\sin \\beta_2 \\cdot \\sin \\gamma_2}{\\sin \\alpha_2 \\cdot \\sin \\beta_1 \\cdot \\sin \\gamma_1} = 1\n$$\nas far as $AL$, $BL$ and $CL$ intersect in one single point, and for them Cheva's theorem in its trigonometrical form is fulfilled. But then from Cheva's theorem in its standard form, it follows that $BK$, $NQ$ and $PT$ intersect in one single point, that is what had to be proven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45456,
    "subject": "Mathematics (Multi-modal)",
    "question": "The country Plato has the shape of a convex polygon in vertices of which there are border towers. Trucker needs to drive through all towers, and for every kilometre he is ought to pay one puylyk (state currency) to swindlers from the government (route does not have to be closed, the only condition is to visit each tower, and one can go in any direction without going beyond the state border). The perimeter of the state is 3000 kilometres and diameter is 1000 kilometres. Which guaranteed amount of puylyks will the trucker pay to swindlers?\n\n*The diameter of a polygon is the largest distance between any pair of vertices.*",
    "options": [],
    "answer": "2000",
    "solution": "If the state has the shape of an equilateral triangle, then obviously, trucker can do the job and pay only 2000 puylyks (in this case the route of the trucker lies along the two sides of the triangle).\n\nNow let us prove that in any case the trucker will have to pay at least 2000 puylyks. Suppose that there is a state that has a shape of a convex $n$-gon for which trucker manages to do the job paying less than 2000 puylyks. Consider his shortest way. Let us close it (i.e. draw a segment from the starting point to the ending one). The length of the new route\n\n![](attached_image_1.png)\n\nFig. 44\n\nwill increase by not less than 1000 km, so it would be less than the perimeter. Let us prove that this is impossible.\n\nThe proof will be based on a few simple statements.\n\n**Statement 1.** Any part of the way between the two points should be a segment.\nIf this is not the segment, then by connecting them with the segment we get a shorter route, which is impossible according to our choosing of the shortest closed route.\n\n![](attached_image_2.png)\n\nFig. 45\n\n**Statement 2.** All turns on the route must be the vertices of the boundaries of the polygon.\nIf there is a turn in the point $A$ that is not the vertex on the border, then we have two segments $CA$ and $AB$ included in the route. If we replace them with the segment $CB$, the route will shorten.\n\n**Statement 3.** The route can not cross itself.\nIf the route $AB...X...CD...Y...A$ contains two segments $AB$ and $CD$ that intersect in some point $O$, then we will consider the route $AC...X...BD...Y...A$ instead. They differ only in pairs of segments $AB$, $CD$ and $AC$, $BD$. Since (fig. 44)\n$$\nAB + CD = AO + OB + CO + OD > AC + BD,\n$$\nthe length of the route decreases. Since the number of all different routes is finite, such decrease in length will end when the route will have no intersections with itself.\n\nFrom these statements it follows that the route runs along the perimeter, because if somewhere not consecutive vertices are connected with the segment $AB$ (fig. 45), it is no longer possible to connect the vertices $C$ and $D$ without intersections with the route. Thus, the shortest route must be the perimeter, and this contradiction with the assumption completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45457,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is given. The circle $w$ with the centre at the point $Q$ touches the side $BC$ and is internally tangent to the circumscribed circle of $\\triangle ABC$ in the point $A$. Let $M$ be the midpoint of the side $BC$, and $N$ be the middle of the arc $BAC$ of the circumscribed circle of $\\triangle ABC$. On the side $BC$ the point $S$ is selected such that $\\angle BAM = \\angle SAC$. Prove that the points $N$, $Q$ and $S$ are collinear.\n(Mykhaylo Plotnykov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $\\Gamma$ the circumscribed circle of $\\triangle ABC$. Let $W$ be the middle of the smaller arc $BC$ of the circle $\\Gamma$, and $O$ be the centre of $\\Gamma$ (Fig. 46). Obviously, points $N$, $M$, $W$, $O$ lie on the median perpendicular to the segment $BC$. By Lemma of Archimedes points of contact of $w$ to $BC$, to $\\Gamma$ and the point $W$ are collinear. As $w$ is tangent to the circle $\\Gamma$ in the point $A$, then the circle $w$ touches $BC$ in the point $L$, which is the intersection of $BC$ and $AW$. Note that\n![](attached_image_1.png)\nFig. 46\n$L$ is the base of the bisector of the angle $\\angle BAC$. Let us draw the rays $AM$ and $AS$ until they intersect with the circle $\\Gamma$ in the points $X$ and $P$ respectively. By the statement of the problem, $\\angle BAX = \\angle PAC$, hence $P$ and $X$ are symmetrical with respect to $WN$. Also $\\angle XAW = \\angle WAP$. $WN$ is the diameter of $\\Gamma$, therefore $\\angle NML = \\angle NAL = 90^\\circ$, and thus the quadrangle $ALMN$ is inscribed. Then $\\angle WNP = \\angle WAP = \\angle XAW = \\angle MNL$, and hence $N$, $L$, $P$ are collinear. Note that $LM$, $NA$, $WP$ are the heights of $\\triangle NLW$, their intersection point is the orthocentre $K$. Let $T$ be the intersection point of the ray $AN$ and $w$. There exists a homothetic transformation transforming $\\Gamma$ into $w$. By this homothetic transformation point $N$ transforms into $T$, point $W$ into $L$, i.e. $TL$ is the diameter of $w$, because $WN$ is the diameter of $\\Gamma$. Hence, $Q$ is the midpoint of the segment $TL$. Let us use Menelaus' theorem to $\\triangle TLK$ and points $N$, $Q$, $S$. Then to prove $N$, $Q$, $S$ are collinear, it is sufficient to show that\n$$\n\\frac{TN}{NK} \\cdot \\frac{KS}{SL} \\cdot \\frac{LQ}{QT} = 1, \\text{ i.e. (as } TQ = QL) \\frac{TN}{NK} = \\frac{SL}{KS}.\n$$\nLet us prove this equality. It is clear that the lines $AL$ and $AK$ are the internal and the extremal bisectors of the angle $\\angle MAS$, hence $\\frac{ML}{LS} = \\frac{MK}{SK}$. Then $\\frac{ML}{MK} = \\frac{SL}{SK}$. Obviously, $TL \\parallel NW$, hence by Thales' theorem $\\frac{TN}{NK} = \\frac{LM}{MK}$, therefore $\\frac{TN}{NK} = \\frac{SL}{SK}$, Q.F.D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(n)$ and $Q(n)$ be two polynomials (not constant) with positive integer or zero coefficients. For $n \\ge 1$ we determine the sequence $x_n = 2016^{P(n)} + Q(n)$. Prove that there are infinitely many prime numbers $p$ for which there is a positive integer $m$ not divisible by the square of any prime, and for which the condition $x_m \\nmid p$ holds.\n(Oleksiy Klurman)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that it is not so. Consider the sequence $(x_n)$ of all members, the indices of which do not contain in their decomposition on factors the squares of prime numbers. Then in this subsequence there is only a finite number of prime divisors $p_1, p_2, \\dots, p_k$ other than $2$, $3$, $7$ (prime factors of the number $2016$). Let $n_0 = \\prod_{i=1}^{k} p_i^{\\alpha_i}$, where $\\alpha_i \\ge x_1$. Consider the following sequence:\n\n$a_n = 1 + n n_0 \\phi(n_0)$, where $\\phi(n)$ is Euler function, i.e. the function that counts the positive integers up to a given integer $n$ that are relatively prime to $n$. Euler function can be represented in the form of the so-called Euler product: $\\phi(n) = n \\prod_{p|n} (1 - \\frac{1}{p})$, where $p$ is a prime number.\n\nIt is clear that $Q(a_n) \\equiv Q(1) \\pmod{n_0}$ and $P(a_n) \\equiv P(1) \\pmod{\\phi(n_0)}$. As by the construction $(n_0, 2016) = 1$, hence $x_{a_n} = 2016^{P(a_n)} + Q(a_n) \\equiv x_1 \\pmod{n_0}$.\n\nLater we will show that for large $n \\ge 1$ and for each $p_i$, $i = 1, k$ the sequence of values $\\text{ord}_{p_i}(x_{a_n})$ is limited. We can choose an infinite number of primes from the sequence $(a_n)$, which follows from Dirichlet's theorem or from using simple arguments about the progression with the first member $1$. Next, let us consider only those $n$ for which corresponding $a_n$ is a prime number. Then we have that there are infinitely many values $n \\ge 1$:\n\n$$\n2016^{P(n)} + Q(n) = 2^a \\cdot 3^b \\cdot 7^c \\cdot M, \\quad (1)\n$$\nfor a fixed limited $M$. But then\n$$\nM \\cdot (\\max\\{2^a; 3^b; 7^c\\})^3 \\ge M \\cdot 2^a \\cdot 3^b \\cdot 7^c = 2016^{P(n)} + Q(n) > 2016^n.\n$$\nThen from one of these prime numbers $A$ (which is the largest in the previous transition) in equation (1) we have that is $2016^{P(n)}$ divisible by $A^{\\frac{1}{3}n-C_0}$. The right-hand side is also divisible by $A^{\\frac{1}{3}n-C_0}$, hence $Q(n)$ should be divisible by the same power. However, for sufficiently large values $n$ it is not possible, because $Q(n)$ is a power function, while $A^{\\frac{1}{3}n-C_0}$ is exponential.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45459,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute-angled triangle $ABC$ is given, where $AB < BC$. Its incircle with the centre in point $I$ is touching the side $BC$ in point $K$. The line $AK$ intersects a second time the circumscribed circle of triangle $ABC$ in point $T$. Let $M$ be the midpoint of $BC$, and $N$ be the midpoint of arc $BAC$ of the circumscribed circle of triangle $ABC$. Line segment $NT$ intersects the circumscribed circle of triangle $BIC$ in point $P$. Prove that $PM \\parallel AK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's assume $W$ is the midpoint of arc $BC$. As is known, points $W$, $M$, and $N$ are collinear, and $W$ is the midpoint of the circumscribed circle of triangle $BIC$ (see fig. 47). Also, it is known that the center of the excircle of triangle $ABC$ that touches $BC$ belongs to this circle. Let $IK$ cross $TN$ at point $Q$. Since $\\angle IAT = \\angle TAW = \\angle TNW = \\angle TQI$, the quadrilateral $AITQ$ is inscribed. Then, according to the theorem about the multiplication of the line segments of chords, $IK \\cdot KQ = AK \\cdot KT = BK \\cdot KC$, so the quadrilateral $IBQC$ is also inscribed, i.e., $Q$ belongs to the circumscribed circle of triangle $BIC$. $\\angle NBW = \\angle NCW = 90^\\circ$, so $NB$ and $NC$ are tangent lines to the circumscribed circle of triangle $BIC$. In triangle $BPC$, the line $PN$ is the symmedian, so $\\angle BPQ = \\angle CPM$. Let's assume $PM$ crosses the circumscribed circle of triangle $BIC$ a second time at point $I_a$. Let's prove that $I_a$ is the center of the excircle of triangle $ABC$. As already noted, $\\angle BPQ = \\angle CPI_a$, so the arcs $BQ$ and $CI_a$ of the circumscribed circle of triangle $BIC$ are equal. Consider the symmetry relative to $NW$. Point $B$ maps to point $C$, and point $Q$ maps to point $I_a$, since the arcs $BQ$ and $CI_a$ are symmetrical relative to $NW$. Let the perpendicular from $I_a$ to $BC$ cross $BC$ at point $X$, then according to symmetry the line segments $MK$ and $MX$ are equal, so $BK = CX$, i.e., $X$ is the touchpoint of the excircle to the $BC$ side. So, $I_a$ belongs to the circumscribed circle of triangle $BIC$, lies with $A$ in a different half-space relative to $NW$, and lies on the perpendicular that passes through the touchpoint of the excircle of triangle $ABC$.\n\nObviously, there is only one such point, and the center of the excircle of triangle $ABC$ that touches the $BC$ side satisfies such conditions. So, $I_a$ is the center of the excircle.\n\nThus, to finish the solution, it's enough to show that $AK \\parallel PI_a$. This statement follows from the fact that $\\angle AI_aP = \\angle IQP = \\angle TQI = \\angle TAI = \\angle KAI_a$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all $x$, $y$, $z$ which satisfy:\n$$\n\\begin{cases} x^2 + x y + x z = y, \\\\ y^2 + y z + y x = z, \\\\ z^2 + z x + z y = x. \\end{cases}\n$$",
    "options": [],
    "answer": "x = y = z = 0 or x = y = z = 1/3",
    "solution": "Summing up the equalities we get\n$$\nx^2 + y^2 + z^2 + 2 x y + 2 x z + 2 y z = x + y + z \\text{ or } (x + y + z)^2 = x + y + z.\n$$\nNow we face two cases.\n\nCase 1. $x + y + z = 1$. Then from the first equality we get\n$$\nx^2 + x y + x z = x(x + y + z) = x = y,\n$$\nAnalogously, $x = y = z$, so the only solution is $x = y = z = \\frac{1}{3}$.\n\nCase 2. $x + y + z = 0$. Again from the first equality\n$$\nx^2 + x y + x z = x(x + y + z) = 0 = y,\n$$\nhence $x = y = z = 0$ – the second solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45461,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisector of an angle $BAC$ of acute angled triangle $ABC$ ($AC \\neq AB$) intersects its circumcircle second time at the point $W$. Let $O$ be the circumcenter of $\\triangle ABC$. Line $AW$ intersects second time the circumcircles of triangles $OWB$ and $OWC$ at points $N$ and $M$ respectively. Prove that $BN + MC = AW$.\n\n(V.Mitrofanov, D.Khilko)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality we can consider that $AC < AB$. From the conditions (fig. 18) $2\\angle CAW = \\angle COW = \\angle CMW$, hence $\\angle ACM = \\angle CAM$. Thus $AM = CM$. Similarly $AN = NB$. Then it is enough to prove that $AM + AN = AW$, thus $AM = NW$. Obviously, $CO = OB$, $CW = WB$. Thus $\\angle OBW = \\angle OCW$. Hence $\\angle OBW = \\angle OCW$. Therefore,\n\n$\\angle ONM = \\angle OBW = \\angle OCW = \\angle OMN$, hence $\\triangle ONM$ is isosceles. It is also obvious that $\\triangle OAW$ is isosceles, hence it is clear that $\\triangle ONW = \\triangle OMA$, thus $AM = NW$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45462,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is number in every cell of the table $5 \\times 5$ written under such conditions: not all number should be different, however, there is no row or column where all five numbers are equal; the middle number (the third one) in every row and column equals the mean of the numbers in its row or column. What is the minimum amount of numbers that are less than the number written in the middle of the table?",
    "options": [],
    "answer": "3",
    "solution": "Let us prove that there cannot be less than three such numbers. Let the middle number of the table be $a$. Then there exists at least one number that is less than $a$ in the third row and column. If there is no such number then all the numbers in the row or column are the same, which contradicts the conditions. Without loss of generality let us consider that the number $b < a$ is situated in the middle row in the fifth column. But then in the fifth column exists the number smaller than $b$, and smaller than $a$. Therefore, at least three such numbers exist.\n\nThe example where answer 3 is achievable is shown on the picture (fig. 17).\n\n| 4 | 4 | 3 | 4 | 0 |\n|---|---|---|---|---|\n| 4 | 4 | 3 | 4 | 0 |\n| 3 | 3 | 0 | 3 | -9 |\n| 4 | 4 | 3 | 4 | 0 |\n| 0 | 0 | -9 | 0 | -36 |\n\nFig. 17",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45463,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $22$ cards, where the numbers $1, 2, \\ldots, 22$ are written. Using these cards one formed $11$ fractions. What is the greatest possible number of integer numbers among the fractions?",
    "options": [],
    "answer": "10",
    "solution": "The numbers $13$, $17$, $19$ may form an integer only if they stand at the numerator position and $1$ stands in the denominator position. Hence, at least one fraction cannot be integer. However, $10$ numbers may occur to be integer:\n$$\n\\frac{22}{11}, \\frac{14}{7}, \\frac{15}{5}, \\frac{21}{3}, \\frac{20}{10}, \\frac{18}{9}, \\frac{16}{8}, \\frac{12}{6}, \\frac{19}{1}, \\frac{4}{2}, \\frac{13}{17}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find at least one pair of positive integers $(x, y)$, which satisfies an equality:\n$$\n\\frac{1}{2}(x^2 - y^3) = 2016.\n$$",
    "options": [],
    "answer": "(64, 4)",
    "solution": "Rewrite the given condition in the following way:\n$$\nx^2 - y^3 = 2 \\cdot 2016 = 2^6 \\cdot 3^2 \\cdot 7 = 2^6 \\cdot (64 - 1) = 2^{12} - 2^6.\n$$\nNow it is easy to see how to find the solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45465,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 22 cards, where the numbers $1, 2, \\ldots, 22$ are written. Using these cards one formed 11 fractions. What is the greatest possible number of integer numbers among the fractions?",
    "options": [],
    "answer": "10",
    "solution": "The numbers $13$, $17$, $19$ may form an integer only if they stand at the numerator position and $1$ stands in the denominator position. Hence, at least one fraction cannot be integer. However, $10$ numbers may occur to be integer:\n$$\n\\frac{22}{11}, \\frac{14}{7}, \\frac{15}{5}, \\frac{21}{3}, \\frac{20}{10}, \\frac{18}{9}, \\frac{16}{8}, \\frac{12}{6}, \\frac{19}{1}, \\frac{4}{2}, \\frac{13}{17}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45466,
    "subject": "Mathematics (Multi-modal)",
    "question": "In acute-angled triangle $ABC$ bisector $AL$, height $BH$ and the perpendicular bisector of line $AB$ intersect at one point. Find the angle $BAC$.\n\n**Answer:** $\\angle BAC = 60^\\circ$.",
    "options": [],
    "answer": "60 degrees",
    "solution": "Let the angle $BAC$ be $2\\alpha$ (fig. 15). Hence $\\triangle APB$ is isosceles, thus $\\angle PBA = \\alpha$. Since $\\triangle AHB$ is right triangle, $3\\alpha = 90^\\circ$, thus $\\angle BAC = 2\\alpha = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$, such that $11^n - 1$ is divisible by $10^n - 1$.",
    "options": [],
    "answer": "no positive integers",
    "solution": "Since $10^n - 1 = 9 \\cdot (10^{n-1} + \\dots + 10 + 1)$ we obtain that $11^n - 1$ is divisible by $9$. Considering the residues modulo $9$: $11^n - 1 \\equiv 2^n - 1 \\pmod{9}$ we get $n = 6k$. But then $10^{6k} - 1$ is divisible by $10^6 - 1$. Hence it is divisible by $10^3 + 1$ and $10 + 1 = 11$ which is not possible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive real numbers $a, b, c$, which satisfy the equality:\n$$\nab\\left(1 - \\frac{c^2}{(a+b)^2}\\right) = bc\\left(1 - \\frac{a^2}{(b+c)^2}\\right) = ca\\left(1 - \\frac{b^2}{(c+a)^2}\\right).\n$$",
    "options": [],
    "answer": "(t, t, t) for any t > 0",
    "solution": "Rewrite the given equality in the following way:\n$$\n\\frac{a((a+b)^2-c^2)}{(a+b)^2} = \\frac{c((b+c)^2-a^2)}{(b+c)^2} \\quad \\text{or} \\quad \\frac{a(a+b-c)(a+b+c)}{(a+b)^2} = \\frac{c(b+c-a)(b+c+a)}{(b+c)^2}.\n$$\nAfter dividing by $a+b+c$ we obtain\n$$\n\\frac{a(a+b-c)}{(a+b)^2} = \\frac{c(b+c-a)}{(b+c)^2}.\n$$\nSuppose that $c \\ge a+b$. Then $b+c > a$, so the left side of the equality is not positive while the right side is positive. This contradiction means that $b+c > a$, $a+c > b$ and $b+a > c$ so we can treat the sides $a, b, c$ as sides of a triangle.\nBut the given equalities literally mean that the angle bisectors of this triangle are equal. Indeed: $l_c^2 = ab - a_1b_1$, where $a_1, b_1$ are the lengths of the two parts of the opposite side which are derived after drawing the angle bisector. Then we have: $a_1 + b_1 = c$ and $\\frac{a_1}{b_1} = \\frac{a}{b}$. It is easy to see that $a_1 = \\frac{ac}{a+b}$ and $b_1 = \\frac{bc}{a+b}$. Then $l_c^2 = ab - a_1b_1 = ab - \\frac{abc^2}{(a+b)^2} = ab\\left(1 - \\frac{c^2}{(a+b)^2}\\right)$. Since if three angle bisectors are equal the triangle is regular, we obtain that the solutions are $(t, t, t)$, where $t > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Petryk programmed calculator in such a way that if there is number $x$ on the screen and then the button «=>» is pressed, number $\\frac{x-1}{x+1}$ appears on the screen. Petryk pressed the button «=>» 2014 times, and then 2016 appeared on the screen. What number was on the screen at the beginning? The screen can show not only integer numbers.",
    "options": [],
    "answer": "-1/2016",
    "solution": "Let us look how the number changes after the button «=>» is pressed. Let the first number be $x$. Then after the first pressing we have $\\frac{x-1}{x+1}$.\n\nAfter the second we have\n$$\n\\frac{\\frac{x-1}{x+1}-1}{\\frac{x-1}{x+1}+1} = \\frac{x-1-x-1}{x-1+x+1} = -\\frac{1}{x}.\n$$\nAfter the third we have\n$$\n\\frac{-\\frac{1}{x}-1}{-\\frac{1}{x}+1} = \\frac{-1-x}{-1+x}.\n$$\nAfter the fourth we have\n$$\n\\frac{-\\frac{1-x}{-1+x}-1}{\\frac{-1-x}{-1+x}+1} = \\frac{-1-x+1-x}{-1-x-1+x} = x.\n$$\nThus, after every fourth pressing the number that was at the beginning appears on the screen. Therefore, after 2012 pressings we have $x$. After 2 more pressings we obtain $-\\frac{1}{x} = 2016$, thus $x = -\\frac{1}{2016}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real nonzero numbers $a$, $b$, $c$, $d$ satisfy the conditions $a^3 + b^3 + c^3 + d^3 = 0$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\neq 0$. Prove that $a + b + c + d \\neq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution.** We will use method from contradiction. Suppose $a + b + c + d = 0$. Thus $a + b = -(c + d)$. Therefore, we can obtain such equalities:\n$$\n(a + b)^3 = -(c + d)^3 \\Leftrightarrow a^3 + b^3 + 3ab(a + b) = -c^3 - d^3 - 3cd(c + d) \\Leftrightarrow \\\\\n3ab(a + b) + 3cd(c + d) = 0 \\Leftrightarrow -3ab(c + d) - 3cd(a + b) = 0 \\Leftrightarrow \\\\\nab(c + d) + cd(a + b) = 0 \\Leftrightarrow abc + abd + acd + bcd = 0.\n$$\nThat contradicts the conditions since:\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\neq 0 \\Leftrightarrow \\frac{abc + abd + acd + bcd}{abcd} \\neq 0 \\Leftrightarrow abc + abd + acd + bcd \\neq 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45471,
    "subject": "Mathematics (Multi-modal)",
    "question": "30 children – boys and girls – formed a circle. It occurred that there is no child such that both its neighbors are boys. What is the least possible number of girls there?",
    "options": [],
    "answer": "16",
    "solution": "Consider a group of consecutive boys. There are less than 3 boys in this group. Also consider a group of consecutive girls. There are at least 2 girls in the group. These groups follow each other hence their amounts are equal. Assume that it may occur that there are 15 boys and 15 girls. Then there are at least 8 groups of boys, hence there are at least 8 groups of girls and there are at least 16 girls in total. So there are at most 14 boys and at least 16 girls.\n\nIt is easy to see that the desired arrangement exists. Consider 7 groups each consisting of 2 boys which are alternating by 6 groups of girls consisting of 2 and one group consisting of 4.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy, Bogdan and Olesia were walking by the same road from home to the school. Andriy was walking with velocity equal to $a$ km/h for $(2-b)$ hours, Bogdan was walking with velocity equal to $b$ km/h for $(2-c)$ hours, Olesia was walking with velocity equal to $c$ km/h for $(2-a)$ hours, where $a, b, c$ are some real numbers. What is the distance between home and school if it is known that it is equal to an integer number?",
    "options": [],
    "answer": "1",
    "solution": "Analyzing the problem we get:\n$$\nS = a(2-b), \\quad S = b(2-c), \\quad S = c(2-a),\n$$\nwhere $S$ -- positive integer which is equal to the distance.\n\nWe may assume that $a \\ge b$. If $a > b$, then $2-b < 2-c$ or $b > c$. Analogously, $2-c < 2-a$ or $c > a$. Contradiction. So $a = b = c$. Then we have that $S = a(2-a)$. Let us show that $S = a(2-a) \\le 1$. Indeed, $2a - a^2 \\le 1 \\Leftrightarrow (a-1)^2 \\ge 0$. Hence, since $S$ is a positive integer which is less or equal than $1$, $S=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all strictly increasing functions $f: \\mathbb{R} \\to \\mathbb{R}$ that satisfies the conditions:\n$$\n\\bullet \\quad f(x^3 + y^2 + f(y)) = x^2 f(x) + y f(y) + y;\n$$\n$$ \\bullet \\quad \\text{Equality } y^2 + f(y) = t \\text{ has a solution for every } t \\in \\mathbb{R}. $$",
    "options": [],
    "answer": "no such function",
    "solution": "**Answer.** Solution does not exist.\n\nLet us substitute $x = y = 0$ at the first condition $\\Rightarrow f(f(0)) = 0$.\nThen when $x=0$ and when $x=f(0)$ we obtain such equations:\n$$\nf(y^2 + f(y)) = y f(y) + y \\quad \\text{ta} \\quad f(f^3(0) + y^2 + f(y)) = y f(y) + y.\n$$\nSince $f$ is strictly increasing:\n$$\nf(y^2 + f(y)) = f(f^3(0) + y^2 + f(y)) \\Rightarrow f(0) = 0.\n$$\nIf we substitute $x=0$ and $y=0$, we obtain\n$$\n\\begin{aligned}\nf(y^2 + f(y)) &= y f(y) + y \\quad \\text{ta} \\quad f(x^3) = x^2 f(x) \\\\\nf(x^3 + y^2 + f(y)) &= f(x^3) + f(y^2 + f(y)).\n\\end{aligned}\n$$\nSince $y^2 + f(y)$ can equal any positive value (from the second condition), the last equation can be rewritten in such a way: $f(x+t) = f(x) + f(t)$, where $x \\in \\mathbb{R}, t \\ge 0$.\n\nTherefore, $f$ is increasing and satisfies Cauchy's functional equation, thus $f(x) = kx$ for some real $k$. After checking the first condition it is clear that only case $k=1$ is correct, but such a function does not satisfy the second condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: N \\to Z$ be the function that for any two positive integers $a, b$ $f(ab) = f(a) + f(b)$. Find all possible values of $f(1000)$.",
    "options": [],
    "answer": "3k for any integer k",
    "solution": "Let us look on such an equality:\n$$\nf(1000) = f(10^3) = 3 \\cdot f(10)\n$$\nLet us show now that for any integer $k$ $f(1000)$ can equal $3k$. Let $\\operatorname{ord}(a)$ be a maximum degree of $2$, by what $a$ is divisible. Let\n$$\na = 2^i(2a_0 + 1), \\quad b = 2^m(2b_0 + 1). \\text{ Thus}\n$$\n$$\n\\operatorname{ord}(ab) = \\operatorname{ord}(2^{i+m}(2a_0+1)(2b_0+1)) = i+m = \\operatorname{ord}(a)+\\operatorname{ord}(b). \\text{ Let } f(a)=k \\cdot \\operatorname{ord}(a), \\text{ then}\n$$\n$$\nf(ab) = k \\cdot \\operatorname{ord}(ab) = k \\cdot (\\operatorname{ord}(a) + \\operatorname{ord}(b)) = k \\cdot \\operatorname{ord}(a) + k \\cdot \\operatorname{ord}(b) = f(a) + f(b)\n$$\n$$\n\\Rightarrow f(1000) = k \\cdot \\operatorname{ord}(2^3 \\cdot 125) = 3k,\n$$\nThat is what we had to prove.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Grasshopper is sitting at point $O$ of coordinate line. He makes $2016$ jumps in positive direction. Firstly his jump has length $1$, every next jump is $k \\in \\mathbb{N}$ times longer than the previous one. There are holes in every point with coordinate $2016l$, $l \\in \\mathbb{N}$. Determine $k$ for which the grasshopper will make all the jumps and will not jump into hole.",
    "options": [],
    "answer": "All natural numbers k with gcd(k, 2016) > 1",
    "solution": "$$\na_1 = 1, \\quad a_n = 1 + k + k^2 + \\dots + k^{n-1}, \\quad n = 2, \\ldots, 2016.\n$$\nWe have to find such $k$, for which $a_n$ is not divisible by $2016$.\n\nLet us suggest that $(k, 2016) = d > 1$. As we see, every coordinate after the jump has residue $1$ modulo $d$ thus cannot be divisible by $2016$. So all such $k$ are the answers.\n\nIf $(k, 2016) = 1$, let us look at all the residues of $a_n$ modulo $2016$. If at least one value has residue $0$ then $k$ does not satisfy the condition. Let us suggest that there are no zeroes, then at least two residues are the same. Let them be $a_m$ and $a_l$, $m > l$. Their difference $a_m - a_l$ is divisible by $2016$, but then\n$$\na_m - a_l = (1 + k + k^2 + \\dots + k^{m-1}) - (1 + k + k^2 + \\dots + k^{l-1}) =\n$$\n$$\nk^l (1 + k + k^2 + \\dots + k^{m-l-1}) = k^l a_{m-l}.\n$$\nThus $a_{m-l}$ is divisible by $2016$, and we have a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = ax^2 + bx + c$ be a polynomial with integer coefficients. For every integer $x$ $f(x)$ is divisible by $N$ where $N$ is a positive integer. Is it true that $N$ necessarily divides all the coefficients of $f(x)$ if\n$$\n\\text{a) } N = 2016; \\quad \\text{b) } N = 2017?\n$$",
    "options": [],
    "answer": "a) No. b) Yes.",
    "solution": "a) Note that for any integer $x$ the product $x(x+1)$ is even. This suggests the following example:\n$$\n1008x(x+1)+2016=1008x^2+1008x+2016.\n$$\nFor every integer $x$, $f(x)$ is divisible by $2016$, but $2016$ does not divide all the coefficients ($1008$ is not divisible by $2016$). Thus, the answer is no.\n\nb) We have $f(x) = ax^2 + bx + c$. Let us do the following substitutions:\n$$\n\\begin{aligned}\nx &= 0 \\Rightarrow f(0) = c \\implies 2017 \\,|\\, c; \\\\\nx &= 1 \\Rightarrow f(1) = a + b + c \\implies 2017 \\,|\\, a + b + c; \\\\\nx = -1 \\Rightarrow f(-1) = a - b + c \\implies 2017 \\,|\\, a - b + c.\n\\end{aligned}\n$$\nThen both $a + b$ and $a - b$ are divisible by $2017$, so are $2a$ and $2b$. Since $2017$ is odd, all the coefficients are divisible by $2017$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a polygon with $2016$ vertices. Alisa and Basilio play the following game. In each turn a player draws a diagonal of the polygon which intersects the other drawn diagonals or the sides only at the vertices. When the polygon is cut into triangles the game is finished. For each triangle having exactly zero sides among the sides of the initial polygon Alisa is paid $1$ cent. For each triangle having exactly two sides among the sides of the initial polygon Basilio is paid $1$ cent. Who will get more money and what would be the difference if both are clever players?\n\n![](attached_image_1.png)\n\nFig. 02",
    "options": [],
    "answer": "Basilio by 2 cents",
    "solution": "Let $a$ be the number of the triangles which have $0$ sides among the sides of the initial polygon, $b$ be the number of the triangles having $1$ such side, and $c$ be the number of the triangles with $2$ such sides (fig. 02). Then $b + 2c = 2016$ since the polygon has $2016$ sides. Also, since we will get $2014$ triangles, $2014 = a + b + c$. Hence $c = a + 2$. So Basilio will get $2$ cents more than Alisa with no dependence on how they are playing.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45478,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Does there exist a set $A$ that has $2016$ pairwise different positive integers and such a property: for every non-empty subset $B \\subset A$ that is not equal to $A$ and every non-empty subset $C \\subset (A \\setminus B)$ the sum of the elements of subset $B$ is not divisible by the sum of the elements of subset $C$?\n\nb) Does there exist a set $A$ that has $2016$ pairwise different positive integers and such a property: for every non-empty subset $B \\subset A$ that is not equal to $A$ and contains at least $2$ elements and every non-empty subset $C \\subset (A \\setminus B)$ the product of the elements of subset $B$ is divisible by the sum of the elements of subset $C$?",
    "options": [],
    "answer": "a) Yes; b) Yes",
    "solution": "a)\nLet us firstly choose numbers in such a way: $2, 3, 5, \\ldots, p_{2016}$, where $p_i$ is a prime number in increasing order. Then we will write down all the possible combinations of subsets $B$ and $C$. Its amount is finite. Let there are $N$ such pairs. Successively we will change the elements in a particular way. And we will build ordered set $A$, at every step we will change some of its elements comparing with the previous step. We will identify the elements as $a_1, a_2, \\ldots, a_{2016}$ (at the $0$ step (at the beginning) $a_i = p_i$, $i=1, \\ldots, 2016$). Elements of the sets $B$ and $C$ will be determined by their indices. For instance, $B = \\{1, 4, 2000\\}$ means that the set $B$ includes $a_1, a_4$ and $a_{2000}$ elements of $A$, after their change at the previous step.\n\nLet us look at the sets $A, B, C$ in $k$-th row. Let $p = p_{2016+k}$. Now we multiply all the elements of the set $C$ by $p$, and from the elements of the set $B$ - the first element (with the lowest index $a_j$) we do not multiply by $p$, and the rest of the elements (if the set $B$ contains only one element, we do not have the rest) multiply by $p$. Thus the sum of the elements of the set $B$ is not divisible by $p$, and the sum of the elements of the set $C$ is divisible by $p$, hence the sum of the elements of $B$ is not divisible by the sum of the elements of $C$.\n\nIn such a way we will make $N$ steps and show that the obtained set satisfies the conditions.\n\nLet us choose sets $B$ and $C$ arbitrarily (the sets are determined by the sets of indices), let it correspond to $k$-th row of the table. Then from the elements of set $B$ all but one are divisible by $p_{2016+k}$, and among elements of set $C$ all are divisible by it, as is the sum.\n\nIt is clear that similarly we can build a corresponding set, that contains arbitrarily finite amount of pairwise distinct elements.\n\nb)\nLet us show how such a set can be built for arbitrarily finite amount of elements $n$. Let $N = \\left(\\frac{n(n+1)}{2}\\right)!$. Let us build the corresponding set: $a_i = i \\cdot N$, $i=1, \\ldots, n$. Let us show that it satisfies the conditions. Consider any subset $C$, sum of its elements $L$ is not greater than $L \\le M = (1+2+\\ldots+n) \\cdot N = \\frac{n(n+1)}{2} N$, thus it can be written as $K \\cdot N$, where $K \\le \\frac{n(n+1)}{2}$. The product of any two numbers of the set $A$ is divisible by $N^2$, thus it is divisible by $L$, since $N$ is divisible by any number less than $\\frac{n(n+1)}{2}$, thus it is divisible by $K$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45479,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are four numbers on the board: $1$, $3$, $6$ and $10$. Each time we can erase any two numbers $a$, $b$ written on the board and write numbers $a+b$, $ab$ instead. Can we obtain such four numbers\n\na) $2015$, $2016$, $2017$, $2018$; after several moves?\n\nb) $2016$, $2017$, $2019$, $2022$",
    "options": [],
    "answer": "a) no; b) no",
    "solution": "**Answer:** a), b) that is not possible.\n\na) Let us look at the numbers modulo $3$. Obviously, the amount of numbers divisible by $3$ cannot decrease. Because if both $a$, $b$ are divisible by $3$, then both $a+b$, $ab$ are also divisible by $3$. If one of the numbers is divisible by $3$, then $ab$ is also divisible by $3$. Thus, at the beginning we had only one number divisible by $3$, and in the end only one, thus such situation is impossible.\n\nb) Let us look at the situation now when exactly three numbers are divisible by $3$. Thus those numbers equal $0$, $0$, $0$, $k$, where $k \\in \\{1, 2\\}$ modulo $3$. Thus these four numbers will never change. Let us check now when the amount of numbers divisible by $3$ can increase. Then $a$, $b$ should be $1$, $2$ modulo $3$. However, then numbers $0$, $2$ appear. Thus in the situation where exactly three numbers are divisible by $3$, they should equal $0$, $0$, $0$, $2$, and four numbers from condition equal $0$, $0$, $0$, $1$. Thus we get a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45480,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive numbers $a, b, c$ prove the inequality :\n$$\n\\sqrt{a^2 + bc} + \\sqrt{b^2 + ca} + \\sqrt{c^2 + ab} \\ge \\sqrt{ab + bc} + \\sqrt{bc + ca} + \\sqrt{ca + ab}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It should be noted, that the inequality is symmetrical, that is, the interchange in places of any two variables does not change its appearance. So, without any loss of generality we can assume that $a \\ge b \\ge c$. If we move all summands to one side and group them accordingly, the inequality becomes like this:\n$$\n(\\sqrt{a^2 + bc} - \\sqrt{ca + ab}) + (\\sqrt{b^2 + ca} - \\sqrt{ab + bc}) + (\\sqrt{c^2 + ab} - \\sqrt{bc + ca}) \\ge 0.\n$$\nAfter multiplying every summand on the conjugate, we obtain that the following inequality should be proven:\n$$\n\\frac{(a-b)(a-c)}{\\sqrt{a^2+bc}+\\sqrt{ca+ab}} + \\frac{(b-c)(b-a)}{\\sqrt{b^2+ca}+\\sqrt{ab+bc}} + \\frac{(c-a)(c-b)}{\\sqrt{c^2+ab}+\\sqrt{bc+ca}} \\ge 0.\n$$\nSince $\\frac{(a-b)(a-c)}{\\sqrt{a^2+bc}+\\sqrt{ca+ab}} \\ge 0$ it will be enough to prove that:\n$$\n\\frac{(a-c)(b-c)}{\\sqrt{c^2 + ab} + \\sqrt{bc + ca}} \\ge \\frac{(b-c)(a-b)}{\\sqrt{b^2 + ca} + \\sqrt{ab + bc}}.\n$$\nIf $b-c=0$, hence there are zeros in both parts of the inequality. Otherwise, we can divide both parts by $b-c>0$. Further, since $a-c \\ge b-c \\ge 0$ it should be enough to prove the inequality:\n$$\n\\frac{1}{\\sqrt{c^2 + ab} + \\sqrt{bc + ca}} \\ge \\frac{1}{\\sqrt{b^2 + ca} + \\sqrt{ab + bc}} \\Leftrightarrow \\\\\n\\sqrt{b^2 + ca} + \\sqrt{ab + bc} \\ge \\sqrt{c^2 + ab} + \\sqrt{bc + ca}.\n$$\nThe last inequality is true, since\n\n$b^2 + ca \\geq bc + ca$ and $ab + bc \\geq c^2 + ab$.\n$$\nThe inequality has been proven.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45481,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive $a$, $b$, $c$, that satisfy the condition $ab + bc + ca = 3$, prove an inequality:\n$$\n\\frac{1}{2a^3+1} + \\frac{1}{2b^3+1} + \\frac{1}{2c^3+1} \\ge 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us make such transformation:\n$$\n1 = \\frac{ab+bc+ca}{3} \\ge \\sqrt[3]{(abc)^2} \\Leftrightarrow abc \\le a \\le \\frac{1}{bc},\\ b \\le \\frac{1}{ac},\\ c \\le \\frac{1}{ab},\n$$\nhence $a+b+c \\le \\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$. Then we use well-known inequality:\n$$\n\\frac{a_1^2}{b_1} + \\frac{a_2^2}{b_2} + \\dots + \\frac{a_n^2}{b_n} \\ge \\frac{(a_1+a_2+\\dots+a_n)^2}{b_1+b_2+\\dots+b_n}.\n$$\nThen we will make such transformation:\n$$\n\\begin{align*}\n\\frac{1}{2a^3+1} + \\frac{1}{2b^3+1} + \\frac{1}{2c^3+1} &= \\frac{\\frac{1}{a^2}}{2a+\\frac{1}{a^2}} + \\frac{\\frac{1}{b^2}}{2b+\\frac{1}{b^2}} + \\frac{\\frac{1}{c^2}}{2c+\\frac{1}{c^2}} \\\\\n&\\ge \\frac{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})^2}{2a+2b+2c+\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}} \\\\\n&\\ge \\frac{\\frac{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})^2}{2}}{\\frac{1}{bc}+\\frac{1}{ca}+\\frac{1}{ab}+\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}} \\\\\n&= \\frac{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})^2}{(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})^2} = 1,\n\\end{align*}\n$$\nThat is what we had to prove.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AM$ be a median in an acute triangle $ABC$. Its extension intersect the circumcircle $w$ of $ABC$ at $P$. Let $AH_1$ be an altitude of $\\triangle ABC$, $H$ - its orthocenter. The rays $MH$ and $PH_1$ intersect $w$ at $K$ and $T$ respectively. Prove that the circumcircle of $\\triangle AKTH_1$ is tangent to $BC$.\n\n(Khilko Danylo)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to show that $\\angle TKH_1 = \\angle TH_1B$ (Fig. 14). Let us extend $KH_1$ and intersect if with $w$ at $S$. Then\n$$\n\\angle TKS = \\angle TAB + \\angle BAS, \\quad \\angle TH_1B = \\angle TAB + \\angle PAC,\n$$\nSo it is sufficient to show that $\\angle PAC = \\angle BAS$.\nDenote by $A_1$ the point such that $AA_1$ is the diameter of $w$. Then $\\angle A_1CA = \\angle ABA_1 = 90^\\circ$, so $BH \\parallel A_1C$ and $CH \\parallel A_1B$. Then $BHCA_1$ is a parallelogram, so $HA_1$ passes through the point $M$.\n\nThen $K$ lies on $HA_1$, hence $\\angle A_1KA = 90^\\circ$. Then the quadrilateral $AKH_1M$ is inscribed. So $\\angle KH_1A = \\angle KMA$. Suppose $AH_1$ intersect $w$ secondly at $F$. Then\n$$\n\\angle KH_1A = \\angle KCA + \\angle FAS, \\quad \\angle KMA = \\angle KCA + \\angle PAA_1.\n$$\nHence, $\\angle FAS = \\angle PAA_1$, also $\\angle ABC = \\angle AA_1C$, so we derive $\\angle BAF = 90^\\circ - \\angle ABC = \\angle A_1AC$. Then\n$$\n\\angle BAS = \\angle BAF + \\angle FAS = \\angle PAA_1 + \\angle A_1AC = \\angle PAC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45483,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertices of a cube are enumerated by the numbers $1$, $2$, $\\ldots$, $8$. Someone chose three sides of the cube and told the numbers which are written on them to Pete: $\\{1, 4, 6, 8\\}$, $\\{1, 2, 6, 7\\}$, $\\{1, 2, 5, 8\\}$. Is it possible to determine which number has the vertex which is opposite to the one numbered $5$?",
    "options": [],
    "answer": "6",
    "solution": "Three edges go from each vertex of a cube. Every edge is a part of two sides. Let us look at the vertex $1$. It lies inside all \"known\" sides. Hence these three sides form an angle with vertex in $1$. Then three edges are $1$-$2$, $1$-$6$, and $1$-$8$. Now it is easy to understand that $6$ is opposite to $5$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to cut a regular triangle into:\na) three equal quadrilaterals;\nb) three equal pentagons?\nConvexity of quadrilaterals and pentagons is not required.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Consider a regular triangle $\\triangle ABC$, denote its midpoints by $M, N, K$ respectively. (fig. 04). The segments $AN$, $BK$ and $CM$ intersect at $O$. It is easy to see that the quadrilaterals $BMON$, $CKON$ and $AMOK$ satisfy the condition.\n\nb) Now let us denote the midpoints of $OM$, $ON$ and $OK$ by $P$, $R$ and $Q$ respectively. The pentagons which we are looking for are $ABROP$, $ROQCB$ and $APOQC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a stripe $1 \\times n$, $n \\ge 4$. One wrote a positive integer in each cell of it (the numbers are not obliged to be equal). After under each number one writes a positive integer which is equal to the quantity of such integers in the previous row. (i.e. if $10$ was written in the previous row three times then under each of the numbers $10$ one writes $3$ in the next row). After the procedure repeats.\n\na) Prove that after a finite number of steps the rows will not change.\n\nb) How long can the next row differ from the previous one if $n=2016$ and if $n$ is arbitrary?",
    "options": [],
    "answer": "For n = 2016, the maximum number of steps before stabilization is 11. In general, the maximum is floor(log2 n) + 1.",
    "solution": "Obviously, it is possible to rearrange the numbers in each row so that they increase. Denote the initial row as $0^{\\text{th}}$, the next - $1^{\\text{st}}$ and so on.\n\nLet us describe the properties of the numbers inside rows beginning from the $1^{\\text{st}}$ one.\n\n**Lemma 1.** If $n$ occur in $k$ th row, then the amount of numbers \"n\" is divisible by $n$.\n\n**Proof.** Number $n$ occurs in the $1^{\\text{st}}$ row if there are $n$ equal numbers in the $0^{\\text{th}}$ row, say $m$. Then under these $n$ numbers \"$m$\" one writes a group of numbers $n$. If another number $l$ in the $0^{\\text{th}}$ row occurs $n$ times too, then a new group which contains $n$ numbers equal to $n$ will emerge in the $1^{\\text{st}}$ row. The number $n$ may occur in the rows after $1^{\\text{st}}$ one if there were numbers $n$ in the previous row and one wrote $n$ under them once again or there was a group of exactly $n$ equal numbers different from $n$. Hence, their amount is divisible by $n$.\n\n**Lemma 2.** If there is a group of numbers \"n\" in the $k$ th row then the same number will be written under these numbers but this number is not obliged to be equal to $n$.\n\n**Proof.** It is an easy consequence of the property that if in the previous row there were two equal numbers then under them one should write the same numbers too.\n\nConsider the first row. Let $l$ be the smallest number inside it. That means the amount of the numbers \"$l$\" will be equal to $sl$, $s \\ge 1$. The other numbers are greater than $l$.\n\nIf $l$ occur exactly $l$ times, then under them only one can write only numbers $l$. They cannot decrease since there are exactly $l$ of them and they cannot increase because in this case we need $l$ equal numbers but there is no such group. So we can erase these numbers and consider a row with $n-l$ numbers. Now we can look for the smallest number in the new row.\n\nIf there is a group of $l$ consisting of $sl$ numbers, $s > 1$, then in the next group one has to write $sl$ under them. Consider some $m$ in the 1st row such that: $l < m < 2l$. Analogously, if there are exactly $m$ of them then they cannot change in the lower rows. The numbers which are less then $m$ either do not change (and does not impact on the changes of the numbers $m$) or at least two times increase, so $m$ cannot occur under them. So we can ignore such numbers. If there are $tm$ of them then they are replaced by $tm$ in the consecutive rows.\n\nHence in the 2nd row we need only consider the numbers which are greater or equal to $2l$.\n\nAnalogously, if the least number in the 2nd row is $l_2$ (among the numbers which we need to consider), then in the 3rd row the least such number will be at least $2l_2$ and so on.\n\nIt is easy to see that a number greater then $n$ cannot emerge in the rows. So the greatest possible number of steps is $k$, for which $2^k \\le n$. But we have to count the move from 0th to 1st row. So at most we can have $k+1$ steps.\n\nLet the 0th row has the following appearance:\n$$\n(n-1), (1), (2, 2), (4, 4, 4), \\dots, \\underbrace{(2^{k-1}, 2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, m, \\dots, m)}_{m} \\text{ (0th row)}\n$$\n\n$$\n(1, 1), (2, 2), (4, 4, 4), \\dots, \\underbrace{(2^{k-1}, 2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, m, \\dots, m)}_{m} \\text{ (1st row)}\n$$\n$$\n(2, 2, 2, 2), (4, 4, 4, 4), \\dots, \\underbrace{(2^{k-1}, 2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, m, \\dots, m)}_{m} \\text{ (2nd row)}\n$$\n$$\n(4, 4, 4, 4, 4, 4, 4), \\dots, \\underbrace{(2^{k-1}, 2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, m, \\dots, m)}_{m} \\text{ (3rd row)}\n$$\n$$\n\\underbrace{(2^{k-1}, 2^{k-1}, \\dots, 2^{k-1})}_{2^k}, \\underbrace{(m, m, \\dots, m)}_{m} \\text{ (k-1\\textsuperscript{th} row)}\n$$\n$$\n\\underbrace{(2^k, 2^k, \\dots, 2^k)}_{2^k}, \\underbrace{(m, m, \\dots, m)}_{m} \\text{ (k\\textsuperscript{th} row)}\n$$\n\nNow let us understand why the condition $m = 2^k - n = 2^i$ is important. If there is a group consisting of $2^i$ then there is a group of the numbers $2^i$, which will give a group of $2^{i+1}$ in the next row. So in the described process the numbers change as: $2^0 \\to 2^1 \\to 2^2 \\to \\dots \\to 2^i$, in the next step the numbers will not change and the changes will last shorter.\n\nFor instance if there are 5 numbers in the beginning then we will have the following estimates: $2^2 = 4 < 5$, so the answer is at most 3 steps. But if we consider the 0th row as 1; 4; 2; 2; 5 we will get the following 1st row: 1; 1; 2; 2; 1, 2nd row: 3; 3; 2; 2; 3. The same situation is if $n = 2^s + 1$.\n\nAnalogously if $n = 2^s + 2$, because in this case either a group of two equal numbers will emerge or there will be two distinct numbers in the 0th row. The last case when we cannot increase the amount of numbers - $n = 2^s + 4$. Here we have that if 4 decomposed into several integer summands then one of this summands would be equal to some power of 2. In this case there will emerge a group of three one's in the first row. So we do not need $2^k \\le n$, but $3 \\cdot 2^{k-1} \\le n$. Now we are going to determine for which $n$ it is possible. $3 \\cdot 2^{k-1} \\le n = 2^k + 4$. Then $2^{k-1} \\le 4$, or $k \\le 3$. So for $n=5$, $n=6$ and $n=12$. For $n=5$ we have already seen that it is impossible, for $n=6$ we have the following situation in 0th row: $1; 2; 2; 2; 5; 6 \\rightarrow 1; 3; 3; 3; 1; 1 \\rightarrow 3; 3; 3; 3; 3; 3 \\rightarrow 6; 6; 6; 6; 6; 6$, Analogously if $n=12$.\n\nIn case of $2^i \\ge 8$ it suffices to take instead of $(\\underbrace{m, m, ..., m}_{m})$ in the basic example two groups: $(\\underbrace{s, s, ..., s}_{s})$ and $(\\underbrace{t, t, ..., t}_{t})$, where $s = 2^{i-1}-1$ and $t = 2^{i-1}+1$, which will not change during the whole procedure and will not interfere with the basic example.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be real numbers from segment $[0; 1]$. Prove that\n$$\n(x^4 + y^4 + z^4) + (x^5 + y^5 + z^5) + (x - y)^6 + (y - z)^6 + (z - x)^6 \\le 6.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly, we will prove that if $\\alpha$, $\\beta \\in [0, 1]$, then $\\alpha^4 + \\beta^5 + (\\alpha - \\beta)^6 \\le 2$. Indeed, $t^n \\le t$ for all real $t \\in [0, 1]$ and any positive integer $n$. Suppose that $\\alpha \\ge \\beta$ (case $\\alpha \\le \\beta$ is considered analogously). Then $0 \\le \\alpha - \\beta \\le 1$, so $(\\alpha - \\beta)^6 \\le \\alpha - \\beta$. Also, $\\alpha^4 \\le \\alpha$ and $\\beta^5 \\le \\beta$. Summing up the previous inequalities we obtain:\n$$\n\\alpha^4 + \\beta^5 + (\\alpha - \\beta)^6 \\le \\alpha + \\beta + \\alpha - \\beta = 2\\alpha \\le 2.\n$$\nNow let $x$, $y$, $z \\in [0, 1]$. By the latter claim:\n$$\nx^4 + y^5 + (x-y)^6 \\le 2, \\quad y^4 + z^5 + (y-z)^6 \\le 2, \\quad z^4 + x^5 + (z-x)^6 \\le 2.\n$$\nAfter summing up we get the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45487,
    "subject": "Mathematics (Multi-modal)",
    "question": "For six integers $a, b, c$ and $A, B, C$ the following correlations are true:\n$$\nb + c = A^2, \\quad c + a = B^2, \\quad a + b = C^2, \\quad C > B > A \\geq 0.\n$$\nFind numbers $a, b, c$, for which the sum $A^2 + B^2 + C^2$ takes the smallest possible value.",
    "options": [],
    "answer": "a = 5, b = 4, c = -4",
    "solution": "Let us solve the given system of equations for numbers $a, b, c$. From the first two equations we obtain: $a - b = B^2 - A^2$. Let us add this to the third equation and find out, that\n$$\na = \\frac{1}{2}(B^2 + C^2 - A^2).\n$$\nSimilarly, or thinking symmetrically, we can find out that\n$$\nb = \\frac{1}{2}(C^2 + A^2 - B^2) \\text{ and } c = \\frac{1}{2}(A^2 + B^2 - C^2).\n$$\nThe smallest possible sum $A^2 + B^2 + C^2$ can be received for three smallest various squares of integers:\n$A^2 = 0, B^2 = 1, C^2 = 4$. But for these three squares, numbers $a, b, c$ will not be integers, for example:\n$a = \\frac{1}{2}(1 + 4 - 0) = \\frac{5}{2}$. The following three squares with the next bigger sum are the numbers:\n$A^2 = 0, B^2 = 1, C^2 = 9$. For this triple, the values of $a, b, c$ are integers, hence desired:\n$$\na = \\frac{1}{2}(1 + 9 - 0) = 5, \\quad b = \\frac{1}{2}(9 + 0 - 1) = 4, \\quad c = \\frac{1}{2}(0 + 1 - 9) = -4.\n$$\n**Answer:** $a = 5,\\ b = 4,\\ c = -4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45488,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points $A_1$ and $C_1$ are chosen on the sides $BC$ and $AB$ of the triangle $ABC$ so that the segments $AA_1$ and $CC_1$ are equal and perpendicular. Prove that if $\\angle ABC = 45^\\circ$, then $AC = AA_1$.\n\n(Andrei Gogolev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly we will show that the triangle is acute. Indeed, suppose that $\\angle BAC \\ge 90^\\circ$. Then $\\angle ACB \\le 45^\\circ = \\angle ABC < \\angle AA_1C$. From $\\triangle AA_1C$ we get that $AC > AA_1$. From the other side\n\n$\\angle C_1AC \\ge 90^\\circ$, then in $\\triangle AC_1C$ the side $CC_1$ is the longest, in particular, $CC_1 > AC$. Hence $CC_1 > AC$; $CC_1 > AC > AA_1$, contradiction.\nLet $AA_1$ and $CC_1$ intersect at $O$. Consider a circle with diameter $AC$, then $O$ lies on it (fig. 10).\n\n![](attached_image_1.png)\nFig. 10\n\nThis circle intersect $AB, BC$ at $P, Q$ respectively. The points $P$ and $Q$ lies inside the sides of $\\triangle ABC$ because the triangle is acute. Notice also that since $AA_1$ and $CC_1$ are cevians, $O$ lies inside $\\triangle ABC$, in other words, on the arc $PQ$ of the latter circle. We will prove that $\\angle A_1AQ = \\angle QAC$. If it is true $\\triangle CAA_1$ is isosceles, because $AQ$ is angle bisector and altitude in it and so $A_1A = AC$. Indeed, suppose that $\\angle A_1AQ = \\beta > \\alpha = \\angle QAC$. Then $A_1A > AC$. Obviously, $\\angle PBC = \\angle BAQ = 45^\\circ$. Then\n$$\n\\angle PCC_1 = 45^\\circ - \\angle OCQ = 45^\\circ - \\angle OAQ = 45^\\circ - \\beta.\n$$\nAlso $\\angle PCA = 90^\\circ - \\angle PAC = 45^\\circ - \\alpha$. Since $\\beta > \\alpha$, we have $45^\\circ - \\beta < 45^\\circ - \\alpha$, so $\\angle C_1CP < \\angle PCA$, hence $CC_1 < AC$. But $A_1A > AC$, so $CC_1 < AA_1$ - contradiction. The case $\\beta < \\alpha$ is considered analogously.\n\n\n**Alternative solution.** Denote $\\angle C_1CB = \\alpha$, $\\angle A_1AB = \\beta$. Then since $\\angle OCA + \\alpha + \\angle OAC + \\beta = 135^\\circ$ and $\\angle OCA + \\angle OAC = 90^\\circ$, so $\\alpha + \\beta = 45^\\circ$ (fig. 11).\n\n![](attached_image_2.png)\nFig. 11\n\nWe can assume that $AA_1 = CC_1 = 1$. Denote $OC_1 = p$, $OA_1 = q$, then $OC = 1-p$, $OA = 1-q$. Since $\\tan \\alpha = \\frac{p}{1-p}$ and $\\tan \\beta = \\frac{q}{1-q}$, rewrite the condition $\\alpha + \\beta = 45^\\circ$ as $\\tan(\\alpha + \\beta) = 1$ or\n$$\n\\begin{aligned}\n\\tan(\\alpha + \\beta) &= \\frac{\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta} = \\frac{\\frac{p}{1-p} + \\frac{q}{1-q}}{1 - \\frac{p}{1-p} \\frac{q}{1-q}} = \\\\\n&= \\frac{q - q^2 + p - p^2}{(1-p)(1-q) - pq} = \\frac{q + p - (q^2 + p^2)}{1 - p - q} = 1,\n\\end{aligned}\n$$\nso $p^2 + q^2 - 2p - 2q + 1 = 0$. Now we have that\n$$\nAC^2 = (1-p)^2 + (1-q)^2 = 2 - 2p - 2q + p^2 + q^2 = 1 = AA_1^2\n$$\nand we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Suppose that $AD$ and $BE$ are its angle bisectors. Prove that $\\angle ACB = 60^\\circ$ if and only if $AE + BD = AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $I$ the incenter of the triangle $\\triangle ABC$ (fig. 03). Also denote the point $D_1$, which is the symmetric reflection of $D$ with respect to $BE$. Then $DB = BD_1$ and $DI = ID_1$.\n\n![](attached_image_1.png)\n\nWithout any dependence on the given conditions we have $D_1 \\in AB$, since in $\\triangle DBD_1$ the line $BE$ contains the altitude, so it is the bisector of $\\angle DBD_1$.\n\nIf $\\angle ACB = 60^\\circ$ we get $\\angle AIB = 90^\\circ + \\frac{1}{2}\\angle ACB = 120^\\circ$. Then $\\angle DIB = 60^\\circ = \\angle D_1IB$, because $\\triangle DBD_1$ is isosceles. Hence, $\\angle EIA = 60^\\circ = \\angle AIB - \\angle D_1IB$. Then $\\triangle AIE = \\triangle AID_1$, so $EA = AD_1$ and $AE + BD = AB$.\n\nConversely, if $AE + BD = AB$, then $EA = AD_1$, so $\\triangle AIE = \\triangle AID_1$. Then $\\angle EIA = \\angle AID_1$. Since $\\triangle DBD_1$ is isosceles we have $\\angle DIB = \\angle D_1IB$. Then $\\angle EIA + \\angle DIB = \\angle AIB = \\angle EID$, so $120^\\circ = \\angle AIB = 90^\\circ + \\frac{1}{2}\\angle ACB \\Rightarrow \\angle ACB = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45490,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of real numbers: $a_0, a_1, a_2, \\dots, a_{2012}$ satisfies the following conditions:\n$$\n|a_0 - a_1| = \\frac{3^1}{2^0} |a_1 - a_2| = \\frac{3^2}{2^1} |a_2 - a_3| = \\dots = \\frac{3^{2011}}{2^{2010}} |a_{2011} - a_{2012}| = \\frac{3^{2012}}{2^{2011}} |a_{2012} - a_0|.\n$$\nWhich values the subtraction: $a_0 - a_{1006}$ can take?",
    "options": [],
    "answer": "0",
    "solution": "Let $|a_0 - a_1| = k$. We can write down the equations:\n$$\n\\begin{align*}\n a_0 - a_1 &= \\pm k, \\\\\n a_1 - a_2 &= \\pm \\frac{2^0}{3^1} k, \\\\\n a_2 - a_3 &= \\pm \\frac{2^1}{3^2} k, \\\\\n \\dots, \\\\\n a_{2011} - a_{2012} &= \\pm \\frac{2^{2010}}{3^{2011}} k, \\\\\n a_{2012} - a_0 &= \\pm \\frac{2^{2011}}{3^{2012}} k.\n\\end{align*}\n$$\nIf $k \\neq 0$, let us add these equations and reduce everything by $k$, as a result, we will get an equation:\n$$\n0 = \\pm 1 \\pm \\frac{2^0}{3^1} \\pm \\frac{2^1}{3^2} \\pm \\dots \\pm \\frac{2^{2010}}{3^{2011}} \\pm \\frac{2^{2011}}{3^{2012}}.\n$$\nBut irrespectively of sign before the 1 we cannot obtain 0 as a result, because module of sum of the summands left is less than 1. It is easy to prove using the formula for the sum of members of geometric progression:\n$$\n\\left| \\pm \\frac{2^0}{3^1} \\pm \\frac{2^1}{3^2} \\pm \\dots \\pm \\frac{2^{2010}}{3^{2011}} \\pm \\frac{2^{2011}}{3^{2012}} \\right| \\le \\frac{2^0}{3^1} + \\frac{2^1}{3^2} + \\dots + \\frac{2^{2010}}{3^{2011}} + \\frac{2^{2011}}{3^{2012}} = \\frac{\\frac{1}{3}\\left(1 - \\left(\\frac{2}{3}\\right)^{2012}\\right)}{1 - \\frac{2}{3}} = 1 - \\left(\\frac{2}{3}\\right)^{2012} < 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45491,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ with sides $BC > AC > AB$, consider the angles between the height and median, being built from one top. Find out, on which top this angle is the largest of the three.",
    "options": [],
    "answer": "Vertex B",
    "solution": "Let us denote in a standard way the sides of triangle $a, b, c$. Then under condition, $a > b > c$. Let us denote the height and median from the top $B$ as $h_b = BH$ and $m_b = BM$ accordingly (fig. 48). Then the interesting for us angle is $\\angle MBH$, while $\\cos\\angle MBH = \\frac{h_b}{m_b}$. Similarly, we can determine cosines of the other studied angles. In order to prove that the angle at the top $B$ is\n\n![](attached_image_1.png)\nFig. 48\n\nthe biggest, it should be enough to show that cosine of this angle is the smallest, that is:\n$$\n\\frac{h_b}{m_b} < \\min \\left\\{ \\frac{h_a}{m_a}, \\frac{h_c}{m_c} \\right\\} . \\text{ For this, let us prove that}\\\n$$\n$$\n\\frac{m_b^2}{h_b^2} > \\frac{m_a^2}{h_a^2} \\quad \\text{ta}\\quad \\frac{m_b^2}{h_b^2} > \\frac{m_c^2}{h_c^2}.\n$$\nTo do this, let us use the following formula to calculate the height and median:\n$$\nh_a = \\frac{2S_{ABC}}{a} \\text{ ta } m_a^2 = \\frac{2b^2 + 2c^2 - a^2}{4}.\n$$\nThe first inequation is equivalent to the following:\n$$\n\\frac{2a^2 + 2c^2 - b^2}{4} \\cdot \\frac{b^2}{4S^2} > \\frac{2b^2 + 2c^2 - a^2}{4} \\cdot \\frac{a^2}{4S^2} \\Leftrightarrow (2a^2 + 2c^2 - b^2)b^2 > (2b^2 + 2c^2 - a^2)a^2 \\Leftrightarrow 2b^2c^2 - b^4 > 2a^2c^2 - a^4 \\Leftrightarrow (a^2 + b^2)(a^2 - b^2) > 2c^2(a^2 - b^2).\n$$\nThe last inequation is true, since under the condition: $a > b > c$.\nSimilarly, for the second ineqation we have:\n$$\n(2a^2 + 2c^2 - b^2)b^2 > (2b^2 + 2a^2 - c^2)c^2 \\Leftrightarrow 2b^2a^2 - b^4 > 2a^2c^2 - c^4 \\Leftrightarrow 2a^2(b^2 - c^2) > (b^2 + c^2)(b^2 - c^2).\n$$\nThe last inequation, again, is true since $a > b > c$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Olesya chose 5 numbers from the set $\\{1; 2; 3; 4; 5; 6; 7\\}$. She told Pavlik the product of these numbers and asked whether the sum of these numbers is odd or even. Pavlik replied that he could not determine it for sure. What product might have had Olesya?",
    "options": [],
    "answer": "420",
    "solution": "If Pavlik knows the product, he could determine the product of the last two numbers that were not chosen. Since he could not determine the pairing of the product, then he could not determine the numbers that were not chosen although he knows their product. Consider all products of two numbers in the given set:\n$$\n\\begin{aligned}\n1 \\cdot 2 &= 2, \\quad 1 \\cdot 3 = 3, \\quad 1 \\cdot 4 = 4, \\quad 1 \\cdot 5 = 5, \\quad 1 \\cdot 6 = 6, \\quad 1 \\cdot 7 = 7, \\quad 2 \\cdot 3 = 6, \\quad 2 \\cdot 4 = 8, \\\\\n2 \\cdot 5 &= 10, \\quad 2 \\cdot 6 = 12, \\quad 2 \\cdot 7 = 14, \\quad 3 \\cdot 4 = 12, \\quad 3 \\cdot 5 = 15, \\quad 3 \\cdot 6 = 18, \\quad 3 \\cdot 7 = 21, \\\\\n4 \\cdot 5 &= 20, \\quad 4 \\cdot 6 = 24, \\quad 4 \\cdot 7 = 28, \\quad 5 \\cdot 6 = 30, \\quad 5 \\cdot 7 = 35, \\quad 6 \\cdot 7 = 42.\n\\end{aligned}\n$$\nThere are only 2 products that appear more than once. This is $1 \\cdot 6 = 6 = 2 \\cdot 3$ and $2 \\cdot 6 = 12 = 3 \\cdot 4$. Suppose that the product of the two numbers that were not chosen equals $6$. In both cases the sum of the numbers is odd: $1+6=7$ and $2+3=5$. Thus the sum of chosen numbers is also odd. This contradicts Pavlik's claim. Therefore, the product of the two numbers that were not chosen might equal $12$. In this case their sum might be even ($2+6=8$) or odd ($3+4=7$). Then the product of 5 chosen numbers is equal to $\\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7}{12} = 420$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute angled triangle and $\\angle ACB = 60^\\circ$. Let $BL$ be a bisector line and $BH$ be an altitude. Let $LD$ be a perpendicular from $L$ on the side $BC$. Determine the angles of $\\triangle ABC$ in case if $AB \\parallel HD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "∠A = 40°, ∠B = 80°, ∠C = 60°",
    "solution": "Firstly, we consider the case when point $L$ belongs to the segment $AH$ (fig. 21). Since $\\angle LHB = \\angle LDB = 90^\\circ$, then $BLHD$ is cyclic quadrilateral. Thus, $\\angle LBD = \\angle DHC$. On the other hand $\\angle LBD = \\angle LBA$, due to the fact that $BL$ is a bisector and $\\angle DHC = \\angle BAC$, because $AB \\parallel HD$. It follows that $\\angle ABC = 2\\angle BAC$ and $\\angle ABC + \\angle BAC = 180^\\circ - 60^\\circ = 120^\\circ$. And we get $\\angle BAC = 40^\\circ$ as $\\angle ABC = 80^\\circ$.\n\n![](attached_image_2.png)\n\nWe point out that $\\angle LBD = \\angle DHL$ and again $\\angle LBD = \\angle LBA$, due to the fact that $BL$ is a bisector of the triangle and $\\angle DHL = \\angle BAC$ since $AB \\parallel HD$. Similarly we get $\\angle BAC = 40^\\circ$ and $\\angle ABC = 80^\\circ$. But, if $\\angle ACB = 60^\\circ > \\angle BAC = 40^\\circ$ then the point $H$ can not belong to the segment $AL$. Thus, this case is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that number $N$ is also integer, where\n$$\nN = \\frac{2 \\cdot 1}{\\sqrt{1^2 + 1 + 4 + \\sqrt{1^2 - 1 + 4}}} + \\frac{2 \\cdot 2}{\\sqrt{2^2 + 2 + 4 + \\sqrt{2^2 - 2 + 4}}} + \\dots + \\frac{2n}{\\sqrt{n^2 + n + 4 + \\sqrt{n^2 - n + 4}}}\n$$\n(Anikushyn A., Rublyov B.)",
    "options": [],
    "answer": "3",
    "solution": "Let us note that\n$$\n\\frac{2k}{\\sqrt{k^2 + k + 4} + \\sqrt{k^2 - k + 4}} = \\frac{(\\sqrt{k^2 + k + 4} - \\sqrt{k^2 - k + 4})2k}{(k^2 + k + 4) - (k^2 - k + 4)} = \\\\\n= \\sqrt{k^2 + k + 4} - \\sqrt{k^2 - k + 4} = \\sqrt{k^2 + k + 4} - \\sqrt{(k-1)^2 + (k-1) + 4}.\n$$\nAnd thus\n$$\nN = (\\sqrt{6} - \\sqrt{4}) + (\\sqrt{10} - \\sqrt{6}) + (\\sqrt{16} - \\sqrt{10}) + \\dots + (\\sqrt{n^2 + n + 4} - \\sqrt{(n-1)^2 + (n-1) + 4}) = \\\\\n= \\sqrt{n^2 + n + 4} - \\sqrt{4} = \\sqrt{n^2 + n + 4} - 2.\n$$\nIt means that $N$ is integer if and only if $n^2 + n + 4$ is a perfect square.\nSince $n^2 < n^2 + n + 4 < n^2 + 4n + 4 = (n + 2)^2$, then for every positive integer $n$ we obtain $n^2 + n + 4 = (n + 1)^2 = n^2 + 2n + 1$. And thus $n = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45495,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square $n \\times n$ is divided into $n^2$ cells. In total $n^2$ tokens are placed at some cell. During every round, a player can move one token from cell $A$ to cell $B$, and one token from cell $A$ to cell $C$, provided that cell $A$ contained at least two tokens, $B$ and $C$ are symmetric with respect to $A$, and finally $B$ and $C$ are adjacent to $A$. Is it possible that after a few such rounds every cell on the board contains exactly one token, in case:\na) $n = 2016$;\nb) $n = 2017$?",
    "options": [],
    "answer": "a) impossible; b) possible",
    "solution": "**Answer:** a) impossible; b) possible.\n\na) Consider square $n \\times n$, for positive integer $n$. Let us denote the most left column by $1$, adjacent column by $2$, and so on, the most right column by $n$. For any cell $c$ we denote function $w(c) = b$, provided that cell $c$ is located in the column $b$. For every token $T$ in the same way let $w(T) = b$ provided that token $T$ is placed on the cell in the column $b$ at the moment. Finally, by $W$ we denote the sum of $w(T)$ for all tokens $T$ on the board. It is clear that after every round $W$ is not changed. At the end of our process we must have\n$$\nW = n + 2n + \\dots + n^2 = n \\cdot (1 + 2 + \\dots + n) = \\frac{n^2(n+1)}{2}.\n$$\nAt the beginning of our process $W = l \\cdot n^2$ provided that all tokens are placed at the cell in $l$. And for $n = 2016$ we get a contradiction.\n\nb) Let us consider the case $n = 2017$. It follows from previous arguments that the statement of the problem could be true only in the case when all tokens are placed on the central cell at the beginning. In this case we provide the algorithm. Let us note that in every cell we write the number of tokens placed there. Using induction, we prove the following statements\n\n**Statement 1.** For every positive integer $n$ in a row of length $2n+1$ from the position\n$0; 0; 0; \\dots; 0; 2n+1; 0; \\dots; 0; 0$\nwe can get the position\n$1; 1; 1; \\dots; 1; 1; 1; \\dots; 1; 1$\n\n**Statement 2.** For every positive integer $n$ in a row of length $2n+1$ from the position\n$0; 1; 1; \\dots; 1; 3; 1; \\dots; 1; 1$\nwe can get the position\n$1; 1; 1; \\dots; 1; 1; 1; \\dots; 1; 1$\n\n*Proof.* It is clear for $n=1$. Suppose that for $n=k-1$ statements are true, let us prove them for $n=k$. At the beginning, we have $2k+1$ tokens on the central cell. Two of them we will not move. The rest $2k-1$ tokens we can move (according to the induction hypothesis) to the position\n$0; 1; 1; \\dots; 1; 1; 3; 1; 1; \\dots; 1; 1$\nNow consider the following replacing\n$0; 1; 1; \\dots; 1; 1; 2; 1; 2; 1; 1; \\dots; 1; 1$ \n$0; 1; 1; \\dots; 1; 2; 0; 3; 0; 2; 1; \\dots; 1; 1$ \n$0; 1; 1; \\dots; 2; 0; 1; 3; 1; 0; 2; \\dots; 1; 1$ \n$\\dots$ \n$0; 1; 2; 0; 1; \\dots; 1; 1; 1; 3; 1; 1; 1; \\dots; 1; 0; 2; 1$ \n$0; 2; 0; 1; 1; \\dots; 1; 1; 1; 3; 1; 1; 1; \\dots; 1; 1; 0; 2$ \n$1; 0; 1; 1; 1; \\dots; 1; 1; 1; 3; 1; 1; 1; \\dots; 1; 1; 1; 0$\nWe can apply the induction hypothesis now, and get\n$1; 1; 1; \\dots; 1; 1; 1; \\dots; 1; 1$\nBoth statements are proved now.\n\nWe can easily prove the statement of the problem now. Divide all tokens into equal groups with $2017$ tokens in every group. Using our statements, we can replace tokens such that in every cell of the central column exactly one group will be placed. And then just use the statements again to replace tokens such that in every cell of every row exactly one token will be placed.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $a$, $b$, $c$ satisfy the conditions:\n$$\na^2 + 2 = b^4, \\quad b^2 + 2 = c^4, \\quad c^2 + 2 = a^4.\n$$\nWhat values can take expression\n$(a^2 - 1)(b^2 - 1)(c^2 - 1)$?",
    "options": [],
    "answer": "1",
    "solution": "Let us subtract $1$ from the left and right parts of equality and obtain:\n$$\na^2 + 1 = b^4 - 1 = (b^2 - 1)(b^2 + 1),\n$$\n\nand analogously from other $2$ equalities. Next multiply the resulting equalities:\n\n$$\n(a^2 + 1)(b^2 + 1)(c^2 + 1) = (a^2 - 1)(a^2 + 1)(b^2 - 1)(b^2 + 1)(c^2 - 1)(c^2 + 1)\n$$\n$$\n(a^2 - 1)(b^2 - 1)(c^2 - 1) = 1.\n$$\nNote that this value is achieved when $a = b = c = \\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45497,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. A convex heptagon is divided into triangles by drawing its diagonals. Prove that in such case you can obtain 5 or 7 triangles, but can not get 6 triangles.\n\nb. Prove that there exists nonconvex heptagon that can be divided by internal diagonals into 6 triangles. Internal diagonal of a polygon $M$ is a segment connecting the two non-neighbouring vertices of $M$ and does not go beyond the figure.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. If all vertices of resulting triangles coincide with vertices of the heptagon, the sum of the angles of the triangles equals the sum of the angles of the heptagon and equals $5 \\cdot 180^\\circ$. Hence, there are only 5 triangles. In case if two diagonals intersect not at the vertex of the heptagon, then this intersection point $A$ lies inside the heptagon because of its convexity. Then the angles of triangles adjacent to point $A$ sum up to $360^\\circ$, and the sum of the angles of resulting triangles is not less than $5 \\cdot 180^\\circ + 360^\\circ = 7 \\cdot 180^\\circ$. Thus, in this case will have at least 7 triangles.\n\n![](attached_image_1.png)\n\nb. The heptagon and corresponding division are shown in fig. 38. Note that the horizontal diagonal lies through the vertex of the heptagon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45498,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each pair of integers $a, b$ a non-negative integer $a*b$ is defined such that it satisfies the following two conditions:\n1) $(a+b)*b = a*b+1$;\n2) $(a*b) \\cdot (b*a) = 0$.\nFind values of the expressions $2016*121$ and $2016*144$.",
    "options": [],
    "answer": "2016*121 = 16, 2016*144 = 13",
    "solution": "**Answer:** $2016 * 121 = 16$, $2016 * 144 = 13$.\n\nSuppose there are two positive integers $a, b$. Let us rewrite them as follows: $a = bq + r$, where $q$ is a non-negative integer, $r$ is a positive integer and $r \\le b$. We prove that under such conditions $a*b = q$.\n\nIf in the second condition $\\forall a \\in \\mathbb{N}$ we put $a = b$, then we will obtain $a*a = 0$. If we suppose that $a*b = 0$, and in condition 1) put $a_1 + b_1 = a$, $b_1 = b$, then we obtain\n$$\n(a_1 + b_1)*b_1 = a_1*b_1 + 1 \\text{ or } a*b = (a-b)*b + 1.\n$$\nBut this contradicts the definition of the operation, because in such case for positive integers $a-b$ and $b$ the operation is not defined, because then $(a-b)*b = -1$, which is impossible. Thus, for $a > b$ $b*a = 0$.\n\nNow suppose $a > b$ and $a = bq + r$, $q, r \\in \\mathbb{N}$, $r \\le b$. Then\n$$\n\\begin{align*}\n& r*b = 0 \\Rightarrow (r + b)*b = r*b + 1 = 1 \\Rightarrow ((r + b) + b)*b = (r + b)*b + 1 = 2 \\Rightarrow \\dots \\\\\n& \\qquad (r + qb)*b = ((r + (q-1)b) + b)*b + 1 = q-1 + 1 = q,\n\\end{align*}\n$$\n\nFinally we obtain\n$$\n\\begin{aligned}\n2016 &= 121 \\cdot 16 + 80 &\\Rightarrow 2016 * 121 &= 16, \\\\\n2016 &= 144 \\cdot 14 = 144 \\cdot 13 + 144 &\\Rightarrow 2016 * 144 &= 13.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle **ABC** is such that $\\angle ABC = \\angle ACB = 30^\\circ$. On the side **BC** point **D** is selected. The point **K** is such that **D** is the midpoint of **AK**. It turned out that $\\angle BKA > 60^\\circ$. Prove that $3AD < CB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us select the points $X$ and $Y$ on the side $BC$ such that $BX = AX$ and $CY = YA$. Then $\\angle AXY = \\angle ABC + \\angle BAX = 2\\angle ABC = 60^\\circ$ (fig. 39).\n\nAnalogously, $\\angle XYA = 60^\\circ$ and then $\\triangle XYA$ is equilateral. Then $BX = AX = XY = AY = YC$, i.e. the points $X$, $Y$ divide $BC$ into three equal parts and $BC = 3AX$.\n\nLet $M$ be the midpoint of $BC$. $\\triangle ABC$ is an isosceles triangle, therefore $\\angle AMB = 90^\\circ$.\n\nConsider the points $B_1, T, K, N, C_1$, where $B$ is the middle of $AB_1$, $X$ is the middle of $AT$, $M$ is the middle of $AN$, $C$ is the middle of $AC_1$. By Thales' theorem, the points $T, K, N$ lie on the segment $B_1C_1$.\n\nWe have that $BX = AX = XT$, and therefore $\\angle ABT = 90^\\circ$. Obviously, $\\angle TNA = 90^\\circ$. Thus, the quadrangle $ABTN$ is inscribed in a circle $w$. Hence, $\\angle ANB = \\angle ATB = 90^\\circ - \\angle BAT = 60^\\circ$.\n\nPoint $K$ lies on the segment $B_1C_1$, i.e. the points $T, K, N$ lie in one half-plane with respect to the line $AB$. Then, as the arc $\\cup BTA$ of the circle $w$ has length $60^\\circ$, then from $\\angle AKB > 60^\\circ$ it follows that the point $K$ lies inside the circle $w$.\n\nOn the other hand, the point $K$ lies on the segment $B_1C_1$, hence $K$ lies inside the segment $TN$. Then $D$ lies inside the segment $XM$. We obtain that $AM \\perp XM$ and the point $D$ is closer to $M$ than to $X$. Then $AD < AX$, and therefore $3AD < 3AX = BC$, Q.E.D.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45500,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Determine if there exist positive integer numbers $x, y, z$ such that\n$$\n2016 = x^3 + y^3 + z^3?\n$$\n\nb) Determine if there exist positive integer numbers $x, y, z, t$ such that\n$$\n2016 = x^3 + y^3 + z^3 + t^3?\n$$",
    "options": [],
    "answer": "a) No.\nb) Yes, for example x = 10, y = 10, z = 2, t = 2 (2016 = 10^3 + 10^3 + 2^3 + 2^3).",
    "solution": "b) It is enough to provide an example: $2016 = 1000 + 1000 + 8 + 8$.\n\na) Let us first note that $2016 = 2^5 \\cdot 3^2 \\cdot 7$. And consider remainders of $a^3, b^3, c^3$ modulo $7$. The remainders could equal to $0$ or $\\pm 1$. Thus, if $2016 = x^3 + y^3 + z^3$ then at least one of the numbers is divisible by $7$. It is easy to see that it must be $7$, because $7^3 = 343 < 2016$, and the next number $14^3 = 2744 > 2016$. Thus two other numbers must satisfy the condition $y^3 + z^3 = 1673$.\n\nNow, suppose that $y \\le z$. Since $12^3 = 1728 > 1673$ then $z \\le 11$. Moreover, $1673 = y^3 + z^3 \\le 2z^3$, then $837 \\le z^3$, and it means that $10 \\le z$. To finish the solution we just need to check cases $z=10,11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bisector of the angle $\\angle ABC$ of the triangle $ABC$ intersects the circumcircle of triangle $ABC$ at the point $K$. Point $N$ belongs to the segment $AB$ and $NK \\perp AB$. Let $P$ be a midpoint of the segment $NB$. Consider the line through $P$ that is parallel to $BC$ and intersects line $BK$ at the point $T$. Prove that the line $NT$ passes through the midpoint of the segment $AC$.\n\n(Nagel Igor)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M = NT \\cap AC$ (fig. 23). We note that since $BK$ is a bisector of $ABC$ then $\\angle KBC = \\angle KBA = \\alpha$, and since $PT$ is parallel to $BC$ then $\\angle KBC = \\angle PTB$. Thus, $PT = PB = PN$. It follows that $\\triangle BNT$ is right-angled with hypotenuse $BN$. Since triangle $\\triangle BNK$ is also right-angled then $\\angle KNT = \\alpha$. Moreover, we get $\\angle KAC = \\angle KBC = \\alpha$. Thus, quadrilateral $KANM$ is cyclic. Since $\\angle ANK = 90^\\circ$, then $AK$ is a diameter of the circumcircle of the $KANM$. Hence, $\\angle AMK = 90^\\circ$. Since $\\triangle AKC$ is isosceles we get that altitude $KM$ also is a median. Hence, $AM = MC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_n$ be positive numbers such that:\n$$\nx_1^{n-1} + x_2^{n-1} + \\dots + x_n^{n-1} = x_1 x_2 \\dots x_n\n$$\nProve the inequality:\n$$\n(x_1 - n + 1)(x_2 - n + 1)\\dots(x_n - n + 1) \\ge 1. \\quad (\\text{Serdiuk Nazar})\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using Cauchy inequality we get $\\forall i=1, n$:\n$$\n\\begin{align*}\nx_1 x_2 \\dots x_n &= x_1^{n-1} + x_2^{n-1} + \\dots + x_n^{n-1} \\ge x_1^{n-1} + (n-1) x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n \\\\\n&\\Rightarrow x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n (x_i - n + 1) \\ge x_i^{n-1} \\Rightarrow x_i - n + 1 \\ge \\frac{x_i^{n-1}}{x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n} \\\\\n&\\Rightarrow \\prod_{i=1}^n (x_i - n + 1) \\ge \\prod_{i=1}^n \\frac{x_i^{n-1}}{x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n} = 1,\n\\end{align*}\n$$\nas needed.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $a$, $b$, $c$ satisfy the conditions:\n$$\n\\frac{a+c}{a+1} = b, \\quad \\frac{c+b}{c+1} = a, \\quad \\frac{b+a}{b+1} = c.\n$$\n\nWhat values can the expression $(a+1)(b+1)(c+1)$ take?",
    "options": [],
    "answer": "1 and 8",
    "solution": "Let us subtract $1$ from the left and right parts of each equality and obtain:\n\n$$\nb - 1 = \\frac{a+c}{a+1} - 1 = \\frac{a+c - a - 1}{a+1} = \\frac{c-1}{a+1},\n$$\n\nand analogously from the other two equalities. Because of the conditions of existence of expressions, none of the variables equals $-1$, so we can multiply these equalities by the corresponding denominator:\n$$\n(b-1)(a+1) = c-1,\n$$\n\nand similarly for the other variables. Then we multiply all the equalities:\n$$\n(a-1)(a+1)(b-1)(b+1)(c-1)(c+1) = (a-1)(b-1)(c-1).\n$$\n\nIf, e.g., $c=1$, we immediately get that $b=1$, hence $a=1$. In this case $(a+1)(b+1)(c+1)=8$. If none of the variables is equal to $1$, then we can divide by $(a-1)(b-1)(c-1)$, and therefore obtain $(a+1)(b+1)(c+1)=1$. Note that the value is $1$ when $a=b=c=0$, and it is $8$ for $a=b=c=1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $1, 2, \\ldots, n$ are written in line. Numbers $1$ and $n$ are painted blue, and the rest are painted yellow. Two players - Olesya and Andriy - in turns repaint one of the yellow numbers blue by the following rules: with the first move Olesya (she starts) repaints blue any of the yellow numbers (let us denote it by $k$). Then Andriy chooses from the intervals of numbers $(1, 2, \\ldots, k)$ or $(k, k+1, \\ldots, n)$ the one that contains more yellow numbers. If these intervals have equal amount of yellow numbers, then he selects any of the two. If, e.g., the interval $(1, 2, \\ldots, k)$ is larger, then another interval is not involved in the game anymore. After that Andriy repaints any yellow number from the new interval blue. Now a new interval is also divided into two smaller ones. Then Olesya chooses one of the two new intervals, the one that contains more yellow numbers for her move, and another interval goes out of the game. And so on. The winner is the one who paints blue a number, both (left and right) neighbours of which are blue. Who wins this game, if both players play perfectly?\n\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Andriy wins if n is a power of two; otherwise Olesya wins.",
    "solution": "We will solve this problem by finding the winning and losing positions. Recall that the position is called a losing position for a player, if after his move he instantly loses, or puts the game in a winning position for another player. Position is a winning one, if the player makes a move and immediately wins, or can put the game in a losing position for another player.\n\nIt is clear that the game is being played in the range $l, l+1, \\ldots, m$, where numbers $l, m$ are blue and the rest are yellow. The values of these numbers are not significant, what is important is the amount of numbers in the interval. We will evaluate positions by the amount of yellow numbers. It is clear that the position containing $1$ yellow number is a winning one, so the position $a_1 = 2$ is a losing one. Then positions $3, 4, 5$ are winning ones, because from each of them a player can put the opponent in a losing position $a_1 = 2$. Hence, position $a_2 = 6$ is a losing one.\n\nLet us show by Mathematical Induction that all losing positions satisfy the condition: $a_{n+1} = 2a_n + 2$. Base of induction was tested. Let all positions from $a_{n-1} + 1$ to $a_n - 1$ be winning, and $a_{n-1}$ and $a_n$ be losing ones. Consider the position $b$ between $a_n + 1$ and $a_{n+1} - 1 = 2a_n + 1$. Then the next player repaints blue the number that splits the interval of the length $b$ into $a_n$ and $b - a_n - 1$. We have that\n$$\na_n \\geq b - a_n - 1 \\Leftrightarrow b \\leq 2a_n + 1,\n$$\nhence the opponent turns out to be in a losing position $a_n$.\n\nIf the player is in position $a_{n+1} = 2a_n + 2$, then after the division this interval into two parts the larger of them will surely be not less than $a_n + 1$, i.e. is a winning one.\n\nWhat is left is to find the explicit form of losing positions. Having calculated the first few members, it is not hard to guess and then prove by Mathematical Induction that $a_n = 2^{n+1} - 2$ (or simply by solving a linear recursive equation).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45505,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ $AB > AC$. A tangent to the circumcircle of the triangle $ABC$ has been drawn through the point $A$. This tangent intersects the line $BC$ at the point $P$. In continuation of the side $BA$ after the point $A$ point $Q$ was selected such that $AQ = AC$. Let $X$ and $Y$ be the midpoints of the segments $CQ$ and $AP$, respectively, and let $R$ belong to the segment $AP$ such that $AR = CP$. Prove that $CR = 2XY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the point $M$, which is the midpoint of the segment $PQ$, then $MX$ and $MY$ are the mid-segments of $\\triangle CPQ$ and $\\triangle APQ$ respectively. It follows that $MX = y$ and $MY = x$, and also $MX \\parallel CP$ and $MY \\parallel AQ$. This parallel gives that $\\angle XMY = \\angle PBQ = \\beta$ (as the angles of respectively parallel sides).\n![](attached_image_1.png)\nFig. 40\nThus, $\\frac{MY}{MX} = \\frac{x}{y} = \\frac{2x}{2y} = \\frac{AC}{AR}$ and $\\angle XMY = \\angle PAC = \\beta$. Which means $\\triangle XMY \\sim \\triangle RAC$, due to the proportionality of the two sides and the angle between them. From the similarity of triangles it follows that\n$$\n\\frac{CR}{XY} = \\frac{AC}{MY} = \\frac{2x}{x} = 2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45506,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which positive integers $n$ there exist $2n$ pairwise different positive integers $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$ satisfying the equations:\n$$\na_1 + a_2 + \\dots + a_n = b_1 + b_2 + \\dots + b_n \\text{ and } a_1 a_2 \\dots a_n = b_1 b_2 \\dots b_n?\n$$",
    "options": [],
    "answer": "All positive integers n greater than or equal to 3",
    "solution": "**Answer:** for all $n \\ge 3$.\n\nAssume $n \\ge 3$. Let us introduce such denotations:\n$$\na_1 + a_2 + \\dots + a_{n-1} = a, \\quad a_1 a_2 \\dots a_{n-1} = A, \\quad b_1 + b_2 + \\dots + b_{n-1} = b, \\quad b_1 b_2 \\dots b_{n-1} = B. \\quad (1)\n$$\nThen we can write that $a + a_n = b + b_n$, $A a_n = B b_n$, and hence,\n$$\n\\begin{aligned}\n\\frac{A a_n}{B} &= b_n \\Rightarrow a + a_n = b + \\frac{A a_n}{B} \\Rightarrow B a_n - A a_n = b B - a B \\Rightarrow \\\\\na_n &= \\frac{B(b-a)}{B-A} \\Rightarrow b_n = \\frac{A(b-a)}{B-A}.\n\\end{aligned} \\quad (2)\n$$\nThus, having a set of $2n-2$ numbers that satisfy conditions (1), one can always add numbers $a_n$ and $b_n$ determined by formulas (2). Then a set of $2n$ numbers satisfies the condition if and only if the signs of the numbers $(b-a)$ and $(B-A)$ are the same.\nAs an example of the opposite, consider the numbers $a_1 = 3, a_2 = 5, b_1 = 2, b_2 = 7$. Then $a=8, b=9, A=15, B=14$. Then $a_3 = \\frac{B(b-a)}{B-A} = -14, b_3 = -15$. Indeed, the sets of numbers\n3; 5; -14 and 2; 7; -15 have the same sums and products, but are constituted by not only positive integers.\n\nThe easiest way is to choose numbers such that the first $2n-2$ numbers had products differing by 1, i.e. $|B-A|=1$. Then numbers found by formulas (2) will be integer, and if they satisfy other conditions, they will meet the conditions of the problem. Thus let us find two numbers, each of which has a different decomposition into at least $n-1$ factors, and which are neighbouring positive integers. As the basis let us choose $2^N - 1$ and $2^N$.\nLet us choose numbers so that:\n$$\na_1 = 2^{2^1} - 1 = 3;\n$$\n$$\na_2 = 2^{2^1} + 1 = 5, \\text{ then } a_1 a_2 = 2^{2^2} - 1;\n$$\n$$\na_3 = 2^{2^2} + 1 = 17, \\text{ then } a_1 a_2 a_3 = 2^{2^3} - 1; \\dots\n$$\n$$\na_{n-2} = 2^{2^{n-3}} + 1, \\text{ then } a_1 a_2 \\dots a_{n-2} = 2^{2^{n-2}} - 1;\n$$\n$$\na_{n-1} = 2^{2^{n-2}} + 1, \\text{ then } a_1 a_2 \\dots a_{n-1} = 2^{2^{n-1}} - 1;\n$$\n$$\nb_1 = 2^{2^0} = 2;\n$$\n$$\nb_2 = 2^{2^1} = 4, \\text{ then } b_1 b_2 = 2^{2^0 + 2^1} = 2^{2^2 - 1};\n$$\n$$\nb_3 = 2^{2^2} = 16, \\text{ then } b_1 b_2 b_3 = 2^{2^3 - 1}; \\dots\n$$\n$$\nb_{n-2} = 2^{2^{n-3}}, \\text{ then } b_1 b_2 \\dots b_{n-2} = 2^{2^{n-2}-1};\n$$\n$$\nb_{n-1} = 2^{2^{n-2}+1}, \\text{ then } b_1 b_2 \\dots b_{n-1} = 2^{2^{n-1}}.\n$$\nDue to the construction we have that $b_1 b_2 \\dots b_{n-1} = B = 2^{2^{n-1}}$, $a_1 a_2 \\dots a_{n-1} = A = 2^{2^{n-1}} - 1$. What is left to check other conditions. $B = A+1 > A$, hence it should be true that $b > a$.\n$$\na = a_1 + a_2 + \\dots + a_{n-1} = 2^{2^1} - 1 + 2^{2^1} + 1 + 2^{2^2} + 1 + \\dots + 2^{2^{n-3}} + 1 + 2^{2^{n-2}} + 1,\n$$\n$$\nb = b_1 + b_2 + \\dots + b_{n-1} = 2^{2^0} + 2^{2^1} + \\dots + 2^{2^{n-3}} + 2^{2^{n-2}+1}.\n$$\n$$\nb-a = 2^{2^0} + 2^{2^1} + \\dots + 2^{2^{n-3}} + 2^{2^{n-2}+1} - 2^{2^1} - 2^{2^1} - 2^{2^2} - \\dots - 2^{2^{n-3}} - 2^{2^{n-2}} - (n-3) = \\\\\n= 2^{2^0} - 2^{2^1} + 2^{2^{n-2}+1} - 2^{2^{n-2}} - (n-3) = 2^{2^{n-2}} - (n-1) > 0\n$$\nThe last inequality can be proved easily by Mathematical Induction.\nIt remains to check that the numbers are pairwise different. I.e., that the added numbers $a_n$ and $b_n$ do not equal any of the numbers constructed earlier. Provided that $B-A=1$, we obtain\n$$\na_n = \\frac{B(b-a)}{B-A} = B(b-a) > b_n = \\frac{A(b-a)}{B-A} = A(b-a) = (2^{2^{n-1}} - 1)(2^{2^{n-2}} - (n-1)) > \\\\\nb_{n-1} = 2^{2^{n-2}+1} > a_{n-1} = 2^{2^{n-2}} + 1.\n$$\nLet us prove that $2^{2^{n-1}} - 1 > 2^{2^{n-2}+1}$, and even a stronger inequality: $2^{2^{n-1}} \\ge 2^{2^{n-2}+2}$. Then we have:\n$$\n2^{2^{n-1}} \\ge 2^{2^{n-2}+2} \\Leftrightarrow 2^{n-1} \\ge 2^{n-2} + 2 \\Leftrightarrow 2^{n-2} \\ge 2.\n$$\nFor $n=1$ it is clear that such numbers do not exist.\n\nFor $n=2$, if $a_1 a_2 = b_1 b_2 = q$ and $a_1 + a_2 = b_1 + b_2 = p$, then the numbers $a_1, a_2$ and $b_1, b_2$ are the roots of the square equation $t^2 - p t + q = 0$, and therefore these pairs of numbers are the same up to the order.\n\nAlternative solution.\nSuppose we have $2n-2$ pairwise different numbers $a_1, a_2, \\dots, a_{n-1}$, $b_1, b_2, \\dots, b_{n-1}$, which satisfy the conditions:\n$$\na_1 + a_2 + \\dots + a_{n-1} = a = b_1 + b_2 + \\dots + b_{n-1} - 1, \\quad a_1 a_2 \\dots a_{n-1} = A = b_1 b_2 \\dots b_{n-1} - 1. \\quad (3)\n$$\nThus, for the formulas (2) we have $b-a=1$ and $B-A=1$, therefore we can make a set of $2n$ pairwise different numbers $a_1, a_2, \\dots, a_{n-1}, a_n, b_1, b_2, \\dots, b_{n-1}, b_n$ by adding two numbers obtained by the formulas (2), $a_n = B = A+1$ and $b_n = A$.\nWhat is left is to understand how for every positive integer $n \\ge 2$ obtain two sets of $n$ numbers, sums and products of which differ by 1.\nSuppose pairwise different numbers $a_1, a_2, \\dots, a_{n-1}, b_1, b_2, \\dots, b_{n-1}$ satisfy conditions (3). Let us add to them numbers: $a_n = A$ and $b_n = A-1$. Then\n$$\na_1 + a_2 + \\dots + a_{n-1} + a_n = a + A, \\quad b_1 + b_2 + \\dots + b_{n-1} + b_n = a + 1 + A - 1 = a + A, \\\\\na_1 a_2 \\dots a_{n-1} a_n = A A = A^2, \\quad b_1 b_2 \\dots b_{n-1} b_n = (A+1)(A-1) = A^2 - 1.\n$$\nNow let us add $a_{n+1} = A^2 - 2$ and $b_{n+1} = A^2 - 1$ to these numbers and obtain:\n$$\na_1 + a_2 + \\dots + a_{n-1} + a_n + a_{n+1} = a + A + A^2 - 2, \\\\\nb_1 + b_2 + \\dots + b_{n-1} + b_n + b_{n+1} = a + A + A^2 - 1, \\\\\na_1 a_2 \\dots a_{n-1} a_n a_{n+1} = A^2 \\cdot (A^2 - 2) = A^4 - 2A^2, \\\\\nb_1 b_2 \\dots b_{n-1} b_n b_{n+1} = (A^2 - 1)(A^2 - 1) = A^4 - 2A^2 + 1.\n$$\nAs you can see, we made sets of $2n+2$ pairwise different numbers $a_1, a_2, \\dots, a_{n+1}, b_1, b_2, \\dots, b_{n+1}$, that satisfy conditions (3), therefore from them one can obtain the desired sets of $2n+4$ numbers. Thus, from the existence of sets satisfying conditions (3) for $n=2$, we obtain that there are desired sets of numbers for any odd $n \\ge 3$, and from existence of sets satisfying conditions (3) for $n=3$ it follows that there are sets for any even $n \\ge 4$. What is left is to state these numbers.\nFor $n=2$ we can take $a_1 = 4, a_2 = 5, b_1 = 3, b_2 = 7$,\nFor $n=3$ we can take $a_1 = 2, a_2 = 3, a_3 = 15, b_1 = 1, b_2 = 7, b_3 = 13$.\nThen, e.g., we make sets as follows:\nFor $n=3$: $a_1 = 4, a_2 = 5, a_3 = 21$ and $b_1 = 3, b_2 = 7, b_3 = 20$.\nWith the set for $n=2$ satisfying conditions (3), we now make a set for $n=4$ that also satisfies the conditions (3):\n$$\na_1 = 4, a_2 = 5, a_3 = 20, a_4 = 398 \\text{ and } b_1 = 3, b_2 = 7, b_3 = 19, b_4 = 399. \\text{ Then we have that} \\\\\na_1 + a_2 + a_3 + a_4 = 427, b_1 + b_2 + b_3 + b_4 = 428, \\text{ and} \\\\\na_1 a_2 a_3 a_4 = 159200, b_1 b_2 b_3 b_4 = 159201.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three pairwise distinct numbers $a$, $b$, $c$ are such that their product is $80$.\nDetermine the least possible prime sum of these numbers.",
    "options": [],
    "answer": "19",
    "solution": "Clearly, their sum is greater than $2$, thus their sum is odd. Since all three numbers cannot be odd (since their product is $80$), then two of the numbers are even and one is odd. There are only two odd divisors of $80 = 2^4 \\cdot 5$: $1$ and $5$. Consider these cases.\n\n$c=1$, the following is possible:\n\n$a = 2$, $b = 40$, $a+b+c = 43$ is prime.\n$a = 4$, $b = 20$, $a+b+c = 25$ is not prime.\n$a = 8$, $b = 10$, $a+b+c = 19$ is prime and less than $43$.\n\n$c=5$, the following is possible:\n$a = 2$, $b = 8$, $a+b+c = 15$ is not prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given three pairwise distinct positive integers $a$, $b$, $c$, whose product is $320$. Determine the smallest possible prime sum of these numbers.",
    "options": [],
    "answer": "37",
    "solution": "Clearly, their sum is greater than $2$, so the prime sum has to be odd. Since all three numbers can't be odd simultaneously, since their product is $320$, then two numbers are even and one is odd. There are exactly two odd divisors of $320 = 2^6 \\cdot 5$: $1$ and $5$. Consider these cases.\n\n$c = 1$, the following is possible:\n\n$a = 2$, $b = 160$, $a+b+c = 163$ is prime.\n$a = 4$, $b = 80$, $a+b+c = 85$ is not prime.\n$a = 8$, $b = 40$, $a+b+c = 49$ is not prime.\n$a = 16$, $b = 20$, $a+b+c = 37$ is prime and less than $163$.\n$a = 32$, $b = 10$, $a+b+c = 43 > 37 \\Rightarrow a+b+c > 37$.\n\n$c = 5$, the following is possible:\n$a = 2$, $b = 32$, $a+b+c = 39$ is not prime.\n$a = 4$, $b = 16$, $a+b+c = 25$ is not prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all possible pairs of integers $a, b$ so that exactly one of them is even and so that there are non-integer $x, y$, such that both $x+y$ and $ax+by$ are integer?",
    "options": [],
    "answer": "All integer pairs with exactly one even and |a−b|>1; pairs with |a−b|=1 do not work.",
    "solution": "Since $ax+by = a(x+y) + (b-a)y$, the value of $(b-a)y$ is integer. If $|a-b|=1$, then $y$ has to be integer – contradiction.\n\nFor $|a-b|>1$ we can let $y = \\frac{1}{b-a}$ and $x = -y$. Then $x+y=0$, $(b-a)y$ is integer, hence $ax+by$ is also integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exist non-integer $x, y$, so that for any integer $a, b$, both $x+y$ and $ax+by$ are integers.",
    "options": [],
    "answer": "No such numbers exist.",
    "solution": "Suppose such numbers exist. Then $ax+by = a(x+y) + (b-a)y$. Hence $(b-a)y$ is an integer. Let $b=2, a=1$, then $y$ is an integer – contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exist noninteger $x$, $y$ such that for any integer $a$, $b$ that are either both odd or both even, numbers $x + y$ and $a x + b y$ are integers?",
    "options": [],
    "answer": "Yes; for example x = 1/2 and y = 1/2.",
    "solution": "Let $x = y = \\frac{1}{2}$, then $x + y = 1$, $a x + b y = \\frac{1}{2}(a + b) \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45512,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum length of a sequence of positive integers $a_1, a_2, \\dots, a_n$, if following conditions hold:\n* $a_1 > 1$ is a prime number;\n* for any $i: 2 \\le i \\le n$, $a_i \\vdash a_1 a_2 \\dots a_{i-1}$ holds;\n* $a_n = 2^2 \\cdot 3^3 \\cdot 5^5 \\cdot 7^7 \\cdot 11^{11} \\cdot 13^{13} \\cdot 17^{17}$.",
    "options": [],
    "answer": "6",
    "solution": "Let $a_1 = p$ is prime. Then it is clear that\n$$\n\\begin{array}{ccc}\na_2 \\vdash p, & a_3 \\vdash a_2 a_1 \\vdash p^2, & a_4 \\vdash a_3 a_2 a_1 \\vdash p^4, \\\\\na_5 \\vdash a_4 a_3 a_2 a_1 \\vdash p^8, & a_6 \\vdash a_5 a_4 a_3 a_2 a_1 \\vdash p^{16}.\n\\end{array}\n$$\nAssume the length of the sequence is greater than 6, thus $a_7 \\vdash a_6 a_5 a_4 a_3 a_2 a_1 \\vdash p^{32}$, there is a prime number that divides $a_7$ and it is included with degree not less than 32. By conditions on $a_n$ that is not possible. Therefore, the maximum length is $n=6$. The only prime number that is included with degree 16 in $a_6 = 2^2 \\cdot 3^3 \\cdot 5^5 \\cdot 7^7 \\cdot 11^{11} \\cdot 13^{13} \\cdot 17^{17}$ is $p=17$. This is the maximum value of $a_1$.\n\nIt suffices to show now, that such sequence exists. Take\n$$\na_1 = 17,\\ a_2 = 17,\\ a_3 = 17^2,\\ a_4 = 17^4,\\ a_5 = 17^8,\\ a_6 = 2^2 \\cdot 3^3 \\cdot 5^5 \\cdot 7^7 \\cdot 11^{11} \\cdot 13^{13} \\cdot 17^{17}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find possible positive solutions $(x, y, z)$ for the given system of equations:\n$$\n\\begin{cases}\nx + y^2 = 2z^3, \\\\\ny + z^2 = 2x^3, \\\\\nz + x^2 = 2y^3.\n\\end{cases}\n$$",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "Without loss of generality let $z \\ge \\max\\{x, y\\}$. Consider the following cases.\n\nCase 1. $z > 1$. Then due to the first equation $x + y^2 \\le z + z^2 < 2z^3$ – contradiction.\n\nCase 2. $z = 1$. Due to the first equation, $2 = x + y^2$.\nSince $\\max\\{x, y\\} \\le 1$, it is possible only if $x = y = 1$. It is easy to check that $(1; 1; 1)$ is a solution.\n\nCase 3. $z < 1$. Then for the smallest variable, say $x$, $x < 1$ holds, then by the second equation:\n$x + y^2 \\le z + z^2 < 2z^3$ – contradiction.\n\nSimilar contradiction holds if $y$ is the smallest variable.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system of equations for positive integer numbers $x, y, z$:\n$$\n\\begin{cases} x^3 - 6y^2 + 27z = 132, \\\\ y^3 - 9z^2 + 3x = 125, \\\\ z^3 - 3x^2 + 12y = -68. \\end{cases}\n$$",
    "options": [],
    "answer": "No positive integer solutions",
    "solution": "Let's add up all three equations:\n\n$$\n(x^3 - 3x^2 + 3x) + (y^3 - 6y^2 + 12y) + (z^3 - 9z^2 + 27z) = 189 \\Leftrightarrow \\\\\n(x^3 - 3x^2 + 3x - 1) + (y^3 - 6y^2 + 3y - 27) + (z^3 - 9z^2 + 27z - 27) = 153 \\Leftrightarrow \\\\\n(x-1)^3 + (y-2)^3 + (z-3)^3 = 153.\n$$\n\nBy a simple exhaustive search, we can see that number 153 has a unique representation in the form of the sum of three cubes: $153 = 5^3 + 3^3 + 1^3$. Therefore, in the last equation, values of $(x-1)$, $(y-2)$ and $(z-3)$ must be selected from 1; 3; 5. Again, simply by trying out the variants, we see that no solution is possible.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Cut right triangle with one angle $30^\\circ$ into three pairwise distinct isosceles triangles so that each has a non acute angle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We can cut $\\triangle ABC$ with $\\angle BAC = 30^\\circ$ into triangles as shown in Fig. 25.\n\n![](attached_image_1.png)\n\nHere $BC = CD$ and $\\triangle DBC$ is isosceles with right angle. Then $BN = MD$ and $\\triangle DBM$ is isosceles with angle $\\angle DMB = 150^\\circ$. Thus $\\angle DMA = 30^\\circ$, $\\triangle ADM$ is isosceles with $\\angle ADM = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exist a function $f : (0; 1) \\rightarrow (2018; +\\infty)$, so that the following conditions hold:\n* $f(x \\cdot y) = f(x) \\cdot f(y)$ for any $x, y \\in (0; 1)$;\n* for any $y \\in (2018; +\\infty)$ there exists $x \\in (0; 1)$ such that $f(x) = y$?",
    "options": [],
    "answer": "No such function exists.",
    "solution": "Suppose such a function exists. Since for any $x \\in (0; 1)$ equation $x = \\sqrt{x} \\cdot \\sqrt{x}$ holds, then $f(x) = f(\\sqrt{x} \\cdot \\sqrt{x}) = f(\\sqrt{x}) \\cdot f(\\sqrt{x}) > 2018^2$, that contradicts the fact that $E_f = (2018, +\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy wrote a 4-digit number. Olesya crossed out the last digit and it turned out that the difference between the initial number and the obtained number equals $2018$. Which number did Andriy write? Provide all possible answers.",
    "options": [],
    "answer": "2242",
    "solution": "Let $abcd$ denote the number Andriy wrote, then $\\overline{abcd} - \\overline{abc} = 2018$. Now we only need to pick corresponding digits. Obviously, $a = 2$ or $a = 3$.\n\nWith $a = 3$, $\\overline{3bcd} - \\overline{3bc} > 3000 - 399 > 2018$, hence, $a = 2$ and $\\overline{2bcd} - \\overline{2bc} = 2018$. Analogously, $b = 2$ or $b = 3$.\n\nWith $b = 3$, $\\overline{33cd} - \\overline{33c} > 3300 - 339 > 2018$, therefore, $b = 2$ and $\\overline{22cd} - \\overline{22c} = 2018$. Analogously, $c = 3$ or $c = 4$.\n\nWhen $c = 4$, we get $\\overline{224d} - \\overline{224} > 2240 - 224 = 2016$, therefore, the first solution would be in the form: $2242 - 224 = 2018$.\n\nWhen $c = 3$, we get $\\overline{223d} - \\overline{223} < 2239 - 223 = 2016$, which means previously found solution is unique possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist such value of $x \\in (0, \\frac{\\pi}{2})$, for which numbers $\\sin x$, $\\cos x$ and $\\operatorname{tg} x$ form a geometric progression?",
    "options": [],
    "answer": "Yes",
    "solution": "Let us write the equation that is necessary and sufficient for these three numbers to form geometric progression:\n$$\n\\sin x \\cdot \\operatorname{tg} x = \\cos^2 x.\n$$\nThen, $\\sin^2 x = \\cos^3 x$ and if we denote by $t = \\cos x \\in (0; 1)$, then we obtain the equation:\n$$\n1 - t^2 = t^3 \\Rightarrow f(t) = t^3 + t^2 - 1 = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the acute $\\triangle ABC$ and point $D$ on the side $AB$. Let's denote the center of the circumscribed circle around the $\\triangle ACD$ as $P$ and the center of the circumscribed circle around the $\\triangle BDC$ as $Q$. Prove that triangles $ABC$ and $DPQ$ are similar.\n\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's denote the radius of the circumscribed circle around the $\\triangle ACD$ as $R_1$, the radius of the circumscribed circle around the $\\triangle BDC$ as $R_2$. So, if $CD = l$ then by the law of sines (Fig.47):\n$$\n\\frac{l}{\\sin \\beta} = 2R_1 \\Rightarrow \\sin \\beta = \\frac{l}{2R_1}.\n$$\nHerewith, $\\cos \\varphi = \\frac{DE}{DQ} = \\frac{l}{2R_1} = \\sin \\beta$.\nTherefore, from the acute triangle $\\triangle ABC$ $\\varphi = \\frac{\\pi}{2} - \\beta$, and similarly $\\psi = \\frac{\\pi}{2} - \\alpha$.\nTherefore $\\angle PDQ = \\varphi + \\psi = \\frac{\\pi}{2} - \\alpha + \\frac{\\pi}{2} - \\beta = \\pi - \\alpha - \\beta = \\gamma = \\angle ACB$.\n\n![](attached_image_1.png)\n**Fig. 47**\n\nMoreover, from the following equation:\n$$\n\\frac{QD}{PD} = \\frac{R_1}{R_2} = \\frac{l}{2\\sin\\beta} \\cdot \\frac{2\\sin\\alpha}{l} = \\frac{\\sin\\alpha}{\\sin\\beta} = \\frac{a}{b} = \\frac{CB}{AC}\n$$\n\nthe sides about the equal angles of our triangles are proportional, so triangles $ABC$ and $DPQ$ are similar for any point $D \\in AB$. Points $A, B$ do not satisfy the condition, otherwise one of the $\\triangle ACD$ or $\\triangle BDC$ degenerates to a line segment.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $[x_1, x_2]$ and $[y_1, y_2, y_3]$ be the least common multiple of $x_1, x_2$ and $y_1, y_2, y_3$ respectively. For any positive integers $a, b, c, d$ let $A$ and $B$ be such that:\n$$\nA = [a, b, c] \\cdot [a, b, d] \\cdot [a, c, d] \\cdot [b, c, d] \\text{ and } B = [a, b] \\cdot [a, c] \\cdot [a, d] \\cdot [b, c] \\cdot [b, d] \\cdot [c, d].\n$$\n\nShow that $A^6 \\geq B^4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Take prime $p$ that divides $abcd$. Without loss of generality, let the prime number be a factor of $a, b, c, d$ of degree $a_1 \\geq b_1 \\geq c_1 \\geq d_1$ respectively. We can find the biggest degree of $p$—$p_A$ and $p_B$, that divide $A$ and $B$ respectively.\n\n$$\np_A = 3a_1 + b_1, \\quad p_B = 3a_1 + 2b_1 + c_1.\n$$\n\nThen degrees $p'_A$ and $p'_B$ for numbers $A^6$ and $B^4$ equal:\n\n$$\np'_A = 6p_A = 18a_1 + 6b_1 \\geq p'_B = 4p_B = 12a_1 + 8b_1 + 4c_1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45521,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is isosceles obtuse triangle $ABC$ with vertex in point $B$ given. Perpendicular bisector to side $BC$ intersects lines $AC$ and $AB$ in points $K$ and $M$ respectively. Prove, that point, symmetric to point $A$ with respect to line $BK$, is on line $CM$.\n\n(Anton Trigub)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1$ be point, symmetric to $A$ with respect to $BK$ (Fig. 37). At first, $\\angle BA_1K = \\angle BAK = \\angle BCK$, thus quadrilateral $BA_1CK$ is cyclic. Then,\n$$\n\\angle A_1CB = \\angle A_1KB = \\angle AKB = 2\\angle BCA = \\angle MBC = \\angle MCB,\n$$\nSo we get, that points $C$, $A_1$, $M$ are on one line.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players – Andriy and Olesya play the following game. On a table, there is a rounded cake, which is cut by one of them into $4n$ different in weight sectors (pieces). Weight of every part is known by each player. After that they choose pieces for themselves upon following rules. At first Andriy chooses 1 piece, then Olesya chooses 2 pieces, but so that the pieces, that are left after her turn, form a sector. Then they take in turns 2 pieces, so that after every step pieces that are on the table form a sector. Making the last step Andriy takes the last piece. Each player aims that the total weight of the part of the cake, he or she took, is bigger, than opponent's one. Is it possible that someone surely takes more than half of the whole cake if:\na) Olesya cuts the cake into sectors;\nb) Andriy cuts the cake into sectors, but at his first step he is prohibited to take the biggest on weight piece?",
    "options": [],
    "answer": "a) Yes. Olesya can cut and play so she surely gets more than half. b) Yes. Andriy can cut and play so he surely gets more than half even if he cannot take the heaviest piece on his first move.",
    "solution": "a) Let's show how Olesya can cut the cake to win. Let $M$ denote the weight of the cake. She cuts it into 3 neighboring pieces weighting $\\frac{1}{3}M$ (heavy), and all the rest have zero weight (light). Let's number the pieces clockwise 1; 2; ...; $4n$. Piece number $4n$ is neighboring to 1. Let pieces $4n-3$, $4n-2$ and $4n-1$ be heavy. Andriy can not begin with any of 5 pieces: $4n-4$, ..., $4n$, because otherwise Olesya by her first step will choose 2 heavy pieces and collect in total $\\frac{2}{3}M > \\frac{1}{2}M$. So Andriy chooses any other piece. Then the one who takes at least one of two pieces $4n-4$ or $4n$ (even if together with one of heavy pieces), loses, because the opponent definitely can take 2 heavy pieces and collect more than half of the cake.\n\nThus, Andriy and Olesya have to make steps in sector between pieces 1 and $4n-5$. After Andriy's turns $4n+1$ pieces will be taken, and after Olesya's turns $-4n+3$. Thus, after completing the full sector between pieces 1 and $4n-5$ Andriy will make his step and so he will lose.\n\nb) Let's show how Andriy should cut the cake to win. Let's number pieces from 1 to $4n$. Let there be 3 heavy pieces weighting $\\frac{1}{3}M$ each, and the rest – light ones of zero weight. Let pieces 1, 3 and 5 be heavy ones. Then Andriy by his first step takes piece 3, then by his second turn he can take one more heavy piece and takes in total $\\frac{2}{3}M > \\frac{1}{2}M$.\n\nIt remains to satisfy the conditions of the task precisely. All the pieces must have different non-zero weight. For this in a) make the small pieces weight, for example, $\\frac{i}{5(4n-3)(4n-2)}M$, $i=1; 4n-3$, then their sum\n$$\n\\frac{1}{5(4n-3)(4n-2)}M + \\frac{2}{5(4n-3)(4n-2)}M + \\dots + \\frac{4n-3}{5(4n-3)(4n-2)}M = \\frac{(4n-3)(4n-2)}{10(4n-3)(4n-2)}M = \\frac{1}{10}M.\n$$\nChoose the big pieces weighting $\\frac{29}{100}M$, $\\frac{30}{100}M$ and $\\frac{31}{100}M$. Order of pieces does not play role. The same distribution of weights in b), but with condition, that 3rd piece has weight of $\\frac{30}{100}M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: [0, +\\infty) \\to [0, +\\infty)$, which for all not-negative $x, y$ satisfy equality:\n$$\nf(f(x) + f(y)) = xyf(x + y).\n$$",
    "options": [],
    "answer": "f(x) = 0 for all x ≥ 0",
    "solution": "For $y=0$ from the condition we have, that $f(f(x)+f(0))=0$ for any $x$, and there is such $a$ that $f(a)=0$. Thus, for $x=a$, $f(f(0))=0$. Then, for $x=y=0$ it follows, that $f(2f(0))=0$, and for $x=y=f(0)$ we have, that\n$$\nf(f(f(0)) + f(f(0))) = f^2(0)f(2f(0)) = 0,\n$$\ni.e. $f(0)=0$ and $f(f(x))=0$ for any $x$.\n\nLet $y=f(y)$ in the condition, then\n$$\nxf(y)f(x+f(y)) = f(f(x)+f(f(y))) = f(f(x)) = 0,\n$$\nso $xf(y)f(x+f(y))=0$ for arbitrary $x$ and $y$.\n\nLet $f(y_0) \\neq 0$ for some $y_0 > 0$. Let $y = y_0 - x$, then\n$$\nf(f(x) + f(y_0 - x)) = x(y_0 - x)f(y_0).\n$$\nExpression in right part takes arbitrary value in segment $[0; \\frac{1}{4}y_0^2 f(y_0)]$, so range of function $f$ includes this segment. Thus, there is some $x_0$ such, that $f(x_0) = \\min\\{\\frac{1}{2}y_0, \\frac{1}{8}y_0^2 f(y_0)\\}$. From equality $xf(y)f(x+f(y))=0$ for $y=x_0$ and $x=y_0-f(x_0)>0$ we have, that $f(x_0)f(y_0)=0$, that is impossible. This contradiction proves, that $f(x)=0$ for arbitrary $x \\in [0, +\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45524,
    "subject": "Mathematics (Multi-modal)",
    "question": "30 military ships are approaching to the island: 10 destroyers and 20 small ships. All ships are arranged in a circle and all distances between neighboring ships are equal. Two battle ships are defending the island. Each of them has exactly 10 rockets. The first battle ship can launch all 10 rockets at the same time, however all 10 targets must be neighboring. The second battle ship can launch all 10 rockets at the same time, however all 10 targets must alternate. Both battle ships launch rockets at the same time (it means that rockets might hit one target). How many destroyers can be saved regardless of how battle ships will launch their rockets?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "3",
    "solution": "Consider all possible tens of neighboring military ships. There exist ten neighboring military ships such that there are at least 4 destroyers among them. The first battle ship can hit these 10 targets. Consider the rest 20 targets and divide them into two groups: with odd position and with even position. One of these groups includes at least half of the destroyers that are saved after first rocket launch. So the second battle ship can hit this group. Therefore, if the first battle ship hits $k \\ge 4$ destroyers, then the second hits at least $\\frac{1}{2}(10-k)$. And in total battle ships hit $\\frac{1}{2}(10-k)+k = 5+\\frac{1}{2}k \\ge 7$ destroyers.\n\nNow we build an example how to place military ships in a way that at least 3 destroyers will be saved. For this purpose we place ships such that every third is a destroyer. Then the first battle ship will hit at most 4 destroyers. If it is exactly 4, then the second battle ship could not hit more than 3, because there is the same amount of destroyers on odd and on even positions. If the first battle ship hits 3 destroyers, then the second could not hit more than 4.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive numbers $a$, $b$, $c$ satisfy the condition $a^2 + b^2 + c^2 + abc = 4$. Prove that the inequality $c + ab \\le 2$ holds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We consider equality from the problem as a quadratic equation with respect to the variable $c$:\n$$\nc^2 + abc + (a^2 + b^2 - 4) = 0 \\Rightarrow c = \\frac{-ab \\pm \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2}}{2}.\n$$\nSince $c$ is a positive number, then $c = \\frac{-ab + \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2}}{2}$. Now we can solve the problem in the following way:\n$$\n\\frac{-ab + \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2}}{2} + ab \\le 2 \\Leftrightarrow \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2} \\le 4 - ab \\Leftrightarrow \\\\\na^2b^2 + 16 - 4a^2 - 4b^2 \\le 16 - 8ab + a^2b^2 \\Leftrightarrow 4a^2 + 4b^2 \\ge 8ab \\Leftrightarrow 4(a-b)^2 \\ge 0.\n$$\n\nNote that $ab \\le 4$, because otherwise $a^2 + b^2 \\ge 2ab \\ge 8$, and this contradicts to the problem statement.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45526,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ positive integer numbers are given, such that greatest common divisors of all nonempty sets of these numbers are pairwise distinct. Determine the smallest possible number of distinct prime divisors of the product of these $N$ numbers.\n(OleksandrGolovanov)",
    "options": [],
    "answer": "N",
    "solution": "Firstly we provide an example of such $N$ numbers. Consider numbers $a_k = p_k p_{k+1} p_{k+2} \\dots p_{k+N-1}$, where $p_1, p_2, \\dots, p_N$ are $N$ distinct prime numbers and $p_{N+i} = p_i$ ($1 \\le i \\le N-1$). Indeed, GCD of any set will include $p_k$ exactly in a power 1 if and only if $a_k$ belongs to this set. It means that GCD of all sets will be distinct.\n\nLet us prove that it could not be less than $N$ prime divisors. Assume that numbers $a_1, a_2, ..., a_N$ satisfy the condition and their product has only $m < N$ distinct prime divisors $p_1, p_2, ..., p_m$. For every $p_i$ we choose $a_{k_i}$ that includes $p_i$ in the smallest possible power. Consider the set of all such $a_{k_i}$ taken once (it might be that $a_{k_i} = a_{k_j}$ with $i \\neq j$, in this case still our set will include $a_{k_i}$ exactly one time). This set will consist of not more than $m < N$ numbers. GCD of all numbers of this set will include $p_i$ in the power that is equal to the smallest power in which $p_i$ belongs to $a_1, a_2, ..., a_N$. Thus if we add any of the $a_i$ to our set, GCD of all numbers in the set will be the same. And we have a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle **ABC** and a line that does not coincide with the triangle sides and passes through the point **A**. This line meets altitudes $BH_2$ and $CH_3$ at the points $D_1$ and $E_1$, respectively. By $D_2$ and $E_2$ we denote points that are symmetrical to $D_1$ and $E_1$ with respect to the sides **AB** and **AC**, respectively. Prove that the circumcircle of the triangle $D_2AB$ is tangent to the circumcircle of the triangle $E_2AC$.\n\n(Myhailo Plotnikov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By $E_3, D_3$ we denote points that are symmetrical to $E_1, D_1$ with respect to **AB** and **AC**, respectively (Fig. 45). It follows from symmetry that $AE_1 = AE_2 = AE_3$, and **CA** is a bisector of an angle that is created by the lines $CE_3$ and $CE_2$. Then we have\n$$\n\\angle(CE_3, E_3A) = \\angle(E_1E_3, E_3A) = \\angle(AE_1, E_1E_3) = \\angle(AE_1, E_1C) = \\angle(CE_2, E_2A).\n$$\n\nThus points **A**, $E_2$, $E_3$ and **C** are cyclic.\nIn the same way points **A**, $D_2$, $D_3$ and **B** are cyclic. Then the circumcircle of the triangle $\\triangle ABD_2$ coincides with the circumcircle of $\\triangle AD_3D_2$. In the same way, the circumcircle of $\\triangle ACE_2$ coincides with the circumcircle of $\\triangle AE_3E_2$. Since $AE_3$ and $AD_2$ are symmetrical to the same line with respect to **AB**, then points **A**, $E_3$, $D_2$ are collinear. In the same way, **A**, $E_2$, $D_3$ are collinear. Moreover,\n$$\nk = \\frac{AD_2}{AE_3} = \\frac{AD_1}{AE_1} = \\frac{AD_3}{AE_2}.\n$$\nThus $\\triangle AE_2E_3 \\sim \\triangle AD_3D_2$, and their circumcircles are tangent to each other, because under homothety $H_A^k$ one of the circles maps to the other circle and they have a common point **A**.\n\n![](attached_image_1.png)\n**Pic. 45**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy has counted the sum of the squares of positive integers from $1$ to $2016$, Vitaliy has counted the sum of the squares of natural numbers from $1$ to $2016$. Yuriy has added Andriy's and Vitaliy's numbers, multiplied by $3$ and added $2016$. What number has Yuriy got?\n\n(Andriy Anikushin)",
    "options": [],
    "answer": "2017^3 - 1",
    "solution": "Let's write down the number that Yuriy had got:\n$$\n\\begin{aligned}\n& 3 \\cdot (1^2 + 2^2 + \\dots + 2016^2) + 3 \\cdot (1 + 2 + \\dots + 2016) + 2016 = \\\\\n& = (3 \\cdot 1^2 + 3 \\cdot 1 + 1) + (3 \\cdot 2^2 + 3 \\cdot 2 + 1) + \\dots + (3 \\cdot 2016^2 + 3 \\cdot 2016 + 1) = \\\\\n& = (1^3 + 3 \\cdot 1^2 + 3 \\cdot 1 + 1) + (2^3 + 3 \\cdot 2^2 + 3 \\cdot 2 + 1) + \\dots + (2016^3 + 3 \\cdot 2016^2 + 3 \\cdot 2016 + 1) - \\\\\n& \\quad - (1^3 + 2^3 + \\dots + 2016^3) = ((1+1)^3 + (2+1)^3 + \\dots + (2016+1)^3) \\\\\n& \\quad - (1^3 + 2^3 + \\dots + 2016^3) = \\\\\n& = (2^3 + 3^3 + \\dots + 2017^3) - (1^3 + 2^3 + \\dots + 2016^3) = 2017^3 - 1.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integer numbers $x, y, p, n, k$ such that:\n$$\n\\begin{cases}\n5x + y = p^k, \\\\\n5y + x = p^{k+n}.\n\\end{cases}\n$$",
    "options": [],
    "answer": "All solutions are p = 2, n = 1, k ≥ 3 with x = 2^{k-3} and y = 3·2^{k-3}. No other positive integer solutions exist.",
    "solution": "Rewrite first equation in the form $25x + 5y = 5p^k$ and subtract the second equation: $24x = p^k(5 - p^n)$. Thus, it is obvious, that $5 - p^n > 0$, so there are only the following cases.\n\nCase 1. $p = 1$, $n \\in \\mathbb{N}$. It is obvious, that $5x + y \\ge 6$, so there are no solutions.\n\nCase 2. $p=2$, $n=1$. Then we have an equality $24x = 2^k \\cdot 3$ or $x = 2^{k-3}$. Thus, $k \\ge 3$.\nThen from the first equality $y = 2^k - 5 \\cdot 2^{k-3} = 3 \\cdot 2^{k-3}$. It is not hard to see, that these values satisfy also the second equality. Thus, for any $k \\ge 3$ the answer is a set of numbers: $x = 2^{k-3}$, $y = 3 \\cdot 2^{k-3}$, $p=2$ and $n=1$.\n\nCase 3. $p=2$, $n=2$ or $p=4$, $n=1$. Then we have equality $24x = 2^k$ and so there are no answers.\n\nCase 4. $p=3$, $n=1$. Then we have an equality $24x = 3^k \\cdot 2$ and then there are no answers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples pairwise distinct positive integer numbers $(a, b, c)$ that satisfy the condition: number $2a-1$ is divisible by $b$, number $2b-1$ is divisible by $c$ and number $2c-1$ is divisible by $a$.",
    "options": [],
    "answer": "(25, 7, 13) and its cyclic permutations: (7, 13, 25) and (13, 25, 7)",
    "solution": "Let's rewrite the conditions in form of a system: there exist natural numbers $k, m, n$, for which equalities are true:\n$$\n2a-1=kb, \\quad 2b-1=nc \\text{ and } 2c-1=ma.\n$$\n\nIt is obvious, that all the numbers $k, m, n$ and $a, b, c$ are odd.\nThen we have, that\n$$\nb = \\frac{2a-1}{k} \\Rightarrow 2b-1 = \\frac{4a-2}{k}-1 = nc \\Rightarrow c = \\frac{4a-2-k}{kn} \\Rightarrow 2c-1 = \\frac{8a-4-2k}{kn}-1 = ma \\Rightarrow \\\\\n8a-4-2k-kn = knma \\Rightarrow a = \\frac{4+2k+kn}{8-knm}.\n$$\nNow we see, that an equality must be true: $knm<8$. Then from symmetry of conditions and oddness of numbers $k, m, n$ it follows, that the following variants are possible (with accuracy to cycle).\n\nVariant 1: $k=m=n=1$. Then we have $c=2b-1 \\Rightarrow a=2c-1=4b-3 \\Rightarrow b=2a-1=8b-7 \\Rightarrow a=b=c=1$, but numbers $a, b, c$ must be different.\n\nVariant 2: $k=3, m=n=1$. Then we have $c=2b-1 \\Rightarrow a=2c-1=4b-3 \\Rightarrow 3b=2a-1=8b-7 \\Rightarrow 5b=7$ – contradiction, because $b$ is not integer.\n\nVariant 3: $k=5, m=n=1$. Then we have $c=2b-1 \\Rightarrow a=2c-1=4b-3 \\Rightarrow 5b=2a-1=8b-7 \\Rightarrow 3b=7$ – contradiction, because $b$ is not integer.\n\nVariant 4: $k=7, m=n=1$. Then we have $c=2b-1 \\Rightarrow a=2c-1=4b-3 \\Rightarrow 7b=2a-1=8b-7 \\Rightarrow b=7 \\Rightarrow c=13, a=25$ and this set satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45531,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ there is an altitude $AH$ and median $AM$. On lines $AB$ and $AC$ there are points $X$ and $Y$ so that $AX = XC$ and $AY = YB$. Prove that the midpoint of segment $XY$ is equidistant from points $H$ and $M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Z$ be the midpoint of segment $XY$, and $N$ and $T$ be the midpoints of segments $AB$ and $AC$ respectively (see the figure).\n\nAs, according to the condition, $AX = XC$ and $AY = YB$, $XT$ and $YN$ are bisectors to segments $AC$ and $AB$. Then triangles $XTY$ and $XNY$ are right triangles, so $XZ = ZN = ZT = ZY$, because the median of a right triangle, drawn to the hypotenuse, equals its half. So, point $Z$ is equidistant from $N$ and $T$.\n\n![](attached_image_1.png)\n\nOn the other side, $NT \\parallel BC$, because $NT$ is the midline of $\\triangle ABC$. $MT$ is also a center line of $\\triangle ABC$, so $MT = \\frac{1}{2}AB$. $NH = \\frac{1}{2}AB$ as the median, drawn to the hypotenuse in a right triangle. So, $HNTM$ is an isosceles trapezoid. Then its bases have a common bisector. We have proved that point $Z$ belongs to bisector $NT$. Then it belongs to bisector $HM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players – Andriy and Olesya play the following game. On a table there is a rounded cake, which is cut by one of them into $2n$, $n>1$ different in weight sectors (pieces). Weight of every piece is known by each player. After that they choose pieces according the following rules. At first Olesya chooses 1 piece, then Andriy chooses 2 pieces, but so, that pieces, that are left on table after her turn, form a sector. Then they choose in turns 2 pieces each so, that after every step pieces, that are left on the table, form a sector. By his last turn one of players takes the last piece. Each of players aims, that total weight of part of the cake, he or she took, is bigger, than opponent's one. For which $n$ Olesya can cut the cake so, that if by her first step she takes the smallest piece?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Exactly for odd n greater than one; for even n, Olesya cannot guarantee a win.",
    "solution": "Let's consider, which values $n$ can take.\n\nIf $n = 2k+1 > 1$, let's consider the following distribution of weights. Pieces $n-1$, $n$ and $n+1$ have weight 1, and other ones – 0. Then after Olesya's first step Andriy can not take non-zero pieces. Until Andriy takes by his turn piece $n-2$ or $n+2$, Olesya just makes her step so, that sector of taken pieces is symmetrical to sector with number 0. If Andriy takes by his turn one of pieces $n-2$ or $n+2$, then it does not matter which other piece he takes, Olesya can take two pieces weighting 1 and win.\n\nIf Andriy by his first step takes pieces of zero weight 1 and $2n-1$, Olesya chooses two big pieces with numbers 2 and 3. After this Andriy by his step can take at maximum two more big pieces, and Olesya by her turn – two more big pieces and win.\n\nSo, we assume, that Andriy by his first step chooses pieces 1 and 2 and we have the total weight $\\frac{99}{100}$. After that Olesya takes pieces 3 and 4 and we have the total weight $\\frac{201}{100}$. Andriy by any his step can not take piece $2n-1$, because Olesya takes two more big pieces and wins. So Andriy has\n---\nto go in direction 4; 5; 6; ..., Olesya takes two neighboring in this direction. No body is beneficial to take a piece with number $2n-1$, because then opponent can take two big pieces at once. So, they take zero pieces until they come to big pieces number $2n-4$, $2n-3$ and $2n-2$. According to the order of steps, Andriy takes big pieces with numbers $2n-3$ and $2n-2$. Olesya by her last turn takes piece number $2n-1$, and by last but one – two pieces with numbers $2n-5$ and $2n-4$, the last of which is big. In total we have, that Olesya's pieces' weight will be: $\\frac{201}{100} + \\frac{101}{100} = \\frac{302}{100}$, and Andriy's: $\\frac{99}{100} + \\frac{199}{100} = \\frac{298}{100}$. So, Olesya wins.\n\nIf $n=2$, so by his first step Andriy will choose two the biggest pieces and win.\nIf $n=4$, we have in total $2n=8$ pieces. Let's show, that Andriy always wins. Here number, which denotes the number of piece, also means its weight in comparisons. Let $(k-1, k, k+1)$ denote the smallest in weight piece, that has corresponding number, $[k-1, k, k+1]$ – the biggest, and $\\{k-1, k, k+1\\}$ – middle.\nWe can assume that $1+2 \\ge 6+7$. Let Andriy take by his first step $1+2$ and does not win. This means, that Olesya could take set $6+7, 3+7$ or $3+4$ and win. Let's consider all these cases. Case $6+7$ has contradiction. Really, as $1+2 \\ge 6+7$, and Andriy chooses $[3, 4, 5]+\\{3, 4, 5\\}$, and Olesya has $(3, 4, 5)$, so Olesya loses.\nIf Olesya gets win after step $3+7$, this means, that inequality is true:\n$$\n3+7+(4,5,6) > \\frac{1}{2}M > 1+2+[4,5,6]+\\{4,5,6\\}. \\quad (1)\n$$\nThen Andriy by his first step takes not $1+2$, but $6+7$. Then he can take by his next step pair of pieces, including 3 (Olesya will not reach it). Then Andriy will have at least set: $6+7+3+(2,4)$. As\n$$\n6+7+3+(2,4) > 3+7+(4,5,6) > \\frac{1}{2}M,\n$$\nAndriy wins.\nIf Olesya gets win after step $3+4$, this means, that inequality is true:\n$$\n3+4+(5,6,7) > \\frac{1}{2}M > 1+2+[5,6,7]+\\{5,6,7\\}. \\quad (2)\n$$\nIf Andriy by his first step plays $6+7$, Olesya can not take $1+2$ and $1+5$, because Andriy will take winning combination $3+4+6+7 > 3+4+(5,6,7) > \\frac{1}{2}M$. Thus, Olesya can win with combination $4+5$, thus inequality must be true:\n$$\n4+5+(1,2,3) > \\frac{1}{2}M > 6+7+[1,2,3]+\\{1,2,3\\}. \\quad (3)\n$$\nIf Andriy by his first step will play $1+7$, so Olesya can not play $5+6$ and $2+6$, because then Andriy takes winning combination $3+4+1+7 > 3+4+(5,6,7) > \\frac{1}{2}M$. Thus, Olesya must win with combination $2+3$, and thus inequality must be true:\n$$\n2+3+(4,5,6) > \\frac{1}{2}M > 1+7+[4,5,6]+\\{4,5,6\\}. \\quad (4)\n$$\nIf to add two last ratios, we have contradiction:\n$$\n2+3+4+5+(1,2,3)+(4,5,6) > 1+6+7+7+[1,2,3]+\\{1,2,3\\}+[4,5,6]+\\{4,5,6\\}\n$$\nas $2+3 \\le [1, 2, 3] + \\{1, 2, 3\\}$, $4+5 \\le [4, 5, 6] + \\{4, 5, 6\\}$, $(1, 2, 3) \\le 1$, $(4, 5, 6) \\le 6$, $0 < 7+7$. Thus, Andriy has winning strategy, knowing ratio of weights of pieces.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45533,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $(x_n)$ satisfies the following conditions: $x_1 = a$, $x_{n+1} = \\frac{1}{2}\\left(x_n - \\frac{1}{x_n}\\right)$, $n \\in \\mathbb{N}$. Prove that there exists a number $a$ such that the sequence $(x_n)$ has exactly 2018 pairwise distinct elements.\n(If one of the elements of the sequence equals 0, then the sequence stops on that element.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote $x_1 = a = \\operatorname{ctg} \\alpha$. Then\n$$\nx_2 = \\frac{1}{2}\\left(x_1 - \\frac{1}{x_1}\\right) = \\frac{1}{2}(\\operatorname{ctg} \\alpha - \\operatorname{tg} \\alpha) = \\frac{1}{2} \\cdot \\frac{\\cos^2 \\alpha - \\sin^2 \\alpha}{\\sin \\alpha \\cdot \\cos \\alpha} = \\frac{\\cos 2\\alpha}{\\sin 2\\alpha} = \\operatorname{ctg} 2\\alpha,\n$$\nIn the same way we easily prove that\n$$\nx_{n+1} = \\frac{1}{2}\\left(x_n - \\frac{1}{x_n}\\right) = \\frac{1}{2}\\left(\\operatorname{ctg} 2^{n-1}\\alpha - \\operatorname{tg} 2^{n-1}\\alpha\\right) = \\frac{1}{2} \\cdot \\frac{\\cos^2 2^{n-1}\\alpha - \\sin^2 2^{n-1}\\alpha}{\\sin 2^{n-1}\\alpha \\cdot \\cos 2^{n-1}\\alpha} = \\operatorname{ctg} 2^n \\alpha, \\forall n \\in \\mathbb{N}.\n$$\nThe statement of the problem will be satisfied if $x_i \\ne 0$, $i = \\overline{1, 2017}$, and $x_{2018} = 0$. Therefore, it is sufficient to choose $\\alpha$ such that $x_{2018} = \\operatorname{ctg}(2^{2017}\\alpha) = 0$. Thus we have $2^{2017}\\alpha = \\frac{\\pi}{2}$, and $\\alpha = \\frac{\\pi}{2^{2018}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute angled triangle, $AA_1$ and $CC_1$ be its bisector, $I$ be the incenter of $ABC$, $M$ and $N$ be the midpoints of $AI$ and $CI$, respectively. Inside the triangles $AC_1I$ and $A_1CI$ we choose points $K$ and $L$, such that\n\n$\\angle AKI = \\angle CLI = \\angle AIC$, $\\angle AKM = \\angle ICA$, $\\angle CLN = \\angle IAC$. Prove that radiuses of the circumcircles of the triangles $KIL$ and $ABC$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Anton Trigub)\n![](attached_image_1.png)\n**Fig. 46**\nSuppose that lines $AI$ and $CI$ meet the circumcircle of $\\triangle ABC$ (second time) in the points $W_A$ and $W_C$, respectively (Fig. 46). Since $\\angle AKI = \\angle AIC$, then circumcircle of the triangle $AKI$ tangent to the line $W_C I$. Consider the triangle $AW_C I$. According to well-known fact this triangle is isosceles. Then the circumcircle of the triangle $\\triangle AKI$ tangents to $W_C A$ as well. Thus a symmedian of the $\\triangle AKI$ belongs to the line $W_C K$. Then $\\angle W_C KI = 180^\\circ - \\angle ICA = 180^\\circ - \\angle W_C W_A I$. It means that $K$ belongs to the circumcircle of the $\\triangle W_C W_A I$. In the same way $L$ belongs to this circumcircle as well. Let us note that $\\triangle W_C B W_A = \\triangle W_C I W_A$ (they have equal sides), and thus the radius of the circumcircle of $\\triangle KIL$ equal to the radius of the circumcircle of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45535,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cube is cut into $2010$ smaller cubes. $2008$ of them are unit cubes (cubes of side $1$) and edges of another $2$ cubes are assigned integer numbers (different from $1$). Find the volume of the big cube.\n\n(Serhiy Zhydkov)",
    "options": [],
    "answer": "2197",
    "solution": "Let's denote the edge of the big cube as $a$ and the edges of two not unit as $b$ and $c$.\nThen we have an equation:\n$$\nb^3 + c^3 + 2008 = a^3.\n$$\nBecause $12^3 = 1728 < 2008$, then $a > 12$, so $a \\ge 13$.\nNow let's find an upper bound of $a$. It is clear that $a \\ge b+c$,\n$$\nb^3 + 3b^2c + 3bc^2 + c^3 \\le a^3 = b^3 + c^3 + 2008,\n$$\nwhich means that $b^2c + bc^2 \\le \\frac{2008}{3}$ or $b^2c + bc^2 \\le 669$. Then $2c^3 \\le b^2c + bc^2 \\le 669$ so $c^3 \\le 334$.\n\nMoreover,\n$$\n2b^2 + 8 \\le b^2c + bc^2 \\le 669 \\Rightarrow 2b^2 \\le 661 \\Rightarrow b^2 \\le 330 \\Rightarrow b \\le 18 \\Rightarrow b^3 \\le 5832 \\Rightarrow \\\\\nb^3 + c^3 \\le 6166 \\Rightarrow a^3 = b^3 + c^3 + 2008 \\le 6166 + 2008 = 8174 \\Rightarrow a \\le 20.\n$$\nThe only thing that remains is to test on cube of integer numbers $b \\ge c \\ge 2$, which satisfy an equation $b^3 + c^3 + 2008 = a^3$, where $a$ is from $13$ to $20$. As a result, we have the unique solution: $5^3 + 4^3 + 2008 = 13^3$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a foundation that is in a form of a rectangle $2m \\times 2n$ that is divided into smaller $1 \\times 1$ squares. There is a gap of length $1$ between any two adjacent squares. The foundation is covered by several layers of bricks of size $2 \\times 1$. Every layer consists of $2mn$ bricks and each brick fully covers exactly one gap of length $1$. Such cover is called *strong*, if every gap is covered by a brick at least in one of the layers. What is the minimum amount of layers that make a strong cover?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Minimum number of layers = 2 if m = n = 1; 3 if exactly one of m, n equals 1 and the other is greater than 1; 4 otherwise.",
    "solution": "Without loss of generality let $2 \\le m \\le n$. Consider a square $A$ of size $1 \\times 1$, that does not touch the sides of a big rectangle. All 4 sides of it have to be covered with different bricks, since half of each brick covers the square $A$ itself. Therefore, covering with less than 4 layers is not possible. It suffices to show that covering with 4 layers is possible (Fig. 19). First two layers are shown on the picture, the other two layers are given by dividing rectangle on $2 \\times 2$ squares. Then two different options of dividing such squares into bricks give two layers as in Fig. 19.\n\nClearly, in case of $m=n=1$ it suffices to have 2 layers.\n\nIf $m=1<n$ covering with 3 layers exists. Example is shown in Fig. 20. Covering with less layers is not possible, since for those squares whose vertices does not coincide with a vertices of a big rectangle, 3 sides have to be covered, that is possible only with at least three layers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find pairs of positive integers $x, y$, which satisfy the system of equations:\n$$\n\\begin{cases}\n[x, y] + (x, y) = 2018, \\\\\nx + y = 2018,\n\\end{cases}\n$$\nWhere $[x, y]$ and $(x, y)$ are LCM and GCD of numbers $x, y$.",
    "options": [],
    "answer": "(x, y) ∈ {(1, 2017), (2017, 1), (2, 2016), (2016, 2), (1009, 1009)}",
    "solution": "Since $(x, y) \\cdot [x, y] = xy$, then the first equation of the system can be re-written as\n$$ \\frac{xy}{(x, y)} + (x, y) = 2018, $$\nwhich gives us quadratic equation with respect to $(x, y)$:\n$$\n(x, y)^2 - 2018(x, y) + xy = 0.\n$$\nIts discriminant, also taking into account the second equation of the system, is\n$$\nD = 2018^2 - 4xy = 2018^2 - 4x(2018 - x) = (2x - 2018)^2.\n$$\nWhich gives $x = \\frac{2018 \\pm (2x - 2018)}{2} \\Rightarrow x_1 = x$ and $x_2 = 2018 - x = y$.\n\nHence, one of the numbers equals GCD, which means it divides the other number. E.g., let $x \\le y$, i.e. $y = kx$, then $(k+1)x = 2018$. Hence, these are the possible cases:\nCase 1. $k+1=2018 \\Rightarrow x=1$ and $y=2017$.\nCase 2. $k+1=1009 \\Rightarrow x=2$ and $y=2016$.\nCase 3. $k+1=2 \\Rightarrow x=1009=y$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given quadrilateral $ABCD$, such that $AB = BC$. $K$ is the midpoint of $CD$, rays $BK$ and $AD$ intersect at $M$, the circumcircle of $\\triangle ABM$ intersects line $AC$ for the second time at point $P$. Show that $\\angle BKP = 90^\\circ$.\n\n(Anton Trygub)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the midpoint of diagonal $AC$, then $NK$ is the midline of $\\triangle ACD$ (Fig. 31). Thus, $\\angle (KN, NP) = \\angle (AM, AP) = \\angle (BM, BP) = \\angle (KB, BP)$, hence $BNKP$ is inscribed. Thus, since $\\triangle ABC$ is equilateral, $\\angle BNP = 90^\\circ$, therefore, $\\angle BKP = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all possible real pairs $(x, y)$ that satisfy the following:\n$$\n4x + 3y = 2x \\cdot \\left[ \\frac{x^2 + y^2}{x^2} \\right].\n$$",
    "options": [],
    "answer": "All real pairs with x ≠ 0 and either y = -2x/3 or y = 2x.",
    "solution": "It follows from the given condition that $x \\neq 0$, therefore, equation can be rewritten as $2 + \\frac{3y}{2x} = \\left[ 1 + \\frac{y^2}{x^2} \\right]$. Let $k = \\frac{3y}{2x}$ is an integer. Since $[1 + \\frac{y^2}{x^2}] = 1 + \\frac{y^2}{x^2}$, the following holds:\n$$\n1 + k = \\left[ \\frac{4k^2}{9} \\right] \\Leftrightarrow 1 + k \\le \\frac{4k^2}{9} < 2 + k.\n$$\n$$\n\\begin{cases} 4k^2 - 9k - 9 \\ge 0, \\\\ 4k^2 - 9k - 18 < 0, \\end{cases}\n$$\n$$\n\\left( \\frac{9 - 3\\sqrt{41}}{8}; -\\frac{3}{4} \\right] \\cup \\left[ 3; \\frac{9 + 3\\sqrt{41}}{8} \\right).\n$$\nTherefore, $k = -1$ or $k = 3$. So, $\\frac{3y}{2x} = -1$ or $\\frac{3y}{2x} = 3$, and thus, all possible solutions are in the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45540,
    "subject": "Mathematics (Multi-modal)",
    "question": "The teacher wrote digits $123\\ldots9123\\ldots9123\\ldots$ on the board until a $2018$-digit number was formed. After that Andriy and Olesya played a game as follows. Alternately (after Andriy begins) they cross out $2$ digits as follows: either the first two digits of the number remaining after the previous move, or the last two digits, or the first and last digits of that number. The game ends when the two-digit number is left. Olesya wins, if this number is a multiple of $3$, otherwise Andriy wins. Who will win if both players play according to the best strategy?\n\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Olesya",
    "solution": "Since we are only interested in divisibility by $3$, we can change all the digits as follows: $1, 4, 7 \\to 1$, $2, 5, 8 \\to 2$ and $3, 6, 9 \\to 3$, which will give us the equivalent problem. Then, after $2014 \\div 2 = 1007$ moves, a $4$-digit number will be left on the board, and Olesya will be making the last move. Obviously, these $4$ digits left will be the ones that were initially together (following one another) on the board. Therefore, the possible options are: $1231$, $2312$ or $3123$. In either of these cases there is a pair of digits that comprises number $12$, and this is exactly the pair which Olesya should leave on the board for her to win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45541,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given that the bookshelf can fit $9$ of the same thick books, but the $10$th one will not fit anymore. Similarly, it can hold $15$ of the same thin books, but $16$th will not fit anymore. Is it possible for that shelf to hold simultaneously:\n\na) $6$ thick and $5$ thin books?\n\nb) $7$ thick and $5$ thin books?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "Let us re-write the statement of the problem as follows. Let us denote the length of the shelf by $S$, the width of the thick book by $x$ and the width of the thin book by $y$. Then, we have the conditions:\n$$\n9x \\leq S < 10x \\text{ and } 15y \\leq S < 16y \\Leftrightarrow \\frac{1}{10}S < x \\leq \\frac{1}{9}S \\text{ and } \\frac{1}{16}S < y \\leq \\frac{1}{15}S.\n$$\n\na) Let's see what is the maximum width of the space these books would take:\n$$\n6x + 5y \\leq \\frac{6}{9}S + \\frac{5}{15}S = S.\n$$\n\nb) Analogously, we show that this set of books cannot fit on the shelf:\n$$\n7x + 5y > \\frac{7}{10}S + \\frac{5}{16}S = \\frac{56+25}{80}S > S.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a foundation that is in form of a square $4 \\times 4$, that is divided into smaller $1 \\times 1$ squares. There is a gap of length $1$ between any two adjacent squares. The foundation is covered with several layers of bricks of size $2 \\times 1$. Every layer consists of $8$ bricks and each brick fully covers exactly one gap of length $1$. Such cover is called *strong*, if every gap is covered by a brick at least in one of the layers. What is the minimum amount of layers that make a strong cover?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "4",
    "solution": "Consider a square $A$ of size $1 \\times 1$, that does not touch the borders of $4 \\times 4$. All $4$ sides of it have to be covered by bricks. Moreover, these bricks have to be different for different\n![](attached_image_1.png)\n**Fig. 16**\nsides, since one half of each brick covers the square $A$. Thus, there have to be at least $4$ layers in order for cover to be strong. It suffices to show that it is possible to have a strong cover with $4$ layers (Fig. 16).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest positive integer that has pairwise distinct digits and is divisible by each of its digits.",
    "options": [],
    "answer": "9867312",
    "solution": "Clearly, $0$ is not one of the digits. In order to be divisible by $5$ last digit has to be $5$, but the number will not be divisible by $2$, $4$, $6$ and $8$, so it will consist of more digits if $5$ is not one of them. But then if all other digits are present, it is not divisible by $3$, $6$ and $9$. Therefore, one more digit has to be taken from the number in order for it to be divisible by $3$. If so, the following cases are possible:\n\nCase 1. If either $1$ or $7$ are taken, then number is divisible by $3$ and $6$, but not by $9$, thus digit $9$ is not present and the number has $6$ digits;\n\nCase 2. If $4$ is taken, then it is possible to arrange digits that are left so that each digit divides the number, so the number has $7$ digits (all except $0$, $4$ and $5$).\n\nTherefore, the number has $7$ digits: $1$, $2$, $3$, $6$, $7$, $8$, $9$. Our goal is to make it the largest possible.\n\nIf the number starts with $987$, then digits $1$, $2$, $3$, $6$ can make numbers divisible by $8$, those are $6312$, $1632$, $2136$ to $3216$. But then any of the numbers is not divisible by $7$.\nIf the number starts with $3216$, then digits $1$, $2$, $3$ form a number $312$ that is divisible by $8$.\nMoreover, $9867312$ is divisible by $7$, therefore, it is the largest number that satisfies given conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45544,
    "subject": "Mathematics (Multi-modal)",
    "question": "We know that a certain number $A$ has exactly 2018 positive integer factors (including 1 and number $A$ itself), and it is divisible by 2018. Prove that $A$ is not divisible by $2018^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A = p_1^{k_1} p_2^{k_2} \\dots p_n^{k_n}$ be a canonical prime factorization of $A$, where $p_i$ are different prime factors, and $k_i$ are positive integers, $i = 1, n$. Knowing the total number of positive integer factors of $A$ we can conclude that\n$$\n(k_1 + 1)(k_2 + 1)\\dots(k_n + 1) = 2018.\n$$\n\nSince the left-hand side comprises factors each of which is greater or equal to 2, we have only two cases:\n\n1) $k_1 + 1 = 2018$.\n2) $(k_1 + 1)(k_2 + 1) = 2 \\cdot 1009$.\n\nFrom the first case, $A$ has a canonical prime factorization of the form: $A = p_1^{2017}$, which contradicts the fact that $A \\neq 2018$, since in that case $A$ can't be divisible by both 2 and the prime number 1009.\n\nHence, $(k_1 + 1)(k_2 + 1) = 2 \\cdot 1009$, which yields, without loss of generality, $k_1 = 1$, $k_2 = 1008$. Therefore, $A = p_1 p_2^{1008}$, moreover from divisibility by 2 and 1009 we obtain $A = 2 \\cdot 1009^{1008}$ or $A = 1009 \\cdot 2^{1008}$. However, none of these numbers is divisible by $2018^2$, Q.E.D.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be an orthocenter of an acute triangle $ABC$, and let $O$ be its circumcenter. The line $AO$ intersects segment $BC$ at $D$. Perpendicular to $BC$ with a foot $D$ intersects altitudes of $ABC$ through $B$ and $C$ at points $X$ and $Y$ respectively. Show that the circumcenter of $\\triangle HXY$ is equally distant from points $B$ and $C$.\n\n(Danylo Khilko)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AH_1$, $BH_2$, $CH_3$ be the altitudes of $\\triangle ABC$ (Fig. 29), $M$ is the midpoint of $BC$, and let $O_1$ be the circumcenter of $\\triangle YHX$. We want to show that $\\triangle YHX \\sim \\triangle ABC$.\n\nSince $\\angle BH_3C = \\angle YDC = 90^\\circ$, $\\angle ABC = \\angle HYX$. Similarly, $\\angle ACB = \\angle HXY$. Therefore, $\\triangle YHX \\sim \\triangle ABC$.\n\nLet $HH'$ be the altitude of $\\triangle YHX$, and let $M'$ be the midpoint of $XY$.\n\nClearly, $O_1M' \\perp XY$, $OM \\perp BC$. Thus, $HH'$ and $AH_1$ are the corresponding elements in similar triangles. Similarly, $O_1M'$ and $OM$ are the corresponding elements in similar triangles. Thus,\n$$\n\\frac{HH'}{AH_1} = \\frac{O_1M'}{OM}.\n$$\nFrom the other side, it is clear that $\\triangle OMD \\sim \\triangle AH_1D$. Hence,\n$$\n\\frac{H_1D}{AH_1} = \\frac{MD}{OM}.\n$$\nThese two equalities, together with $HH' = H_1D$, since $HH_1DH'$ is a rectangle, give $O_1M' = MD$. Thus, $O_1M'DM$ is a rectangle. Therefore, $O_1M \\perp BC$, so $O_1$ belongs to the perpendicular bisector of $BC$. Hence, $O_1B = O_1C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45546,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which positive integers $n$, square $n \\times n$ can be completely covered (without overlaps) by rectangles $k \\times 1$ and one square $1 \\times 1$, where:\na) $k=4$;\nb) $k=8$?",
    "options": [],
    "answer": "a) Exactly the odd n with n ≥ 5.\nb) Exactly the odd n with n > 8 and n ≡ 1 or 7 (mod 8), equivalently n = 8m + 9 or n = 8m + 15 for integers m ≥ 0.",
    "solution": "For both a) and b) it is obvious that $n$ has to be odd and greater than $k$.\n\na) Hence, $n$ is odd and greater than $4$. Let us show that any such odd $n$ satisfies the condition. It is clear that any stripe $4 \\times l$ can be covered by rectangles $4 \\times 1$. For $5 \\times 5$ and $7 \\times 7$ the required coverage is shown on Fig. 7–8. For any odd $n = 4m + 5$ and $n = 4m + 7$ it is sufficient to cut the square $n \\times n$ into a square $5 \\times 5$ or $7 \\times 7$ and several stripes $4 \\times l$ (Fig. 9).\n\n![](attached_image_1.png)\nFig. 9\n\nb) Let us show that by a similar scheme we can cover squares $n = 8m + 9$ and $n = 8m + 15$. For this, it is sufficient to cover $9 \\times 9$ and $15 \\times 15$. Then, each $n \\times n$ of the above size can be cut into a square $9 \\times 9$ or $15 \\times 15$ and several stripes $8 \\times l$ (Fig. 10–11).\n\nNow, let us show that squares $n = 8m + 11$ and $n = 8m + 13$ cannot be covered according to the condition. For the square $11 \\times 11$, consider this coloring in black and white squares, as shown in Fig. 12. Out of $121$ squares, $65$ are black and $56$ are white, which means there are $9$ more black cells than white ones. Each rectangle $8 \\times 1$ inside of this square covers the same number of black and white cells. Therefore, if there existed such coverage, the number of black and white cells would differ only by $1$. For the general case square $n \\times n$ when $n = 8m + 11$, analogous coloring yields $9$ more black cells than white cells.\n\n![](attached_image_2.png)\nFig. 10\n\nAnalogously, for square $13 \\times 13$, we show the coloring (Fig. 13) with $89$ black and $80$ white cells, which again means there is no such coverage. And for the case $n \\times n$ when $n = 8m + 13$, the same holds.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to draw a square with area of $50$ units on a graphing paper using a ruler that can connect two edges with a line?",
    "options": [],
    "answer": "Yes",
    "solution": "Consider a $10 \\times 10$ square and connect the midpoints of the sides as shown in Fig. 17. The square obtained clearly has area that is a half of the original square's area, that is $\\frac{1}{2} \\cdot 10 \\cdot 10 = 50$.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a foundation that is in a form of a $6\\times 6$ square, that is divided into smaller $1\\times 1$ squares. There is a gap of length $1$ between any two adjacent squares. The foundation is covered by several layers of bricks of size $3\\times 1$. Every layer consists of $12$ bricks and each brick fully covers exactly two gaps of length $1$. Such covering is called *strong*, if every gap is covered by a brick at least in one of the layers. Determine the minimum amount of layers required for a strong covering.\n(Bogdan Rublyov)",
    "options": [],
    "answer": "4",
    "solution": "By contradiction, suppose cover can have three layers. Consider a grey square and four marked gaps as in Fig. 26. Each gap can be covered only by a brick that will also cover a grey square. Any such brick can't cover two of the marked gaps simultaneously. Thus, the grey square has to be covered by at least four bricks. Therefore, $4$ is the least possible amount of layers for a strong cover.\n\nIt suffices to find an example for four layers, that is shown in Fig. 27.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several representatives of competing companies that produce the game \"Overwatch\" arrived at the conference. Consider all representatives of different companies to be competitors. It is known that each participant of the conference has exactly $2018$ competitors among all other participants. What is the largest possible number of participants who took part in the conference?",
    "options": [],
    "answer": "4036",
    "solution": "Notice that from the problem statement it follows that the same number of participants arrived from each company. Indeed, if there were two different numbers of participants from some two companies, they would have a different number of competitors. Let $m$ be the number of companies participating in the conference, each of them having $k$ members. Then, the number of competitors of each participant of the conference equals $(m-1)k$. It is equally easy to devise that the total number of participants in the conference is $mk$. Therefore, it should hold that $(m-1)k = 2018$, or $mk = 2018 + k$. Thus, the largest number of participants will be at the highest value of $k$, and this value is $k = 2018$ when $m = 2$. Thus, the maximum number of participants of the conference is $4036$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45550,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive integers, prove the inequality:\n$$\n\\frac{y}{2x + y} + \\frac{z}{2y + z} + \\frac{x}{2z + x} \\geq 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove it by using Cauchy–Schwarz inequality:\n$$\n\\left(\\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x}\\right) (x+y+z)^2 = \\left(\\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x}\\right) \\left(y(2x+y) + z(2y+z) + x(2z+x)\\right) \n\\geq \\left(\\sqrt{\\frac{y}{2x+y}} \\cdot \\sqrt{y(2x+y)} + \\sqrt{\\frac{z}{2y+z}} \\cdot \\sqrt{z(2y+z)} + \\sqrt{\\frac{x}{2z+x}} \\cdot \\sqrt{x(2z+x)}\\right)^2 = (x+y+z)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45551,
    "subject": "Mathematics (Multi-modal)",
    "question": "In isosceles triangle $ABC$ with the vertex in $B$ there are altitudes $BH$ and $CL$. Point $D$ is such that $BDCH$ is a rectangle. Find the angle $DLH$.\n(Bogdan Rublyov)",
    "options": [],
    "answer": "90°",
    "solution": "Let us denote by $O$ the intersection of diagonals of rectangle $HBDC$ (Fig. 3). Since $\\triangle CBL$ has right angle, $BO = LO = DO$, therefore, $HO = LO = DO$. Hence, $\\triangle DHL$ is the right triangle with the hypotenuse $DH$, which yields $\\angle DLH = 90^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$ with $\\angle ACB > 90^\\circ$, $\\angle CBA > 45^\\circ$. Points $P$ and $T$ belong to the sides $AC$ and $AB$, respectively, so that $PT = BC$ and $PT \\perp BC$. Points $P_1$ and $T_1$ belong to the sides $AC$ and $AB$ so that $AP = CP_1$ and $AT = BT_1$. Show that $\\angle CBA - \\angle P_1T_1A = 45^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be such that $\\triangle ABC$ and $\\triangle BCS$ are in a different half planes with respect to $BC$, and $\\triangle BCS$ is isosceles and has right angle $CBS$. Then segments $BS$ and $PT$ are equal and parallel ($TP = BC = BS$, $PS \\parallel AB$), hence $TPSB$ is a parallelogram (Fig. 21). Thus, $PS \\parallel AB$, and also $PS = TB$, hence $AT_1 = PS$. Thus $AT_1SP$ is a parallelogram. Thus, $AP \\parallel ST_1$ and $AP = ST_1$. Therefore, $P_1C \\parallel T_1S$ and $P_1C = T_1S$ is a parallelogram. Hence $P_1T_1 \\parallel CS$ and $\\angle P_1T_1A = 180^\\circ - \\angle P_1T_1S - \\angle ST_1B = \\angle CST_1 - \\angle ST_1B$.\n\n![](attached_image_1.png)\n**Fig. 21**\nFinally,\n$$\n\\begin{aligned}\n\\angle CBA - \\angle P_1T_1A &= \\angle CBA - \\angle CST_1 + \\angle ST_1B = \\\\\n&= 90^\\circ - \\angle T_1SB - \\angle CST_1 = 90^\\circ - \\angle CSB = 45^\\circ.\n\\end{aligned}\n$$\nLet $PT$ and $BC$ intersect at $K$, $M_c$ is a midpoint of $CP$, $M_b$ is a midpoint of $BT$ (Fig. 22). Thus, medians in right triangles $CKP$ and $BKT$ equal half of hypotenuses. Therefore:\n$$\n\\angle M_c KM_b = \\angle TKM_b - \\angle PKM_c = \\angle M_b TK - \\angle M_c PK = \\angle CAB.\n$$\nAlso, $\\frac{AR}{AT_1} = \\frac{CP}{BT} = \\frac{KM_c}{KM_b}$, hence $\\triangle AP_1T_1 \\sim \\triangle KM_cM_b$. We can show that $\\angle CBA - \\angle P_1T_1A = 45^\\circ$. Also, $\\angle CBA - \\angle P_1T_1A = \\angle M_bKB - \\angle M_cM_bK$, that is an angle between $M_cM_b$ and $BC$.\nThus, it suffices to show that $M_cM_b$ is parallel to bisector of $\\angle TKB$.\n\nLet $X$ is a midpoint of $PB$, $Y$ is a midpoint of $TC$ (Fig. 23) where $M_c M_b$ is a bisector of $\\angle XM_c Y$. $XM_c \\parallel KB$ and $YM_c \\parallel KT$ finish the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Can one draw $6$ circles on a plane, such that each one passes through the centers of exactly three other circles?",
    "options": [],
    "answer": "Yes",
    "solution": "Possible example is shown on Fig. 18 where each segment has length $1$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45554,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ students arrived to the summer camp. Every child (boy or girl) in the camp knows exactly one boy and one girl. Is it possible if\n\na) $n = 2000$;\n\nb) $n = 2018$.",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "a) Suppose $1000$ boys and $1000$ girls arrived to the camp. We can split them into groups with $4$ students in each: $2$ boys and $2$ girls. Moreover, they know each other in the following way: $B_1 \\leftrightarrow B_2 \\leftrightarrow G_2 \\leftrightarrow G_1 \\leftrightarrow B_1$. One can check that all conditions are satisfied.\n\nb) Suppose by contradiction that it is possible. Then every boy can be put into a pair with a girl if they know each other. Thus, there has to be exactly $1009$ boys and $1009$ girls. Similarly, all the boys can be divided in pairs if they know each other, so the amount of boys has to be even. That leads to a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$, $D$ and $K$ belong to the sides $AB$, $BC$ and $CA$ of isosceles triangle $ABC$ with apex $B$ such that $AM = 2DC$ and $\\angle AMD = \\angle KDC$. Show that $MD = KD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $FD \\parallel AC$ (Fig. 24).\n\n![](attached_image_1.png)\n\nThen $AF = FM = DC$, hence $\\triangle FMD = \\triangle KDC$ by the side and adjacent angles. Thus, $MD = KD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45556,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are several gentlemen in the club.\nEvery two are either friends, or enemies. It is known that each of the gentlemen has exactly $4$ enemies. In addition, for each of them, the enemy of his friend is his enemy. How many gentlemen can be present at the club?",
    "options": [],
    "answer": "5, 6, 8",
    "solution": "Note that the condition implies that each of the gentlemen has equal number of friends. Let $n$ denote the number of gentlemen in the club. Each of them has exactly $4$ enemies, therefore exactly $n-5$ friends.\n\nConsider a single gentleman $A_1$. Let $B_1, B_2, B_3, B_4$ denote his enemies. Since each friend of $A_1$ is also an enemy of $B_1$, then $A_1$ has no more than three friends, since $B_1$ has exactly $4$ enemies and no more than that. Consider the following cases.\n\n**Case 1.** $A_1$ has $3$ friends. Let us denote them by $A_2, A_3, A_4$. In this case, there are total $8$ gentlemen in the club, and this situation is possible when each two of gentlemen $A_1, A_2, A_3, A_4$ are friends, each two of gentlemen $B_1, B_2, B_3, B_4$ are also friends, and each pair $A_i$ and $B_j$ are enemies.\n\n$A_2, A_3$ already have $4$ enemies, which means they must be friends with each other. Consider gentleman $B_1$. Without loss of generality, we can state that his friends are $B_2, B_3$. Analogous to the above, we show that $B_2, B_3$ are friends. Then, $B_4$ does not have any friends. Which makes this case impossible.\n\n**Case 2.** $A_1$ has $2$ friends. Let us denote them by $A_2, A_3$. Then the club has the total of $7$ gentlemen. Notice that each of the friends of $A_1$ is the enemy of $B_j$. Then each of the gentlemen ...\n\n**Case 3.** $A_1$ has $1$ friend. Then the total number of gentlemen in the club is $6$. This case is possible. For example, let the pairs $(A_1, A_2)$, $(B_1, B_2)$ and $(C_1, C_2)$ be pairs of friends. All the rest are enemies to each other. It is easy to see that this example satisfies the condition.\n\n**Case 4.** $A_1$ has no friends. Then $5$ gentlemen, all of which are the enemies of one another, satisfy the condition of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45557,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triple of numbers $(x, y, p)$, where $x, y$ are positive integers and $p$ is a prime number, which satisfy the condition:\n$$\ny(x^2 + p) - x(y^2 + p) = p.\n$$",
    "options": [],
    "answer": "All triples (x, y, p) with p prime given by (x, y) = (p + 1, 1) for any prime p, and the exceptional triple (3, 2, 3).",
    "solution": "Factor the left-hand side of the equation:\n$$\n(yx^2 - xy^2) + (yp - xp) = p \\Rightarrow yx(x - y) - p(x - y) = p \\Rightarrow (yx - p)(x - y) = p.\n$$\nThe last equation is possible in several cases.\n\n**Case 1.** $yx - p = 1$, $x - y = p$. Then, we get a quadratic equation: $x = y + p$\n$$\ny(y + p) - p = 1 \\Rightarrow y^2 + yp - p - 1 = 0.\n$$\nSince $y=1$ is the solution to this equation, then another solution is $y=-p-1$, but it is not a positive integer. Hence, we obtain that $x = p+1$. Putting these values to the initial equation, we see that a tuple $(p+1; 1; p)$ satisfies the statement for arbitrary prime number $p$.\n\nCase 2. $yx - p = -1$, $x - y = -p$. Then, we get a quadratic equation: $y = x + p$\n$$\nx(x + p) - p = -1 \\Rightarrow x^2 + xp - p + 1 = 0.\n$$\nSince $x \\in \\mathbb{N}$, the discriminant of the last equation must be the square of an integer number:\n$$D = p^2 - 4(-p+1) = p^2 + 4p - 4.$$ Let us see which $p$ satisfies the conditions:\n$$\n(p+1)^2 < p^2 + 4p - 4 < (p+2)^2.\n$$\nRight inequality holds for any $p$, while the left one can be re-written:\n$$\np^2 + 2p + 1 < p^2 + 4p - 4 \\Leftrightarrow 2p > 5 \\Leftrightarrow p \\ge 3.\n$$\nTherefore, we must check separately the case $p=2$, whereas for any other prime $p$ the discriminant is not the square of an integer number. For $p=2$, we get $D=8$, which is also not the square of an integer, which means such positive integer $x$ does not exist.\n\nCase 3. $yx - p = p$, $x - y = 1$. Then, we obtain the equations: $x = y+1$ and $yx = 2p$. From the 1st equation, $x > y$, from the 2nd, $y=1$ and $x = 2p$ or $y = 2$ and $x = p$. If we combine these conditions, then we obtain $y=1, x=2, p=1$ – contradiction, or $y=2, x=p=3$ – solution.\n\nCase 4. $yx - p = -p$, $x - y = -1$. Obviously, the first equation yields the contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-isosceles triangle $ABC$ is given, in which $2AC = AB + BC$. Let $I$ be the incenter of the inscribed circle in $ABC$, $K$ be the middle of the sector $ABC$ of the circumscribed circle. Let $T$ be such point on the line $AC$ that $\\angle TIB = 90^\\circ$. Prove that line $TB$ is tangent to the circumcircle of $\\triangle KBI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, suppose $AB < BC$. Let angular bisector of $\\angle ABC$ intersect the circumcircle of the triangle second time in point $W$, and let point $M$ be the middle of $AC$. Let us prove that $BI = IW$. For that, from $I$ and $W$, we draw perpendiculars $II_1$ and $WW_1$ on line $BC$ (Fig. 14). It is known that\n$$\nBW_1 = \\frac{1}{2}(AB + BC) = AC, \\quad \\text{and}\n$$\n$$\nBI_1 = \\frac{1}{2}(AB + BC - AC) = \\frac{1}{2}AC, \\quad \\text{hence, } BI_1 = I_1W_1 \\quad \\text{and}\n$$\n$$\nBI = WI.\n$$\n\nNotice also that quadrilateral $TIMW$ is inscribed with diameter $TW$.\nLet us show that $\\angle BKI = \\angle IMA$. From the trillium theorem, $WA = WC = WI$, i.e.\n$$WI^2 = WM \\cdot WK. \\quad \\text{Then,}$$\n$$\\Delta WIM \\sim \\Delta KIW, \\quad \\text{which means}$$\n$$\\angle WMI = \\angle WIK \\Rightarrow \\angle BKI = \\angle IMA.$$ \nNow, in $\\triangle BTW$, $TI$ is altitude and median, which makes $\\triangle BTW$ isosceles, and so\n$$\\angle TBI = \\angle TWI = \\angle TMI = \\angle BKI,$$\nwhich yields the statement of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real numbers $a$, $b$, $c$ such that: $ab - c = 1000$, $bc - a = 1018$ and $ca - b = -2018$. Show that $a + b + c \\neq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By contradiction, assume that $a + b + c = 0$. Thus,\n$$\nab + bc + ca = a + b + c = 0.\n$$\n\nSuppose that $abc = 0$, so without loss of generality let $c = 0$. Hence $ab = 0$. Therefore, two variables are zeros. By the first condition on numbers in the problem, we obtain a contradiction. Therefore, $abc \\neq 0$.\n\nPlug $c = -a - b$ in equation $ab + bc + ca = 0$. Hence\n$$\n\\begin{aligned}\n(a + b)b + (a + b)a - ab &= 0 \\\\\na^2 + ab + b^2 &= 0 \\\\\na^2 + ab + \\frac{1}{4}b^2 + \\frac{3}{4}b^2 &= 0 \\\\\n(a + \\frac{1}{2}b)^2 + \\frac{3}{4}b^2 &= 0 \\\\\na = b = 0.\n\\end{aligned}\n$$\n\nContradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, ..., x_{14}$ be non-negative numbers, sum of which is equal to $1$. Prove that:\n$$\nx_1x_2x_3x_4 + x_2x_3x_4x_5 + \\dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_{14}x_1 + x_{13}x_{14}x_1x_2 + x_{14}x_1x_2x_3 \\le \\frac{1}{4^4}. \\quad \\text{(Serhiy Zhyliaiev)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's denote the left side of inequality that should be proven as $S$. Let $x_{14}x_1$ be the smallest product among all pairwise products of adjacent numbers $x_kx_{k+1}$, $k = 1, 14$ (cyclically adjacent numbers: $x_{15} = x_1$). Then we can bound from above all the summands containing this pair like this:\n$$\nx_{12}x_{13}x_{14}x_1 \\le x_{12}x_{13}x_6x_7, \\quad x_{13}x_{14}x_1x_2 \\le x_{13}x_7x_8x_2, \\quad x_{14}x_1x_2x_3 \\le x_8x_9x_2x_3,\n$$\nBut then\n$$\n\\begin{aligned}\nS &= x_1x_2x_3x_4 + x_2x_3x_4x_5 + \\dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_{14}x_1 + x_{13}x_{14}x_1x_2 + x_{14}x_1x_2x_3 \\\\ &\\le x_1x_2x_3x_4 + x_2x_3x_4x_5 + \\dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_6x_7 + x_{13}x_7x_8x_2 + x_8x_9x_2x_3 = P.\n\\end{aligned}\n$$\n\nIn every group of $4$ summands from $P$ every multiplier has different remainder modulo $4$, that's why\n$$\n\\begin{aligned}\n& (x_1 + x_5 + x_9 + x_{13})(x_2 + x_6 + x_{10} + x_{14})(x_3 + x_7 + x_{11})(x_4 + x_8 + x_{12}) \\le \\\\\n& \\le \\frac{1}{4^4} \\left( \\sum_{k=1}^{14} x_k \\right)^4 = \\frac{1}{4^4},\n\\end{aligned}\n$$\nwhat was to be demonstrated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ was chosen inside the triangle $ABC$ so that $BC = AP$ and $\\angle APC = 180^\\circ - \\angle ABC$. On side $AB$, there exists point $K$ such that $AK = KB + PC$. Prove $\\angle AKC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We extend the ray $AB$ further after $B$ to find the point $T$ such that $BT = PC$ (Fig. 1). Then, $\\Delta TBC = \\Delta CPA$ due to two equal length sides and the same angle between them. Hence, $TC = CA$. Analogously,\n$$\nAK = KB + PC = KB + BT = KT.\n$$\nTherefore, in the isosceles triangle $ATC$ segment $KC$ is a median, which also makes it the altitude. From which follows that $KC \\perp AB$, Q.E.D.\n\n![](attached_image_1.png)\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45562,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the acute triangle $ABC$ there are altitudes $BP$ and $CQ$, point $T$ is the intersection of altitudes of $\\triangle PAQ$. It turns out that $\\angle CTB = 90^\\circ$. Find the value of $\\angle BAC$.",
    "options": [],
    "answer": "45°",
    "solution": "The statement of the problem yields $\\angle BQC = \\angle BTC = \\angle BPC = 90^\\circ$, hence, points $Q, T, P$ lie on the circle with diameter $BC$, and in this exact order: $B, Q, T, P, C$ (Fig. 6), since $\\triangle ABC$ is acute. Then,\n$$\n\\angle QTP = 180^\\circ - \\angle ABP = 180^\\circ - (90^\\circ - \\angle BAC) = 90^\\circ + \\angle BAC.\n$$\nOn the other hand, $QT \\perp AC$, $PT \\perp AB$, $\\angle QTP = 180^\\circ - \\angle BAC$. Hence,\n$$\n90^\\circ + \\angle BAC = 180^\\circ - \\angle BAC \\Rightarrow \\angle BAC = 45^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Country \"U\" has cities $A_1, A_2, \\dots, A_n$. They are connected with flights so that one can get from any city to any other city with layovers. Let $d_{i,j}$ be the smallest amount of flights needed to get from city $A_i$ to city $A_j$, and let $d$ be the total number of flights in country \"U\" and let\n$$\nD = d_{1,1} + d_{1,2} + \\dots + d_{1,n} + d_{2,3} + d_{2,4} + \\dots + d_{2,n} + \\dots + d_{n-1,n}.\n$$\n\nShow that:\n$$\nd + \\frac{D}{n-1} \\le \\frac{1}{2}(n+1)(n+2).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Show that:\n$$\nd + \\frac{D}{n-1} \\le \\frac{1}{2}(n+1)(n+2).\n$$\n\nLet $i \\ne j$, and let the shortest path between $A_i$ and $A_j$ be the following: $A_i = A_{\\gamma_0}, A_{\\gamma_1}, \\dots, A_{\\gamma_k} = A_j$. Let $X$ be the set of such vertices ($k+1$ vertices), and let $Y$ be the set of all other vertices ($n-k-1$ vertices total). The number of edges that connects vertices inside $Y$ is not greater than $C_{n-k-1}^2$. All edges that connect vertices inside $X$ are the edges of the shortest path, otherwise there exists shorter path between $A_i$ and $A_j$. Therefore, there are exactly $k$ edges.\n\nIf $A \\in Y$, then there are 3 or less edges that connect $A$ with vertices from $X$. Otherwise, there exists shorter path than the one that has $k$ edges. Therefore, there exists not more than $3(n-k-1)$ edges that connect vertices from $X$ and vertices from $Y$. Thus,\n$$\n\\begin{aligned}\nd \\le C_{n-k-1}^2 + k + 3(n-k-1) &= \\frac{1}{2}(n^2 + 3n - 4 + k^2 - k - 2kn) \\le \\\\\n& \\quad (k < n, \\text{ hence } kn < n^2) \\\\\n\\le \\frac{1}{2}(n^2 + 3n - 4) - \\frac{n+1}{2}k &\\Rightarrow d + \\frac{n+1}{2}d_{i,j} \\le \\frac{1}{2}(n^2 + 3n - 4), \\text{ thus} \\\\\n\\sum_{1 \\le i < j \\le n} \\left(d + \\frac{n+1}{2}d_{i,j}\\right) &\\le \\frac{1}{2}C_n^2(n^2 + 3n - 4), \\text{ so} \\\\\nC_n^2d + \\frac{n+1}{2}D &\\le \\frac{1}{2}C_n^2(n^2 + 3n - 4).\n\\end{aligned}\n$$\n\nThe given inequality can be obtained by dividing inequality by $C_n^2$ and using\n$$\n\\frac{n+1}{n(n-1)} \\ge \\frac{1}{n-1} \\quad \\text{and} \\quad n^2 + 3n - 4 \\le n^2 + 3n + 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45564,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are numbers $1, 2, 3, \\ldots, 100$ on the board. Each is written exactly once. Petryk and Ivasyk are playing in the following game (Petryk starts): each player erases one number from the board. If after player's turn the sum of all erased numbers (by both players) cannot be represented as a difference of squares of integers, the player loses. Who will win in this game if both players want to win?",
    "options": [],
    "answer": "Petryk",
    "solution": "Notice that $n$ cannot be represented as a difference of squares of integers iff $n \\equiv 2 \\pmod{4}$. Indeed, if $n \\equiv 2 \\pmod{4}$, then suppose $n = (x + y)(x - y)$. If $x, y$ are either both odd or both even, then $n \\equiv 0 \\pmod{4}$, otherwise $n \\equiv \\pm 1 \\pmod{4}$. If $n = 4k$, then let $x = k + 1, y = k - 1$, if $n \\equiv \\pm 1 \\pmod{4}$, then let $x = \\frac{n+1}{2}, y = \\frac{n-1}{2}$.\n\nThe strategy for Petryk is the following: he erases $100$ first, then if Ivasyk chooses $m$, then Petryk erases $100-m$. Since choice of number $50$ leads to loss, then Petryk wins, because if Ivasyk hasn't lost yet then Petryk has a number to erase that will not make him lose.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45565,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the quadrilateral $ABCD$ point $E$ is the middle of the side $AB$, point $F$ is the middle of the side $BC$, point $G$ is the middle of the side $AD$. Turns out that the segment $GE$ is perpendicular to $AB$, and segment $GF$ is perpendicular to $BC$. Find the angle $GCD$ given $∠ADC = 70°$.",
    "options": [],
    "answer": "70°",
    "solution": "Since the segment $GE$ is both altitude and median in $ΔABG$ (Fig. 2), this triangle is isosceles, therefore $AG = GB$. Analogously, $ΔCBG$ is isosceles, hence $BG = GC$. Moreover, from the statement of the problem, point $G$ is the middle of the segment $AD$. Hence, we obtain that $DG = AG = BG = CG$, i.e. $ΔCGD$ is also isosceles, which means $∠GCD = ∠GDC = 70°$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45566,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 30 cities in the country, some of them are connected by flights. The total amount of flights satisfies the following: if one does not consider any 26 cities with all flights that connect one of these cities and any other, then one can get from any of the 4 cities that is left to any other city of the four, maybe with layovers, only using flights that are left. Determine the smallest amount of flights for which the condition holds.",
    "options": [],
    "answer": "405",
    "solution": "Let $A$ is a vertex with the smallest degree $k(A)$. If $k(A) < 27$, then there exist at least 3 vertices that are not connected with a chosen vertex. Then 3 of those vertices together with $A$\n\nmake a not connected graph, so it contradicts the condition. Thus, the smallest size of graph is $\\frac{1}{2} \\cdot 27 \\cdot 30 = 405$.\nWe want to show that such a graph exists. Call vertices $A_1, A_2, ..., A_{29}, A_{30}$ and $A_1 = A_{31}$. Connect every vertex $A_i$ with 27 vertices, all except $A_{i-1}$ and $A_{i+1}$, $i = 1, 30$. We want to show that such graph satisfies the condition. Suppose by contradiction that, condition is not satisfied for vertices $A_i, A_j, A_k$ and $A_l$, where, $i < j < k < l$, moreover the biggest number of vertices is between $A_i$ and $A_l$. Then there are edges between $A_i \\leftrightarrow A_k, A_i \\leftrightarrow A_l$ and $A_j \\leftrightarrow A_l$, since they are not neighbors. Obtained contradiction finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ are such that $\\frac{7a^3b^3}{a^6-8b^6} = 1$. Determine what the value of $\\frac{a^2-b^2}{a^2+b^2}$ can be.",
    "options": [],
    "answer": "0 or 3/5",
    "solution": "Rewrite given equation as $a^6 - 7a^3b^3 - 8b^6 = 0$, hence $(a^3 + b^3)(a^3 - 8b^3) = 0$. We can show that $x^2 \\pm xy + y^2 > 0$ holds iff $x, y \\neq 0$, since\n$$\nx^2 \\pm xy + y^2 = (x \\pm \\frac{1}{2}y)^2 + \\frac{3}{4}y^2 = 0 \\Leftrightarrow y = 0 \\text{ and } x \\pm \\frac{1}{2}y = 0 \\Leftrightarrow x = y = 0.\n$$\n\nHence given equation can be written as $a^3 + b^3 = 0$ or $a^3 - 8b^3 = 0$. And at the same time values $a, b$ cannot both be zero.\n\n**Case 1.** $a^3 + b^3 = 0 \\Rightarrow (a+b)(a^2 - ab + b^2) = 0 \\Rightarrow a = -b$, since $a^2 - ab + b^2 \\neq 0$. Hence\n$$\n\\frac{a^2 - b^2}{a^2 + b^2} = \\frac{b^2 - b^2}{b^2 + b^2} = 0.\n$$\n\n**Case 2.** $a^3 - 8b^3 = 0 \\Rightarrow (a - 2b)(a^2 + 2ab + 4b^2) = 0 \\Rightarrow a = 2b$, since $a^2 + 2ab + 4b^2 \\neq 0$. Hence\n$$\n\\frac{a^2 - b^2}{a^2 + b^2} = \\frac{4b^2 - b^2}{4b^2 + b^2} = \\frac{3}{5}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Capablanca and Alyokhin decided to play a match of 16 games according to the following rules. The winner of the first game received $1 = 3^0$ peso, the winner of the second one got $3 = 3^1$ peso, the winner of the third game received $9 = 3^2$ peso and so on.\n\nIf the game ended in a draw, then they split the prize pool of the game in half. It turned out that at the end of the match, Alyokhin earned 2018 pesos more than Capablanca. How many games has each of the players won?",
    "options": [],
    "answer": "Each player won 3 games.",
    "solution": "Without loss of generality, we can say that after each draw they both got 0 peso. Then Alyokhin in $k^{th}$ game could get $a_k \\cdot 3^{k-1}$, where $a_k \\in \\{-1; 0; 1\\}$. Let us show that his gain after the $k^{th}$ game could be from $A_k = 3^0 + 3^1 + \\dots + 3^{k-1}$ to $(-A_k)$. Moreover, each possible result is yielded by the unique variant of the results of the games. Indeed, for $k=1, 2$ everything is obvious from the simple search. Suppose after $k$ games he could obtain by one variant from $(-A_k)$ to $A_k$. Let us add to each of the results the result of the $(k+1)^{th}$ game. If it was a draw, then no result would change. If Alyokhin wins, he will get all the results from $3^k - A_k$ to $3^k + A_k$.\n\n$$\n3^k - A_k = 3^k - 3^{k-1} - 3^{k-2} - \\dots - 3 - 1 = A_k + 1 = 3^{k-1} + 3^{k-2} + \\dots + 3 + 1 + 1 \\Leftrightarrow\n$$\n$$\n3^k = 2 \\cdot 3^{k-1} + \\dots + 2 \\cdot 3^2 + 2 \\cdot 3 + 2 \\cdot 1 + 1 = 2 \\cdot 3^{k-1} + \\dots + 2 \\cdot 3^2 + 2 \\cdot 3 + 3 =\n$$\n$$\n3^k = 2 \\cdot 3^{k-1} + \\dots + 2 \\cdot 3^3 + 3 \\cdot 3^2 = 2 \\cdot 3^{k-1} + \\dots + 3 \\cdot 3^3 = 2 \\cdot 3^{k-1} + 3 \\cdot 3^{k-2} = 3^k,\n$$\nThen the result after $(k+1)^{th}$ game will also will every value from $(-A_{k+1})$ to $A_{k+1}$, and each result follows from a unique combination of results of the games.\n\nNow we can create a table of the results, which Alyokhin can achieve:\n1 game: $[-1; 1]$;\n3 game: $[-13; 13]$;\n5 game: $[-161; 161]$;\n7 game: $[-1133; 1133]$;\n2 game: $[-4; 4]$;\n4 game: $[-40; 40]$;\n6 game: $[-404; 404]$;\n8 game: $[-3320; 3320]$.\n\nTherefore, after 8 games, 8 other games resulted in a draw. From the obtained ranges it is easy to find the corresponding set of the results:\n$$\n2018 = 2187 - 169 = 2187 - 243 + 74 = 2187 - 243 + 81 - 7 = 2187 - 243 + 81 - 9 + 3 - 1\n$$\n$$\n\\Rightarrow 2018 = 3^7 - 3^5 + 3^4 - 3^2 + 3^1 - 3^0.\n$$\nHence, they both won 3 games each, and two games finished in a draw.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the angles of at least one triangle, one bisector of which is twice bigger than another.",
    "options": [],
    "answer": "36°, 36°, 108°",
    "solution": "Let's find an isosceles triangle. Let's pretend that the vertex angle is obtuse. Then let's denote the base angle as $2\\alpha$ (Fig. 49). Then we can easily find values of some angles (Fig. 49):\n$$\n\\angle ALC = \\pi - 3\\alpha, \\quad \\angle ALB = 3\\alpha, \\quad \\angle ABC = \\pi - 4\\alpha.\n$$\n\nLet's denote the bisection and the sides as follows: $AL = l$, $AD = h$, $AB = b$, then $2h = l$.\nThen $h = b \\sin 2\\alpha$. By the law of sines for $\\triangle ABL$: $\\frac{b}{\\sin 3\\alpha} = \\frac{l}{\\sin 4\\alpha}$. So, $2h = 2b \\sin 2\\alpha = \\frac{b \\sin 4\\alpha}{\\sin 3\\alpha}$\n$\\Rightarrow 2 \\sin 2\\alpha \\sin 3\\alpha = 2 \\sin 2\\alpha \\cos 2\\alpha \\Rightarrow \\sin 3\\alpha = \\cos 2\\alpha$.\nThe answer of this equality is pretty clear: $\\alpha = 18^\\circ$.\n\n**Answer:** $36^\\circ$, $36^\\circ$, $108^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45570,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, perpendicular bisector of the side $AC$ intersects the angle bisector $AK$ in point $P$, $M$ is such point that $\\angle MAC = \\angle PCB$, $\\angle MPA = \\angle CPK$, and points $M$ and $K$ lie on different sides from the segment $AC$. Prove that the line $AK$ divides the segment $BM$ in two equal segments.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the point, symmetrical to $M$ with respect to $AK$ (Fig. 4). Obviously, to prove the statement, it suffices to prove that $BT \\parallel AK$. Notice that points $C, P, T$ lie on the same line. Also,\n$$\n\\angle TAB = \\angle TAK - \\angle BAK = \\angle MAK - \\angle KAC = \\angle CAM = \\angle TCB,\n$$\nWhich means the quadrilateral $BCAT$ is inscribed, and\n$$\n\\angle TBA = \\angle TCA = \\angle PAC = \\angle PAB,\n$$\nwhich proves the parallelism.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45571,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive numbers $x$, $y$, $z$ prove the inequality:\n$$\n2 \\cdot \\left( \\frac{x}{2x+y} \\right)^2 + 2 \\cdot \\left( \\frac{y}{2y+z} \\right)^2 + 2 \\cdot \\left( \\frac{z}{2z+x} \\right)^2 + \\frac{9xyz}{(2x+y)(2y+z)(2z+x)} \\le 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We make a substitution: $a = \\frac{2x}{2x+y}$, $b = \\frac{2y}{2y+z}$, $c = \\frac{2z}{2z+x}$, then the inequality will be as follows:\n$$\n2 \\cdot \\left(\\frac{x}{2x+y}\\right)^2 + 2 \\cdot \\left(\\frac{y}{2y+z}\\right)^2 + 2 \\cdot \\left(\\frac{z}{2z+x}\\right)^2 + \\frac{9xyz}{(2x+y)(2y+z)(2z+x)} = \\frac{a^2+b^2+c^2}{2} + \\frac{9}{8}abc \\le 1, \\text{ or}\n$$\n$$\na^2 + b^2 + c^2 + \\frac{9}{4}abc \\le 2.\n$$\n\nConsider the expression:\n$$\n\\begin{aligned} \\left(\\frac{1}{a}-1\\right)\\left(\\frac{1}{b}-1\\right)\\left(\\frac{1}{c}-1\\right) &= \\left(\\frac{2x+y}{2x}-1\\right)\\left(\\frac{2y+z}{2y}-1\\right)\\left(\\frac{2z+x}{2z}-1\\right) = \\frac{y}{2x} \\cdot \\frac{z}{2y} \\cdot \\frac{x}{2z} = \\frac{1}{8}, \\text{ or} \\\\ \\frac{(1-a)}{a} \\cdot \\frac{(1-b)}{b} \\cdot \\frac{(1-c)}{c} &= \\frac{1}{8} \\Leftrightarrow 8(1-a)(1-b)(1-c) = abc \\Leftrightarrow 8(1-a)(1-b)(1-c) = abc \\\\ &\\Leftrightarrow 8-8(a+b+c)+8(ab+bc+ca)-8abc = abc \\Rightarrow \\\\ &\\frac{9}{4}abc = 2-2(a+b+c)+2(ab+bc+ca). \\end{aligned}\n$$\nContinue the proof using the obtained conditions:\n$$\n\\begin{aligned} & a^2 + b^2 + c^2 + \\frac{9}{4}abc - 2 = \\\\ & = (a+b+c)^2 - 2(ab+bc+ca) + 2 - 2(a+b+c) + 2(ab+bc+ca) - 2 = \\\\ & = (a+b+c)^2 - 2(a+b+c) \\le 0. \\end{aligned}\n$$\n\nThe last inequality holds when $a+b+c \\le 2$.\n\nPerform transformations:\n$$\n\\begin{aligned} a+b+c &= \\frac{2x}{2x+y} + \\frac{2y}{2y+z} + \\frac{2z}{2z+x} = 1 - \\frac{y}{2x+y} + 1 - \\frac{z}{2y+z} + 1 - \\frac{x}{2z+x} = 2 + \\left(1 - \\frac{y}{2x+y} - \\frac{z}{2y+z} - \\frac{x}{2z+x}\\right) \\le 2 \\\\ &\\Leftrightarrow \\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x} \\ge 1. \\end{aligned}\n$$\n\nProve the last inequality using Cauchy–Schwarz inequality:\n$$\n\\left( \\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x} \\right) (x+y+z)^2 = \\left( \\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x} \\right) (y(2x+y) + z(2y+z) + x(2z+x)) \\ge\n$$\n$$\n\\geq \\left( \\sqrt{\\frac{y}{2x+y}} \\cdot \\sqrt{y(2x+y)} + \\sqrt{\\frac{z}{2y+z}} \\cdot \\sqrt{z(2y+z)} + \\sqrt{\\frac{x}{2z+x}} \\cdot \\sqrt{x(2z+x)} \\right)^2 = (x+y+z)^2,\n$$\nQ.E.D.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45572,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive integers $m$ and $n$, compare the numbers $A = m^{525} + n^{525}$ and $B = (m+n)(m^2+n^2)(m^4+n^4)(m^8+n^8)\\dots(m^{128}+n^{128})$.",
    "options": [],
    "answer": "B > A only when both integers equal one; in all other cases A > B.",
    "solution": "If $m = n$, then $A = 2m^{525}$, $B = 2m \\cdot 2m^2 \\cdot 2m^4 \\dots \\cdot 2m^{128} = 2^8 m^{255} > A = 2m^{525}$\n$\\Leftrightarrow m = n = 1$, while for $m = n > 1$ we get $A > B$.\n\nNow, without the loss of generality, $m > n$. Then we can do the transformations:\n$$\n\\begin{aligned}\nB \\le (m-n)B &= (m-n)(m+n)(m^2+n^2)(m^4+n^4)\\dots(m^{128}+n^{128}) = \\\\\n&= (m^2-n^2)(m^2+n^2)(m^4+n^4)\\dots(m^{128}+n^{128}) = (m^4-n^4)(m^4+n^4)\\dots(m^{128}+n^{128}) = \\\\\n&= m^{256}-n^{256} < m^{525} + n^{525} = A \\Rightarrow B < A.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is circle $\\Gamma$ with center in point $O$ and diameter $AB$. $OBDE$ is a square, $F$ is the second point of intersection of $AD$ and circle $\\Gamma$, $C$ is the middle of the segment $AF$. Find the value of the angle $OCB$.",
    "options": [],
    "answer": "45°",
    "solution": "Since $AB$ is a diameter, then $\\angle AFB = 90^\\circ$, and $CO \\parallel FB$ as the middle segment. Therefore, $CO \\perp CD$ and quadrilateral $OBDC$ is inscribed. Hence, $\\angle OCB = \\angle ODB = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y, z$ be positive real numbers such that $x + y + z = \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}$. Prove that $xy + yz + zx \\ge 3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using this equation and the inequality between the arithmetic mean and the geometric mean, we have\n$$\nxy + yz + zx = xyz\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) = \\frac{xyz\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right)^2}{x + y + z} = \\frac{xyz\\left(\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}\\right) + 2x + 2y + 2z}{x + y + z} \\ge \\frac{xyz\\left(\\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx}\\right)}{x + y + z} + 2 = 3\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that following inequality is true for any triangle:\n$$\n2R + (3\\sqrt{3} - 4)r \\ge p,\n$$\n\nwhere $R$ and $r$ are the radii of the circumcircle and incircle of a triangle, and $p$ is the semiperimeter of the triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's use $u$, $v$, $w$ - method. You can get more information about this method in numerous articles, e.g. ($x = p - a$, $y = p - b$, $z = p - c$):\n$$\nR = \\frac{abc}{4S} = \\frac{(x+y)(y+z)(z+x)}{4\\sqrt{xyz}(x+y+z)} \\quad \\text{and} \\quad r = \\frac{S}{p} = \\sqrt{\\frac{xyz}{x+y+z}}\n$$\n\nThen we have to prove equivalent inequality:\n$$\n(x+y)(y+z)(z+x) + 2(3\\sqrt{3}-4)xyz \\geq 2(x+y+z)\\sqrt{xyz}(x+y+z).\n$$\nLet's denote:\n$$\n\\begin{cases}\nx + y + z = 3u, \\\\\nxy + yz + zx = 3v^2, \\quad \\text{Then } 9uv^2 + (6\\sqrt{3}-9)w^3 \\ge 6\\sqrt{3}\\sqrt{uw^3}u \\\\\nxyz = w^3.\n\\end{cases}\n$$\n\n$$\n(3uv^2 + (2\\sqrt{3}-3)w^3)^2 \\ge 12u^3w^3 \\Leftrightarrow \\\\\nf(w^3) = (2\\sqrt{3}-3)^2w^6 + w^3(6(2\\sqrt{3}-3)uv^2 - 12u^3) + 9u^2v^4 \\ge 0.\n$$\n\nIf $u$, $v^2$ are fixed then $f$ is convex. That's why $f$ has a minimum value when 2 variables $\\{x, y, z\\}$ coincide. Without loss of generality $y = z$. Let's check our inequality. It's equivalent to the following:\n$$\n(x + y)((2x + y) - xy) + 2(3\\sqrt{3} - 4)xy \\geq 2\\sqrt{x(x + 2y)^3}.\n$$\nLet's denote: $t = \\frac{x}{y} \\Rightarrow$\n$$\n(t+2)(t+1) + 2(3\\sqrt{3}-4)t \\ge 2\\sqrt{t(t+2)^3} \\Leftrightarrow t^2 + 3(3\\sqrt{3}-2)t + 1 \\ge \\sqrt{t(t+2)^3} \\Leftrightarrow \\\\\nt^3(6\\sqrt{3}-10) + t^2(21-12\\sqrt{3}) + t(6\\sqrt{3}-12) + 1 \\ge 0 \\Leftrightarrow (t-1)^2((6\\sqrt{3}-10)t+1) \\ge 0.\n$$\nWhat was to be demonstrated. The equation is true for $x = y = z \\Leftrightarrow a = b = c$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45576,
    "subject": "Mathematics (Multi-modal)",
    "question": "All the diagonals of the lateral faces of an $n$-sided prism are colored in yellow and blue colors, so that single-colored diagonals do not have common points. Prove that the sum of the squares of lengths of all yellow diagonals is equal to the sum of the squares of lengths of the blue ones.\n(Oleg Kryzhanovsky.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n**Fig. 50**\nAccording to the condition, single-colored diagonals don't have common points. That's why without loss of generality let's consider that the following diagonals are colored in blue: $B_1A_2, B_2A_3, \\ldots, B_{n-1}A_n, B_nA_1$, the others are colored in yellow. As $\\overrightarrow{B_1A_2} = \\vec{h} + \\overrightarrow{A_1A_2}$, so $(\\overrightarrow{B_1A_2})^2 = (\\vec{h})^2 + (\\overrightarrow{A_1A_2})^2 + 2 \\cdot \\vec{h} \\cdot \\overrightarrow{A_1A_2}$. Similarly, $(\\overrightarrow{B_2A_3})^2 = (\\vec{h})^2 + (\\overrightarrow{A_2A_3})^2 + 2 \\cdot \\vec{h} \\cdot \\overrightarrow{A_2A_3}, \\ldots, (\\overrightarrow{B_{n-1}A_n})^2 = (\\vec{h})^2 + (\\overrightarrow{A_{n-1}A_n})^2 + 2 \\cdot \\vec{h} \\cdot \\overrightarrow{A_{n-1}A_n}$. Let's sum up our equalities:\n\n$$\n\\begin{aligned}\nB_1 A_2^2 + \\dots + B_n A_1^2 &= (\\overrightarrow{B_1 A_2})^2 + \\dots + (\\overrightarrow{B_n A_1})^2 = \\\\\n&= n(\\vec{h})^2 + (\\overrightarrow{A_1 A_2})^2 + \\dots + (\\overrightarrow{A_n A_1})^2 + 2 \\cdot \\vec{h} \\cdot (\\overrightarrow{A_1 A_2} + \\dots + \\overrightarrow{A_n A_1}) = nh^2 + A_1 A_2^2 + \\dots + A_n A_1^2\n\\end{aligned}\n$$\nIf you make similar calculations for yellow diagonals, then you'll get the same expression.\nWhat was to be demonstrated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45577,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known, that for some value $a$ the equality: $a^4 - \\frac{1}{a^2} = 4$ is true. Is it possible, that number $x = a^4 + \\frac{1}{a^2}$ is integer?",
    "options": [],
    "answer": "No",
    "solution": "If we add these two equalities, we have equality $4 + x = 2a^4$, if we subtract, $x - 4 = \\frac{2}{a^2}$. Then we have, that $(x+4)(x-4)^2 = \\frac{4}{a^4} \\cdot 2a^4 = 8$. As we are interested only in integer $x$, each of expressions $x+4$ and $x-4$ has to be integer. As these two numbers are of the same parity and $(x-4)^2$ is a square of an integer, equalities $x+4=2$ and $(x-4)^2 = 4$ must be true. But it is impossible for both to be true at the same time. Thus, there is no such value $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is there a rectangle, that can be cut into 5 squares, among which there are two same ones, and all the other ones are distinct and differ from the same ones, and the same squares are\n![](attached_image_1.png)\nFig. 33\na) the smallest of all;\nb) the biggest of all?",
    "options": [],
    "answer": "a) yes; b) no",
    "solution": "a) Let's show the corresponding example (Fig. 32).\n\nb) Consider a square of the smallest size. It can not be on the border of the rectangle (Fig. 33), as well as it can not be inside of the border of one of the other squares. This means, that only the following arrangement of the other four squares around the smallest one (Fig. 34). But then, if $AB$ is a side of the biggest square, then for the point $B$ not to be a vertex of the outer rectangle, it has to be inside the side, but then there must be a square with a side $CB > AB$ – this contradiction ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45579,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the biggest three-digit number $n$, for which the following condition is true: there exist exactly 16 pairs of natural numbers $(a, b)$, where $a < b$, for which $n$ is the least common multiple.",
    "options": [],
    "answer": "992",
    "solution": "At first suppose, that number $n = p_1^{m_1} \\dots p_k^{m_k}$, where $p_i$, $i = 1, \\overline{k}$ – are distinct prime numbers and $m_i$ – are natural numbers, $i = 1, \\overline{k}$. Then let's choose all ordered pairs of numbers $(a, b)$, for which $[a, b] = n$. Here we do not have condition $a < b$, so pairs $(a, b)$ and $(b, a)$ for $a \\neq b$ are considered different. If $a = p_1^{a_1} \\dots p_k^{a_k}$ and $b = p_1^{b_1} \\dots p_k^{b_k}$, so for satisfying condition $[a, b] = n$ it is necessary and sufficient that $\\forall i = 1, \\overline{k} \\max\\{a_i, b_i\\} = m_i$. Then possible variants of pairs $(a_i, b_i)$:\n$(0, m_i), (1, m_i), \\dots, (m_i - 1, m_i), (m_i, m_i), (m_i, m_i - 1), (m_i, m_i - 2), \\dots, (m_i, 0).$\nIn total there are $2m_i + 1$ variants. So, in total there will be exactly $N = (2m_1 + 1) \\dots (2m_k + 1)$ ordered pairs of numbers. And there are exactly $\\frac{1}{2}(N-1)$ pairs, for which $a < b$. Really, for all computed pairs we need to delete pair $(n, n)$, and for every two pairs $(a, b)$ and $(b, a)$ we need to leave one. So, we need to find the biggest three-digit natural number, for which $\\frac{1}{2}(N-1) = 16$.\n\nThen $N = 33$, so there are two possible variants: $N = 2 \\cdot 16 + 1$ or $N = (2 \\cdot 5 + 1)(2 \\cdot 1 + 1)$. In first case, for some prime $p$ condition is true $n = p^{16} \\ge 2^{16} > 999$. In second case, we have, that $n = p^5 q$, where $p, q$ – are distinct prime numbers. As $5^5 > 999$, then $p = 2$ or $p = 3$.\nFor $p=3$ we have, that $3^5 = 243$, so the number can only be $3^5 \\cdot 2 = 486$, as $3^5 \\cdot 5 > 999$.\nFor $p=2$ we have, that $2^5 = 32$, so the number can be maximum three-digit number of form: $2^5 \\cdot p$, where $p > 2$ – is prime. As $1000: 2^5 = 31.25$, it is not hard to see, that the number is $2^5 \\cdot 31 = 992$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy and Olesya write a natural number each on a chalkboard. It turns out, that number, written by Olesya, has sum of digits $2018$ and has precisely $1$ digit less than Andriy's number. It is also known, that difference of numbers, written by him, equals to one-digit number. What can be the number, written by Andriy?",
    "options": [],
    "answer": "10^225 or 10^225 + 1",
    "solution": "It is not hard to see, that Andriy's number can only be $\\overline{100...0a}$, and Olesya's – only: $\\overline{99...9b}$. Otherwise, the difference will not be a one-digit number. Really, if not all Olesya's digits, except for the last, are $9$, then after adding a one-digit number, the number of digits will not change. So this is the presentation of Andriy's number. The sum of digits of Olesya's number equals $2018$, so it is $\\overline{99...92}$ (as $2018 = 224 \\cdot 9 + 2$). So Andriy's number has to have the last digit less than $2$, because otherwise the difference of written numbers will not be less than $10$. So, this number can be $0$ or $1$. So, he wrote number $\\overline{100...0}$, or $\\overline{100...01}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45581,
    "subject": "Mathematics (Multi-modal)",
    "question": "On sides $AB$, $BC$, $AC$ of triangle $ABC$ with $\\angle BAC = 120°$ there are points $M$, $K$, $N$ respectively so, that $\\triangle MKN$ is equilateral, and $AM = 2,017$, $AN = 2,018$. Baron Munchausen assures, that $\\triangle MKN$ has the smallest perimeter of all equilateral triangles, that have exactly one vertex on each of sides of $\\triangle ABC$. Is baron right?",
    "options": [],
    "answer": "No",
    "solution": "According to the condition, all the angles of $\\triangle MKN$ equal to $60°$ (Fig. 39). As $\\angle BAC + \\angle MKN = 180°$, quadrilateral $AMKN$ is cyclic, so\n$$\n\\angle KAC = \\angle KMN = 60° = \\angle MNK = \\angle BAK.\n$$\nLet's consider points $M_1$ and $N_1$ – projections of point $K$ on sides $AB$ and $AC$ respectively. So $\\triangle AKM_1 = \\triangle AKN_1$, so $AM_1 = AN_1$. Apart from that, $\\angle M_1KN_1 = 60°$ and $KM_1 = KN_1$. So, $\\triangle KM_1N_1$ is also equilateral. Case $N = N_1$, $M = M_1$ is impossible, as otherwise we would have\n\n$AM = AN$. So $KM_1 < KM$, so perimeter of $\\triangle KM_1N_1$ is strictly less than perimeter of $\\triangle KMN$. So, baron Munchausen is not right.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, x, y$ be non-negative numbers, $k > 0$ and for these numbers the following inequalities are true:\n$$\na + kx \\le 1; \\quad a + ky \\le 1; \\quad b + \\frac{x}{k} \\le 1; \\quad b + \\frac{y}{k} \\le 1.\n$$\n\nProve that the following equation is true:\n$$\n(a + x)^2 + (b + y)^2 \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Edward Turkevich)\n**Answer:** the equation is true when $a = b, x = y = 1 - a, k = 1$.\n\nFrom $a + kx \\le 1$: $x \\le \\frac{1-a}{k}$ and from $b + \\frac{x}{k} \\le 1$: $x \\le k(1-b)$.\nSo, $x^2 \\le (1-a)(1-b)$ and $x \\le \\frac{1}{2}(2-a-b)$ (based on the inequality of arithmetic and geometric means of $1-a$ and $1-b$).\nSimilarly, $y^2 \\le (1-a)(1-b)$ and $y \\le \\frac{1}{2}(2-a-b)$.\n\n$$\n\\begin{aligned}\n(a+x)^2 + (b+y)^2 &= a^2 + 2ax + x^2 + b^2 + 2by + y^2 \\\\ &\\le a^2 + 2a \\cdot \\frac{1}{2}(2-a-b) + (1-a)(1-b) + b^2 + 2b \\cdot \\frac{1}{2}(2-a-b) + (1-a)(1-b) \\\\ &= a^2 + b^2 + (a+b)(2-a-b) + 2(1-a)(1-b) = 2,\n\\end{aligned}\n$$\nWhat was to be proven.\n\nThe equality is true when $1-a=1-b$, so $a=b, x=y=1-a, k=1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45583,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, $H$ is the orthocenter and $AK$ is an altitude. Circle $w$ goes through points $A$ and $K$ and intersects sides $AB$ and $AC$ at points $M$ and $N$ respectively. The line through point $A$ parallel to $BC$ intersects the circumscribed circles of triangles $AHM$ and $AHN$ a second time at points $X$ and $Y$ respectively. Prove that $XY = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Z$ be the point of intersection of circle $w$ with line $BC$, then $AZ$ is the diameter of $w$ (see Fig. 35).\n\nReally, if $K = Z$, $w$ is tangent to $BC$, so, as $AK \\perp BC$, the center of $w$ is on $AK$. If $K \\neq Z$, then $\\angle AKZ = 90^\\circ$ and $AZ$ is the diameter of $w$. Then $\\angle AMZ = 90^\\circ$.\n\nLet line $MZ$ intersect the circumscribed circle of $\\triangle XAH$ a second time at point $S$, then $\\angle AMS = 90^\\circ$. Also $\\angle SHA = \\angle SMA$, then $SH \\perp AH$. So $AX \\parallel SH \\parallel BC$, it also follows that $XAHS$ is a rectangle. So, $AX = SH$.\n\nAlso notice that $CH \\perp AB$. Apart from that, $ZM \\perp AB$. So $HC \\parallel SZ$. Then $SHCZ$ is a parallelogram, and so $SH = ZC$. Then $AX = SZ$. Also, $AY = BZ$, so $XY = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45584,
    "subject": "Mathematics (Multi-modal)",
    "question": "60 participants took part in the Olympiad. They were offered 8 tasks, evaluated from 0 to 7 points each. Prove, that in total there are 3 participants, whose results differ in not more than 1 point. Would the statement be true, if 58 took part in the Olympiad?\n\n*Result of a participant at Olympiad is the total amount of the points he got.*",
    "options": [],
    "answer": "Yes for sixty participants; no for fifty-eight participants. A counterexample for fifty-eight is two participants at each even total from zero to fifty-six.",
    "solution": "Minimal amount of points, that was possible to earn equals to $0$, maximum – to $56$. Consider such segments in points: $[0; 1]$, $[2; 3]$, $[4; 5]$, ..., $[54; 55]$ and $56$ points. If at least 3 students are in at least one of these segments, the statement is proved. If not, then for each of these segments there are not more than two students. There are $29$ segments, so not more than $58$ might have taken part in the Olympiad. This contradiction ends the proof of the first part.\n\nFor the second part. If exactly $2$ participants got each of points $0$, $2$, $4$, ..., $56$. Then there are no three, difference of whose results is not more than $1$ point. And in total there are exactly $58$ participants.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andriy read a big book for a month. He was reading the book according to a schedule: from 1 until 20 April he read in average 20 pages per day, from 6 until 25 April he read in average of 30 pages per day, and from 11 until 30 April he read in average of 40 pages per day. What are maximum and minimum amounts of pages, that this book could contain?",
    "options": [],
    "answer": "maximum 1200, minimum 800",
    "solution": "Let's split April into 5 segments. Let from 1 until 5 April he read in average $x$ pages, from 6 until 10 April – in average $a$ pages, from 11 until 20 April – in average $c$ pages, from 21 until 25 April – in average $b$ pages, from 26 until 30 – in average $y$ pages.\nThen equalities are true:\n$$\n\\begin{gathered}\n\\frac{5x+5a+10c}{20} = 20, \\quad \\frac{5a+10c+5b}{20} = 30, \\quad \\frac{10c+5b+5y}{20} = 40, \\text{ are} \\\\\n x+a+2c=80, \\quad a+2c+b=120, \\quad 2c+b+y=160.\n\\end{gathered}\n$$\n$$\n\\begin{aligned}\nS &= 5x + 5a + 10c + 5b + 5y = 5(x + a + 2c + b + y) = \\\\\n&= 5((x + a + 2c) + (2c + b + y) - 2c) = 5(240 - 2c).\n\\end{aligned}\n$$\nSo, the book has the maximum amount of pages when value $2c$ is the least, and minimal amount of pages, when $2c$ is the biggest.\n\nCase of maximum $S$, or minimal $2c$. As $c \\ge 0$, let's try to find possible values of variables for $c = 0$. So we have conditions:\n$$\nx+a=80, \\quad a+b=120, \\quad b+y=160.\n$$\nOne of possible cases, when $a=0$, then $x=80$, $b=120$ and $y=40$. Then maximum amount of pages equals to $S_{\\max} = 1200$.\n\nCase of minimal $S$, or maximum $2c$. As $x+a+2c=80$, then $2c \\le 80$. Let's try to find possible values of variables for $c = 40$. Then we have conditions:\n$$\nx+a=0, \\quad a+b=40, \\quad b+y=80.\n$$\nOne of possible cases, when $a = x = 0$, then $b = y = 40$. Then minimal amount of pages equals to $S_{\\min} = 800$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $1 < d_1 < d_2 < \\ldots < d_{l-1} < d_l < n$ denote divisors of composite natural number $n$, that is not a square of prime number, $l \\ge 2$. For which $n$ are there natural numbers $a$, $b$ and $N$, that satisfy the conditions: $d_1 + d_2 = N^a$ and $d_{l-1} + d_l = N^b$?",
    "options": [],
    "answer": "Exactly the following n: (i) n has exactly two proper divisors greater than one, i.e., n is either the product of two primes or the cube of a prime (then taking both exponents equal to one works); and (ii) for any pair of twin primes p and p plus two and any integer b at least one, n = 2 · p · (p + 2)^(b − 1) with N = p + 2 and a = 1 (this also includes the case b = 1 giving n = 2p). No other n satisfy the conditions.",
    "solution": "Notice, that from conditions of the task, it follows, that $d_1 \\cdot d_l = d_2 \\cdot d_{l-1} = n$. Using this equality and first equality from conditions, it is possible to rewrite the second equality from the task:\n$$\nN^b = d_{l-1} + d_l = \\frac{n}{d_2} + \\frac{n}{d_l} = \\frac{n(d_1+d_2)}{d_1d_2} = \\frac{nN^a}{d_1d_2} \\text{ or } n = d_1d_2N^{b-a}.\n$$\n\nApart from that, if $b=a$, then $d_1+d_2 = N^a = N^b = d_{l-1} + d_l$, so $d_1 = d_{l-1}$; $d_2 = d_l$. Then the number needed has four divisors, or exactly two proper divisors. There are two forms of such numbers: $n = pq$, where $p, q$ are prime numbers, or $n = p^3$, where $p$ is prime. So, we can consider only case $b > a$.\n\nConsider two cases.\n\n1. Let $N$ be odd number. Then from equality $d_1 + d_2 = N^a$ we have, that sum $d_1 + d_2$ is odd number, then from these divisors there is exactly one even, another is odd. So it is obvious, that $d_1 = 2$, as it is the least even divisor. Then $d_2 = N^a - 2$ must be prime number. Let $d_2 = N^a - 2 = p$, then $N^a = p + 2$. So, $n = 2 \\cdot p \\cdot N^{b-a}$.\n\na) If $N$ is not prime number, then as $N \\le p+2$, it has prime divisor $q \\le p+2$. Obviously, $q \\ne p$, and so $q < p$. But this means, that for $b > a$ number $n = 2 \\cdot p \\cdot N^{b-a}$ is divisible by $q < p = d_2$, and then number $p$ can not be $d_2$. We get contradiction.\n\nb) If $N$ is prime number, then for $a > 1$ we get, that it follows from equality $N^a = p+2$, that $N < p+2$, then it is proper divisor of number $n$, less than $p$, that is impossible. So, $a=1$, i.e. numbers $p$ and $N = p+2$ is pair of twin primes. And then $\\forall b \\ge 1$ we have, that number $n = 2 \\cdot p \\cdot (p+2)^{b-1}$ is solution. Really, $d_{l-1} = 2(p+2)^{b-1}$, $d_l = p(p+2)^{b-1}$ and $d_{l-1} + d_l = 2(p+2)^{b-1} + p(p+2)^{b-1} = (p+2)^b = N^b$.\n\n2. Let $N$ be even number. As $n = d_1d_2N^{a-b}$, then $n$ is also even, i.e. $d_1 = 2$.\nThen as $d_1 = 2$, then $d_2 = 4$, as sum of two the least divisors is even. Then $N^a = 6$. Then $N = 6$, $a = 1$, thus $n = 2 \\cdot 4 \\cdot 6^{b-1}$. If $b > a = 1$, then number 3 is divisor of number $n$, then $d_2 = 3$ – contradiction. Thus $b = a = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45587,
    "subject": "Mathematics (Multi-modal)",
    "question": "For natural number $n$ let $S(n)$ denote sum of its digits. From all pairs of natural numbers $(n, m)$, that satisfy equality $S(n) \\cdot S(n+1) \\cdots S(n+m) = 2018$, find such, for which sum $n+m$ takes the least possible value.",
    "options": [],
    "answer": "n = 1 followed by 112 nines; m = 1",
    "solution": "Numbers $S(n)$ and $S(n+1)$ have the following ratios: $S(n+1) = S(n)+1$, or $S(n) > S(n+1)$, and the last inequality can not be true for two numbers in a row. As $2018 = 2018 \\cdot 1 = 1009 \\cdot 2 = 1009 \\cdot 1 \\cdot 2$, and there are no other factor decompositions, that satisfy the condition, so the following cases are possible:\n* $m=1$, $S(n)=2018$ and $S(n+1)=1$;\n* $m=1$, $S(n)=1009$ and $S(n+1)=2$;\n* $m=2$, $S(n)=1009$, $S(n+1)=1$ and $S(n+2)=2$.\n\nIf $S(n)=1009$ and $S(n+1)=1$, so we get contradiction as in the previous case. If $S(n)=1009$ and $S(n+1)=2$, so $n+1=100\\dots0100\\dots0$ or $n+1=200\\dots0$. Then $n=100\\dots0099\\dots9$ and $n=199\\dots9$. As $1009=112\\cdot9+1$, so these variants are possible, and number $n$ has to have in the end 112 digits 9. The least value for $n$ is $n=\\underbrace{199\\dots9}_{112}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45588,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ there are points $M_1, M_2, M_3$ – midpoints of sides $BC$, $AC$, $AB$ correspondingly. Point $K$ is symmetrical to $M_2$ with respect to line $BC$, $AH$ is the altitude of triangle $ABC$. Prove that line $KM_3$ splits the segment $HM_1$ in half.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From right-angled triangles $ABH$ and $ACH$ we get that $M_3H = \\frac{1}{2}AB$, $M_2H = \\frac{1}{2}AC$ (Fig. 42).\n\nAlso $M_1M_2 = \\frac{1}{2}AB$, $M_1M_3 = \\frac{1}{2}AC$ as centerlines. From symmetry $M_2M_1 = M_1K$ and $M_2H = HK$. So $M_3H = M_1K$ and $M_3M_1 = HK$, so quadrilateral $HM_3M_1K$ is a parallelogram, so line $KM_3$ splits segment $HM_1$ in half.\n\n![](attached_image_1.png)\n**Fig. 42**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45589,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive numbers $x$, $y$, $z$ prove inequality:\n$$\n\\frac{1}{3}(x^3 + y^3 + z^3) \\geq xyz + \\frac{2}{9}(x + y + z)(x - z)^2 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From known identity\n$$\nx^3 + y^3 + z^3 - 3xyz = (x + y + z) \\cdot \\frac{(x-y)^2 + (y-z)^2 + (z-x)^2}{2} .\n$$\n\nFrom inequality between $n$ arithmetic mean and root-means-square we have, that\n$$\n\\begin{aligned} (x - y)^2 + (y - z)^2 + (z - x)^2 &\\ge \\frac{1}{3}(|x - y| + |y - z| + |z - x|)^2 \\ge \\\\ &\\ge \\frac{1}{3}(|x - y| + |y - z| + |x - z|)^2 = \\frac{4}{3}(x - z)^2. \\end{aligned}\n$$\n\nFrom this follows the inequality needed.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Rounded tower has 16 doors, behind each there is a chest with gold of captain Flint. These doors are at equal distances to the neighboring ones, and are numbered clockwise from 1 to 16. 16 pirates come to the tower, each having a key, all keys are numbered from 1 to 16. It is known that key with number $n$ opens doors with number $m$ if and only if $m \\neq n$. Pirates stand one next to each door, but they do not know the number of the door they stand in front of. Jim Hawkins knows which pirate has which key and wants them to take as small an amount of chests with gold as possible. Jim can turn the tower so that doors are situated in front of pirates as he wants – but still all the numbers go clockwise 1-16 starting from some door. What is the maximum amount of chests with gold pirates can definitely take in such conditions?\n\n(Bogdan Rublyov)",
    "options": [],
    "answer": "3",
    "solution": "Consider some arrangement of keys. Then for every column there is a corresponding key. Paint gray all the cells of the column that can be opened by the corresponding key. In the column where there is key 2, 8 cells will be painted, where there is key 5 – 3 cells. In total there will be painted\n\n$$\n16 + 8 + 5 + 4 + 3 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 50 \\text{ cells.}\n$$\n\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |\n|---|---|---|---|---|---|---|---|---|----|----|----|----|----|----|----|\n| 16| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9  | 10 | 11 | 12 | 13 | 14 | 15 |\n| 15| 16| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8  | 9  | 10 | 11 | 12 | 13 | 14 |\n| 14| 15| 16| 1 | 2 | 3 | 4 | 5 | 6 | 7  | 8  | 9  | 10 | 11 | 12 | 13 |\n| 13| 14| 15| 16| 1 | 2 | 3 | 4 | 5 | 6  | 7  | 8  | 9  | 10 | 11 | 12 |\n| 12| 13| 14| 15| 16| 1 | 2 | 3 | 4 | 5  | 6  | 7  | 8  | 9  | 10 | 11 |\n| 11| 12| 13| 14| 15| 16| 1 | 2 | 3 | 4  | 5  | 6  | 7  | 8  | 9  | 10 |\n| 10| 11| 12| 13| 14| 15| 16| 1 | 2 | 3  | 4  | 5  | 6  | 7  | 8  | 9  |\n| 9 | 10| 11| 12| 13| 14| 15| 16| 1 | 2  | 3  | 4  | 5  | 6  | 7  | 8  |\n| 8 | 9 | 10| 11| 12| 13| 14| 15| 16| 1  | 2  | 3  | 4  | 5  | 6  | 7  |\n| 7 | 8 | 9 | 10| 11| 12| 13| 14| 15| 16 | 1  | 2  | 3  | 4  | 5  | 6  |\n| 6 | 7 | 8 | 9 | 10| 11| 12| 13| 14| 15 | 16 | 1  | 2  | 3  | 4  | 5  |\n| 5 | 6 | 7 | 8 | 9 | 10| 11| 12| 13| 14 | 15 | 16 | 1  | 2  | 3  | 4  |\n| 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12| 13 | 14 | 15 | 16 | 1  | 2  | 3  |\n| 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12 | 13 | 14 | 15 | 16 | 1  | 2  |\n| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11 | 12 | 13 | 14 | 15 | 16 | 1  |\n\n**Fig. 43**\n\nAs there are exactly 16 rows, so according to the Dirichlet's drawer principle there will be such that has not more than three gray cells. Then Jim can turn the tower in a way that corresponds to this row. In this case pirates can open at maximum 3 doors.\n\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |\n|---|---|---|---|---|---|---|---|---|----|----|----|----|----|----|----|\n| 16| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9  | 10 | 11 | 12 | 13 | 14 | 15 |\n| 15| 16| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8  | 9  | 10 | 11 | 12 | 13 | 14 |\n| 14| 15| 16| 1 | 2 | 3 | 4 | 5 | 6 | 7  | 8  | 9  | 10 | 11 | 12 | 13 |\n| 13| 14| 15| 16| 1 | 2 | 3 | 4 | 5 | 6  | 7  | 8  | 9  | 10 | 11 | 12 |\n| 12| 13| 14| 15| 16| 1 | 2 | 3 | 4 | 5  | 6  | 7  | 8  | 9  | 10 | 11 |\n| 11| 12| 13| 14| 15| 16| 1 | 2 | 3 | 4  | 5  | 6  | 7  | 8  | 9  | 10 |\n| 10| 11| 12| 13| 14| 15| 16| 1 | 2 | 3  | 4  | 5  | 6  | 7  | 8  | 9  |\n| 9 | 10| 11| 12| 13| 14| 15| 16| 1 | 2  | 3  | 4  | 5  | 6  | 7  | 8  |\n| 8 | 9 | 10| 11| 12| 13| 14| 15| 16| 1  | 2  | 3  | 4  | 5  | 6  | 7  |\n| 7 | 8 | 9 | 10| 11| 12| 13| 14| 15| 16 | 1  | 2  | 3  | 4  | 5  | 6  |\n| 6 | 7 | 8 | 9 | 10| 11| 12| 13| 14| 15 | 16 | 1  | 2  | 3  | 4  | 5  |\n| 5 | 6 | 7 | 8 | 9 | 10| 11| 12| 13| 14 | 15 | 16 | 1  | 2  | 3  | 4  |\n| 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12| 13 | 14 | 15 | 16 | 1  | 2  | 3  |\n| 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12 | 13 | 14 | 15 | 16 | 1  | 2  |\n| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11 | 12 | 13 | 14 | 15 | 16 | 1  |\n| No of key | 2 | 4 | 12 | 8 | 3 | 6 | 1 | 16 | 14 | 9 | 7 |   | 5 |   | 15 | 11 |\n\n**Fig. 44**\n\nIf pirates stand with the keys, as shown in the last row of Fig. 44, they can always open at least three locks.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does the triangle, sides of which can be expressed as a positive integer in centimeters and 2 of its medians are perpendicular, exist?",
    "options": [],
    "answer": "Yes; for example, a triangle with sides 13 cm, 19 cm, and 22 cm has two perpendicular medians.",
    "solution": "Let's find the condition on the sides of the triangle, under which the medians are perpendicular. Without loss of generality let's consider that medians $AM$ (from the edge $A$) and $BN$ (from the edge $B$) are perpendicular. Let's use vectors: $\\vec{c} = \\overrightarrow{AB}$, $\\vec{b} = \\overrightarrow{AC}$. Then $\\overrightarrow{AM} = \\frac{1}{2}(\\overrightarrow{AB} + \\overrightarrow{AC}) = \\frac{1}{2}(\\vec{c} + \\vec{b})$, $\\overrightarrow{BN} = \\frac{1}{2}(\\overrightarrow{BA} + \\overrightarrow{BC}) = \\frac{1}{2}(2\\overrightarrow{BA} + \\overrightarrow{AC}) = \\frac{1}{2}(\\vec{b} - 2\\vec{c})$. These vectors are perpendicular, so, their dot product is equal to $0$. So,\n$$\n\\begin{gathered}\n0 = (\\overrightarrow{AM}, \\overrightarrow{BN}) = \\frac{1}{4}(\\vec{c} + \\vec{b}, \\vec{b} - 2\\vec{c}) = \\frac{1}{4}(|\\vec{b}|^2 - 2|\\vec{c}|^2 - (\\vec{c}, \\vec{b})), \\quad (*) \\\\\n|\\overrightarrow{BC}|^2 = (\\vec{c} + \\vec{b}, \\vec{c} + \\vec{b}) = |\\vec{c}|^2 + |\\vec{b}|^2 + 2(\\vec{c}, \\vec{b}). \\text{ But then: } (\\vec{c}, \\vec{b}) = \\frac{1}{2}(BC^2 - |\\vec{b}|^2 - 2|\\vec{c}|^2).\n\\end{gathered}\n$$\n\nLet's substitute (*): $0 = |\\vec{b}|^2 - 2|\\vec{c}|^2 - \\frac{1}{2}(BC^2 - |\\vec{b}|^2 - |\\vec{c}|^2)$.\nLet's notice, that $|\\vec{c}|^2 = AB^2, |\\vec{b}|^2 = AC^2$.\n\nSo, we can rewrite the equation like this: $5AB^2 = BC^2 + AC^2$. Let's notice, that this condition is necessary and sufficient for medians to be perpendicular. That's why the only thing that remains is to check if this equation has the solution in integers, that is true for triangle inequality. Let's find $AB$ in this form: $d^2 + e^2$. Then,\n$$\n(1^2 + 2^2)(d^2 + e^2)(d^2 + e^2) = ((d-2e)^2 + (2d+e)^2)(d^2 + e^2) \\\\ = (4de - d^2 + e^2)^2 + (2d^2 + 2de - 2e^2)^2\n$$\nIf $e=2$ and $d=3$, the triangle with the sides $AB=13$, $BC=19$ and $AC=22$ satisfy the condition. So, we've found the triangle, 2 medians of which are perpendicular and the sides are integers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve $\\frac{2 \\cos 2x}{6 - 3 \\cos 3x} = \\frac{\\cos 2x + 1}{\\cos 3x + 2}$ for $-\\pi \\le x \\le \\pi$.",
    "options": [],
    "answer": "0",
    "solution": "The equation can be written as:\n$$\n\\frac{2 \\cos 2x}{\\cos 2x + 1} = \\frac{6 - 3 \\cos 3x}{\\cos 3x + 2}.\n$$\nConsider the following two functions: $f(x) = \\frac{2x}{x+1}$ and $g(x) = \\frac{6-3x}{x+2}$, $x \\in (-1; 1]$. If $x \\in (-1; 1)$, then\n$$\nf(x) = \\frac{2x}{x+1} < 1 \\Leftrightarrow 2x < x+1 \\Leftrightarrow x < 1, \\quad g(x) = \\frac{6-3x}{x+2} > 1 \\Leftrightarrow 6-3x > x+2 \\Leftrightarrow x < 1.\n$$\n\nThus, the equality holds only if $\\cos 2x = \\cos 3x = 1$. A sketch of the graphs of the functions can lead to the same conclusion.\n\nIn order for the last equation to be true, the following has to hold:\n$2x = 2\\pi k$, and $3x = 2\\pi l$, $k, l \\in \\mathbb{Z}$. Due to the first equation $x \\in \\{-\\pi, 0, \\pi\\}$, with the help of the second equation it is easy to check that the only answer is $x = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $\\frac{\\sqrt{x+2}}{\\cos 2x+3} = \\frac{\\sqrt{x+1}}{\\cos 2x+1}$.",
    "options": [],
    "answer": "x = 0",
    "solution": "The equation can be written the following way:\n$$\n\\frac{\\sqrt{x+2}}{\\sqrt{x+1}} = \\frac{\\cos 2x + 3}{\\cos 2x + 1}.\n$$\nConsider two functions: $f(t) = \\frac{t+2}{t+1}$, $t \\ge 0$ and $g(y) = \\frac{y+3}{y+1}$, $y \\in (-1; 1]$.\n\nIf $t \\ge 0$, then $f(t) = \\frac{t+2}{t+1} < 2 \\Leftrightarrow t+2 < 2t+2 \\Leftrightarrow t > 0$. Also, $f(t) = \\frac{t+2}{t+1} = 2$ only if $t = 0$.\n\nIf $y \\in (-1; 1]$, then $g(y) = \\frac{y+3}{y+1} > 2 \\Leftrightarrow y+3 > 2y+2 \\Leftrightarrow y < 1$. Also, $g(1) = \\frac{1+3}{1+1} = 2$ only if $y = 1$.\n\nThus, given equation holds only if $f(\\sqrt{x}) = 2 = g(\\cos 2x)$, that is possible only if $\\sqrt{x} = 0$ and $\\cos 2x = 1$. Therefore, the only solution is $x = 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two cyclists left towns $A$ and $B$, went towards one another with the speeds $v_1$ and $v_2$, where $v_1 \\ge v_2$, and met for the first time after 1 hour. After having met, they both continued their journeys without stopping until their destination town. If one of them reached his/her final point, he/she turned around and went in the opposite direction.\nHow much time after their first meeting did their second meeting take place?\n\n![](attached_image_1.png)\nFig. 22",
    "options": [],
    "answer": "t2 = 2 hours if v1 < 2*v2; otherwise t2 = 2*v2/(v1 - v2) hours",
    "solution": "Until the first meeting point, the first cyclist travelled the distance $S_1 = v_1$, while the second cyclist travelled $S_2 = v_2$. Therefore, the distance between towns $A$ and $B$ equals $v_1 + v_2$. There are two possible cases.\n\nCase 1. Before their second meeting, both cyclists reached their destination town, and turned around. Suppose their second meeting occurred at the distance $S_3$ from town $B$ and time $t_2$ later after the first meeting (fig. 22). Then, we get the equations\n![](attached_image_2.png)\nFig. 23\n$$\nv_2 + S_3 = v_1 t_2 \\text{ and } v_1 + (v_1 + v_2 - S_3) = v_2 t_2.\n$$\nBy adding these equations, we obtain\n$$\nv_2 + S_3 + 2v_1 + v_2 - S_3 = (v_2 + v_1)t_2 \\Rightarrow t_2 = 2.\n$$\n\nCase 2. Before their second meeting, the first cyclist reached destination town $B$, turned around and reached the second cyclist before he/she reached town $A$ (fig. 23). Suppose their second meeting occurred at the distance $S_3$ from town $B$ and time $t_2$ later after the first meeting (fig. 4). Then, we get the equations\n$$\nS_3 - v_2 = v_2 t_2 \\text{ and } v_2 + S_3 = v_1 t_2.\n$$\nTaking the difference of these equations, we obtain\n$$\nv_2 + S_3 + v_2 - S_3 = (v_1 - v_2)t_2 \\Rightarrow t_2 = \\frac{2v_2}{v_1 - v_2}.\n$$\n\nNow, the only thing left is to figure out which of the two cases occurs for which $v_1, v_2$. Case 1 happens when the first cyclist reaches $A$ later than the second cyclist, i.e. $\\frac{v_2+v_2+v_1}{v_1} > \\frac{v_1}{v_2} \\Rightarrow 2v_2^2 + v_1v_2 > v_1^2$. Let us denote $x = \\frac{v_1}{v_2}$, then it must satisfy the equation\n$$\nx^2 - x - 2 < 0 \\Rightarrow (x+1)(x-2) < 0 \\Rightarrow x < 2, \\text{ i.e. } v_1 < 2v_2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(a, b)$, which satisfy the equation:\n$$\nab^3 + a^3 + b + 1 = 2019.\n$$",
    "options": [],
    "answer": "a = 2, b = 10",
    "solution": "**Answer:** $a = 2, b = 10$.\n\nIt can be seen from the problem statement that both unknowns are even, because otherwise we would have an equality of an even and an odd number. Let us denote $a = 2n$ and $b = 2k \\Rightarrow$\n$$\n2n \\cdot 8k^3 + 8n^3 + 2k = 2018 \\text{ or } 8nk^3 + 4n^3 + k = 1009.\n$$\nHere, it can be seen that $k$ must be odd, and now a simple exhaustive search will be enough to find the solution $nk^3 \\le \\frac{1009}{8}$, hence, $k^3 \\le 125$.\nTherefore, there are only three options.\n\n$k = 5 \\Rightarrow 1000n + 4n^3 = 1004 \\Rightarrow n = 1$ satisfies the inequality. For larger values of $n$, the left-hand side becomes larger than $1004$.\n\n$k = 3 \\Rightarrow 64n + 4n^3 = 1006 \\Rightarrow$ no solution exists, since 4 divides the left-hand side but not the right-hand side.\n\n$k = 1 \\Rightarrow 8n + 4n^3 = 1008 \\Rightarrow 2n + n^3 = 252 \\Rightarrow n = 2m \\Rightarrow 4m + 8m^3 = 252 \\Rightarrow m + 2m^3 = 63$. Now, we need to check only three cases for $m$, since, from the last equation, $m \\le 3$. $m = 1, m = 2$ and $m = 3$ do not satisfy the equation.\n\nHence, there is only a single pair $k = 5$ and $n = 1 \\Rightarrow b = 10$ and $a = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45596,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is an equilateral trapezoid with bases $BC$ and $AD$ and known angles: $\\angle BDC = 10^\\circ$ and $\\angle BDA = 70^\\circ$. Prove that the following equality holds true: $AD^2 = BC(AD + AB)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is clear that $\\angle AMD = 20^\\circ$, $\\angle DBM = 150^\\circ$. Let's construct equilateral triangle $\\triangle KMD$, then points $M$, $B$, $D$ are on the circle centered at $K$ (fig. 8.76). After that $KM = KB$, $\\angle KMB = 80^\\circ$, from where $\\angle MBK = 80^\\circ$, but where $\\angle MBC = 80^\\circ$ as well, which implies that points $B$, $C$, $K$ are on the same line.\n\nThen $\\triangle AMD = \\triangle BKM$, as isosceles with equal sides and angles at the base. Therefore $MB = AD$. From similarity of $\\triangle MBC \\sim \\triangle MAD$ we have that $\\frac{BC}{AD} = \\frac{MB}{MA}$ $\\Rightarrow$ $AD \\cdot BM = BC \\cdot MA$, hence we obtain that\n\n![](attached_image_1.png)\nFig. 45\n\n$$\nAD^2 = BC(MB + AB) = BC(AD + AB).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real values of $k$, such that all solutions of the equation $k(2-k)x^2 - (k+4)x + 6 = 0$ are positive integers.",
    "options": [],
    "answer": "k = 1/2, 1, 2",
    "solution": "First, we go through cases when this equation is not quadratic.\nFor $k=0$, equation becomes $-4x+6=0$, and has a non-integer solution.\nFor $k=2$, equation becomes $-6x+6=0$, and has a single solution $x=1$, which satisfies the statement.\n\nNow, suppose $k(2-k) \\neq 0$. We find the discriminant of this quadratic equation:\n$$\nD = (k^2 + 8k + 16) - 24(2k - k^2) = 25k^2 - 40k + 16 = (5k - 4)^2\n$$\nSo the roots are:\n$$\nx_1, x_2 = \\frac{(k+4) \\pm (5k-4)}{2k(2-k)}\n$$\nCalculating both roots:\n$$\nx_1 = \\frac{(k+4) + (5k-4)}{2k(2-k)} = \\frac{6k}{2k(2-k)} = \\frac{3}{2-k}\n$$\n$$\nx_2 = \\frac{(k+4) - (5k-4)}{2k(2-k)} = \\frac{-4k+8}{2k(2-k)} = \\frac{2}{k}\n$$\nWe must find all cases when this is a positive integer. Let $\\frac{2}{k} = m$ be a positive integer, then $k = \\frac{2}{m}$.\n\nThen:\n$$\n\\frac{3}{2-k} = \\frac{3}{2-\\frac{2}{m}} = \\frac{3m}{2(m-1)}\n$$\nWe require $\\frac{3m}{2(m-1)}$ to be a positive integer. Since $m$ is a positive integer, $m$ and $m-1$ are coprime.\nHence, the last condition is only possible for $m-1=1 \\Rightarrow k=1$ or $m-1=3 \\Rightarrow k=\\frac{1}{2}$.\n\nTherefore, the real values of $k$ are $k=2$, $k=1$, and $k=\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find positive integers $a_1, a_2, \\dots, a_{2019}$, which satisfy the equation\n$$\na_1 + a_2 + \\dots + a_{2019} = a_1 a_2 \\dots a_{2019} = \\sqrt[2018]{2019^{2019}}.\n$$",
    "options": [],
    "answer": "No positive integers exist",
    "solution": "From the Arithmetic mean - Geometric mean Inequality,\n$$\n\\frac{1}{2019}(a_1 + a_2 + \\dots + a_{2019}) \\ge \\sqrt[2019]{a_1 a_2 \\dots a_{2019}}, \\text{ or} \\\\\n(a_1 + a_2 + \\dots + a_{2019})^{2019} \\ge 2019^{2019} a_1 a_2 \\dots a_{2019}.\n$$\nFrom problem statement,\n$$\n(a_1 + a_2 + \\dots + a_{2019})^{2019} = (\\sqrt[2018]{2019^{2019}})^{2019} = (2019 \\cdot \\sqrt[2018]{2019})^{2019} = \\\\\n= 2019^{2019} \\cdot \\sqrt[2018]{2019^{2019}} = 2019^{2019} a_1 a_2 \\dots a_{2019},\n$$\nWhich means that in this Arithmetic mean - Geometric mean Inequality, equality holds, which is possible iff\n$$\na_1 = a_2 = \\dots = a_{2019} = \\frac{1}{\\sqrt[2019]{2019 \\cdot 2019^{2019}}} = \\sqrt[2019]{2019^{2019}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists a pair of proper fractions that cannot be simplified so that their difference equals their product and one of denominators is $2019$. If such pair exists, find at least two of such pairs of fractions.",
    "options": [],
    "answer": "(1/2019, 1/2020) and (1/2018, 1/2019)",
    "solution": "The solution is based on a simple fact: $\\frac{1}{n} - \\frac{1}{n+1} = \\frac{1}{n(n+1)} = \\frac{1}{n} \\cdot \\frac{1}{n+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45600,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given that real roots of a quadratic polynomial $g(x) = x^2 - 3x + a$ are also roots of polynomial $f(x) = x^3 - x^2 + c x + 4$. Analogously, both real roots of a quadratic polynomial $h(x) = x^2 + x + b$ are also roots of $f(x)$. What values can $f(1)$ take?",
    "options": [],
    "answer": "0",
    "solution": "Since $f(x)$ is a cubic polynomial, it has no more than three real roots, hence, quadratic polynomials $g(x)$ and $h(x)$ have the same root. Let us denote this root by $t$. Then,\n$$\ng(t) = t^2 - 3t + a = 0 \\text{ and } h(t) = t^2 + t + b = 0 \\Rightarrow 4t + b - a = 0 \\Rightarrow t = \\frac{1}{4}(a-b).\n$$\n\nThen, the equality must hold: $f(x)(x-t) = g(x)h(x)$.\n$$\n(x^3 - x^2 + c x + 4)(x - t) = (x^2 - 3x + a)(x^2 + x + 2).\n$$\nBy collecting coefficients of $x^3$, we obtain that the equation must be satisfied:\n$$\n-1-t = -3+1 \\Rightarrow t=1, \\text{ hence, } f(1) = f(t) = 0.\n$$\n\nIt is easy to find an explicit form of polynomials $f$, $g$ and $h$, which satisfy the given statement:\n$$\nf(x) = x^3 - x^2 - 4x + 4, \\quad g(x) = x^2 - 3x + 2, \\quad h(x) = x^2 + x - 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the collection of lines of the form $y = (k+n)x + (k-n)$ on a plane, where $k, n$ are any integers. Is there a point with integer coordinates that doesn't belong to any of such lines?\n\n**Answer:** yes, there is.",
    "options": [],
    "answer": "yes, there is",
    "solution": "Let $x=1$, then the second coordinate of all the points that belong to the lines is $y = (k+n) + (k-n) = 2k$ which is even. Therefore, the point $(1, 1)$ doesn't belong to any of the lines.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the family of parabolas of the form $y = kx^2 + (k-n)x + (k+n)$ on a plane, where $k, n$ are any integer numbers. Is there a point with integer coordinates that doesn't belong to any parabola?",
    "options": [],
    "answer": "(1, 1)",
    "solution": "Let $x=1$, then the second coordinate is $y = k + (k-n) + (k+n) = 3k$, which is divisible by three. Therefore, $(1; 1)$ doesn't belong to any parabola.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45603,
    "subject": "Mathematics (Multi-modal)",
    "question": "From natural numbers $2, 3, 4, \\ldots, 2019$ one constructs $1009$ fractions, and chooses the maximum of these. What is the minimum possible value of this maximum fraction?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "1010/2019",
    "solution": "First of all, note that this value can be achieved with the following fractions choice:\n$$\n\\frac{1}{1011}, \\frac{2}{1012}, \\frac{3}{1013}, \\dots, \\frac{1010}{2019}.\n$$\n\nFor the sake of contradiction, let's suppose that the smaller value of the maximum fraction can be obtained in some other way. Clearly, $2019$ cannot be a numerator, hence we can consider the fraction with denominator $2019$. Its numerator should be smaller than $1010$. Let's denote it as $a < 1010$.\nThere are at least $1009$ numbers in the set\n$$\nM = \\{a, a+1, \\dots, 1010, 1011, \\dots, 2018\\}.\n$$\nFrom the pigeonhole principle it follows that some of these numbers form one of the rest $1008$ (excluding $\\frac{a}{2019}$) fractions. If we denote them $b < c$ then we will have $\\frac{1010}{2019} < \\frac{b}{2019} < \\frac{b}{c}$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andrew and Olesya in turn cut some squares from the rectangle $4000 \\times 2019$, following the lines in such a way that after every turn the remaining figure stays connected. The one who cannot make a move loses. Who is going to win if both children play the best they can, and Olesya is first?\n\n(Bogdan Rublyov)\n\n![](attached_image_1.png)\n**Fig. 34**",
    "options": [],
    "answer": "Olesya",
    "solution": "In her first turn Olesya cuts the square $2018 \\times 2018$ as shown in fig. 34.\n\nAfter this move Olesya can follow the symmetric strategy, with the only not trivial moves regarding the black unit squares.\n\nHowever, if Andrew is able to cut (w.l.o.g.) the left black square then it implies that all squares to the left were cut as well. Because of the symmetric play it also implies that all squares to the right were cut, meaning that Olesya can cut the right black square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45605,
    "subject": "Mathematics (Multi-modal)",
    "question": "The inscribed circle of an acute triangle **ABC** touches sides **AB** and **BC** in points $C_1$ and $A_1$ respectively. Let **M** be the midpoint of the side **AC**, and **N** be the midpoint of an arc **ABC** of the circumcircle of **ABC**. Also, let **P** be the projection of a point **M** to the segment $A_1C_1$. Prove the points **I**, **P**, and **N** lie on the same line.\n(Anton Trygub)\n\n![](attached_image_1.png)\n**Fig. 35**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's denote the second intersection point of the line **BI** with the circumcircle of $\\triangle ABC$ as **W** (fig. 35).\nClearly, **W** is a midpoint of the smaller arc **AC**, and points **M**, **W**, and **N** lie on the same line.\n\nHence, it is enough to prove that triangles $PMN$ and $IWN$ are similar (which will imply that $\\angle PNM = \\angle INW$ and the desired collinearity). In order to prove the similarity, we are going to show that $\\frac{NM}{NW} = \\frac{PM}{IW}$.\nLet $\\angle B = 2\\beta$, then $\\angle MNA = \\beta$ and\n$$\n\\frac{NM}{NW} = \\frac{PM}{IW} \\cdot \\frac{NA}{NW} = \\cos^2 \\beta.\n$$\n\nConsidering a segment **PM** yields that **PM** is a midline of a trapezoid **AXYC**, where **X**, **Y** are the projections of **A** and **C** onto the line $A_1C_1$ respectively. Going further, we get\n$$\n2\\mathrm{PM} = \\mathrm{CY} + \\mathrm{XA} = \\mathrm{CA}_1 \\sin \\mathrm{YA}_1\\mathrm{C} + \\mathrm{C}_1 \\sin \\mathrm{XC}_1\\mathrm{A} = (\\mathrm{CA}_1 + \\mathrm{AC}_1)\\cos \\beta = \\mathrm{AC}\\cos \\beta,\n$$\nbecause $\\angle \\mathrm{YA}_1\\mathrm{C} = \\angle \\mathrm{XC}_1\\mathrm{A} = 90^\\circ - \\beta$, and $\\mathrm{CA}_1 + \\mathrm{AC}_1 = \\mathrm{AC}$, because $A_1$ and $C_1$ are the points in which inscribed circle touches the sides of a triangle $\\triangle ABC$.\nFinally, from the trillium theorem, $IW = CW$, meaning that\n$$\n\\frac{PM}{IW} = \\frac{AC}{2CW} \\cos \\beta = \\frac{CM}{CW} \\cos \\beta = \\cos^2 \\beta,\n$$\nfinishing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two distinct natural numbers $a$ and $b$ greater than $1$,\n\na) Prove that there are infinitely many natural numbers $n$ such that $s_n = a^n + b^{n+1}$ is composite.\n\nb) Prove that there are infinitely many prime numbers $p$ such that $s_n$ is divisible by $p$ for some $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If $s_n = p$ is prime for some $n$ larger than $a$ and $b$, then for some $k$ and $l$ the numbers $a^k - 1$ and $b^l - 1$ are divisible by $p$. If we set $m = kl$, then both $a^{mt} - 1$ and $b^{mt} - 1$ are divisible by $p$ for all natural numbers $t$. This implies that the number\n$$\ns_{n+mt} = a^{n+mt} + b^{n+1+mt} = a^n + b^n + a^n(a^{mt} - 1) + b^{n+1}(b^{mt} - 1)\n$$\nis divisible by $p$ for all $t$, which gives an infinite amount of composite members in the sequence of $s_n$.\n\nb) All prime divisors of $s_n$ either divide both $a$ and $b$ or none of them.\nLet $q$ be a common prime divisor of $a$ and $b$, and $q^k$ and $q^l$ are the largest powers of $q$ that are divisors for $a$ and $b$ respectively. If $k \\le l$ then $kn < l(n+1)$, and if $k > l$ then $kn > l(n+1)$ for all large enough $n$. Therefore, for sufficiently large $n$, the prime divisor $q$ raised to one of the powers $kn$ or $l(n+1)$ divides $s_n$, implying that the greatest common divisor of $a^n$ and $b^{n+1}$ cannot be larger than $d^{n+1}$, where $d = (a, b)$ – GCD of $a$ and $b$.\nLet $p$ be a prime number dividing neither $a$ nor $b$. Let $p^k$ be the largest power of $p$ that is divisor of $b+1$ (maybe, $k=0$). For some natural number $m$ both $a^m - 1$ and $b^m - 1$ are divisible by $p^{k+1}$. Then, for some $n$ divisible by $m$, the number $s_n = (b+1) + (a^n - 1) + b(b^n - 1)$ is divisible by $p$ raised to the same power as $b+1$.\nNow we can finally get back to the original problem. For the sake of contradiction, we assume that there are only finitely many primes dividing some $s_n$. In particular, this means that there are only finitely many prime numbers $p_1, p_2, \\dots, p_j$ which do not divide $a$ or $b$, but divide $s_n$ for some $n$.\nWe have just shown that for all such $p_i$ there is $m_i$ such that for all $n$ divisible by $m_i$, the largest number, such that $p_i$ raised to the power of this number is a divisor of, $s_n$ is smaller than the largest number, such that $p_i$ raised to the power of this number, is a divisor of $b+1$.\nHowever, if we now choose $n$ divisible by all $m_i$, then we will get that $s_n = a^n + b^{n+1}$ is not greater than $d^{n+1}(b+1)$. However, this cannot be the case for all $n$ because of the fact that one of $a$ or $b$ is bigger than $d$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45607,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $a$ and $b$, for which the number $2^{a!} + 2^{b!}$ is a cube of a natural number.\n*Recall that for a natural number $n$, $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot n$.*\n\n(Arseniy Nicolaev)",
    "options": [],
    "answer": "a = b = 2",
    "solution": "It is clear that if $n \\ge 3$, $n!$ is divisible by $3$, that is $n! = 3k$, for some natural number $k$. But then $2^{n!} = 2^{3k} = 8^k \\equiv 1 \\pmod{7}$. It's easy to see that the cubes of integer numbers are equal to $0$ or $\\pm 1$ modulo $7$. Without loss of generality, we may assume that $a \\ge b$.\n\nCase 1. $a \\ge 3$. If $b \\ge 3$, then $2^{a!} + 2^{b!} \\equiv 2 \\pmod{7}$ is not a cube of an integer number. If $b=2$, then $2^{a!} + 2^{b!} \\equiv 1+4 \\equiv 5 \\pmod{7}$ is not a cube of an integer number. If $b=1$, then $2^{a!} + 2^{b!} \\equiv 1+2 \\equiv 3 \\pmod{7}$ is not a cube of an integer number.\n\nCase 2. $2 \\ge a \\ge b$, here we have three cases, from which we simply find a single answer in a simple overview.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram. The circle through $A$ and $D$ intersects the lines $AB$, $BD$, $AC$ and $CD$ in points $B_1$, $B_2$, $C_1$ and $C_2$ respectively. Let $K$ be the intersection point of the lines $B_1B_2$ and $C_1C_2$. Prove that $K$ is equidistant from the lines $AB$ and $CD$.\n\n![](attached_image_1.png)\n\nFig. 41",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the intersection point of the diagonals of the parallelogram. Let's prove that the points $C_1$, $O$, $B_2$ and $K$ lie on the same circle. In fact, with the cyclicity of points $A$, $B_1$, $B_2$, $C_1$, $C_2$ and $D$ and parallelism of $AB$ and $CD$ we have (Fig. 41):\n$$\n\\angle(KC_1, C_1O) = \\angle(CD, DA) = \\angle(BA, AD) = \\angle(B_1B_2, B_2D) = \\angle(B_2K, B_2O).\n$$\nNow $\\angle(BO, OK) = \\angle(B_2C_1, C_1K) = \\angle(B_2D, DC)$, thus $OK \\parallel DC$. As known, diagonals $AC$ and $BD$ are divided by a point $O$ in half. By the Thales' theorem the line $OK$ (and accordingly, the point $K$) is equidistant from the lines $AB$ and $CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45609,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country, the government has decided to build a transport link between cities by railway or airline in such a way that no more than four other cities can be reached directly from each city. Prove that the government can always choose the type of connection of the corresponding pairs of cities in such a way that there are no three cities, each pair of which is connected by one type of transport.\n(Arseniy Nicolaev)\n\n![](attached_image_1.png)\nFig. 42\n\n![](attached_image_2.png)\nFig. 43",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's rewrite the problem statement in terms of graphs: in the graph the degree of each vertex is not more than four. Prove that all edges can be painted in two colors so that there are no single-color triangles.\nLet our graph have $n$ vertices. We prove the statement of the problem by induction on the number of vertices.\n\nBase case: if $n=1, 2$ everything is obvious\nAssume the statement is true for any graph with $n \\le k$ vertices. Consider a graph with $n=k+1$ vertices. Select the vertex $A$ arbitrarily. Remove this vertex from the graph as well as all edges from $A$. Thus we have a graph with $n=k$ vertices, that, by the induction hypothesis, can be painted as needed. Now paint vertex $A$ with its edges. Consider three cases.\n\n**I.** The degree of each vertex $A$ is not more than 2. Then paint all edges from $A$ in different colors, which proves the problem statement.\n\n**II.** The degree vertex $A$ is equal to 3, that is there is exactly 3 edges $AX, AY$ and $AZ$ (Fig. 42); the dotted line means it is not significant whether there is a corresponding edge. We know that the edges of $\\Delta XYZ$ have different colors. Let, for instance, $XY$ be painted in the first color $YZ$ – the second color. Let's paint the edges as shown on Fig. 42: $AX$ and $AY$ – in the second color, and $AZ$ – first. By the induction hypothesis the statement is proved.\n\n![](attached_image_1.png)\nFig. 42\n\n**III.** The degree vertex $A$ is equal to 4, that is there are the following edges from $A$: $AP, AQ, AR$ and $AS$.\n\na) If the graph on the vertices $P, Q, R, S$ is complete, we have a separate connected component on the vertices $A, P, Q, R, S$. Wherein by the induction hypothesis it is a complete graph. Then let's paint its edges as shown on Fig. 43 and the statement is proved. Since all other vertices form a graph with less then $k$ vertices, by the induction hypothesis, it is painted as needed.\n\n![](attached_image_2.png)\nFig. 43\n\nb) If the graph on the vertices $P, Q, R, S$ is not complete, suppose there that there is no edge $QR$. Let's paint the edges $AP, AQ$ in a color, opposite to the edge $PQ$, and the edges $AR, AS$ – in a color, opposite to the edge $RS$. It is straightforward to check that there are no single-color triangles. Hence, by the induction, the problem statement is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45610,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive numbers $x$, $y$, $z$, satisfying equality $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$, prove that the following inequality is true:\n$$(x-1)(y-1)(z-1) \\le \\frac{1}{4}(xyz-1).$$\n(Vadim Mitrofanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the description of the problem we have that $xy + yz + zx = 3xyz$. The given inequality is equivalent to\n$$\n(x-1)(y-1)(z-1) = (xyz-1) + (x+y+z-xy-yz-xz) = (xyz-1) + (x+y+z-3xyz) = x+y+z-1-2xyz \\le \\frac{1}{4}(xyz-1) \\Leftrightarrow \\\\\n x+y+z \\le \\frac{9}{4}xyz + \\frac{3}{4}.\n$$\n\nLet us get rid of the dependence between the variables $x$, $y$, $z$, choose the positive ones $a$, $b$, $c$, for which the following equality is true:\n$$\nx = \\frac{a+b+c}{3a}, \\quad y = \\frac{a+b+c}{3b}, \\quad z = \\frac{a+b+c}{3c}.\n$$\n\nIt is clear that for any positive $a$, $b$, $c$ the condition $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$ is fulfilled, and the inequality that must be proved is rewritten as follows:\n$$\n\\left(\\frac{a+b+c}{3}\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\le \\frac{9}{4} \\cdot \\frac{(a+b+c)^3}{27abc} + \\frac{3}{4} \\Leftrightarrow 4(a+b+c)(ab+bc+ca) \\le (a+b+c)^3 + 9abc.\n$$\nAfter opening the brackets and reducing fraction we get:\n$$\n\\begin{aligned}\n& 4(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) + 12abc \\le \\\\\n& \\le a^3 + b^3 + c^3 + 3(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) + 15abc \\Leftrightarrow \\\\\n& a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 3abc \\le a^3 + b^3 + c^3,\n\\end{aligned}\n$$\n\nwhich is a case of Schur's inequality",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Are there integers $a < b < c < d$ such that\n$$\n\\frac{a}{a} + \\frac{a}{b} + \\frac{a}{c} + \\frac{a}{d} = \\frac{b}{a} + \\frac{b}{b} + \\frac{b}{c} + \\frac{b}{d} ?\n$$",
    "options": [],
    "answer": "a = -28, b = -14, c = -7, d = 4",
    "solution": "**Answer.** Yes, for example $a = -28$, $b = -14$, $c = -7$ and $d = 4$.\n\nClearly, all numbers cannot be positive. Let's take $a = -4$, $b = -2$, $c = -1$ and find the corresponding $d$:\n$$\n\\begin{aligned}\n\\frac{-4}{-4} + \\frac{-4}{-2} + \\frac{-4}{-1} + \\frac{-4}{d} &= \\frac{-2}{-4} + \\frac{-2}{-2} + \\frac{-2}{-1} + \\frac{-2}{d} \\\\\n1 + 2 + 4 + \\frac{-4}{d} &= \\frac{1}{2} + 1 + 2 + \\frac{-2}{d} \\\\\n7 - \\frac{4}{d} &= \\frac{7}{2} - \\frac{2}{d} \\\\\n7 - \\frac{4}{d} + \\frac{2}{d} &= \\frac{7}{2} \\\\\n7 - \\frac{2}{d} = \\frac{7}{2} \\\\\n7 - \\frac{7}{2} = \\frac{2}{d} \\\\\n\\frac{7}{2} = \\frac{2}{d} \\\\\nd = \\frac{4}{7}.\n\\end{aligned}\n$$\n\nThe equality is homogeneous, meaning that we can simply multiply all numbers by $7$ to obtain the aforementioned integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andrew and Olesya in turn cut some squares $2 \\times 2$ or $1 \\times 1$ from the rectangle $2 \\times 2n$, following the lines in such a way that after every turn the remaining figure stays connected. The one who cannot make a move loses. Who is going to win if both children play the best they can, and Olesya is the first?\n\n(Bogdan Rublyov)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Olesya wins when n is odd; Andrew wins when n is even.",
    "solution": "We proceed by induction on $n$. One can verify that if the field is $2 \\times 2$ Olesya wins (by cutting entire $2 \\times 2$ square), and if the field is $2 \\times 4$ then Andrew wins (casework). If $n = 2m+1$, then Olesya can cut $2 \\times 2$ square from the end of a strip and place herself in the position of a second player for even $n$. Hence, it remains to prove only that second player wins for even $n$.\n\nPrior to and including the first $1 \\times 1$ square cut by the first player in the bordering column (i.e. the first or the last column of the “active” (see fig. 36) part of the playing field if we assume that the playing strip is placed horizontally) second player can follow the symmetric strategy. Clearly, he will be able to do so. Also, note that the first $1 \\times 1$ square will be cut when active playing field has dimensions $2 \\times k$, where $k$ is even (fig. 37).\n\nAfter that point, the only thing he has to do is to prevent the cutting of $2 \\times 2$ square, because it will mean that the game will end after the even number of moves (because both players can cut only $1 \\times 1$).\n\nMoreover, it is clear that the only places from which the $2 \\times 2$ square can be cut are the ends of the playing strip. It means that the second player does not have much to do at all: he cuts a unit square from one of the bordering (for the active part of the field) $2 \\times 2$ squares that has no unit squares already cut out (\"free\" ends).\n\nClearly, there is no greater than one free end, because Olesya cannot create two free ends in one move, and we \"start\" (after the first two bordering $1 \\times 1$ squares are cut) with zero free ends.\n\nIf, however, there are no free ends, then one of the following holds:\n1. bordering column has no squares cut out, then Andrew can simply cut one unit square from it;\n2. bordering column has one square cut out,\n   a) and the second to border column has a square cut out, then Andrew can simply cut the second unit square from the bordering column;\n   b) and the second to border column has no squares cut out,\n      i. and the third to border column has no squares cut out, or it has a square cut out in the same row as for the bordering column, then Andrew can cut a unit square from the second to border column in the same row;\n      ii. and the third to border column has a square cut out in the different row from that of the bordering column, then Andrew can cut a unit square from the bordering column, effectively decreasing the \"active\" playing field, but without creating free ends.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $B$ and $C$ are chosen on the circle with diameter $AD$ in such a way that $AB = AC$. Point $P$ is an arbitrary point of the segment $BC$, and points $M$ and $N$ are chosen on the segments $AB$ and $AC$ respectively in such a way that $PMAN$ is a parallelogram. Let $PL$ be a bisector in triangle $MPN$. Line $PD$ intersects $MN$ in point $Q$. Prove that points $B$, $Q$, $L$, and $C$ are cyclic.\n\n(Mykhaylo Plotnikov, Danylo Khilko)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we are going to prove that $\\angle BDP = \\angle AMN$ and $\\angle PDC = ANM$. Indeed, $MP \\parallel AC$, $NP \\parallel AB$, because $PMAN$ is a parallelogram (fig. 38). Then $\\angle MPB = \\angle ABC = \\angle ACB = \\angle NPC$, meaning that $\\triangle BMP \\sim \\triangle PNC$. We also note that $BC$ is a bisector of interior angle $\\triangle MPN$. Then $\\frac{BP}{PC} = \\frac{MP}{NC}$. Using $MP = AN$ one gets $NC = NP = AM \\cdot \\frac{PB}{PC} = \\frac{AN}{AM}$.\n\nNotice that $\\triangle BDC$ is isosceles, and $\\angle CBD = \\angle BCD$. By cosine law for triangles $BPD$ and $CPD$:\n$$\n\\frac{BP}{\\sin \\angle BDP} = \\frac{PD}{\\sin \\angle CBD} = \\frac{PD}{\\sin \\angle BCD} = \\frac{PC}{\\sin \\angle PDC}.\n$$\nFrom here it follows that $\\frac{\\sin \\angle PDC}{\\sin \\angle BDP} = \\frac{PC}{BP}$ and $\\frac{\\sin \\angle PDC}{\\sin \\angle BDP} = \\frac{AM}{AN}$. By sine law for $\\triangle AMN$\nwe have $\\frac{\\sin \\angle ANM}{\\sin \\angle AMN} = \\frac{AM}{AN}$. Finally we get $\\frac{\\sin \\angle PDC}{\\sin \\angle BDP} = \\frac{\\sin \\angle ANM}{\\sin \\angle AMN}$.\n\nNoticing\n$$\n\\angle PDC + \\angle BDP = \\angle BDC = 180^\\circ - \\angle MAN = \\angle MNA + \\angle AMN\n$$\nyields $\\angle PDC = \\angle MNA$ and $\\angle PDB = \\angle AMN$.\n\nGoing further, $\\angle QMD = \\angle AMN = \\angle QDB$, hence $MQBD$ is cyclic. Considering triangle $CPD$ and $PLM$ gives\n$$\n\\angle CDP = \\angle NMP = \\angle ANM, \\quad \\angle LPM = 90^\\circ - \\angle MPB = 90^\\circ - \\angle ACB = \\angle PCD,\n$$\nmeaning that $\\triangle CPD \\sim \\triangle PLM$.\n\nTherefore $\\frac{CP}{LP} = \\frac{CD}{MP}$. Using $MP = MB$ and $CD = DB$, one gets $\\frac{CP}{LP} = \\frac{BD}{MB}$.\n\nConsidering triangles $LPC$ and $MBD$ gives $\\angle MBD = \\angle LPC = 90^\\circ$, implying (together with the previous equality) $\\triangle LPC \\sim \\triangle MBD$.\n\nFinally,\n$\\angle LPC = \\angle MDB = 180^\\circ - \\angle BQM$, meaning that $\\angle BQL + \\angle LCB = 180^\\circ$, i.e. that the quadrilateral $BQLC$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $a$, $b$ and $c$, such that the number $2^{a!} + 2^{b!} + 2^{c!}$ is a cube of a natural number.\n(Arseniy Nicolaev)",
    "options": [],
    "answer": "All permutations of (1,1,2).",
    "solution": "It is clear that if $n \\ge 3$, $n! \\nmid 3$, that is $n! \\ne 3k$, for some natural number $k$. But then $2^{n!} = 2^{3k} = 8^k \\equiv 1 \\pmod{7}$. Analogously modulo 9: if $n \\ge 3$ we have that $n! \\nmid 6$, that is $n! \\ne 6l$, for some natural $l$. But then $2^{n!} = 2^{6l} = 64^l \\equiv 1 \\pmod{9}$.\n\nIt's easy to see, that the cubes of integer numbers are equal $0$ or $\\pm 1$ modulo $7$ and modulo $9$. Without loss of generality, we may assume that $a \\ge b \\ge c$.\n\n**Case 1.** $a \\ge 3$ and $b \\ge 3$. If $c \\ge 3$, then $2^{a!} + 2^{b!} + 2^{c!} \\equiv 3 \\pmod{7}$ – is not a cube of an integer.\nIf $c=2$ then $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1+1+4 \\equiv -1 \\pmod{7}$ – can be a cube of an integer.\nCheck modulo 9. $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1+1+4 \\equiv 6 \\pmod{9}$ – is not a cube of an integer.\nIf $c=1$ then $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1+1+2 \\equiv 4 \\pmod{7}$ – is not a cube of an integer.\n\n**Case 2.** $a \\ge 3 \\Rightarrow b \\le 2$. Hence we have three cases.\nIf $b=c=2$ $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1+4+4 \\equiv 2 \\pmod{7}$ – is not a cube of an integer.\nIf $b=2, c=1$ $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1+4+2 \\equiv 0 \\pmod{7}$ – can be a cube of an integer.\nThen analogously modulo 9 $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1+4+2 \\equiv 7 \\pmod{9}$ – is not a cube of an integer.\nIf $b=c=1$ $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1+2+2 \\equiv 5 \\pmod{7}$ – is not a cube of an integer.\n\n---\n\nCase 3. $2 \\ge a \\ge b \\ge c$, here we have 4 cases overall, from which we simply find a single answer in a simple overview.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose there are $2019$ points $A_1, A_2, \\dots, A_{2019}$ in a circle, that form the regular $2019$-gon $A_1A_2\\dots A_{2019}$. Olesya and Andrew take turns, Olesya is the first to go. According to certain rules, Olesya makes a move to form an obtuse triangle, and Andrew – an acute triangle. The rules for forming triangles are as follows. At the beginning vertices $A_1$ and $A_2$ are supposed to be marked. In the first step Olesya marks an unmarked vertex $A_i$ so that $\\Delta A_1A_2A_i$ is formed and vertex $A_i$ becomes marked. Let in the next step Olesya (Andrew) marked an unmarked earlier vertex $A_j$ and formed an obtuse (acute) $\\Delta A_kA_lA_j$. Then in the next step Andrew (Olesya) can mark such an unmarked vertex $A_m$, for which for an acute (obtuse) triangle $\\Delta A_nA_jA_m$ is formed, where $n=k$ or $n=l$. One loses if not able to follow the rules to make another move, that is to mark the vertex to form a proper triangle. Who wins the game if both players play correctly?",
    "options": [],
    "answer": "Olesya",
    "solution": "Let's place the circle so that the diameter through the vertex $A_1$ is vertical and this vertex itself is located below. Let's renumber all the vertices as shown on Fig. 44:\n\n$A_i = B_{i-1},\\ i = 1, 1010$ and\n$A_i = B_{1010-i},\\ i = 1011, 2019$.\n\nThe strategy of Olesya is as follows: she always marks a vertex, that is symmetrical to the one Andrew marked with respect to the vertical diameter. In her first move she marks $B_{-1}$. As we see, if vertex $B_i$ is not marked, then the vertex $B_{-i}$ is unmarked too and vice versa.\n\nIf Andrew is able to make a move, he marks some point $B_{-k}$ and $\\Delta B_m B_l B_{-k}$ (Fig. 44). Since this triangle is acute, then the center of the circle is inside this triangle. Therefore, we will have a diameter $B_{-k}O$, then vertices $B_m$ and $B_l$ lie on different sides of that diameter. One of them is located in one half-plane with a vertex $B_k$, let's say, $B_m$. Thus Olesya can mark vertex $B_k$ and $\\Delta B_m B_{-k} B_k$, it is obtuse, because it doesn't have the circle center inside.\n\nThus, Olesya will always be able to answer Andrew's move, even if all the vertices are marked, then Andrew will not be the first to make a move and lose.\n\n![](attached_image_1.png)\n**Fig. 44**",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f:(0, +\\infty) \\rightarrow (0, +\\infty)$ for which for any positive numbers $x, y$ the following equality is true:\n$$\nf(f(x)+y) = f(f(x))+2yf(x)-f(y)+2y^2+1. \\qquad (\\text{Ihor Voronovych})\n$$",
    "options": [],
    "answer": "f(x) = x^2 + 1 for all x > 0",
    "solution": "From the task of the problem when substituting $y = f(y)$ we have that\n$$\n\\begin{aligned}\nf(f(x)+f(y)) &= f(f(x))+2f(y)f(x)-f(f(y))+2f^2(y)+1 \\Rightarrow \\\\\nf(f(x)+f(y))-f(f(x))-2f(y)f(x)-f(f(y)) &= -2f(f(y))+2f^2(y)+1.\n\\end{aligned}\n$$\nTherefore, due to the symmetry of the left part relative to the variables $x, y$ the following equality must be obtained:\n$$\n\\begin{aligned}\n&f(f(y)) = f^2(y)+c, \\text{ where } c = \\text{const} \\Rightarrow \\\\\n&f(f(x)+f(y)) = f^2(x)+c+2f(y)f(x)-f^2(y)-c+2f^2(y)+1 \\text{ and} \\\\\n&f(f(x)+f(y)) = (f(x)+f(y))^2 +1.\n\\end{aligned} \\qquad (1)\n$$\n\nLet's fix some $x_0$, and let $y_0 = f(x_0)$. Then if $x = x_0$ we get that\n$$\n\\begin{aligned}\n&f(f(x_0)+y)+f(y) = f(f(x_0))+2yf(x_0)+2y^2+1, \\text{ or} \\\\\n&f(y_0+y)+f(y) = f(y_0)+2yy_0+2y^2+1 \\Leftrightarrow \\\\\n&f(y_0+y)+f(y) = 2yy_0+2y^2+f(y_0)+1.\n\\end{aligned}\n$$\n\nTherefore for any $t > f(y_0)+1$ exists $y_t$ such that $2yy_0 + 2y^2 + f(y_0) + 1 = t$, thus $f(y_0 + y_t) + f(y_t) = t$. Taking into account (1), substituting $x = y_0 + y_t$ and $y = y_t$ in the given equation we obtain:\n$$\n\\begin{aligned}\n&f(f(y_0 + y_t) + f(y_t)) = (f(y_0 + y_t) + f(y_t))^2 + 1 \\text{ or} \\\\\n&f(t) = t^2 + 1\n\\end{aligned} \\qquad (2)\n$$\n\nfor any $t > f(y_0)+1 = a$.\nFor any $x > a$ we have that $x^2 + 1 \\ge 2x > x > a$, from the statement of the problem if $x > a$ applying equality (2) several times, we get that\n$$\n\\begin{aligned}\n&f(x^2+1+y) = f(x^2+1)+2y(x^2+1)-f(y)+2y^2+1 \\text{ or} \\\\\n&\\phantom{f(x^2+1+y) = } (x^2+1+y)^2+1 = (x^2+1)^2+1+2y(x^2+1)-f(y)+2y^2+1.\n\\end{aligned}\n$$\nAfter opening all brackets we get that for any positive $y$ we have equality:\n$$\n\\begin{aligned}\n&f(y) = -x^4 -1 -y^2 -2x^2 -2x^2y -2y + x^4 + 2x^2 +1+2yx^2 +2y +2y^2 +1 \\Leftrightarrow \\\\\n&f(y) = y^2 +1.\n\\end{aligned}\n$$\n\nIt is straightforward to check that $f(x) = x^2 + 1$ satisfies the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45617,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a group of $2n$ people, among whom there are couples of friends. It is known that every person in this group has exactly $k \\ge 1$ friends (if \"A\" is familiar with \"B\", then conversely, \"B\" is familiar with \"A\"). Find such a number $k$ for which this group can always be divided into two subgroups of $n$ people so that in both subgroups each person has at least one friend?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "k ≥ 3",
    "solution": "For $k=1$ everything it is obvious. All friends are divided into pairs. In order to be able to split everyone properly, each pair must fall into one subgroup. But each subgroup must have the same number of people. Therefore, if there is an even number of such pairs, that is $n=2m$, it can be done by adding exactly $m$ pairs to each of the subgroups. If $n=2m-1$, then for any division of all people into two equal parts at least in one of the couples people will fall into different subgroups, and therefore they will not have the same friend in their subgroup.\n\nNow, let $k \\ge 2$ and, for example, there is a division in which it is impossible to properly divide the group into two subgroups. Let's consider an arbitrary splitting into two subgroups of $n$ people. We denote people with points that are broken up by a vertical line into two subgroups - the left (L) and the right (R).\n\nDraw the line segments between all pairs of acquaintances. The segments between people who fall into one subgroup with such a division will be called *important*. Now among all possible division, choose the one in which the number of important segments is *maximal*. We call such a division (or any of these, if there may be several) maximal. Presumably it is impossible to divide the group into two subgroups as needed. Therefore, for a maximal division there exists, for example, in the left subgroup, some person $X$ whose all his/her $k$ friends $A_1, \\dots, A_k$ are in the right subgroup. Consider two cases.\n\nSuppose there are two people $B \\in L$ and $C \\in R$, that are friends with each other. Clearly, according to the rules the line segment between them is not important. Suppose $C$ has $l \\ge 1$ friends in the left side, and $k-l$ friends in the right side. Then let's swap people $X$ and $C$ in subgroups. Let's count how much the number of important segments has changed. The left subgroup has $l$ segments more, and the right side $-k-(k-l)=l$ more. Thus their number increased, and therefore we got a contradiction with the assumption that this is the maximum division.\n\nBy the hypothesis for the maximal division we have that there is a person $X \\in L$ with all his/her friends are in the right subgroup. At the same time, all other people have friends only in their subgroup.\n\nConsider the case $k=2$. Then all people divide into chains of 3 or more people. Let's denote them as follows: $B_1B_2B_3\\dots B_j, j \\ge 3$. The first $B_1$ and the last $B_j$ as well as $B_t$ and $B_{t+1}$ are acquaintances.\n\n**Case 1.** Suppose a chain with element $X$ has 3 elements, that is $XX_2X_3$, then all these three elements have to fall into the same subgroup. Let's move them to the left subgroup. If there is at least one chain of 4 or more people $B_1B_2B_3\\dots B_j$ in the left subgroup, $j \\ge 4$, then it's enough to move $B_1$ and $B_2$ to the right subgroup the required distribution is done. It is impossible to do it this way, if only all chains in the left side consist of 3 elements. Thus the right subgroup has element $X$ and a few chains of 3 elements. Hence the right subgroup has elements $X_2, X_3$ and at least one chain that does not consist of 3 elements.\nIf there is exactly one cycle of 5 elements with the other cycles of 3 elements, then the required distribution does not exist. Clearly, all elements of cycles of 3 elements have to be in one subgroup. Then the cycle of 5 elements can be divided in two parts $3+2$ and two equal groups does not exist. Hence for $n=2+6m, m \\ge 2$ the needed distribution may not exist.\nShow, that for all other cases the distribution does exists. If there is at least 1 cycle $B_1B_2B_3\\dots B_j$, $j=4$ or $j \\ge 6$, which has exactly 4 or more then 5 elements, then we do the following rearrangements: $X$ and some cycle of 3 elements must be moved to the right subgroup, and $B_1B_2B_3B_4$ -- left one.\n\nIf there is at least two cycles of exactly 5 elements $B_1B_2B_3B_4B_5$ and $C_1C_2C_3C_4C_5$, then we move elements $X_2, X_3, B_1, B_2$ and $C_1, C_2$, from left to right, and two cycles of length 3 - from right to left.\n\n**Case 2.** Let chain with the element $X$ has 5 elements $XX_2X_3X_4X_5$. If all other chains have 3 elements each, the distribution may not exist. An example is similar to the one given above for $n=2+6m$, but now $m \\in \\mathbb{N}$. For $n=8$ we have two cycles of 3 and 5 elements, which, obviously, cannot be divided into two equal subgroups as needed.\nLet's show that for all other cases distribution is possible. That is, there are cycles not only of 3 elements. If there is a cycle of 4 elements in the left subgroup, move it to the right side, and $X_2, X_3, X_4, X_5$ - to the left side. If there is a cycle of at least 5 elements $B_1B_2B_3...$, do the following displacement: $B_1, B_2, B_3$ у праву, $X_2, X_3$ у ліву. Thus the left subgroup has only 3 - element cycles. If the right subgroup has a cycle of at least 4 elements $B_1B_2B_3...$, then move $B_1, B_2$ and $X$ to the left subgroup, and the cycle 3 elements to the right side.\n\n**Case 3.** Let chain with the element $X$ has 4 elements $XX_2X_3X_4$. If in the left side there are cycles of 3 elements $B_1B_2B_3$ or at least 5 elements $B_1B_2B_3...$, then move $X_2, X_3, X_4$ to the left side, and $B_1, B_2, B_3$ - to the right side. The only possibility left is when there are only cycles of 4 elements in the left side. Suppose one of them consists of elements $C_1C_2C_3C_4$. Then the left side cannot have all cycles of 4 elements each. If in the right subgroup there is a cycle of 3 elements $B_1B_2B_3$ or of at least 5 elements $B_1B_2B_3...$, then move $C_1, C_2$ and $X$ to the right subgroup, and $B_1, B_2, B_3$ to the left subgroup.\n\n**Case 4.** Let chain with the element $X$ has at least 6 elements $XX_2X_3X_4...$. If in the left subgroup each cycle has 3 elements $B_1B_2B_3$, move $B_1, B_2, B_3$ to the right side, and $X_2, X_3, X_4$ - to the left side. If there is a cycle of at least 4 elements $B_1B_2B_3B_4...$, move $B_1, B_2$ to the right side, and $X_2, X_3$ - to the left side.\n\nConsider the case when $k \\ge 3$. Then all people are divided into connected components, which have no less than 4 elements each and the degree of each vertex of a component is $k \\ge 3$, moreover element $X$ is connected to elements $X_2X_3X_4...$. If the left side has a connected component of at least 5 elements $B_1B_2B_3...$, it's enough to move $X_2, X_3, X_4$ to the left side, and $B_1, B_2, B_3$ - to the right side, if the whole connected component consists of 4 elements $XX_2X_3X_4$, otherwise move $X_2, X_3$ to the left side, and $B_1, B_2$ - to the right side.\nOtherwise, each connected component in the left side has exactly 4 elements. Clearly, that under such conditions the only possibility is $k=3$. Then the right side cannot have connected components of 4 elements. Then since there is a component in the right subgroup, that has at least 5 elements $B_1B_2B_3...$, move $B_1, B_2, B_3$ to the left side, $X$ and $C_1, C_2$ to the right side, where $C_1, C_2$ - two elements of one connected component.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose a country has a system of roads such that the roads intersect in towns only, and do not intersect in-between the towns. Furthermore, from any town one can reach any other town, if one can go in either of the two directions on each road. For no pair of towns, there is more than one direct road connecting them. The government decided to make each road a one-way road, i.e. if towns $A$ and $B$ are connected by a road, then one can use it to get from $A$ to $B$, or from $B$ to $A$. Moreover, for each town, there must be at least one road for coming in that town and at least one road for leaving it. Will it always be the case that in this country, there exists a town from which one can get to any other town (even if going through some other towns on the way), or to which one can get from any other town (even if going through some other towns on the way)?",
    "options": [],
    "answer": "No",
    "solution": "Let us show such system of roads where such town does not exist. Denote by $A, B, C, D$ 3-tuples of towns, where the roads form a cycle, e.g. $A_1 \\to A_2 \\to A_3 \\to A_1$ (fig. 24). In this way, the condition that each town has one incoming and one outcoming road is satisfied. Now, we place additional roads with such directions: $A_1 \\to B_1$, $C_1 \\to B_1$ and $C_1 \\to D_1$. Then, one cannot get to any town of groups $A$ and $C$ from other groups, one cannot get to any town in $B$ from group $D$, and vice versa. Analogously, from any town of any group one cannot get to any town from other groups.\n\n![](attached_image_1.png)\n\nFig. 24",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45619,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number $2019$ is written on the board. Katia and Mykola are playing the following game: one by one (starting with Katia) they choose any divisor $d$ of the number $N$ written on the board and change the number on the board $N$ to the number $N - (2d - 1)$, if it is positive integer. Whoever writes number $1$ loses. Who will win in this game and what is the strategy, considering both players want to win?",
    "options": [],
    "answer": "Katia wins; because the number decreases and parity alternates each turn, the number one is written on Mykola’s turn, so he loses. Katia can simply make any legal move each turn.",
    "solution": "Firstly, we will show that the number on the board decreases with every turn. Clearly, it will be smaller and integer. It will be positive, since: $N = dD \\Rightarrow M = N - (2d - 1) = dD - 2d + 1 = d(D - 2) + 1 \\ge 1$, since if $d < N$ then $D \\ge 2$. Therefore, number $1$ will be written on the board after finite number of turns.\n\nIt isn't hard to notice, that every turn changes the parity of the number on the board. Since Katia starts the game, there will be even number after her turn, and odd number after Mykola's turn. Therefore, number $1$ will be written on the board after Mykola's turn, so he will lose.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45620,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ rays $OA_1, OA_2, ..., OA_n$ that start at point $O$ and are enumerated counterclockwise on a plane so that $\\angle A_1OA_n < 180^\\circ$. Determine the smallest $n$ such that there exists a pair of $60^\\circ$ angles, a pair of $45^\\circ$ angles and a pair of $30^\\circ$ angles among angles $\\angle A_iOA_j$, $1 \\le i < j \\le n$.\n(Bogdan Rublyov)\n\n![](attached_image_1.png)\nFig. 1",
    "options": [],
    "answer": "5",
    "solution": "Firstly, we will show that $n = 5$ satisfies the conditions (Fig. 1).\n\nSuppose there exists a configuration that satisfies the conditions with 4 rays (Fig. 2). Then these rays form 6 different angles. Thus, if conditions were satisfied then all the pairs of angles would be among these 6 angles. But if the greatest angle $\\angle A_1OA_4$ equals $60^\\circ$, then another $60^\\circ$ angle will not exist. If $\\angle A_1OA_4 < 60^\\circ$, then there are no $60^\\circ$ angles. If $\\angle A_1OA_4 > 60^\\circ$, then there won't be enough angles to satisfy all the conditions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let **ABC** be a right triangle such that its legs satisfy the following: $BC = \\sqrt{2}AC$. Show that the medians **AN** and **CM** are perpendicular to each other.\n\n(Danylo Khilko)\n\n![](attached_image_1.png)\n**Fig. 6**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since **AB** is a hypothenuse, then $\\angle MCB = \\angle MBC$ (Fig. 6).\n\nOn the other hand, $AC^2 = CN \\cdot CB$, since $BC = \\sqrt{2}AC$ and $CN = \\frac{1}{2}BC = \\frac{\\sqrt{2}}{2}AC$.\n\nThen $\\triangle ACN \\sim \\triangle ABC$. Hence $\\angle CAN = \\angle ABC$.\n\nThus, $\\angle AKC = \\angle ACK + \\angle NAC = \\angle ACK + \\angle KCB = 90^\\circ$, that finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45622,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is an infinite sequence of letters $a$ and $b$. In this sequence, one can perform a substitution $abb \\rightarrow baa$. It is known that regardless of the order of these substitutions, one can only make a finite number of them. Prove that, in that case, substitutions $aabb \\rightarrow bbaa$ can also be made only a finite number of times.\n(Arsenii Nikolaiev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since there can only be made a finite number of substitutions, there is a number $N$, starting from which substitutions are not possible. That means that there are no pairs of subsequent letters $b$. Because, otherwise, on the left from them there is letter $a$, and substitution is possible. But then the second type of substitution is not possible, either, because starting from $N$, there is no pair of letters $b$. And all other pairs of $b$ get moved to the left by each second-type substitution. And there is only a finite number of such moves possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute angle triangle, where $AB < AC$, and let $M$ be the midpoint of $BC$, and $K$ be the midpoint of the polygonal chain $BAC$. Show that $\\sqrt{2} KM > AB$.\n\n(Heorhii Naumenko)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the midpoint of $AC$. Since $K$ is the midpoint of polygonal chain $BAC$, the following holds (Fig. 15):\n$$\n\\frac{1}{2}(AB + AC) = KC = KN + NC = KN + \\frac{1}{2}AC \\Rightarrow KN = \\frac{1}{2}AB = NM.\n$$\nBy the cosine theorem for $\\triangle KNM$:\n$$\n\\begin{aligned}\n& KM^2 = KN^2 + NM^2 - 2 \\cdot KN \\cdot NM \\cdot \\cos \\angle KNM \\\\\n& \\Rightarrow KM^2 = 2KN^2(1 - \\cos \\angle KNM) = \\frac{1}{2} AB^2 (1 + \\cos \\angle CNM)\n\\end{aligned}\n$$\nSince $\\triangle ABC$ is acute angled, $\\angle CNM = \\angle CAB < 90^\\circ \\Rightarrow KM^2 > \\frac{1}{2} AB^2$, that completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45624,
    "subject": "Mathematics (Multi-modal)",
    "question": "A right triangle $ABC$ is called *special*, if the lengths of its sides $AB$, $BC$ and $CA$ are integers, and there is a point $X$ on each side (different from the vertices of $\\triangle ABC$), for which lengths of $AX$, $BX$ and $CX$ are integers. Find at least one special triangle.\n\n![](attached_image_1.png)\n**Fig. 11**",
    "options": [],
    "answer": "A special triangle is one with side lengths 36, 48, and 60.",
    "solution": "Let triangle $\\triangle ABC$ have a right angle at $C$. Then for any triangle with even integer hypotenuse, the midpoint of hypotenuse is the point of interest (Fig. 11).\n\nLet $BC = an$, $AC = bn$, $AB = cn$, $n \\in \\mathbb{N}$ and $a^2 + b^2 = c^2$. Let $K \\in AC$, $N \\in BC$, $CN = bk$, $CK = ak$, $k \\in \\mathbb{N}$. It suffices for lengths $AN = b\\sqrt{n^2 + k^2}$ and $BK = a\\sqrt{n^2 + k^2}$ to be integer. Thus, it suffices that $n^2 + k^2 = m^2$ for some $m \\in \\mathbb{N}$, $bk < an$ and $ak < bn$. Let $a = 3$, $b = 4$, $c = 5$, $k = 5$, $n = 12$, $m = 13$, hence $an = 36$, $bn = 48$, $ak = 15$, $bk = 20$. Thus, we obtain the following segments:\n$$\nAC = 48, \\ CK = 15, \\ BC = 36, \\ CN = 20, \\ AB = 60.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three numbers $2^{100}$, $3^{100}$ and $5^{100}$ are written on a long paper strip without any space in-between, thus, creating one big number $N$. Arsenii claims that he can change the last digit of number $N$ so that the new number is a power of $13$. Is he right?",
    "options": [],
    "answer": "No",
    "solution": "Suppose Arsenii's claim is correct and by changing the last digit of number $N$ he obtained $13^k$, where $k$ is a positive integer. Clearly, he had to change the last digit, because it is $5$. Since $2^{100} \\equiv 1 \\pmod{3}$, $3^{100} \\equiv 0 \\pmod{3}$ and $5^{100} \\equiv 1 \\pmod{3}$, the sums of their digits have the same remainders modulo $3$. The change of the last digit $5$ to digit $l$ means that we subtracted $2$ (or added $1$) and added $l$ modulo $3$. Thus, the new number (denoted by $M$) equals $l \\equiv 0 \\pmod{3}$. Since $13^k \\equiv 1 \\pmod{3}$, we could add one of the following digits: $1$, $4$ or $7$. Since the last digit of $13^k$ can be $1$, $3$, $7$ or $9$, we need to consider two cases.\nCase 1. $13^k$ ends with $1$, it is possible in the case $k \\equiv 0 \\pmod{4}$, then the following is true modulo $8$: $M = 13^k = 13^{4j} = 169^{2j} \\equiv 1 \\pmod{8}$. Note that $M = N - 4$. $N \\equiv 5^{100} = 25^{50} \\equiv 1 \\pmod{8} \\Rightarrow M = N - 4 \\equiv 5 \\pmod{8}$, so we get a contradiction.\nCase 2. $13^k$ ends with $7$, it is possible if $k \\equiv 3 \\pmod{4}$, then the following is true modulo $8$: $M = 13^k = 13^{4j+3} = 169^{2j} \\cdot 13 \\equiv 5 \\pmod{8}$. Note that $M = N + 2 \\Rightarrow M \\equiv 3 \\pmod{8}$ – contradiction.\nThese contradictions complete the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the average of all 5-digit numbers that satisfy the following:\n* The number is of the form $\\overline{ab0cd}$, that is, its third digit is zero;\n* Digits are pairwise distinct;\n* Both numbers $\\overline{ab0cd}$, and $\\overline{dc0ba}$ are divisible by 7.",
    "options": [],
    "answer": "55055",
    "solution": "Let us rewrite the problem the following way: since $1001$ is divisible by $7$, then\n$$\n\\begin{align*}\n\\overline{ab0cd} &= 1000 \\cdot \\overline{ab} + \\overline{cd} = 1001 \\cdot \\overline{ab} + (\\overline{cd} - \\overline{ab}) \\\\\n&\\Rightarrow (\\overline{cd} - \\overline{ab}) \\text{ is divisible by } 7 \\\\\n&\\Rightarrow 10(c-a) + (d-b) \\text{ is divisible by } 7 \\\\\n&\\Rightarrow 3(c-a) + (d-b) \\text{ is divisible by } 7.\n\\end{align*}\n$$\n\nSimilarly, $3(d-b)+(a-c)$ must be divisible by $7$. Thus, $3x+y=7k$ and $3y+x=7l \\Rightarrow 9x+3y=21k \\Rightarrow 8x=7(3k-l) \\Rightarrow x$ is divisible by $7$, and similarly $y$ is divisible by $7$. Therefore, there exist two pairs of nonzero digits, difference of which is divisible by $7$. There are two such pairs: $(8, 1)$ and $(9, 2)$. Clearly, the pair $(7, 0)$ is not considered, since it contains zero, that already is one of the digits of the number. So, there are eight numbers that satisfy the conditions: $12089$, $19082$, $21098$, $28091$, $89012$, $82019$, $98021$ and $91028$. Their average equals $55055$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45627,
    "subject": "Mathematics (Multi-modal)",
    "question": "In expression $1\\ 2\\ 3\\ 4\\ 5\\ 6\\ 7\\ 8\\ 9$ you should arrange parentheses and four arithmetic symbols $+, -, \\times, :$ between some of the digits to obtain the largest possible number.\n(Bogdan Rublyov)",
    "options": [],
    "answer": "123456789",
    "solution": "It is clear that the largest number you can get if you use only the actions of multiplication and addition. We will show that there is actually no sense to put any characters and is the largest nine-digit number $123456789$.\n\nFirst remove one \"+\" sign, that is write $\\overline{XY}$ instead of expression $X + Y$, as a result, the expression will obviously increase, because\n\n$$\n\\overline{XY} \\ge 10X + Y > X + Y.\n$$\n\nNow the expression contains only the multiplication signs. Then, we will substitute $X \\cdot Y$ with $\\overline{XY}$, and then we will show that the new expression will increase. Let $Y$ have $k$ digits, i.e. $10^{k-1} < Y < 10^k$, then\n$$\n\\overline{XY} \\ge 10^k X + Y > X \\cdot Y \\Leftrightarrow (10^k - Y)X + Y > 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all tuples of positive integers $(m, n, k)$ that satisfy the equation\n$$\n(m! + m)(n! + n) = (k! + k).\n$$",
    "options": [],
    "answer": "(1, 1, 2)",
    "solution": "From the problem statement, it is obvious that $k > m$ and $k > n$. We rewrite the given equation as\n$$\nm n ((m-1)! + 1)((n-1)! + 1) = k ((k-1)! + 1).\n$$\n\nSince $(k-1)!$ is divisible by $m$ and $n$, $(k-1)! + 1$ is not divisible by any factor of these numbers. Hence, $k \\nmid m n$. Suppose $n \\ge m$. If $m \\ge 2$, then\n$$\n\\begin{aligned}\n4(n!)^2 &\\ge 4 m! n! = (2 m!) (2 n!) \\ge (m! + m)(n! + n) = k! + k > (m n)! \\ge (2 n)! \\\\\n4 n! n! &> (2 n) \\cdot (2 n - 1) \\cdots (n + 1) \\cdot n! \\Rightarrow 4 n! > (2 n) \\cdot (2 n - 1) \\cdots (n + 1) \\\\\n4 n \\cdot (n - 1) \\cdot (n - 2) \\cdots 2 \\cdot 1 &> (2 n) \\cdot (2 n - 1) \\cdots (2 n - 2) \\cdots (n + 2) \\cdot (n + 1) \\\\\n4 &> \\frac{2 n}{n} \\cdot \\frac{2 n - 1}{n - 1} \\cdots \\frac{n + 1}{1} \\Rightarrow \\text{contradictions for } n \\ge 2.\n\\end{aligned}\n$$\nLet us show that for $n \\ge 2$ this yields a contradiction. On the right-hand side, each factor except the first and last one is greater than the ones on the left. Now, let us show that the product of the first and last factors is still greater on the right-hand side: $4 n \\cdot 1 > (2 n) \\cdot (n + 1) > 6 n$.\n\nTherefore, there is only one option $m = 1$, hence,\n$$\n2 \\cdot n! + 2 n = k! + k \\ge (n + 1)! + (n + 1) > (n + 1)! + n \\Rightarrow n > n! (n - 1) \\Rightarrow n = 1 \\Rightarrow m = 1.\n$$\n\nThen, we easily get $k = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the triangle **ABC** be such that $2AC = AB$ and $\\angle A = 2\\angle B$. Let **AL** be its bisector and let **M** be the midpoint of **AB**. It turns out that $CL = ML$. Show that $\\angle B = 30^\\circ$.\n\n(Danylo Khilko)\n\n![](attached_image_1.png)\n**Fig. 4**",
    "options": [],
    "answer": "30°",
    "solution": "Since **AL** is a bisector, then $\\angle CAL = \\angle LAB = \\angle CBA$ (Fig. 4). Then $\\triangle ALB$ is isosceles, so **LM** is its altitude and a median. Thus $\\angle LMA = 90^\\circ$. Consider the triangles **AML** and **ALC**. Let **C**' be the projection of **L** on **AC**. Then right triangles **AML** and **AC'L** are equal by hypothenuse and the angle. Thus, $LC' = LM = LC$. Therefore, $C = C'$, since there exists only one projection.\n\nThus $\\triangle ABC$ is a right triangle for which $2AC = AB$, hence $\\angle ABC = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45630,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a quadrilateral $ABCD$, $\\angle ABD = \\angle DBC$ and $AD = CD$. Let $DH$ be the height of $\\triangle ABD$. Prove that $|BC - BH| = HA$.\n\n(Danylo Khilko)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On the ray $BA$, we put down a segment $BE = BC$. If point $E$ belongs to $AB$ (fig. 19), then $\\triangle BCD = \\triangle BED$ due to two pairs of equal sides and the angle between them. Then, $AD = CD = ED$, which yields that $\\triangle ADE$ is isosceles. There, $HD$ is the height and, hence, the median. Therefore,\n\n$$\nAH = HE = HB - BE = HB - BC \\Rightarrow BC = BH - HA.\n$$\n\nIf point $A$ belongs to the segment $BE$ (fig. 20), then, analogously, $\\triangle BCD = \\triangle BED$ and\n\n$$\nAH = HE = BE - HB = BC - HB \\Rightarrow BC = BH + HA.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the $m \\times n$ table, $m, n \\ge 2$ ($m$ rows are enumerated $1, 2, \\ldots, m$ and $n$ columns are enumerated $1, 2, \\ldots, n$), which is filled with positive integers. Let $b_i$ be the $lcm$ (least common multiple) of all numbers in the $i^{th}$ row, $1 \\le i \\le m$, and let $B$ be the $gcd$ (greatest common divisor) of numbers $(b_1, b_2, \\ldots, b_m)$. Also, let $c_j$ be the $gcd$ of all numbers in $j^{th}$ column, $1 \\le j \\le n$, and let $C$ be the $lcm$ of numbers $(c_1, c_2, \\ldots, c_n)$. Is it true that $B$ is divisible by $C$, or is $C$ divisible by $B$?",
    "options": [],
    "answer": "B is divisible by C",
    "solution": "**Answer:** $B$ is divisible by $C$.\n\nConsider any prime number $p$, and its power for each number in the table. Replace all the numbers in the table with a power of the chosen prime number. Let the $m \\times n$ table is filled with $\\alpha_{i,j}$, $i=1, \\overline{m}$, $j=1, \\overline{n}$. Now $\\beta_i$ is the greatest number from the corresponding row, and $B$ is the smallest of $\\beta_i$. Similarly, $\\gamma_j$ is the smallest number of the corresponding column, and $\\Gamma$ is the greatest of $\\gamma_j$. Thus both $B$ and $\\Gamma$ are in the table. If they belong to the same row or column, then $B \\ge \\Gamma$ by construction. If they are in different rows and columns, find $\\Delta$, which is the number on the intersection of the column of $\\Gamma$ and the row of $B$. Then by construction $\\Gamma \\le \\Delta \\le B$. Thus, the power of $p$ in $B$ is not less than its power in $C$. Since $p$ is arbitrary, $C|B$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Line $l$ is perpendicular to the side $AC$ of acute triangle $ABC$, and intersects $AC$ at $K$. $l$ intersects circumscribed circle of $\\triangle ABC$ at $P$ and $T$ (point $P$ in the same half-plane w.r.t. $AC$ as vertex $B$). By $P_1$ and $T_1$ are denoted projections of points $P$ and $T$ to line $AB$. Furthermore, vertices $A, B$ belong to the segment $P_1T_1$. Prove that the center of circumscribed circle of $\\triangle P_1KT_1$ lies on the line, which contains midsegment of $\\triangle ABC$, parallel to the side $AC$.\n(Anton Tryhub)\n\n![](attached_image_1.png)\nFig. 31",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We denote by $B_1$ a projection of vertex $B$ to line $l$, then quadrilaterals $BB_1PP_1$ and $TKAT_1$ are inscribed, with diameters $BP$ and $AT$. Then, we have the following equalities of angles:\n\n$$\n\\angle P_1B_1K = \\pi - \\angle P_1B_1P = \\pi - \\angle P_1BP = \\\\\n\\angle ABP = \\pi - \\angle ATP = \\pi - \\angle ATK = \\\\\n= \\pi - \\angle AT_1K = \\pi - \\angle P_1T_1K.\n$$\n\nTherefore, quadrilateral $P_1B_1KT_1$ is inscribed, hence, center of circumscribed circle of $\\triangle P_1KT_1$ lies on the perpendicular bisector to the segment $B_1K$. And, clearly, midsegment of $\\triangle ABC$ parallel to $AC$ lies on this exact perpendicular bisector. Q.E.D.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45633,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, by $T_A, T_B, T_C$ are denoted tangency points of excircles of $\\triangle ABC$ tangent to sides $BC, AC$ and $AB$, respectively. Let $O$ be the center of circumscribed circle of $\\triangle ABC$, and $I$ be the center of its inscribed circle. It is known that $OI \\parallel AC$. Prove that $\\angle T_A T_B T_C = 90^\\circ - \\frac{1}{2} \\angle ABC$.\n\n![](attached_image_1.png)\n\n**Fig.29**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I_A, I_B, I_C$ be the centers of excircles, $I_1$ be the point, symmetrical to $I$ with respect to point $O$ (fig. 29). Let us look at $\\Delta I_A I_B I_C$. $I$ is its orthocenter, since bisectors of outer and inner angles are perpendicular. It is also clear that $O$ is the center of Euler's circle of this triangle. Then, point $I_1$ is the center point of circumscribed circle of $\\Delta I_A I_B I_C$. Then,\n\n$\\angle I_1 I_B C = \\angle I_C I_B B$, which yields $I_1 I_B \\perp AC$, since $\\angle B I_C A = 90^\\circ - \\frac{1}{2} \\angle BCA = \\angle AC I_B$.\n\nHence, lines $I_B T_B$, $I_A T_A$ and $I_C T_C$ intersect at point $I_1$.\n\nClearly, $AII_1C$ is a trapezoid. Points $I, I_1$ are symmetrical with respect to the median perpendicular of $AC$, since $II_1 \\parallel AC$. Then, $AII_1C$ is an isosceles trapezoid, and $\\angle A I_1 C = \\angle A I C = 90^\\circ + \\frac{1}{2} \\angle ABC$. Let us also note that quadrilaterals $I_1 T_C A T_B$ and $I_1 T_A C T_B$ are inscribed. Then,\n\n$$\n\\angle T_A T_B T_C = \\angle T_C T_B I_1 + \\angle I_1 T_B T_A = \\angle I_1 A B + \\angle I_1 C B = \\angle A I_1 AC - \\angle ABC = 90^\\circ - \\frac{\\angle ABC}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Equilateral triangle $ABC$ is inscribed in circle $w$. Points $F$ and $E$ are chosen on sides $AB$ and $AC$, respectively, so that $\\angle ABE + \\angle ACF = 60^\\circ$. Circumscribed circle of $\\triangle AFE$ intersects circle $w$ at point $D$. Rays $DE$ and $DF$ intersect line $BC$ at points $X$ and $Y$, respectively. Prove that the center point of inscribed circle of $\\triangle DXY$ does not depend on the choice of points $F$ and $E$.\n(Danylo Khilko)\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote by $P$ intersection point of $CF$ and $BE$. Then, $\\angle CBE = 60^\\circ - \\angle ABE = \\angle FCA$. Analogously, $\\angle FCB = \\angle ABE$ (fig. 27). Then, $\\angle FKE = \\angle CKB = 180^\\circ - \\angle CBE - \\angle FCB = 120^\\circ$.\n\nThus, quadrilateral $AFPE$ is inscribed in circle $\\gamma$, which point $D$ lies on. We draw a bisector of $\\angle CAB$ until its intersection with $\\gamma$ at some point $O$. Obviously, point $O$ is the middle of the smaller arc $EF$ of circle $\\gamma$. Let us prove that $O$ is the center of equilateral $\\triangle ABC$. $\\triangle AFC = \\triangle BEC$ by the side and two adjacent angles. Hence, $FC = BE$. Naturally, $\\angle OAFC = \\angle BEO$ and $OF = OE$. Thus, $\\triangle OFC = \\triangle BEO$ by two pairs of sides and an angle between them, which yields $\\angle PBO = \\angle PCO$, i.e. quadrilateral $BPOC$ is inscribed, hence, $\\angle BOC = 120^\\circ$. Therefore, point $O$ lies on the bisector of $\\angle CAB$ in the same half-plane as point $P$ with respect to line $BC$, and $\\angle BOC = 120^\\circ$. Obviously, then point $O$ is the center of $\\triangle ABC$.\n\nNow, we prove that $O$ is the incenter of $\\triangle DXY$. Clearly, then this center does not depend on the points $E$ and $F$. Obviously, $\\angle YDO = \\angle XDO = 30^\\circ$, i.e. $DO$ is a bisector of $\\angle XDY$. Suffices to show that $YO$ is a bisector of $\\angle DYX$. $\\angle YDO = \\angle OBY = 30^\\circ$, hence, quadrilateral $DOYB$ is inscribed, and $DO = OB$, as radii. Therefore,\n$$\n\\angle OYX = \\angle ODB = \\angle DBO = \\angle DYO,\n$$\ni.e. $YO$ is a bisector of $\\angle DYX$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circle $k$ of radius $r$ is inscribed in $\\triangle ABC$. Tangent lines of $k$, that are parallel to sides $AB$, $BC$ and $CA$, intersect other sides of $\\triangle ABC$ at points $M, N$; $P, Q$ and $L, T$ ($P, T \\in AB$, $L, N \\in BC$ and $M, Q \\in AC$). Denote by $r_1, r_2, r_3$ radii of circles inscribed in triangles $MNC, PQA$ and $LTB$, respectively. Prove that $r_1 + r_2 + r_3 = r$.\n\n![](attached_image_1.png)\n**Fig. 28**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since all these triangles are similar, we get\n$$\n\\frac{r_1 + r_2 + r_3}{r} = \\frac{p_1 + p_2 + p_3}{p}\n$$\nIt is not hard to see that (fig. 28)\n$$\n\\begin{aligned}\n2p_1 &= CM + CN + MN = \\\\\n&= CM + CN + MZ + ZN = \\\\\n&= CM + MX + CN + NY = 2p - (AX + AB + BY) = 2p - 2c.\n\\end{aligned}\n$$\nThus,\n$$\n\\frac{r_1 + r_2 + r_3}{r} = \\frac{p_1 + p_2 + p_3}{p} = \\frac{p-a+p-b+p-c}{p} = \\frac{3p-2p}{p} = 1.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $a$, $b$, $c$, $d$ such that $a < b < c < d$. Is it possible for the least common multiple of $a$ and $b$ to be greater than the least common multiple of $c$ and $d$?",
    "options": [],
    "answer": "Yes; for example, a=8, b=9, c=10, d=25 gives lcm(a,b)=72 and lcm(c,d)=50.",
    "solution": "Let $a = 8$, $b = 9$, then $[a; b] = 72$. We will choose the other two numbers such that $[c; d] = 50$: $c = 10$, $d = 25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer of the form $\\overline{30x070y03}$, which is divisible by $37$, where $x$, $y$ are digits.",
    "options": [],
    "answer": "300070703",
    "solution": "Let us rewrite this number as follows:\n$$\n\\overline{30x070y03} = 300070003 + 10^6x + 10^2y = 37 \\cdot (8110000 + 27027x + 3y) + (3 + x - 11y).\n$$\nThus, $3 + x - 11y$ must be divisible by $37$ for the smallest $x$. Since $x$, $y$ are digits, then this expression can only take values $0$, $-37$, $-74$. For each, we have the following answers.\n$$\n3 + x - 11y = 0 \\Rightarrow y = 1 \\text{ and } x = 8.\n$$\n$$\n3 + x - 11y = -37 \\Rightarrow y = 4 \\text{ and } x = 4.\n$$\n$$\n3 + x - 11y = -74 \\Rightarrow y = 7 \\text{ and } x = 0.\n$$\nThus, the smallest number of these is $300070703$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$, and prime numbers $p, q$, which satisfy the equation\n$$\nn^3 = p^3 + 2p^2q + 2pq^2 + q^3.\n$$",
    "options": [],
    "answer": "No solutions",
    "solution": "We rewrite the equation as follows.\n$$\nn^3 = p^3 + 2p^2q + 2pq^2 + q^3 = (p+q)(p^2 + pq + q^2)\n$$\nClearly, $p \\neq q$, because, otherwise, the equation $n^3 = 6p^3$ would hold, which is not possible for positive integers.\nSuppose there exists such $s > 1$, which is a factor of both $p+q$ and $(p^2 + pq + q^2)$. But then $s \\mid (p+q)^2 - (p^2 + pq + q^2) = pq$. If, for example, $s \\mid p$, then from $s \\mid (p+q)$ follows $s \\mid q \\Rightarrow$ from $p, q$ being prime, it must be that $p = q$, which leads to contradiction.\nIf, on the other hand, $((p+q), (p^2 + pq + q^2)) = 1$, then each factor must be a cube of a positive integer. Suppose $p+q = x^3$, $p^2 + pq + q^2 = y^3$, and $n = xy$. Then,\n$$\npq = (p+q)^2 - (p^2 + pq + q^2) = x^6 - y^3 = (x^2 - y)(x^4 + x^2y + y^2).\n$$\nSince $x^4 + x^2y + y^2 > x^3 = p+q$, we obtain $x^4 + x^2y + y^2 = pq$ and $x^2 - y = 1 \\Rightarrow pq = x^4 + x^2(x^2-1) + (x^2-1)^2 = 3x^4 - 3x^2 + 1 = 3x^2(x-1)(x+1) + 1 \\equiv 1 \\pmod 9$, since $x(x-1)(x+1) \\not\\equiv 3 \\pmod 9$.\nWhat is left is to search through the cases modulo 9. Cube of an integer modulo 9 can be equal to 0, \\pm 1. Let us list all possible cases for remainders modulo 9 of primes $p, q$ so that condition $pq \\equiv 1 \\pmod 9$ is satisfied.\n$$\np \\equiv 1 \\pmod 9 \\Rightarrow q \\equiv 1 \\pmod 9 \\Rightarrow x^3 = p+q \\equiv 2 \\pmod 9 \\{\\text{ -- contradiction.}\\}\n$$\n$$\np \\equiv 2 \\pmod 9 \\Rightarrow q \\equiv 5 \\pmod 9 \\Rightarrow x^3 = p+q \\equiv 7 \\pmod 9 \\{\\text{ -- contradiction.}\\}\n$$\n$$\np \\equiv 4 \\pmod 9 \\Rightarrow q \\equiv 7 \\pmod 9 \\Rightarrow x^3 = p+q \\equiv 2 \\pmod 9 \\{\\text{ -- contradiction.}\\}\n$$\n$$\np \\equiv 8 \\pmod 9 \\Rightarrow q \\equiv 8 \\pmod 9 \\Rightarrow x^3 = p+q \\equiv 7 \\pmod 9 \\{\\text{ -- contradiction.}\\}\n$$\nAll the cases were checked, which concludes the proof that such numbers do not exist.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for the positive $x$, $y$, $z$ the following inequality holds:\n$$\n\\frac{x^8+1}{x^4} + \\frac{y^8+1}{y^4} + \\frac{z^8+1}{z^4} \\geq 2 \\cdot \\left( \\frac{x}{z} + \\frac{z}{y} + \\frac{y}{x} \\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Due to inequality of arithmetic and geometric means applied several times:\n$$\n\\left( x^4 + \\frac{1}{y^4} \\right) + \\left( y^4 + \\frac{1}{z^4} \\right) + \\left( z^4 + \\frac{1}{x^4} \\right) \\geq 2 \\cdot \\left( \\frac{x^2}{y^2} + \\frac{y^2}{z^2} + \\frac{z^2}{x^2} \\right) = \\\\\n= \\left( \\frac{x^2}{y^2} + \\frac{y^2}{z^2} \\right) + \\left( \\frac{y^2}{z^2} + \\frac{z^2}{x^2} \\right) + \\left( \\frac{z^2}{x^2} + \\frac{x^2}{y^2} \\right) \\geq 2 \\cdot \\left( \\frac{x}{z} + \\frac{y}{x} + \\frac{z}{y} \\right),\n$$\nthis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45640,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which of the sets $A = \\mathbb{R}$ or $A = \\mathbb{Q}$ is there a function $f : A \\to (0, +\\infty)$ such that $\\forall x_1, x_2 \\in A,\\ x_1 \\neq x_2$ the condition $\\min\\{f(x_1), f(x_2)\\} \\leq |x_1 - x_2|$ is satisfied?\n(Andrii Anikushin)",
    "options": [],
    "answer": "Only for A = Q; no such function exists for A = R.",
    "solution": "Suppose that the set $A$ is more than countable set.\nDivide half-plane $\\mathbb{R} \\times (0, +\\infty)$ to the counted number of rectangles:\n$$\nD_{n,m} = [n, n+1) \\times \\left[\\frac{1}{m+1}, \\frac{1}{m}\\right), \\ n \\in \\mathbb{Z},\\ m \\in \\mathbb{N},\\ D_{n,0} = [n, n+1) \\times [1, +\\infty).\n$$\nBy construction, $\\bigcup_{n \\in \\mathbb{Z}} \\bigcup_{m \\in \\mathbb{Z}^+} D_{n,m} = \\mathbb{R} \\times (0, +\\infty)$, in addition, any two of determined earlier rectangles do not intersect. Consider the graph $G = \\{(x, f(x)) \\mid x \\in A\\}$ of the function $f$.\nBecause $A$ is more than countable, the set $G$ is also more than countable ($A \\sim G$).\nSuppose that $\\forall n \\in \\mathbb{Z}, m \\in \\mathbb{Z}^+$ (i.e. the set $G \\cap D_{n,m}$) is finite, but then the whole set $G$ is at most countable, that contradicts its construction. Therefore, $\\exists n_0 \\in \\mathbb{Z}, m_0 \\in \\mathbb{Z}^+ : G \\cap D_{n_0,m_0}$ is infinite set. This means that there is an infinite set $X \\subset [n_0, n_0+1) \\cap A$ such that $\\forall x \\in X$ $f(x) > \\frac{1}{m_0+1}$. Then due to Bolzano-Weierstrass theorem there is a sequence of different elements that converges to some number. As a result, this sequence is fundamental and $\\forall k \\in \\mathbb{N}$ $f(x_k) > \\frac{1}{m_0+1}$. But then we have the following relationship: $0 = \\lim_{k \\to \\infty} |x_{k+1} - x_k|$ and $\\min\\{f(x_k), f(x_{k+1})\\} \\geq \\frac{1}{m_0+1}$, which contradicts the problem's condition.\n\nThereby, $A$ should be no more than countable. Now determine the required function. Let $A = \\{x_1, x_2, ...\\}$. Define the function by induction: $f(x_1) = 1$. Now let us know $f(x_1), ..., f(x_n)$. Let us make the following notation: $r = \\min_{k=1,n} |x_{n+1} - x_k| > 0$. Let's set $f(x_{n+1}) = \\frac{r}{2}$.\nWe should verify that the function which is defined for $A$ satisfies the conditions of the problem.\n\nExample: for $A = \\mathbb{Q}$: if $x = \\frac{p}{q}$, where $(p, q) = 1$, $p \\in \\mathbb{Z}$, $q \\in \\mathbb{N}$ ($0 = \\frac{0}{1}$), we will set $f\\left(\\frac{p}{q}\\right) = \\frac{1}{q^2}$. Then if $\\frac{p_1}{q_1} \\neq \\frac{p_2}{q_2}$ we have:\n$$\n|x_1 - x_2| = \\frac{|p_1 q_2 - q_1 p_2|}{q_1 q_2} \\geq \\frac{1}{q_1 q_2} \\geq \\frac{1}{\\max\\{q_1^2, q_2^2\\}} = \\min\\{f(x_1), f(x_2)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(m, n)$ that satisfy the equation\n$$\nm! + n! = m^n + 1.\n$$",
    "options": [],
    "answer": "(1, 1), (2, 1), (5, 3)",
    "solution": "**Answer:** $(1, 1)$, $(2, 1)$, $(5, 3)$.\n\nBy checking trivial cases $m = 1$, $n = 1$ and $n = 2$, we find the first two of the solutions listed. From now on, we assume $m \\ge 2$ and $n \\ge 3$.\n\n**Step 1.** We prove that $m > n$.\nSuppose $n \\ge m$. Then, numbers $m!$, $n!$ and $m^n$ are divisible by $m$, hence, $m$ also divides $m! + n! - m^n = 1$ – contradiction. In particular, it follows that $m \\ge 4$.\n\n**Step 2.** Now, we show that $n > \\frac{1}{2}m$.\nFrom well-known inequality $m! > m^{\\frac{1}{2}m}$, which follows from the Root-Mean Square-Arithmetic-Mean-Geometric-Mean-Harmonic-Mean Inequality, we obtain $m^n + 1 = m! + n! > m^{\\frac{1}{2}n} + 1$.\n\n**Step 3.** We prove that $m$ is prime.\nOtherwise, $m$ has a prime factor $p$ not larger than $\\frac{1}{2}m$ and $n$. In that case, $p$ is a factor of $m! + n! - m^n = 1$ – contradiction.\n\n**Step 4.** We prove that $m-1$ is either prime or a square of a prime. Suppose the contrary. Then, $m-1$ can be represented as a product of two factors $a \\ne b$, both smaller than $\\frac{1}{2}m$ and $n$. Therefore, $m!$ and $n!$ are divisible by $n-1 = ab$, since $n!$ contains both these factors. $m^n + 1 = ((m-1) + 1)^n + 1$ gives a remainder of 2 when divided by $m-1$. Hence, $m-1$ is a factor of 2, which is impossible, since $m \\ge 4$.\n\nFrom the results of Steps 3 and 4 and from parity, we get that the only possible case is $m = 5$. Indeed, since $m \\ge 4$ is prime, it is odd, hence, $m-1$ is even and can be a square of a prime, if that prime is 2. From the results of Steps 1 and 2, we get that possible solutions are only $(5, 3)$ and $(5, 4)$. By checking, we see that only the first pair satisfies the initial equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the positive integer $n$, for which the following holds:\n\n$$\nn^2 = 2 \\cdot (20^4 + 19^4 + 39^4).\n$$",
    "options": [],
    "answer": "2282",
    "solution": "Consider the following expression:\n$$\nx^4 + y^4 + (x + y)^4 = 2x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 2y^4,\n$$\nthus\n$$\n2(x^4 + y^4 + (x+y)^4) = 4(x^4 + 2x^3y + 3x^2y^2 + 2xy^3 + y^4) = (2(x^2 + xy + y^2))^2.\n$$\nHence for $x = 20$, $y = 19$ the following holds:\n\n$$\nn^2 = (2(x^2 + xy + y^2))^2 \\text{ or } n = 2(20^2 + 20 \\cdot 19 + 19^2) = 2282.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest fraction out of 1010 fractions given below:\n$$\n\\begin{array}{ccccccc}\n\\frac{1}{2019}, & \\frac{1+2018}{2019+2}, & \\frac{1+2018+3}{2019+2+2017}, & \\frac{1+2018+3+2016}{2019+2+2017+4}, & \\\\\n& \\frac{1+2018+3+2016+5}{2019+2+2017+4+2015}, & \\dots, & & \\\\\n& & \\frac{1+2018+3+2016+...+1009}{2019+2+2017+4+...+1011}, & \\frac{1+2018+3+2016+...+1009+1010}{2019+2+2017+4+...+1011+1010}, & \\\\\n& & & \\frac{1+2018+3+2016+...+1009+...+1008+1010}{2019+2+2017+4+...+1011+...+1010+1010}. & \\frac{1+2018+3+2016+...+1009+...+1008+1010}{2019+2+2017+4+...+1011+...+1010+1010}. & \\frac{1}{2019} &\n\\end{array}\n$$\n(Bohdan Rublyov)",
    "options": [],
    "answer": "2019/2021",
    "solution": "Let us split all fractions into two natural groups: those on the odd position and those on the even.\n$$\n\\begin{align*}\na_1 &= \\frac{1}{2019}, \\\\\na_2 &= \\frac{1+2018+3}{2019+2+2017} = \\frac{1+2021}{2019+2019}, \\\\\na_3 &= \\frac{1+2018+3+2016+5}{2019+2+2017+4+2015} = \\frac{1+2021+2}{2019+2019+2}, \\dots, \\\\\na_{505} &= \\frac{1+2018+3+2016+5+\\dots+1012+1009}{2019+2+2017+4+2015+\\dots+1008+1011} = \\frac{1+2021+504}{2019+2019+504}. \\\\\nb_1 &= \\frac{1+2018}{2019+2} = \\frac{2019}{2021}, \\\\\nb_2 &= \\frac{1+2018+3+2016}{2019+2+2017+4} = \\frac{2019+2}{2021+2} = \\frac{2019}{2021}, \\\\\nb_3 &= \\frac{1+2018+3+2016+5+2014}{2019+2+2017+4+2015+6} = \\frac{2019+3}{2021+3} = \\frac{2019}{2021}, \\dots, \\\\\nb_{505} &= \\frac{1+2018+3+2016+5+\\dots+1012+1009+1010}{2019+2+2017+4+2015+\\dots+1008+1011+1010} = \\frac{2019+505}{2021+505} = \\frac{2019}{2021}.\n\\end{align*}\n$$\n\nIn this way, all fractions denoted by $b_i$, $i = 1, 505$ are of the same value, $\\frac{2019}{2021}$. Let us now compare values of $a_k$ and $a_{k+1}$:\n$$\na_{k+1} - a_k = \\frac{1+2021k}{2019+2019k} - \\frac{1+2021(k-1)}{2019+2019(k-1)} = \\frac{(1+2021k)(2019+2021(k-1)) - (1+2021k)(2019+2021k)}{(2019+2019(k-1))(2019+2019k)} > 0 \\Leftrightarrow\n$$\n\n2021 + 2021 \\cdot (k-1) + 2021^2 k + 2021^2 k(k-1) - 2021 - 2021 \\cdot k - 2021^2(k-1) - 2021^2 k(k-1) = 2021^2 - 2021 > 0\n$$\nTherefore, the largest fraction would be $\\frac{2019}{2021}$ or $a_{505} = \\frac{1+2021\\cdot504}{2019+2019\\cdot504}$. Let us compare them.\n$$\n\\frac{1+2021\\cdot504}{2019+2019\\cdot504} - \\frac{2019}{2021} = \\frac{(1+2021\\cdot504) \\cdot 2021 - (2019+2019\\cdot504) \\cdot 2019}{(2019+2019\\cdot504) \\cdot 2021} < 0 \\Leftrightarrow \\\\ 2021 + 2021 \\cdot 2021 \\cdot 504 - 2019^2 - 2019^2 \\cdot 504 = \\\\ = 2021 + (2021^2 - 2019^2) \\cdot 504 - 2019^2 = 2021 + 8080 \\cdot 504 - 2019^2 = -2020 < 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45644,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers, neither is divisible by $6$. A rectangle $m \\times n$ is filled with $2 \\times 2$ and $3 \\times 3$ squares. Show that such rectangle can be filled with squares of one of the types: $2 \\times 2$ or $3 \\times 3$.\n\n(Edvard Turkevych)\n\n![](attached_image_1.png)\n\n**Fig. 12**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the case when one of the numbers is not divisible by two and another one is not divisible by three. Without loss of generality, let $m$ be not divisible by $2$, and let $n$ be not divisible by $3$, and let $m$ be the width and $n$ be the height of the rectangle.\n\nWe will color in black all the rows of the rectangle $m \\times n$ of the type $3k$ and $3k+2$, $k \\in \\mathbb{N}$ (Fig. 12). The number of black rows is odd: $2\\left[\\frac{1}{3}n\\right] + 1$, hence the odd number of black cells, since there are odd ($m$) number of cells in each row. Since every square $2 \\times 2$ and $3 \\times 3$ covers even number of black cells, it is not possible to cover the rectangle with $2 \\times 2$ and $3 \\times 3$ squares.\n\nIt suffices to consider the case when both $m$ and $n$ are divisible by $2$, or $3$. Then it is clear how to divide the rectangle into $2 \\times 2$ squares (in the first case) or $3 \\times 3$ squares (in the second case).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a white $6 \\times 2019$ table. Andrii and Arsenii are playing the following game: one after another (starting with Andrii) a player colors one of the $1 \\times 1$ cells in black. Moreover, one's turn cannot create a $3 \\times 3$ square that contains two black cells. Whoever doesn't have a turn loses. Who will win if both want to win the game? And what is the strategy?\n\n![](attached_image_1.png)\n\n**Fig. 3**",
    "options": [],
    "answer": "Arsenii (the second player) wins. Strategy: partition each column into friendly pairs of cells three rows apart, and after Andrii colors a cell, Arsenii colors its paired cell in the same column.",
    "solution": "We will show that there exists a strategy such that Arsenii always has a turn after Andrii's turn. We will split all the cells into *friendly pairs*. Two $1 \\times 1$ cells make a friendly pair, if they are in the same column and there are exactly two $1 \\times 1$ cells in-between. Then Arsenii colors the cell that makes a friendly pair with a cell that Andrii colored during his last turn.\n\nWe want to show that Arsenii always has such a turn. If Andrii colored some cell $P$, then a cell $V$, that makes a friendly pair with $P$ is white. Moreover, suppose the cell $V$ cannot be colored in black because there already is another black cell $Y$ in some $3 \\times 3$ square that is located either in the very top or the very bottom of the table. Then there also exists another similar $3 \\times 3$ square either in the very top or the very bottom that contains cells $P$ and $Y$, both of which are black, where $Y$ is a friendly pair of $V$ (Fig. 3). This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45646,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle passes through vertices **A**, **B** of a parallelogram $ABCD$, and intersects diagonals $BD$ and $AC$ in points $X$ and $Y$, respectively. Circumscribed circle of $\\angle ADX$ intersects diagonal $AC$ in point $Z$. Prove that $AY = CZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $A$, $B$, $Z$, $D$ lie on one circle, and $A$, $B$, $X$, $Y$ lie on another circle (fig. 26),\n\n$$\n\\angle AYB = \\angle AXB = 180^\\circ - \\angle AXD = 180^\\circ - \\angle AZD = \\angle DZC.\n$$\n\nSince $ABCD$ is a parallelogram, $\\angle BAY = \\angle DCZ$, hence, $\\angle ABY = \\angle CDZ$, which yields $AY = CZ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45647,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many different ways to cover the $4 \\times 4$ square with five $3 \\times 1$ rectangles are there, so that exactly one $1 \\times 1$ cell is left uncovered?\n![](attached_image_1.png)",
    "options": [],
    "answer": "16",
    "solution": "We will cover the $1 \\times 1$ cells two ways as shown in Fig. 13. Since any $3 \\times 1$ rectangle covers exactly one cell of each color, only white cell can be left uncovered (since there are 6 white and 5 of grey and black cells). Same positions of white cells are only on the edges of $4 \\times 4$ square. Thus, there are 4 kinds of uncovered cell. Clearly, the number of covers for each of these types is the same. It is easy to find that there are four covers (puc. 14).\nThere are three ways with one horizontal $3 \\times 1$ rectangle on the bottom, and one way without it. Thus, there are 16 ways in total.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for positive numbers $x, y, z, a, b, c$ so that $x + y + z = a + b + c$ the following inequality holds:\n$$\n\\frac{x}{a+b} + \\frac{y}{b+c} + \\frac{z}{c+a} + \\frac{a}{x+z} + \\frac{b}{x+y} + \\frac{c}{y+z} > 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The following completes the proof:\n$$\n\\frac{x}{a+b} + \\frac{y}{b+c} + \\frac{z}{c+a} + \\frac{a}{x+z} + \\frac{b}{x+y} + \\frac{c}{y+z} > \\\\\n\\frac{x}{a+b+c} + \\frac{y}{a+b+c} + \\frac{z}{a+b+c} + \\frac{a}{x+y+z} + \\frac{b}{x+y+z} + \\frac{c}{x+y+z} = \\\\\n= \\frac{x+y+z}{a+b+c} + \\frac{a+b+c}{x+y+z} = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45649,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which positive integers $n \\ge 2$, there exist $n$ odd (not necessarily different) numbers $a_1, a_2, \\dots, a_n$ such that\n$a_1^2 + a_2^2 + \\dots + a_n^2$ is a square of some positive integer?",
    "options": [],
    "answer": "Exactly those n with n ≡ 0, 1, or 4 (mod 8).",
    "solution": "Clearly, square of an integer number can give a remainder of $0$, $1$ or $4$ modulo $8$. Therefore, only for $n$ of the form $8k + r$, where $r \\in \\{0, 1, 4\\}$, such numbers can exist. Let us show how they can be constructed.\n\n$$\nn = 4t,\\ a_1 = \\dots = a_{n-1} = 1,\\ a_n = (2t-1):\n$$\n$$\na_1^2 + a_2^2 + \\dots + a_n^2 = (4t-1) \\cdot 1^2 + (2t-1)^2 = (2t)^2.\n$$\n\n$$\nn = 8t + 1,\\ a_1 = \\dots = a_{n-1} = 1,\\ a_n = (2t-1):\n$$\n$$\na_1^2 + a_2^2 + \\dots + a_n^2 = (8t) \\cdot 1^2 + (2t-1)^2 = (2t+1)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequence of positive integers $a_1, a_2, a_3, \\dots$ is defined by $a_{n+1} = a_n^2 + 2018$, where $a_1$ is some positive integer. Prove that in this sequence no more than one number can be a cube of a positive integer number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose there is more than one number, which is a cube of a positive integer. Let $a_k$ be the smallest of all possible cubes. Then, it gives $0, \\pm 1$ modulo $9$, hence, $a_k^2 \\equiv 0, 1 \\pmod{9}$. From now on, we write all remainders modulo $9$. Let us list all the cases.\n\n$a_k^2 \\equiv 0$ and $a_{k+1} \\equiv 2 \\Rightarrow a_{k+1}^2 \\equiv 4$ and $a_{k+2} \\equiv 6 \\Rightarrow a_{k+2}^2 \\equiv 0$ and $a_{k+3} \\equiv 2$, and there cannot be any more cubes.\n\n$a_k^2 \\equiv 1$ and $a_{k+1} \\equiv 3 \\Rightarrow a_{k+1}^2 \\equiv 0$ and $a_{k+2} \\equiv 2$, and there cannot be any more cubes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Number $1000$ was split into $9$ (not necessarily different) positive integer additive terms. After that, we list all different numbers that can be obtained from adding some of these terms (from one to eight). What is the minimum number of numbers listed?",
    "options": [],
    "answer": "9",
    "solution": "Let us split $1000$ into $8$ numbers $100$ and one number $200$. In this case, the sum of some terms can have $9$ different values: $100 \\cdot 1$, $100 \\cdot 2$, $\\ldots$, $100 \\cdot 8$, and $200 + 100 \\cdot 7 = 900$.\n\nLet us prove that it is impossible to obtain less than $9$ different numbers. Since $9$ does not divide $1000$, a split of $1000$ results in at least $2$ different terms. Let us sort the terms in ascending order. For the list, we first pick only the $1$st term, then $1$st and $2$nd, then $1$st, $2$nd and $3$rd, and so on, until we pick the first $8$ numbers – in this way, we get $8$ different numbers. Now we take the last $8$ terms: this new sum is larger than any previous sum, therefore, there cannot be fewer than $9$ different numbers listed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andrii chooses a sign in front of each number in the expression $\\pm 1 \\pm 2 \\pm 3 \\pm \\dots \\pm 2018$. How many different positive values can Andrii obtain as a result of the computed expression?",
    "options": [],
    "answer": "1018586",
    "solution": "Clearly, the expression cannot equal $0$, since it contains $1009$ odd numbers. We want to show that the expression can be any odd number between $1$ and $1+2+3+\\dots+2018$.\n\nTake some configuration. Find the first from the left consecutive numbers with signs \"-\" and \"+\". The switching of these signs decreases the value of the expression by $2$. We will start with the following expression: \"+1+2+3+\\dots+2018\" and will change it to expression \"-1+2+3+\\dots+2018\" that is less by $2$ as described above. By following the described algorithm, we will obtain all the numbers decreased by $2$.\n\nWhen we get the expression\n\n+1 - 2 - 3 - ... - 2018, for which the algorithm doesn't work, the last step will be to switch to the smallest expression -1 - 2 - 3 - ... - 2018.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest possible value of $x^6 + x^4y^2 + x^2y^4 + y^6$, given that the product of real numbers $x, y$ is $1$?",
    "options": [],
    "answer": "4",
    "solution": "Without loss of generality, let $x, y$ be positive. Then we can factor the expression as $(x^4 + y^4)(x^2 + y^2)$. Since $(x^2 - y^2)^2 = x^4 - 2x^2y^2 + y^4 \\ge 0$, then $x^4 + y^4 \\ge 2x^2y^2 = 2$, and similarly $x^2 + y^2 \\ge 2xy = 2$. Therefore, the smallest number is $4$, that can be obtained when $x = y = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive $x, y, z$, that satisfy the following system of inequalities:\n$$\n\\begin{cases} (x+1)(y+1) \\le (z+1)^2, \\\\ \\left(\\frac{1}{x}+1\\right)\\left(\\frac{1}{y}+1\\right) \\le \\left(\\frac{1}{z}+1\\right)^2. \\end{cases}\n$$",
    "options": [],
    "answer": "All positive triples with x = y = z",
    "solution": "Consider the following substitutions: $a = x+1 > 1$, $b = y+1 > 1$ and $c = z+1 > 1$. Then the first inequality is $ab \\le c^2$. The second inequality can be rewritten the following way (using the first one):\n$$\n\\frac{ab}{(a-1)(b-1)} \\le \\left(\\frac{c}{c-1}\\right)^2 = \\left(1+\\frac{1}{c-1}\\right)^2 \\le \\left(1+\\frac{1}{\\sqrt{ab}-1}\\right)^2 = \\frac{ab}{(\\sqrt{ab}-1)^2} \\Leftrightarrow \\\\\n(\\sqrt{ab}-1)^2 \\le (a-1)(b-1) \\Leftrightarrow a - 2\\sqrt{ab} + b \\le 0 \\Leftrightarrow (\\sqrt{a}-\\sqrt{b})^2 \\le 0 \\Leftrightarrow a = b,\n$$\nMoreover, equality is possible only if all the transitions satisfy the equality. So, the following equality has to hold: $ab = c^2 \\Leftrightarrow a = b = c \\Leftrightarrow x = y = z$.\nConsider the following substitution: $x = \\operatorname{tg}^2 \\alpha$, $y = \\operatorname{tg}^2 \\beta$, $\\alpha, \\beta \\in (0, \\frac{\\pi}{2})$. Let the triple $(x, y, z)$ satisfy the conditions, then\n$$\n\\begin{cases} \n\\frac{1}{\\cos^2 \\alpha \\cos^2 \\beta} \\le (z+1)^2, \\\\ \n\\frac{1}{\\sin^2 \\alpha \\sin^2 \\beta} \\le \\left(\\frac{1}{z} + 1\\right)^2, \\\\ \n1 \\ge \\left(\\frac{1}{\\cos \\alpha \\cos \\beta} - 1\\right) \\left(\\frac{1}{\\sin \\alpha \\sin \\beta} - 1\\right) \\\\ \n\\frac{1}{\\cos \\alpha \\cos \\beta} + \\frac{1}{\\sin \\alpha \\sin \\beta} \\ge \\frac{1}{\\cos \\alpha \\cos \\beta \\sin \\alpha \\sin \\beta} \\\\ \n\\sin \\alpha \\sin \\beta + \\cos \\alpha \\cos \\beta = \\cos(\\alpha - \\beta) \\ge 1 \\Rightarrow \\alpha = \\beta \\Rightarrow x = y. \n\\end{cases} \n\\Rightarrow \n\\begin{cases} \nz \\ge \\frac{1}{\\cos \\alpha \\cos \\beta} - 1, \\\\ \n\\frac{1}{z} \\ge \\frac{1}{\\sin \\alpha \\sin \\beta} - 1, \\\\ \n1 \\ge \\left(\\frac{1}{\\cos \\alpha \\cos \\beta} - 1\\right) \\left(\\frac{1}{\\sin \\alpha \\sin \\beta} - 1\\right) \\\\ \n\\frac{1}{\\cos \\alpha \\cos \\beta} + \\frac{1}{\\sin \\alpha \\sin \\beta} \\ge \\frac{1}{\\cos \\alpha \\cos \\beta \\sin \\alpha \\sin \\beta} \\\\ \n\\sin \\alpha \\sin \\beta + \\cos \\alpha \\cos \\beta = \\cos(\\alpha - \\beta) \\ge 1 \\Rightarrow \\alpha = \\beta \\Rightarrow x = y. \n\\end{cases}\n$$\n\n$$\n\\begin{cases} \n(x+1)^2 \\le (z+1)^2, \\\\ \n\\left(\\frac{1}{x}+1\\right)^2 \\le \\left(\\frac{1}{z}+1\\right)^2, \n\\end{cases} \n\\Rightarrow \n\\begin{cases} \nx \\le z, \\\\ \n\\frac{1}{x} \\le \\frac{1}{z}, \n\\end{cases} \n\\Rightarrow \n\\begin{cases} \nx \\le z, \\\\ \nz \\le x, \n\\end{cases} \n\\Rightarrow x = z,\n$$\n\nthat leads to the solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45655,
    "subject": "Mathematics (Multi-modal)",
    "question": "<table><tr><td>$a_1$</td><td>$a_2$</td><td>$a_3$</td><td>$a_4$</td></tr><tr><td>$a_5$</td><td>$a_6$</td><td>$a_7$</td><td>$a_8$</td></tr><tr><td>$a_9$</td><td>$a_{10}$</td><td>$a_{11}$</td><td>$a_{12}$</td></tr><tr><td>$a_{13}$</td><td>$a_{14}$</td><td>$a_{15}$</td><td>$a_{16}$</td></tr></table>\n**Fig. 7**\n\nA $4 \\times 4$ table has positive integers in its cells so that the sum of any two cells that share a side is a factorial of some positive integer. Show that there are at least 4 equal numbers in this table.\n(Arsenii Nikolaiev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will start with the following lemma.\n\n**Lemma 1.** There exists a diagonal that has two equal numbers for any $2 \\times 2$ square.\n**Proof.** Let $X$ be the greatest number in $2 \\times 2$ square. Let his neighbors in this $2 \\times 2$ square be $A$ and $B$. Clearly, they are located on a diagonal, so if $A = B$, the lemma is proven. Suppose $A > B$. Then $A + X = k! > B + X = m! > 1$, thus $k > m$. It is also clear that $m > 1, k > 2$. But then\n$$\n2(B + X) = 2m! > 2X \\geq A + X = k! \\geq k \\cdot m! > 2m!,\n$$\nthat leads to a contradiction. This finishes the proof of Lemma 1.\n\nBy contradiction, let the table consist of the numbers $a_1, a_2, ..., a_{16}$, that satisfy the conditions (Fig. 7).\nUse Lemma 1 for a square with $a_1, a_2, a_5, a_6$. Without loss of generality, let $a_2 = a_5$. Use Lemma 1 for squares with $a_2, a_3, a_6, a_7$ and $a_5, a_6, a_9, a_{10}$.\n\n**Case I.** $a_2 = a_7$ or $a_5 = a_{10}$. Since these cases are similar, let $a_2 = a_5 = a_{10} = x$.\n\n<table><tr><td>$a_1$</td><td>$x$</td><td>$a_3$</td><td>$a_4$</td></tr><tr><td>$x$</td><td>$a_6$</td><td>$a_7$</td><td>$a_8$</td></tr><tr><td>$a_9$</td><td>$x$</td><td>$a_{11}$</td><td>$a_{12}$</td></tr><tr><td>$a_{13}$</td><td>$a_{14}$</td><td>$a_{15}$</td><td>$a_{16}$</td></tr></table>\n**Fig. 8**\n\nFrom Lemma 1 for the square $a_9$, $a_{10} = x, a_{13}, a_{14}$, if $a_{13} = x$, then we have four equal numbers that lead to a contradiction. Then $a_9 = a_{14}$. Similarly, from Lemma 1 for a square with $a_{10} = x, a_{11}, a_{14}, a_{15}$ we have that $a_9 = a_{14} = a_{11} = y$\n\n<table><tr><td>$a_1$</td><td>$x$</td><td>$a_3$</td><td>$a_4$</td></tr><tr><td>$x$</td><td>$a_6$</td><td>$a_7$</td><td>$a_8$</td></tr><tr><td>$y$</td><td>$x$</td><td>$y$</td><td>$a_{12}$</td></tr><tr><td>$a_{13}$</td><td>$y$</td><td>$a_{15}$</td><td>$a_{16}$</td></tr></table>\n**Fig. 9**\n\nIt suffices to use the Lemma 1 for a square with $a_6, a_7, a_{10} = x, a_{11} = y$ thus, the table has either four numbers $x$ or four numbers $y$. The contradiction completes the proof.\n\n**Case II.** $a_3 = a_6 = a_9 = t$.\n\n<table><tr><td>$a_1$</td><td>$a_2$</td><td>$t$</td><td>$a_4$</td></tr><tr><td>$a_5$</td><td>$t$</td><td>$a_7$</td><td>$a_8$</td></tr><tr><td>$t$</td><td>$a_{10}$</td><td>$a_{11}$</td><td>$a_{12}$</td></tr><tr><td>$a_{13}$</td><td>$a_{14}$</td><td>$a_{15}$</td><td>$a_{16}$</td></tr></table>\n**Fig. 10**\n\nBy assumption, there is no other number $t$ among the rest of the values, so by Lemma 1 the following holds: $a_4 = a_7 = a_{10} = a_{13}$, that leads to a contradiction and completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45656,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $3 \\times 3$ table has positive integers in its cells so that the sum of numbers in any two cells that share common side is a factorial of a positive integer. Show that there are at least $3$ equal numbers in the table.\nThe factorial of a positive integer $n$ is a product $1 \\cdot 2 \\cdot 3 \\cdots n$.\n\n![](attached_image_1.png)\n**Fig. 5**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will start with the following lemma.\n**Lemma 1.** There exists a diagonal that has two equal numbers for any $2 \\times 2$ square.\n**Proof.** Let $X$ be the greatest number in $2 \\times 2$ square. Let his neighbors in this $2 \\times 2$ square be $A$ and $B$. Clearly, they are located on a diagonal, so if $A = B$, the lemma is proven. Suppose $A > B$. Then $A + X = k! > B + X = m! > 1$, thus $k > m$. It is also clear that $m > 1, k > 2$. But then\n$$\n2(B + X) = 2m! > 2X \\ge A + X = k! \\ge k \\cdot m! > 2m!,\n$$\nthat leads to a contradiction. This finishes the proof of Lemma 1.\n\nBy contradiction, suppose $a, b, c, d, e, f, g, h, i$ are in the table (Fig. 5). By Lemma 1 for the square that consists of $a, b, d, e$ either $b = d$ or $a = e$. In the first case, we will use Lemma 1 for the square with $d, e, g, h$, and then for $b, c, e, f$. Thus, either $b = d = h$, or $b = d = f$, or $g = e = c$.\n\nIn the second case, we will use Lemma 1 for $e, f, h, i$, and then for $b, c, e, f$. Thus, either $a = e = i$ or $a = e = c$, or $h = f = b$. In all cases we obtain three equal numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45657,
    "subject": "Mathematics (Multi-modal)",
    "question": "The convex polygon $M$ is given. There is a square $K$, that contains $M$ inside and has the minimum possible area. Is it obligatory for at least one of the sides of a square $K$ to contain one of the sides of the polygon $M$?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "No",
    "solution": "Let's consider the smallest tangential square, that is circumscribed around the equilateral triangle $\\triangle ABC$ (Fig. 46). And the side of the square contains the side of the triangle $AC$. It's clear, that it will be a square $ADEC$ and the edge $B$ is located inside the square. Then let's make a small turn of the square around the edge $A$. Let it be the square $ATUV$. So, the edge $C$ will be inside the new square. It's clear from the Fig. 46, if to continue the line $AC$ till the intersection with $UV$ at some point $C'$, then at $\\triangle AVC'$ $AC'$ is a hypotenuse, so, $AC' > AV = AC$. That's why we can make a small compression of the $ATUV$ (homothety with the center at the point $A$ and coefficient $k < 1$, but close enough to $1$), after which the area of the square will become smaller, but it will still cover the triangle $\\triangle ABC$.\n\n![](attached_image_1.png)\nFig. 46",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists a connected figure $F$ that consists of $1\\times 1$ squares and satisfies the following conditions:\n* it consists of $s$ $1\\times 1$ squares and is not a rectangle;\n* any $k\\times s$ rectangle, where $k \\ge s$ can be split into figures that coincide with $F$?\nA figure that consists of $1\\times 1$ squares is called *connected*, if one can get from any cell to any cell by moving to adjacent cells.\n\n![](attached_image_1.png)\n\n**Fig. 16**\n\n**Fig. 17**",
    "options": [],
    "answer": "Yes",
    "solution": "Yes, such figure exists.\n\nConsider an L-shaped tromino that consists of three $1\\times 1$ cells. $3\\times 2$, $2\\times 3$ rectangles can be easily split into such figures, as well as $9\\times 5$ rectangle (Fig. 16). Now let every $1\\times 1$ square be a $2\\times 3$ rectangle. Then the tromino became a new figure that will be denoted as $F$ (Fig. 17), then $S=18$.\n\nAn $m\\times n$ rectangle is called *nice*, if it can be split into figures $F$. Clearly, if a rectangle can be split into nice rectangles, it is nice itself. Since $3\\times 2$, $2\\times 3$ and $9\\times 5$ rectangles are nice, then so are $18\\times 15$, $6\\times 6$ and $4\\times 9$ rectangles.\n\nWe will continue the proof with induction. Consider $k\\times 18$ rectangle.\n\n$k=18$, we can split $18\\times 18$ into six $6\\times 6$.\n\n$k=19$, firstly, from $19\\times 18$ we are cutting $18\\times 15$ off, then the remainder $4\\times 18$ can be split into two $4\\times 9$.\n\n$k=20$, can split into ten $9\\times 4$.\n\n$k=21$, firstly, from $21\\times 18$ we are cutting $18\\times 15$ off, then the remainder $6\\times 18$ can be split into three $6\\times 6$.\n\nSuppose all rectangles of the type $k\\times 18$ for $k \\ge 21$ are nice.\nConsider $(k+1)\\times 18$ rectangle. Split it into $(k-3)\\times 18$ and $4\\times 18$.\n\n![](attached_image_2.png)\n\n**Fig. 17**\nBoth are nice by the induction hypothesis and by an easy split into other nice rectangles. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for positive numbers $x, y, z, t$ the following inequality holds:\n$$\n\\frac{x^8+1}{x^4} + \\frac{y^8+1}{y^4} + \\frac{z^8+1}{z^4} + \\frac{t^8+1}{t^4} \\ge 2 \\cdot \\left( \\frac{x}{y} + \\frac{y}{z} + \\frac{z}{t} + \\frac{t}{x} \\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality can be rewritten as follows:\n$$\n\\begin{aligned}\n& (x^4 + \\frac{1}{z^4} + z^4 + \\frac{1}{y^4}) + (y^4 + \\frac{1}{t^4} + t^4 + \\frac{1}{z^4}) + (z^4 + \\frac{1}{x^4} + x^4 + \\frac{1}{t^4}) + (t^4 + \\frac{1}{y^4} + y^4 + \\frac{1}{x^4}) \\\\ & \\ge 4 \\cdot \\left( \\frac{x}{y} + \\frac{y}{z} + \\frac{z}{t} + \\frac{t}{x} \\right).\n\\end{aligned}\n$$\n\n$(x^4 + \\frac{1}{z^4}) + (z^4 + \\frac{1}{y^4}) \\geq 2\\sqrt{x^4 \\cdot \\frac{1}{z^4}} + 2\\sqrt{z^4 \\cdot \\frac{1}{y^4}} = \\frac{2x^2}{z^2} + \\frac{2z^2}{y^2} \\geq 4\\sqrt{\\frac{x^2}{z^2} \\cdot \\frac{z^2}{y^2}} = \\frac{4x}{y z},$\nthat finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45660,
    "subject": "Mathematics (Multi-modal)",
    "question": "A figure is called *polyomino* if it is formed by joining one or more $1 \\times 1$ squares edge to edge. It is known that a rectangle which is not a square can be split into 8 pairwise distinct polyominos. Polyominos are considered equal if one can be transformed into another by translations and rotations. What is the smallest area of such a rectangle?",
    "options": [],
    "answer": "26",
    "solution": "We will start with counting polyominos of smallest area. There is only 1 polyomino of area 1, 1 polyomino of area 2, and 2 polyominos of area 3. Thus, the smallest area of a rectangle that consists of 8 polyominos is $1 \\cdot 1 + 1 \\cdot 2 + 2 \\cdot 3 + 4 \\cdot 4 = 25$. But then it has to be of a size $25 \\times 1$ (since $5 \\times 5$ square doesn't satisfy the condition). It is clear for such a rectangle that it can only be split into rectangles of size $k \\times 1$ and, but such a rectangle has an area at least $1+2+...+8=36 > 25$. Thus, the smallest possible area is 26. An example of splitting a $13 \\times 2$ rectangle is shown on Fig. 18.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45661,
    "subject": "Mathematics (Multi-modal)",
    "question": "The convex quadrilateral $ABCD$ is given.\n$\\angle BAD = \\angle BCD = 120^\\circ$, $\\angle ACD = 80^\\circ$, $BC = CD$, $O$ is the intersection point of its diagonals.\nProve, that $BO^2 = AO \\cdot AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's draw a circle with the center at the point $C$ and radius $BC = CD$. Points $B$ and $D$ are located on this circle, but $\\angle BCD = 120^\\circ$, so, arc $AD$, that doesn't contain point $A$, is $240^\\circ$. That's why its inscribed angle is $120^\\circ$. So, point $A$ is located on the circle too. Then it's easy to calculate angles (Fig. 47).\n\n$\\angle ACB = 40^\\circ$, $\\angle DBA = 40^\\circ$ -- is subtended by the central angle $\\angle ACD = 80^\\circ$, because $\\triangle ACB$ is isosceles and the apex angle is $40^\\circ$. So,\n\n$\\angle CAB = \\angle ABC = 70^\\circ$. So, $\\angle AOB = 70^\\circ$ that's why $\\triangle AOB$ is also isosceles and similar to $\\triangle ACB$, that's why we can write the following equation:\n\n![](attached_image_1.png)\nFig. 47\n\n$$\n\\frac{AO}{AB} = \\frac{AB}{AC} \\Rightarrow AO \\cdot AC = AB^2 = BO^2,\n$$\n\nWhat had to be demonstrated.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's consider the sequence of positive integers $(x_n)$, that is given by the formula:\n$x_n = 5 \\cdot 2^n - 1$, $n \\in N$. Prove, that there is an infinite number of pairs $(x_i, x_j)$ of elements that are mutually-prime, and at the same time none of the elements $x_k$ in this infinite sequence of pairs is included at the initial sequence infinite number of times.\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Everything follows from this equation: $\\forall n \\in N$\n$$\nx_{n+1} - 2x_n = 5 \\cdot 2^{n+1} - 1 - 2 \\cdot (5 \\cdot 2^n - 1) = 10 \\cdot 2^n - 1 - 10 \\cdot 2^n + 2 = 1.\n$$\nSo, GCF of $(x_{n+1}, x_n)$ is a divisor of $1$, what means that they're coprime. What had to be shown.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some natural numbers are placed around a circle in such a way that the product of any two neighboring numbers is a perfect square. Prove that the product of any (not necessarily neighboring) two numbers is also a perfect square.\n(Arseniy Nikolaev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote numbers as $a_1, a_2, \\ldots, a_k$. From the problem statement it follows that for any $1 \\le i < j \\le k$ the product $(a_i a_{i+1})(a_{i+1} a_{i+2}) \\ldots (a_{j-1} a_j) = n^2$ is a perfect square. Hence the product $a_i a_j = \\frac{n^2}{a_{i+1}^2 a_{i+2}^2 \\ldots a_{j-1}^2}$ is also a perfect square, as the natural ratio of two perfect squares, Q.E.D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bogdan drawn $2017$ vertical and $2018$ horizontal lines in the rectangle $Q$. These lines divide $Q$ into $2018 \\times 2019$ smaller not necessarily equal rectangles. Two children want to determine the perimeter of $Q$. Andrew says that he can do it by choosing some $2019$ smaller rectangles and getting to know their perimeters. On the contrary, Olesya says that she can do it by choosing $4036$ smaller rectangles. Who is right about the number of smaller rectangles, perimeters of which one should know in order to determine the perimeter of $Q$?\n\n(BogdanRublyov)",
    "options": [],
    "answer": "Neither",
    "solution": "Let us consider two rectangles constructed from $2018 \\times 2019$ smaller rectangles, of sizes $2 \\times 2$ and $3 \\times 1$ respectively. It is clear that all small rectangles in both of them have equal perimeters, namely $8$. However, the first one is a rectangle $4036 \\times 4038$ with the perimeter $16148$, not equaling $16146$, which is the perimeter of the second, a $6054 \\times 2019$ rectangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45665,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the company there are no three people who know each other, but for any group of five there are three who have the common acquaintance (not necessarily from that group). Prove that this company can be split into two groups of pairwise unfamiliar people.\n(ViktorBogdanskyi)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We are going to show that the acquaintances graph has no odd cycles. For the sake of contradiction, let us take the smallest odd cycle. Clearly, there are no edges connecting vertices of this cycle (as it would divide the cycle into two smaller cycles, at least one of which would be odd). Next, let us enumerate people from $1$ to $2k+1$ along the cycle and use the statement for people with numbers $1$, $2$, $3$, $k+2$, $k+3$ (one can easily check that they are pairwise distinct). Note that\nthere cannot be edges between any three people having the common acquaintance, because it would lead to the existence of a triangle. Hence, if $2k+1=5$, there would be such edges between people $1$, $2$, $3$, $k+2$, $k+3$, so $k>2$.\nThen there must be a common acquaintance for $1$, $3$, and one of $k+2$, $k+3$, w.l.o.g. let $1$, $3$, $k+2$ have the common acquaintance.\nFrom here one can easily construct a smaller odd cycle by using the common acquaintance $X$ of $1$ and $k+2$ ($X$ doesn't belong to the original cycle), namely $1$, $2$, $3$, $\\ldots$, $k+2$, $X$ and $k+2$, $k+3$, $\\ldots$, $2k+1$, $1$, $X$. Both of them are smaller than the original cycle and have lengths of different parity, leading to the contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45666,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any natural number $n \\ge 3$ find integers $a_1 < a_2 < \\dots < a_n$ such that the following equality holds:\n$$\n\\frac{a_1}{a_1} + \\frac{a_1}{a_2} + \\frac{a_1}{a_3} + \\dots + \\frac{a_1}{a_n} = \\frac{a_2}{a_1} + \\frac{a_2}{a_2} + \\frac{a_2}{a_3} + \\dots + \\frac{a_2}{a_n}. \\qquad (\\text{BogdanRublyov})\n$$",
    "options": [],
    "answer": "One valid construction is: for k from 1 to n−1, set a_k = −2^{n−k}(2^{n−1}−1), and set a_n = 2^{n−1}.",
    "solution": "Let us first find rational numbers satisfying the equality, and then multiply them by the least common multiplier of the denominators in order to obtain integers.\nOne can fix $b_1 = -2^{n-1}$, $b_2 = -2^{n-2}$, ..., $b_{n-1} = -2^1$. Let us rewrite the equation as follows:\n$$\n\\frac{b_1}{b_1} + \\frac{b_1}{b_2} + \\frac{b_1}{b_3} + \\dots + \\frac{b_1}{b_{n-1}} - \\frac{b_2}{b_1} - \\frac{b_2}{b_2} - \\frac{b_2}{b_3} - \\dots - \\frac{b_2}{b_{n-1}} = \\frac{b_2}{b_n} - \\frac{b_1}{b_n} \\quad \\text{or}\n$$\n$$\nX = \\frac{2^{n-1}}{2^{n-2}} - \\frac{2^{n-2}}{2^{n-1}} + \\frac{2^{n-1}}{2^{n-3}} - \\frac{2^{n-2}}{2^{n-3}} + \\frac{2^{n-1}}{2^{n-4}} - \\frac{2^{n-2}}{2^{n-4}} + \\dots + \\frac{2^{n-1}}{2^1} - \\frac{2^{n-2}}{2^1} = \\frac{2^{n-1}}{b_n} - \\frac{2^{n-2}}{b_n} \\Rightarrow\n$$\n$$\nX = 2 - \\frac{1}{2} + 2^2 - 2 + 2^3 - 2^2 + \\dots + 2^{n-2} - 2^{n-3} = 2^{n-2} - \\frac{1}{2} = \\frac{2^{n-1}-1}{2} = \\frac{2^{n-2}}{b_n} \\Rightarrow\n$$\n$$\nb_n = \\frac{2^{n-1}}{2^{n-1}-1} > 0.\n$$\n\nAll $b_k$, $k = \\overline{1, n-1}$ are negative, and from the last inequality it follows that\n$b_1 < b_2 < \\dots < b_{n-1} < b_n$. Finally, we can get aforementioned integers as follows:\n$$\na_k = b_k \\cdot (2^{n-1} - 1), \\quad k = \\overline{1, n}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45667,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that nonzero real numbers $x$, $y$, $z$ satisfy the condition $xy + yz + zx = 0$. What value can the expression\n$$\n\\frac{1}{x^2+2yz} + \\frac{1}{y^2+2zx} + \\frac{1}{z^2+2xy}\n$$\nbe equal to?",
    "options": [],
    "answer": "0",
    "solution": "Since $xyz \\neq 0$, we can do the following transformation of the given expression:\n$$\n\\begin{aligned}\n\\frac{1}{x^2+2yz} + \\frac{1}{y^2+2zx} + \\frac{1}{z^2+2xy} &= \\frac{1}{x^2+2yz-xy-yz-zx} + \\frac{1}{y^2+2zx-xy-yz-zx} + \\frac{1}{z^2+2xy-xy-yz-zx} \\\\\n&= \\frac{1}{x^2+yz-xy-zx} + \\frac{1}{y^2+zx-xy-yz} + \\frac{1}{z^2+xy-yz-zx} \\\\\n&= \\frac{1}{(x-y)(x-z)} + \\frac{1}{(y-x)(y-z)} + \\frac{1}{(z-x)(z-y)} \\\\\n&= -\\frac{1}{(x-y)(z-x)} - \\frac{1}{(x-y)(y-z)} - \\frac{1}{(z-x)(y-z)} \\\\\n&= -\\frac{(y-z)+(z-x)+(x-y)}{(x-y)(z-x)(y-z)} = 0.\n\\end{aligned}\n$$\n\nThus, there are no zeros among the three terms on the left-hand side, and it turned out that the sum of two is equal to the third modulo, that is, they can not be the sides of a non-degenerate triangle. The resulting contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each player - Andriy and Olesya - has a set of 2019 cards, which contains numbers $1, 2, \\ldots, 2019$ (each number is exactly one time for each player). The game follows the following rules. At the beginning of the game there is a card with a number $k \\in \\{1, 2, \\ldots, 2019\\}$ on the table. After that, the players take turns (Andriy begins) to change one of their cards to the one that is currently on the table. Besides that, Andriy can change the card on the table to the one he holds, if it has a number that is greater than the number written on the card on the table, and Olesya - on her card with a smaller written number than on the card on the table. A person who cannot make a move is considered to be the loser. Who will win in this game, if everyone wants to win?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "Olesya",
    "solution": "For each time Andriy plays a card $n > k$, Olesya plays a card $n - 1 < n$. Since Andriy has a set of cards $\\{1, 2, \\ldots, k, k, k+1, \\ldots, n-1, n+1, n+2, \\ldots, 2018\\}$, he must put one of the cards that is greater or equal to $n+1$. Olesya plays the card $n$ back. Thus, after each move, Andriy should play a greater card, while Olesya will always have a card with a number what is 1 less than the one Andriy plays. When he plays the card with the number 2018, he will have a set of cards that will not contain the number 2018, and Olesya will have a card that contains the number 2017. She will change the card with the number 2018 to the card with the number 2017 and Andriy will not be able to make his move.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Let $C_1$, $A_1$ and $B_1$ be the points of $AB$, $BC$ and $AC$, respectively. Let $K$ be the projection of $B_1$ on the line $A_1C_1$. Let the points $M$ and $N$ lie on the rays $B_1A$ and $B_1C$ respectively, so that $\\angle B_1A_1C_1 = 2\\angle KNB_1$ and $\\angle B_1C_1A_1 = 2\\angle KMB_1$. Prove that the length of the segment $MN$ is not greater than the perimeter of the triangle $\\Delta A_1B_1C_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M_1$ be the point of the ray $A_1C_1$ such that $M_1C_1 = B_1C_1$, $N_1$ be the point of the ray $C_1A_1$ such that $N_1A_1 = B_1A_1$ (Fig. 39). Then\n$\\angle B_1M_1K = \\frac{1}{2} \\angle A_1C_1B_1 = \\angle KMB_1$,\nwhence $KM_1MB_1$ is inscribed and $M$ is the projection of $M_1$ on the line $AC$.\nAnalogously, $N$ is the projection of $N_1$. Then, the segment $MN$ is the projection $M_1N_1$ on the line $AC$, hence\n$$\nMN \\leq M_1N_1 = B_1C_1 + A_1C_1 + B_1A_1,\n$$\nwhich proves the problem statement.\n\n![](attached_image_1.png)\n**Fig. 39**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be natural numbers. Prove that there exists an integer nonnegative number $k$, such that $\\text{GCD}(a^k + bc, b^k + ca, c^k + ab) > 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p$ be some prime divisor of the number $abc + 1$. Let's prove that $k = p - 2$ satisfies the statement of the problem. Since $p$ doesn't divide any of the numbers $a$, $b$, $c$, by Fermat's little theorem: $a^{p-1} \\equiv b^{p-1} \\equiv c^{p-1} \\equiv 1 \\pmod{p}$. Hence $p$ divides each number $a^{p-1} + abc$, $b^{p-1} + abc$ and $c^{p-1} + abc$. Since $p$ doesn't divide any of the numbers $a$, $b$, $c$ and $p$ is prime, then prime number $p$ divides each number $a^k + bc$, $b^k + ca$ and $c^k + ab$. The latter is the consequence of the following equality: $a^k + bc = a^{p-2} + bc = \\frac{1}{a}(a^{p-1} + abc)$. The statement is proved.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a fixed natural number $n > 1$, 2019 natural numbers are placed around the circle in such a way that the product of any two neighboring numbers is a perfect $n$-th power. Is it always the case that the product of any (not necessarily neighboring) two numbers is also a perfect $n$-th power?\n\n(A. Nikolaev, B. Rublyov)",
    "options": [],
    "answer": "Yes",
    "solution": "For the odd $n = 2l + 1$ one can prove the statement as follows:\n$$\nP^2 = (a_1a_2)(a_2a_3)(a_3a_4)\\dots(a_{2018}a_{2019})(a_{2019}a_1) = m^{2l+1},\n$$\nwhere $P = a_1a_2\\dots a_{2019}$. From here it follows that $P = a_1a_2\\dots a_{2019} = s^{2l+1}$. Therefore,\n$$\na_1 = \\frac{P}{(a_2a_3)(a_4a_5)\\dots(a_{2018}a_{2019})} = \\frac{s^{2l+1}}{u^{2l+1}} = v_1^{2l+1},\n$$\nand the same holds for all $a_i = v_i^{2l+1}$, $i = 2, \\ldots, 2019$. From here it directly follows that\n$$\na_i a_j = v_i^{2l+1} v_j^{2l+1} = v^{2l+1}.\n$$\n\nFor even $n = 2l$ one can write\n$$\nP^2 = (a_1a_2)(a_2a_3)(a_3a_4)\\dots(a_{2018}a_{2019})(a_{2019}a_1) = m^{2l}, \\text{ where } P = a_1a_2\\dots a_{2019}.\n$$\nFrom here it follows that $P = m^l$. Therefore,\n$$\na_1^2 = \\frac{P^2}{(a_2a_3)(a_4a_5)\\dots(a_{2018}a_{2019})} = \\frac{m^{2l}}{u^{2l}} = v_1^{2l},\n$$\nand the same holds for all $a_i^2 = v_i^{2l}$, $i = 2, \\ldots, 2019$. From here it directly follows that\n$$\nr^{2l} = (a_1a_2)(a_2a_3)(a_3a_4)\\dots(a_{k-1}a_k) = a_1a_2^2a_3^2\\dots a_{k-1}a_k = a_1a_k w^{2l} \\Rightarrow a_1a_k = \\frac{r^{2l}}{w^{2l}} = z^{2l}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a right triangle $ABC$, let $M$ be the midpoint of its hypotenuse $AB$. A perpendicular bisector of $AB$ intersects side $BC$ in point $K$. A perpendicular from $K$ to $CM$ intersects the ray $AC$ in point $P$, which happens not to lie on the segment $AC$. Lines $CM$ and $BP$ intersect in point $T$. Prove that $AC = TB$.\n\n(Danylo Khilko)\n\n![](attached_image_1.png)\nFig.33",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From $KP \\perp CM$ it follows that $\\angle CPK = \\angle MBC = \\angle ABC$ (fig. 33), hence the quadrilateral $PAKB$ is cyclic. Therefore, $PK$ is a bisector in the triangle $\\triangle CPT$. Clearly, $PK$ is also a height and a median of this triangle.\n\nTherefore, $\\triangle CPT$ is isosceles. In particular, it implies that $PK$ is a perpendicular bisector of $CT$.\n\nFurther, $CK = KT$, and $\\triangle PCK = \\triangle PTK$, meaning that $\\angle PTK = 90^\\circ$. Finally, using that $KB = KA$, we get $\\triangle KTB = \\triangle KCA$, implying $TB = AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45673,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polynomial $x^2 + 1$ is written on the blackboard. Every day Kate wipes off currently written polynomial $F(x)$ and writes one of the $F^2(x) + 1$ and $F(x^2 + 1)$ instead, choosing at her discretion. Prove that the constant (free) term will exceed $2^{2^{222}}$ in one year.\n\n(Arseniy Nikolaev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We are going to prove a lemma prior to the problem itself: the polynomial which will be written after $k$ days does not depend on Kate's choices and equals to\n$$\nF_k = \\underbrace{(\\dots((x^2 + 1)^2 + 1)^2 + \\dots)^2}_{k} + 1.\n$$\n\nUsing induction one can directly verify this lemma. Alternatively, one can use the following equality:\n$$\ng^2(x) + 1 = (h^2(x) + 1)^2 + 1 = (h((x^2 + 1))^2 + 1 = h^2(x^2 + 1) + 1 = h^2(t) + 1 = g(t) = g(x^2 + 1).\n$$\nto get the aforementioned result.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Natural numbers $a < b$ are written on the board. At each step, two numbers written on the board are wiped out, and their sum and the modulus of difference are written down instead. At some point the number 2019 appeared on the board. What is the smallest possible value of $b$?",
    "options": [],
    "answer": "1010",
    "solution": "It is easy to write down all the pairs of the numbers, that will successively appear on the board:\n$$\na, b \\rightarrow b+a, b-a \\rightarrow 2a, 2b \\rightarrow 2(b+a), 2(b-a) \\rightarrow 2a^2, 2b^2 \\dots\n$$\n\nAs we see, after the appearance of the first four numbers: $a, b, b+a, b-a$ – all the other numbers that may appear on the board are even. Since 2019 is an odd number, then the number 2019 must appear among these first four numbers.\n\nCase 1: $b = 2019$.\n\nCase 2: $a = 2019 < b$.\n\nCase 3: $b-a = 2019 < b$.\n\nCase 4: $b+a = 2019$, then $2b > b+a = 2019 \\Rightarrow 2b \\ge 2020 \\Rightarrow b \\ge 1010$. Thus the smallest possible value of $b$ is $b=1010$. Then if $b=1010$ and $a=1009$ the number 2019 can appear on the board.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any real numbers $x, y, z$ the following inequality is true:\n$$\nx^2(3y^2 + 3z^2 - 2yz) \\geq yz(2xy + 2xz - yz).\n$$\nFind the triples that turn it into equality.",
    "options": [],
    "answer": "(x, y, z) = (t, 2t, 2t), (t, 0, 0), (0, t, 0), (0, 0, t) for any real t",
    "solution": "**Answer:** $(x, y, z) = (t, 2t, 2t), (t, 0, 0), (0, t, 0), (0, 0, t)$.\n\nConsider the given inequality as quadratic relative to the variable $x$:\n$$\nx^2(3y^2 + 3z^2 - 2yz) - 2x(y^2z + z^2y) + y^2z^2 \\geq 0.\n$$\nSince $3y^2 + 3z^2 \\geq 6|yz| \\geq 2yz$, consider the case that $3y^2 + 3z^2 = 2yz$. Then both variables should be equal to zero, which gives a solution in which equality is achieved. In all other cases it is a quadratic trinomial with branches up. Let's find its discriminant:\n$$\n\\frac{1}{4} D = (y^2z + z^2y)^2 - (3y^2 + 3z^2 - 2yz)y^2z^2 = y^2z^2((y+z)^2 - (3y^2 + 3z^2 - 2yz)) = -2y^2z^2(y-z)^2 \\leq 0.\n$$\n\nThus it is proved that the left part is non-negative, because the discriminant is not positive.\n\nConsider the conditions under which equality is possible, that is, when the discriminant is zero.\n\n1 case. $y=0$. Then the equality transforms into $3x^2z^2 = 0$. Thus all triples $(0, 0, z), (0, y, 0)$ and $(x, 0, 0)$ are sought ones.\n\n2 case: $y = z \\neq 0$. Then $4x^2y^2 = 4xy^3 - y^4 \\Rightarrow 4x^2 - 4xy + y^2 = 0 \\Rightarrow (2x - y)^2 = 0$. Thus the triples $(t, 2t, 2t)$ for any real $t$ are the answer too.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with circumcircle $w$ with center $O$. Let $A_0$ and $C_0$ be the second intersection points of $w$ with the continuation of the altitudes from $A$ and $C$. Let $A_1$ and $C_1$ be the intersection points of the line $A_0C_0$ with the sides of the triangle $AB$ and $BC$ respectively. Let $A_2$ and $C_2$ be the points of $AC$ such that $A_2O \\parallel BC$, and $C_2O \\parallel AB$. Let $H$ – orthocenter of $\\triangle ABC$, $T$ is the intersection point $A_1A_2$ and $C_1C_2$. Prove that $HT \\parallel AC$. Suppose that the tangents to $w$ at $A$ and at $B$ meet at a point $K$.\n(Anton Trygub)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For now let's consider the points $A_0$, $A_1$ and $A_2$. Since $\\triangle ABC$ is an acute triangle, then the point $O$ lies inside $\\triangle ABC$, hence the point $A_2$ lies on the line $AC$ (Fig. 40). Let's prove that the quadrilateral $OA_0CA_2$ is inscribed. Indeed, since $A_2O \\parallel BC$ we have\n$$\n\\angle OA_2A = \\angle BCA = 90^\\circ - \\angle A_0AC = 90^\\circ - \\frac{1}{2} \\angle A_0OC = \\angle OA_0C.\n$$\n\nNow let $X$ be the second intersection point of circumscribed circle of quadrilateral $OA_0CA_2$ with the line $CH$. Let's prove that the points $A_0$, $B$, $A_1$ and $X$ lie on the same circle. Indeed,\n$$\n\\angle A_1BH = 90^\\circ - \\angle A = \\angle C_0CA = \\angle AA_0C_0,\n$$\nthat is quadrilateral $A_0BA_1H$ is inscribed. Also $\\angle A_1BH = 90^\\circ - \\angle A = \\angle C_0CA = \\angle AA_0C_0$\n\n$$\n\\angle HXA_0 = 180^\\circ - \\angle A_0XC = \\angle A_0OC = 2\\angle A_0AC = 2\\angle A_0BC = \\angle A_0BH,\n$$\nsince because of the fact that the points $H$ and $A_0$ are symmetric with respect to the line $BC$.\n\nNow let's prove that the line $A_1A_2$ through a point $X$. This follows from that fact that\n$$\n\\angle A_1XA_0 = 180^\\circ - \\angle A_1BA_0 = \\angle A_0CA_2 = 180^\\circ - \\angle A_2XA_0.\n$$\nNow we have\n$$\n\\angle OXH = 180^\\circ - \\angle OXC = \\angle OA_2A = \\angle C, \\text{ and also } \\angle OXT = \\angle OCA = 90^\\circ - \\angle B.\n$$\nSo we proved that if $X$ is an intersection point of $CC_0$ and $A_1A_2$, then $\\angle OXT = 90^\\circ - \\angle B$, $\\angle OXH = \\angle C$.\n\n![](attached_image_1.png)\n**Fig. 40**\n\nNow let $Y$ be the intersection point of $AA_0$ and $C_1C_2$. Analogously one can prove that $\\angle OYT = 90^\\circ - \\angle B$, and $\\angle OYH = \\angle A$. Consider the points $X, H, Y, O$. Clearly $\\angle YHX = \\angle A$. Then, since $\\angle OXH + \\angle XHY + \\angle HYO = 180^\\circ$, the points $X, Y, O$ lie on one line. Hence, we have $\\triangle XHY$ with the point $O$ on the side $XY$. This triangle is similar to $\\triangle ABC$, moreover since $\\angle TXY = \\angle TYX = 90^\\circ - \\angle B$, then in this similarity the ray $XT$ corresponds to the ray $CO$, and the ray $YT$ to $AO$. Hence, the point $T$ corresponds to the point $O$, so $T$ is the center of circumscribed circle $\\triangle YHX$. Thus we have $\\angle THX = 90^\\circ - \\angle XYH = 90^\\circ - \\angle A = \\angle HCA$, that is $HT \\parallel AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all possible pairs of positive integers $(m, n)$ that satisfy the following:\n$$\nm^{10} + 23mn + n! = 2020,\n$$\nwhere $n!$ denotes the product of all positive integers from $1$ to $n$.",
    "options": [],
    "answer": "(2, 6)",
    "solution": "When $m \\ge 3$, $m^{10} \\ge 3^{10} > 3^7 = 2187 > 2020$, so the given equality doesn't hold. Thus, there are two possible values for $m$: $m=1$ or $m=2$.\n\nIf $m=1$, the condition to be satisfied is $23n + n! = 2019$. Since $7! = 5040 > 2019$, $n \\le 6$. For $n \\le 6$, $23n + n! \\le 138 + 720 = 850 < 2019$, so, considered case has no solutions.\n\nIf $m=2$, the condition to be satisfied is $46n + n! = 996$. Similarly, $n \\le 6$. If $n=6$, then $46n + n! = 276 + 720 = 996$, so the equation holds. Clearly, if $n$ is smaller, then so is the left side of equation, and thus, there are no more solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine which of the following numbers is greater: $2$ or $\\operatorname{tg} 1$?",
    "options": [],
    "answer": "2 > \\tan(1)",
    "solution": "Consider the following inequalities: $2 > \\sqrt{3} = \\operatorname{tg} \\frac{\\pi}{3} > \\operatorname{tg} 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Out of three expressions $\\frac{x}{y}$, $\\frac{x^2+x}{y^2+y}$ and $\\frac{x^2+2}{y^2+2}$, for some integer $x, y$ all three are defined, two take the same integer value, and the remaining one takes a different integer value. For which pairs of integers $x, y$ is this possible?\n\n(Bohdan Rublyov)",
    "options": [],
    "answer": "(2,1) and (-4,1)",
    "solution": "Obviously, $|x| > |y|$, for $x \\neq 0$, since otherwise, the first fraction will not be an integer. If the first two expressions are equal, then we have the equality:\n$$\n\\frac{x}{y} = \\frac{x^2+x}{y^2+y} \\Rightarrow xy^2 + xy = yx^2 + xy \\Rightarrow xy(y-x) = 0.\n$$\nFor $x \\neq y$ and possible range of values, the given equality is satisfied only by pairs $(x, y) = (0, t)$, $t \\neq 0; -1$. The third expression will obviously take a non-zero, but non-integer value, since $0 < \\frac{2}{t^2+2} < 1$ for $t \\neq 0$.\n\n\nIf the first and third expressions are equal, we have the equality:\n$$\n\\frac{x}{y} = \\frac{x^2+2}{y^2+2} \\Rightarrow xy^2 + 2x = yx^2 + 2y \\Rightarrow xy(y-x) = 2(y-x) \\Rightarrow xy = 2,\n$$\nThe possible pairs of $(x, y)$ are (2, 1) and (-2, -1). Here, the second expression in the first case will take on a value 3, hence, different from the other two. In the second case, it is not defined and, therefore, not possible.\n\n\nIf the second and third expressions are equal, we have the equality:\n$$\n\\frac{x^2+x}{y^2+y} = \\frac{x^2+2}{y^2+2} \\Rightarrow x^2y^2 + xy^2 + 2x^2 + 2x = x^2y^2 + yx^2 + 2y^2 + 2y \\Rightarrow \\\\\nxy(y-x) = 2(y-x)(y+x) + 2(y-x) \\Rightarrow xy - 2y - 2x = 2 \\Rightarrow (x-2)(y-2) = 6.\n$$\n\n$x-2=6, y-2=1 \\Rightarrow x=8, y=3 \\Rightarrow \\frac{x^2+2}{y^2+2} = \\frac{66}{11} = 6$ and $\\frac{x}{y} = \\frac{8}{3}$ does not satisfy the conditions.\n\n$x-2=3, y-2=2 \\Rightarrow x=5, y=4 \\Rightarrow \\frac{x^2+2}{y^2+2} = \\frac{27}{18}$ is not integer, so does not satisfy the conditions.\n$x-2=-2, y-2=-3 \\Rightarrow x=0, y=-1$ does not lie in the allowed range of values.\n$x-2=-6, y-2=-1 \\Rightarrow x=-4, y=1 \\Rightarrow \\frac{x^2+2}{y^2+2} = \\frac{18}{3} = 6$ and $\\frac{x}{y} = -4$ satisfies the conditions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no two distinct positive integers $a$ and $b$, such that $\\{\\frac{a}{b}\\} + \\{\\frac{b}{a}\\} = 0$.\nHere $\\{x\\}$ denotes the difference between $x$ and the greatest integer that does not exceed $x$, for example, $\\{\\frac{7}{5}\\} = \\frac{2}{5}$, $\\{\\frac{2019}{3}\\} = 0$ and $\\{\\frac{2020}{3}\\} = \\frac{1}{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose such two numbers $a$ and $b$ exist. Without loss of generality, one can assume they are relatively prime. The condition of the problem implies that the number $\\frac{a+b}{a} = \\frac{a^2+b^2}{ab}$ is an integer. But then $a^2 + b^2$ has to be divisible by both $a$ and $b$. Hence $b^2 \\mid a$ and $a^2 \\mid b$, which contradicts $a$ and $b$ being relatively prime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist pairwise distinct positive integers $a$, $b$ and $c$, such that $\\left\\{ \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} \\right\\} = 0$? The fractions are not necessarily irreducible.\n\nHere $\\{x\\}$ denotes the difference between $x$ and the greatest integer that does not exceed $x$, for example, $\\left\\{ \\frac{7}{5} \\right\\} = \\frac{2}{5}$, $\\left\\{ \\frac{2019}{3} \\right\\} = 0$ and $\\left\\{ \\frac{2020}{3} \\right\\} = \\frac{1}{3}$.",
    "options": [],
    "answer": "Yes; for example a=2, b=12, c=9.",
    "solution": "A source for an example is the following equation: $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6} = 1$. Clearly, irreducible fractions would not work, thus we will choose $\\left\\{ \\frac{2}{12} \\right\\} = \\left\\{ \\frac{1}{6} \\right\\} = \\frac{1}{6}$. Then $\\left\\{ \\frac{12}{9} \\right\\} = \\left\\{ \\frac{4}{3} \\right\\} = \\frac{1}{3}$ and $\\left\\{ \\frac{9}{2} \\right\\} = \\frac{1}{2}$. Thus, $\\left\\{ \\frac{2}{12} \\right\\} + \\left\\{ \\frac{12}{9} \\right\\} + \\left\\{ \\frac{9}{2} \\right\\} = 1$ and $\\{1\\} = 0$.\n\nAnother example is based on the equality $\\frac{1}{2} + \\frac{1}{2} + 0 = 1 \\Rightarrow \\left\\{ \\frac{1}{2} \\right\\} + \\left\\{ \\frac{2}{4} \\right\\} + \\left\\{ \\frac{4}{1} \\right\\} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Square $ABCD$ of size $2019 \\times 2019$ is divided by two lines into four rectangles with integer side lengths, and some of the rectangles might be squares. Turns out, the area of the rectangle containing vertex $A$ equals the perimeter of the rectangle containing vertex $B$. What is the area of the smallest of the four rectangles (or squares)?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "16",
    "solution": "Let us denote the sides of the rectangle containing vertex $A$ as $a$ and $b$ (Fig. 21). Then the sides of the rectangle containing vertex $B$ are equal to $a$ and $2019-b$. The equality of the area of the first rectangle and perimeter of the second yields:\n$$\nab = 2 \\cdot (a + 2019 - b) \\Rightarrow ab - 2a + 2b = 4038 \\Rightarrow \\\\ (a+2)(b-2) = 4034 = 2017 \\cdot 2.\n$$\n\nSince, on the left-hand side, one of the multipliers is greater than $2$, and $2017$ is prime, the only solution is: $a+2=2017$ and $b-2=2$. Hence, $a=2015$ and $b=4$. Clearly, the smallest rectangle of the four would be the square with the area $4 \\cdot 4 = 16$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Out of three expressions $\\frac{x}{y}$, $\\frac{x^2+4}{y^2+4}$ and $\\frac{x^3+8}{y^3+8}$, for some integer $x, y$, two take the same value, and the remaining one takes a different value. For which pairs of integers $x, y$ is this possible?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "[(1, 4), (4, 1)]",
    "solution": "Obviously, $x \\neq y$, because otherwise, all three expressions would take the same value, which contradicts problem statement.\nIf the first two expressions are equal, then we have the equality:\n$$\n\\frac{x}{y} = \\frac{x^2+4}{y^2+4} \\Rightarrow xy^2 + 4x = yx^2 + 4y \\Rightarrow xy(y-x) = 4(y-x).\n$$\nSince $x \\neq y$, we can divide by $y-x \\Rightarrow xy=4$. There are two pairs of positive integer numbers that satisfy the last equality: (1, 4) and (4, 1). Substitute them into the third equality, and we get\n$$\n\\frac{x^3+8}{y^3+8} = \\frac{9}{72} = \\frac{1}{8} \\neq \\frac{x}{y} \\quad \\text{and, analogously,} \\quad \\frac{x^3+8}{y^3+8} = 8 \\neq \\frac{x}{y}, \\quad \\text{so these pairs of numbers satisfy the}\n$$\ncondition.\n\nIf the first and third expressions are equal, then\n$$\n\\frac{x}{y} = \\frac{x^3 + 8}{y^3 + 8} \\Rightarrow xy^3 + 8x = yx^3 + 8y \\Rightarrow xy(y-x)(y+x) = 8(y-x) \\Rightarrow xy(y+x) = 8,\n$$\nwhich has no solution, as one can confirm by a simple search.\nIf the second and third expressions are equal, then we have the equality:\n$$\n\\begin{aligned}\n\\frac{x^2+4}{y^2+4} &= \\frac{x^3+8}{y^3+8} \\Rightarrow x^2y^3 + 8x^2 + 4y^3 = x^3y^2 + 8y^2 + 4x^3 \\\\\nx^2y^2(y-x) + 4(y-x)(y^2 + xy + x^2) &= 8(y-x)(y+x) \\Rightarrow x^2y^2 + 4(y^2 + xy + x^2) = 8(y+x) \\Rightarrow 4xy + x^2y^2 = 4(2y-y^2) + 4(2x-x^2).\n\\end{aligned}\n$$\n\nSo, $x=1$ or $y=1$. If $x=1$, then $4y+y^2 = 4(2y-y^2)+4$ or $5y^2-3y+4=0$ -\nequality is obviously not possible, because $5y^2 > 3y$ for positive integer $y$. Analogously for $y=1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45684,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the smallest possible value the expression $ab + a + b$ can take, if real numbers $a, b$ satisfy the condition $a^2 + b^2 = 25$.",
    "options": [],
    "answer": "-13",
    "solution": "$$(a+b+1)^2 \\ge 0$$\n$$a^2 + b^2 + 1 + 2ab + 2a + 2b \\ge 0 \\Rightarrow ab + a + b \\ge -\\frac{1}{2}(a^2 + b^2 + 1) = -13.$$ \nBy choosing $a = -4$ and $b = 3$, we obtain: $ab + a + b = -12 - 4 + 3 = -13$, i.e. the smallest possible value is reached.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ which are less than 1% of the number 2020, and such that $n+1$ is more than 1% of the number 2019.",
    "options": [],
    "answer": "20",
    "solution": "The conditions of the problem can be written the following way: $\\frac{n}{2020} < \\frac{1}{100} < \\frac{n+1}{2019}$.\n\nThe left inequality implies that $\\frac{n}{2020} < \\frac{1}{100}$, so $n < \\frac{2020}{100} = 20.2$, hence $n \\le 20$.\n\nIf $n=20$ is substituted into the right inequality, one can see that it holds: $\\frac{21}{2019} > \\frac{1}{100} \\Leftrightarrow 2100 > 2019$.\n\nThe right inequality doesn't hold for $n \\le 19$, since $\\frac{n+1}{2019} \\le \\frac{20}{2019} < \\frac{20}{2000} = \\frac{1}{100}$.\n\nThus, the only possible solution is $n=20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45686,
    "subject": "Mathematics (Multi-modal)",
    "question": "For arbitrary integer $a, b, c$, we define a function $f(x) = a x^2 + b x + c$. Prove that the expression\n$$\nf(2020) + f(2019) + \\dots + f(1011) - f(1010) - f(1009) - \\dots - f(1)\n$$\nis divisible by $2020$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote the expressions as follows:\n$$\n\\begin{aligned}\nS &= f(2020) + f(2019) + \\dots + f(1011) - f(1010) - f(1009) - \\dots - f(1) \\\\\n&= a(2020^2 + 2019^2 + \\dots + 1011^2 - 1010^2 - 1009^2 - \\dots - 1^2) \\\\\n&\\quad + b(2020 + 2019 + \\dots + 1011 - 1010 - 1009 - \\dots - 1) \\\\\n&\\quad + c(1010 - 1010) \\\\\n&= S_1 + S_2.\n\\end{aligned}\n$$\n\n\\begin{aligned}\nS_1 &= a(2020^2 + 2019^2 + \\dots + 1011^2 - 1010^2 - 1009^2 - \\dots - 1^2) \\\\\nS_2 &= b(2020 + 2019 + \\dots + 1011 - 1010 - 1009 - \\dots - 1)\n\\end{aligned}\n\nLet us analyze $S_1$ and $S_2$ separately.\n\nFirst, note that $c$ cancels out because there are $1010$ positive and $1010$ negative terms, so $c(1010 - 1010) = 0$.\n\nConsider $S_2$:\n\nThe sum $2020 + 2019 + \\dots + 1011$ has $1010$ terms, and $1010 + 1009 + \\dots + 1$ also has $1010$ terms. Pairing each term:\n$$\n(2020 - 1010) + (2019 - 1009) + \\dots + (1011 - 1)\n$$\nEach pair is $1010$, and there are $1010$ such pairs, so\n$$\nS_2 = b \\cdot 1010 \\cdot 1010 = b \\cdot 1010^2.\n$$\n\nNow, consider $S_1$:\n\nWe can pair the squares similarly:\n$$\n(2020^2 - 1010^2) + (2019^2 - 1009^2) + \\dots + (1011^2 - 1^2)\n$$\nThere are $1010$ such pairs. Each pair is:\n$$\nn^2 - m^2 = (n - m)(n + m)\n$$\nFor the $k$-th pair:\n$$\n(2020 - (k-1))^2 - (1010 - (k-1))^2 = [2020 - 1010] \\cdot [2020 + 1010 - 2(k-1)] = 1010 \\cdot [3030 - 2(k-1)]\n$$\nfor $k = 1, 2, \\dots, 1010$.\n\nSo,\n$$\nS_1 = a \\sum_{k=1}^{1010} 1010 \\cdot [3030 - 2(k-1)] = a \\cdot 1010 \\sum_{k=1}^{1010} [3030 - 2(k-1)]\n$$\n\nCalculate the sum:\n$$\n\\sum_{k=1}^{1010} [3030 - 2(k-1)] = 3030 \\cdot 1010 - 2 \\sum_{k=1}^{1010} (k-1)\n$$\nBut $\\sum_{k=1}^{1010} (k-1) = \\sum_{j=0}^{1009} j = \\frac{1009 \\cdot 1010}{2}$.\n\nSo,\n$$\nS_1 = a \\cdot 1010 [3030 \\cdot 1010 - 2 \\cdot \\frac{1009 \\cdot 1010}{2}] = a \\cdot 1010 [3030 \\cdot 1010 - 1009 \\cdot 1010]\n$$\n$$\n= a \\cdot 1010^2 (3030 - 1009) = a \\cdot 1010^2 \\cdot 2021\n$$\n\nTherefore,\n$$\nS = S_1 + S_2 = a \\cdot 1010^2 \\cdot 2021 + b \\cdot 1010^2\n$$\n$$\n= 1010^2 (a \\cdot 2021 + b)\n$$\n\nNow, $1010^2 = (2 \\cdot 5 \\cdot 101)^2 = 2^2 \\cdot 5^2 \\cdot 101^2$. Note that $2020 = 2^2 \\cdot 5 \\cdot 101$.\n\nThus, $1010^2$ is divisible by $2020$ (since $1010^2 = 1010 \\cdot 1010 = (2 \\cdot 5 \\cdot 101)^2$ contains all the prime factors of $2020$ at least once).\n\nTherefore, $S$ is divisible by $2020$ for any integers $a, b, c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45687,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that $n^n + 1$ is divisible by $n+1$.",
    "options": [],
    "answer": "All odd positive integers",
    "solution": "For odd integer $n$ it is enough to decompose:\n$$\nn^n + 1 = (n+1)(n^{n-1} - n^{n-2} + n^{n-3} - \\dots + 1).\n$$\nLet's assume that there exists an even number $n = 2k$ such that $K = (2k)^{2k} + 1$ is divisible by $2k + 1$. Then $2k + 1$ is also a divisor of $2kK = (2k)^{2k+1} + 2k = ((2k)^{2k+1} + 1) + (2k-1)$. Since $(2k)^{2k+1} + 1 = (2k+1)((2k)^{2k} - (2k)^{2k-1} + (2k)^{2k-2} - \\dots + 1)$, then $2k-1$ must also be divisible by $2k+1$. This is impossible and therefore no even number $n$ satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define $H_a$, $H_b$, $H_c$ as projections of $A$, $B$, $C$ on $BC$, $AC$, $AB$ respectively. $H$ is the orthocenter of $\\triangle ABC$ and $K$ is a reflection of $A$ over $BC$. The line parallel to $H_bH_c$ through $H$ meets $AB$ and $AC$ in points $X$ and $Y$. Prove that inscribed circles of $\\triangle ABC$ and $\\triangle XYK$ touch.\n(Michael Bondarenko)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define new points: $M$ and $N$ - points symmetric to $H$ over $AB$ and $AC$ respectively, $S$ and $L$ - midpoints of $XH$ and $HY$ respectively. Let's prove a few facts (Fig 34).\n\n![](attached_image_1.png)\nFig. 34\n\n1. $H_c$, $S$, $H_a$ are collinear.\nIt can be seen from $\\angle SH_cH = \\angle SHH_c = \\angle HH_cH_b = \\angle H_bBC = \\angle HBH_a = \\angle HH_cH_a$, where the first equality follows from the fact that $S$ is a midpoint of the hypotenuse of $\\Delta XH_cH$, the second from $XY \\parallel H_cH_b$, others from properties of inscribed quadrilaterals $BH_cH_bC$ and $BH_cHH_a$.\n\n2. $M$, $X$, $K$ are collinear.\nIf we make a homothety at $H$ with a coefficient $2$, we map $H_c \\to M$, $S \\to X$ and $H_a \\to K$, which gives the desired result.\nSimilarly can be shown collinearity of $N$, $Y$, $K$.\n\n3. $XY \\parallel MN$.\nIt can be deduced from $XY \\parallel H_cH_b$ and $MN \\parallel H_cH_b$ because $H_cH_b$ is a midline of $\\Delta HMN$.\n\nTherefore, if we make a homothety at $K$ which maps $\\Delta KXY \\to \\Delta KMN$, we map their circumscribed circles into each other. Since $K$ is their common point, they touch in that point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Increasing geometric progression (sequence) of 5 natural numbers satisfies the following condition: the square of the sum of first and fourth elements is 100 times greater than the sum of first, fifth, and doubled third elements. Find the largest 3-digit number that can be a member of such progression.\n\n(Bogdan Rublyov)",
    "options": [],
    "answer": "625",
    "solution": "Let $b_1 = b$, $b_2 = bq$, $b_3 = bq^2$, $b_4 = bq^3$, $b_5 = bq^4$ denote elements of the progression. Rewriting the problem by using our notation, we get:\n$$\n(b_2 + b_4)^2 = 100(b_1 + 2b_3 + b_5)\n$$\nSubstitute the terms:\n$$\n(bq + bq^3)^2 = 100(b + 2bq^2 + bq^4)\n$$\n$$\nb^2q^2(1 + q^2)^2 = 100b(1 + 2q^2 + q^4)\n$$\nDivide both sides by $b$ (since $b > 0$):\n$$\nbq^2(1 + q^2)^2 = 100(1 + 2q^2 + q^4)\n$$\nLet $bq^2 = 100$, then $b = \\frac{100}{q^2}$.\n\nNow, since the sequence consists of natural numbers, $b$ is a natural number and $q = \\frac{m}{n}$ is rational, where $(m, n) = 1$. Therefore $b m^2 = 100 n^2$. Hence we have $m \\in \\{2, 5, 10\\}$ as $m > n$ and $m^2 \\mid 100$.\n\nLet us recall that $b_3 = 100$ for all cases.\n\nFor $q = 2$:\n$$\nb_4 = bq^3 = 100 \\cdot 2 / 4 = 200,\\quad b_5 = bq^4 = 100 \\cdot 4 / 4 = 400.\n$$\nFor $q = 5$:\n$$\nb_4 = bq^3 = 100 \\cdot 125 / 25 = 500,\\quad b_5 = bq^4 = 100 \\cdot 625 / 25 = 2500.\n$$\nFor $q = 10$:\n$$\nb_4 = bq^3 = 100 \\cdot 1000 / 100 = 1000,\\quad b_5 = bq^4 = 100 \\cdot 10000 / 100 = 10000.\n$$\nFor $q = \\frac{5}{2}$:\n$$\nb_4 = bq^3 = 100 \\cdot \\left(\\frac{5}{2}\\right)^3 / \\left(\\frac{5}{2}\\right)^2 = 100 \\cdot \\frac{125}{8} / \\frac{25}{4} = 100 \\cdot 5 = 250,\\quad b_5 = bq^4 = 100 \\cdot \\left(\\frac{5}{2}\\right)^4 / \\left(\\frac{5}{2}\\right)^2 = 100 \\cdot \\frac{625}{16} / \\frac{25}{4} = 100 \\cdot 25 = 625.\n$$\n\nSo the desired value is $625$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45690,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an isosceles triangle $ABC$, where $AC$ is a base, $P$ is an arbitrary point on $AC$, $T$ is a projection of $P$ onto $BC$. Determine in the ratio in which a symmedian drawn from a vertex $C$ of a triangle $\\triangle PBC$ divides $AT$? Symmedian $CS$, $S \\in BP$ of a $\\triangle PBC$ is a reflection of median $CF$ over the angle bisector $CL$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "2:1",
    "solution": "Let $BH$ be an altitude of $\\triangle ABC$, then $H$ is a midpoint of $AC$, $BH$ and $PT$ are altitudes of $\\triangle PBC$ (Fig. 39). Recall that symmedian bisects a line segment whose endpoints are feet of the altitudes. Therefore, a symmedian of $\\triangle PBC$ is passing through $M$ which is a midpoint of $HT$. Let $CM$ intersect $AT$ at $K$, then applying the Menelaus theorem on the triangle $\\triangle ATH$ we get:\n$$\n\\frac{AK}{KT} \\cdot \\frac{TM}{MH} \\cdot \\frac{HC}{CA} = 1 \\Rightarrow \\frac{AK}{KT} = \\frac{AC}{CH} = 2.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45691,
    "subject": "Mathematics (Multi-modal)",
    "question": "The chessboard was split into domino tiles, meaning it was split into $1 \\times 2$ and $2 \\times 1$ rectangles. Each tile has a number written on it equal to the number of tiles that it has a common line segment with, without taking into account the tile itself. What is a least possible sum of all numbers that are written on the chessboard?\n(Arsenii Nikolaiev)",
    "options": [],
    "answer": "104",
    "solution": "It is clear that there are $32$ domino tiles, hence there are exactly $32$ unit intervals that are covered by domino tiles thus they are internal for tiles and therefore they cannot be common for two tiles. The rest of unit intervals that are not at the edge of the chessboard are common for two tiles. Let their number be equal to $7 \\cdot 8 \\cdot 2 - 32 = 80$.\nIt is obvious that $2F + N = 80$, since the common segment for them has a length of $2$. Hence,\n$$\nS = 2F + 2N = (2F + N) + N = 80 + N = 160 - 2F.\n$$\n\nA set of tiles $D_1, \\dots, D_k$, where $D_i$ has a tile $D_{i+1}$ are friendly for $i=1, k$, or a separate tile that has no friendly tiles, we call a *chain*. Each tile belongs to exactly one chain. Thus we have $C$ chains. Let it be $k_1$ tiles in the $l$-th chain. Then it has $(k_l-1)$ friendly pairs in total. Then there are $(k_1-1)+\\dots+(k_C-1)=32-C=F$ friendly pairs in total. Therefore $S=160-2F=96+2C$. Thus in order to minimize $S$ we have to minimize the number of chains $C$. It is clear that there are no more than $8$ tiles in the chain, hence $C \\ge 4$. Hence we have that the least possible sum is $S=96+2C=104$.\nIt remains to give an example when $C=4$. It is enough to place all the tiles in the same manner, for instance, the bigger side in all to be vertical.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two different natural numbers such that $a^2 + b^2 + 1$ is divisible by $2ab + 1$. Prove that $2ab + 1$ is a perfect square (is a square of an integer).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume the contrary. Let $(a, b)$ be a pair of natural numbers for which this does not hold, such that $a < b$ and this pair has a minimal sum of $a+b$.\n\n**Lemma 1.**  $a^2 + b^2 + 1$ is divisible by $2ab + 1$ if and only if $(2a^2 + 1)^2$ is divisible by $2ab + 1$.\n\n*Proof.* Since $(a^2, 2ab+1)=1$, then $a^2+b^2+1$ is divisible by $2ab+1$ if and only if $a^4+b^2a^2+a^2$ is divisible by $2ab+1$ if and only if $4a^4+4b^2a^2+4a^2$ is divisible by $2ab+1$, since $2ab+1$ is odd. The rest follows from:\n$$\n4a^4+4b^2a^2+4a^2 = (4a^4+4a^2+1)+(4b^2a^2-1) = (4a^4+4a^2+1)+(2ab-1)(2ab+1)\n$$\nSo the lemma is proved.\n\nThus for the fixed pair of natural numbers $(a, b)$ we have $a^2 + b^2 + 1$ is divisible by $2ab + 1$, therefore $(2a^2 + 1)^2$ is divisible by $2ab + 1$. So $(2a^2 + 1)^2 = (2ab + 1)(2ac + 1)$. It follows from the condition $a < b$ that $0 < c < a < b$, otherwise, the right-hand side is greater than the left-hand side. If $2ab + 1$ is not a perfect square then neither is $2ac + 1$. It follows from the lemma that if $(2a^2 + 1)^2$ is divisible by $2ac + 1$, then $a^2 + c^2 + 1$ is divisible by $2ac + 1$ as well. Thus we have found a pair of natural numbers that satisfies the problem statement and has a sum of components $c+a < a+b$, which yields the contradiction with the choice of $(a, b)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 < a_2 < \\dots < a_k$ be positive integers. For each $a_i$ Zarina has written down all the positive divisors of $a_i$ in the notebook (some numbers might be written several times). Then, Marina split all the numbers in the notebook into several groups. It turned out that the numbers in each of these groups form a set of all the positive divisors of some positive integer, which Marina has decided to also write down on the desk. Prove that the numbers on the desk are $a_1, a_2, \\dots, a_k$ in some order.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that number $a_k$ is written only once in the notebook. Then there exists some Marina's group that contains all divisors of $a_k$. Therefore $a_k$ is also on the desk. Then we cross out from the notebook all divisors of $a_k$ one time. Now, by similar thoughts it follows that $a_{k-1}$ is also on the desk, we can cross out it and all of its divisors and so on. Therefore, all the numbers $a_1, a_2, \\dots, a_k$ are present on the desk.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $a$, $b$, $n$ such that $a + b = n - 1$. In some school each student has at most $n$ friends from this school. Prove that one can split all the students from this school into two groups: $A$ and $B$, in such a way that every student from the group $A$ will know at most $a$ students from the group $A$, and every student from the group $B$ will know at most $b$ students from the group $B$.\n\n(Anton Trygub)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a graph, where nodes represent pupils of this school and two nodes are connected by an edge if corresponding pupils are friends. Among all possible partitions of nodes on two sets $A$ and $B$ we choose a partition where $S = b \\cdot S_A + a \\cdot S_B$ is the smallest, where $S_A$ and $S_B$ denote the amount of edges inside groups $A$ and $B$. We show that this partition satisfies the required condition.\n\nIndeed, let's assume that there exists a node $X$ with a degree at least $a+1$ in the group $A$, then all edges coming from $X$ add at least $b(a+1)$ to the sum $S$. Also, there are no more than $b$ nodes in the group $B$ that are adjacent to $X$. Thus if we shift $X$ to the group $B$, the sum $S$ will decrease by at least $b(a+1)$ and increase by $ba$. Therefore, $S$ decreases and we have a contradiction with our assumption of choosing the partition. Similarly, there is no node in the group $B$ with a degree $\\ge b$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute and non-isosceles triangle. Its angle bisectors $AL_1$ and $BL_2$ intersect at the point $I$. Points $D$ and $E$ are chosen on the segments $AL_1$ and $BL_2$ in such a way that $\\angle DBC = \\frac{1}{2}\\angle A$ and $\\angle EAC = \\frac{1}{2}\\angle B$. The lines $AE$ and $BD$ intersect at a point $P$. Let $K$ be the point that is symmetric to the point $I$ with respect to the line $DE$. Prove that the lines $KP$ and $DE$ intersect at a point on the circumcircle of $\\triangle ABC$.\n(Danylo Khilko)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$. Let $N$ be the midpoint of arc $ACB$ of the circumcircle of $\\triangle ABC$ (Fig. 35). We are going to show that $N$ belongs to the line $DE$.\n\n![](attached_image_1.png)\n\nFirstly, observe that both points $D$ and $E$ lie inside $\\triangle ANB$. Indeed, since both of them are inside of $\\triangle ABC$, $\\frac{1}{2}\\beta = \\angle EAC < \\angle BAC = \\alpha$ and $\\frac{1}{2}\\alpha = \\angle DBC < \\angle ABC = \\beta$.\n\nAlso $\\angle NAB = \\angle NBA = 90^\\circ - \\frac{1}{2}\\angle ACB = 90^\\circ - \\frac{1}{2}\\gamma = \\frac{1}{2}(\\alpha + \\beta)$,\n$$\n\\angle EBA = \\frac{1}{2}\\beta < \\frac{1}{2}(\\alpha + \\beta) = \\angle NBA \\text{ and } \\angle EAB = \\alpha - \\frac{1}{2}\\beta < \\frac{1}{2}(\\alpha + \\beta) = \\angle NAB,\n$$\nbecause $\\frac{1}{2}\\alpha < \\beta$. So $E$ lies inside $\\triangle ANB$, similarly $D$ lies there. Now we have that\n$$\n\\begin{aligned}\n\\angle EAB &= \\alpha - \\frac{1}{2}\\beta, \\\\\n\\angle NAE &= \\frac{1}{2}(\\alpha + \\beta) - (\\alpha - \\frac{1}{2}\\beta) = \\beta - \\frac{1}{2}\\alpha, \\\\\n\\angle NBI &= \\frac{1}{2}\\alpha, \\quad \\angle IBA = \\frac{1}{2}\\beta. \\text{ By Ceva's Theorem from the triangle } ABN \\text{ and the point } E\n\\end{aligned}\n$$\nTheorem from the triangle $ABN$ and the point $E$\n$$\n1 = \\frac{\\sin \\angle ANE}{\\sin \\angle ENB} \\cdot \\frac{\\sin \\angle NBE}{\\sin \\angle EBA} \\cdot \\frac{\\sin \\angle BAE}{\\sin \\angle EAN}\n$$\nAfter substitutions we get\n$$\n\\frac{\\sin \\angle ENB}{\\sin \\angle ANE} = \\frac{\\sin \\frac{1}{2}\\alpha}{\\sin \\frac{1}{2}\\beta} \\cdot \\frac{\\sin(\\alpha - \\frac{1}{2}\\beta)}{\\sin(\\beta - \\frac{1}{2}\\alpha)}\n$$\nAnalogously, $\\angle DAB = \\frac{1}{2}\\alpha$,\n$$\n\\begin{aligned}\n\\angle DAN &= \\frac{1}{2}\\beta, & \\angle DBA = \\beta - \\frac{1}{2}\\alpha, \\\\\n\\angle NBD &= \\alpha - \\frac{1}{2}\\beta. \\text{ Using Ceva's Theorem}\n\\end{aligned}\n$$\n$$\n1 = \\frac{\\sin \\angle AND \\cdot \\sin \\angle NBD \\cdot \\sin \\angle BAD}{\\sin \\angle DNB \\cdot \\sin \\angle DBA \\cdot \\sin \\angle DAN} \\text{ or } \\frac{\\sin \\angle DNB}{\\sin \\angle AND} = \\frac{\\sin \\frac{1}{2}\\alpha \\cdot \\sin(\\alpha - \\frac{1}{2}\\beta)}{\\sin \\frac{1}{2}\\beta \\cdot \\sin(\\beta - \\frac{1}{2}\\alpha)} \\Rightarrow \\\\\n\\frac{\\sin \\angle ENB}{\\sin \\angle ANE} = \\frac{\\sin \\angle DNB}{\\sin \\angle AND}\n$$\nTherefore the rays $ND$ and $NE$ coincide since\n$$\n\\angle BNA = \\angle ANE + \\angle ENB = \\angle AND + \\angle DNB\n$$\nNow let $L$ be the second intersection point of $DE$ and the circumcircle of $\\triangle ABC$. We claim that $KP$ passes through $L$. Obviously\n$$\n\\angle (AL, LN) = \\angle (AB, BN) = \\frac{1}{2}(\\alpha + \\beta).\n$$\nOn the other hand,\n$$\n\\angle (AP, PB) = \\angle (PA, AB) + \\angle (AB, BP) = \\alpha - \\frac{1}{2}\\beta + \\beta - \\frac{1}{2}\\alpha = \\frac{1}{2}(\\alpha + \\beta).\n$$\nHence $\\angle (AL, LN) = \\angle (AP, PB)$ and the quadrilateral $APDL$ is inscribed. Also we have $\\angle (DI, IE) = \\frac{1}{2}(\\alpha + \\beta) = \\angle (EP, PB)$. By symmetry $\\angle (EK, KD) = \\angle (DI, IE)$ and $\\angle (KD, DE) = \\angle (ED, DI)$, so $\\angle (EK, KD) = \\angle (EP, PD)$ and $PKED$ is inscribed.\nFinally\n$$\n\\angle (KP, PE) = \\angle (KD, DE) = \\angle (ED, DI) = \\angle (LD, DA) = \\angle (LP, PA).\n$$\nWe obtain $\\angle (KP, PE) = \\angle (LP, PA)$ and since $AP$ and $EP$ coincide we get that the line $KP$ passes through $L$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45696,
    "subject": "Mathematics (Multi-modal)",
    "question": "A palindromic number is a number whose digits stand symmetrically with respect to the center of the number's decimal notation, for instance, $7$, $1221$ and $57575$ are palindromic, while $1212$ and $3330$ are not. Prove that for any number of pairwise distinct palindromic numbers the sum of their reciprocals is not greater than $11$.\n(Arsenii Nikolaiev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let an arbitrary natural number $n$ and consider all palindromic numbers that have exactly $n$ digits. We consider $n$ odd and $n$ even separately.\n\nWhen $n = 2m$, palindrome becomes $a_1a_2\\dots a_{m-1}a_m a_m a_{m-1}\\dots a_2a_1$, the only restriction on its digits is $a_1 \\neq 0$. Hence there are $9 \\cdot 10^{m-1}$ of them and each is greater than $10^{2m-1}$. Therefore the sum of reciprocals is not greater than $\\frac{9 \\cdot 10^{m-1}}{10^{2m-1}} = \\frac{9}{10^m}$. For all $m$ the sum is not greater than\n$$\n\\sum_{m=1}^{\\infty} \\frac{9}{10^m} = 9 \\cdot \\sum_{m=1}^{\\infty} \\frac{1}{10^m} = 9 \\cdot \\frac{\\frac{1}{10}}{1-\\frac{1}{10}} = 9 \\cdot \\frac{1}{9} = 1.\n$$\n\nWhen $n = 2m-1$, palindrome becomes $a_1a_2\\dots a_{m-1}a_m a_{m-1}\\dots a_2a_1$, and again the only restriction on its digits is $a_1 \\neq 0$. Hence there are $9 \\cdot 10^{m-1}$ of them and each is greater than $10^{2m-2}$. Therefore the sum of reciprocals is not greater than $\\frac{9 \\cdot 10^{m-1}}{10^{2m-2}} = \\frac{9}{10^{m-1}}$. For all $m$ the sum is not greater than\n$$\n\\sum_{m=1}^{\\infty} \\frac{9}{10^{m-1}} = 90 \\cdot \\sum_{m=1}^{\\infty} \\frac{1}{10^m} = 90 \\cdot \\frac{\\frac{1}{10}}{1-\\frac{1}{10}} = 90 \\cdot \\frac{1}{9} = 10.\n$$\n\nTherefore, the sum does not exceed $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Equation $\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = \\frac{577}{408}$ has an infinite number of natural solutions $x_1, x_2, \\dots, x_m, y_1, y_2, \\dots, y_n$ for some non-negative integers $m, n$. Prove that an equation $\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_k} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_l} = \\frac{577}{408}$ has a solution in natural numbers $x_1, x_2, \\dots, x_k, y_1, y_2, \\dots, y_l$ for some non-negative integers $k < m$ and $l < n$.\n(Olexander Golovanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove the claim by induction on $m+n$. When $m+n=1$, if $a>0$ we have $m=1$ and $n=0$, therefore given equation becomes $\\frac{1}{x_1} = a$ and has no more than one solution. Similarly $a<0$.\n\nSuppose our claim is true for $m+n<t$, and that equation $\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = a$ where $m+n=t$, has an infinite number of solutions. Without loss of generality, we may assume $a>0$; then, we can consider only those solutions, for which $x_1 \\le x_2 \\le \\dots \\le x_m$ (there is an infinite number of these too, obviously). Since\n$a \\le \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_m} \\le \\frac{m}{a}$, then $x_1 \\le \\frac{m}{a}$. Therefore, $x_1$ takes a finite number of values, and at least when $x_1 = c$ equation $\\frac{1}{x_2} + \\frac{1}{x_3} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = a - \\frac{1}{c}$ has an infinite number of solutions. If $a - \\frac{1}{c} \\ne 0$, then by the inductive assumption equation\n$\\frac{1}{x_2} + \\frac{1}{x_3} + \\dots + \\frac{1}{x_k} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_l} = a - \\frac{1}{c}$ has a solution for some $k < m$ and $l < n$. Then the following equation\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_k} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_l} = a\n$$\nhas a solution as well.\n\nIt remains to consider the case when $a - \\frac{1}{c} = 0$. For this case, a single value of $x_1$, for which the equation $\\frac{1}{x_2} + \\frac{1}{x_3} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = a - \\frac{1}{x_1} = 0$ has an infinite number of solutions, equals $\\frac{1}{a}$.\nIt is clear that $m > 0$ and $n > 0$, and the solution $x_1 = \\frac{1}{a}$ is the desired one.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45698,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have to place natural numbers $a_1, a_2, \\dots, a_n$ that are not all equal at the vertices of a regular $n$-sided polygon $A_1A_2\\dots A_n$, $n \\ge 6$ with center at $O$ in such a way that for every vertex $A_i$ there exist two vertices $A_k$ and $A_l$, that are symmetrical with respect to the line $OA_i$ and the equation $a_i = \\frac{1}{2}(a_k + a_l)$ holds. For which $n \\ge 6$ can we do this?\n(Bogdan Rublyov)",
    "options": [],
    "answer": "all integers n ≥ 6 with n ≠ 7",
    "solution": "Let us assume we managed to place numbers properly. Let $m = \\min_{j=1, n} a_j$, $M = \\max_{j=1, n} a_j$. If $m = a_i = \\frac{1}{2}(a_k + a_l)$, then $m = a_k = a_l$, hence there have to be at least 3 of the smallest numbers, similarly for the biggest. It is easy to fill in numbers when\n\n![](attached_image_1.png)\nFig. 40\n\n$n \\ge 6$ is composite. Let $n = pq$, $p, q > 1$. Then we place $m$ in vertices $A_p, A_{2p}, \\dots, A_{qp}$, and we place $M$ in the rest of the vertices. It is clear that we satisfy conditions since we split vertices of our polygon into $q$ regular $p$-sided polygons, with the same number at each vertex, and a number in each vertex of a polygon is the arithmetic mean of numbers in the rest of the vertices of the same polygon, hence we satisfy the condition for $n$-sided polygon.\n\nNow we consider the case when $n$ is prime. Let us represent this number as $n = 4s + r$, where $r \\in \\{3, 5\\}$. Next we propose the following arrangement of numbers in vertices of $n$-sided polygon.\n\nFirst for $n=4s+3$. For the sake of simplification, on Fig. 40 we denote vertex $A_i$ as $i$, $m$ as black disk, and $M$ as a white disk. Now for every disk $i$ we have to indicate a pair of disks of the same color that are located symmetrically (on equal distance from $i$, hence with index $i+u$ and $i-u$ (mod $n$)). In case there are 3 or more consecutive disks of the same color, then we have to consider only those on the sides of the group since for the rest of them the desired pair is two adjacent disks. Therefore for the group of black vertices we have to find a pair for the disk with index $s+3$ (due to symmetry we have a similar answer for $3s+2$). It is obvious that a pair for $s+3$ will be 1 and a symmetric disk. We can calculate that its index is $2s+5$, since it has to satisfy $s+3 = \\frac{1}{2}(1+2s+5)$. Hence it will be black if $2s+5 \\le 3s+2$ or $s \\ge 3$. Due to symmetry, for the white group it is enough to consider indices $2, 3, \\dots, s+3$.\n\nFor 2 a pair is $4s+3$ and 4. Now we consider $s+2$. One could be a disk with the least possible index of $3s+3$. The other one has to have an index of\n$$\n(s+2)-(2s+1)=1-s=1-s+4s+3=3s+4.\n$$\nA distinct black vertex with index of 1 clearly satisfies conditions, for instance, its pair of symmetric vertices could be $2s+2$ and $2s+3$. Therefore this construction will do when $s \\ge 3$, meaning $n=4s+3 \\ge 15$.\n\nNow we consider $n=4s+5$ (Fig. 41). For the group of black vertices we have to find a pair for the disk with index $s+4$. The desired pair is 1 and a symmetric disk with index of $2s+7$, since it has to satisfy $s+4 = \\frac{1}{2}(1+2s+7)$. Hence it will be black when $2s+7 \\le 3s+3$ or $s \\ge 4$.\n\nFor the white group for 2 the pair is $4s+5$ and 4. Now we consider $s+3$. One could be a disk with the least possible index of $3s+4$. The other one has to have an index of\n$$\n(s+3)-(2s+1)=2-s=2-s+4s+5=3s+7.\n$$\nA distinct black vertex with index of 1 clearly satisfies conditions, its pair of symmetric vertices could be $2s+3$ and $2s+4$. Therefore this construction will do when $s \\ge 3$, meaning $n=4s+5 \\ge 21$.\n\nNow we have to consider the following prime numbers: 17, 13, 11 and 7.\nLet's present construction for the first three of them.\n\nHave a look at Fig. 42 for $n=17$. Non-obvious pairs are for the following vertices:\n$$\n\\begin{aligned}\n13 &= \\frac{1}{2}(5+21) \\equiv \\frac{1}{2}(5+4), \\\\\n5 &= \\frac{1}{2}(13-3) \\equiv \\frac{1}{2}(13+14), \\\\\n12 &= \\frac{1}{2}(18+6) \\equiv \\frac{1}{2}(1+6), \\\\\n6 &= \\frac{1}{2}(1+11).\n\\end{aligned}\n$$\n\nHave a look at Fig. 43 for $n=13$. Non-obvious pairs are for the following vertices:\n$$\n\\begin{aligned}\n3 &= \\frac{1}{2}(-1+7) \\equiv \\frac{1}{2}(12+7), \\\\\n6 &= \\frac{1}{2}(9+3), \\\\\n4 &= \\frac{1}{2}(10-2) \\equiv \\frac{1}{2}(10+11), \\\\\n5 &= \\frac{1}{2}(10+0) \\equiv \\frac{1}{2}(10+13).\n\\end{aligned}\n$$\n\nHave a look at Fig. 44 for $n=11$. Black and white disks are located symmetrically thus it is enough to indicate pairs for one of the colors.\n\n$$\n1 = \\frac{1}{2}(5-3) = \\frac{1}{2}(5+8), \\quad 3 = \\frac{1}{2}(1+5), \\quad 8 = \\frac{1}{2}(2+14) = \\frac{1}{2}(2+3).\n$$\n\nThere is a distinct gray disk, therefore for number $x$ the following equation $x = \\frac{1}{2}(M+m)$ has to be satisfied, for instance, $x = 2$, $m = 1$, $M = 3$.\n\nIt is impossible to satisfy conditions for $n=7$. Assume there are no more minimal than maximal. Then there are no more than 3 of them. It is clear that there cannot be 2 of them. Same goes for their location, since in case they are consecutive, then the ones on the sides do not have a pair, in case two are adjacent, then each of them does not have a pair. In case they are all distinct, we have an alternating placement and two of them do not have a pair.\n\nTherefore, the answer is: for all $n \\ne 7$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be the sides of a triangle. How many numbers (maximum) out of $\\frac{a+b}{a+b-c}$, $\\frac{b+c}{b+c-a}$ and $\\frac{c+a}{c+a-b}$ can be greater than $2$?",
    "options": [],
    "answer": "2",
    "solution": "First, we show an example where two numbers are, indeed, greater than $2$:\n$$\na = b = 5,\\ c = 1,\\ \\text{ then } \\frac{a+b}{a+b-c} = \\frac{10}{9} < 2,\\ \\text{ and } \\frac{b+c}{b+c-a} = \\frac{c+a}{c+a-b} = \\frac{6}{1} > 2.\n$$\n\nSuppose all of these fractions are greater than $2$. Then, clearly, this triangle is not equilateral. Let $a$, $b$ be two bigger sides of the triangle, then $a + b > 2c$. From our assumption,\n$$\n\\frac{a+b}{a+b-c} > 2 \\Rightarrow \\frac{a+b+2c-2a-2b}{a+b-c} > 0 \\Rightarrow \\frac{2c-a-b}{a+b-c} > 0,\n$$\nsince $a + b > 2c$, meaning the fraction at hand is negative – contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that one can choose 7 pairwise distinct numbers from $1, 2, \\ldots, 10000$, such that none of them is a perfect square and no sum of the several numbers out of the 7 chosen ones is a perfect square either.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the following 7 numbers: $2^1$, $2^3$, $\\ldots$, $2^{13} = 8192 < 10000$. It is clear that the sum of any subset of them is such that the highest power of $2$ that divides the sum is odd, thus, it is not a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p > 2$. There are $N$ people in a circle, each one came up with a positive integer and then wrote down the remainder when divided by $p$ on a piece of paper. Then everyone showed their piece of paper to the neighbor on the right, computed the product of their number on a piece of paper and their neighbor's number, and wrote down the second number, which is the remainder of a product divided by $p$. What is the largest value that $N$ could be, if it is known that everyone's first numbers are pairwise distinct, and each piece of paper has two distinct numbers written down?",
    "options": [],
    "answer": "p-2",
    "solution": "Clearly, there cannot be more than $p$ different numbers. It follows from the condition of the problem, that $0$ could not be someone's number on the first step, since then that person will have it on the second step as well. Also, no one can have $1$ as a first number, since then the neighbor on the left will have two equal numbers. We will show that the conditions hold for $N = p-2$.\n\nIndeed, let the following consecutive numbers be written down on the first step from left to right: $2, 3, 4, \\ldots, p-1$. They are all pairwise distinct. Assume that after the second step someone's two numbers are equal. If it's one of the first $N-1$ people, it means that the product $k(k+1)$, where $2 \\le k \\le p-2$, has the same remainder when divided by $p$, as the number $k$. But then their difference is divisible by $p$, that is, $k(k+1)-k=k^2$ is divisible by a prime number $p$, thus, $k \\nmid p$, which is not possible. If the $N$-th person has two equal numbers, then the number $p-1$ will be multiplied by $2$, but then two numbers will be different, since $2(p-1)-(p-1)=p-1$ is not divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the middle point of the side $AC$ of triangle $ABC$. Inside $\\triangle BMC$, there is such point $P$ that $\\angle BMP = 90^\\circ$, $\\angle ABC + \\angle APC = 180^\\circ$. Prove that $\\angle PBM + \\angle CBM = \\angle PCA$.\n\n(Anton Tryhub)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We construct point $B'$ such that $ABCB'$ is a parallelogram. Then (Fig. 26)\n$$\n\\angle ABC + \\angle APC = 180^\\circ = \\angle AB'C + \\angle APC = 180^\\circ,\n$$\nhence, quadrilateral $APCB'$ is inscribed. Obviously, $\\triangle BPB'$ is isosceles, which yields\n$$\n\\angle PCA = \\angle PB'A = \\angle PB'M + \\angle MB'A = \\\\ = \\angle PBM + \\angle MBC, \\text{ Q.E.D.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ is chosen on the smaller arc $BC$ of the circumscribed circle of an acute-angled triangle $ABC$. Points $R$ and $S$ on sides $AB$ and $AC$ respectively are chosen such that $CPRS$ is a parallelogram. Point $T$ on arc $AC$ of the circumscribed circle of $\\triangle ABC$ is such that $BT \\parallel CP$. Prove that $\\angle TSC = \\angle BAC$.\n\n*(Anton Tryhub)*",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let line $PR$ intersect the circumscribed circle of $\\triangle ABC$ at point $K$ (Fig. 27).\n\nThen, $AKPC$ and $BPCT$ are isosceles trapezoids, hence, $\\angle RSA = \\angle PCA = 180^\\circ - \\angle RKA$, so $AKRS$ is inscribed, which makes it an isosceles trapezoid.\n\nThen, $\\angle RKS = \\angle RAS = \\angle BAC = \\angle PKT = \\angle RKT$. Hence, points $K$, $S$, $T$ lie on the same line.\n\nThen, $\\angle TSC = \\angle TKP = \\angle BAC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45704,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex quadrilateral $ABCD$ is given, and $\\angle CBD = 90^\\circ$, $\\angle BCD = \\angle CAD$ and $AD = 2BC$. Prove that $CA = CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We denote by $C_1$ the point symmetric to $C$ with regards to (w.r.t) point $B$ (Fig. 23). Then $\\angle BCD = \\angle CAD = \\angle DC_1C$, hence, $ADCC_1$ is inscribed. Since $AD = 2BC = CC_1$,\n\n![](attached_image_1.png)\n\nFig. 23\n\nthen $\\angle CDC_1 = \\angle DC_1A$, since they are based on the arcs of the same size, formed by chords of the same length. Hence, $C_1A \\parallel CD$, and inscribed quadrilateral $ADCC_1$ is either an isosceles trapezoid or a rectangle.\n\nQuadrilateral $ADCC_1$ cannot be a rectangle, since then from point $D$ onto line $BC$ would be drawn two different perpendiculars – $DB$ and $DC$. Hence, $ADCC_1$ is an isosceles trapezoid, which yields $AC = C_1D = DC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that one can choose 8 pairwise distinct numbers among $1, 2, \\ldots, 10000$, such that none of them is a perfect square and that no sum of the several of them is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the following 7 numbers: $2^1, 2^3, \\ldots, 2^{13} = 8192 < 10000$. Clearly, the sum of any subset of them is such that the highest power of $2$ that divides the sum is odd, thus, it is not a perfect square.\n\nAdd number $3$ to the chosen numbers. Suppose it is possible to choose several numbers such that their sum is a perfect square. Then $3$ is one of such numbers, otherwise we get a contradiction as shown above. Since a perfect square can only have a remainder $0$, $1$ or $4$ when divided by $8$, and since the sum is odd, then the only possible remainder is $1$. However, only numbers $2$ and $3$ have a non-zero remainder when divided by $8$, thus, it is not possible to obtain a sum with remainder $1$ when divided by $8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vasia wrote down all seven-digit numbers that contain every digit between $1$ and $7$ exactly once. Prove that neither of the numbers Vasia wrote down divides another such number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose one of such numbers $a$ is divisible by $b$, that is, there is an integer $n > 1$ such that $a = n b$. Since both $a$ and $b$ have remainder $1$ when divided by $9$, then the number $n$ also has remainder $1$ when divided by $9$. Since $n \\ne 1$, then $n \\ge 10$, thus the number $a$ has at least one digit more than the number $b$, which is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vika chose a 20-letter word that consists only of letters $A$ and $B$. Oleksii wants to know what Vika's word is. He can ask Vika if there are more $A$'s or $B$'s among several (possibly one) consecutive letters of her word. If there are as many $A$'s as there are $B$'s, Vika's answer may be any of the two letters. What is the least number of questions after which Oleksii can guaranteed determine the word Vika chose?",
    "options": [],
    "answer": "20",
    "solution": "It is clear how Oleksii can determine the word in 20 questions – it suffices to ask about each letter separately.\n\nWe want to show that smaller number of questions would not be enough. Suppose Oleksii determined the word in no more than 19 questions. Suppose Vika chose a word that consists of 20 letters $A$. Clearly, for every question Oleksii asked, Vika's answer was that there are more letters $A$. Since there were less than 20 questions, there is a letter that wasn't asked about separately. Let it be the $t$-th letter. Then consider a word that consists of letters $A$ everywhere, except for the $t$-th letter, which is $B$. Clearly, for both such words Vika could have had the same answers. Thus, Oleksii couldn't have determined which one of these two words Vika chose.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45708,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the round table, 50 inhabitants of the island \"Loud Mouths\" are sitting at equal distance from one another and discussing something. Each of them is either a knight who always tells the truth, or a liar who lies every time, and people of each type are present at the table. During the discussion, everyone said that the islander sitting directly opposite of him/her across the table and the two neighbours of that person were not all of one type, i.e., not all three knights and not all three liars. Prove that there are no fewer than 10 and not over 25 liars at the table.",
    "options": [],
    "answer": "Between 10 and 25 liars (inclusive).",
    "solution": "Lemma 1. There is at least one knight among any two diametrically opposite islanders.\n\n*Proof.* Suppose the opposite is true: let $C$ and $X$ be two opposite sitting islanders, which are both liars (Fig. 24). Then the neighbours of $X$, islanders $Y$ and $W$, must be liars, since $C$ must have lied, which meant that \"an islander sitting directly in front of him and his two neighbours are all of one type\". Analogously, both neighbours of $C$, islanders $B$ and $D$, must be liars. However, then there are two more opposite pairs of liars: $D$, $Y$ and $B$, $W$. Applying the same reasoning to them, we get that all the people at the table are liars. This contradicts the statement, so our assumption is false, and the lemma is proved.\n\n![](attached_image_1.png)\n\nFrom here, the estimate of the maximum number of liars becomes clear. Since there are 25 pairs of opposite-sitting people, the number of knights is not less than 25, hence, the number of liars is not greater than 25.\n\nLemma 2. There is at least one liar among any five islanders sitting in a row.\n\n*Proof.* Suppose this does not hold, and there are five knights sitting in a row: $L_1, \\dots, L_5$. Then people opposite to $L_2, L_3$ and $L_4$ are liars, because they lied about three people of the same type sitting together. But then $L_3$ must have lied, because all three opposite of him/her are of the same type. This contradiction *proves the lemma*.\n\nNow we divide all 50 islanders sitting at the table into 10 groups of 5, each of them must contain at least 1 liar, hence, in total there are at least 10 liars.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all tuples of positive integers $(a, b, c)$, which satisfy the equation:\n$$\na + b + c^2 = abc.\n$$",
    "options": [],
    "answer": "(2,2,2), (5,1,2), (1,5,2), (5,1,3), (1,5,3), (3,2,1), (2,3,1), (3,2,5), (2,3,5)",
    "solution": "Consider this equation as a square equation with respect to $c$. Then,\n$$\nc^2 - abc + (a + b) = 0 \\Rightarrow D = (ab)^2 - 4(a + b) = m^2.\n$$\nIf $b = 1$, $D_1 = a^2 - 4a - 4 = (a - 2)^2 - 8 = x^2 - 8$, where $x = a - 2$.\nIf $x = 3$, $D_1 = 1$ is a perfect square, so $a = 5$, $c^2 - 5c + 6 = 0$, which yields the answers: $(5, 1, 2)$ and $(5, 1, 3)$, and corresponding symmetric tuples: $(1, 5, 2)$ and $(1, 5, 3)$.\nIf $x = 4$, $D_1 = 8$ is not a perfect square.\nIf $x \\ge 5$, $x^2 > D_1 > (x - 1)^2 = x^2 - 2x + 1$, so $D_1$ is not a perfect square.\n\nNow, suppose $b \\ge 2$. Then, $2ab \\ge a + b + 4 \\Leftrightarrow a(b - 1) + b(a - 1) \\ge 4$. Hence,\n$$\n(ab)^2 > D = (ab)^2 - 4(a + b) \\ge (ab - 4)^2 = (ab)^2 - 8ab + 16.\n$$\nThus, 4 cases remain.\n$D = (ab - 4)^2$, then, from previous estimates, this is only possible if $a = b = 2$, thus, $c = 2$, and we have the solution $(2, 2, 2)$.\n$D = (ab - 3)^2$ and $D = (ab - 1)^2$ are impossible, since then $D = (ab - n)^2 = (ab)^2 - 4(a + b) \\Leftrightarrow 2abn = n^2 + 4(a + b)$, which gives a contradiction for an odd $n$.\n\n$D = (ab - 2)^2$, then $4ab = 4 + 4(a + b) \\Rightarrow ab - a - b - 1 = 0 \\Rightarrow a(b - 1) - (b - 1) = 2 \\Rightarrow (a - 1)(b - 1) = 2 \\Rightarrow a - 1 = 2$ and $b - 1 = 1$. Thus, $a = 3$ and $b = 2$, and we find $c$\n\nfrom the equation: $c^2 - 6c + 5 = 0$. This gives us the solutions $(3, 2, 1)$ and $(3, 2, 5)$, and their symmetric counterparts $(2, 3, 1)$ and $(2, 3, 5)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let quasi-square be a figure that consists of an $n \\times n$-square with one more $1 \\times 1$ square attached along one of its sides, so that the unit square shares a side with one of the unit squares of an $n \\times n$-square as well as shares a vertex with one of the corner squares of the $n \\times n$-square. Thus, the upper two figures on Fig. 3 are quasi-squares, whereas the lower two are not. Determine all possible integer $n \\ge 3$ for which a plane can be filled with the identical quasi-squares. Quasi-squares can be rotated and reflected but are not allowed to overlap.\n\n![](attached_image_1.png)\nFig. 3",
    "options": [],
    "answer": "All integers n ≥ 3",
    "solution": "Fig. 4 shows an example of filling the plane with quasi-squares in which an extra $1 \\times 1$-square shares a side with an edge square. Fig. 5 shows an example for quasi-squares which have unit square sharing the vertex but not the side with the edge square of an $n \\times n$-square.\n\n![](attached_image_2.png)\nFig. 4",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Give an example of a hexagon (not necessarily convex) that can be cut with one straight line into a triangle and a quadrilateral (not necessarily convex), but which cannot be cut into two triangles or two quadrilaterals.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On Fig. 21, the dashed line shows how to cut the hexagon – one has to draw a segment $CF$ or $BD$.\n\nLet us now see where the line of separation of $ABCDEF$ can be drawn.\nIf it passes through a vertex of a hexagon and is different from lines $AC$ and $BE$, e.g. $AL$, then\non the side, a new point ($L$) appears, meaning the resulting polygons must have 7 vertices, two of\nwhich are counted twice (in this case, $A$ and $L$), which means that in total, these polygons must\nhave 9 vertices, which is not possible for both two triangles and two quadrilaterals. Analogously,\nif the line does not pass through the vertex (e.g., $MN$), then in total, there must be 10 vertices,\nwhich is also impossible for two triangles and two quadrilaterals. The only case left is when the\nsegment connects two vertices of a hexagon, e.g. $BE$, then in total this yields 8 vertices, which\ncould be formed by two quadrilaterals. But it suffices\nto check all such segments to see that none such\nsegment partitions the hexagon into two\nquadrilaterals. $BE$ and $FD$ are the only such\nsegments, and each of them partitions the hexagon\ninto a triangle and a pentagon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45712,
    "subject": "Mathematics (Multi-modal)",
    "question": "After $x$ matches in a championship a team had exactly $n\\%$ of wins, where $x, n$ are positive integers. What is the least $x$ for which it is possible that after $x+1$-st match the team had exactly $(n+1)\\%$ of wins?",
    "options": [],
    "answer": "24",
    "solution": "It follows from the conditions of the problem that after $x$ matches the team had $y$ wins, such that $\\frac{y}{x} = \\frac{n}{100}$. Then, the following has to hold: $\\frac{y+1}{x+1} = \\frac{n+1}{100}$.\n\nThe first equation implies $100y = nx$, and so $100$ has to divide $nx$.\nThe second equation implies $100y + 100 = nx + x + n + 1 \\Rightarrow 100 = x + n + 1$.\nHence $x = 100 - n - 1 \\le 98$. Moreover, $n = 100 - x - 1 \\Rightarrow xn = 100x - x^2 - x$. Therefore, $100$ has to divide $x^2 + x$. Since $x^2 + x = x(x+1)$ and the numbers $x$ and $x+1$ are relatively prime and less than $100$, one of the numbers has to be divisible by $4$, and the other by $25$. Clearly, the smallest such number is $x = 24$. Thus, $n = 75$ and $y = 18$. It is easy to check that\n$$\n\\frac{y}{x} = \\frac{18}{24} = 75\\% \\text{ and } \\frac{y+1}{x+1} = \\frac{19}{25} = 76\\%\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a white square $ABCD$ of size $8 \\times 8$, that consists of $64$ unit squares of size $1 \\times 1$. In a turn one can choose any\n\na) square;\nb) rectangle,\n\nthat consists of a whole number of unit squares and contains at least one of the vertices of the square $ABCD$, and change a color of every unit square in it to the opposite one (white is changed to black and vice versa). Is it possible to obtain arbitrary coloring of the square $ABCD$ using the turns described above?",
    "options": [],
    "answer": "a) No. b) Yes.",
    "solution": "a) Clearly, choosing the same square and acting in it twice is equivalent to not acting at all. Thus, we can assume that every square can be chosen not more than once. There are $32$ squares for which one can make a turn. Thus, there are not more than $2^{32}$ different colorings, which is less than $2^{64}$ – the amount of all possible colorings of unit squares in two colors.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nb) Choose any unit square (depicted in black) and change color of every highlighted rectangle into the opposite one. There are either two (Fig. 20a), three (Fig. 20b) or four (Fig. 20c) such rectangles. After that change the color of the whole square $ABCD$ into the opposite one. This way, only the color of the chosen unit square is changed. Thus, it is possible to obtain any coloring of the original square in two colors.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Petryk and Vasyl' are playing a game with the numbers written on the board. In a single one move – Petryk goes first – the player chooses two co-prime numbers out of the ones written on the board, erases them, and writes down their sum instead. The one who can't make the move loses. Who will win if both players play correctly and if the numbers written initially are:\na) $2019$ digits $1$; b) $2020$ digits $1$?",
    "options": [],
    "answer": "a) Vasyl wins. b) Vasyl wins.",
    "solution": "a) Let us show that Vasyl' can achieve the situation that after each his move, the board contains some odd number $n$ and $2019-n$ numbers $1$. Then, after $1009$ pairs of moves, only number $2019$ remains on the board, and Petryk will not be able to make a move and will lose. As his first move, Petryk wipes two numbers $1$ and writes down number $2$. Vasyl' makes number $3$. If at some point an odd number $n$ is written on the board, and $2019-n$ numbers $1$, then Petryk can either erase two $1$ and write $2$, after which Vasyl' erases $n$ and $2$ and writes $n+2$, or he exchanges $n$ and $1$ for $n+1$, after which Vasyl' erases $n+1$ and $1$ and writes $n+2$, and wins.\n\nb) Here Vasyl' follows a strategy similar to a) except for the last move. Before the last move of Petryk, four numbers are written on the board: $2017$, $1$, $1$ and $1$. If Petryk leaves the numbers $2018$, $1$ and $1$, then Vasyl' turns it into $2018$ and $2$, and Petryk cannot make a move. If Petryk leaves the numbers $2017$, $2$ and $1$, then Vasyl' again turns it into $2018$ and $2$, and Petryk cannot make a move.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45715,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many distinct prime divisors does the number $11^8 + 11^7 - 132$ have?",
    "options": [],
    "answer": "7",
    "solution": "Consider the following transformation:\n$$\n11^8 + 11^7 - 132 = 11^6 \\cdot (11^2 + 11) - 132 = 11^6 \\cdot 132 - 132 = 132 \\cdot (11^6 - 1).\n$$\nSince $132 = 11 \\cdot 12 = 2^2 \\cdot 3 \\cdot 11$, it has three prime divisors: $2$, $3$, $11$.\n$$\n\\begin{aligned}\n11^6 - 1 &= (11^3 - 1)(11^3 + 1) = (11-1)(11^2 + 11+1)(11+1)(11^2 - 11+1) \\\\\n&= 10 \\cdot 133 \\cdot 12 \\cdot 111 = 2^3 \\cdot 3^2 \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 37.\n\\end{aligned}\n$$\nTherefore, we have four more prime divisors: $5$, $7$, $19$, $37$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45716,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given that there exists such prime $p$, for which $10^{17} \\le p \\le 10^{17} + 10$. Find $p$.",
    "options": [],
    "answer": "10^17 + 3",
    "solution": "We should find the last digit of this number. It cannot be even or equal to $5$.\n\nNumber $10^{17} + 1$ is divisible by $11$.\n\nConsider number $10^{17} + 9$ modulo $7$. Then, $10 \\equiv 3 \\pmod{7} \\Rightarrow 10^2 \\equiv 2 \\pmod{7} \\Rightarrow 10^3 \\equiv -1 \\pmod{7} \\Rightarrow 10^{15} \\equiv -1 \\pmod{7} \\Rightarrow 10^{17} \\equiv -2 \\pmod{7} \\Rightarrow 10^{17} + 9 \\equiv 0 \\pmod{7}$.\n\nConsider number $10^{17} + 7$ modulo $17$. $10 \\equiv 10 \\pmod{17} \\Rightarrow 10^2 \\equiv -2 \\pmod{17} \\Rightarrow 10^8 \\equiv 16 \\equiv -1 \\pmod{17} \\Rightarrow 10^{16} \\equiv 1 \\pmod{17} \\Rightarrow 10^{17} \\equiv 10 \\pmod{17} \\Rightarrow 10^{17} + 7 \\equiv 0 \\pmod{17}$.\n\nThus, the only candidate for a prime number left is $10^{17} + 3$, and since it is given that such a number exists, this must be it.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45717,
    "subject": "Mathematics (Multi-modal)",
    "question": "299 digits $0$ and one digit $1$ are written in a circle. The following moves are allowed:\n* from each digit, subtract the sum of the adjacent digits;\n* select two digits with exactly two digits between them and increase both by $1$ or decrease both by $1$.\n\nIs it possible to obtain such an arrangement of numbers (after a finite number of such moves), in which there are two adjacent digits $1$, and the rest of the digits are $0$?",
    "options": [],
    "answer": "No",
    "solution": "Let us analyse how the recorded moves affect the sum of the digits written in a circle. Let us denote the numbers by $a_1, a_2, \\dots, a_{300}$. The following numbers will be written after the move of the first type: $b_k = a_k - a_{k-1} - a_{k+1}$, $k = 1, \\dots, 300$ ($a_{301} \\equiv a_1$). Let $S = a_1 + a_2 + \\dots + a_{300}$,\n\nthen $b_1 + b_2 + \\dots + b_{300} = S - S - S = -S$. After the move of the second type, the sum becomes $S+2$ or $S-2$. Since the parity of the sum after each move does not change, it is impossible to get an even sum from the original odd sum.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45718,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which greatest value of $n$, there exist integers $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$ such that:\n* all numbers $b_1, b_2, ..., b_n$ are different and lie in the range $[0; 99]$;\n* $1 \\le a_1 < a_2 < ... < a_n \\le 100$;\n* for all $i: 1 \\le i \\le n$, $b_i = a_i - i$ or $b_i = a_i - i + n$ holds?",
    "options": [],
    "answer": "67",
    "solution": "We denote $c_i = a_i - i$, then $0 \\le c_1 \\le c_2 \\le ... \\le c_n \\le 100 - n$, $b_i = c_i$ or $b_i = c_i + n$ for all $i = 1, n$. Then, $c_i + n \\le c_n + n = a_n \\le 100$. Note that among the three consecutive numbers $c_i, c_{i+1}$ and $c_{i+2}$, at least two are different, because otherwise, at least two out of three numbers $b_i, b_{i+1}$ and $b_{i+2}$ would be the same, which contradicts problem statement. Hence, $c_{i+2} \\ge c_i + 1$.\n\nThus, $c_n \\ge c_{n-2} + 1 \\ge \\dots \\ge \\frac{1}{2}(n-1)$. On the other hand, $c_n \\le 100 - n$. Thus, $\\frac{1}{2}(n-1) \\le c_n \\le 100 - n \\Rightarrow n \\le 67$. Now we show an example for $n=67$:\n\n$$\nc_1 = c_2 = 0,\\ c_3 = c_4 = 1,\\ \\dots,\\ c_{65} = c_{66} = 32\\ \\text{ and }\\ c_{67} = 33,\n$$\n$$\nb_1 = 0,\\ b_2 = 67,\\ b_3 = 1,\\ b_4 = 68,\\ \\dots,\\ b_{65} = 32,\\ b_{66} = 99\\ \\text{ and }\\ b_{67} = 33,\n$$\ntherefore, $a_i = c_i + i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the point on the arc $AC$ of the circumcircle of the triangle $ABC$ ($AB < BC$), that doesn't contain point $B$. Let $X$ and $X'$ be any two points on the side $AC$ such that $\\angle ABX = \\angle CBX'$. Show that regardless of the choice of point $X$, the circumcircle of $\\triangle DXX'$ passes through a fixed point different from $D$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $W$ be the circumcircle of $\\triangle ABC$, let $w$ be the circumcircle of $\\triangle BXX'$, and let $v$ be the circumcircle of $\\triangle DXX'$. Let $TB$ be a tangent line to the circle $W$, where $T$ belongs to the line $AC$. Then $TB^2 = TA \\cdot TC$, as well as (Fig. 12)\n$$\n\\angle TBX = \\angle TBA + \\angle ABX = \\angle ACB + \\angle CBX' = \\angle BX'X.\n$$\nThus, the line $TB$ is also tangent to the circle $w$, thus $TB^2 = TX \\cdot TC$. Let $w \\cap TD = F \\neq D$. Then $TX \\cdot TC = TF \\cdot TD = TB^2$. Thus, the points $X, X', D, F$ lie on the circle $v$ (due to the property of inscribed quadrilateral). Since $T, B$ and $D$ are fixed points, then so is the point $W \\cap TD = F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45720,
    "subject": "Mathematics (Multi-modal)",
    "question": "We enumerate all prime numbers in ascending order: $p_1 = 2, p_2 = 3, p_3 = 5, \\dots$. Find all positive integer $n$, for which $p_1!+p_2!+\\dots+p_n!=a^b$, for some positive integer $a, b>1$, where $k!$ denotes the product of all integers from 1 to $k$.",
    "options": [],
    "answer": "n = 2, 3",
    "solution": "Directly checking yields:\n$$\np_1! = 2, \\quad p_1! + p_2! = 8 = 2^3, \\quad p_1! + p_2! + p_3! = 128 = 2^7\n$$\nFor $n \\ge 4$, $p_1! + p_2! + \\dots + p_n! = 128 + 7! + 11! + \\dots + p_n!$ is a number in which all summands are divisible by $2^5$, except $p_4! = 7! = 2^4 \\cdot 315$, which is divisible by $2^4$ but not by $2^5$. Therefore, this number cannot be a power of 2. If it is a power of another number, then since $2^4$ divides it, it can only be $2^4$ or $x^2$. Let us show that it cannot be a square. Indeed, $2! + 3! + \\dots + p_n!$ gives a remainder 2 modulo 3, which is impossible for squares.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square with the side-length $10$. Let $E$ and $F$ be the points on the sides $BC$ and $AD$ respectively, so that $ABEF$ is a rectangle. The rectangle $KLMN$ is such that its vertices $K$, $L$, $M$ and $N$ belong to the segments $CD$, $DF$, $FE$ and $EC$ respectively. It turns out that the rectangles $ABEF$ and $KLMN$ are equal, and $AB = MN$. Determine the length of the segment $AL$.\n\n![](attached_image_1.png)\nFig. 1",
    "options": [],
    "answer": "5",
    "solution": "Consider the diagonals $AE$ and $LN$ of the equal rectangles $ABEF$ and $KLMN$. Clearly, $AE = LN$. Let $NX$ be the perpendicular to the segment $AD$ (Fig. 1). Clearly, $NX = AB = 10$. Then $\\triangle AEF = \\triangle NXL$, since they are right triangles with equal hypotenuses and sides. Thus, $AF = LX$. Let $LY \\perp BC$, then the rectangle $LXNY$ is equal to the rectangles $ABEF$ and $KLMN$. Thus, $\\triangle LNY = \\triangle NXL$, since they are right triangles with equal hypotenuses and sides. Hence\n$$\n\\angle MNE = \\angle LNY - \\angle LNM = \\angle NLX - \\angle NLK = \\angle KLD,\n$$\nso $\\triangle MNE = \\triangle KLD$, since they are right triangles with equal hypotenuses and angles. Therefore, $LD = NE = BY = AL$, hence $L$ is a midpoint of $AD$, thus, $AL = 5$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n$ pairwise distinct positive numbers. Vasia wrote on the board all possible numbers of the form $a + \\frac{b}{c}$, where $a, b, c$ are distinct numbers from the given set. Can one always find two numbers on the board such that the larger one is not more than twice the other one, if\n\na) $n=4$?\n\nb) $n=3$?",
    "options": [],
    "answer": "a) Yes. b) No; for example, the set {ten to the minus one, ten to the minus three, ten to the minus nine}.",
    "solution": "a) We will show that for four given numbers one can always find two numbers on the board such that one is not more than twice the other one. Suppose by contradiction, it is not true. Let the given numbers be $a < b < c < d$. Then $d + \\frac{b}{a} > c + \\frac{b}{a} \\Rightarrow d + \\frac{b}{a} > 2(c + \\frac{b}{a}) \\Rightarrow d + \\frac{b}{a} > 1$.\n\nOn the other hand, $d + \\frac{b}{c} > d + \\frac{a}{c} \\Rightarrow d + \\frac{b}{c} > 2(d + \\frac{a}{c}) \\Rightarrow d < \\frac{b}{c} < 1$. Thus, we get a contradiction.\n\n\nb) We will give an example of three numbers for which the condition doesn't hold: $a = 10^{-1}$, $b = 10^{-3}$ and $c = 10^{-9}$. Then, the following numbers are written on the board:\n\n$10^{-1} + 10^{6}$, $10^{-1} + 10^{-6}$, $10^{-3} + 10^{8}$, $10^{-3} + 10^{-8}$, $10^{-9} + 10^{2}$, $10^{-9} + 10^{-2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integer $n$ has no more than $2020$ digits. Prove that there exist $m$ such palindromes that their sum equals $n$, where $m \\le 13$.\n\nA palindrome is a number which is read the same both left to right and right to left, i.e. $1001$, $9$ and $767$ are palindromes, while $1212$ and $110$ are not.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we prove a lemma:\n\n**Lemma 1.** Suppose there is a $k$-digit positive integer $X$, $k > 2$. Then such a palindrome can be subtracted from it, that the remaining number consists of no more than $\\lfloor \\frac{1}{2}k+1 \\rfloor$ digits.\n\n*Proof.* Suppose the first $\\lfloor \\frac{k+1}{2} \\rfloor$ digits of $X$ make up the number $A \\ne 100\\dots0$. Then $B$, a number formed from $(A-1)$ by writing the digits in the opposite order (i.e. $B$ can begin with several zeros). Then, for an even $k$, we subtract number $\\overline{(A-1)B}$, and for an odd $k$, number $\\overline{(A-1)B'}$, where $B'$ is number $B$ but without the first digit. Then after such subtraction, the first digit on the left, which may be non-zero, is $\\lfloor \\frac{1}{2}k+1 \\rfloor$-th digit. In case if $A=100\\dots0$, we simply subtract from $X$ number $99\\dots9$, which has one fewer digits.\n\n*Lemma is proved.*\n\n**Lemma 2.** A number with no more than three digits can be represented as a sum of no more than three palindromes.\n\n*Proof.* For two-digit and single-digit numbers, this is obvious. For the three-digit number $abc$, we consider the following cases.\n\nFor $a \\le c$, $abc = aba + (c-a)$.\n\nFor $a = c+1$, $abc = cbc + 99 + 1$.\n\nFor $a > c + 1$, $abc = cbc + (a-c)00 = cbc + (a-c-1)9(a-c-1) + (11-a+c)$.\n\n*Lemma is proved.*\n\nNow let's calculate how the number of digits will change if we first take a $2020$-digit number, then apply lemma 1 ten times and then lemma 2:\n$$\n2020 \\rightarrow 1011 \\rightarrow 506 \\rightarrow 254 \\rightarrow 128 \\rightarrow 65 \\rightarrow 33 \\rightarrow 17 \\rightarrow 9 \\rightarrow 5 \\rightarrow 3.\n$$\nThus, we get no more than $13$ palindromes. If the initial number has fewer digits, then the number of palindromes can only decrease.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45724,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) For which positive integer $l$ there exists a pair of positive integers $(m, n)$, which consist of no more than $l$ digits, have different last digits and which satisfy the condition: the last $l$ digits of $m^2$ form the number $n$, and analogously, the last $l$ digits of $n^2$ form number $m$ (discarding any zeros that the last $l$ digits may begin with).\n\nb) For which positive integer $l$ there exists a pair of positive integers $(m, n)$, which consist of no more than $l$ digits, have different last digits and which satisfy the condition: the last $l$ digits of $m^3$ form the number $n$, and analogously, the last $l$ digits of $n^3$ form number $m$ (discarding any zeros that the last $l$ digits may begin with).",
    "options": [],
    "answer": "a) No positive integer l. b) Every positive integer l.",
    "solution": "a) Consider the last digits of numbers satisfying the conditions of the problem. We follow the last digit of their squares, so we put the matching the last digit to each square:\n\n$1^2 \\rightarrow 1, 2^2 \\rightarrow 4, 3^2 \\rightarrow 9, 4^2 \\rightarrow 6, 5^2 \\rightarrow 5, 6^2 \\rightarrow 6, 7^2 \\rightarrow 9, 8^2 \\rightarrow 4, 9^2 \\rightarrow 1$\n\nb) Analogously to a), we find those single-digit numbers that satisfy the conditions and write down the last digit for each third power:\n\n$1^3 \\rightarrow 1, 2^3 \\rightarrow 8, 3^3 \\rightarrow 7, 4^3 \\rightarrow 4, 5^3 \\rightarrow 5, 6^3 \\rightarrow 6, 7^3 \\rightarrow 3, 8^3 \\rightarrow 2, 9^3 \\rightarrow 9$\n\n- pairs $(3; 7)$ and $(2; 8)$ satisfy the conditions.\n\nFor example, we already found a pair of $l$-digit numbers $(m, n)$, satisfying the condition: $m$'s last digit is $2$ its third power ends with $n$, and analogously, number $n$'s last digit is $8$ and its third power ends with $m$. Let us try to find $l+1$-st digits of these numbers. In all equalities, all digits to the left of $l+1$-st one are discarded. We denote the unknown numbers by $\\overline{xm}$ and $\\overline{yn}$.\n\nThen let $m^3 = 10' A + n$, $n^3 = 10' B + m$, which yields:\n\n$$\n\\begin{align*}\n\\overline{xm}^3 &= (10' x + m)^3 \\rightarrow 3 \\cdot 10' x \\cdot m^2 + m^3 = 3 \\cdot 10' x \\cdot 4 + 10' A + n \\rightarrow \\\\\n&\\rightarrow 10' (2x + A) + n = \\overline{yn} = 10' y + n, \\\\\n\\overline{yn}^3 &= (10' y + n)^3 \\rightarrow 3 \\cdot 10' y \\cdot n^2 + n^3 = 3 \\cdot 10' y \\cdot 4 + 10' B + m \\rightarrow \\\\\n&\\rightarrow 10' (2y + B) + m = \\overline{xm} = 10' x + m.\n\\end{align*}\n$$\n\nAfter simplifications and reductions, get the system:\n\n$$\n\\begin{cases}\n2x + A = y, \\\\\n2y + B = x.\n\\end{cases}\n$$\n\nFrom which, $x = 2y + B = 2(2x + A) + B = 4x + (2A + B)$, hence, $\\begin{cases} 3x = -(2A + B), \\\\ 3y = -(2B + A). \\end{cases}$\n\nSince for arbitrary $A, B$ there exist corresponding digits $x, y$ (recall that the equality only applies to the last digits here), and $3x$ and $3y$ take all possible values.\n\nThus, corresponding pairs of numbers exist for any value of $l$.\n\nLet us show by examples: the transition from a single-digit number to a two-digit number:\n\n$$\n\\begin{align*}\nm_1 = 2, m_1^3 = 8 &\\rightarrow 10 \\cdot 0 + 8 \\Rightarrow A = 0, n_1 = 8, n_1^3 = 512 \\rightarrow 10 \\cdot 1 + 2 \\Rightarrow B = 1. \\\\\n3x \\rightarrow -(2A+B) = -1 &\\rightarrow 9 \\Rightarrow x = 3, 3x \\rightarrow -(2B+A) = -2 \\rightarrow 8 \\Rightarrow y = 6.\n\\end{align*}\n$$\n\nWe obtain the numbers:\n\n$$\nm_2 = \\overline{xm_1} = 32, m_2^3 = 32^3 = 32768, n_2 = \\overline{yn_1} = 68, n_2^3 = 68^3 = 314432,\n$$\n\nwhich satisfy the statement.\n\nThe transition from a two-digit number to a three-digit number:\n\n$$\n\\begin{align*}\nm_2 = 32, m_2^3 = 32768 &\\rightarrow 100 \\cdot 7 + 68 \\Rightarrow A = 7. n_2 = 68, n_2^3 = 314432 \\rightarrow 100 \\cdot 4 + 32 \\\\\n&\\Rightarrow B = 4.\n\\end{align*}\n$$\n\n$$\nm_3 = \\overline{xm_2} = 432, m_3^3 = 432^3 = 80621568, n_3 = \\overline{yn_2} = 568, n_3^3 = 568^3 = 183250432,\n$$\n\nwhich, too, satisfy the statement.\nb) Let us prove that such pair exists for every value of $l$. We show that there is such positive integer $x$ that $2^l \\mid x^2 - 1$ and $5^l \\mid x^2 + 1$.\n\nFor the first divisibility, we set $x \\equiv 1 \\pmod{2^l}$. Using Method of Mathematical Induction (MMI) for $l$, we show that for any $l$ there exists $x$ such that $5^l \\mid x^2 + 1$. If $l = 1$, then we set $x = 2$.\n\nSuppose for some $l$, the statement holds. Consider $l+1$: if $x$ works for $l$, i.e. $x^2+1=t \\cdot 5^l$, then we set $x_1 = x+k \\cdot 5^l$; so\n\n$$\nx_1^2 + 1 = x^2 + k^2 \\cdot 5^{2l} + 2xk \\cdot 5^l + 1 = t \\cdot 5^l + k^2 \\cdot 5^{2l} + 2xk \\cdot 5^l = 5^l(t + k^2 \\cdot 5^l + 2xk).\n$$\n\nObviously, there exists such $k$ that the expression in parentheses is divisible by $5$, we choose such $x_1$ for this $k$, and the statement is proved.\n\nThen, by the Chinese remainder theorem, we select $x$ which satisfies both conditions. Then, we show that the last $l$ digits of the pair $x$ and $x^3$ satisfy the statement. Firstly, by the construction, both numbers have no more than $l$ digits. Secondly, they have different last digits: otherwise, $x^3 - x$ would be divisible by $5$, but $x^3 - x = x(x^2 - 1) \\equiv -2x \\pmod 5$ hence, $x$ must have been divisible by $5$, which, obviously, is impossible if $x^2 + 1 = t \\cdot 5^l$.\n\nIt is left to show that $x - x^3 \\equiv 0 \\pmod{10^l}$, which is obvious by construction, and also\n\n$$\n(x^3)^3 - x = x^9 - x = x(x^4 + 1)(x^2 + 1)(x^2 - 1) \\equiv 0 \\pmod{10^l},\n$$\n\nwhich is obvious, as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral with equal sides $AB$ and $BC$. Let $E$ be the point on the line $AB$, such that $BD = BE$ and $AD \\perp DE$. Show that the perpendicular bisectors of segments $AD$, $CD$ and $CE$ intersect in precisely one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\triangle BAD$ is isosceles with the base $DE$, $\\angle BDE = \\angle BED < 90^\\circ$. Since $\\angle AED < 90^\\circ$, then the points $B$ and $E$ are on the same side from point $E$. Moreover, $\\angle BED < 90^\\circ = \\angle ADE$, so the point $B$ belongs to the segment $AE$ (Fig. 2). Then\n$$\n\\angle BAD = \\angle EAD = 180^\\circ - \\angle ADE - \\angle AED = 90^\\circ - \\angle BED = 90^\\circ - \\angle BDE = \\angle ADE - \\angle BDE = \\angle BDA.\n$$\nThus, $\\triangle BAD$ is isosceles with the base $AD$. Then $BA = BD$. Thus, $BC = BA = BD = BE$. Thus, from the property of the perpendicular bisector of the base of the isosceles triangle, perpendicular bisectors of the segments $AD$, $CD$ and $CE$ pass through the point $B$. Q.E.D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all possible positive integers $m$ and $n$, that satisfy the following:\n$$\n(m+n)! = 2m! \\cdot n!\n$$\nwhere $k!$ denotes the product $1 \\cdot 2 \\cdot \\dots \\cdot k$, where $k$ is a positive integer.",
    "options": [],
    "answer": "m = 1, n = 1",
    "solution": "Without loss of generality, assume that $m \\ge n$. Then, if $n > 1$ equation can be written the following way:\n$$\n1 \\cdot 2 \\cdot 3 \\cdots m \\cdot (m+1) \\cdot (m+2) \\cdots (m+n) = 2 \\cdot 1 \\cdot 2 \\cdot 3 \\cdots m \\cdot 1 \\cdot 2 \\cdot 3 \\cdots n \\Rightarrow \\\\ (m+1) \\cdot (m+2) \\cdots (m+n) = 2 \\cdot 1 \\cdot 2 \\cdot 3 \\cdots n.\n$$\nThe factors on the left side are not less than the factors on the right side, since the factors can be paired the following way:\n$$\nm+1 > 1,\\ m+2 > 2,\\ \\dots,\\ m+n-1 > n-1,\\ m+n \\ge 2n.\n$$\nSince the factor $m+1$ always exists and $m+1 > 1$, equation doesn't hold. If $n=1$, equation $(m+1)! = 2m!$ holds only if $m+1=2$. Thus, the only solution is $m=n=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $(x, y)$ that satisfy the equality: $x - y = \\frac{x}{y}$.",
    "options": [],
    "answer": "(4, 2)",
    "solution": "Let us rewrite the equality as follows: $xy - y^2 = x$, or $x(y-1) = y^2$. Since numbers $y^2$ and $y-1$ are co-prime, i.e. don't have common divisors greater than 1, then from the last equality, $x \\mid y^2$ and $y^2 \\mid x \\Rightarrow x = \\pm y^2$. Note that $y \\neq 0$.\nIf $x = y^2$, then $y^3 - y^2 = y^2 \\Rightarrow y = 2$ and $x = 4$.\nIf $x = -y^2$, then we obtain that $-y^3 - y^2 = -y^2 \\Rightarrow y = 0$, which doesn't give any new solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45728,
    "subject": "Mathematics (Multi-modal)",
    "question": "The managers of two companies are 1000 km apart. The segment that connects them contains $n$ points, in each of which sits a mathematician. Every second each mathematician moves to a point which is a middle of a segment that connects him/her and the closest mathematician or manager. If there is more than one such point, mathematician chooses any of them. If two mathematicians arrive to the same point according to the rules above, younger of them leaves the game. Show that in finitely many seconds every mathematician which is still in the game can shake hands with one of the managers. A mathematician can shake hands when the distance from the manager does not exceed 1 m.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S_0 = \\{0, x_1, x_2, \\dots, x_n, 1000\\}$ be the initial locations of mathematicians and managers, where $0$ and $1000$ represent the locations of managers, and $x_1 < x_2 < \\dots < x_n$ represent the locations of mathematicians. Similarly, let $S_i$ denote the positions of everyone in $i$ seconds. Note that the numbers of elements of $S_i$ will decrease if there are two points closest to each other among the locations of mathematicians. Since there are finitely many elements in $S_0$, there are finitely many decreases, thus, there exists an integer non-negative $N$ such that $|S_k| \\le |S_N|$, $\\forall k \\ge N$.\n\nLet $S_N = \\{0, y_1, y_2, \\dots, y_m, 1000\\}$, $0 < y_1 < y_2 < \\dots < y_m < 1000$, $m \\le n$. Since $|S_{N+1}| \\le |S_N|$, then\n$$\ny_1 - 0 < y_2 - y_1 < \\dots < y_t - y_{t-1} < y_{t+1} - y_1 > y_{t+2} - y_{t+1} > \\dots > y_m - y_{m-1} > 1000 - y_m,\n$$\nthat is, for the first $t$ mathematicians, their closest neighbor is on the left, and for everyone else – on the right.\n\nThis way, both groups of mathematicians will move in their direction and will end up at a distance less than 1 meter from a manager in a finite number of steps, with whom they will shake hands.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all four-digit positive integers $\\overline{abcd}$, which satisfy the conditions:\n1) $a \\le b \\le c \\le d$;\n2) $a^2 + b^2 + c^2 + d^2$ is divisible by 4;\n3) $\\overline{abcd}$ divided by $c$ gives the remainder 7.",
    "options": [],
    "answer": "7999",
    "solution": "From the second condition, it becomes clear that all the digits are either even or odd, since the square of the integer gives the remainder 0 or 1 modulo 4. From the third condition, digit $c$ equals 8 or 9.\nIf $c=8$, then $abcd$ is even, but then it cannot give an odd remainder modulo 8, because then the equality would hold: $abcd = 8r + 7$.\nHence, $c=9$. Thus, all the digits are odd and the remainder of the number modulo 9 equals the remainder of the sum of the digits modulo 9. So, $a+b+d=9r+7$. Then, from the first condition, $d=9$. Consider the cases.\n$r=0$, then $a+b+d=7$, and we immediately have a contradiction.\n$r=1$, then $a+b+d=16$, and $a+b=7$ is impossible, since $a$ and $b$ are odd.\n$r=2$, then $a+b+d=25$, and $a+b=16$. Then, the number we are looking for is 7999.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest positive integer $n$ such that it has exactly 4 positive integer divisors (including $1$ and $n$), the sum $S$ of which satisfies the condition $40 \\le S \\le 42$.",
    "options": [],
    "answer": "27",
    "solution": "In the case $n = pq$, $S = 1 + p + q + pq = (1+p)(1+q)$. Suppose $p < q$.\nLet us consider possible cases.\nIf $S = 40$, then $(p+1)(q+1) = 40$. Since $p+1 \\ge 3$, the following cases are possible:\n$$\np+1=4, q+1=10 \\Rightarrow p=3, q=9 \\Rightarrow q \\text{ is not prime.}\n$$\n$$\np+1=5, q+1=8 \\Rightarrow p=4, q=7 \\Rightarrow p \\text{ is not prime.}\n$$\nIf $S = 41$, then $(p+1)(q+1) = 41$, since $41$ is prime, there are no solutions.\nIf $S = 42$, then $(p+1)(q+1) = 42$. The options are:\n$$\np+1=3, q+1=14 \\Rightarrow p=2, q=13 \\Rightarrow n=26.\n$$\n$$\np+1=6, q+1=7 \\Rightarrow p=5, q=6 \\Rightarrow q \\text{ is not prime.}\n$$\nSo, in this case, we have a single number $n=26$, which satisfies the conditions.\nIn case $n = p^3$, $S = 1 + p + p^2 + p^3$. Note that if $p=2$, then $1 + p + p^2 + p^3 = 15$; if $p=3$, $1 + p + p^2 + p^3 = 40$; and if $p \\ge 4$, $1 + p + p^2 + p^3 \\ge 85$.\nIn this case, there is also a single number $n=27$, which satisfies the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45731,
    "subject": "Mathematics (Multi-modal)",
    "question": "In how many ways can a chessboard $n \\times n$, $n \\ge 3$, from which two diagonally opposite corner cells $1 \\times 1$ were cut out, accommodate $n$ rooks, neither of which attacks one other? A rook is a chess piece that attacks all the cells adjacent horizontally or vertically to the cell it is located in.",
    "options": [],
    "answer": "(n-2)!(n^2 - 3n + 3)",
    "solution": "Without limitation of generality, suppose that the left lower cell $A$ and the right upper cell $Z$ were cut out. First, let us see in how many ways rooks can be placed on a $n \\times n$ chessboard. There are $n!$ ways, since for a rook, there are $n$ options in the first column, $n-1$ option in the second column, and so on. Consider the arrangements of rooks, in which one of them is located in cell $A$. There are $(n-1)!$ such arrangements, and analogously, $(n-1)!$ ways, for when a rook occupies cell $Z$. In the expression $n!-2(n-1)!$ positions where rooks occupy cells $A$ and $Z$ are rejected twice. There are $(n-2)!$ such arrangements. Thus, the correct expression would be:\n$$\nn!-2(n-1)!+(n-2)!=(n-2)!(n(n-1)-2(n-1)+1)=(n-2)!(n^2-3n+3).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the following inequality for positive $a, b, c$:\n$$\n\\frac{a^3 + b^3 + c^3}{abc} + 6 \\ge 9 \\cdot \\frac{a^2 + b^2 + c^2}{ab + bc + ca}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Subtract $9$ from both sides and use the well-known identity\n$$\na^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).\n$$\n\n$$\n\\frac{a^3 + b^3 + c^3}{abc} - 3 \\ge 9 \\cdot \\frac{a^2 + b^2 + c^2}{ab + bc + ca} - 9 \\Leftrightarrow \\\\\n\\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc}{abc} - 3 \\ge 9 \\cdot \\frac{a^2+b^2+c^2}{ab+bc+ca} - 9 \\Leftrightarrow\n$$\n\n$$\n\\begin{aligned}\n\\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc-3abc}{abc} &\\ge \\frac{9(a^2+b^2+c^2-ab-bc-ca)}{ab+bc+ca} \\\\\n&\\iff \\\\\n\\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{abc} &\\ge \\frac{9(a^2+b^2+c^2-ab-bc-ca)}{ab+bc+ca} \\\\\n&\\iff \\\\\n\\frac{(a^2+b^2+c^2-ab-bc-ca)((a+b+c)(ab+bc+ca)-9abc)}{abc(ab+bc+ca)} &\\ge 0.\n\\end{aligned}\n$$\n\nThe last inequality holds, since\n$$\na^2 + b^2 + c^2 - ab - bc - ca \\ge 0 \\quad \\text{and} \\quad (a+b+c)(ab+bc+ca) - 9abc \\ge 0.\n$$\nBy the inequality with the sum of three squares:\n$$\n\\begin{aligned}\n(a-b)^2 + (b-c)^2 + (c-a)^2 \\ge 0 &\\iff a^2 + b^2 + c^2 \\ge ab + bc + ca \\\\\n(a+b+c)^2 \\ge 3(ab+bc+ca) &\\Rightarrow (a+b+c)^6 \\ge 3^3(ab+bc+ca)^3 \\ge 27 \\cdot 27(abc)^2 \\\\\n(a+b+c)^3 &\\ge 27abc.\n\\end{aligned}\n$$\n$$\n\\text{Moreover, } \\left(\\frac{1}{3}(x+y+z)\\right)^3 \\ge xyz \\Rightarrow (ab+bc+ca)^3 \\ge 27a^2b^2c^2.\n$$\n$$\n\\text{Hence } (a+b+c)^3(ab+bc+ca)^3 \\ge (9abc)^3.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive numbers $a, b, c$ that satisfy the condition:\n$$\na^2 + b^2 + c^2 + abc = 4,\n$$\nprove the following inequality:\n$$\n(4 - a^2)(4 - b^2)(4 - c^2)a^2 b^2 c^2 \\leq (2a + bc)(2b + ca)(2c + ab).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we will prove the following lemmas:\n\n**Lemma 1.** If $\\alpha, \\beta, \\gamma \\in (0^\\circ, 90^\\circ)$ and $\\cos^2 \\alpha + \\cos^2 \\beta + \\cos^2 \\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma = 1$, then $\\alpha, \\beta, \\gamma$ are the angles of an acute triangle.\n\n*Proof.* Consider the equation in the lemma statement as quadratic in terms of $\\cos\\gamma$. Its discriminant equals\n$$\nD = 4\\cos^2 \\alpha \\cos^2 \\beta - 4\\cos^2 \\alpha - 4\\cos^2 \\beta + 4 = 4\\sin^2 \\alpha \\sin^2 \\beta.\n$$\nHence\n$$\n\\cos\\gamma = -\\cos\\alpha\\cos\\beta + \\sin\\alpha\\sin\\beta = -\\cos(\\alpha \\pm \\beta).\n$$\nAssuming that $|\\alpha - \\beta| < 90^\\circ$, we obtain $\\cos(\\alpha - \\beta) > 0$, thus, equality $\\cos\\gamma = -\\cos(\\alpha - \\beta)$ is not possible. Thus,\n$$\n\\cos\\gamma = -\\cos(\\alpha + \\beta) = \\cos(180^\\circ - \\alpha - \\beta) \\Rightarrow \\alpha + \\beta + \\gamma = 180^\\circ, \\text{ since } \\alpha + \\beta < 180^\\circ.\n$$\n\n**Lemma 2.** If $\\alpha, \\beta, \\gamma$ are the angles of an acute triangle, then the following inequality holds:\n$$\n8\\cos\\alpha\\cos\\beta\\cos\\gamma \\le 1.\n$$\n*Proof.* This lemma is well-known, and below is one of its proofs. Consider three vectors $\\vec{AB}, \\vec{BC}$ and $\\vec{CA}$, as well as unit vectors $e_a, e_b$ and $e_c$, that are co-directed and collinear to vectors $\\vec{AB}, \\vec{BC}$ and $\\vec{CA}$ respectively. Angles between them are $180^\\circ - \\alpha, 180^\\circ - \\beta$ and $180^\\circ - \\gamma$ in some order. Then\n$$\n\\begin{align*}\n(e_a + e_b + e_c)(e_a + e_b + e_c) \\ge 0 &\\Leftrightarrow e_a^2 + e_b^2 + e_c^2 + 2e_a e_b + 2e_a e_c + 2e_b e_c \\ge 0 \\\\\n&\\Leftrightarrow 3 - 2(\\cos\\alpha + \\cos\\beta + \\cos\\gamma) \\ge 0 \\\\\n&\\Leftrightarrow \\cos\\alpha + \\cos\\beta + \\cos\\gamma \\le \\frac{3}{2} \\\\\n&\\Leftrightarrow \\sqrt[3]{\\cos\\alpha\\cos\\beta\\cos\\gamma} \\le \\frac{1}{3}(\\cos\\alpha + \\cos\\beta + \\cos\\gamma) \\le \\frac{1}{2}.\n\\end{align*}\n$$\n\nSince $a, b, c < 2$, there exist $\\alpha, \\beta, \\gamma \\in (0^\\circ, 90^\\circ)$ so that $a = 2\\cos\\alpha, b = 2\\cos\\beta$ and $c = 2\\cos\\gamma$. From the conditions of the problem\n$$\na^2 + b^2 + c^2 + abc = 4 \\Rightarrow \\cos^2 \\alpha + \\cos^2 \\beta + \\cos^2 \\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma = 1,\n$$\nthus, by Lemma 1 $\\alpha, \\beta, \\gamma$ are the angles of an acute triangle.\n\n$$\n4 - a^2 = 4(1 - \\cos^2 \\alpha) = 4\\sin^2 \\alpha \\text{ and}\n$$\n$$2a + bc = 4(\\cos\\alpha + \\cos\\beta\\cos\\gamma) = 4(-\\cos(\\beta + \\gamma) + \\cos\\beta\\cos\\gamma) = 4\\sin\\beta\\sin\\gamma,$$\n\nas well as similar equalities, the original inequality can be written the following way:\n$$\n4^6 \\cdot \\cos^2 \\alpha \\cos^2 \\beta \\cos^2 \\gamma \\sin^2 \\alpha \\sin^2 \\beta \\sin^2 \\gamma \\le 4^3 \\cdot \\sin^2 \\alpha \\sin^2 \\beta \\sin^2 \\gamma \\Leftrightarrow\n$$\n$$\n2^6 \\cdot \\cos^2 \\alpha \\cos^2 \\beta \\cos^2 \\gamma \\le 1 \\Leftrightarrow 8\\cos\\alpha\\cos\\beta\\cos\\gamma \\le 1, \\text{ which is proven in Lemma 2 above.}\n$$\n\n\nAlternative solution:\n\nFirst, we will prove the following lemma.\n\n**Lemma 1.** If positive numbers $a, b, c$ satisfy $a^2 + b^2 + c^2 + abc = 4$, then the variables $x, y$ and $z$, determined by the following equations: $a = \\frac{2}{\\sqrt{(3y-1)(3z-1)}}$, $b = \\frac{2}{\\sqrt{(3z-1)(3x-1)}}$, $c = \\frac{2}{\\sqrt{(3x-1)(3y-1)}}$ satisfy $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$.\n\n*Proof.* By plugging in variables into Lemma's condition, we obtain\n$$\n4 = \\frac{4}{(3y-1)(3z-1)} + \\frac{4}{(3z-1)(3x-1)} + \\frac{4}{(3x-1)(3y-1)} + \\frac{8}{(3x-1)(3y-1)(3z-1)} \\Rightarrow\n$$\n$$\n(3x-1)(3y-1)(3z-1) = (3x-1) + (3y-1) + (3z-1) + 2 = 3x+3y+3z-1 \\Rightarrow 27xyz - 9xy - 9yz - 9zx + (3x+3y+3z-1) = 3x+3y+3z-1 \\Rightarrow 3xyz = xy + yz + zx\n$$\nwhich implies needed equality.\n\nApply Lemma 1. Then:\n$$\n\\begin{aligned}\n(4-a^2)(4-b^2)(4-c^2) &= \\left(4-\\frac{4}{(3y-1)(3z-1)}\\right) \\cdot \\left(4-\\frac{4}{(3z-1)(3x-1)}\\right) \\cdot \\left(4-\\frac{4}{(3x-1)(3y-1)}\\right) = \\\\\n&= 64 \\cdot 27 \\cdot \\frac{(3yz-y-z)(3zx-z-x)(3xy-x-y)}{(3x-1)^2(3y-1)^2(3z-1)^2} \\le \\frac{8}{27} \\\\\n\\text{(since by Lemma 1, we obtain } (3x-1)(3y-1)(3z-1) = 3x+3y+3z-1 \\ge 8) \\\\\n&\\le 27 \\cdot (3yz-y-z)(3zx-z-x)(3xy-x-y) \\le \\\\\n&\\le 27 \\cdot (27(xyz)^2 - 9(x^2yz(y+z)+\\dots) + 3(yz(y+x)(z+x)+\\dots) - \\\\\n&\\qquad -(x+y)(y+z)(z+x)) = \\\\\n&\\le 27 \\cdot (27(xyz)^2 - 18xyz(xy + yz + zx) + 9xyz(x+y+x) + \\\\\n&\\qquad 3((xy)^2 + (yz)^2 + (zx)^2) - (xy + yz + zx)(x+y+x) + xyz) = \\\\\n\\text{(since } 3xyz = xy + yz + zx) \\\\\n&= 27 \\cdot (-27(xyz)^2 + 6xyz(x+y+x) + 3((xy)^2 + (yz)^2 + (zx)^2) + xyz) = \\\\\n\\text{(since } 9(xyz)^2 = (xy)^2 + (yz)^2 + (zx)^2 + 2x^2yz + 2xy^2z + 2xyz^2) \\\\\n&= 27 \\cdot (3(9(xyz)^2 - 2xyz(x+y+z)) - 27(xyz)^2 + 6xyz(x+y+x) + xyz) = 27xyz \\Rightarrow \\\\\n&\\qquad (4-a^2)(4-b^2)(4-c^2) \\le 27xyz.\n\\end{aligned}\n$$\n\n**Lemma 2.** If positive numbers $x, y, z$ satisfy $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$, then variables $a, b$ and $c$, which are determined by equations $x = \\frac{2a+bc}{3bc}$, $y = \\frac{2b+ca}{3ca}$, $z = \\frac{2c+ab}{3ab}$, satisfy the following:\n$$\na^2 + b^2 + c^2 + abc = 4.\n$$\n*Proof.* By plugging in variables into Lemma's condition we obtain\n$$\n\\begin{aligned}\n3 &= \\frac{3bc}{2a+bc} + \\frac{3ca}{2b+ca} + \\frac{3ab}{2c+ab} \\Rightarrow \\\\\n(2a+bc)(2b+ca)(2c+ab) &= \\\\\n&= 3(abc)^2 + 4abc(a^2 + b^2 + c^2) + 4a^2b^2 + 4b^2c^2 + 4c^2a^2.\n\\end{aligned}\n$$\nOn the other hand,\n$$\n\\begin{aligned}\n(2a+bc)(2b+ca)(2c+ab) &= (abc)^2 + 2abc(a^2 + b^2 + c^2) + 4a^2b^2 + 4b^2c^2 + 4c^2a^2 + 8abc \\Rightarrow \\\\\na^2 + b^2 + c^2 + abc &= 4, \\text{ Q.E.D.}\n\\end{aligned}\n$$\n\nNow by Lemma 2 we obtain that\n$$\n(4-a^2)(4-b^2)(4-c^2) \\le 27xyz = \\frac{(2a+bc)(2b+ca)(2c+ab)}{(abc)^2} \\Rightarrow \\\\\n(4-a^2)(4-b^2)(4-c^2)a^2b^2c^2 \\le (2a+bc)(2b+ca)(2c+ab).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute non-isosceles triangle $ABC$, let $AK$ and $CN$ be its bisectors, let $I$ be the point of their intersection. Let $X$ be the second point of intersection of the circumcircles of $\\triangle ABC$ and $\\triangle KBN$. Let $M$ be the midpoint of $AC$. Show that the Euler line of $\\triangle ABC$ is perpendicular to $BI$ if and only if points $X$, $I$ and $M$ are collinear.\n\n*The Euler line of a non-isosceles triangle is a line that passes through the orthocenter, centroid and a circumcircle.*",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we will show that the Euler line of the triangle $ABC$ is perpendicular to $BI$ if and only if $\\angle ABC = 60^\\circ$. Indeed, let $H$ and $O$ be the orthocenter and the circumcircle of $\\triangle ABC$ respectively, then $\\angle HBI = \\angle IBO$, and $BI \\perp OH \\Leftrightarrow BO = BH \\Leftrightarrow \\sin \\angle BAH = \\frac{1}{2}$, which, due to the triangle $\\triangle ABC$ being acute-angled, is equivalent to $\\angle ABC = 60^\\circ$.\n\n**Lemma.** If $O$ and $H$ are the circumcircle and the orthocenter of $\\triangle ABC$ respectively, then\n$$\nBO = BH \\Leftrightarrow \\angle ABC = 60^\\circ \\text{ or } \\sin \\angle BAH = \\frac{1}{2}.\n$$\n\n![](attached_image_1.png)\nFig. 16\n**Proof.** Consider a triangle whose sides are parallel to the sides of $\\triangle ABC$ and pass through the vertices of $\\triangle ABC$ (Fig. 16), then the new triangle is similar to $\\triangle ABC$ with coefficient 2. Moreover, point $H$ becomes a circumcircle of the larger triangle, and point $B$ has a corresponding point $K$, that is a midpoint of $AC$. Then $BH = 2OK = 2OC \\cos \\angle COK = 2R \\cos \\angle ABC$. Hence by the conditions of the problem $BH = BO = R$, which is equivalent to $\\cos \\angle ABC = \\frac{1}{2} \\Leftrightarrow \\angle ABC = 60^\\circ$.\n\nIt suffices now to show that $\\angle ABC = 60^\\circ$ if and only if points $X$, $I$ and $M$ are collinear. Let the line $XI$ intersect the side at $M_0$. Then by the Law of Sines for triangles $AIM_0$ and $CIM_0$ (Fig. 17):\n$$\n\\frac{\\sin \\angle AIM_0}{\\sin \\angle M_0 IC} = \\frac{AM_0}{M_0 C} \\cdot \\frac{M_0 I}{\\sin \\angle ICM_0} \\cdot \\frac{\\sin \\angle IAM_0}{M_0 I} = \\frac{AM_0}{M_0 C} \\cdot \\frac{\\sin \\angle IAM_0}{\\sin \\angle ICM_0},\n$$\nhence $\\frac{\\sin \\angle AIM_0}{\\sin \\angle CIM_0} = \\frac{\\sin \\angle KIX}{\\sin \\angle NIX}$. By the Law of Sines for triangles $XIK$ and $XIN$ we obtain that $\\frac{\\sin \\angle KIX}{\\sin \\angle NIX} = \\frac{XK}{XN} \\cdot \\frac{NI}{KI} \\cdot \\frac{\\sin \\angle KXI}{\\sin \\angle NXI}$.\n\nNote that $\\triangle XNA \\sim \\triangle XKC$. Indeed, $\\angle XAB = \\angle XCB$ and $\\angle AXN = \\angle CXK$, since $\\angle AXC = \\angle ABC = \\angle NBK = \\angle NXX$. Thus, $\\frac{KX}{NX} = \\frac{KC}{NA}$. Hence\n$$\n\\frac{\\sin \\angle AIM_0}{\\sin \\angle CIM_0} = \\frac{\\sin \\angle KIX}{\\sin \\angle NIX} = \\frac{XK}{XN} \\cdot \\frac{NI}{KI} \\cdot \\frac{\\sin \\angle KXI}{\\sin \\angle NXI} = \\frac{KC}{NA} \\cdot \\frac{NI}{KI} \\cdot \\frac{\\sin \\angle KXI}{\\sin \\angle NXI}.\n$$\nPlugging it into the equality above we obtain that $\\frac{KC}{NA} \\cdot \\frac{NI}{KI} \\cdot \\frac{\\sin \\angle KXI}{\\sin \\angle NXI} = \\frac{AM_0}{CM_0} \\cdot \\frac{\\sin \\angle IAM_0}{\\sin \\angle ICM_0}$, which is equivalent to the following:\n$$\n\\frac{AM_0}{CM_0} = \\frac{KC}{NA} \\cdot \\frac{NI}{KI} \\cdot \\frac{\\sin \\angle KXI}{\\sin \\angle NXI} \\cdot \\frac{\\sin \\angle ICA}{\\sin \\angle IAC} =\n$$\n\n$= \\frac{\\sin \\angle KXI}{\\sin \\angle NXI} \\cdot \\left( \\frac{\\sin \\angle ICB}{KI} \\cdot \\frac{KC}{\\sin \\angle KIC} \\right) \\cdot \\left( \\frac{\\sin \\angle KIC}{NA} \\cdot \\frac{NI}{\\sin \\angle IAB} \\right)$.\n$$\nBut from the Law of Sines for the triangles $ANI$ and $KIC$ we obtain the following:\n$$\n\\frac{\\sin \\angle ICB}{KI} \\cdot \\frac{KC}{\\sin \\angle KIC} = 1, \\quad \\frac{\\sin \\angle KIC}{NA} \\cdot \\frac{NI}{\\sin \\angle IAB} = 1 \\Rightarrow \\frac{AM_0}{CM_0} = \\frac{\\sin \\angle KXI}{\\sin \\angle NXI}.\n$$\n\nThus, the condition that the points $X$, $I$ and $M$ are collinear is equivalent to the equality of acute angles $\\angle KXI = \\angle NXI$.\nLet $Y$ and $Z$ be the points of intersection of $XI$ and $BI$ respectively with the circumcircle of $\\triangle BNK$. Then, since $BI$ is a bisector, $Z$ is a midpoint of the arc $NK$, thus, it is necessary for an equality $\\angle KXI = \\angle NXI$ that $Y \\equiv Z$, but then either $X \\equiv B$, which contradicts non-isoscelesness of $\\triangle ABC$, or $I$ belongs to the circumcircle of $\\triangle KBN$.\n![](attached_image_2.png)\nFig. 17\nThus, the condition that points $X$, $I$ and $M$ are collinear is equivalent to the points $B$, $N$, $K$ and $I$ lying on one circle, which is equivalent to $\\angle ABC = 60^\\circ \\Leftrightarrow 180^\\circ - \\angle ABC = 90^\\circ + \\frac{1}{2} \\angle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$, where $O$ is its circumcenter, $M$ is a midpoint of $BC$, and $W$ is a point of the second intersection of bisector of $C$ with the circumcircle. The line parallel to $BC$ that passes through $W$ intersects $AB$ at point $K$, so that $BK = BO$. Find the angle $WMB$.",
    "options": [],
    "answer": "45°",
    "solution": "Let the line that passes through $W$ parallel to $AB$ intersect the line $BC$ at point $T$ (Fig. 6). Then, $KWTB$ is a parallelogram and:\n$$WT = BK = BO = WO.$$\nNotice that $WO \\perp AB$, since $\\triangle ABO$ is isosceles, and $CW$ is a bisector of $\\angle BCA$, thus $\\angle OWT = 90^\\circ$. It is also clear that $\\angle OMT = 90^\\circ$, thus, $OWTM$ is inscribed with diameter $OT$. Therefore, since $WT = WO$, $MW$ is a bisector of $\\angle OMT$, hence $\\angle WMT = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma$ be a half circle with diameter $AB$. Let $C$ be the point on the diameter, and let $D$ and $E$ be the points on the half circle so that $E$ lies between $B$ and $D$. It turns out that $\\angle ACD = \\angle ECB$. Let $F$ be the point of intersection of the lines tangent to $\\Gamma$ at the points $D$ and $E$. Show that $\\angle EFD = \\angle ACD + \\angle ECB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFig. 18\n\nLet $O$ be the center of the half circle $\\Gamma$, then by the properties of the tangent lines $\\angle OEF = 90^\\circ = \\angle FDO$ (Fig. 18). Thus, $O, E, F, D$ are cyclic.\n\nLet $\\angle ACD = \\angle ECB = \\theta$. By the Law of Sines for $\\triangle COD$ we obtain that\n$$\n\\frac{\\sin \\angle ODC}{OD} = \\frac{\\sin \\angle DCO}{OD} = \\frac{\\sin(180^\\circ - \\theta)}{OD}.\n$$\nSimilarly, for $\\triangle COE$ : $\\frac{\\sin \\angle OEC}{CO} = \\frac{\\sin \\angle ECO}{OE} = \\frac{\\sin \\theta}{OE}$.\nSince $OD = OE$ as radii of the half circle,\n$\\sin \\angle ODC = \\sin \\angle OEC$. Since\n$\\angle ODC < \\angle BDA = 90^\\circ$, then the angle is acute, as well\nas $\\angle OEC$. Thus, the angles are equal and five points\n$O, E, F, D, C$ are cyclic. Thus, by the property of the\ninscribed quadrilateral\n$$\n\\angle EFD = 180^\\circ - \\angle DCE = \\angle ACD + \\angle ECB.\n$$\n**Alternative solution.** Let $E'$ be the point symmetric to $E$ with respect to $AB$, then the point $E'$ lies on a circumcircle of $\\triangle AEB$ and $\\angle ACD = \\angle BCE = \\angle BCE'$ (Fig. 19), hence\n![](attached_image_2.png)\nFig. 19\n$$\n\\begin{align*}\n\\angle ACD &= \\angle BCE = \\frac{1}{2}(\\cup AD + \\cup BE) = \\\\\n\\phantom{\\angle ACD} &= \\frac{1}{2}(\\cup AD + \\cup BE) = \\frac{1}{2}(180^\\circ - \\cup DE).\n\\end{align*}\n$$\nOn the other hand, $DF$ and $FE$ are tangent to the circle $\\Gamma$, thus,\n$$\n\\begin{align*}\n\\angle FED = \\angle FDE &= \\frac{1}{2} \\cup DE \\Rightarrow \\angle ACD + \\angle BCE = 180^\\circ - \\cup DE = \\\\\n&= 180^\\circ - \\angle FED - \\angle FDE = \\angle DFE, \\text{ Q.E.D.}\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be an inscribed hexagon, such that $AB = BC$, $CD = DE$ and $EF = FA$. Show that the lines $AD$, $BE$ and $CF$ intersect at exactly one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Equality $AB = BC$ implies that $\\angle AEB = \\angle BEC$, hence $EB$ is a bisector of $\\triangle ACE$, similarly, one can conclude the same about $AD$ and $CF$ (Fig. 7). Therefore, they intersect at one point.\n\n![](attached_image_1.png)\nFig. 7",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integers from $1$ to $100$ inclusive are written on the blackboard. Andrew wants to cross out some numbers in such a way, that the product of the remaining numbers is not divisible by $250$. What is the smallest number of numbers that he can cross?",
    "options": [],
    "answer": "18",
    "solution": "Since $250 = 2 \\cdot 5^3$, Andrew has to cross out from the product $1 \\cdot 2 \\cdots 100$ all numbers which are divisible by $5$, except for two numbers that are not divisible by $25$ (for example, we can leave $5$ and $10$). The resulting product satisfies the condition because it is not divisible by $5^3 = 125$. Meanwhile, Andrew crossed out $18$ numbers, because among the factors $1, 2, \\ldots, 100$ exactly $20$ are divisible by $5$.\n\nLet us assume that it is possible to cross out no more than $17$ numbers. Thus, among $20$ numbers that are divisible by $5$, Andrew left at least three, then the product is divisible by $5^3$. In order for the product not to be divisible by $250$, all factors must be odd. However, since $83$ numbers are left and only $50$ initial numbers were odd, at least one of the factors is even. Contradiction completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eight children, who did not know each other, came to the dance class. To introduce them to each other, teacher decided to choose four children each minute and let them dance in a circle. During this minute, each of the children in a circle will get to know a neighbor on the left and on the right. What is the minimal number of minutes the teacher needs to make all eight children know each other?\n(Arsenii Nikolaiev)",
    "options": [],
    "answer": "8",
    "solution": "Consider any child from this group. They need to get to know seven other children, and participating once in a dancing circle they meet maximum two new acquaintances. Thus, each child should participate in at least four dancing circles. In each dancing circle, there are no more than four children participating. Then, there were at least $(8 \\cdot 4) = 8$ circles. Now we are going to give an example in which eight circles are enough for everybody to know each other. We enumerate children from 1 to 8 and make the next ordered groups: 1234, 5678, 1357, 2468, 1526, 3748, 2736 and 1845. Here we assume that the last child stands in a circle near the first one.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45740,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) For non-zero integers $a, b, c$ it holds that: $\\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{a} = \\frac{a^2}{c} + \\frac{c^2}{b} + \\frac{b^2}{a}$.\nDoes it follow then that: $\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} = \\frac{a}{c} + \\frac{c}{b} + \\frac{b}{a}$?\n\n6) Consider all triples of non-zero integers $(a, b, c)$, which belong to the interval $[-2020, 2020]$. Find for how many of them the conditions $a+b+c=0$ and $\\frac{a^2}{b}+\\frac{b^2}{c}+\\frac{c^2}{a}=\\frac{a^2}{c}+\\frac{c^2}{b}+\\frac{b^2}{a}$ are equivalent? In other words, how many such triples are there, such that either both conditions hold, or none of the two holds?",
    "options": [],
    "answer": "a) No. 6) 65890313340",
    "solution": "Consider the following transformation:\n$$\n\\left( \\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{a} \\right) - \\left( \\frac{a^2}{c} + \\frac{c^2}{b} + \\frac{b^2}{a} \\right) = \\frac{a^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c}{abc}\n$$\nNow we transform only numerator:\n$$\n\\begin{aligned} a^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c &= (b-a)(ab^2 + a^2 b + c^2 - ca^2 - abc - cb^2) = \\\\ &= (b-a)(ab^2 + a^2 b + c^2 - ca^2 - abc - cb^2) = (b-a)(a-c)(b^2 + ab - c^2 - ca) = \\\\ &= (b-a)(a-c)(b-c)(a+b+c). \\end{aligned}\n$$\nTherefore the condition $\\frac{a^2}{b}+\\frac{b^2}{c}+\\frac{c^2}{a}=\\frac{a^2}{c}+\\frac{c^2}{b}+\\frac{b^2}{a}$ for non-zero numbers $a, b, c$ is equivalent to condition: $(b-a)(a-c)(b-c)(a+b+c)=0$.\n\na) Similarly we conduct following transformation:\n$$\n\\left(\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}\\right) - \\left(\\frac{a}{c} + \\frac{c}{b} + \\frac{b}{a}\\right) = \\frac{a^2 c - a^2 b + b^2 a - a^2 b + c^2 b - b^2 c}{abc} = \\\\ \\frac{(b-a)(c-a)(c-b)}{abc} = 0.\n$$\nNow, as a counter example let's choose a triple $a=1, b=2, c=-3$. It satisfies the first equation because $a+b+c=0$. However, the second equation is not correct:\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} = \\frac{1}{2} - \\frac{2}{3} + \\frac{3}{1} = \\frac{3-4+18}{6} = \\frac{17}{6}, \\quad \\frac{a}{c} + \\frac{c}{b} + \\frac{b}{a} = -\\frac{1}{3} + \\frac{3}{2} + \\frac{2}{1} = \\frac{-2+9+12}{6} = \\frac{19}{6}.\n$$\n\n6) From transformations above we conclude that the required triples can be two types\nType 1. Integers $a, b, c$ are pairwise distinct.\nType 2. If at least two integers are equal, for example $a=b \\neq c$, then second equation is correct, while second is correct only if $c = -(a+b)$.\nNow we count the number of such triples. The amount of pairwise different non-zero triples is $4040 \\cdot 4039 \\cdot 4038$. For the Type 2, if $a=b \\neq c$, then there are $2020$ numbers to choose $a$ and then $b$ and $c$ are defined. Therefore, there are $3 \\cdot 2020$ such triples.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45741,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ $\\angle A=75^\\circ$ and $\\angle C=45^\\circ$. Points $P$ and $T$ are chosen on the segments $AB$ and $BC$ in such a way that quadrilateral $APTC$ is cyclic and $CT = 2AP$. Point $O$ is the circumcenter of $\\triangle ABC$. The ray $TO$ crosses side $AC$ in a point $K$. Prove that $TO=OK$.\n\n(Anton Trygub)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $CD$ be a diameter of $(ABC)$. Then $\\triangle ADC$ is a right triangle with an angle $60^\\circ$. Hence, $CD = 2AD$ (fig. 30) and $\\triangle DTC \\sim \\triangle DPA$ by two proportional sides and equal included angles. Therefore, $\\angle BPD = \\angle BTD$ and $PDBT$ is cyclic quadrilateral. We point out that $\\angle BTD = \\angle BPD = \\angle DPT - \\angle BPT = 180^\\circ - \\angle DBT - 45^\\circ = 180^\\circ - 90^\\circ - 45^\\circ = 45^\\circ$, from which $DT \\parallel AC$. Since $DO = OC$ and $DT \\parallel CK$, $DTCK$ is parallelogram and $TO = OK$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $2n$ points on the plane, prove that it is possible to split them into $n$ pairs such that: if for each pair of points we construct a circle as on diameter, then the obtained $n$ circles will have a common point which does not necessarily belong to given ones.\n(Anton Trygub)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a line $l$ such that there are $n$ points on one side of the line and $n$ points on the other. Clearly such a line exists. Without loss of generality, assume $l$ to be the horizontal axis. Then arrange points in increasing order of abscissa. In the upper half-plane we have: $A_1, A_2, \\ldots, A_n$ and $B_1, B_2, \\ldots, B_n$ are in the lower half-plane. Then we group them in pairs $(A_1, B_n), (A_2, B_{n-1}), \\ldots, (A_n, B_1)$. We will show that circles constructed on those pairs of points as on diameters have a common point that belongs to $l$.\n\nLet us show that projections of two arbitrary line segments onto line $l$ do intersect. Now, we will show that projections onto line $l$ of any two segments with endpoints in pairs listed above intersect. Indeed, when $i < j$, projections of line segments $A_i B_{n+1-i}$ and $A_j B_{n+1-j}$ are such that $A_i$ is not projected to the right of $A_j$ and $B_i$ is not projected to the left of $B_j$. Now, among all line segments $A_i B_n, \\ldots, A_n B_1$, we choose the last one that has a positive slope. Let it be $A_i B_{n+1-i}$, hence the desired segment is an overlap of projections of $A_i B_{n+1-i}$ and $A_{i+1} B_n$. Obviously, if a point belongs to the projection of a segment onto a line, then it is contained inside a disk that is constructed on segment's endpoints as on diameter (Fig. 36). This completes the proof.\n\n![](attached_image_1.png)\nFig. 36",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Say a pair of natural numbers $(a, b)$ is *interesting* if there exists natural number $n$, such that minimal prime divisor of $a + n$ equals maximal prime divisor of $b + n$. Find all pairs of natural numbers that are interesting.",
    "options": [],
    "answer": "All pairs (a, b) with |a − b| ≠ 1",
    "solution": "Let us now assume that $|a-b| \\neq 1$. We will show that the pair $(a, b)$ is interesting. Let's consider the smallest prime number $p$ dividing $|a-b|$. Denote by $p_1 < p_2 < \\ldots < p_s$ all prime numbers less than $p$. It is clear that they may not exist. From the choice of $p$ it follows that none of $p_1, \\ldots, p_s$ divides $|a-b|$. Let's pick a large enough natural number $k$, such that $p^k > b$ and set $n = p^k p_1 \\ldots p_s - b$. Clearly, greatest common divisor of $b+n$ is $p$. Now let us consider a number $n+a = p^k p_1 \\ldots p_s + (a-b)$. It follows that $p^k p_1 \\ldots p_s + (a-b)$ is divisible by $p$, and is not divisible by any of $p_1, \\ldots, p_s$, which implies that the smallest prime divisor of $a+n$ is $p$, which is the same as greatest prime divisor of $b+n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45744,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the size of the largest set of numbers we can choose among $1$, $2$, ..., $2n$ in such a way that any two of them have a common divisor greater than $1$?",
    "options": [],
    "answer": "n",
    "solution": "If we choose all even numbers, they follow required condition and there are exactly $n$ of them.\n\nNow let's consider next pairs of integers: $(1, 2)$, $(3, 4)$, ..., $(2n-1, 2n)$. In each pair, numbers are relatively prime and we cannot choose both of them to our set. Therefore, it is impossible to choose more than $n$ integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Equilateral triangle with the side length $n$ is divided into $n^2$ smaller equilateral triangles with the side length $1$ (Fig. 31 shows this division for $n=10$). Initially, one of these triangles is blue and the rest are yellow. The blue triangle is guaranteed not to have any common points with the boundary of the big triangle. It is allowed to consider any of $n^2$ equilateral triangles and change its color together with the colors of its neighbors (blue changes to yellow, yellow changes to blue). Is there any $n$ for which it is possible to make all $n^2$ equilateral triangles of the same color?\n![](attached_image_1.png)\n(Arsenii Nikolaiev)",
    "options": [],
    "answer": "No",
    "solution": "![](attached_image_2.png)\nLet's consider one step and assume that some triangle $a$ and his neighbors changed color on this step. Pairs that include $a$ do not influence $X$, because if they had the same color - they will stay the same, and if they had different colors, after the recoloring they will also have different colors. Now, we point out that each of neighboring triangles\nto *a* is always adjacent with two more small triangles. If both of them had the same color then after repainting they will increase/decrease *X* by $2$. If they had the same color, then they do not change *X*. If it is possible to make all triangles unicolor, then initial $X = 3$ will become equal to $0$, which contradicts the invariant.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any integer $n > 1$ there exists such a sequence of positive integers $a_1 \\le a_2 \\le \\dots \\le a_m = n$, $m > 1$ such that\n$$\n5(a_1^2 + a_2^2 + \\dots + a_m^2) - 4(a_1a_2 + a_2a_3 + \\dots + a_{m-1}a_m) \\le 4n^2 + \\frac{1}{2}(m+1).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Rewrite the given inequality in the following way:\n$$\n(a_m - 2a_{m-1})^2 + \\dots + (a_2 - 2a_1)^2 + a_1^2 \\le \\frac{1}{2}(m+1)\n$$\nBuild a sequence $(b_n)$ such that: $b_1 = n$, if $b_i$ is even, then set $b_{i+1}$ to be $\\frac{1}{2}b_i$, otherwise define it as one of the numbers $\\frac{1}{2}(b_i + 1)$, $\\frac{1}{2}(b_i - 1)$ which is even.\nBy the algorithm exists $k$, for which $b_k = 1$. Consider the smallest such index for which that occurred. Set $a_{k-i+1} = b_i$, $i = 1, \\overline{k}$. Hence, among $(a_{i+1} - 2a_i)^2$ will be either zero or ones. In addition, according to the construction no consecutive ones will appear because we choose an even value among two possible which suffices the next number to be zero. Overall, we sum over $k$ elements, therefore,\n$$\n(a_k - 2a_{k-1})^2 + \\dots + (a_2 - 2a_1)^2 + a_1^2 \\le \\frac{1}{2}(k+1),\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45747,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, $H$ is a midpoint of the altitude $AD$ and $O$ is a centre of the circumscribed circle. A line perpendicular to the line $HO$ passing through $H$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the midpoints of $BP$, $CQ$ and $O$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define $M_C$ and $M_B$ as midpoints of $AB$ and $AC$ respectively. Then $H$ lies on the line $M_B M_C$ due to its initial position. Note that projections of $O$ on the lines $AB$, $AC$ and $PQ$ are points $M_C$, $M_B$ and $H$ which lie on the Simson line (Fig 33). From this can be concluded that $O$ lies on the circumscribed circle of the $\\triangle PAQ$.\n\nIntersect a ray $HO$ with the line $BC$ in a point $S$. Observe that $\\triangle OM_B M_C \\sim \\triangle HBC$ as they have the same angles. Also, $\\angle OHM_B = \\angle HSB$, so $S$ and $H$ are respective vertices of the similar triangles. Since $BC = 2M_B M_C$, we have that $SC = 2M_C H = BD$. Hence, points $S$ and $D$ are symmetric over $BC$.\n\nName midpoints of $BP$ and $CQ$ as $P_1$ and $Q_1$ respectively. Define $C_1$ as a symmetric point of $C$ over $D$. Therefore, can deduce that\n$$\n\\angle OQC = \\angle OHM_B = \\angle OSC_1 = \\angle HSC_1 \\text{ and } \\angle OCA = 90^\\circ - \\angle B = \\angle HCB = \\angle HC_1S,\n$$\nwhich means that $\\triangle HSC_1 \\sim \\triangle OQC$. Let $X$ be a midpoint of the $SC_1$. From $C_1D = DC = BS$, we have that the midpoints of the $SC_1$ and $BD$ coincide, so $X$ is also the midpoint of the $BD$. From the previously described similarity $\\triangle HSC_1 \\sim \\triangle OQC$ we get that $\\angle OQ_1A = \\angle HXS$ (an angle between a median and a side) and $\\angle OQ_1A = \\angle HXS = \\angle B$ (midline). Similarly,\n\n$\\angle OP_1A = \\angle C$. From the angle sum $\\angle A + \\angle B + \\angle C = 180^\\circ$ we conclude that $P_1, O$ and $Q_1$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for natural numbers $a \\ge b \\ge c \\ge d$ the inequality:\n$$\nab+bc+cd-b^{2}-c^{2}-d^{2} \\ge a-d.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have:\n$$\nab+bc+cd-b^{2}-c^{2}-d^{2} = b(a-b)+c(b-c)-d(c-d) \\ge \\\\\n\\ge (a-b)+(b-c)+(c-d)=a-d.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral circumscribed around a circle with center $O$. We construct equal segments $AA_1$, $AA_2$, $CC_1$ and $CC_2$ on rays $AB$, $AD$, $CB$ and $CD$ respectively, so that their lengths are greater than that of any side of $ABCD$, and $A_1$, $A_2$, $C_1$ and $C_2$ are located on the circle with center $O$. Prove that lines $A_1A_2$, $C_1C_2$ and $BD$ either intersect at one point or are parallel.\n\n(Olena Artemchuk, Mykola Moroz)\n\n![](attached_image_1.png)\nFig. 38",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ and $P$ be tangent points of an inscribed circle of $ABCD$ to $AB$ and $BC$ respectively (Fig. 38). It is that $\\triangle C_1PO = \\triangle A_1KO$, since they are right triangles with equal legs $KO = PO$ and hypotenuses $C_1O = A_1O$ as radii of respectful circles. Thus we have $C_1P = A_1K$. Since $BK = BP$, as two tangent segments that are drawn from the same external point, then $C_1B = A_1B$. Similarly, $C_2D = A_2D$. Moreover, from the discussion above we find that $AB = BC$ and $AD = CD$. Therefore, $ABCD$ is a kite, whose diagonal $BD$ is a line of symmetry, hence $B$, $D$ and $O$ lie on the same line.\n\nLine $BD$ passes through the points $B$ and $O$. Those points are equidistant from the endpoints of segment $A_1C_1$, hence $B$ and $O$ belong to its perpendicular bisector. Similarly, line $BD$ is a perpendicular bisector of $A_2C_2$. Since $BD \\perp A_1C_1$ and $BD \\perp A_2C_2$, then $A_1C_1 \\parallel A_2C_2$.\n\nNext, we consider two cases.\n\nIf $A_1A_2 \\parallel C_1C_2$, then the quadrilateral $A_1A_2C_2C_1$ is a parallelogram, then $A_1A_2C_2C_1$ is a rectangle. All three lines $A_1A_2$, $C_1C_2$ and $BD$ are perpendicular to $A_1C_1$, hence they are parallel.\n\nIf $A_1A_2$ and $C_1C_2$ are not parallel, then the quadrilateral $A_1A_2C_2C_1$ is a trapezoid. Since a line $BD$ passes through midpoints of trapezoid bases, then it contains a point of intersection of extensions of legs $A_1A_2$ and $C_1C_2$. Hence lines $A_1A_2$, $C_1C_2$ and $BD$ intersect at one point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45750,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are four countries, each consists of several cities. The cities of any two countries are connected by at least $\\frac{5}{6}$ of a number of all possible roads between these two countries. Prove that it is possible to choose one city from each country so that any two of them are connected. Any pair of cities can be connected by no more than one road.\n(Nazar Serdyuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote countries as $X, Y, Z, T$. We randomly choose one city $x, y, z, t$ from each country respectively. In case there are all 6 roads between them, then the statement has been proved.\n\nIn case there is quadruple that is connected via less than 5 roads, then there has to exist a quadruple that is connected via all 6 roads, otherwise, the number of possible roads is less than $\\frac{5}{6}$. Therefore, in order to comply with the problem statement, each four cities have to be connected with exactly 5 roads. Same every two countries have to be connected via exactly $\\frac{5}{6}$ from the total number of possible roads between those countries, hence there are unconnected cities between countries.\n\nChoose cities $x \\in X$, $y \\in Y$ that are not connected. Consider arbitrary $z \\in Z$ and $t \\in T$. Since quadruple $x, y, z, t$ is connected with exactly 5 roads, then $z, t$ have to be connected. Since we chose arbitrary cities $z \\in Z$ and $t \\in T$, then every two cities of $Z$ and $T$ are connected, which yields a contradiction with connection between countries.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45751,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which values of $n > 1$ is there a set of pairwise distinct natural numbers $(a_1, a_2, ..., a_n)$ such that the following condition is satisfied:\n$$\n(a_1+1)!+(a_2+1)!+...+(a_n+1)! : a_1!+a_2!+...+a_n!\n$$",
    "options": [],
    "answer": "no values of n greater than 1",
    "solution": "$a_1, a_2, ..., a_n$ Without the loss of generality, let us assume that $a_1 < a_2 < ... < a_n$. Then $(a_i + 1)! \\le a_i!(a_n + 1)$, and equality is reached only when $i = n$.\n\nLet $(a_1 + 1)! + (a_2 + 1)! + ... + (a_n + 1)! = N(a_1! + a_2! + ... + a_n!)$ for some natural number $N$. Then the condition above yields $N \\le a_n + 1$, moreover, if $n > 1$, then $N < a_n + 1$.\n\nLet now $n > 1$, then we have that $N \\le a_n$. Let us prove that it is impossible by showing that:\n$$\n(a_1+1)!+(a_2+1)!+...+(a_n+1)! > a_n(a_1!+a_2!+...+a_n!)\n$$\n\nLet's consider the following expression:\n$$\n(a_1+1)!+(a_2+1)!+...+(a_n+1)! - a_n(a_1!+a_2!+...+a_n!) = \\sum_{i=1}^{n} a_i!(a_i+1-a_n).\n$$\nThe last term is equal to $a_n!$. If $a_{n-1} = a_n - 1$, then the penultimate term is zero. It is enough to consider the following:\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} a_i!(a_i+1-a_n) &\\le \\sum_{i=1}^{a_n-1} a_i!(a_n-1-i) = \\sum_{i=1}^{a_n-2} a_i!(a_n-1-i) \\\n&\\le \\sum_{i=1}^{a_n-2} i!a_n = a_n \\sum_{i=1}^{a_n-2} i! \\le \\\\\na_n(a_n-2)(a_n-2)! < a_n! \\Rightarrow \\\\\n(a_1+1)!+(a_2+1)!+...+(a_n+1)! - a_n(a_1!+a_2!+...+a_n!) > 0,\n\\end{aligned}\n$$\nwhich yields the desired result. Hence, there are no such sets for $n > 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do positive real numbers $x, y, z$ have to be equal, if they satisfy\n$$\n\\frac{xy + 1}{x + 1} = \\frac{yz + 1}{y + 1} = \\frac{zx + 1}{z + 1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the statement, we have $(xy + 1)(y + 1) = (yz + 1)(x + 1)$, so $xy^2 + xy + y = xyz + x + yz$. Similarly $yz^2 + yz + z = xyz + y + zx$, and also $zx^2 + zx + x = xyz + z + xy$. After adding these three equations, we get $xy^2 + yz^2 + zx^2 = 3xyz$, or $\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y} = 3$. But for positive real numbers $x, y, z$, by inequality between the arithmetic mean and the geometric mean, we get $\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y} \\ge 3$, and the equality is achieved only when $\\frac{x}{y} = \\frac{y}{z} = \\frac{z}{x} = 1$, implying $x = y = z$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of positive integers $a, b, c$, such that\n$$\na + [a, b] = b + [b, c] = c + [c, a],\n$$\nwhere by $[x, y]$ we denote the smallest common multiple of $x, y$.",
    "options": [],
    "answer": "(n, n, n) for any positive integer n",
    "solution": "Note that $b$, $[a, b]$ and $[b, c]$ are divisible by $b$, so $a$ is divisible by $b$. Similarly, $c$ is divisible by $a$, and $b$ is divisible by $c$, so $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$ with $\\angle A = 120^\\circ$. Let $P$ and $Q$ denote the projections of $B$ onto $CO$ and $AO$, respectively. Let $M$ be the midpoint of $AO$. Prove that the circumcircle of $\\triangle MPQ$ touches $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be the projection of $B$ onto $AC$ (fig. 23). Then $P, Q, R$ are the projections of $B$ onto the sides of $\\triangle AOC$. Let $B_1$ be the isogonal conjugate of $B$ with respect to this triangle. Since $\\angle BAC = \\angle BOC = 120^\\circ$, we have $\\angle B_1AO = \\angle B_1OA$. Then the projection of $B_1$ onto $AC$ is $M$. Then the points $P, Q, R$, and $M$ are concyclic. Thus, it suffices to prove $\\angle QRA = \\angle QPR$. We will use the cyclicity of $ARBQ, OQBP,$ and $CPBR$. Indeed,\n$$\n\\angle RPQ = \\angle RPC - \\angle QPC = \\angle RBC - \\angle QBO = (90^\\circ - \\angle BCA) - (90^\\circ - \\angle BOA) = \\angle BCA,\n$$\n$$\n\\angle QRA = \\angle ABQ = 90^\\circ - \\angle BAO = \\angle BCA,\n$$\ncompleting the proof.\n\n![](attached_image_1.png)\nFig. 23",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of positive integers $a$, $b$, $c$, such that\n$$\na + (a, b) = b + (b, c) = c + (c, a),\n$$\nwhere by $(x, y)$ we denote the largest common divisor of integers $x$, $y$.",
    "options": [],
    "answer": "(n, n, n) for any positive integer n",
    "solution": "If numbers $a$, $b$, $c$ have a common divisor, we can divide by it and get a triple of integers $a_1$, $b_1$, $c_1$, whose largest common divisor is $1$. As $a_1 + (a_1, b_1) = b_1 + (b_1, c_1)$, $(b_1, c_1)$ is divisible by $(a_1, b_1)$. But $((a_1, b_1), (b_1, c_1)) = 1$, as explained above, so $(a_1, b_1) = 1$. Similarly we get that our numbers are pairwise coprime, so $a_1 + 1 = b_1 + 1 = c_1 + 1$, implying $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible for some positive integers $a$ and $d$ to satisfy:\n$$\na) [a, a + d] = [a, a + 2d];\n$$\n$$\nb) [a, a + d] = [a, a + 4d];\n$$\nwhere by $[x, y]$ we denote the least common multiple of integer $x, y$?",
    "options": [],
    "answer": "a) No, impossible. b) Yes; for example a = 4 and d = 2.",
    "solution": "a) As $a + 2d > a$, there exists some power of a prime $p^k$, that $a + 2d$ is divisible by $p^k$ and $a$ isn't divisible by $p^k$. From the given equality it follows that $a + d$ must be divisible by $p^k$. But then $2(a + d) - (a + 2d) = a$ is divisible by $p^k$, contradicting the choice of $p^k$. This contradiction completes the proof.\n\nb) It's enough to provide an example: $a = 4, d = 2$, then $a + d = 6$ and $a + 4d = 12$. Checking:\n$$\n[a, a + d] = [4, 6] = 12 = [4, 12] = [a, a + 4d].\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45757,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the beginning of work, the memory of computer contained a single polynomial $x^2 - 1$. Each minute, the computer can do one of two things: choose any polynomial $f(x)$ from its memory and memorize also polynomials $f^2(x) - 1$ and $f(x^2 - 1)$, or chooses any two different polynomials $g(x)$ and $h(x)$ from its memory and memorizes the polynomial $\\frac{1}{2}(g(x) + h(x))$. Is it possible that after some time the memory of computer contains polynomial $P(x) = \\frac{1}{1024}(x^2 - 1)^{2048} - 1$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45758,
    "subject": "Mathematics (Multi-modal)",
    "question": "$AH_a$, $BH_b$ and $CH_c$ are the altitudes of the triangle $ABC$. Prove that if $\\frac{H_b C}{AC} = \\frac{H_c A}{AB}$, then the line, symmetric to the line $BC$ with respect to the line $H_b H_c$, is tangent to the circumscribed circle of $\\triangle H_b H_c A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45759,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2022$ points on the circle, one of which is painted black, and the other $2021$ — white. In one move Hedgehog is allowed to do one of the following operations:\n* repaint in the opposite color two consecutive points of the same color;\n* repaint in the opposite colors two points of different colors, between which there is exactly one other point.\nWill Hedgehog be able to do such operations so that each point changes its color to the opposite (compared to the initial coloring)?",
    "options": [],
    "answer": "No",
    "solution": "Let's enumerate the points in clockwise order (starting, for example, from the black one), by $1, 2, \\ldots, 2022$. Note that from the very beginning, there was one more black point in the positions with odd numbers than the black points in the positions with even numbers. After the first operation, we either add one black point to each group or subtract one black point from each group, so the difference between even and odd black points remains constant. After the second operation, we do not change the number of black points in both groups, so their difference remains constant. Thus, the number of black points in odd places will always be one more than in even places. But this contradicts the fact that in the desired configuration, $1010$ points stand in even places and $1009$ points stand in odd places.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Monica and Bogdan are playing a game that depends on two positive integers $n$ and $k$. First, Monica chooses and writes $k$ positive numbers. Bogdan wins if he manages to mark $n$ points on the plane so that for each number $m$ written by Monica, there are two marked points at a distance precisely $m$, otherwise Monica wins. Find out, depending on $n$ and $k$, who wins, if both players play optimally.\n(Fedir Yudin)",
    "options": [],
    "answer": "Bogdan wins if k < n; Monica wins if k >= n.",
    "solution": "If $k < n$, Bogdan can choose $n$ points on a line so that the distance between the first and second points is equal to the first written number, between the second and third to the second written number, and so on.\n\nIf $k \\ge n$, Monica can choose numbers $2^0, 2^1, \\dots, 2^{k-1}$. Suppose that Bogdan can win. Then for each $i = 0, k-1$ connect with a segment two points at a distance $2^i$. As $k \\ge n$, this graph contains a cycle. As the length of the longest segment in this cycle is larger than the sum of the lengths of other segments, we get a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45761,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a central train station at point $O$, which is connected to other train stations $A_1, A_2, \\dots, A_8$ with tracks. There is also a track between stations $A_i$ and $A_{i+1}$ for each $1 \\le i \\le 8$ (here $A_1 = A_9$). The length of each track $A_iA_{i+1}$ is equal to $1$, and the length of each track $OA_i$ is equal to $2$, for each $1 \\le i \\le 8$.\n\nThere are also $8$ trains $B_1, B_2, \\dots, B_8$, the speed of the train $B_j$ is $j$. Trains can move only by the tracks above, in both directions. No time is wasted on changing directions. If two or more trains meet at some point, they will move together from now on, with the speed equal to that of the fastest of them.\n\nIs it possible to arrange trains into stations $A_1, A_2, \\dots, A_8$ (each station has to contain one train initially), and to organize their movement in such a way, that all trains arrive at $O$ in time $t < \\frac{1}{2}$?",
    "options": [],
    "answer": "yes",
    "solution": "**Answer:** yes.\n\nArrange the trains clockwise as following: $B_8, B_1, B_7, B_2, B_6, B_3, B_5$ and $B_4$. Now let's show how they should move to meet in time.\n\nLet $B_8$ move through the station of $B_1$ (at this time $B_1$ stays at place and is waiting for $B_8$), and they arrive to $O$ in time $t_8 = \\frac{3}{8}$.\n\nLet train $B_7$ move through the station of $B_2$ (at this time $B_2$ stays at place and is waiting for $B_7$), and they arrive to $O$ in time $t_7 = \\frac{3}{7}$.\n\nLet train $B_6$ move through the station of $B_3$ (at this time $B_3$ stays at place and is waiting for $B_6$) and they arrive to $O$ in time $t_7 = \\frac{1}{2}$.\n\n$B_5$ goes to $O$ right away in time $t_5 = \\frac{2}{5}$.\n\n$B_4$ goes to $O$ right away in time $t_4 = \\frac{1}{2}$.\n\nNow we see that two groups — $B_6 + B_3$ and $B_4$ — arrive at the time $\\frac{1}{2}$, but the groups $B_8 + B_1$ and $B_7 + B_2$ arrive faster than in $\\frac{1}{2}$, and their speeds are larger than those of first groups. So, the group $B_8 + B_1$ upon its arrival to $O$ goes to the group $B_6 + B_3$, unites with it, and goes to the point $O$ with its speed. Similarly, group $B_7 + B_2$ will get $B_4$ to $O$ faster.\n\nIt's possible to find the time of arrival of each group to $O$. For example, for the last group this time is $\\frac{37}{77} < \\frac{1}{2}$. Indeed, group $B_7 + B_2$ arrives at $O$ at $\\frac{3}{7}$, at this time $B_4$ has travelled $\\frac{12}{7}$ in the direction of point $O$. The time before their meet is $\\frac{2}{4+7} = \\frac{2}{77}$, and it's equal to the time spent to arrive back to $O$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral with equal angles $B$ and $D$. Circles $\\omega_1, \\omega_2$ are symmetric with respect to $AC$; $\\omega_1$ passes through $B$ and intersects $AB$ and $BC$ for the second time at $K$ and $L$, respectively, and $\\omega_2$ passes through $D$ and intersects $CD$ and $AD$ for the second time at $M$ and $N$, respectively.\nProve that $KM$ and $LN$ intersect on $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D'$ be symmetric to $D$ with respect to $AC$. If $D'$ coincides with $B$, then $N$ and $M$ are symmetric to $K$ and $L$, respectively, with respect to $AC$, and the problem clearly holds. Let $D' \\neq B$. Since $\\omega_2$ is symmetric to $\\omega_1$ with respect to $AC$, If we denote by $M'$ and $N'$ the points symmetric to $M$ and $N$, respectively, with respect to $AC$ we will have that $B, D', M', L, N', K$ all lie on $\\omega_1$ (fig. 24). Since $\\angle AD'C = \\angle ADC = \\angle ABC$, points $A, B, D', C$ are concyclic. We get $\\angle (CA, AD') = \\angle (CB, BD') = \\angle (LN', N'D')$, so $LN' \\parallel AC$, and similarly $M'K \\parallel AC$. Then $LM'KN'$ is a cyclic trapezoid, so it's isosceles, and $KL = M'N' = MN$. We also get $\\angle (MN, AC) = \\angle (AC, M'N') = \\angle (N'L, M'N') = \\angle (KL, N'L) = \\angle (KL, AC)$, so $MN \\parallel KL$. Since segments $KL$ and $MN$ are equal, $KLMN$ is a parallelogram, so $KM$ intersects $NL$ at its midpoint, and since $N'L \\parallel AC$ and $N$ is symmetric to $N'$ with respect to $AC$, the midpoint of $NL$ lies on $AC$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nonzero real numbers $x_1, x_2, \\dots, x_n$ satisfy the following conditions:\n$$\nx_1 - \\frac{1}{x_2} = x_2 - \\frac{1}{x_3} = \\dots = x_{n-1} - \\frac{1}{x_n} = x_n - \\frac{1}{x_1}.\n$$\n\nFor which $n$ do the numbers $x_1, x_2, \\dots, x_n$ all have to be equal?",
    "options": [],
    "answer": "All odd positive integers n",
    "solution": "Suppose that $x_i = x_{i+1}$ for some $i$ (from now on we denote $x_{n+k} = x_k$.) Then from the equality $x_i - \\frac{1}{x_{i+1}} = x_{i+1} - \\frac{1}{x_{i+2}}$ we get $x_{i+1} = x_{i+2}$, then $x_{i+2} = x_{i+3}$ and so on. So we will get that all numbers are equal.\n\nSuppose now that we don't have equal adjacent numbers. From the statement, we get\n$$\nx_i - x_{i+1} = \\frac{1}{x_{i+1}} - \\frac{1}{x_{i+2}} = -\\frac{x_{i+1} - x_{i+2}}{x_{i+1}x_{i+2}}, \\quad i = \\underline{1, n}.\n$$\nIf we multiply all such equalities, the product of the differences will cancel out, and we will get $\\frac{1}{(x_1x_2\\dots x_n)^2} = (-1)^n$. For an odd $n$ it's impossible, and for even $n$ one possible array is $(2, \\frac{1}{2}, 2, \\frac{1}{2}, \\dots, 2, \\frac{1}{2})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45764,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given a triangle $ABC$, $\\omega$ is its circumscribed circle, $I$ is its incenter. Let $K$ be any point on the arc $AC$ of $\\omega$, not containing point $B$. Point $P$ is symmetrical to the point $I$ with respect to the point $K$. Point $T$ on the arc $AC$ of the circle $\\omega$, which contains point $B$, is such that $\\angle KCT = \\angle PCI$. Prove that the bisectors of angles $AKC$ and $ATC$ meet at the line $CI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all bijections $f: (0, +\\infty) \\to (0, +\\infty)$ such that, for any $x, y > 0$ the following is satisfied:\n\n$$f(xf(x) + yf(y)) = f^2(x) + f^2(y).$$\n(Oleksii Masalitin, Fedir Yudin)",
    "options": [],
    "answer": "All functions f(x) = c x with c > 0.",
    "solution": "Denote the statement by $P$, and let $P(x, y)$ denote the substitutions into it.\nLet $a$ be the real number such that $f(a) = 1$. Let $S$ denote the set of $x > 0$ such that $f(ax) = x$. Then $1 \\in S$. We prove several lemmas:\n\n**Lemma 1.** If $x \\in S$, then $2x^2 \\in S$.\n*Proof.* Use $P(ax, ax)$: $f(a2x^2) = 2x^2 \\Rightarrow 2x^2 \\in S$.\n\n**Lemma 2.** If $x \\in S$, then $\\sqrt{\\frac{x}{2}} \\in S$.\n*Proof.* Denote by $x_0$ the real number that satisfies $f(x_0) = \\sqrt{\\frac{x}{2}}$. Use $P(x_0, x_0)$: $f(x_0\\sqrt{2}x) = x = f(ax)$, so $x_0\\sqrt{2}x = ax$. We get $x_0 = a\\sqrt{\\frac{x}{2}}$ and $f(a\\sqrt{\\frac{x}{2}}) = \\sqrt{\\frac{x}{2}}$, so $\\sqrt{\\frac{x}{2}} \\in S$.\n\n**Lemma 3.** If $x, y \\in S$, then $\\frac{1}{2}(x + y) \\in S$.\n*Proof.* From Lemma 2 we have $\\sqrt{\\frac{x}{2}}, \\sqrt{\\frac{y}{2}} \\in S$. Use $P(a\\sqrt{\\frac{x}{2}}, a\\sqrt{\\frac{y}{2}}) \\Rightarrow f(\\frac{1}{2}a(x + y)) = \\frac{1}{2}(x + y)$.\n\n**Lemma 4.** The function $x \\mapsto xf(x)$ is a surjection.\n*Proof.* Use $P(x, x)$ and get $f(2xf(x)) = 2f^2(x)$. Consider arbitrary $x > 0$. Take $y$ such that $f(y) = \\sqrt{\\frac{1}{2}f(2x)}$. Substitute it into the last equality to get $f(2yf(y)) = f(2x) \\Rightarrow yf(y) = x$, as desired.\n\n**Lemma 5.** If $xf(x) > yf(y)$, then $f(x) > f(y)$.\n*Proof.* Suppose the contrary, let $xf(x) > yf(y)$, but $f(x) \\le f(y)$. Since $x \\ne y$, we have $f(x) < f(y)$. Choose real $t$ with $xf(x) > t > yf(y)$. Then there is a real $u$ such that $uf(u) + yf(y) = t$. Then $f(uf(u) + yf(y)) = f(t) = f^2(u) + f^2(y) > f^2(y) > f^2(x)$. Then there is a real $v$ such that $f(t) = f^2(v) + f^2(x) = f(xf(x) + vf(v))$. From here we get $t = xf(x) + vf(v) > xf(x)$, which is a contradiction, and the proof of the lemma is complete.\n\n**Lemma 6.** The function $x \\mapsto f(x)$ is increasing on $(0, +\\infty)$.\n*Proof.* It suffices to prove that for any positive $a, b, c$ with $b > c$ we have $f(a + b) > f(a + c)$. Consider $x, y, z$ such that $a = xf(x), b = yf(y)$, and $c = zf(z)$. Since $b > c$, we have $f(y) > f(z)$. Then\n$$\n\\begin{aligned}\nf(a + b) &= f(xf(x) + yf(y)) = f^2(x) + f^2(y) > f^2(x) + f^2(z) \\\\\n&= f(xf(x) + zf(z)) = f(a + c),\n\\end{aligned}\n$$\nas desired.\n\nFrom Lemma 1, $2^1 \\in S$, $2^3 \\in S$, $2^7 \\in S$, etc., so $S$ contains arbitrary large reals. Let us prove that $S$ contains arbitrary small (close to zero) reals. Consider arbitrary $\\varepsilon > 0$. It suffices to prove that $\\exists x \\in S: 0 < x < \\varepsilon$. From Lemma 2, we have that $2^{\\frac{n-1}{2}} \\in S$ if $2^n \\in S$. Since $2^0 \\in S$, we have $2^{\\frac{1}{2}-1} \\in S$, $2^{\\frac{1}{4}-1} \\in S$, $2^{\\frac{1}{8}-1} \\in S$, etc., so there is $x \\in S: \\frac{1}{2} < x < \\frac{1}{2} + \\varepsilon^2$. Denote by $x_0$ the real such that $f(x_0) = \\sqrt{x - \\frac{1}{2}}$. Use $P(x_0, \\frac{1}{\\sqrt{2}}a)$. Since $\\frac{1}{\\sqrt{2}} \\in S$, we have $f(\\frac{a}{2} + x_0\\sqrt{x - \\frac{1}{2}}) = x = f(ax)$. This means $\\frac{a}{2} + x_0\\sqrt{x - \\frac{1}{2}} = x$.\n\nTherefore $\\sqrt{x - \\frac{1}{2}} \\in S$ and $\\sqrt{x - \\frac{1}{2}} < \\varepsilon$. We have proven that $S$ contains arbitrarily large and arbitrarily small numbers.\n\nSuppose that $x \\notin S$. Let us prove that there is $y \\in S$, lying between $x$ and $f(ax)$. Suppose otherwise. We've proven that there are $y_1, y_2 \\in S$ such that $y_1 < x, f(ax) < y_2$. Consider the following process: given $z_1, z_2 \\in S$ such that $z_1 < x, f(ax) < z_2$, consider $z_3$, the midpoint of $[z_1, z_2]$. By Lemma 3, it lies in $S$ and by our assumption is not between $x$ and $f(ax)$. This means that the closed interval between $x$ and $f(ax)$ lies entirely either in $[z_1, z_3]$ or $[z_3, z_2]$. Then we can choose one of these intervals and repeat the operation. Since after each step the length of the interval halves, it will eventually be smaller than the length of the interval between $x$ and $f(ax)$, leading to a contradiction.\n\nThus, there is $y \\in S$ between $x$ and $f(ax)$. Since $f$ is increasing, $0 > (x - y)(f(ax) - y) = (x - y)(f(ax) - f(ay)) > 0$. This contradiction proves that $S = (0, +\\infty)$, so $f(x) = \\frac{x}{a}$. It's easy to check that all functions of the form $f(x) = cx$ with $c > 0$ satisfy the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45766,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which largest $k$ does there exist a permutation $(a_1, a_2, \\ldots, a_{2022})$ of integers $(1, 2, \\ldots, 2022)$ such that for some $k$ integers $1 \\le i \\le 2022$ the fraction $\\frac{a_1 + a_2 + \\cdots + a_i}{1 + 2 + \\cdots + i}$ is an integer larger than $1$?\n\n(Oleksii Masalitin)",
    "options": [],
    "answer": "1011",
    "solution": "Denote by $s_i = a_1 + a_2 + \\dots + a_i$, $t_i = 1 + 2 + \\dots + i$. We will show that there exists at most one $i \\ge 1011$, for which $s_i \\ne t_i$ is divisible by $t_i$. Indeed, note that $s_i \\le 2022 + 2021 + \\dots + (2023 - i) = 2023i - 1011$. Also, for $i \\ge 1011$ we have $2023i - 1011 < 3t_i$, as $4t_i = 2i(i+1) \\ge 2024i$, so if for $i \\ge 1011$ $t_i$ is an integer larger than $1$, then $s_i = 2t_i$.\n\nSuppose that there exist two such $i > j \\ge 1011$, that $s_i = 2t_i, s_j = 2t_j$, then $s_i - s_j = 2(t_i - t_j) = i^2 + j^2 - i - j = (i-j)(i+j+1) \\ge 2023(i-j)$, but $s_i - s_j = a_{j+1} + a_{j+2} + \\dots + a_i \\le 2022(i-j)$, contradiction. So, there can't be more than $1011$.\n\nLet it be $i$. It's enough to note that for a permutation $(2, 4, \\dots, 2022, 1, 3, \\dots, 2021)$ the number of such $i$ is precisely $1011$, as for $i = 1, \\ldots, 1011$ we have $s_i = 2+4+\\dots+2i = 2t_i$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 45767,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ the median $BM$ is equal to the half of the side $BC$. Prove, that\n$$\n\\angle ABM = \\angle BCA + \\angle BAC.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Mark on the extension of median $BM$ by point $M$ such point $D$, that $BM = MD$ (fig. 11). Then triangles $ABM$, $CDM$ are equal by two sides and the angle between them, and also $BD = 2BM = BC$. Then triangle $BDC$ is isosceles, and\n$$\n\\begin{align*}\n\\angle ABM &= \\angle CDM = \\angle BCD = \\angle BCA + \\angle DCA \\\\\n&= \\angle BCA + \\angle BAC,\n\\end{align*}\n$$\nas desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Petrik uses the computer program \"Three\", which converts the numbers written on the display. For one application of this program Petrik chooses 5 numbers from the written ones, and the program increases each of these 5 numbers in 3 times. At the beginning, the following 20 numbers are written on the display: 1, $3^1$, $3^2$, ..., $3^{19}$. What smallest number of times does Petrik have to use the program to be able to get a set of equal numbers on the display?",
    "options": [],
    "answer": "38",
    "solution": "In one operation the product of all written numbers increases in $3^5$ times. At the beginning, this product equals $3^{0+1+2+...+19} = 3^{190}$. So, after using it $n$ times, the product will be equal to $3^{190+5n}$. By that time all numbers would have to become equal, and therefore at least $3^{19}$, so in the end the product is at least $3^{20p}$, where $p \\ge 19$. Then $3^{20p} = 3^{190+5n}$, $5n + 190 = 20p$ and $n + 38 = 4p$, and as $p \\ge 19$, we get $n \\ge 38$. Now let's show how to achieve this by using the program 38 times.\n\n$$\n\\begin{aligned}\n3^0, 3^1, 3^2, 3^3, 3^4 \\quad (15) &\\rightarrow 3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}; \\\\\n3^5, 3^6, 3^7, 3^8, 3^9 \\quad (10) &\\rightarrow 3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}; \\\\\n3^{10}, 3^{11}, 3^{12}, 3^{13}, 3^{14} \\quad (5) &\\rightarrow 3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}.\n\\end{aligned}\n$$\nSo after 30 uses we have 4 of each of $3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}$.\n\n$$\n\\begin{aligned}\n3^{15}, 3^{16}, 3^{17}, 3^{18} \\quad (3) &\\rightarrow 3^{18}, 3^{19}, 3^{19}, 3^{19}; \\\\\n3^{16}, 3^{17}, 3^{18}, 3^{19} \\quad (2) &\\rightarrow 3^{18}, 3^{19}, 3^{19}, 3^{19}; \\\\\n3^{17}, 3^{18}, 3^{19}, 3^{20} \\quad (1) &\\rightarrow 3^{18}, 3^{19}, 3^{20}, 3^{20}.\n\\end{aligned}\n$$\nSo after 36 uses we have 10 of $3^{18}, 3^{19}$. In the last two moves we make them all equal to $3^{19}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45769,
    "subject": "Mathematics (Multi-modal)",
    "question": "The inscribed circle of triangle $ABC$ is tangent to its sides $BC$, $AC$ and $AB$ at points $K$, $L$ and $M$ correspondently. Let $P$ be the point of intersection of the bisector of $\\angle BCA$ with the line $MK$. Prove that $AP \\parallel LK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the inscribed circle by $k$, its center by $S$ (fig. 19), and its angles by $\\alpha, \\beta, \\gamma$. From symmetry, $KL \\perp CP$ and $\\angle LPC = \\angle KPC$. Then with simple calculations we get $\\angle MKB = 90^\\circ - \\frac{1}{2}\\beta$ and $\\angle LKC = 90^\\circ - \\frac{1}{2}\\gamma \\Rightarrow \\angle MKL = 90^\\circ - \\frac{1}{2}\\alpha$. Similarly for other angles.\n\n![](attached_image_1.png)\n\nFig. 19\n\n$$\n\\text{As } \\angle KPC + \\frac{1}{2}\\gamma = \\angle BKP = 90^\\circ - \\frac{1}{2}\\beta,\n$$\n$$\n\\angle LPC = \\angle KPC = 90^\\circ - \\frac{1}{2}(\\beta + \\gamma) = \\frac{1}{2}\\alpha.\n$$\n\n$$\n\\angle LPC = \\angle LPS = \\angle LAS = \\frac{1}{2}\\alpha,\n$$\n\nthe quadrilateral $LSPA$ is inscribed. As $\\angle ALS = 90^\\circ$, we get $AP \\perp CP \\Rightarrow AP \\parallel KL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45770,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given some positive integer $n$. Prove that for any real numbers $a_1, a_2, ..., a_n$ there exists some number of form $k\\sqrt{2}$, where $k$ is some positive integer, that all numbers $k\\sqrt{2} + a_1, k\\sqrt{2} + a_2, ..., k\\sqrt{2} + a_n$ are irrational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider numbers $x_1 = \\sqrt{2}, x_2 = 2\\sqrt{2}, ..., x_{n+1} = (n + 1)\\sqrt{2}$. Suppose that for each $k = 1, n + 1$ at least one of the numbers $x_k + a_1, x_k + a_2, ..., x_k + a_n$ is rational. As we have $n$ numbers, and try $n + 1$ options, from the Dirichlet principle some two numbers of form $x_i + a_i$ and $x_j + a_j$ will be rational. But then their difference also will be rational, so the number\n$$\n(x_j + a_i) - (x_i + a_j) = x_j - x_i = (j - i)\\sqrt{2}\n$$\nwill be rational, which isn't true. This contradiction finishes the proof, as for some $x_k = k\\sqrt{2}$ all integers $x_k + a_1, x_k + a_2, ..., x_k + a_n$ are irrational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prime number $p > 2$ and the polynomial $Q$ with integer coefficients are such that there do not exist two positive integers $i, j$, such that $i < j < p$ and the number $(Q(j) - Q(i))(jQ(j) - iQ(i))$ is divisible by $p$. What's the smallest possible degree of $Q$?",
    "options": [],
    "answer": "p-2",
    "solution": "Example: $Q(x) = x^{p-2} - 1$. For any $i < j < p$:\n$$\nQ(j) - Q(i) = y^{p-2} - x^{p-2} \\equiv \\frac{1}{y} - \\frac{1}{x} \\pmod{p},\n$$\nwhich clearly isn't divisible by $p$, and\n$$\njQ(j) - iQ(i) = (y^{p-1} - x^{p-1}) - (y - x) \\equiv (x - y) \\pmod{p}\n$$\nalso isn't divisible by $p$.\n\nSuppose that there exists such a polynomial of smaller degree. It follows from the statement that all of $Q(1), Q(2), \\dots, Q(p-1)$ give different remainders modulo $p$, and also $1Q(1), 2Q(2), \\dots, (p-1)Q(p-1)$ give different remainders modulo $p$. Let's use the following fact: $1^k + 2^k + \\dots + (p-1)^k \\equiv 0 \\pmod{p}$ for any $k$ from $1$ to $p-2$. Then consider $1Q(1) + 2Q(2) + \\dots + (p-1)Q(p-1)$. From the fact above, this number is $0$ modulo $p$ (as each degree in the polynomial $xQ(x)$ is in the range $[1, p-2]$). Suppose that among the numbers $1P(1), \\dots, (p-1)P(p-1)$ there are all remainders modulo $p$, except for $x$. Then:\n$$\n0 \\equiv 0 + 1 + \\dots + (p-1) \\equiv 1Q(1) + 2Q(2) + \\dots + (p-1)Q(p-1) + x \\equiv x \\pmod{p},\n$$\nso $1Q(1), 2Q(2), \\dots, (p-1)Q(p-1)$ modulo $p$ give remainders $1, 2, \\dots, p-1$ in some order. Then none of $Q(1), Q(2), \\dots, Q(p-1)$ is divisible by $p$, and they also give remainders $1, 2, \\dots, p-1$ in some order.\nBut then:\n$$\n(p-1)! \\equiv 1Q(1) \\times 2Q(2) \\times \\dots \\times (p-1)Q(p-1) \\equiv (p-1)!^2 \\pmod{p},\n$$\nwhich, by the Wilson theorem, gives, $-1 \\equiv 1 \\pmod{p}$, this contradiction completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any integer $n$ there is a monic quadratic polynomial $x^2 + bx + c$ with integer coefficients which attains values $n, n^2, n^3$ at some three integer points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\nf(x_1) = n,\\ f(x_2) = n^2,\\ f(x_3) = n^3.\n$$\nThen for a polynomial $g(x) = f(x) - n$ we have $g(x_1) = 0$, $g(x_2) = n^2 - n$, $g(x_3) = n^3 - n$, so $g(x) = (x - x_1)(x - t)$ for some integer $t$. Then we need to find $x_1, x_2, x_3, t$, such that $g(x_2) = (x_2 - x_1)(x_2 - t) = n^2 - n$, $g(x_3) = (x_3 - x_1)(x_3 - t) = n^3 - n$. It suffices to have\n$$\nx_2 - x_1 = 1,\\ x_2 - t = n^2 - n,\\ x_3 - x_1 = n,\\ x_3 - t = (n^2 - 1).\n$$\nFor example, $x_1 = 0$, $x_2 = 1$, $x_3 = n$ and $t = -n^2 + n + 1$ satisfy these relations. It's easy to check that the polynomial $f(x) = x(x + n^2 - n - 1) + n$ attains the desired values at $x = 0$, $x = 1$, and $x = n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fedir and Mykhailo have three piles of stones: the first contains $100$ stones, the second $101$, the third $102$. They are playing a game with the following rules: they move in turns, in his turn a player chooses any two piles of stones, containing, say, $a$ and $b$ stones correspondingly, and takes from each of them the number of stones equal to the greatest common divisor of numbers $a$ and $b$. The winner is the player, after whose move some pile becomes empty for the first time. Who wins if Fedir moves first, and both try to win?",
    "options": [],
    "answer": "Mykhailo (the second player) wins.",
    "solution": "Let's provide the winning strategy for Mykhailo. Suppose that before Fedir's move, the piles contained $(2n, 2n+1, 2n+2)$ stones for some integer $n > 1$. Then there are 3 possible moves.\n\nIf he takes first two piles, then, as $(2n, 2n+1) = 1$, after his move we will have the following numbers of stones: $(2n-1, 2n, 2n+2)$. After this Mykhailo chooses the last two piles, and as $(2n, 2n+2) = 2$, after his move the numbers will be: $(2n-2, 2n-1, 2n)$. This is the initial configuration.\n\nIf Fedir chooses first and the third piles, then, as $(2n, 2n+2) = 2$, after his move we will have the following configuration: $(2n-2, 2n, 2n+1)$. Mykhailo chooses last two piles once again, and gets the configuration $(2n-2, 2n-1, 2n)$ again.\n\nIf Fedir chooses the second and the third piles, then from $(2n+1, 2n+2) = 1$, we see that after his move we will have the following configuration: $(2n, 2n, 2n+1)$. In this situation Mykhailo just takes first two equal piles and wins, as after his move there will be two empty piles: $(0, 0, 2n+1)$.\n\nThe initial configuration is just $n = 50$. After the moves of Fedir and Mykhailo we will get the configuration for $n = 49$ (or Mykhailo will win). And so on. After $48$ moves, unless Mykhailo has already won, we will have $(4, 5, 6)$. Fedir will change it to one of: $(3, 4, 6)$, $(2, 5, 4)$ or $(4, 4, 5)$. It's easy to see that in all of them Mykhailo wins in the next move.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45774,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a right triangle $ABC$ with right angle $C$ on the sides $BC$, $AC$ and $AB$ points $D, E$ and $F$ correspondingly were chosen so that $\\angle DAB = \\angle CBE$ and $\\angle BEC = \\angle AEF$. Prove that $DB = DF$.\n(Mykhailo Shtandenko)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the point $K$, symmetric to the point $B$ with respect to point $C$ (fig. 13). Then $\\Delta BEC = \\Delta KCE$, and $\\angle KEC = \\angle BEC = \\angle AEF$, so points $K, E, F$ lie on the same line. Then from the statement it follows, that\n![](attached_image_1.png)\nFig. 13\n$$\n\\angle FAD = \\angle EBC = \\angle EKC = \\angle FKD,\n$$\nso points $A, K, D, F$ are concyclic. Therefore\n$$\n\\angle BFD = 180^\\circ - \\angle AFD = \\angle AKD.\n$$\nAs $\\Delta ABC = \\Delta AKC$, and $\\angle AKC = \\angle ABC$, the equality above is rewritten as $\\angle BFD = \\angle ABC = \\angle FBD$. From this in $\\Delta FBD$ it follows that $DB = DF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Suppose that there exists a line $l \\parallel BD$, which is tangent to the inscribed circles of triangles $ABC$ and $CDA$. Prove that the line $l$ contains the incenter of one of $\\triangle ABCD$ and $\\triangle DAB$.\n(Fedir Yudin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose it's false, wlog $BC < CD$. Let $CK$ and $CL$ be the bisectors of the correspondent angles in triangles $\\triangle ABC$ and $\\triangle CDA$, and $B'$ and $D'$ be the points symmetric to the points $B$ and $D$ with respect to the lines $CK$ and $CL$ correspondingly (fig. 20). Then lines $KB$ and $KB'$ are symmetric with respect to the line $CK$, so $KB'$ is tangent to the inscribed circle of $\\triangle ABC$, similarly $LD'$ is tangent to the inscribed circle of $\\triangle CDA$.\n\n![](attached_image_1.png)\nFig. 20\n\nDenote by $d(X)$ the oriented distance from the point $X$ to the line $BD$, where $d(A) > 0$.\nNote that $\\angle DBC > \\frac{1}{2}\\angle BAD$, implying\n$$\n\\begin{aligned}\n\\angle AKB' &= 180^\\circ - 2\\angle BKC \\\\\n&= 2\\angle ABC + \\angle ACB - 180^\\circ \\\\\n&> 2\\angle ABD + \\angle BAD + \\angle ADB \\\\\n&\\ge -180^\\circ = \\angle ABD,\n\\end{aligned}\n$$\nso $d(B') < d(K)$. Similarly, $\\angle ALD' < \\angle ADB$, so $d(D') > d(L)$. Also note that\n$$\n\\frac{AK}{KB} = \\frac{AC}{CB} > \\frac{AC}{CD} = \\frac{AL}{LD} \\Rightarrow d(K) < d(L).\n$$\nConsider the line $k \\parallel BD$, with respect to which the points $K$ and $L$ lie on different sides. Then the inscribed circle of $\\triangle ABC$ lies on the same side from the line $k$, as the segment $BD$ (as this circle lies in the quadrilateral $BCB'K$, which lies on this side with respect to $k$) and the inscribed circle of $\\triangle CDA$ has a point, lying from the different side with respect to $k$ (as it's tangent to the segment $LD'$, which lies from the different side with respect to $k$). So these inscribed circles can't have a common tangent, parallel to $BD$, which lies on the same side from $BD$, as point $A$, this contradiction completes the proof.\n\nAlso note that when $C$ is the midpoint of arc $BD$, we have $CB' = CB = CD = CD'$, so $B' = D'$ is the incenter of $\\angle DAB$ (fig. 21). Furthermore, $\\angle AKB' = 180^\\circ - 2\\angle BKC = 2\\angle ABC + \\angle ACB - 180^\\circ = 2\\angle ABD + \\angle BAD + \\angle ADB - 180^\\circ = \\angle ABD$, so $KB' \\parallel BD$. Similarly $LD' \\parallel BD$, so the line $KL$ is the common tangent.\n\n![](attached_image_2.png)\nFig. 21",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45776,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides $AB$ and $BC$ of the square $ABCD$ we marked points $K$ and $L$ correspondingly so that $KB = LC$. Segments $AL$ and $CK$ intersect at point $P$. Prove that lines $DP$ and $KL$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By construction $\\Delta ADK = \\Delta BAL$. Then $\\angle BAL + \\angle ALB = \\angle BAL + \\angle AKD = 90^\\circ$, and $\\Delta AMK$ has a right angle $M$, so $AL \\perp DK$ (fig. 14).\n\nSimilarly $DL \\perp CK$. Then lines $AL$ and $CK$, which intersect at point $P$, contain two altitudes of the $\\Delta DKL$, and therefore the line $DP$ has to contain the third altitude, so $KL \\perp DP$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with hypothenuse $BC$ and altitude $AD$. Let's denote the midpoints of $AD$ and $AC$ by $E$ and $F$ correspondingly. Let point $M$ be the circumcenter of $\\triangle BEF$. Prove that $AC||BM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $\\triangle ADB \\sim \\triangle CAB$, we get $\\frac{AD}{AB} = \\frac{CA}{CB}$. As $AE = \\frac{1}{2}AD$, $CF = \\frac{1}{2}CA$, we get $\\frac{AF}{AB} = \\frac{CF}{CB}$. From this similarity we get that (fig. 16):\n$$\n\\angle BAE = \\angle BAD = \\angle BCA = \\angle BCF\n$$\nThen by angle and the ratio of the sides we get that $\\triangle AEB \\sim \\triangle CFB \\Rightarrow \\angle ABE = \\angle CBF$.\nAs $EF$ is the midline of $\\triangle CAD$, we also get $EF||BC \\Rightarrow \\angle BFE = \\angle CBF$. Next, we get the following equalities of angles: $\\angle BFE = \\frac{1}{2}\\angle BME = 90^\\circ - \\angle EBM \\Rightarrow \\angle ABE = \\angle CBF = 90^\\circ - \\angle EBM$. Therefore $\\angle ABM = \\angle ABE + \\angle EBM = 90^\\circ, AC||BM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45778,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle *ABC* the point *N* is the midpoint of the median *CM*. The point *X* satisfies the following conditions: $\\angle XMC = \\angle MBC$, $\\angle XCM = \\angle MCB$ and points *X* and *B* lie from different sides with respect to the line *CM*. Let $\\omega$ be the circumscribed circle of $\\triangle AMX$. Prove that:\n\na) *CM* is tangent to $\\omega$;\n\nb) Lines *NX* and *AC* intersect on the circle $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) as from the statement $\\triangle XMC \\sim \\triangle MBC$, we get\n$$\n\\begin{aligned} \n\\angle AMX &= 180^\\circ - \\angle XMC - \\angle BMC = \\\\ \n&= 180^\\circ - \\angle XMC - \\angle MXC = \\angle MCX \n\\end{aligned}\n$$\n\nand $\\frac{AM}{MX} = \\frac{BM}{MX} = \\frac{MC}{CX} \\Rightarrow \\triangle AXM \\sim \\triangle MXC \\Rightarrow \\angle XAM = \\angle XMC$, so the claim follows from the equality of the inscribed angles.\n\n![](attached_image_1.png)\n**Fig. 22**\n\nb) Let *S* be the second point of the intersection of *AC* and $\\omega$. We need to show that the point *S* lies on *NX*. From the properties of inscribed angles, we get $\\angle SMC = \\angle SAM = \\angle CAM$. It follows that $\\triangle CSM \\sim \\triangle CMA \\Rightarrow \\frac{CM}{CA} = \\frac{SM}{MA}$. Then, as $CM = 2MN$ and $MA = \\frac{1}{2}AB \\Rightarrow \\frac{MN}{CA} = \\frac{SM}{AB}$. Together with the condition $\\angle SMN = \\angle SMC = \\angle CAM = \\angle CAB$ we get that $\\triangle SNM \\sim \\triangle BCA$. Then $\\angle MSN = \\angle ABC$. Furthermore, $\\angle XSM = 180^\\circ - \\angle XAM$. From the claim in a), we also get $\\angle XAM = \\angle XMC = \\angle MBC = \\angle ABC$, so $\\angle XSM = 180^\\circ - \\angle ABC = 180^\\circ - \\angle MSN$, and $S \\in MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45779,
    "subject": "Mathematics (Multi-modal)",
    "question": "**Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that the following condition is fulfilled for arbitrary real numbers $x$ and $y$**\n$$\nf^2(x + y) = f^2(x) + 2f(xy) + f^2(y).\n$$",
    "options": [],
    "answer": "f(x) = 0; f(x) = -2; f(x) = x; f(x) = x - 2",
    "solution": "Let us denote the equation given in the statement of a problem by $(*)$ and substitute $y = -x$ into it. We obtain\n$$\nf^2(x) + f^2(-x) + 2f(-x^2) = f^2(0) \\text{\\{ (**).\\}}\n$$\nSubstituting $x = x + y, y = -x$ into $(*)$, we get\n$$\nf^2(y) = f^2(x + y) + f^2(-x) + 2f(-x^2 - xy).\n$$\nAdding $(*)$ and $(**)$ to the last-mentioned equality, we obtain\n$$\nf(xy) + f(-x^2 - xy) = f(-x^2) - f^2(0)/2.\n$$\nLet us denote $g(x) = f(x) + f^2(0)/2$. Let $a = -x^2 \\le 0, b = xy \\in \\mathbb{R}$ when $a \\ne 0$. The last-mentioned equality assumes the form\n$$\ng(a) + g(b) = g(a + b) \\text{\\{ (***)\\}.}\n$$\nIf $a, b < 0$, then we obtain the Cauchy's equation. Let us prove that $g(x)$ is bounded above when $x < 0$. Then from the Cauchy's equation we get that $g(x) = kx$ for arbitrary negative number $x$ and real number $k$. From $(**)$ we get\n\nThus, $g(x) = kx \\quad \\forall x < 0$. Let us assume that $x \\ge 0$. Substituting $y = -1 - x$ into $(***)$, we obtain $g(x) = g(-1) - g(-1 - x) = kx$ for every positive number $x$. Thus, we proved that\n$$\nf(x) = g(x) - f^{2}(0)/2 = kx - f^{2}(0)/2 = kx + c\n$$\nfor every real $x$. Substituting the explicit form of the function $f$ into $(*)$, we obtain:\n$$\n\\begin{align*}\nk^2 (x + y)^2 + 2kc(x + y) + c^2 &= k^2 x^2 + 2kcx + c^2 + 2kxy + 2c + k^2 y^2 + 2kcy + c^2, \\\\\n(k^2 - k)xy &= c^2 + 2c.\n\\end{align*}\n$$\nConsidering that $xy$ is able to possess arbitrary real value, from the last-mentioned equality we get that $k \\in \\{0,1\\}$, $c \\in \\{0,-2\\}$.\nThus, we obtain that the condition can be fulfilled only by the functions $f(x) = 0$, $f(x) = -2$, $f(x) = x$, and $f(x) = x - 2$. We verify that all these functions fulfill the condition of the problem by checking.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a polynomial with integer coefficients of degree $d$. For the set $A = \\{a_1, a_2, \\dots, a_k\\}$ of positive integers denote $S(A) = P(a_1) + P(a_2) + \\dots + P(a_k)$. The positive integers $m, n$ are such that $m^{d+1} \\mid n$. Prove that the set $\\{1, 2, \\dots, n\\}$ can be split into $m$ disjoint subsets $A_1, A_2, \\dots, A_m$ of equal size such that $S(A_1) = S(A_2) = \\dots = S(A_m)$.\n(Klurman Oleksiy)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = k m^{d+1}$. We will construct the desired splitting in the following way. To determine which of the subsets each number $x \\in \\{1; 2; \\dots n\\}$ belongs to, let us write $x - 1$ in the form:\n$$\nx - 1 = c m^{d+1} + x_0 + x_1 m + x_2 m^2 + \\dots + x_d m^d,\n$$\nwhere $c \\ge 0$ is an integer, $x_i \\in \\{0; 1; \\dots m-1\\}$, $i = 0, \\dots, d$, i.e. we write the remainder of $x-1$ divided by $m^{d+1}$ in base $m$ numeral system. We include the number $x$ to subset $A_j$, $j \\in \\{1; 2; \\dots m\\}$ if $x_0 + x_1 + x_2 + \\dots + x_d \\equiv j \\pmod m$. Let us show that such splitting satisfies the condition. Clearly, it is enough to prove our statement for polynomials of the form $P(x) = x^t$, $t = 0, \\dots, d$.\n\nWe have\n$$\nS(A_j) = \\sum_{c=0}^{k-1} \\sum_j (c m^{d+1} + x_0 + x_1 m + x_2 m^2 + \\dots + x_d m^d)^t,\n$$\nwhere $\\sum_j$ denotes the sum over all tuples $x_0, \\dots, x_d$ of numbers from the set $\\{0; 1; \\dots m-1\\}$ such that $x_0 + x_1 + x_2 + \\dots + x_d \\equiv j \\pmod m$. Let us show that the inner sum does not depend on $j$. Denoting $z = 1 + c m^{d+1}$, we can transform it by expanding the brackets:\n$$\n\\sum_j (c m^{d+1} + x_0 + x_1 m + x_2 m^2 + \\dots + x_d m^d)^t = \\sum_d \\frac{t!}{u! t_0! t_1! \\dots t_d!} z^u m^{t_1 + 2 t_2 + \\dots + d t_d} \\sum_j x_0^{t_0} x_1^{t_1} \\dots x_d^{t_d},\n$$\nwhere $\\sum_d$ denotes the sum over all tuples of non-negative integers $u, t_0, \\dots, t_d$ such that $u + t_0 + \\dots + t_d = t$. Since $t_0 + \\dots + t_d \\le t < d+1$, there is at least one number among $t_0, \\dots, t_d$ which equals zero. Suppose that $t_0 = 0$, then\n$$\n\\sum_j x_0^{t_0} x_1^{t_1} \\dots x_d^{t_d} = \\sum_j x_1^{t_1} \\dots x_d^{t_d} = \\sum_{x_1, \\dots, x_d = 0}^{m-1} x_1^{t_1} \\dots x_d^{t_d},\n$$\nbecause for any tuple $x_1, \\dots, x_d$ of numbers from $\\{0; 1; \\dots m-1\\}$ there is a unique number $x_0 \\in \\{0; 1; \\dots m-1\\}$ for which $x_0 + x_1 + x_2 + \\dots + x_d \\equiv j \\pmod m$. The last sum is clearly independent of $j$, which is what we wanted to prove.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45781,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that the triple $a, b, c$ from the segment $[-1; 1]$ is *worthy*, if these numbers satisfy the inequality $1 + 2abc > a^2 + b^2 + c^2$. Prove that if the triples $a, b, c$ and $x, y, z$ are worthy, then $ax, by, cz$ is worthy as well.\n\n(Voloshyn Denys)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We are given the following inequalities:\n$$\n1 + 2abc \\ge a^2 + b^2 + c^2 \\text{ and } 1 + 2xyz \\ge x^2 + y^2 + z^2, \\quad (1)\n$$\nand we need to show that the following inequality holds:\n$$\n1 + 2abcxyz \\ge (ax)^2 + (by)^2 + (cz)^2. \\quad (2)\n$$\nFirst observation: we can assume that all numbers are non-negative. Indeed, if $abcxyz \\ge 0$, then after taking the absolute values the inequality (2) does not change, while the inequalities (1) can only become stronger. If $abcxyz < 0$, then either $abc < 0$ or $xyz < 0$; without loss of generality, assume the former, then\n$$\n1 + 2abcxyz \\ge 1 + 2abc \\ge a^2 + b^2 + c^2 \\ge (ax)^2 + (by)^2 + (cz)^2.\n$$\nNext, we can assume without loss of generality that $a \\le b \\le c$. Inequalities (1) and the left-hand side of (2) are invariant to the permutation of $x, y, z$. From the inequality for ordered tuples we get that the right-hand side of (2) is maximized when $x \\le y \\le z$, hence, it is enough to prove the statement only for this case.\n\nSolving the inequality (2) as quadratic w.r.t. $cz$, we get the equivalent inequality:\n$$\nabxy - \\sqrt{(1 - a^2x^2)(1 - b^2y^2)} \\le cz \\le abxy + \\sqrt{(1 - a^2x^2)(1 - b^2y^2)}.\n$$\nThe inequality on the left is obvious since $cz \\ge ax \\ge abxy$. Let us prove the inequality on the right. By solving the inequalities (1) as quadratic w.r.t. $c$ and $z$, we get\n$$\nc \\le ab + \\sqrt{(1 - a^2)(1 - b^2)} \\text{ and } z \\le xy + \\sqrt{(1 - x^2)(1 - y^2)}.\n$$\nThus, it is enough to show that\n$$\n(ab + \\sqrt{(1 - a^2)(1 - b^2)}) (xy + \\sqrt{(1 - x^2)(1 - y^2)}) \\le abxy + \\sqrt{(1 - a^2x^2)(1 - b^2y^2)}. \\quad (3)\n$$\nSuppose that some of the numbers $a, b, x, y$ are equal to 1. Taking into account that $a \\le b$ and $x \\le y$, it is enough to consider the case $b=1$ (the case $y=1$ is similar). The inequality (3) becomes\n$$\na\\sqrt{(1 - x^2)(1 - y^2)} \\le \\sqrt{(1 - a^2x^2)(1 - y^2)},\n$$\nwhich is now obvious since $a \\le 1, 1 - a^2x^2 \\ge 1 - x^2$.\nNow, suppose that none of the numbers is equal to 1. By expanding the brackets in the left-hand side and dividing by $\\sqrt{(1 - a^2)(1 - b^2)(1 - x^2)(1 - y^2)}$, we get after minor simplifications the equivalent inequality:\n$$\n1 + AB + XY \\le \\sqrt{(1 + A^2 + X^2)(1 + B^2 + Y^2)},\n$$\nwhere $A = \\frac{a}{\\sqrt{1-a^2}}, B = \\frac{b}{\\sqrt{1-b^2}}, X = \\frac{x}{\\sqrt{1-x^2}}, Y = \\frac{y}{\\sqrt{1-y^2}}$. And the last inequality is a partial case of the Cauchy-Schwarz inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $F$, $F_A$, $F_B$ and $F_C$ be the Feuerbach points of non-equilateral triangle $ABC$, and point $I$ be its incenter. Prove that lines $AF_A$, $BF_B$, $CF_C$ and $IF$ pass through one point. Feuerbach points are the tangent points of the nine-point circle with incircle (touches internally) and three excircles (touch externally) of the triangle: $F$ — tangent point to incircle, $F_A$ — tangent point to excircle that touches $BC$, $F_B$ — tangent point to excircle that touches $AC$, $F_C$ — tangent point to excircle that touches $AB$.\n\nThe nine-point circle is the circle that passes through the midpoint of each side of the triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $O_B$ the center of the excircle that touches $AC$, and by $E$ the center of the nine-point circle (fig. 23). Points $E$, $I$ and $F$, as well as the points $E$, $F_B$ and $O_B$ lie on the same line since $F$ and $F_B$ are the tangent points of the circles with the corresponding centers. Besides, $E \\neq I$, since otherwise the circles either would not have any common points or would coincide. Taking this into account, if the point $E$ lies on the line $IO_B$, i.e. on a bisector of the angle $B$, then the points $F$ and $F_B$ lie on this line as well. Then the lines $IF$ and $BF_B$ both coincide with the bisector.\n\nNow, suppose $E$ lies outside the line $IO_B$. Point $F_B$ is on the side $EO_B$ of triangle $EIO_B$ since the circles touch externally. In contrast, $B$ lies on the continuation of the side $O_B I$ of this triangle. Hence, the line $BF_B$ intersects the line $EI = IF$ at some point $X_B$, moreover, $X_B$ is the inner point of segment $EI$. Now we can apply Menelaus's theorem to the triangle $EIO_B$ and line passing through the points $B$, $X_B$ and $F_B$:\n\n![](attached_image_1.png)\nFig. 23\n\n$$\n\\frac{EX_B}{X_B I} \\cdot \\frac{IB}{B O_B} \\cdot \\frac{O_B F_B}{F_B E} = 1.\n$$\n\nDenote by $r$ the radius of the incircle of triangle $ABC$, by $r_B$ the radius of its excircle that touches side $AC$, and by $R_9$ the radius of the nine-point circle. If we drop the perpendiculars-radiuses $II'$ and $O_B O'_B$ from the points $I$ and $O_B$ to the line $AB$, it will form similar right triangles $BII'$ and $BO_B O'_B$ with the similarity coefficient $\\frac{BO_B}{BI} = \\frac{O_B O'_B}{II'} = \\frac{r_B}{r}$. It means that\n\n$$\n\\frac{EX_B}{X_B I} = \\frac{B O_B}{I B} \\cdot \\frac{F_B E}{O_B F_B} = \\frac{r_B}{r} \\cdot \\frac{R_9}{r_B} = \\frac{R_9}{r}.\n$$\n\nThus, line $BF_B$ passes through the unique point $X$ from the segment $EI$ which satisfies $\\frac{EX}{XI} = \\frac{R_9}{r}$. The same is also true in the case when $E$ lies on the line $IO_B$, because, as shown above, in this case all points of the segment $EI$ belong to the line $BF_B$.\n\nBy similar reasoning it follows that the lines $AF_A$ and $CF_C$ pass through the same point $X \\in EI$, which satisfies $\\frac{EX}{XI} = \\frac{R_9}{r}$. To finish the proof, it remains to mention that point $X$, as well as the whole segment $EI$, also belongs to the line $IF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45783,
    "subject": "Mathematics (Multi-modal)",
    "question": "What smallest value can be attained by the expression\n\n$$\n\\frac{(x + y + |x - y|)^2}{xy}\n$$\n\nfor positive $x$, $y$?",
    "options": [],
    "answer": "4",
    "solution": "We assume $x \\ge y$, then we can rewrite as:\n$$\n\\frac{(x + y + |x - y|)^2}{xy} = \\frac{(x + y + x - y)^2}{xy} = \\frac{(2x)^2}{xy} = \\frac{4x^2}{xy} = \\frac{4x}{y} \\ge 4\n$$\nWhen $x = y$, we get equality.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Represent the fraction $\\frac{1}{2022}$ as a difference of two regular fractions with smaller denominators.",
    "options": [],
    "answer": "1/674 - 1/1011",
    "solution": "One of the possible representations is\n$$\n\\frac{1}{2022} = \\frac{1}{2 \\cdot 3 \\cdot 337} = \\frac{3-2}{2 \\cdot 3 \\cdot 337} = \\frac{3}{2 \\cdot 3 \\cdot 337} - \\frac{2}{2 \\cdot 3 \\cdot 337} = \\frac{1}{2 \\cdot 337} - \\frac{1}{3 \\cdot 237} = \\frac{1}{674} - \\frac{1}{1011}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Teacher wrote on the board 5 distinct numbers. After that Petrik counted the sums of each two of these numbers and wrote them on the left half of the board, Vasyl did the same for the sums of each three of these numbers and wrote them on the right half of the board. Could the teacher write such numbers so that the sets of numbers written on the left and right halves of the board are the same (counting multiplicity)?",
    "options": [],
    "answer": "Yes; for example, −2, −1, 0, 1, 2.",
    "solution": "It's enough to choose the following numbers: $-2, -1, 0, 1, 2$. Then we can write down the sets of integers, but we can also apply the following reasoning: for any two numbers, say, $a, b$, selected by Petrik, from one side exists pair of numbers $(-a, -b)$, whose sum is the opposite to the initial, and, from another side, there exists a triple of numbers except $a, b$, and their sum also is $(-a - b)$, as the sum of all five numbers is zero. So we get a correspondence between the numbers from the left and from the right parts of the board.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45786,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given 5 distinct positive integers. Can their arithmetic mean be:\na) exactly 3 times larger than their largest common divisor;\nb) exactly 2 times larger than their largest common divisor;",
    "options": [],
    "answer": "a) Yes; for example, 1, 2, 3, 4, 5. b) No.",
    "solution": "a) It's enough to provide an example of such 5 integers. One example is the set $1, 2, 3, 4, 5$, whose arithmetic mean is $3$, and the largest common divisor is $1$.\n\nb) Suppose that such numbers $a_1, a_2, a_3, a_4$ and $a_5$ exist, let $d$ be their largest common divisor, then these 5 integers can be rewritten as $a_i = d b_i$, $i = 1, 5$ and the equality in the statement is rewritten as:\n$$\n\\frac{a_1 + \\cdots + a_5}{5} = 2d \\Leftrightarrow d(b_1 + \\cdots + b_5) = 10d \\Leftrightarrow b_1 + \\cdots + b_5 = 10.\n$$\nClearly numbers $b_1, \\dots, b_5$ are distinct, so their smallest possible sum is $1 + 2 + 3 + 4 + 5 = 15 > 10$, this contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45787,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any real numbers $x, y$ prove the inequality\n$$\n\\sqrt{(x + 4)^2 + (y + 2)^2} + \\sqrt{(x - 5)^2 + (y + 4)^2} \\le \\sqrt{(x - 2)^2 + (y - 6)^2} + \\sqrt{(x - 5)^2 + (y - 6)^2} + 20.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On a coordinate plane consider points $A(-4, -2)$, $B(2, 6)$, $C(5, 6)$ and $D(5, -4)$ (fig. 2). For any point $M(x, y)$ of the plane the inequality is rewritten as: $MA + MD - MB - MC \\le 20$. Let's find the largest possible value of the expression\n\n![](attached_image_1.png)\nFig. 2\n$$\nS(M) = MA + MD - MB - MC.\n$$\nFor any point $M$ of the plane from the triangle inequality we get:\n$$\nS(M) = (MA - MB) + (MD - MC) \\le AB + CD.\n$$\nIn addition, for the point $X = AB \\cap CD$ we get\n$$\nS(X) = (XA - XB) + (XD - XC) = AB + CD = 10 + 10 = 20.\n$$\nSo, the largest value is achieved for point $X$, and for all others the inequality is strict.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the circumscribed circle of an obtuse triangle $ABC$ with an obtuse angle $B$. Tangents to this circle at points $A$ and $B$ meet at point $P$, and the perpendicular to the line $BC$ at point $B$ intersects $AC$ at point $K$. Prove, that $PA = PK$.\n\n(Danylo Khilko)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45789,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\ge 3$ segments, their lengths in centimeters are distinct positive integers. It's known that it's possible to form a nondegenerate triangle from any three of these $n$ segments. Suppose that among these segments there are segments with lengths $5$ cm and $12$ cm. What's the largest value $n$ can attain?",
    "options": [],
    "answer": "6",
    "solution": "Reorder the segments by their lengths, so that $a_1 < a_2 < \\dots < a_n$. Clearly, any three segments form a triangle if and only if the sum of the lengths of the smallest two segments is larger than the length of the longest segment. So, the smallest segment except from the given two can't have a length smaller than $8$, as in that case we wouldn't be able to form a triangle from segments $5$, $a \\le 7$ and $12$, as $5 + 7 \\le 12$.\n\nSo, $a_1 = 5$ and $a_2 \\ge 8$. Clearly, $a_2 \\le 12$ and the number of segments can't exceed the number of elements of the set $\\{5, a_2, a_2 + 1, a_2 + 2, a_2 + 3, a_2 + 4\\}$, so $n \\le 6$.\n\nNote that the set of segments with lengths $\\{5, 8, 9, 10, 11, 12\\}$ satisfies the conditions. Therefore, the answer is $n = 6$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to write five integers on a board so that for any two numbers, there exists a pair of numbers among the remaining three whose sum is equal to the sum of the original two numbers?",
    "options": [],
    "answer": "Yes; for example: −2, −1, 0, 1, 2.",
    "solution": "It's enough to choose the following numbers: $-2$, $-1$, $0$, $1$, $2$. Then we can write down the sets of integers, but we can also apply the following reasoning: for any two numbers, say, $a$, $b$, selected by Petrik, from one side exists pair of numbers $(-a, -b)$, whose sum is the opposite to the initial, and, from another side, there exists a triple of numbers except $a$, $b$, and their sum also is $(-a - b)$, as the sum of all five numbers is zero. So we get a correspondence between the numbers from the left and from the right parts of the board.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45791,
    "subject": "Mathematics (Multi-modal)",
    "question": "$2n$ distinct positive integers are given. What's the largest number of pairs that can always be formed from these numbers, so that each number belongs to at most one pair, and the sum of integers in each pair is a composite number?",
    "options": [],
    "answer": "n-1",
    "solution": "Let $p_1, p_2, \\dots, p_{2n-1}$ be distinct prime integers larger than $2$. Then for a set $(1, p_1-1, p_2-1, \\dots, p_{2n-1}-1)$ it's not possible to form $n$ such pairs, as no number can be paired with number $1$.\n\nFrom the other side, we can always form at least $n-1$ pairs from the numbers of the same parity. In each such pair the sum will be even and not less than $4$, therefore composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45792,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point of the intersection of diagonals of the cyclic quadrilateral $ABCD$. The circumscribed circles of $\\triangle APD$ and $\\triangle BPC$ intersect the line $AB$ at points $E$ and $F$ correspondingly. $Q$ and $R$ are projections of the point $P$ onto the lines $FC$ and $DE$. Prove that $AB \\parallel QR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $d(Z, XY)$ the distance from the point $Z$ to the line $XY$. From the equality of inscribed angles it follows that (fig. 6)\n$$\n\\angle PEB = \\angle ADB = \\angle ACB = \\angle PFA \\Rightarrow PE = PF.\n$$\nFurthermore,\n$$\n\\angle EDP = \\angle BAP = \\angle BDC, \\quad \\angle FCP = \\angle PBA = \\angle DCP,\n$$\nso $P$ is the intersection of bisectors of angles $\\angle EDC$ and $\\angle DCF$, so $d(P, DE) = d(P, DC) = d(P, CF)$, so $PQ = PR$. Then $\\triangle FPQ \\cong \\triangle EPR$ as right triangles with equal cathetus and hypotenuse. From the equality $PQ = PR$ it follows that $\\angle PQR = \\angle PRQ$, so $\\angle FQR = \\angle QRE$. Similarly, $\\angle QFE = \\angle REF$. This means that $EQRF$ is an isosceles trapezoid, so $QR \\parallel EF$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AL$ be a bisector of triangle $ABC$. The circle centered at $B$ with radius $BL$ meets the ray $AL$ at point $E$, and the circle centered at $C$ with radius $CL$ meets the ray $AL$ at point $D$ (points $E$ and $D$ are different from point $L$). Prove that $AL^2 = AE \\cdot AD$.\n(Mykola Moroz)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, triangles $BEL$ and $CDL$ are isosceles, and angles $\\angle CLD$ and $\\angle BLE$ are vertical (fig. 3). Then $\\angle BEL = \\angle BLE = \\angle CLD = \\angle CDL$. Then $\\angle AEB = \\angle ALC$ as adjacent to equal angles. Also $\\angle CAL = \\angle BAL$, as $AL$ is a bisector.\n\nNote that triangles $CLA$ and $BEA$ are similar by two angles. Then $\\frac{AL}{AE} = \\frac{AC}{AB}$, from where $AL = AE \\cdot \\frac{AC}{AB}$. Also note that triangles $BLA$ and $ACD$ are similar by two angles. Then $\\frac{AL}{AD} =$\n$$\n\\frac{AB}{AC}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the circumscribed circle of an obtuse triangle $ABC$ with an obtuse angle $B$. Tangents to this circle at points $A$ and $B$ meet at point $P$, and the perpendicular to the line $BC$ at point $B$ intersects $AC$ at point $K$. Prove, that $PA = PK$.\n\n(Danylo Khilko)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First note, that as $\\angle ABC > 90^\\circ$, point $K$ lies on $AC$ (fig. 1). Also, it's clear that $PA = PB$. We will show that $K$ lies on the circle $\\omega$ with a center $P$ and radius $PA$. It's enough to prove, that $\\angle APB = 2(180^\\circ - \\angle AKB)$. Indeed, take on the larger arc of circle $\\omega$ any point $X$. Then $\\angle AXB = \\frac{1}{2}\\angle APB$. If this condition holds, then $\\angle AKB + \\angle AXB = 180^\\circ$, so points $A, K, B$ and $X$ lie on a circle $\\omega$. As $P$ is the center of this circle, $PK = PA$.\n\nSo, let's prove that $\\angle APB = 2(180^\\circ - \\angle AKB)$. As $PA = PB$, and also using the theorem about the angle between the chord and a tangent, we get the following: $\\angle APB = 180^\\circ - 2\\angle ABP = 180^\\circ - 2\\angle ACB = 180^\\circ - 2(90^\\circ - \\angle CKB) = 2\\angle CKB = 2(180^\\circ - \\angle AKB)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45795,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given a set of $n$ not necessarily distinct numbers $\\{a_1, a_2, \\dots, a_n\\}$, (meaning that some of them can be equal). Consider all $2^n - 1$ nonempty subsets of this set, and for each such subset, find the sum of its elements. What largest number of these sums could turn out to be equal to $1$? For example, for a set $\\{-1; 2; 2\\}$ we have $7$ nonempty subsets: $\\{-1\\}$, $\\{2\\}$, $\\{2\\}$, $\\{-1; 2\\}$, $\\{-1; 2\\}$, $\\{2; 2\\}$ and $\\{-1; 2; 2\\}$, and among them there are exactly two subsets with sum $1$.\n(Anton Trygub)",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "Example, where we reach equality, is: $(1, 0, 0, \\dots, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45796,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ points $H$ and $O$ are the orthocenter and the circumcenter correspondingly. The line $HO$ intersects the sides $AB$ and $AC$ in points $X$ and $Y$ correspondingly, so that the point $H$ lies on the segment $OX$. It turned out that $XH = HO = OY$. Find the angle $\\angle BAC$.\n\n(Oleksii Masalitin)",
    "options": [],
    "answer": "60°",
    "solution": "In the solution of this problem, we will be using the following well-known fact: in any triangle, the distance from the vertex to the orthocenter is twice larger than the distance from the circumcenter to the opposite side.\n\nDenote by $O_B$ and $H_B$ the projections of $O$ and $H$ correspondingly onto the line $AC$, define points $O_C$ and $H_C$ similarly (fig. 9). From the fact above it follows that $BH = 2OO_B$ and $CH = 2OO_C$. Also note, that $OO_B \\parallel HH_B$ and $HO = OY$, and therefore $OO_B$ is the midline of $\\triangle YHH_B$, so $HH_B = 2OO_B = BH$. Similarly $HH_C$ is the midline of $\\triangle XOO_C$, so $HH_C = \\frac{1}{2}OO_C = \\frac{1}{4}CH$. As the quadrilateral $BHC_HC_B$ is cyclic, $BH \\cdot HH_B = CH \\cdot HH_C$, and $BH^2 = 4HH_C^2$, so $BH = 2HH_C$, and $\\angle HBH_C = 30^\\circ$. Now it's clear that $\\angle BAC = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45797,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of coprime integers $m$ and $n$ equals $90$. What largest possible value can be attained by the product $mn$ of these integers?",
    "options": [],
    "answer": "2021",
    "solution": "Make the following transformations:\n$$\nmn = \\left(\\frac{m+n}{2}\\right)^2 - \\left(\\frac{m-n}{2}\\right)^2 = 45^2 - \\left(\\frac{m-(90-m)}{2}\\right)^2 = 2025 - (m-45)^2 \\rightarrow \\max.\n$$\nSo the max will be achieved, when the value of the expression $(m - 45)^2$ is minimized, but here we have to not forget that $m, n$ are coprime.\n\n$m = 45$, then $n = 45$ and numbers $m, n$ aren't coprime.\n\n$m = 46$, then $n = 44$ and numbers $m, n$ aren't coprime.\n\n$m = 47$, then $n = 43$ and numbers $m, n$ are coprime, as required. So the largest value of the product is $2025 - (47 - 45)^2 = 2021$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45798,
    "subject": "Mathematics (Multi-modal)",
    "question": "In one magic country there are only banknotes of nominal 3, 25 and 80 hryvnyas. Businessman Victor ate in a restaurant of this country for 2024 days in a row, and each day he paid (without change) exactly 1 hryvnya more than the previous one. Is it possible that he paid exactly a million banknotes?",
    "options": [],
    "answer": "No",
    "solution": "Suppose that he paid the sum of $S$ UAH with $k$ banknotes, among which there are $a$ of 3 UAH, $b$ of 25 UAH, and $c$ of 80 UAH. Then\n$$\nS = 3a + 25b + 80c \\equiv 3a + 3b + 3c = 3k \\pmod{11}.\n$$\nIf $n+1$ is the sum that he paid in the first day, then in the $i$-th day he paid a sum of $n+i$ with precisely $k_i$ banknotes. From the above formula it follows that $3(k_1 + k_2 + \\cdots + k_{2024}) \\equiv (n+1) + (n+2) + \\cdots + (n+2024) = 2024n + 1012 \\cdot 2025 \\pmod{11}$.\nAs $1012 \\equiv 11$, the right side is divisible by 11, so the left side, equal to the total number of pair banknotes, also is divisible by 11. But the number 1000000 isn't divisible by 11, so Victor couldn't have used exactly a million banknotes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45799,
    "subject": "Mathematics (Multi-modal)",
    "question": "We chose several numbers among $1, 2, \\ldots, 2022$. It turned out that sum of any two of the chosen numbers isn't divisible by the difference between any two of the chosen numbers. What largest possible number of numbers could be selected?\n\n(Oleksii Masalitin)",
    "options": [],
    "answer": "674",
    "solution": "Suppose that more than $674$ numbers were chosen. Then there exists a triple $3n + 1, 3n + 2, 3n + 3$, among which at least two numbers were chosen, so the absolute difference between some two of the chosen numbers doesn't exceed $2$. Clearly, there exist some two chosen numbers with the same parity, so their sum will be divisible by that difference not exceeding $2$. So, not more than $674$ numbers were chosen.\n\nLet's show that we can choose a given number of numbers. Consider the set $\\{1, 4, 7, \\ldots, 2020\\}$, consisting of $674$ numbers. As the difference between any two of these numbers is divisible by $3$, and the sum of any two of these numbers isn't divisible by $3$, this set satisfies the conditions from the statement.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45800,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given $n \\ge 4$ positive real numbers. It turned out that their $\\frac{n(n-1)}{2}$ pairwise products form an arithmetic progression in some order. Prove that all of these numbers are equal.\n\n(Anton Trygub)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If some two products are equal, then all products are equal, and all numbers are equal. If some two numbers are equal, then some products are equal, so all numbers are equal. Now consider 4 largest numbers $a < b < c < d$. The largest two products are $cd, bd$. Then the difference of the progression is $cd - bd$. But then $ac - ab = a(c - b) < d(c - b)$, so the difference between some two elements of the progression is smaller than the difference of the progression, which is impossible, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's call a positive integer square-free, if it's not divisible by $p^2$ for any prime $p$. You are given a squarefree integer $n > 1$, which has precisely $d$ positive divisors. What largest number of divisors of this number can you choose, so that for any two of them, let's denote them by $a$ and $b$, the number $a^2 + ab - n$ isn't a square of an integer?\n(Oleksii Masalitin)",
    "options": [],
    "answer": "d/2",
    "solution": "**Answer:** $\\frac{d}{2}$\n\nAs $n > 1$ is squarefree, it can't be a square of an integer. Then all divisors of $n$ can be split into pairs $(t_1, t_2)$, $(t_3, t_4)$, ..., $(t_{d-1}, t_d)$ in such a way, that the product of numbers in each pair is $n$. If we choose $(a, b)$ from the same pair, then $a^2 + ab - n = a^2$, contradiction. So, we chose at most one number from each pair, so we selected at most $\\frac{d}{2}$ in total.\n\nLet's show that we can always choose $\\frac{d}{2}$ integers. Consider any prime divisor $p$ of $n$ and choose all divisors of $n$, which are divisible by $p$. It's easy to see that there are $d/2$ of them. Let's show that they satisfy the condition from the statement. Let $a = kp, b = lp$ be any divisors from the chosen group, and $n = pt$. Note that $l, k, t$ aren't divisible by $p$. Then $a^2 + ab - n = p(pk^2 + pkl - t)$ and the number in brackets isn't divisible by $p$, as $t$ isn't divisible by $p$. So, $a^2 + ab - n$ is divisible by $p$, but not by $p^2$, and, therefore, can't be a square of an integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45802,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any non-negative real numbers $x$ and $y$ show the inequality:\n$$\nx^2 y^2 + x^2 y + xy^2 \\leq x^4 y + x + y^4.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $x = 0$ or $y = 0$, the inequality is trivial. Now assume $x, y > 0$. Divide both parts of the inequality by $xy$. We get:\n$$\nxy + x + y \\leq x^3 + \\frac{1}{y} + \\frac{y^3}{x}.\n$$\nApply the famous inequality, which is a consequence of the Cauchy-Schwarz inequality:\n$$\n\\frac{a_1^2}{b_1} + \\frac{a_2^2}{b_2} + \\cdots + \\frac{a_n^2}{b_n} \\geq \\frac{(a_1 + a_2 + \\cdots + a_n)^2}{b_1 + b_2 + \\cdots + b_n}\n$$\n$$\nx^3 + \\frac{1}{y} + \\frac{y^3}{x} \\geq \\frac{(x^2)^2}{x} + \\frac{1^2}{y} + \\frac{(y^2)^2}{xy} \\geq \\frac{(x^2 + 1 + y^2)^2}{x + y + xy} \\geq \\frac{(x + y + xy)^2}{x + y + xy} = xy + x + y,\n$$\nwhich completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45803,
    "subject": "Mathematics (Multi-modal)",
    "question": "When dividing with remainder some four consecutive positive integers by some three-digit integer it turned out, that the sum of these four remainders is equal to $983$. Find the remainder under the division of the smallest of these four numbers by $109$.",
    "options": [],
    "answer": "108",
    "solution": "Denote these four consecutive integers by $n$, $n+1$, $n+2$ and $n+3$. Denote the three-digit number that we were dividing by as $b$. Let $n = bq + r$. Consider possible values of $r$.\n\nIf $r \\le b - 4$, then these remainders are $r$, $r+1$, $r+2$ and $r+3 \\Rightarrow r + r+1 + r+2 + r+3 = 983 \\Rightarrow 4r = 973$, contradiction.\n\nIf $r = b - 3$, then these remainders are $r$, $r+1$, $r+2$ and $0 \\Rightarrow r + r+1 + r+2 + 0 = 983 \\Rightarrow 3r = 980$, contradiction.\n\nIf $r = b - 2$, then these remainders are $r$, $r+1$, $0$ and $1 \\Rightarrow r + r+1 + 0 + 1 = 983 \\Rightarrow 2r = 981$, contradiction.\n\nIf $r = b - 1$, then these remainders are $r$, $0$, $1$ and $2 \\Rightarrow$\n$$\nr + 0 + 1 + 2 = 983 \\Rightarrow r = 980 \\Rightarrow b = r + 1 = 981.\n$$\nIt remains to find the required remainder:\n$$\nn = bq + r = 981q + 980 = 109(9q + 8) + 108.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a trinomial $f(x) = ax^2 + bx + c$ with integer coefficients such that $a$ isn't divisible by $2022$ and all numbers $f(1), f(2), \\ldots, f(2022)$ give different remainders under the division by $2022$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the following trinomial:\n$$\nf(x) = 1011x^2 + 1012x = 1011x(x + 1) + x.\n$$\nThe first term is always divisible by $2022$, so $f(x)$ gives remainder $x$. Proof completed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45805,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive real numbers $a$, $b$, $c$, which satisfy the condition $ab + bc + ca = 1$, prove the following inequality:\n$$\n\\left(\\sqrt{bc} + \\frac{1}{2a + \\sqrt{bc}}\\right) \\cdot \\left(\\sqrt{ca} + \\frac{1}{2b + \\sqrt{ca}}\\right) \\cdot \\left(\\sqrt{ab} + \\frac{1}{2c + \\sqrt{ab}}\\right) \\ge 8abc.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the statement it follows, that\n$$\n\\frac{1}{2a + \\sqrt{bc}} = \\frac{ab + bc + ca}{2a + \\sqrt{bc}} = \\frac{bc + a(b + c)}{2a + \\sqrt{bc}} \\ge \\frac{bc + 2a\\sqrt{bc}}{2a + \\sqrt{bc}} = \\sqrt{bc},\n$$\nimplying\n$$\n\\sqrt{bc} + \\frac{1}{2a + \\sqrt{bc}} \\ge 2\\sqrt{bc}.\n$$\nIf we write two similar inequalities\n$$\n\\sqrt{ca} + \\frac{1}{2b + \\sqrt{ca}} \\ge 2\\sqrt{ca},\n$$\n$$\n\\sqrt{ab} + \\frac{1}{2c + \\sqrt{ab}} \\ge 2\\sqrt{ab},\n$$\nand multiply them, we get the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ for which it's possible to cut the square into $2n$ squares of two different sizes: $n$ squares of one size and $n$ more of other size.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "9",
    "solution": "The example of cutting into $2n = 18$ squares is provided in the fig. 10.\n\nSuppose that there exists an example for $n \\le 8$. Denote the sides of the squares by $a, b$, and of the large square as $N$. Let's first show that it's possible to represent $N$ as the sum $ka + bl$, where $k, l \\le n$, at least in two different ways.\n\nClearly, at least one such representation has to exist. Let $N = k_1a + l_1b$, then any vertical/horizontal line, not passing through any side of any square, intersects precisely $k_1$ squares with side $a$ and precisely $k_2$ squares with side $b$.\n\nChoose any $x$ and draw vertical lines on distances $x, x+a, x+2a, \\ldots$ from the left side of the large square ($x$ has to be such that no such line passes through any side of any of the squares). Each such line will intersect precisely $k_1$ squares with size $a$, so the large square is intersected by precisely $\\frac{n}{k_1}$ lines for any choice of $x$, but it's possible only if $N = \\frac{n}{k_1}a$. Similarly, $N = a\\frac{n}{l_1}$. So, we got two representations of $N$, contradiction.\n\nNow that we know that such $k_2, l_2$ exist, so we get $a(k_1 - k_2) = b(l_2 - l_1)$, so we can zoom in and let $a = l, b = k$ for some $l, k \\le n$. So, from now on, $1 \\le a, b \\le n, a \\ne b, a, b, N$ are positive integers, and their largest common divisor $(a, b) = 1$.\n\nWrite the condition on the area: $n(a^2 + b^2) = N^2$.\n\nFor $n = 3, 6, 7$ this equation doesn't have a solution: as $n$ has a prime divisor of form $p = 4k-1$, we get $N^2 \\mid p$, so $(a^2 + b^2) \\mid p$, so $a$ and $b$ are divisible by $p$, but they are coprime.\n\nFor $n = 1, 2$ we can just check all pairs $(a, b)$ which satisfy the conditions above and see that for no of them the value of $n(a^2 + b^2)$ is a perfect square.\n\nFor $n = 4$ the only such pair is $(a, b) = (3, 4)$. Then $N = 10$, but $10$ can be represented as $3a + 4b$ in a unique way, so it's wrong.\n\nFor $n = 5$ the only such pair is $(a, b) = (1, 2)$, then $N = 5$, but from the square $5 \\times 5$ we can't cut out $5$ squares $2 \\times 2$, as each of them contains at least one of the cells $(2,2), (2,4), (4,2), (4,4)$.\n\nFor $n = 8$ the only such pair – $(1, 7)$, then $N = 20$, but from the square $20 \\times 20$ we can't cut out $8$ squares $7 \\times 7$, as each of them contains at least one of the cells $(7,7), (7,14), (14,7), (14,14)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45807,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the left-lower corner $1 \\times 1$ cell of the $m \\times n$ board ($m, n \\geq 3$) there is a black chip, and in the left-upper and right-lower cells, there are white chips. In his move, Petrik can move the black chip to the adjacent by side cell twice in a row, and Vasyl can either move one of the white chips to the adjacent by side cell twice in a row or move each of the two white chips separately to the adjacent by side cells. Chips can not be placed on fields that have already been visited by a chip of another color. Vasyl wins if after some point both white chips end up in the same cell. If Vasyl wants to win, and Petrik wants not to give him such an opportunity, then under what conditions on $m$ and $n$ does Vasily win? Petrik moves first.\n*(Arseniy Nikolaev)*",
    "options": [],
    "answer": "Vasyl wins if and only if m and n have the same parity; otherwise Petrik wins.",
    "solution": "**Answer:** if $m$ and $n$ have the same parity, Vasyl wins, else Petrik wins.\n\nWlog the number of rows doesn't exceed the number of columns, so $m \\leq n$. Enumerate the rows from bottom to top as 1, 2, ..., $m$, and columns from left to right as 1, 2, ..., $n$, then each cell has two coordinates – numbers of the row and the column. Let's denote the chip that is initially in the left-upper cell with coordinates $(m, 1)$ by A, and the other chip – in the cell $(1, n)$ – by B.\n\n*Let's show that if $m$ and $n$ have the same parity, then Vasyl has a winning strategy.*\n\nVasyl can act as follows: first, he moves $\\frac{1}{2}(n-m)$ times to the right with a chip A twice in a row. After that, the white chips will be in the opposite corners of the square $m \\times m$. After this each time he will make 1 move with chip A to the right, and with chip B – 1 one move up. Let's show that in this case white chips will end up in the same cell with coordinates $(m, n)$, so black chip wouldn't be able to prevent this. For this, we will show that the black chip won't ever be able to get to any cell of the top row and to any cell of the right column. We will show this by contradiction. Suppose it's not true.\n\n**Case 1.** Black chip got to the top row. Chip A at this point has made $x$ double and $y$ unary moves. Then black chip made $(x + y + 1)$ moves, each of which is moving a chip twice. So the sum of its coordinates could change by at most $2(x + y + 1)$. As it ended up in the top row, the first coordinate of its cell is $m$, so it is in the column with number at most $2 + 2(x + y + 1) - m$. A, however, also is in the top row in a column with number $1 + 2x + y$. By assumption the chip A has to stand to the left of the black chip, as otherwise black chip wouldn't be able to get to that cell. So, the following inequality has to hold:\n$$\n1 + 2x + y < 2 + 2(x + y + 1) - m \\Rightarrow y + 3 > m \\Rightarrow m - 2 \\leq y\n$$\nBut this inequality means that the white chip A is at least in the cell $(m, n - 1)$, and the white cell B is at least in the cell $(m - 1, n)$, so the black chip has to be in a cell $(m, n)$. But then at least on one of the cells $(m, n-1)$ and $(m-1, n)$ were both black and white chips, contradiction, which completes the first case.\n\n**Case 2.** The black chip got to the rightmost column. Suppose that it made $k$ double moves. Then the sum of coordinates of the black chip is at most $2+2k$, and as it is in the rightmost column, its first coordinate equals $n$. So the number of its row doesn't exceed $2+2k-n$. Let's count the number of row of **B**. For the first $\\frac{1}{2}(n-m)$ moves it hasn't moved, and for the rest $2k-\\frac{1}{2}(n-m)$ moves it was moving by one cell up at a time. So, its row number is $1+2k-\\frac{1}{2}(n-m)$. By assumption, **chip B** has to be in a row with a smaller number than the black chip. So, the following inequality has to hold:\n$$\n2+2k-n > 1+2k-\\frac{1}{2}(n-m) \\Rightarrow 1 > \\frac{1}{2}(n+m),\n$$\nwhich is impossible.\n\nSo, under any conditions, the white chips will meet at the corner and Vasyl wins.\n\nNow let's show that if $m$ and $n$ have different parity, then Petrik has a winning strategy. With a first move, he will move the first chip one cell to the right and one cell up. Next, he will act by the following algorithm, while possible: if Vasyl moves chip **A(B)** twice, Petrik will move his chip to the right(up) twice, and if he moves both chips, Petrik will move his chip one cell up and one cell to the right.\n\n![](attached_image_1.png)\nFig. 7\n\n**Lemma.** The algorithm of moves of Petrik has the following properties:\nIf Petrik can perform the corresponding move, then after this move the black chip is always to the right of the cell **A** by columns and always higher to the cell **B** by rows, and the white cells haven't yet been to that part of the board (to the right and up from the black chip);\nPetrik can't make a move only if the black chip has reached the edge of the board, or white chips are in the adjacent cells above and to the right of it.\n\n![](attached_image_2.png)\nFig. 8\n\n**Proof.** We will do this by induction. After the first move, the statement is trivial. Suppose that Petrik did $k$ double moves. If Vasyl makes a double move with one chip (fig. 7), then Petrik can make the analogous move, without intersecting the trajectory of any white cell. For example, **B** moved up two cells, then Petrik also moves up two cells in other column, where white cells weren't before. If both white chips move by one cell, Petrik will also be able to go according to the algorithm (fig. 8). It's easy to see that after the $(k+1)$-st move the invariant is saved, so the lemma is proved.\n\nIn the case, when white chips surround the black chip in such a way, the black chip has coordinates $(k, l)$, and white have coordinates $(k, l+1)$ and $(k+1, l)$, and the value of the invariant is:\n$$\n((k+1+l)+(k+l+1)-(k+l)) \\equiv 0 \\pmod{2}.\n$$\nWe see the contradiction, which shows that this situation isn't possible for such values of $(m, n)$, which completes the proof.\n\nFrom this lemma, if Petrik achieved the border, he separated white cells with his trajectory, so they won't be able to meet. Let's show that the situation, when both white cells are adjacent to the black chip is impossible. The parity of the sum of coordinates of white chips minus the coordinates of the black chip is an invariant, as after each move two coordinates change by 1, or one coordinate changes by 2. At the beginning of the game we have the following value of this invariant:\n$$\n((1+n)+(1+m)-(1+1)) = (m+n) \\equiv 1 \\pmod{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which number is larger, $A = \\frac{1}{9} : \\sqrt[3]{\\frac{1}{2023}}$ or $B = \\log_{2023} 91125$?",
    "options": [],
    "answer": "B",
    "solution": "To prove this, we will show that the following inequalities hold: $A < \\frac{3}{2} < B$.\n\n$$\nA = \\frac{1}{9} : \\sqrt[3]{\\frac{1}{2023}} = \\frac{\\sqrt[3]{2023}}{9} < \\frac{13}{9} < \\frac{3}{2}, \\quad B = \\log_{2023} 91125 > \\log_{2023} 45^3 = \\log_{45^2} 45^3 = \\frac{3}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the integer closest to the value of the expression:\n$$\n((7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023})^2 - ((7 + \\sqrt{48})^{2023} - (7 - \\sqrt{48})^{2023})^2.\n$$",
    "options": [],
    "answer": "4",
    "solution": "Let's transform the given expression as follows:\n$$\n((7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023})^2 - ((7 + \\sqrt{48})^{2023} - (7 - \\sqrt{48})^{2023})^2 = \\\\\n= ((7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023} + (7 + \\sqrt{48})^{2023} - (7 - \\sqrt{48})^{2023}) \\cdot \\\\\n\\cdot ((7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023} - (7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023}) = \\\\\n= (2(7 + \\sqrt{48})^{2023}) \\cdot (2(7 - \\sqrt{48})^{2023}) = 2^2(49 - 48)^{2023} = 4.\n$$\nTherefore, the expression equals $4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the integer that is closest to the value of the expression:\n$$\n\\left( (3 + \\sqrt{1})^{2023} - \\left( \\frac{1}{3 - \\sqrt{1}} \\right)^{2023} \\right) \\cdot \\left( (3 + \\sqrt{2})^{2023} - \\left( \\frac{1}{3 - \\sqrt{2}} \\right)^{2023} \\right) \\cdot \\left( (3 + \\sqrt{3})^{2023} - \\left( \\frac{1}{3 - \\sqrt{3}} \\right)^{2023} \\right) \\dots \\left( (3 + \\sqrt{8})^{2023} - \\left( \\frac{1}{3 - \\sqrt{8}} \\right)^{2023} \\right).\n$$",
    "options": [],
    "answer": "0",
    "solution": "Let's consider the last factor:\n$$\n\\frac{1}{3 - \\sqrt{8}} = \\frac{3 + \\sqrt{8}}{9 - 8} = 3 + \\sqrt{8} \\Rightarrow (3 + \\sqrt{8})^{2023} = \\left(\\frac{1}{3 - \\sqrt{8}}\\right)^{2023},\n$$\nwhich means that the last factor equals $0$, and therefore the whole product equals $0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ that satisfy the inequalities:\n$$\n-46 \\le \\frac{2023}{46-n} \\le 46-n.\n$$",
    "options": [],
    "answer": "1 and 90",
    "solution": "Let $x = 46 - n$ and find the corresponding integer values of $x$ that satisfy the condition $-46 \\le \\frac{2023}{x} \\le x$. We consider two cases.\n\nCase 1. $x > 0$. The left inequality holds for all the mentioned values of $x$, and the right one can be rewritten as:\n$$\n2023 \\le x^2 \\Rightarrow x \\ge 45 \\Rightarrow 46 - n \\ge 45 \\Rightarrow n \\le 1 \\Rightarrow n = 1.\n$$\n\nCase 2. $x < 0$. The inequality can be rewritten as:\n$$\nx^2 \\le 2023 \\le -46x \\Rightarrow -44 \\le x \\le 44 \\text{ and } x \\le -44 \\Rightarrow x = -44 \\Rightarrow 46 - n = -44 \\Rightarrow n = 90.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45812,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$ is one of the points of intersection of circles $\\omega_1$ and $\\omega_2$, $T_1T_2$ is the external common tangent to these circles, which tangents $\\omega_1$ at $T_1$ and tangents $\\omega_2$ at $T_2$. $B_1$ is any point on the circle $\\omega_1$, $B_2$ is any point on the circle $\\omega_2$, such that $A$, $B_1$ and $B_2$ are not collinear. Circumcircle of the $\\triangle AB_1B_2$ intersects lines $T_1B_1$, $T_2B_2$ at points $C_1, C_2$ respectively. Prove that $C_1T_2$ and $C_2T_1$ intersect on $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X$ be the second intersection point of $\\omega$ and $(AT_1T_2)$. Observe that $\\angle (AX, XC_1) = \\angle (AB_1, B_1C_1) = \\angle (AB_1, T_1B_1) = \\angle (AT_1, T_1T_2) = \\angle (AX, XT_2)$ and then $X \\in C_1T_2$. Similarly, $X \\in C_2T_1$ and hence $C_1T_2$ and $C_2T_1$ intersect on $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $w_1$ and $w_2$ intersect at the points $P$ and $Q$. Suppose $AB$ and $CD$ are parallel diameters of the circles $w_1$ and $w_2$ respectively. Moreover, none of the points $A, B, C, D$ coincides with $P$ or $Q$, and the points are located on the circles in the following order: $A, B, P, Q$ on $w_1$ and $C, D, P, Q$ on $w_2$. Lines $AP$ and $BQ$ intersect at point $X$ and lines $CP$ and $DQ$ intersect at $Y \\neq X$. Prove that, regardless of the choice of parallel diameters $AB$ and $CD$, all the lines $XY$ intersect at one point or are parallel.\n(Nazar Serdiuk)\n\n![](attached_image_1.png)\nFig. 26",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O_1$ be the center of circle $w_1$ and $X' = BP \\cap AQ$ as on Fig. 26. Since $AP \\perp BP$ and $BQ \\perp AQ$, then $AP$ and $BQ$ are heights of the triangle $ABX'$, and $X$ is the orthocenter. Then $X'X$ is also a height in the triangle, therefore $X'X \\perp AB$ and $\\angle PX'X = 90^\\circ - \\angle ABX' = \\angle PAB$. This implies that triangles $\\triangle APB$ and $\\triangle X'PX$ have the same angles, so there is a rotation and homothety that transform triangle $\\triangle APB$ into $\\triangle X'PX$. The center of transformations is at $P$, the rotation is by $90^\\circ$ and the homothety coefficient is $\\frac{XP}{BP} = \\angle XBP = \\frac{1}{2} \\angle QO_1P$.\n\nDenote by $M_1$ the midpoint of $XX'$. The segments $PO_1$ and $PM_1$ are medians of the triangles $\\triangle APB$ and $\\triangle X'PX$ respectively, therefore $PM_1 \\perp PO_1$, that is, $PM_1$ is tangent to $w_1$. Similarly, we get that $QM_1$ is tangent to $w_1$ as well. Consequently, $M_1$ is the intersection of tangents to $w_1$ drawn from $P$ and $Q$ that are not dependent on the choice of points $A$ and $B$. Note that the length of the segment\n\n$$\nXX' = AB \\cdot \\frac{1}{2} \\angle QO_1P\n$$\ndoes not depend on $A$ and $B$ as well, because the length of diameter $AB$ and the angle $\\angle QO_1P$ are constant.\n\nAnalogously, we construct the point $M_2$ for the circle $w_2$. Assume that $T = XY \\cap M_1M_2$ (the case when $XY \\parallel M_1M_2$ will be considered further). We will show that point $T$ is fixed, which implies that all the lines $XY$ intersect at one point regardless of the choice of diameters $AB$ and $CD$.\n\nNote that $M_1X$ and $M_2Y$ are parallel since both are perpendicular to $AB \\parallel CD$, that is why\n$$\n\\frac{M_1T}{M_2T} = \\frac{M_1X}{M_2Y}.\n$$\nThe points $M_1$ and $M_2$ as well as the ratio $\\frac{M_1X}{M_2Y} = \\frac{\\frac{XX'}{2}}{\\frac{YY'}{2}} = \\frac{XX'}{YY'}$ are fixed, therefore $T$ is also fixed.\n\nThe lines $XY$ and $M_1M_2$ are parallel and do not coincide if and only if segments $M_1X$ and $M_2Y$ are equal and parallel. These properties hold as well with different choices of the diameters $AB$ and $CD$. Hence, if $XY$ and $M_1M_2$ are parallel it will imply that $XY$ is parallel to $M_1M_2$ for all other initializations of diameters $AB$ and $CD$.\n\nThe lines $XY$ and $M_1M_2$ coincide if and only if $AB \\parallel CD \\perp M_1M_2$. In this case one may consider another initialization of the diameters $AB$ and $CD$, that will determine in which of two cases considered above we are. Note, that the initial line $XY = M_1M_2$ satisfies both of the considered cases.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $k$. The product of some $k$ consecutive positive integers ends with the number $k$. What value can the number $k$ attain?",
    "options": [],
    "answer": "1, 2, 4",
    "solution": "**Answer:** $k \\in 1, 2, 4$\n\nSuppose that $k \\ge 5$. It is clear that among any $k$ consecutive numbers, there is one that is divisible by $5$, and one that is divisible by $2$, so their product ends in $0$, hence $k$ is divisible by $10$. It is clear that then $k \\ge 10$, so in the product of $k$ consecutive numbers there are at least two numbers divisible by $5$ and at least two numbers divisible by $2$, and one of them is divisible by $k$. Hence, the product is divisible by $10k$, so there will be more zeros at the end than zeros at the end of $k$, which is where we get the contradiction. Thus, $k \\le 4$.\n\nIf $k = 3$, then the product will be even, but it cannot end in the digit $3$.\n\nFor $k = 1, 2, 4$ it is enough to consider the following examples: $1$, $1 \\cdot 2 = 2$, $1 \\cdot 2 \\cdot 3 \\cdot 4 = 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45815,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a rectangular board $100 \\times 300$ two people take turns coloring the uncolored cells. The first one paints yellow, the second one paints blue. The coloring is completed when every cell on the board is colored. A *sequence of* cells is a set of cells in which two consecutive cells share a common side (all cells in the sequence are different). Consider all possible sequences of yellow cells. The *result* of the first player is the number of cells in the sequence of yellow cells of maximum length. The first player's goal is to maximize the result, and the second player's goal is to make the first player's result as small as possible. Prove that if each player strives to achieve his goal, the first player's result will be no more than $200$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's divide the entire board into vertical dominoes, whose larger sides are parallel to the larger side of the board. Let's consider the second player's strategy: he paints the second square of the domino whose first square was painted by the first player in the previous move. Then it is clear that no sequence of yellow cells crosses the horizontal grid lines of the rectangular board, which divide the dominoes in half. So under such conditions, the sequence with the maximum number of yellow cells can contain only the adjacent two rows of our board, that is, its result will not exceed $200$, which is what we needed to prove.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of the triangle $ABC$, and $P$ be any point on the arc $BAC$ of the circumscribed circle. On the tangent to the circumscribed circle $\\omega$ of the triangle $API$ at the point $I$ points $K$ and $L$ were selected so that $BK = KI$ and $CL = LI$. Prove that the circumcircle of the triangle $PKL$ is tangent to $\\omega$.\n(Mykhailo Shtandenko)\n\n![](attached_image_1.png)\nFig. 13\n\n![](attached_image_2.png)\nFig. 14",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Mark the midpoints of the arcs $AB$, $AC$, $BC$ that do not contain other points, by the points $W_C$, $W_B$, $W_A$ respectively (fig. 13). It is clear that $K$ and $L$ lie on $W_A W_C$ and $W_A W_B$ respectively. Then.\n![](attached_image_1.png)\nFig. 13\n$$\n\\begin{aligned}\n\\angle(KL, PI) &= \\angle(AI, AP) = \\angle(AW_A, AP) = \\angle(W_A W_C, W_C P) = \\\\\n&= \\angle(KW_C, W_C P) = \\angle(W_A W_B, W_B P) = \\angle(LW_B, W_B P),\n\\end{aligned}\n$$\n\n*Archimedes' Lemma.* If a circle is inscribed in a segment of another circle bounded by the chord $BC$ and touches the arc at the point $A_1$ and the chords at the point $A_2$ then the line $A_1A_2$ is a bisector $\\angle BA_1C$ (fig. 14).\n![](attached_image_2.png)\nFig. 14",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(x_n)$ be a sequence of real numbers from the interval $(0; 1)$. The sequence of positive integers $(a_n)$ is defined as follows: $a_1 = 1$, $a_{i+1} = m$, where $m$ is the smallest positive integer for which $[x_1 + x_2 + \\cdots + x_m] = a_i$. Prove that for any indexes $i, j$ the inequality $a_{i+j} \\ge a_i + a_j$ holds.\n(Nazar Serdyuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that we can put $a_0 = 0$, then we would get $a_1 = 1$. First, let's prove the following lemma:\n\n**Lemma.** For all $i \\in \\mathbb{N}$ the condition holds: $a_{i+1} - a_i \\ge a_i - a_{i-1}$.\n\n*Proof:*\n$$\nx_1 + x_2 + \\cdots + x_{2a_i - a_{i-1} - 1} = (x_1 + x_2 + \\cdots + x_{a_i}) + (x_{a_i + 1} + x_{a_i + 2} + \\cdots + x_{2a_i - a_{i-1} - 1}) < \\\\\n< (a_{i-1} + 1) + (1 + 1 + \\cdots + 1)_{\\text{($a_i - a_{i-1} - 1$ terms)}} < (a_{i-1} + 1) + (a_i - a_{i-1} - 1) = a_i.\n$$\nThus, $a_{i+1}$ there must be at least $2a_i - a_{i-1}$.\n\nLet's choose some positive integer $t$. First, write down the inequalities of the form $a_{i-k} - a_{i-k-1} \\ge a_{i-k-1} - a_{i-k-2}$, $k = 0, \\ldots, t-1$.\nAdd up all these inequalities: $a_i - a_{i-t} \\ge a_{i-1} - a_{i-t-1}$ and then we have that\n$$\na_i - a_{i-t} \\ge a_{i-1} - a_{i-t-1} \\ge a_{i-2} - a_{i-t-2} \\ge \\cdots \\ge a_t - a_0.\n$$\nNow we just put $i \\to i+j$, $t \\to j$ and get the desired inequality from the problem: $a_{i+j} \\ge a_i + a_j$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45818,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex pentagon $ABCDE$ the following conditions are satisfied: $AB \\parallel CD$, $BC \\parallel DE$ and $\\angle BAE = \\angle AED$. Prove that $AB + BC = CD + DE$.\n\n![](attached_image_1.png)\nFig. 16",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the rays $AB$ and $DE$ intersect at the point $O$ (fig. 16). Then $BCDO$ is a parallelogram, and $\\Delta AOE$ is isosceles, since $\\angle OAE = \\angle AEO$. So,\n$$\nAB + BC = (OB - OA) + BC = CD - OE + OD = CD + DE.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$, $Q(x)$ and $R(x)$ be polynomials with integer coefficients, such that the equality holds: $P(x) = Q(x)R(x)$. We denote by $a$ and $b$ the maximum of the absolute values of coefficients of the polynomials $P(x)$ and $Q(x)$ respectively. Does the condition $b \\le 2023a$ always hold?",
    "options": [],
    "answer": "No",
    "solution": "**Answer:** No, not necessarily.\n\nAs an example, consider the following two polynomials:\n\n$$\nQ(x) = 1 + 2x + 3x^2 + 4x^3 + \\cdots + 2023x^{2022} + 2024x^{2023} + 2023x^{2024} + \\cdots + x^{4046}, \\quad R(x) = x - 1,\n$$\n$$\n\\text{then } P(x) = Q(x)(x-1) = -1 - x - x^2 - \\cdots - x^{2023} + x^{2024} + x^{2025} + \\cdots + x^{4047}.\n$$\nThen $a = 1$ and $b = 2024$, with $b > 2023a$.\n\n**Alternative solution.** Consider the following two polynomials:\n$$\nP(x) = (x^3 - 1)^N, \\quad Q(x) = (x^2 + x + 1)^N.\n$$\nBy the construction of $P(x) : Q(x)$, that is, the polynomial $R(x)$ exists. Since\n$$\nP(x) = (x^3 - 1)^N = \\sum_{j=0}^{N} C_N^j x^{3j} \\Rightarrow a = C_N^j \\text{ for some } j = 0, N \\Rightarrow a = C_N^j < \\sum_{j=0}^{N} C_N^j = (1+1)^N = 2^N.\n$$\nThe sum of the coefficients of the polynomial $Q(x)$ is equal to $Q(1) = (1 + 1 + 1)^N = 3^N$ and the number of coefficients is $2N + 1$, because it has degree $2N$. Therefore, according to Dirichlet's principle, there is a coefficient at least\n$$\n\\frac{3^N}{2N + 1} \\Rightarrow b \\ge \\frac{3^N}{2N + 1}.\n$$\nIt is clear that with sufficiently large $N$ inequality will hold:\n$$\n\\frac{b}{a} \\ge \\frac{3^N}{2^N(2N + 1)} > 2023.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45820,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given $n \\ge 2$ distinct positive integers. For each pair $a < b$ of these numbers, consider the difference $b - a$. For each of these differences, Vlada writes down the maximum power of two by which this difference is divisible. What is the largest possible number of distinct numbers that Vlada could write?\n(Oleksii Masalitin)",
    "options": [],
    "answer": "n - 1",
    "solution": "We will prove that there are at most $n-1$ different degrees with induction by $n$. The base case for $n=2$ is obvious, let's prove the transition. Let the statement be proved for $k \\le n-1$, let us prove it for $n$ numbers. Suppose that all numbers are divisible by the same power of $2$: $2^k$. If $k > 0$, then we divide these numbers by $2^k$, it is clear that the degree of occurrence of two in all differences has decreased by $k$, so the number of distinct among them has not changed. If all the numbers are odd, then we can add 1, then none of the differences will change, and the numbers will all become even. We will repeat this operation. It is clear that the only odd number that does not decrease after the operations of addition 1 and division by the maximum degree of occurrence of two is 1. Since there are more than one number, the sum of these numbers will decrease after each pair of operations, so sooner or later we will get a set of numbers that are divisible by different degrees of twos.\n\nLet's divide the numbers into groups $A_1, A_2, \\dots, A_m$, according to the degree of occurrence of two (the degrees are equal to $\\alpha_1 < \\alpha_2 < \\dots < \\alpha_m$). It is clear that the difference $a_i - a_j$ for $a_i \\in A_i, a_j \\in A_j$ ($i \\ne j$) is divisible by $2^{\\min(a_i, a_j)}$, therefore, among the differences between different groups there are only $m-1$ different degrees. For each of these groups, the induction assumption can be applied, so the number of different powers of the differences of numbers from $A_i$ does not exceed $|A_i| - 1$. Thus, the number of different degrees of occurrence is no more than\n$$\n|A_1| - 1 + |A_2| - 1 + \\cdots + |A_m| - 1 + m - 1 = |A_1| + |A_2| + \\cdots + |A_m| - 1 = n - 1.\n$$\n\nIt remains to show that $n - 1$ different degrees are possible for any $n \\ge 2$. Indeed, let us choose $a_1 = 1, a_2 = 2^2 + 1, a_3 = 2^3 + 1, ..., a_n = 2^n + 1$. Then, the pairs $(a_1, a_i)$ for $i = 2, n$ have different degrees.\n\nAn alternative solution.\nWe prove that there are at most $n-1$ different degrees of occurrence. Indeed, let there be $n$ pairs in which the degrees of occurrence of 2 in the difference of the pair's numbers are pairwise different. Then consider the following graph with $n$ vertices. Each number is assigned a vertex, and two vertices are connected by an edge if their corresponding numbers form one of the considered $n$ pairs. It is clear that the total number of edges $n$, so the graph in question has a cycle. Then let the numbers corresponding to the vertices of this cycle be $a_1, a_2, ..., a_m$. Then the degrees of occurrence of the two in the difference $a_1 - a_2, a_2 - a_3, ..., a_m - a_1$ are pairwise different. Without loss of generality, let the degree of occurrence of the twos, $k$, in the number $a_m - a_1$ be the smallest of them. Then all other differences are divisible by $2^{k+1}$, so $a_1 \\equiv a_2 \\equiv \\dots \\equiv a_m (\\text{mod } 2^{k+1})$, so $a_m - a_1$ is divisible by $2^{k+1}$, which contradicts the assumption. Example is the same as in previous solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a complete graph with 4046 vertices with edges colored in some colors. We call a graph $k$-good graph if all the vertices of the graph can be divided into 2023 pairs in such a way that among the colors of the 2023 of the edges connecting the vertices in the pairs are exactly $k$ distinct colors. Is it possible that the graph is 999-good and 1001-good, but not 1000-good?\n\n(Anton Trygub)",
    "options": [],
    "answer": "Yes",
    "solution": "Number the vertices of the graph with numbers from 1 to 4046. Consider the following graph: all edges between vertices of different parity are colored 1 and all other edges are colored in their unique colors. Consider any partition of the vertices into 2023 into pairs. Suppose that in $x$ of these pairs the vertices have different parity, and in $(2023 - x)$ the same.\n\nNote that $(2023 - x)$ is an even number, because even numbers that are not in $x$ pairs where the numbers are of different parity must be paired with each other. Thus, $x$ is odd, and we have at least one edge of color 1 we have. Thus, the total number of different colors is\n$$\n1_{\\{\\text{first color}\\}} + (2023 - x)_{\\{\\text{different colors in pairs with same parity}\\}} = 2024 - x,\n$$\nis an odd number, i.e. we cannot get exactly 1000 colors.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume $P_1, P_2, \\dots, P_n$ ($n > 10$) are pairwise different polynomials with coefficients $1$, $0$, or $-1$ and such that do not have integer roots. Additionally, $\\forall i = \\overline{1,n}$ $|P_i(5)| \\le \\frac{n^2}{2}$. Prove that $P_i + P_j = P_k + P_l$ for some $1 \\le i,j,k,l \\le n$ and $\\{i, j\\} \\neq \\{k, l\\}$. Note that $i$ may be equal $j$, and $k$ may be equal $l$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the polynomials $P_1, \\dots, P_n$ do not have integer roots then $P_i(0) \\neq 0$ for any $i = \\overline{1,n}$ and $P_i(0) = \\pm 1$. Let us consider all pairs of polynomials $P_k$ and $P_j$, such that $k \\neq j$ and $P_k(0) = P_j(0)$. For each pair of polynomials we determine $Q_{k,j} = P_k - P_j$. If $a$ and $b$ are the number of polynomials such that $P_i(0) = 1$ and $P_i(0) = -1$ respectively, then the number of polynomials $Q_{k,j}$ will be\n$$\n2C_a^2 + 2C_b^2 = a(a-1) + b(b-1) = a^2 + b^2 - n \\ge \\frac{1}{2}(a+b)^2 - n = \\frac{1}{2}n(n-2).\n$$\nNote that coefficients of polynomials $Q_{k,j}$ may be only equal to $-2, -1, 0, 1, 2$. Hence, the coefficients of $Q_{k,j} - Q_{i,l}$ cannot be bigger than $4$ by absolute value. Therefore $Q_{k,j}(5) = Q_{i,l}(5)$ if and only if $Q_{k,j}$ and $Q_{i,l}$ are the same, because $\\forall n \\in \\mathbb{N}$, $5^n > 4(5^{n-1}+\\dots+5+1)$.\n\nIt is easy to see that $Q_{k,j}(5)$ is divisible by $5$. Moreover, $Q_{k,j}(5) \\neq 0$ (otherwise $Q_{k,j}$ is the zero polynomial which is impossible) and $|Q_{k,j}(5)| \\le |P_i(5)| + |P_j(5)| \\le n^2$. Therefore, $Q_{k,j}(5)$ can take at most $\\frac{2}{5}n^2$ different values. But the overall number of polynomials $Q_{k,j}$ is bigger than $\\frac{2}{5}n^2$, because\n$$\n\\frac{1}{2}n(n-2) > \\frac{2}{5}n^2 \\Leftrightarrow 5(n-2) > 4n \\Leftrightarrow n > 10.\n$$\nThis implies that $Q_{k,j}(5) = Q_{i,l}(5) \\Rightarrow Q_{k,j} = Q_{i,l}$ for some $i,j,k,l$. Consequently, $P_i + P_j = P_k + P_l$ for some $1 \\le i,j,k,l \\le n$, $\\{i, j\\} \\neq \\{k, l\\}$ which finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Set $M$ contains $n \\ge 2$ positive integers. It is known that for any two distinct numbers $a, b \\in M$ the number $a^2 + 1$ is divisible by $b$. Find the largest possible value of $n$.",
    "options": [],
    "answer": "2",
    "solution": "**Answer:** $n = 2$.\n\nSuppose that there are at least three numbers in the set $M$, denote them by $a < b < c$. By the condition $b^2 + 1 \\vdash c$, and therefore the numbers $b, c$ are mutually prime. Since $a^2 + 1 \\vdash b$ and $a^2 + 1 \\vdash c \\Rightarrow$\n$$\na^2 + 1 \\vdash bc \\ge (a+1)(a+2) = a^2 + 3a + 2.\n$$\nWe obtained a contradiction that proves that $n \\le 2$, and for $n = 2$ it is enough to consider, for example, the set $M = \\{1, 2\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45824,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the plane 2022 points $A_1, A_2, \\dots, A_{2022}$ are given, no three of which lie on the same line. Consider all the angles $A_i A_j A_k$ for the triples of distinct points $A_i, A_j, A_k$. What largest number of these angles can be right?",
    "options": [],
    "answer": "2042220",
    "solution": "Consider any point $A_i$ and count the number of pairs of points $(A_j, A_k)$ such that $\\angle A_i A_j A_k = 90^\\circ$. For each point $A_j$ there exists at most one point $A_k$ (because on the line through $A_j$ perpendicular to $A_iA_j$ there can be at most one point other than $A_j$). Also note that if $X$ is the point at the largest distance from $A_i$ then there can be no point $A_k$ with $\\angle A_iXA_k = 90^\\circ$, because then we would have $A_iA_k > A_iX$.\n\nThus, each point can be a vertex of the hypotenuse in at most 2020 right triangles at these points. Since the hypotenuse of each triangle has two vertices, the total number of these triangles does not exceed $\\frac{2022 \\cdot 2020}{2} = 2022 \\cdot 1010$.\n\nThis number can be achieved because, for example, we could take 2022 points on a circle so that they are divided into 1011 pairs so that in each pair the points form a circle diameter. Note that for each diameter there will be exactly 2020 points that form a right triangle with it. Then we have at least $1011 \\cdot 2020$ different right angles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45825,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the quadrilateral $ABCD$ with $\\angle ABC = \\angle CDA = 90^\\circ$. Let us denote $P = AC \\cap BD$, $Q = AB \\cap CD$, $R = AD \\cap BC$. Let $l$ be the midline of the triangle $PQR$ parallel to $QR$. Prove that the circumcircle of the triangle formed by the lines $AB, AD$ and $l$ is tangent to the circumcircle of the triangle formed by the lines $CD, CB$ and $l$.\n\n(Fedir Yudin)\n\n![](attached_image_1.png)\n**Fig. 15**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $l$ intersect $AB, AD, BC, CD$ at the points $M, N, K, L$ and let $E = AC \\cap QR$. Since $C$ is the orthocenter $AQR$, $AC \\perp QR$, and therefore $l$ is the perpendicular bisector of the segment $PE$ (fig. 15). Since $BC$ and $BA$ are the internal and external bisectors of the angle $DBE$, the points $K$ and $M$ are the midpoints of the arcs *PE* of the circumscribed circle of *BPE*. Similarly, the points *L* and *N* are the midpoints of the arcs *PE* of the circumscribed circle of *DPE*. Then $\\angle EKL = \\angle EBM = \\angle ARQ = \\angle ECL$, and hence the points *C*, *L*, *K*, *E* lie on the same circle. Similarly, *A*, *M*, *N*, *E* lie on the same circle. It remains to see that $\\angle LEM = \\angle AEM - \\angle CEL = \\angle ANM - \\angle CKL = \\angle ARQ - \\angle CRQ = 90^\\circ - \\angle QAR$. Similarly, $\\angle KEN = 90^\\circ - \\angle QAR = \\angle LEM$, so the circumcircles of the triangles *AMN* and *CKL* are in contact.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all such functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that for any real $x, y$ the following equality holds:\n$$\nf(x + y f(x + y)) = f(y^2) + x f(y) + f(x).\n$$",
    "options": [],
    "answer": "f(x) = x or f(x) = 0",
    "solution": "**Answer:** $f(x) = x$, $f(x) = 0$.\n\nLet us denote the given equation by (1):\n$$\nf(x + y f(x + y)) = f(y^2) + x f(y) + f(x). \\quad (1)\n$$\n\nSubstitute $x = y = 0$ into (1):\n$$\nf(0) = 2f(0) \\implies f(0) = 0.\n$$\n\nSubstitute $x = 0$ into (1):\n$$\nf(y f(y)) = f(y^2). \\quad (2)\n$$\n\nSubstitute $y = -x$ into (1):\n$$\nf(x + (-x) f(x - x)) = f(x^2) + x f(-x) + f(x) \\implies f(x) = f(x^2) + x f(-x) + f(x)\n$$\nSo,\n$$\nf(x^2) = -x f(-x). \\quad (3)\n$$\n\nNow, substitute $x \\to -x$ in (3):\n$$\nf(x^2) = -(-x) f(-(-x)) = x f(x)\n$$\nSo,\n$$\nf(x^2) = x f(x)\n$$\nComparing with (3):\n$$\nx f(x) = -x f(-x) \\implies f(-x) = -f(x). \\quad (4)\n$$\nSo $f$ is odd.\n\nNow, substitute $y = -y$ in (1):\n$$\nf(x + (-y) f(x - y)) = f(y^2) + x f(-y) + f(x)\n$$\nBut $f(-y) = -f(y)$, so\n$$\nf(x + (-y) f(x - y)) = f(y^2) - x f(y) + f(x)\n$$\nLet us write this as:\n$$\nf(x + y f(y - x)) = f(y^2) - x f(y) + f(x). \\quad (5)\n$$\n\nNow, consider the difference between (1) and (5):\n$$\nf(x + y f(x + y)) - f(x + y f(y - x)) = 2x f(y). \\quad (6)\n$$\n\nSuppose that for some $x_0 \\neq 0$, $f(x_0) = 0$. Substitute $y = x_0 - x$ into (1):\n$$\n\\begin{align*}\nf(x + (x_0 - x) f(x + x_0 - x)) &= f((x_0 - x)^2) + x f(x_0 - x) + f(x) \\\\\nf(x) &= f((x_0 - x)^2) + x f(x_0 - x) + f(x) \\\\\nf((x_0 - x)^2) &= -x f(x_0 - x)\n\\end{align*}\n$$\nBut from (3):\n$$\nf((x_0 - x)^2) = -(x_0 - x) f(x - x_0) = (x_0 - x) f(x_0 - x)\n$$\nSo,\n$$\n-x f(x_0 - x) = (x_0 - x) f(x_0 - x) \\implies f(x_0 - x) = 0, \\forall x \\in \\mathbb{R}.\n$$\nThus, $f$ is identically zero, $f(x) = 0$ for all $x$.\n\nIf $f$ is not identically zero, then such $x_0 \\neq 0$ cannot exist. Thus, from the condition $f(x_0) = 0$ it follows that $x_0 = 0$.\n\nNow, let us prove injectivity. Suppose $\\exists x_1 \\neq x_2$ such that $f(x_1) = f(x_2)$.\nIn (6), substitute $x = \\frac{x_1 - x_2}{2}$, $y = \\frac{x_1 + x_2}{2}$:\n$$\n\\begin{align*}\nf\\left(\\frac{x_1 - x_2}{2} + \\frac{x_1 + x_2}{2} f(x_1)\\right) - f\\left(\\frac{x_1 - x_2}{2} + \\frac{x_1 + x_2}{2} f(x_2)\\right) &= (x_1 - x_2) f\\left(\\frac{x_1 + x_2}{2}\\right) \\\\\n(x_1 - x_2) f\\left(\\frac{x_1 + x_2}{2}\\right) &= 0\n\\end{align*}\n$$\nSo, either $x_1 = x_2$ (contradicts assumption), or $f\\left(\\frac{x_1 + x_2}{2}\\right) = 0$. But as above, the only zero is at $0$, so $x_1 + x_2 = 0 \\implies x_1 = -x_2$.\nBut then\n$$\nf(x_2) = f(-x_1) = -f(x_1) = -f(x_2) \\implies f(x_2) = 0 \\implies x_2 = 0 \\implies x_1 = 0\n$$\nSo, $f$ is injective.\n\nNow, from (2):\n$$\nf(y f(y)) = f(y^2)\n$$\nBut $f$ is injective, so $y f(y) = y^2 \\implies f(y) = y$ for all $y$.\n\nThus, the only solutions are $f(x) = 0$ and $f(x) = x$.\n\nIt is easy to check that both satisfy the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45827,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a polynomial *mixed* if it has both positive and negative coefficients (we do not consider zero to be either positive or negative). Is it true that the product of two mixed polynomials is always a mixed polynomial?",
    "options": [],
    "answer": "No",
    "solution": "Here is a counterexample:\n$$\nf(x) = x^5 + x^4 - x^3 - x^2 + x + 1 = (x^4 - x^2 + 1)(x + 1)\n$$\n$$\ng(x) = x^4 - x^3 + 2x^2 - x + 1 = (x^2 - x + 1)(x^2 + 1).\n$$\n\n\\begin{aligned}\nf(x)g(x) &= \\\\\n&= (x^4 - x^2 + 1)(x + 1)(x^2 - x + 1)(x^2 + 1) = \\\\\n&= (x^6 + 1)(x^3 + 1) = x^9 + x^6 + x^3 + 1.\n\\end{aligned}\n$$\n\nBoth of these polynomials are mixed, but their product is not mixed, since all coefficients are positive.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K$ be the midpoint of the median $AM$ of the triangle $ABC$. On the side $AB$ there is a point $X$ such that $\\angle KXM = \\angle ACB$ and $AX > BX$, and on the side $AC$ there is a point $Y$ such that $\\angle KYM = \\angle ABC$ and $AY > CY$. Prove that the points $B, X, C, Y$ lie on the same circle.\n\n(Mykhailo Shtandenko)\n\n![](attached_image_1.png)\nFig. 19",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the middle of $AB$, and $L$ be the middle of $AC$. Then the points $N$, $K$ and $L$ lie on the same line—the midline of $\\triangle ABC$ which is parallel to $BC$ (fig. 19).\n\nThen we have that $\\angle KXM = \\angle ACB = \\angle KNM$, so the quadrilateral $KNXM$ is inscribed, and similarly $KMYL$ is also inscribed.\n\nTherefore, $AN \\cdot AX = AK \\cdot AM = AL \\cdot AY$, so the points $Y$, $N$, $L$ and $X$ lie on the same circle, hence $\\angle AXY = \\angle ALN = \\angle ACB$, so the quadrilateral $BXYC$ is inscribed, which is what we needed to prove.\n\n**Remarks:** In this configuration, the quadrilateral $AXMY$ is harmonic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45829,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$ we write out all its divisors $1 = d_1 < d_2 < \\cdots < d_k = n$. A divisor $d_i$ is called a good divisor if $d_{i-1}d_{i+1}$ is not divisible by $d_i$, $2 \\le i \\le k-1$. Find all $n$ for which the number of their good divisors is smaller than the number of their different prime divisors.\n\n(Mykhailo Shtandenko)",
    "options": [],
    "answer": "All n of the form p^β and p^α q with q > p^α, where p and q are prime numbers.",
    "solution": "**Answer.** $p^\\beta$ and $p^\\alpha q$, where $q > p^\\alpha$ where $p, q$ are prime numbers.\n\nSolution:\n\nFirst, let's prove that if a number has at least three different prime divisors, it is not suitable for us. Let $m$ be the number of different prime divisors and $p < q$ be the two smallest prime divisors of $n$. Then it is clear that there exists $\\alpha$ such that the first consecutive divisors of $n$ will be the numbers $1 < p < p^2 < \\cdots < p^\\alpha < q$. Note that taking as $d_{i+1}$ all prime divisors of $n$ except for the smallest, we get $m-1$ numbers that satisfy the condition. Indeed, all divisors smaller than a prime number are coprime with that number, so $(d_i, d_{i+1}) = 1$, and since $d_{i-1} < d_i$ we get the conclusion. It remains to find another good number. Let $n : p^{\\alpha+1}$. Then the divisor $q$ is a good divisor. Indeed, it cannot be followed by a divisor divisible by $q$, because the smallest of these not yet chosen divisors is $pq$, but $p^{\\alpha+1} < pq$. So $p^{\\alpha+1}$ is the next composite divisor. Then take the last prime number that comes after $q$ (possibly $q$), after which the next composite number will be $p^{\\alpha+1}$. Then it will be the desired $m$-th good number. Now let $n$ be not divisible by $p^{\\alpha+1}$. Then the next composite divisor is $pq$. If between $q$ and $pq$ there is at least one prime divisor, then we take the last of them as our $m$-th good number. Otherwise, the divisors $p^\\alpha, q, pq$ will be consecutive divisors, and hence $\\frac{n}{pq}, \\frac{n}{q}, \\frac{n}{p^\\alpha}$ will be consecutive divisors, but then the divisor $\\frac{n}{q}$ will be the desired one, because since $\\alpha$ is the degree of occurrence of a prime $p$ in the number $n$, $\\frac{n}{p^\\alpha}$ is not divisible by $p$, and $\\frac{n}{pq}$ is not divisible by $p^\\alpha$, but $\\frac{n}{q} : p^\\alpha$, and $\\frac{n}{p^\\alpha}$ is divisible by more than one prime divisor of $n$, which means that we have not yet taken it into account, which is what we wanted to prove.\n\nIf a number is a power of a prime, then obviously it suits us, because it has no good divisors at all. It remains to consider the case when a number has two different prime divisors, which we denote $p < q$. Then it is clear that there exists $\\alpha$ that the first consecutive divisors of $n$ will be the numbers $1 < p < p^2 < \\cdots < p^\\alpha < q$. The divisor $p^\\alpha$ is a good divisor, which means it is the only good divisor of our number. Hence, the next divisor after $q$ is a divisor that is divisible by $q$, but the smallest such divisor not yet used is $pq$, and therefore it comes after $q$. However, $p^{\\alpha+1} < pq$, and hence our number $n$ is not divisible by $p^{\\alpha+1}$. Then, similarly, the divisors $\\frac{n}{pq}, \\frac{n}{q}, \\frac{n}{p^\\alpha}$ will be consecutive divisors, and the divisor $\\frac{n}{q}$ will be a good divisor. So, $\\frac{n}{q} = p^\\alpha$, so $n = p^\\alpha q$, and $p^\\alpha < q$, and the above answer follows. A simple check shows that both answers do indeed satisfy the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45830,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country there are $2024$ cities, each two of which are connected bidirectionally by exactly one of three modes of transportation: rail, air, or road. A tourist arrives in this country and has the entire transportation scheme. He chooses a travel ticket for one of the modes of transportation and the city from which he starts his trip. He wants to visit as many cities as possible but using only the ticket for the specified type of transportation. How many cities is a tourist guaranteed to be able to visit? During the route, he can return to the cities he has already visited.\n\n(Bogdan Rublov)\n\n![](attached_image_1.png)\n**Fig. 20**",
    "options": [],
    "answer": "1012",
    "solution": "Suppose we have a complete graph with each edge colored in one of three colors. We show that there will always exist a connected component of at least one of the colors of size at least $1012$. First, we show that it is possible to color the edges in such a way that connected components of size more than $1012$ vertices do not exist. Let's label the colors with numbers - 1st, 2nd and 3rd. Let's divide the vertices into groups by $506$ - groups \"A\", \"B\", \"C\" and \"D\". Let all $1012$ vertices of groups \"A\" and \"B\" be connected by edges of color 1, all vertices of group \"C\" - by color 2, group \"D\" - by color 3. Between them, all vertices are connected by edges of those colors, as shown in fig. 20. Therefore, groups \"A\" or \"B\" form a connected component of size $1012$ of color 1. Groups \"A\" and \"C\" are of color 2, groups \"B\" and \"D\" are of color 3. Obviously, there is no larger connected component in terms of the number of vertices.\n\nSuppose that there exists a coloring such that the largest connected component contains fewer than $1012$ vertices. Let's consider the coloring that gives the smallest answer to the problem.\n\nConsider the largest connected component of one of the colors. This component, or rather the number of vertices in it, is the answer to the problem. If there are several largest components, choose any of them. Let this component correspond to color 1. Let's denote this component as I. Let this component contain exactly $m$ vertices. Note that not all the edges between the vertices of this component have the color 1. But if we repaint all the edges of this component with color 1, then the answer $m$ in the problem will not change. In this case, there are no vertices of the component I from which an edge of color 1 can come out outside this component I, that is, to a vertex that does not belong to the component I.\n\nNext, consider the largest connected component among all other vertices, II, with color 2 or 3. It includes all the vertices that are connected by color 2, i.e., also those that can be reached through the vertices of the component I. For example, if $A \\in$ II, $B \\in$ I, and for the vertex $C \\notin$ I the following conditions are true: $A(II)\\ 2 \\leftrightarrow B(I)\\ 2 \\leftrightarrow C$, then we assume that $C \\in$ II, but we do not assume that $B \\in$ II. In other words, a vertex cannot belong to two components at the same time (this principle will be maintained in the future). The number of vertices in the component II is denoted by $k$. Now we can repaint all edges in this component into color 2, and, again, the answer $m$ in the problem will not change.\n\nNow let's denote by $X_2$ the group of vertices of the component I that are connected to at least one vertex of the component II by an edge of color 2, let their number be $x_2$. Then all the edges between the vertices $X_2$ and the vertices of component II can be colored in color 2. They still form the connected component of color 2 before this coloring, but since the component of color 1 was chosen first, the total number of vertices $k + x_2 \\le m$.\n\nThen all the vertices of the component I that are not included in the group $X_2$, we will combine into the group $X_3$ and denote their number by $x_3$. From these vertices, only edges of color 3 can go to each vertex of component II. But then the vertices of the group $X_3$ and the component II form a connected component of color 3. But then from the vertices of the group $X_3$ there cannot be any edges of color 1 or 3 to vertices that are not included in the components I and II. If there is an edge of color 3, then the connected component of color 3 will be larger than the component II, and this contradicts its construction. Therefore, all vertices that are not included in the components I and II are connected by color 2 (since $X_3$ is not empty). We denote all these vertices as a component III. Similarly, the vertices of the group $X_2$ are connected with III only with edges of color 3. But then the components II and III should be connected only by edges of color 1. That is, they form a connected component of color 1. But we initially chose the largest connected component of color 1, which has $m$ vertices. Therefore, groups II and III groups have no more than $m$ vertices, hence $2m \\ge 2024$ and this completes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45831,
    "subject": "Mathematics (Multi-modal)",
    "question": "For arbitrary positive numbers $a, b, c$, solve the system of equations:\n$$\n\\begin{cases} ax^3 + by = cz^5, \\\\ az^3 + bx = cy^5, \\\\ ay^3 + bz = cx^5. \\end{cases}\n$$",
    "options": [],
    "answer": "(0, 0, 0) and (t, t, t), where t = ±sqrt((a + sqrt(a^2 + 4bc)) / (2c))",
    "solution": "**Answer:** $(0, 0, 0)$ and $(t, t, t)$, where $t = \\pm\\sqrt{\\frac{a+\\sqrt{a^2+4bc}}{2c}}$.\n\nSuppose that $x < y < z$ (one of the inequalities may not be strict). Then subtract the third equation from the first: $a(x^3 - y^3) + b(y - z) < 0 \\le c(z^5 - x^5)$ – a contradiction. If we assume that $x < z < y$ (one of the inequalities may not be strict), then subtract the third equation from the second: $a(z^3 - y^3) + b(x - z) < 0 \\le c(y^5 - x^5)$ – a contradiction. Without loss of generality, we can assume that all cases have been considered, since the system of equations is cyclic. Thus, we are left with the condition $x = y = z$.\n\nTo find $x$, we need to solve the equation: $cx^5 - ax^3 - bx = 0$. Obviously, $x_1 = 0$, then we need to solve the equation $cx^4 - ax^2 - b = 0$. Since $a^2 + 4bc > 0$ and\n$$\na - \\sqrt{a^2 + 4bc} < 0, \\text{ we have that } x_{2,3} = \\pm\\sqrt{\\frac{a+\\sqrt{a^2+4bc}}{2c}}, \\text{ otherwise there are no other solutions.}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nonzero real numbers $a, b, c$ satisfy the equation $ab + bc + ac = 0$. Prove that the numbers $a + b + c$ and $\\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a}$ have the same sign.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If we assume that $b + c = 0$, then we have\n$$\nab + bc + ac = a(b + c) + ac = ac = 0,\n$$\nwhich contradicts the condition. Thus $(a+b)(b+c)(c+a) \\neq 0$.\nSince\n$$\n(a + b + c)(a + b) = a^2 + ab + b^2 + ab + bc + ac = a^2 + ab + b^2,\n$$\nwe have $\\frac{1}{a+b} = (a+b+c) \\cdot \\frac{1}{a^2+ab+b^2}$. Then, adding three similar equations, we get\n$$\n\\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{a+c} = (a+b+c) \\left( \\frac{1}{a^2+ab+b^2} + \\frac{1}{b^2+bc+c^2} + \\frac{1}{a^2+ac+c^2} \\right).\n$$\nIt is easy to see that each of the expressions of the form $a^2 + ab + b^2 = \\frac{a^2+(a+b)^2+b^2}{2} > 0$, which implies the desired statement.\nSimilarly to the previous solution, we verify that the given expressions are defined correctly, that is, $(a+b)(b+c)(c+a) \\neq 0$.\nFrom the given equation, we write $c = -\\frac{ab}{a+b}$ and substitute this expression into the product:\n$$\n(a+b+c)\\left(\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{a+c}\\right) = (a+b-\\frac{ab}{a+b})\\left(\\frac{1}{a+b}+\\frac{1}{b-\\frac{ab}{a+b}}+\\frac{1}{a-\\frac{ab}{a+b}}\\right) = \\\\\n= \\left(\\frac{(a+b)^2-ab}{a+b}\\right)\\left(\\frac{1}{a+b}+\\frac{a+b}{b^2}+\\frac{a+b}{a^2}\\right) = \\\\\n= \\frac{a^2+ab+b^2}{a+b} \\cdot \\frac{a^2b^2+(a+b)^2a^2+(a+b)^2b^2}{(a+b)a^2b^2} = \\\\\n= (a^2+ab+b^2) \\cdot \\frac{a^2b^2+(a+b)^2a^2+(a+b)^2b^2}{(a+b)^2a^2b^2} > 0,\n$$\nsince $a^2 + ab + b^2 = \\frac{a^2+(a+b)^2+b^2}{2} > 0$, and the second factor is also obviously positive. If the product of two expressions is always positive, then they have the same sign, which is what we needed to prove.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n \\ge 4$ positive numbers. Consider all $\\frac{n(n-1)}{2}$ pairwise sums of these numbers.\nShow that there exist two sums that differ by no more than a factor of $\\sqrt[4]{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the numbers be arranged in non-increasing order: $x_1 \\ge x_2 \\ge \\dots \\ge x_n$. Consider the sums $2x_1 \\ge x_1 + x_2 \\ge x_1 + x_3 \\ge \\dots \\ge x_1 + x_n > x_1$. Therefore, some two of the sums $x_1 + x_2, x_1 + x_3, \\dots, x_1 + x_n$ differ by no more than a factor of $\\sqrt[4]{2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ and $n$ be arbitrary natural numbers that satisfy the condition $3 \\le k \\le n$. Prove that among any $n$ pairwise distinct real numbers, there are either $k$ numbers with a positive sum or $(k-1)$ numbers with a negative sum.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If there are no positive numbers in the given set, then there is at most one number that is equal to zero, and all other numbers are negative. Therefore, there are $(n-1)$ negative numbers in the set. Hence, we can take any set with $k-1 \\ge 2$ numbers as the desired subset.\n\nOtherwise, if there is at least one positive number in this set, we separate this number. Among the other $(n-1)$ numbers that remain, we take any $k-1$ numbers. If their sum is negative, then the desired subset has been found and the problem is solved. If the sum is non-negative, we add to this subset with a non-negative sum the positive number that was separated. Then we obtain the desired $k$ numbers with a positive sum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45835,
    "subject": "Mathematics (Multi-modal)",
    "question": "$BE$ and $CF$ are the altitudes of the acute scalene $\\triangle ABC$, $O$ is its circumcenter and $M$ is the midpoint of the side $BC$. If the point, which is symmetric to $M$ with respect to $O$, lies on the line $EF$, find all possible values of the ratio $\\frac{AM}{AO}$.\n\n![](attached_image_1.png)\nFig. 21",
    "options": [],
    "answer": "sqrt(2)",
    "solution": "Let $H$ be the orthocenter of $\\triangle ABC$, $A'$ be the antipode of $A$ in $(ABC)$, $S$ be the point, which is symmetric to $M$ with respect to $O$ (fig. 21), and $T$ be the intersection of $BC$ and $EF$. It's well-known that in this construction $H$ is the orthocenter of $\\triangle AMT$, $BHCA'$ is parallelogram (then $M$ is the midpoint of $A'H$) and $AH = 2OM$.\n\n![](attached_image_1.png)\n\nAs $SM = 2OM = AH$ and $AH \\perp BC \\perp SM$, we can obtain that $AHMS$ is parallelogram. Hence, $S$ is the antipode of $T$ in $(AMT)$ and $\\angle AMH = 90^\\circ - \\angle MAT = 90^\\circ - \\angle MST = \\angle STM$. As $OM \\perp BC$ and $AO \\perp EF$, we obtain that $\\angle A'OM = \\angle STM$. So, $\\angle A'OM = \\angle AMH$, hence $\\triangle AA'M \\sim \\triangle AMO$ and then $AA' \\cdot AO = AM^2 \\Rightarrow \\frac{AM}{AO} = \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n \\ge 3$ pairwise distinct real numbers. Prove that there are either 3 numbers with a positive sum or 2 numbers with a negative sum.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If there are no positive numbers in the given set, then there is at most one number equal to zero, and all other numbers are negative, which means that there are $n-1$ negative numbers. Therefore, as the required set of 2 numbers, we can take any two numbers.\n\nOtherwise, if there is at least one positive number in this set, we will isolate this number. Among the other $n-1$ numbers that remained, we take any 2 numbers. If their sum is negative, then the required set is found and the problem is solved. If the sum is non-negative, then we add the previously isolated positive number to this sum. If the new sum is positive, then we have found the required set of 3 numbers. Otherwise, the sum of all three numbers is non-positive, which means that the required set of 2 numbers with a negative sum is also found.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $I$ is the incenter of triangle $ABC$, where $AB < AC$. On the external angle bisector of angle $ABC$, a point $X$ is chosen such that $IC = IX$. Let the tangent to the circumcircle of triangle $BXC$ at point $X$ intersect line $AB$ at point $Y$. Prove that $AC = AY$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the fact that $XY$ is tangent to the circumcircle of $\\triangle BXC$, we have $\\angle BXY = \\angle BCX$. Combining this with the fact that $\\angle YBX = \\angle XBC$, we have the similarity $\\triangle BXY \\sim \\triangle BCX$. Therefore, we have $\\frac{BX}{BC} = \\frac{BY}{BX}$, so $BY = \\frac{BX^2}{BC}$. Notice that $\\angle IBX = 90^\\circ$, so (Fig. 9)\n\n![](attached_image_1.png)\n\n$$\nBY = \\frac{BX^2}{BC} = \\frac{IX^2 - BI^2}{BC} = \\frac{CI^2 - BI^2}{BC}.\n$$\n\nLet $K$ be the point of tangency of the incircle of $\\triangle ABC$ with side $BC$. Then\n$$\nBY = \\frac{CI^2 - BI^2}{BC} = \\frac{(IK^2 + CK^2) - (IK^2 + BK^2)}{BC} = \\frac{CK^2 - BK^2}{BC} = CK - BK.\n$$\n\n$AY = AB + BY = AB + CK - BK = c + (p - c) - (p - b) = b = AC$,\n\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45838,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will call a circle without boundary, i.e., a circle without the points on its circumference, a \"hedgehog\". The diameter of the hedgehog is the diameter of this circle. We will say that the hedgehog \"sits\" at a point where the center of the corresponding circle is located. Let us consider a triangle with sides $a, b, c$, and hedgehogs sitting at its vertices. It is known that there exists a point inside the triangle from which one can reach any side of the triangle along a straight trajectory without touching any of the hedgehogs. What is the largest possible sum of the diameters of these hedgehogs?\n(Oleksii Masalitin)",
    "options": [],
    "answer": "a + b + c",
    "solution": "Let us denote the width of the triangle as $ABC$, and the diameters of the hedgehogs as $d_a, d_b, d_c$, respectively. Then suppose $d_a + d_b > 2c$. This means that any point on the side $AB$ of the triangle is inside one of the hedgehogs, and therefore it is impossible to reach this side. Then we have $d_a + d_b \\le 2c$, and similarly $d_a + d_c \\le 2b$ and $d_b + d_c \\le 2a$, which implies that $d_a + d_b + d_c \\le a + b + c$.\nWe will prove that there exists an example where equality is achieved in this inequality. Let $I$ be the center of the inscribed circle of the triangle and let $A_1, B_1, C_1$ be its points of tangency to the sides of the triangle (Fig. 5). Then it suffices to consider the hedgehogs sitting at the vertices and whose corresponding circles have radii $AB_1 = AC_1$, $BA_1 = BC_1$, and $CA_1 = CB_1$. Since $IA_1 \\perp BC$, $IB_1 \\perp AC$, and $IC_1 \\perp AB$, this means that $IA_1, IB_1, IC_1$ are tangents to the hedgehogs sitting at the corresponding vertices, and therefore they will not touch any hedgehogs, which is what we wanted to prove.\n\n![](attached_image_1.png)\nFig. 5",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's consider on the Cartesian plane all pairs of distinct points $(A, B)$, each of which has both integer coordinates. Among these pairs of points, find all those for which there exist two distinct points $(X, Y)$ with both integer coordinates, such that quadrilateral $AXBY$ is convex and inscribed.\nA quadrilateral is called convex if both of its diagonals lie inside the quadrilateral.",
    "options": [],
    "answer": "All pairs of distinct lattice points except those at unit distance apart; for distance 1 no such points exist, and for any other distance such points can be found.",
    "solution": "First, we will show that for points that are at a distance of $1$ from each other, there are no points $(X, Y)$ that satisfy the condition. Indeed, let us assume that such points exist. Then, $\\angle AXB + \\angle AYB = 180^\\circ$, which means that at least one of these angles is not less than $90^\\circ$. Therefore, at least one of the points $X, Y$ must lie inside or on the circle with diameter $AB$, but this circle does not contain any other integer points except for $A$ and $B$.\n\nNow, let's show that for all other pairs of points, such a pair $(X, Y)$ can be found. Let $A = (a_1, a_2)$ and $B = (b_1, b_2)$. If $a_1 \\ne b_1$ and $a_2 \\ne b_2$, then we can take $X = (a_1, b_2)$ and $Y = (b_1, a_2)$, and $AXBY$\n\nwill be a rectangle, which means that it is inscribed. Otherwise, without loss of generality, we can assume that $a_2 = b_2 = t$, and $A = (a_1, t)$ and $B = (b_1, t)$, where $|a_1 - b_1| > 1$. Without loss of generality, we can also assume that $a_1 < b_1$. Then, it is sufficient to take the following points: $X = (a_1 + 1, t - 1)$ and $Y = (a_1 + 1, t + (b_1 - a_1 - 1))$. It is easy to see that the segments $AB$ and $XY$ intersect at the point $K = (a_1 + 1, t)$, and $AK \\cdot BK = XK \\cdot YK = 1 \\cdot (b_1 - a_1 - 1)$, so the quadrilateral $AXBY$ is indeed inscribed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let its incircle, centred at $I$, touch the side $BC$ at $D$. A line through $A$ intersects the lines $BC$, $BI$ and $CI$ at $X$, $Y$ and $Z$, respectively. The circle $(ABC)$ intersects the circles $(AIY)$ and $(AIZ)$ again at $U$ and $V$, respectively. Prove that the points $D$, $U$, $V$ and $X$ are concyclic.\n\n(Fedir Yudin and Mykhailo Shtandenko)\n\n![](attached_image_1.png)\nFig. 22",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $ABC$ be a triangle with orthocenter $I$, $D$ is the antipode of $I$ in $(BIC)$ (we'll call this circle as $\\omega$). So, $ABCD$ is a parallelogram. A circle, which passes through the points $A$ and $I$, intersects $\\omega$ at $X$ and lines $BI$ and $CI$ at points $Y$ and $Z$ respectively ($X,Y,Z \\neq I$). Lines $AY$ and $AZ$ intersect $(ABC)$ for the second time at points $U$ and $V$ respectively. Prove that the points $D$, $U$, $V$ and $X$ are concyclic. Let line $UV$ meets $BC$ at point $L$. Then it will be sufficient to prove that $DX$ passes through the point $L$ (because then $LU \\cdot LV = LB \\cdot LC = LX \\cdot LD$ and the problem will be solved). Let $K$ be the second intersection point of $(ABC)$ and $AX$. Observe that (fig. 22)\n\n$$\n\\angle(VA, AU) = \\angle(ZA, AY) = \\angle(ZI, IY) = \\angle(CI, BI) = \\angle(BA, AC),\n$$\nSo, arcs $UV$ and $BC$ are equal and then $UV = BC$. Triangles $UVK$ and $BCX$ are similar, because\n$$\n\\angle UVK = \\angle UAK = \\angle YAX = \\angle(XI, IY) = \\angle(XI, IB) = \\angle(XC, CB) = \\angle BCX\n$$\nand, similarly, $\\angle VUK = \\angle CBX$.\n\nSo, points $D$, $L$ and $X$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45841,
    "subject": "Mathematics (Multi-modal)",
    "question": "The problem gives us a right triangle $ABC$ with a right angle at $ACB$. Let $W_A$ and $W_B$ be the midpoints of the smaller arcs $BC$ and $AC$ of the circumcircle of $\\triangle ABC$, and $N_A$ and $N_B$ be the midpoints of the larger arcs $BC$ and $AC$. Let $P$ and $Q$ be the intersection points of segment $AB$ with lines $N_A W_B$ and $N_B W_A$, respectively. Prove that $AP = BQ$.\n\n![](attached_image_1.png)\nFig. 7",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of the hypotenuse $AB$ of triangle $ABC$ (see figure 7). It is clear that $N_A W_B W_A N_B$ is a rectangle with center $M$. Therefore, its sides $N_A W_B$ and $N_B W_A$ are symmetric with respect to $M$. This means that $AP = BQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no natural numbers $n$ and $k$ that satisfy the equation:\n$$\nn^n + (n + 1)^{n+1} + (n + 2)^{n+2} = 2023^k.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the given equation modulo $3$. The right-hand side, $2023^k = 1 \\pmod{3}$. Among any three consecutive natural numbers $n$, $n+1$, $n+2$, one is divisible by $3$, one leaves a remainder of $1$ when divided by $3$, and one leaves a remainder of $2$.\n\nIf $x$ is divisible by $3$, then $x^x$ is also divisible by $3$; if $x$ has a remainder $1$, then $x^x$ also has a remainder $1$. If $x$ has a remainder $2$, then $x^x$ has remainder $1$ or $2$. Then, $n^n + (n+1)^{n+1} + (n+2)^{n+2}$ has a remainder $2$ or $0$ modulo $3$. This contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of natural numbers $(m, n)$ for which $2^n - 13^m$ is a cube of a natural number.",
    "options": [],
    "answer": "(m, n) = (2, 9)",
    "solution": "Let $2^n - 13^m = k^3$. Then $2^n \\equiv k^3 \\pmod{13}$. If $n$ is not divisible by $3$, then $2^n$ can be written as $4l^3$ or $2l^3$. Then either $4$ or $2$ is a cube of some integer modulo $13$. But by simple checking, we can see that only remainders $0$, $\\pm1$, $\\pm5$ can be cubes of integers modulo $13$. Therefore, $n$ is divisible by $3$, so let $n = 3t$. The condition can be rewritten as\n$$\n13^m = (2^t - k)(4^t + k \\cdot 2^t + k^2).\n$$\nThen both factors are powers of $13$, and the left one is smaller than the right one, so $2^t - k$ divides $4^t + k \\cdot 2^t + k^2$. Since $2^t \\equiv k \\pmod{(2^t - k)}$, we have $4^t + k \\cdot 2^t + k^2 \\equiv 3 \\cdot 4^t \\pmod{(2^t - k)}$. Therefore, $3 \\cdot 4^t$ is divisible by $2^t - k$, which is a power of $13$, so $2^t - k = 13^0 = 1$ and $13^m = 3 \\cdot 4^t - 3 \\cdot 2^t + 1$. Suppose that $t \\ge 4$. Then $13^m \\equiv 1 \\pmod{16}$, so $m$ is divisible by Therefore, $m = 4s$ and\n$$\n(13^{2s} + 1)(13^{2s} - 1) = 3 \\cdot 2^t \\cdot (2^t - 1).\n$$\nBut $13^{2s} + 1$ is not divisible by $4$ and $3$, so $13^{2s} + 1$ divides $2 \\cdot (2^t - 1)$. If these numbers are not equal, then\n$$\n13^{2s} + 1 \\le 2^t - 1 < \\sqrt{3 \\cdot 2^t \\cdot (2^t - 1)} < 13^{2s}.\n$$\nOtherwise, $13^{2s} + 1 = 2 \\cdot (2^t - 1)$ and $13^{2s} - 1 = 3 \\cdot 2^{t-1}$. We will subtract the second equation from the first equation and get $2^{t-1} - 2 = 2$, so $t = 3$, which contradicts the assumption that $t > 3$. By checking for $t = 1$, $t = 2$, and $t = 3$, we verify that only $t = 3$ satisfies the condition. Therefore, $n = 9$ and $m = 2$ is the unique solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's call a pair of natural numbers $\\overline{a_1 a_2 \\dots a_k}$ and $\\overline{b_1 b_2 \\dots b_k}$ $k$-similar if all the digits $a_1, a_2, \\dots, a_k, b_1, b_2, \\dots, b_k$ are pairwise distinct and there exist distinct natural numbers $m, n$ such that the following equation holds:\n$$\na_1^m + a_2^m + \\dots + a_k^m = b_1^n + b_2^n + \\dots + b_k^n.\n$$\n\nWhat is the largest value of $k$ for which there exist $k$-similar numbers?",
    "options": [],
    "answer": "4",
    "solution": "Let us consider the numbers $1234$ and $6789$. For them, we see that the following equation holds:\n$$\n6^1 + 7^1 + 8^1 + 9^1 = 30 = 1^2 + 2^2 + 3^2 + 4^2.\n$$\nThus, this pair of numbers is $4$-similar.\n\nObviously, for $k > 5$ there are no $k$-similar numbers, because there are only $10$ different digits. Let us assume that there is a pair of $5$-similar numbers $\\overline{a_1 a_2 a_3 a_4 a_5}$ and $\\overline{b_1 b_2 b_3 b_4 b_5}$, for which the following equation holds:\n$$\na_1^m + a_2^m + a_3^m + a_4^m + a_5^m = b_1^n + b_2^n + b_3^n + b_4^n + b_5^n.\n$$\n\nThe parity of the numbers on both sides of the equation is the same, and therefore their sum is an even number. Since these numbers are all the $10$ digits, their sum must also be even, because the parity of a number does not change when it is raised to a natural power. But the sum of all digits $0 + 1 + 2 + \\cdots + 9 = 45$ is odd. The resulting contradiction shows that the maximum value of $k$ is $4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$, such that for any real $x$, $y$ holds the following:\n$$\nxf(x) + yf(xy) = xf(x + yf(y))\n$$",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x for all real x",
    "solution": "Let $P(x, y)$ be the given assertion,\n$$\nP(0,1): f(0) = 0\n$$\nAssume that there exists $a \\neq 0$ such that $f(a) = 0$. Then $P(x, a)$:\n$$\nxf(x) + af(xa) = xf(x) \\Rightarrow \\forall x \\in \\mathbb{R} \\ f(xa) = 0 \\Rightarrow \\forall x \\in \\mathbb{R} \\ f(x) = 0\n$$\nAnd we found the first solution.\n\nLet us now assume that $f$ is not identically zero, so $f(x) = 0 \\Leftrightarrow x = 0$\n$$\nP(-1,1): 0 = -f(-1) + f(-1) = -f(-1 + f(1)) \\Rightarrow f(1) = 1\n$$\n$$\nP(x, 1): xf(x) + f(x) = xf(x+1) \\Rightarrow (x+1)f(x) = xf(x+1) \\Rightarrow \\forall x \\neq 0, -1\n$$\n$$\n\\frac{f(x)}{x} = \\frac{f(x+1)}{x+1} \\quad (1)\n$$\nWe already have that $f(1) = 1$, so we can get from (1) that $\\forall n \\in \\mathbb{N} \\ f(n) = n$. Also, we can obtain from (1) that $f(-2) = 2f(-1)$.\n$$\nP(-1,2): -f(-1) + 2f(-2) = -f(-1 + 2f(2)) = -f(3) = -3 \\Rightarrow f(-1) = -1\n$$\nBy combining this with (1) we can get that $\\forall n \\in \\mathbb{Z} \\ f(n) = n$.\n\nLet us now show that $\\forall n \\in \\mathbb{Z}$ and $\\forall x \\in \\mathbb{R}$, $(x+n)f(x) = xf(x+n)$. If $x \\in \\mathbb{Z}$ this is obvious, otherwise we can write that $\\frac{f(x)}{x} = \\frac{f(x+1)}{x+1} = \\cdots = \\frac{f(x+n)}{x+n}$ and obtain this equality.\n\n$$\n\\begin{align*}\nP(x, n): \\quad xf(x) + nf(nx) &= xf(x + nf(n)) \\\\\n&\\Rightarrow xf(x) + nf(nx) = xf(x + n^2) = (x + n^2)f(x) \\\\\n&\\Rightarrow xf(x) + nf(nx) = xf(x) + n^2f(x) \\\\\n&\\Rightarrow f(nx) = nf(x). \\tag{2}\n\\end{align*}\n$$\n\nAfter we'll divide $P(x, y)$ by $x \\neq 0$:\n$$\nf(x) + \\frac{yf(xy)}{x} = f(x + yf(y)). \\quad (4)\n$$\n\n$P(1, y): 1 + yf(y) = f(1 + yf(y))$\n\nLet us now put $x = yf(y)$ in (1), so $\\frac{f(yf(y))}{yf(y)} = \\frac{f(yf(y)+1)}{yf(y)+1} = 1 \\Rightarrow$\n$$\nf(yf(y)) = yf(y). \\quad (3)\n$$\n\nPut $x = xf(x)$ there and by symmetry we'll get the following:\n$$\n\\begin{align*}\nf(xf(x)) + \\frac{yf(xyf(x))}{xf(x)} &= f(xf(x) + yf(y)) \\\\\n&= f(yf(y)) + \\frac{xf(xyf(y))}{yf(y)} \\\\\n\\Rightarrow f(xf(x)) - \\frac{xf(xyf(y))}{yf(y)} &= f(yf(y)) - \\frac{yf(xyf(x))}{xf(x)}\n\\end{align*}\n$$\nIf we put here $2x$ instead of $x$ there, then the LHS will increase by 4 times because of (2) and the RHS will not change, so both are zero and hence, $f(xf(x)) = \\frac{xf(xyf(y))}{yf(y)} \\Rightarrow$\n$$\nf(xyf(y)) = f(x)yf(y) \\quad (5)\n$$\n\nLet us put $x = x + 1$ into (5):\n$$\n\\begin{align*}\nf(x+1)yf(y) &= f((x+1)yf(y)) = f(xyf(y) + yf(y)) \\\\\n&\\stackrel{(4)}{=} f(xyf(y)) + \\frac{yf(xy^2f(y))}{xyf(y)} \\\\\n&\\stackrel{(5)}{=} f(x)yf(y) + \\frac{y^2f(y)f(xy)}{xyf(y)} = f(x)yf(y) + \\frac{yf(xy)}{x}\n\\end{align*}\n$$\nSo, $(f(x+1) - f(x))x = \\frac{f(xy)}{f(y)}$ and by substituting $y=1$ here, we are getting that\n$$\n(f(x+1) - f(x))x = f(x) \\Rightarrow f(xy) = f(x)f(y) \\quad (6)\n$$\nBy using (6) in (3) we get that $f(f(y)) = y$. So\n$$\n\\begin{align*}\nP(f(y), y): \\quad f(y)y + yf(f(y)y) &= f(y)f(f(y) + yf(y)) \\\\\n&\\Rightarrow f(y)y + y^2f(y) = yf(y)f(y+1) \\\\\n&\\Rightarrow f(y+1) = y+1 \\Rightarrow \\forall x \\in \\mathbb{R} \\ f(x) = x \\text{ is the second solution.}\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45846,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a natural number $n \\ge 2$, consider an $n \\times n$ board. Let $n^2$ points denote the centers of each of the $1 \\times 1$ squares on this board. What is the largest number of these points that can be marked in such a way that no three marked points form the vertices of a right triangle?\n*(Mykhailo Shtandenko)*",
    "options": [],
    "answer": "2n-2",
    "solution": "We will show that there is an example where the marked number of points satisfies the conditions of the problem. For this, we will denote the centers of all the cells in the first row and the first column, except for the center of the cell at the intersection of the first row and the first column (see Fig. 3). By simple enumeration, it is easy to see that no three marked points are vertices of a right triangle.\n\nSuppose that this is not the maximum possible value for the number of marked points, i.e., it is possible to mark $2n-1$ centers of the marked centers in such a way that no three of them form the vertices of a right triangle. It follows that for each marked point, this is the only marked point in the row or column with this point. Let there be $a_1, a_2, ..., a_n$ marked points in the columns respectively. If $a_i > 1$ for some $i$, then for each of the marked points in the $i$-th column, this point is the only one in its row, otherwise a triangular rectangle will be formed. Therefore, the total number of such points for which $a_i > 1$ is no more than $n$ — the total number of rows. But if there are exactly $n$ of these points, then the total number of marked points is also $n$, which does not exceed $2n-2$. Otherwise, the sum of all $a_i$ greater than one does not exceed $n-1$. On the other hand, the sum of all $a_i$ that are equal to one is also no more than $n$. If there are exactly $n$ of these points, then again all the marked points are $n$, which does not exceed $2n-2$. Thus, the number of both types of marked points is no more than $n-1$, and therefore their total number is no more than $2n-2$. Thus, there cannot be more marked points than this value.\n\n![](attached_image_1.png)\n**Fig. 3**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45847,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 100 cards, each with one of the numbers $1$, $2$, $\\ldots$, $100$ written on it, such that each number appears on exactly one card. The cards are stacked in such a way that the numbers $1$, $2$, $\\ldots$, $100$ are written from top to bottom in order. Petrik rearranges the cards according to the following rules. If, before his $k$-th move, the numbers on the cards from top to bottom are arranged as follows: $a_1, a_2, \\ldots, a_{k-1}, a_k, a_{k+1}, \\ldots, a_{100}$, then after his move they are arranged as follows: $a_k, a_{k-1}, \\ldots, a_2, a_1, a_{k+1}, \\ldots, a_{100}$ (so the order of the cards does not change when $k=1$). Petryk takes turns making moves $1$, $2$, $\\ldots$, $100$, and then makes moves $1$, $2$, $\\ldots$, $100$ again, and so on. Will the initial arrangement of the cards with the numbers $1$, $2$, $\\ldots$, $100$ from top to bottom necessarily be repeated after a finite number of moves?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider two moves: $2k-1$ and $2k$. Suppose that before the first of them there was a layout of cards $a_1, a_2, \\ldots, a_{2k-1}, a_{2k}, a_{2k+1}, \\ldots, a_{100}$, then after two moves we will have the following changes:\n$$\na_{2k-1}, a_{2k-2}, \\dots, a_1, a_{2k}, a_{2k+1}, \\dots, a_{100} \\to a_{2k}, a_1, a_2, \\dots, a_{2k-1}, a_{2k+1}, \\dots, a_{100}.\n$$\nThus, after each pair of moves, the corresponding even number is moved to the first place. Therefore, after 100 moves, we will have such a layout of cards: $100$, $98$, $96$, $\\ldots$, $4$, $2$, $1$, $3$, $5$, $\\ldots$, $97$, $99$. We will call this perturbation a *megamove*. We will call the initial position $A_0$, the position after the first megamove $A_1$, after the second $A_2$, and so on. It is not difficult to understand that if the position $A_m$ arises from the position $A_{m+1}$, then vice versa, it cannot arise from any other position except for $A_m$. Therefore, let us consider the situation after a sufficiently large number of megamoves. They have to start repeating because there is a finite number of them possible. But if the position is repeated, then all previous ones coincide. Thus, the initial position had to repeat as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45848,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ lamps and $2024$ switches in a room. Each lamp is connected to exactly $1000$ switches. When a switch is pressed, the state of each connected lamp changes from \"ON\" to \"OFF\" or vice versa. It is known that by pressing some of the switches, all lamps can be turned on. Prove that this can be achieved by pressing the switches no more than $1012$ times.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Obviously, each switch should not be pressed more than once. Let us divide all switches into two groups: Group I - those switches that must be pressed to turn all lamps on, and Group II - all other switches. Since each lamp is connected to an even number of switches, if all switches in Group II are pressed, then all lamps will also turn on. Indeed, we choose lamp \"A\". If the number of switches connected to \"A\" is even in Group I, then their number is also even in Group II, and pressing them in either group will not change the state of lamp \"A\". Similarly, for an odd number.\n\nIn total, there are $2024$ switches in Groups I and II. Therefore, there are no more than $1012$ switches in at least one of them, which proves the desired answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $x, y, z$ that satisfy the equation: $2^x + 21^y = z^2$.",
    "options": [],
    "answer": "(x, y, z) = (2, 1, 5)",
    "solution": "Let's consider two cases for $x$.\n\nFor odd $x$, we have\n$$\nz^2 \\equiv 2^{2n+1} + 21^y \\equiv 2 + 0 \\equiv 2 \\pmod{3},\n$$\nwhich has no solutions.\n\nLet $x = 2n$. Then\n$$\n21^y = z^2 - 2^{2n} = (z - 2^n)(z + 2^n).\n$$\nSuppose that one of the two prime numbers $3$ or $7$ divides each of the two factors on the right-hand side. Then this prime also divides the number $2z = (z + 2^n) + (z - 2^n)$ and hence $z$. However, this contradicts the equation $2^x = z^2 - 21^y$, in which the right-hand side is divisible by the corresponding prime number, while the left-hand side is not. Therefore, these factors are coprime. Hence, we have the following two cases.\n\nCase 1. $z - 2^n = 1$, $z + 2^n = 21^y \\Rightarrow 2^{n+1} = 21^y - 1 \\Rightarrow 2^{n+1} \\equiv 6 \\pmod 7$, which contradicts the fact that $2^{n+1} \\equiv 1, 2, 4 \\pmod 7$.\n\nCase 2. $z - 2^n = 3^y$, $z + 2^n = 7^y \\Rightarrow z = 2^n + 3^y$ and $2^{n+1} = 7^y - 3^y$. For $y = 1$, we find that $x = 2$ and $z = 5$, which satisfies the equation. For $y \\ge 2$, the expression $7^y - 3^y \\equiv 0 \\pmod 8$ is only possible for even $y$. If $y = 2m$, then $2^{n+1} = (7^m - 3^m)(7^m + 3^m)$. But under these conditions, $7^m + 3^m \\equiv 2 \\pmod 4$ and it is greater than $4$, so it cannot be a power of $2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Petryk solved 33 problems at the exam. For the lesser part of them, including the first problem, he got $a$ points, while for the rest of them he got $b$ points. It is known that the natural numbers $a$ and $b$ satisfy the condition: $1 \\le b < a \\le 10$. After the exam, Petrik calculated the average score for all problems and it turned out to be an integer. For solving how many problems did Petrik get $a$ points?\n**Answer:** 11.",
    "options": [],
    "answer": "11",
    "solution": "Let Petryk solve $n$ problems for $a$ points and $(33-n)$ problems for $b$ points. Then the average result of Petrik is:\n$$\nS = \\frac{na + (33 - n)b}{33} = \\frac{n(a - b) + 33b}{33} = \\frac{n(a - b)}{33} + b.\n$$\nFor this number to be an integer, $n(a-b) \\equiv 0 \\pmod{33}$ is required. Since $1 \\le a-b \\le 9$ and $n < 33$, then $n=11$ or $n=22$. Since for the score of $a$ he solved a smaller part of the problems, there remains only one possible option $n=11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45851,
    "subject": "Mathematics (Multi-modal)",
    "question": "11 points are given on a circle. Petrik numbered them by the numbers $1, 2, \\ldots, 11$. After that, pairs of points were connected by segments: $1$ and $2$, $2$ and $3$, $\\ldots$, $10$ and $11$, $11$ and $1$. What is the largest possible number of intersection points of these segments? The given 11 points are not counted as intersection points.\n\n![](attached_image_1.png)\n**Fig. 4**",
    "options": [],
    "answer": "44",
    "solution": "Consider one of the drawn segments. It cannot intersect itself, nor the adjacent segments on either side, e.g., the segment $3$–$4$ does not intersect the segments $2$–$3$, $3$–$4$, and $4$–$5$. Thus the maximum number of possible intersection points occurs when each of the segments intersects all the other $8$ non-adjacent segments. This can be achieved by numbering the points around the circle in the following way (see Fig. 4):\n$$\n1 - 3 - 5 - 7 - 9 - 11 - 2 - 4 - 6 - 8 - 10 - 1.\n$$\nThe number of intersection points is equal to $\\frac{1}{2} \\cdot 11 \\cdot 8 = 44$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45852,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a far, far galaxy there are 225 inhabited planets. Between some pairs of inhabited planets there is a two-way space connection, and from each planet you can get to any other planet (possibly with several transfers). The influence of a planet is defined as the number of other planets with which this planet has a direct connection. It is known that if two planets are not connected by a direct space flight, then they have different influence. What is the smallest number of connections possible under these conditions?",
    "options": [],
    "answer": "1593",
    "solution": "Let's reformulate this problem in terms of graphs. Planets are vertices of the graph, direct flights are edges. It is known that in any graph, any two vertices that are not connected by an edge have different degrees. The question is how many edges can be in such a graph, the number of vertices being irrelevant. First, let's prove the following lemma.\n\n**Lemma 1.** In a graph, there are no more than $k + 1$ vertices of degree $k$ for any natural number $k$.\n\n*Proof.* By contradiction. Let $M$ be a set of at least $k + 2$ vertices, each of which has degree $k$. If at least two vertices in $M$ are not connected, then the condition of the problem is violated. Thus, every two vertices in $M$ are connected. But then the degree of each vertex is at least $k + 1$, which contradicts the condition. This contradiction completes the proof.\n\n*Lemma proved.*\n\n**Lemma 2.** For any natural number $k \\ge 3$, if the minimum number of edges in a simple graph, then there are at most $k$ vertices of degree $k$.\n\n*Proof.* By contradiction. From the previous lemma, there can be at most $k + 1$ vertices of degree $k$. If there are fewer than $k + 1$, then the statement is proven. Suppose that for some $k$, there exists a vertex of degree $k + 1$. But then they are all connected to each other and there are no other vertices in the graph. Thus, we have a complete graph on ($k + 1$) vertices.\n\nNow consider a complete graph on $k$ vertices and one vertex connected to one of the $k$ vertices. Then, there is one vertex of degree 1, one vertex of degree $k$, and $k - 1$ vertices of degree $k - 1$. Thus, the vertices not connected to the added vertex have different degrees. Let's count the number of edges for both cases. For the complete graph on ($k + 1$) vertices, there are $\\frac{1}{2}k(k + 1)$ edges, and for the second case, a complete graph on $k$ vertices with an additional edge, there are $\\frac{1}{2}(k-1)k + 1$ edges.\n\nThen,\n$$\n\\frac{k(k+1)}{2} > \\frac{(k-1)k}{2} + 1 \\Leftrightarrow k^2 + k > k^2 - k + 2 \\Leftrightarrow k > 1,\n$$\nso the number of edges has decreased. This contradiction completes the *proof of the lemma*.\n\n**Lemma 3.** If we decrease the degree of at least one vertex in a graph, the total number of edges will decrease.\n\n*Proof.* It is sufficient to recall the formula for the sum of the degrees of all vertices $S$ and the number of edges $R$. Clearly, $S = 2R$. Therefore, if $S$ decreases, $R$ also decreases.\n\n*Lemma proved.*\n\nThus, let us assume we have 225 vertices. We will find a value of $k$ for which the following inequality holds:\n$$\n1 + 2 + \\cdots + k = \\frac{k(k+1)}{2} \\le 225 < 1 + 2 + \\cdots + k + (k+1) = \\frac{(k+1)(k+2)}{2}.\n$$\nThis value of $k$ is 20. Let\n$$\nl = 225 - (1 + 2 + \\cdots + k) = 225 - 210 = 15.\n$$\nThen for the minimum number of edges, we should have 1 vertex of degree 1, 2 vertices of degree 2, ..., $k$ vertices of degree $k$. For the remaining $l$ vertices, there are several options. The smallest number of edges would be if all these vertices had degree $k+1$. But then the total number of edges leaving each vertex would be:\n$$\nL = 1 \\cdot 1 + 2 \\cdot 2 + \\cdots + 20 \\cdot 20 + 15 \\cdot 21 = 3185.\n$$\nIn this count, each edge is counted twice, so this number should be even. Since this number is odd, the minimum number of edges should be $\\frac{1}{2}(L+1) = 1593$. For this case, there should be 14 vertices of degree 21 and 1 vertex of degree 22. It remains to show that such a situation is possible.\n\nIn this way, we should have 1 vertex of degree 1, 2 vertices of degree 2, ..., 20 vertices of degree 20, 14 vertices of degree 21, and 1 vertex of degree 22. It remains to construct an example of a graph that satisfies the conditions of the problem. First, we build complete graphs for vertices of degree 2, 3, ..., 20. Then we build a complete graph for the remaining 15 vertices. Thus, the condition of having vertices with the same degree is satisfied.\n\nNext, we need to make the graph connected and ensure that no pair of vertices is connected twice. The following edges remain unconnected: 1 edge for each vertex of degrees 1 through 20, and from the last group of 15 vertices, we have 14 vertices of degree 7 and 1 vertex of degree 8, where the last vertices are already connected. This follows from the fact that they are connected to each other by 14 edges. Thus, the first group has $1 + 2 + \\cdots + 20 = 210$ edges, and the second group has $14 \\cdot 7 + 8 = 106$ edges.\nWe then connect each group from degree 1 to degree 20 with a common edge, and connect the last group to the vertex of degree 8 (Fig. 10).\n\n![](attached_image_1.png)\n\n**Fig. 10**\n\nIn this way, we use 39 edges from the first group of 210 edges and 1 edge from the second group of 106 edges, and achieve a connected graph. We are left with 105 edges that need to be connected to the 105 edges from the first group, but in a way that evenly distributes them, starting from the groups with the maximum number of edges. Obviously, the remaining edges can be connected easily.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n = 2m + 1$ for some integer $m$. $n$ points $A_1, A_2, \\dots, A_n$ are chosen on two parallel lines. What is the largest possible number of acute triangles among the triangles $A_iA_jA_k$ over $1 \\le i < j < k \\le n$?",
    "options": [],
    "answer": "m(m+1)(2m+1)/6",
    "solution": "Let us put points on these two lines in such a way that projections of every two adjacent points from one line to another surround exactly one point on another line and there are no obtuse angles $\\angle ABC$ such that $B$ lies on one line and $A, C$ on another one (clearly, we can achieve this if distances between points will be sufficiently small). It's easy to check that in this case there are exactly $\\frac{1}{6}m(m+1)(2m+1)$ acute triangles.\n\nLet us project all the points on one of the two lines and color projections of the points from one line in white and from another one in black. Then an acute triangle may be obtained only from points $A, B, C$ such that $B$ lies between $A$ and $C$ and $A$ and $C$ have the same color which is different from $B$ (but we should note that the converse is not necessarily true). We will show by induction that there are at most $1^2 + 2^2 + \\cdots + m^2 = \\frac{1}{6}m(m+1)(2m+1)$ such good triples if there are $2m+1$ points.\n\nFor $m=0$ the statement is obvious. Suppose that we proved it for $m-1$. Let us show that this is also true for $m$. We number the points as $A_1, A_2, \\dots, A_{2m+1}$ and consider a coloring with the largest number of such triples.\n\nAssume that $A_1, A_{2m+1}$ have the same color (WLOG it's white). Let among the points from $A_2$ to $A_{2m}$ exactly $w$ are white and $b$ are black points. Observe that there are at most $1^2 + 2^2 + \\cdots + (m-1)^2$ good triples among the points from $A_2$ to $A_{2m}$, $b$ triples which contain $A_1$ and $A_{2m+1}$ and also $bw$ triples which contain exactly one point from $A_1$ and $A_{2m+1}$ (because for every pair of points of different colors, there is exactly one point from $A_1$ and $A_{2m+1}$ which will form a good triple with them). So, we have at most $1^2 + 2^2 + \\cdots + (m-1)^2 + b(w+1)$ good triples. Observe also that $b+w=2m-1$, hence $b(w+1) \\le m^2$ and there are at most $1^2 + 2^2 + \\cdots + (m-1)^2 + m^2$ good triples and in this case the induction proceeds.\n\nFrom now we will assume that $A_1$ and $A_{2m+1}$ have different colors.\n\nConsider any two adjacent points $A_i, A_{i+1}$ such that $A_i$ is white and $A_{i+1}$ is black. Let there are $w_1$ white and $b_1$ black points among the points from $A_1$ to $A_{i-1}$, and $w_2$ white and $b_2$ black points among the points from $A_{i+2}$ to $A_{2m+1}$. Then if we swap $A_i$ and $A_{i+1}$, the number of good triples will increase by $(b_2+w_1)-(b_1+w_2)$. Since our coloring is optimal by assumption then $(b_1+w_2) \\ge (b_2+w_1)$ and so $(w_1-b_1) \\le (w_2-b_2)$.\n\nWLOG $A_1$ is white and $A_{2m+1}$ is black. Let $A_k$ be the first black point and $A_l$ be the last white ($k<l$, because otherwise, we don't have good triples). Let us apply the statement above for the points $A_{k-1}, A_k$: for the points to the left of them the difference (white points)–(black points) = $k-2 \\ge 0$ and so for the points to the right of them this is non-negative. Hence, the difference (white points)–(black points) for all the points is non-negative. On the other hand, let us apply this statement for the points $A_l, A_{l+1}$: for the points to the right of them the difference (white points)–(black points) = $-(2m+1-l-1)=l-2m \\le 0$ and so for the points to the left of them this is non-positive. Hence, the difference (white points)–(black points) for all the points is non-positive. But before we showed that this difference is non-negative, so it's zero and then the number of points is even, so we got a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45854,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square of size $n \\times n$ is given. Some of its $1 \\times 1$ cells are marked. It turned out that there is no convex quadrilateral with vertices at these marked points. For each natural number $n \\ge 3$, find the largest value of $m$ for which this is possible.\nA quadrilateral is called convex if both of its diagonals lie inside the quadrilateral.\n\n![](attached_image_1.png)\nFig. 6",
    "options": [],
    "answer": "n + 2",
    "solution": "We mark two points in the corner cells of the left column, as well as all the points in some non-edge row. Then we have $n + 2$ marked points, none of which form a vertex of a convex quadrilateral (Fig. 6).\n\nWe will show by contradiction that it is not possible to mark more than $n + 2$ points. Suppose at least $n + 3$ cell centers are marked. It is clear that if there are two rows, each with at least 2 marked points, then by taking exactly 2 points from each of these two rows, we obtain a convex quadrilateral. Otherwise, in at least the $(n-1)$-th row, no more than 1 point is marked. Then there must be a row in which at least 4 points are marked. Similarly, there must be a column in which at least 4 points are marked. Let this row and column intersect at point $A$. It is easy to see that there are at least 2 points in this row that lie on one side of point $A$, and similar 2 points can be found for the column. These 4 points form a convex quadrilateral.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a convex 2023-gon on the Cartesian plane with vertices at points whose coordinates are both integers, such that all its side lengths are equal?",
    "options": [],
    "answer": "No, such a polygon does not exist.",
    "solution": "Suppose such a 2023-gon exists.\nLet its side be denoted by $a$, so $a^2$ is an integer, and its vertices as $(x_1, y_1)$, $(x_2, y_2)$, ..., $(x_{2023}, y_{2023})$. Consider the 2023-gon with the smallest value of $a^2$. We have $(x_i - x_{i+1})^2 + (y_i - y_{i+1})^2 = a^2$ for each $i$, where $x_{2024} = x_1$, $y_{2024} = y_1$.\nIf $a^2$ is a multiple of 4, then since if the sum of two squares of integers is a multiple of 4, then both numbers are even, we have $x_i \\equiv x_{i+1} \\pmod{2}$, $y_i \\equiv y_{i+1} \\pmod{2}$ for each $i$. But then we can consider a polygon with half the number of vertices with vertices at $(\\frac{x_i-x_1}{2}, \\frac{y_i-y_1}{2})$, whose vertices are also all integer points, and whose side length is $\\frac{a}{2}$, obtaining a contradiction.\n\nIf $a^2 \\equiv 2 \\pmod 4$, then $x_i$ and $x_{i+1}$ have different parity for each $i$, which leads to a contradiction, since we obtain that $x_1$ and $x_1$ have different parity.\nIf $a^2 \\equiv 1 \\pmod 2$, then $x_i + y_i$ and $x_{i+1} + y_{i+1}$ have different parity for each $i$, which leads to a contradiction, since we obtain that $x_1 + y_1$ and $x_1 + y_1$ have different parity.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of non-negative integers $x \\ge y$ for which $x + 3^y$ and $y + 3^x$ are two consecutive integers.",
    "options": [],
    "answer": "(x, y) = (1, 0)",
    "solution": "Obviously, $x > y$ (since $x + 3^y \\neq y + 3^x$). First, we prove the following lemma.\n\n**Lemma.** For any natural number $n > 1$, the inequality $3^n > n + 2$ holds.\n\n*Proof.* We prove the statement by induction. For $n=2$, we have that $3^2 = 9 > 2 + 2 = 4$. Suppose that the statement holds for some $n \\ge 2$. Then\n$$\n3^{n+1} = 3 \\cdot 3^n > 3 \\cdot (n+2) = 3n + 6 > n + 3 = (n+1) + 2.\n$$\n*Lemma proved.*\n\nNote that the equality $3^n = n + 2$ is only achieved when $n = 1$.\n\nConsider the difference:\n$$\n1 = y + 3^x - x - 3^y = 3^y(3^{x-y} - 1) - (x-y) \\geq 1 \\cdot ((x-y) + 2 - 1) - (x-y) = 1.\n$$\nFor this equality to hold, all intermediate inequalities must also be equalities, so the following conditions must be satisfied: $3^y = 1$ and $x - y = 1$, which gives us the answer stated above.\n\nIf $1 = (x + 3^y) - (y + 3^x)$, then\n$$\n3^x - 3^y = x - y - 1 \\Rightarrow 3^y(3^{x-y} - 1) = x - y - 1 \\Rightarrow 3^{x-y} - 1 \\leq x - y - 1 \\Rightarrow 3^{x-y} \\leq x - y\n$$\nBut we have proved that for all natural numbers $n$, $3^n \\ge n + 2 > n \\Rightarrow 3^n > n$, so we obtain a contradiction in this case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45857,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\ge 3$ children standing in a circle, each of them has two cards, one with the digit $0$ and the other with the digit $1$. At a certain moment, each child raises one of their cards at their discretion. Then every minute, each child whose card number is different from the numbers on both of their neighbors' cards (on the left and on the right) changes their card. Can the situation last indefinitely, when at least one child changes their card?",
    "options": [],
    "answer": "Yes, if the number of children is even (an alternating arrangement yields perpetual flipping); no, if the number is odd (the process eventually stops).",
    "solution": "For even $n$, at the beginning, the kids flip every other sign, starting with $0$ and $1$. Then, every minute, they continue to flip the signs alternately, and this will go on forever.\n\nFor odd $n$, such a distribution is not possible. If two identical digits are next to each other, they will remain so forever. Such signs are called *stable*. All unstable signs change every minute. Let's consider any group of stable digits that are next to each other. We also consider the group of unstable signs that are next to the stable group. It consists of digits that alternate: $0-1-0-1-\\dots$ (or in a different order). Such a group is adjacent to a stable group on each side, and therefore loses two end elements that are added to the stable group. Thus, all unstable groups will disappear after some time, and the process will stop.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45858,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ with an angle $\\angle A = 60^\\circ$, the Euler line intersects the circumcircle of the triangle at the points $U$ and $V$. $AD$ is the height $\\Delta ABC$ and $\\omega$ is the nine-point circle. The circumcircle $\\Omega$ of triangle $\\Delta UVD$ intersects the $\\omega$ for the second time at the point $K$. $S$ is a point such that $KS$ – diameter of $\\omega$. Prove that $AS$ is the symmedian of $\\Delta ABC$.\n\n*An Euler line is a line through the orthocenter, the center of the circumcenter, and the point where the medians of the triangle intersect.*\n*The nine-point circle is the circle on which the bases of the heights, the midpoints of the sides, and the midpoints of the segments between the vertices and the orthocenter lie.*\n*The symmedian of a triangle from a vertex is a line symmetric to the median from the vertex about the bisector from the vertex.*\n\n![](attached_image_1.png)\nFig. 23",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ and $O$ be the orthocentre and the centre of the circumcircle of $\\triangle ABC$ respectively, and $E$ be the centre of $\\omega$. Then, as you know $E$ be the centre of $HO$. Let $M$ be the midpoint of $BC$. Let the circle $\\Omega$ intersect $BC$ for the second time at point $X$. Since $\\angle CAB = 60^\\circ$, then $\\angle BHC = \\angle BOC = 120^\\circ$, so points $B, H, O$ and $C$ are on the same circle (fig. 23). Let $T$ be the point of intersection of the lines $UV$ and $BC$, then, according to the degree of the point, $TD \\cdot TX = TU \\cdot TV = TB \\cdot TC = TH \\cdot TO$. Thus, the points lie on the same circle, and since $\\angle HDX = 90^\\circ$, then $\\angle HOX = 90^\\circ$, whence $XO \\perp HO$. And remembering that $OU = OV$, we get that $XO$ bisects chord $UV$ of the circle $\\Omega$, from where $XO$ passes through the centre of the circle $\\Omega$, and hence $XO$ passes through the point diametrically opposite to the point $X$ on the circle $\\Omega$. However $\\angle XDA = 90^\\circ$, hence $Z$ is the point diametrically opposite to $X$ on the circle $\\omega$. So $XO$ passes through $Z$. In the right $\\triangle BOM$, $\\angle BOM = \\frac{1}{2}\\angle BOC = \\angle BAC = 60^\\circ$, so\n\n$AH = BO = AO$. Thus, the condition $\\angle BAC = 60^\\circ$ implies that $AH = AO$, and therefore $AE \\perp HO$, with $AE$ being the bisector $\\angle HAO$, and hence $AE$ is also the bisector of $\\angle CAB$. The condition $\\angle BAC = 60^\\circ$ implies that the radius of the circle $\\omega$ is $\\frac{1}{2}AO = \\frac{1}{2}AH = OM$. Since $\\angle HOZ = 90^\\circ$ the condition $AH = AO$ implies that $A$ is the midpoint of the segment $HZ$. Denote by $W$ the point that is symmetric to $O$ with respect to $M$. Then $W$ is also symmetric to $O$ with respect to $BC$, since $OM \\perp BC$, so $BW = WC$ and $\\angle BWC = 120^\\circ$, whence $W$ is the midpoint of the smaller arc $BC$ of the circumcircle $\\triangle ABC$, and therefore $W$ lies on $AE$. Moreover, since $OA = OW$ and $OE \\perp AW$, then $E$ is the midpoint of the segment $AW$. In the trapezoid $ZOWH$, $A$ and $M$ are the midpoints of the bases $ZH$ and $OW$, so, by a well-known fact, the point of intersection of its diagonals $ZW \\cap HO$ lies on $AM$, but $AM \\cap HO = G$ is the centroid of $\\triangle ABC$, so $G$ lies on $ZW$. Let us prove the collinearity of the points $Z, K$ and $W$, which implies that $G, K, Z$ and $W$ lie on the same line. To do this, let $K'$ be the projection of point $X$ onto $ZW$. Moreover, since $OA=OW$, and $OE \\perp AW$, $E$ is the midpoint of the segment $AW$. In the trapezoid $ZOWH$, $A$ and $M$ are the midpoints of the bases $ZH$ and $OW$, so, by a well-known fact, the point of intersection of its diagonals $ZW \\cap HO$ lies on $AM$, but $AM \\cap HO=G$ is the centroid of $\\triangle ABC$, so $G$ lies on $ZW$. Let us prove the collinearity of the points $Z, K$ and $W$, which implies that $G, K, Z$ and $W$ lie on the same line. To do this, let $K'$ be the projection of point $X$ onto $ZW$. Then, since, $\\angle XK'Z = \\angle XDZ = 90^\\circ$, the points $X, Z, D$ to $K'$ are cyclic, and therefore $K'$ lies on the circle $\\Omega$. However $K'$ lies on $ZW$, so to prove the collinearity of the points $Z, K$ and $W$, it is enough to show that $K'$ lies on the circle $\\omega$, which implies that $K'$ is the second intersection of the circles $\\Omega$ and $\\omega$, i.e. $K' = K$, and therefore $K$ like $K'$ will lie on the line $ZW$. We have that, $\\angle WK'X = \\angle WMX = 90^\\circ$, so the points $X, M, K'$ and $W$ lie on the same circle, and hence $\\angle MK'X = \\angle MWX = \\angle MOX$ (this equality is obtained from the axial symmetry of points $O$ and $W$ with respect to line $BC$). However, the two lines $ZD$ and $OM$ are perpendicular to $BC$, so they are parallel, so $\\angle MOX = \\angle DZX$, i.e. $\\angle MK'X = \\angle DZX$. Since the points $X, Z, K'$ and $D$ are cyclic, then $\\angle DK'X = 180^\\circ - \\angle DZX$. Therefore\n$$\n\\angle DK'M = \\angle DK'X - \\angle MK'X = 180^\\circ - \\angle DZX - \\angle DZX = 180^\\circ - 2\\angle DZX.\n$$\n\n$$\n\\angle DK'M = 180^\\circ - 2\\angle DZX = 180^\\circ - 2\\angle DAE = 180^\\circ - \\angle DAO.\n$$\nLet $N$ be the midpoint of $AH$, then, as we know, $N$ lies on $\\omega$, and $\\angle MDN$ is right, so $MN$ is the diameter of the circle $\\omega$, so the points $M, E$ and $N$ are collinear. Moreover, since $AN = \\frac{1}{2}AH = OM$ and $AH\\parallel OM$, then $AOMN$ is a parallelogram, hence $MN\\parallel AO$, and therefore $\\angle DAO = \\angle DNM$. Thus, as we proved above, $\\angle DK'M = 180^\\circ - \\angle DAO = 180^\\circ - \\angle DNM$, and hence $\\angle DK'M + \\angle DNM = 180^\\circ$, so the point $K'$ lies on the circumcircle $\\Delta DNM$ – $\\omega$, which proves the collinearity of points $G, K, Z$ and $W$. In particular points $G, K$ and $W$ lie on the same line. Let $P$ be a point symmetric to $M$ with respect to $HO$. Then, since $A$ and $W$ are also symmetric with respect to $HO$, then $APMW$ is an isosceles trapezoid, so the point of intersection of its diagonals $PW \\cap AM$ lies on $HO$. However $G = HO \\cap AM$, so $T$ lies on $GW$, which passes through $K$. So $S$ is the second intersection of $KE$ and the circle $\\omega$. Then $\\angle KSM = \\angle KTM$, and from the isosceles trapezoid $APMW$, $\\angle KTM = \\angle EAM$, so $\\angle EAM = \\angle ESM$, from which points $A, S, E$ and $M$ lie on the same circle, but $ES = EM$, from which $AE$ is the bisector of $\\angle SAM$, that is $AS$ is the symmedian $\\Delta ABC$, which was the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Anna has placed real numbers with sum $S$ in the cells of a row. It turned out that she cannot cut the row into two parts so that the sum of the numbers in one part is positive and in the other part is negative. Prove that the modulus of $S$ is not less than any of Anna's numbers.\n(Oleksii Masalitin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's assume that $S = 0$. Let's choose an arbitrary division of the string into two parts. It is clear that the sum of the numbers in the two parts is $0$, so one of them is not less than $0$, and the other is not greater than $0$. If they are not $0$, we get a contradiction, so the sum of the numbers of any smaller string is $0$, so all the numbers in the string are $0$, which is what we need to prove.\n\nLet $S \\neq 0$, without restriction of generality let $S > 0$. Let $a$ be any number in the string. Consider an arbitrary division of the string into two parts: either both sums are not less than $0$, or not greater than $0$. It is clear that the second option is impossible, because the sum of two nonnegative integers cannot be equal to $S > 0$. Then the sum of the numbers of any lesser row is not less than $0$.\n\nLet $x$ and $y$ denote the sum of the numbers to the left and right of $a$ respectively (if there are no such numbers, then we assume the corresponding variable is $0$). Then, from the above, $x, y \\geq 0$ and $x + a, y + a \\geq 0$, therefore $x + y \\geq 0$ and $x + y + 2a \\geq 0$. By definition, $S = x + y + a$, so $S - a \\geq 0$ and $S + a \\geq 0$, i.e. $S \\leq a \\leq S$, therefore we get $|a| \\leq S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to write positive integers in the cells of the board $2022 \\times 2022$ in such a way that the sum of the numbers in any rectangle $R$ is a perfect square if and only if $R$ is a square?\n(Arsenii Nikolaiev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Yes, you can.\n\nNumber the columns from left to right and the rows from top to bottom with the numbers $1, 2, \\dots, 2022$. Write in the cell $(r, c)$, i.e. the one in the $r$-th row and $c$-th column, the number $x^{2(r+c)}$, where the number $x$ is selected later.\n\nConsider an arbitrary rectangle $R$, which lies in rows with numbers $s, s + 1, \\dots, t$ and columns with numbers $u, u + 1, \\dots, v$. By construction, the sum of all numbers in the rectangle $R$ is equal to\n$$\n(x^{2s} + x^{2s+2} + \\dots + x^{2t}) \\cdot (x^{2u} + x^{2u+2} + \\dots + x^{2v}) = x^{2s+2u} \\cdot \\frac{x^{2a} - 1}{x^2 - 1} \\cdot \\frac{x^{2b} - 1}{x^2 - 1},\n$$\nwhere $a = t - s + 1$ and $b = v - u + 1$. It is immediately clear that if $a = b$, the sum of numbers in $R$ is the perfect square of any real number $x$. Now it remains to find such an $x$ that no matter how $a \\neq b$, where $a, b \\in \\{2, 3, \\dots, 2021\\}$, the number $(x^{2a} - 1)(x^{2b} - 1)$ would not be a perfect square.\n\n*Proving.* Let $P(x) = x^{2n} + a_{2n-1}x^{2n-1} + \\dots + a_1x + a_0$. We can immediately see that starting from some point $P(x)$ is positive, because its highest coefficient is $1$. In the following, we will consider only $x$ from this point onwards. First, let's find the following polynomial $Q(x) = x^n + b_{n-1}x^{n-1} + \\dots + b_1x + b_0$ with rational coefficients such that the degree of the polynomial $P(x) - Q(x)^2$ is less than $n$. To do this, it is enough to restore the coefficients one by one $b_{n-1}, \\dots, b_1, b_0$, using identities:\n$$\n\\sum_{k=n}^{2n-1} \\sum_{i+j=k} b_i b_j x^k = a_k x^k.\n$$\nThus, we can write the equality $P(x) = Q(x)^2 + R(x)$, $\\deg R < n$. Let's multiply both parts of this equality by $M^2$ so that the polynomials $M^2P(x) = P_1(x)$, $MQ(x) = H(x)$ and $M^2R(x) = S(x)$ become integer coefficients. We have that $P_1(x) = H^2(x) + S(x)$, where again $\\deg S < n$ and all polynomials $P_1, H$ and $S$ have integer coefficients. Suppose that what you want to prove is not true. This would mean that $S(x)$ is not identically zero. We know that there are infinitely many $x$ such that $P_1(x) = M^2P(x)$ is a perfect square. We will see that since the coefficients of the polynomial $S(x)$ are integer and nonnegative and $S(x)$ is not exactly zero, starting from some point $S(x) \\neq 0$, and therefore from the same point $H(x)^2 \\neq P_1(x)$. Then consider sufficiently large positive integers $t$, of which there are infinitely many, such that $P_1(t) = w^2$, where $w$ is a positive integer and $H(t)^2 \\neq P_1(t)$. We see\n$$\n|P_1(t) - H^2(t)| \\geq 2w - 1 \\Rightarrow |S(t)| \\geq 2w - 1 = 2\\sqrt{P(t)} - 1 \\geq 2t^n - 1 > t^n.\n$$\nBut since $\\deg S < n$, the inequality $S(t) > t^n$ can only hold for a finite number of different $t$, which leads to a contradiction.\n\nThe lemma implies that for any given $a$ and $b$, there are only a finite number of $x$ such that $(x^{2a} - 1)(x^{2b} - 1)$ is a perfect square. Indeed, otherwise, there would be a polynomial $Q(x)$ with rational coefficients such that $(x^{2a} - 1)(x^{2b} - 1) = Q^2(x)$. But we can see that there is a complex root $e^{\\frac{2\\pi i}{a}}$ on the left-hand side, which has a multiplicity of $1$, and this is where we get the contradiction. Alternatively, we can see that $(x^{2a} - 1)' = 2a x^{2a-1}$, which has a GCD of $1$ with $(x^{2a} - 1)$, and therefore $x^{2a} - 1$ has no roots of multiplicity. Therefore, for any given $a$ and $b$, there are only a finite number of $x$ such that $(x^{2a} - 1)(x^{2b} - 1)$ is a perfect square. And since there are a finite number of ways to choose a pair $(a, b)$, it means that there is an $x_0$, for which $(x_0^{2a} - 1)(x_0^{2b} - 1)$ is not a perfect square for any positive integers $a$ and $b$ not greater than $2019$. Therefore, we can put $x = x_0$ and the statement is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Oleksii placed positive integers in the cells of the chessboard of size $8 \\times 8$. For each pair of adjacent cells, Fedir wrote down the product of the numbers in them and added all the obtained numbers. Oleksii wrote down the sum of the numbers in each pair of adjacent cells and multiplied all the obtained numbers. It turned out that both numbers have the same last digit, $1$. Prove that at least one of the boys made a mistake in the calculation. For example, for the square $3 \\times 3$ and the given arrangement of numbers (fig. 11), Fedir would write the following numbers: $2$, $6$, $8$, $24$, $15$, $35$, $2$, $6$, $8$, $20$, $18$, $42$ and their sum ends with digit $6$, and Oleksii would write the following numbers: $3$, $5$, $6$, $10$, $8$, $12$, $3$, $5$, $6$, $9$, $9$, $13$ and their product ends with digit $0$.\n\n(Oleksii Masalitin, Fedir Yudin)\n\n| 1 | 2 | 3 |\n|---|---|---|\n| 2 | 4 | 6 |\n| 3 | 5 | 7 |\n\nFig. 11",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that this could happen. Since Oleksii's product is odd, all the factors are odd, and therefore the sum of the numbers in any neighboring cells is odd. But then for any neighboring cells, the parity of the numbers in those cells is different, and therefore the product of those numbers is even, and therefore the sum is even. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45862,
    "subject": "Mathematics (Multi-modal)",
    "question": "A country held a one-round tennis tournament. Participants received $1$ point for winning the match, and $0$ points for losing. There are no draws in tennis. At the end of the tournament, Oleksii saw the number of points scored by each participant, as well as the schedule of all the matches in the tournament, which showed the pairs of players, but not the winners. He chooses a match in some particular order (however he wants) and tries to guess the winner, after which he is told if he is correct. Prove that Oleksii can act in such a way that he is guaranteed to guess the winners of more than half of the matches.\n(Oleksii Masalitin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's choose player $A$, who has the lowest number of points. He has at least as many defeats as wins, so by indicating a defeat in each his game, Oleksii will guess at least half of the results of player $A$. Then Oleksii subtracts the results of player $A$ from the results of each participant and again selects player $B$ with the lowest number of points and repeats this procedure. If the corresponding player has an odd number of games, then Oleksii will guess more than half of the results of this player, if the number of games is even, then at least half, but then he guessed more than half for $A$ and $B$ together, since the number of games for them was of different parity. Let's continue to choose the player with the lowest number of points. In total, he will guess more than half of all outcomes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45863,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integers $x$, $y$ satisfy the conditions:\n$$\n\\{\\sqrt{x^2 + 2y}\\} > \\frac{2}{3}, \\{\\sqrt{y^2 + 2x}\\} > \\frac{2}{3}.\n$$\nProve that $x = y$.\nHere, $\\{a\\} \\in [0; 1)$ denotes the fractional part of the number $a$, that is, there exists an integer $n$ for which the equality $a = n + \\{a\\}$ holds. For example, $\\{3.14\\} = 0.14$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that for some positive integers $x < y$ these inequalities are true: $\\{\\sqrt{x^2 + 2y}\\} > \\frac{2}{3}, \\{\\sqrt{y^2 + 2x}\\} > \\frac{2}{3}$. Note that $y^2 < y^2 + 2x < (y + 1)^2$, so we have\n$$\ny^2 + 2x > \\left(y + \\frac{2}{3}\\right)^2 \\Leftrightarrow 2x > \\frac{4}{3}y + \\frac{4}{9} \\Rightarrow x > \\frac{2}{3}y.\n$$\n\n$$\nx^2 + 2y > \\left(x + 1 + \\frac{2}{3}\\right)^2 \\Leftrightarrow 2y > \\frac{10}{3}x + \\frac{25}{9} \\Rightarrow y > \\frac{5}{3}x.\n$$\nBut then, $xy > xy \\cdot \\frac{2}{3} \\cdot \\frac{5}{3}$, a contradiction that completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose for a rhombus $ABCD$ there exists a point $T$ such that the following conditions are satisfied: $\\angle ATC + \\angle BTD = 180^\\circ$ and the circumcircles of the triangles $ATC$ and $BTD$ are tangent to each other. Prove that the point $T$ is equidistant from the diagonals of the rhombus.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the point of intersection of the rhombus diagonals, $O_1$ and $O_2$ be the centers of the described circles $\\Delta ATC$ and $\\Delta BTD$ respectively (Fig. 1). Hence, the following equalities are true:\n$$\n\\angle A O_1 C = 2 \\angle A T C = 2 \\cdot (180^\\circ - \\angle B T D) = \\angle B O_2 D,\n$$\nso the isosceles triangles $AO_1C$ and $BO_2D$ are similar. Then their altitudes are proportional to their sides, i.e. $\\frac{PO_1}{PO_2} = \\frac{O_1A}{O_2B} = \\frac{O_1T}{O_2T}$. Then $PT$ is a bisector (internal or external) of $\\angle O_1PO_2$, exactly what we needed to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45865,
    "subject": "Mathematics (Multi-modal)",
    "question": "The country has $n \\ge 3$ airports, some pairs of which are connected by bidirectional flights. Every day, the government closes the airport from which the largest number of flights is flying. What is the maximum number of days this can continue?\n\n![](attached_image_1.png)\nFig. 17",
    "options": [],
    "answer": "n - 3 for n ≥ 4; 1 for n = 3",
    "solution": "We will use the obvious interpretation in the language of graphs, and for convenience, we will move on to graph complements. From now on, every day the vertex of strictly lowest degree will be removed.\n\nIt is clear that when 2 vertices remain, nothing else will happen. For $n = 3$ this bound is achieved by connecting two of the three vertices with an edge.\n\nWe show that if at some point there were 4 vertices, there will always be at least 3. Indeed, suppose that there were vertices $A$, $B$, $C$, $D$, and first vertex $D$ was removed, and then $C$. Then the vertex $D$ cannot be connected to any of the vertices $A$, $B$. Also note that in a graph of three vertices $A$, $B$, $C$ the vertex $C$ also cannot be connected to any of the vertices $A$, $B$. But then in the graph on these four vertices the degree of $D$ was at least that of $C$, contradiction.\n\nFor example, consider the following graph: a chain of vertices $A_1$, $A_2$, $A_3$, ..., $A_n$, where every two adjacent vertices are connected, and in addition $A_{n-2}$ is connected to $A_n$ (fig. 17). It is clear that the vertices will be removed one by one, in order $A_1$, $A_2$, $A_3$, ..., $A_{n-3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45866,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given a set of $m$ positive integers such that they all give distinct remainders modulo some positive integer $n$. Prove that for any positive integer $k \\leq m$ this set can be partitioned into $k$ nonempty subsets such that the sums of the numbers in these subsets are also distinct modulo $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let these numbers be $a_1, a_2, ..., a_m$. It is enough to show that you can choose some two of these numbers $a_i, a_j$ (with $i < j$) so that all the numbers $a_1, a_2, ..., a_{i-1}, a_{i+1}, ..., a_{j-1}, a_{j+1}, ..., a_m, a_i + a_j$ give distinct remainders when divided by $n$, then we can combine the numbers $(m-k)$ times and get the statement of the problem.\nIf any of the numbers is divisible by $n$, for example, $a_1$ then we can combine the numbers $a_1, a_2$.\nOtherwise, replace these numbers with their remainders when divided by $n$ and sort them. Let $0 < a_1 < a_2 ... < a_m < n$. Then we can combine $a_1, a_m$. Indeed: for any $j$, $a_1 + a_m > a_j$ and $a_1 + a_m < a_j + n$, so this remainder does not occur among the others.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 45867,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum number of points that can be placed on the plane so that there are exactly $2012$ straight lines, which pass through at least two of them?\n\n(Danylo Mysak)\n\n![](attached_image_1.png)\nFig. 24\n\n![](attached_image_2.png)\n**Fig. 25**",
    "options": [],
    "answer": "2012",
    "solution": "It is easy to place $2012$ points for the condition of the problem being fulfilled: we will place $2011$ points on a straight line, and the last one — outside of it (fig. 24).\n\n*Proof.* Let's draw all the lines through every pair of points from the set $M$. We will find a line and a point from $M$ such that the distance between them is minimal (excluding any pair where the point lies on the line). Let's denote the corresponding line as $l$ and the point as $A$. We will prove by contradiction that there are only two points from $M$ lying on line $l$.\nLet's assume that there are at least three points lying on $l$ from the set $M$, and let's denote them as \"from left to right\" as $B_1$, $B_2$ and $B_3$ (fig. 25). Since $AB_1 + AB_3 > B_1B_3 = B_1B_2 + B_2B_3$, then either $AB_1 > B_1B_2$ or $AB_3 > B_2B_3$. Without loss of generality, we assume that $AB_3 > B_2B_3$. Let us denote by $h_1$ – the height of the triangle $AB_2B_3$ corresponding to the vertex $A$, $h_2$ – the height of the same triangle corresponding to the vertex $B_2$. Then the area of the triangle $AB_2B_3$ is given by $S = \\frac{1}{2}h_1 \\cdot B_2B_3 = \\frac{1}{2}h_2 \\cdot AB_3$, where $h_1 > h_2$ when $B_2B_3 < AB_3$. However it implies that the distance between point $B_2$ and line $AB_3$ is less than the distance between $A$ and $l$. We obtain a contradiction which finishes the *proof of the lemma*.\n\nNow we will prove that having $n$ points on the plane implies that either all of them belong to the same line or there exists at least $n$ different lines such that each contains at least two points among $n$ given. This will lead to the answer $2012$ for the question asked in this problem.\nWe will prove this statement by mathematical induction. If $n=3$ then it is obvious. Suppose that the statement of induction is true for some value $n$. Then we will prove the step of induction for $n+1$. Assume that we have arbitrary $n+1$ points on the plane, not all of which lie on the same line. According to the lemma we can choose two points such that the line through them does not contain any other points. Let us delete one of those two points such that all other points are not on the same line. After this operation we have $n$ points that satisfy the conditions of mathematical induction. Therefore for a new set of points we can find at least $n$ different lines such that each contains at least $2$ of given points. Adding the line constructed in the beginning we will get at least $n+1$ different lines for the initial set of $n+1$ points, which finishes the proof of the step of induction.\n\n**Answer:** $2012$ points.\n\n![](attached_image_1.png)\nFig. 24\n\n![](attached_image_2.png)\n**Fig. 25**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45868,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n \\ge 4$ real numbers are placed on the circle. It is known that for any four consecutive numbers $a, b, c, d$ that go around the circle in this specified order, the condition $a + d = b + c$ holds. For which $n$ can we conclude that all the numbers are equal?",
    "options": [],
    "answer": "all odd n",
    "solution": "We denote the numbers in the circle by $a_1, a_2, \\ldots, a_n$. Choose the largest among them (or any of them, if there are several such numbers). Without loss of generality, let this number be $a_1$. From the condition of the problem we have that\n$$\na_1 + a_2 = a_n + a_3, \\quad a_1 + a_n = a_2 + a_{n-1} \\Rightarrow 2a_1 = a_3 + a_{n-1}.\n$$\nSince $a_1$ is the largest number, the last equality implies that $a_1 = a_3 = a_{n-1}$: that is, every second number is the largest, if we continue with the same reasoning. If $n$ is odd, then all the numbers will be equal.\n\nFor even numbers $n$ it is enough to consider the following example with not all equal numbers: $1, 0, 1, 0, \\ldots, 1, 0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integers $a_1, a_2, ..., a_{101}$ are such that $a_i + 1$ are divisible by $a_{i+1}$ for $1 \\le i \\le 101$ (we assume that $a_{102} = a_1$). What is the largest value that the maximum of these numbers can attain?",
    "options": [],
    "answer": "201",
    "solution": "Without loss of generality, let $a_{101}$ be the largest of these numbers (or one of the largest). It is clear that $a_i + 1 \\ge a_{i+1}$ for any $i = 1, 100$. If we add all these 100 inequalities, we get $a_1 + a_2 + ... + a_{100} + 100 \\ge a_2 + a_3 + ... + a_{101}$, that is $a_1 \\ge a_{101} - 100$. From the condition $a_{101} + 1$ is divisible by $a_1$, and from the fact that $a_{101}$ is the largest we have $a_{101} + 1 > a_1$, so $a_{101} + 1 \\ge 2a_1 \\ge 2(a_{101} - 100) = 2a_{101} - 200$, therefore $a_{101} \\le 201$. It remains to show that there is an example where the maximum is equal to 201. Indeed, consider the set 101, 102, ..., 201, for it $a_i + 1 = a_{i+1}$ for $i = 1, 100$ and $a_{101} + 1 = 2a_1$, so this example satisfies the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45870,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute-angled triangle is given $ABC$ with the circumcircle $\\omega$. The points $F$ on $AC$, $E$ on $AB$ and $P, Q$ on $\\omega$ are selected so that $\\angle AFB = \\angle AEC = \\angle APE = \\angle AQF = 90^\\circ$. Prove that the lines $BC, EF$ and $PQ$ intersect at one point.\n\n![](attached_image_1.png)\n**Fig. 12**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A'$ be the point diametrically opposite to $A$ in $\\omega$, and let $R$ be the projection of $A$ onto $EF$ (fig. 12). Since $\\angle BAA' = 90^\\circ - \\angle ACB = 90^\\circ - \\angle AEF$, the line $AR$ passes through $A'$. Lines $PE$ and $QF$ also pass through $A'$. Since the quadrilaterals $AREP$ and $ARFQ$ are inscribed in circles with diameters $AE$ and $AF$ respectively, we have $A'E \\cdot A'P = A'A \\cdot A'R = A'F \\cdot A'Q$, so the quadrilateral $EFPQ$ is cyclic. Then the lines $BC$, $EF$ and $PQ$ intersect in the radial center of the circumscribed circles of the quadrilaterals $EFPQ$, $BCQP$ and $BCEF$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest real number $C$ for which the following condition is true: for any different positive integers $x, y$ the inequality holds\n$$\n\\{\\{\\sqrt{x^2 + 2y}\\}, \\{\\sqrt{y^2 + 2x}\\}\\} < C.\n$$\n\nHere, $\\{a\\} \\in [0; 1)$ denotes the fractional part of the number $a$, that is, there exists an integer $n$ for which the equality $a = n + \\{a\\}$ holds. For example, $\\{3.14\\} = 0.14$.",
    "options": [],
    "answer": "(\\sqrt{5}-1)/2",
    "solution": "We will show that the desired $C$ is the positive root of the equation $x^2 + x = 1$, $C = \\frac{\\sqrt{5}-1}{2}$.\n\nFirst, suppose that for some positive integers $x < y$ the inequalities $\\sqrt{x^2 + 2y} > C$, $\\{\\sqrt{y^2 + 2x}\\} > C$ hold. Note that $y^2 < y^2 + 2x < (y + 1)^2$, so we have $y^2 + 2x > (y + C)^2 \\Leftrightarrow 2x > 2Cy + C^2 \\Rightarrow x > Cy$. Then $(x+1)^2 < x^2 + 2y < x^2 + \\frac{2}{C}x < x^2 + 4x < (x+2)^2$.\n\nSo $x^2 + 2y > (x + 1 + C)^2 \\Leftrightarrow 2y > 2(C + 1)x + (C + 1)^2 \\Rightarrow y > (C + 1)x$.\nBut then $xy > xy \\cdot C(C + 1) = xy$: a contradiction.\n\nNow let's show that any $C_1 < C$ does not satisfy the condition. Consider $y = [(C + 1)x]$ for some sufficiently large $x$.\nWe show that starting from some $x$ we have $\\sqrt{x^2 + 2y} > C_1$, $\\sqrt{y^2 + 2x} > C_1$.\nIt is easy to see that $y^2 < y^2 + 2x < (y + 1)^2$ and also $(x + 1)^2 < x^2 + 2y < (x + 2)^2$ so it suffices to show that for sufficiently large $x$ it holds that $x^2 + 2y > (x + 1 + C_1)^2$, $y^2 + 2x > (y + C_1)^2$. These inequalities are equivalent to the following:\n$2y > 2x(C_1 + 1) + (C_1 + 1)^2$, $2x > 2yC_1 + C_1^2$. Note that $2y - 2x(C_1 + 1) \\ge 2x(C + 1) - 2 - 2x(C_1 + 1) = 2x(C - C_1) - 2$ which is greater than $(C_1 + 1)^2$ for a sufficiently large $x$. Also note that $2x - 2yC_1 < 2x - 2x(C + 1)C_1 = x(2 - (C + 1)C_1)$. Since $(C + 1)C_1 < (C + 1)C = 1$, this value is greater than $C_1^2$ for a sufficiently large $x$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Are there any 10 numbers, not all of which are the same, each of which is equal to the square of the sum of all the other numbers?",
    "options": [],
    "answer": "No",
    "solution": "Suppose that such numbers exist. Since they are equal to some squares, each of these numbers is nonnegative. Let's denote the sum of all these 10 numbers by $S$. Let's pick one of these numbers and denote it by $a$, then $S \\ge a$, and also\n$$\na = (S - a)^2 \\Rightarrow a^2 - (2S + 1)a + S^2 = 0.\n$$\nIf there are two different $a$, satisfying this condition, then by Vieta's theorem their product is equal to $S^2$. However, this means that one of them is greater than $S$, which contradicts what we proved above. Thus, there is only one possible $a$, and it follows that all numbers are equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside the triangle $ABC$ there exists a point $O$ such that $\\angle BOC = 90^\\circ - \\angle BAC$. The rays $BO$ and $CO$ intersect the sides $AC$ and $AB$ at the points $K$ and $L$ respectively. The points $K_1$ and $L_1$ are selected on the segments $LC$ and $BK$ respectively, so that $BK_1 = K_1K$ and $CL_1 = L_1L$. Let $M$ be the middle of the side $BC$. Prove that $\\angle K_1ML_1$ is a right angle.\n\n(Anton Trygub)\n\n![](attached_image_1.png)\n**Fig. 18**",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K_2$ and $L_2$ be the midpoints of the segments $BK$ and $CL$ respectively (fig. 18). Then $MK_2 \\parallel AC$ and $ML_2 \\parallel AB$. Hence, $\\angle K_2ML_2 = \\angle BAC$. From the isosceles triangles $BK_1K$ and $CL_1L$ we have that $\\angle L_1L_2K_1 = \\angle K_1K_2L_1 = 90^\\circ$ and then $L_1, L_2, K_1, K_2$ lie on the same circle with diameter $K_1L_1$. Also note that\n$$ \\angle K_2K_1L_2 = \\angle K_2L_1L_2 = 90^\\circ - \\angle BOC = \\angle BAC = \\angle K_2ML_2. $$\nSo, $M$ also lies on a circle with diameter $K_1L_1$. Hence, $\\angle K_1ML_1 = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45874,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given $n \\ge 2$ distinct positive integers. A pair of integers is called *elegant* if their sum is $2^k$ for some positive integer $k$. For each $n$ find the largest possible number of *elegant* pairs.\n(Oleksii Masalitin)",
    "options": [],
    "answer": "n-1",
    "solution": "**Answer:** $n-1$\n\nWe prove that there exist at most $n-1$ *elegant* pairs by induction by $n$. The base case for $n=2$ is obvious, we prove the transition. Let the statement be proved for $n-1$, let us prove it for $n$ numbers. Let us denote by $a$ the largest of them, suppose that the number $a$ is contained in some two *elegant* pairs. Then $a+b$ and $a+c$ are powers of two. Let us denote by $m$ a number such that $2^m \\le a < 2^{m+1}$. Then $a+b$ and $a+c$ are greater than $2^m$, but less than $2a < 2^{m+2}$, because $a+b = a+c = 2^{m+1}$, which contradicts the fact that the numbers are pairwise distinct. By induction, among other $n-1$ integers there are no more than $n-2$ *elegant* pairs, so there are no more than $n-1$ in total, which completes the proof of the transition.\n\nIt remains to show that there can be $n-1$ *elegant* pairs for any $n \\ge 2$. Indeed, let us choose $a_1 = 1$, $a_2 = 2^2-1$, $a_3 = 2^3-1$, $\\ldots$, $a_n = 2^n-1$. Then the pairs $(a_1, a_i)$ are elegant for $i = \\underline{2, n}$.\n**Alternative solution.** Consider the following graph with $n$ vertices. Each number corresponds to a vertex, and two vertices are connected by an edge if their corresponding numbers form *an elegant* pair. Suppose that the graph in question has a cycle. Then let the numbers corresponding to the vertices of this cycle be $a_1, a_2, \\ldots, a_t$. This means that the numbers $a_1 + a_2, a_2 + a_3, \\ldots, a_t + a_1$ are powers of two. Without loss of generality, let $a_1$ be the largest of these numbers. Let us denote by $m$ a number such that $2^m \\le a_1 < 2^{m+1}$. Then $a_1 + a_2$ and $a_1 + a_t$ are greater than $2^m$, but less than $2a_1 < 2^{m+2}$, because $a_1 + a_2 = a_1 + a_t = 2^{m+1}$, which contradicts the fact that the numbers are pairwise distinct. Thus, this graph has no cycles, i.e. it has no more than $n-1$ edge.\n\nIt remains to show that $n-1$ there can be an elegant pair for any $n \\ge 2$. Indeed, let us choose $a_1 = 1$, $a_2 = 2^2-1$, $a_3 = 2^3-1$, $\\ldots$, $a_n = 2^n-1$. Then the pairs $(a_1, a_i)$ are elegant for $i = \\underline{2, n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45875,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum possible number of edges in a graph with $2n$ vertices, if there is exactly one way to divide its vertices into $n$ pairs such that in each pair the vertices are connected by an edge?",
    "options": [],
    "answer": "n^2",
    "solution": "Consider this partition into pairs, denote the vertices $A_1, A_2, ..., A_{2n}$ where the vertices $A_{2i-1}, A_{2i}$ are connected by an edge for each $i$ from $1$ to $n$. Note that for every two pairs $(A_{2i-1}, A_{2i})$, $(A_{2j-1}, A_{2j})$, there are at most two edges between them: there cannot be two edges $(A_{2i-1}, A_{2j-1})$, $(A_{2i}, A_{2j})$, as then we could replace the chosen edges with these and still get a perfect pairing, and also the edges $(A_{2i-1}, A_{2j})$, $(A_{2i}, A_{2j-1})$. Thus, there can be no more than $n + 2 \\cdot C_n^2 = n^2$ edges in total.\n\nThe example looks like this: we draw all the edges of the form $(A_{2i-1}, A_{2i})$, and also for each $i < j$ draw the edges $(A_{2i-1}, A_{2j-1})$, $(A_{2i-1}, A_{2j})$. Consider any perfect pairing in this graph. Notice that the vertex $A_2$ is connected only to $A_1$ so it must have an edge $(A_1, A_2)$. Among the remaining vertices, the vertex $A_4$ is connected only to $A_3$ so there must be an edge in the pairing $(A_3, A_4)$. Continuing similarly, all edges $(A_{2i-1}, A_{2i})$ will be in this pairing.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn integer is called formidable if it can be written as a sum of distinct powers of $4$, and successful if it can be written as a sum of distinct powers of $6$. Can $2005$ be written as a sum of a formidable number and a successful number? Prove your answer.",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nSuppose that $2005 = a + b$, where $a$ is formidable and $b$ is successful. Then $a$ must be a sum of some of the powers of $4$ less than $2005$, namely $1, 4, 16, 64, 256, 1024$. Similarly, $b$ must be a sum of some of the numbers $1, 6, 36, 216, 1296$. So, $a + b$ must be a sum of some distinct entries from the list\n$$\n1, 1, 4, 6, 16, 36, 64, 216, 256, 1024, 1296.\n$$\nIf we use both $1024$ and $1296$, we get at least $1024 + 1296 = 2320$ which is too big. But if we omit one of them, the most we can get is\n$$\n1 + 1 + 4 + 6 + 16 + 36 + 64 + 216 + 256 + 1296 = 1896\n$$\nwhich is too small. So there is no way to achieve a sum of $2005$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that\n$$\nn(n+1)(2 n+1)\n$$\nis always divisible by $6$, for $n$ a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe number is even, because either $n$ or $n+1$ is even.\nNow we show it is always divisible by three. Assume for contradiction that it isn't. Then neither $n$ nor $n+1$ is divisible by three, so $n+2$ must be. However, $2 n+1 = 2(n+2) - 3$ is then also a multiple of three, which is a contradiction.\n\nIn fact, one can also notice the result from the fact that\n$$\n1^{2} + 2^{2} + \\cdots + n^{2} = \\frac{n(n+1)(2 n+1)}{6} .\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFifty counters are on a table. Two players alternate taking away 1, 2, 3, 4, or 5 of them. Whoever picks up the last counter is the loser. Who has a winning strategy, the first player or the second?",
    "options": [],
    "answer": "first player",
    "solution": "Solution:\n\nNote that if you make a turn and there is 1 counter left, you have won since the other player must pick up that counter.\n\nIf you make a turn and there are 7 counters left, you can win: if your opponent picks up 1, 2, 3, 4, or 5 of them, you can respectively take $5, 4, 3, 2$, or $1$ of them to leave $1$.\n\nLikewise, if you play and there are $13$ counters left, you can in the same way play to leave $7$ on your next turn.\n\nContinuing in this way, we see that the first player can win by removing $1$ counter, leaving $49$, and then playing on the succeeding turns to leave $43, 37, 31, 25, 19, 13, 7$, and $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAt a party with 100 people, everyone is either a knight, who always tells the truth, or a knave, who always lies. Each person says they shook hands with a different number of knights at the party, from 0 to 99. Each pair of people shook hands at most once, and everyone knows whether each other person is a knight or knave. Determine how many knights were at the party.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nCall the person who said they shook hands with $i$ people person $i$ for each $i$ from $0$ to $99$. Now, if person $99$ is telling the truth, person $99$ must have shaken hands with everyone else, and all the other people must be knights. But then person $0$ would be lying, since they said they shook hands with no knights but must have shaken hands with person $99$. This is a contradiction, so person $99$ must be a knave.\n\nSimilarly, if person $98$ is a knight, person $98$ must have shaken hands with people $0$ to $97$, and they must all be knights. This again means person $0$ would be lying, a contradiction, so person $98$ is also a knave. Continuing in this manner, we find that person $97$ is a knave, as is person $96$, and so on down to person $1$. Thus, only person $0$ is telling the truth, so there were $99$ knaves and one knight at the party.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $P_{1} P_{2} \\ldots P_{100}$ is a regular 100-gon, what is the measure of the angle $\\angle P_{20} P_{2} P_{1}$ in degrees?",
    "options": [],
    "answer": "145.8°",
    "solution": "Solution:\nIntroduce the circumcircle of the polygon. The aforementioned angle cuts out an arc which is $\\frac{81}{100}$ of that of the entire circle, i.e. which measures $\\frac{81}{100} \\cdot 360^{\\circ}$. By the inscribed angle theorem, this means an answer of $\\frac{81}{100} \\cdot 180^{\\circ} = 145.8^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider an $8 \\times 8$ chessboard, on which we place some bishops in the 64 squares. Two bishops are said to attack each other if they lie on a common diagonal.\n\na. Prove that we can place 14 bishops in such a way that no two attack each other.\n\nb. Prove that we cannot do so with 15 bishops.",
    "options": [],
    "answer": "14",
    "solution": "Solution:\n\nFor the first part, here is one maximal arrangement, where the location of the bishops are indicated by the letter $B$.\n\n| $B$ |   |   |   |   |   |   |   |\n| :---: | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| $B$ |   |   |   |   |   |   | $B$ |\n| $B$ |   |   |   |   |   |   | $B$ |\n| $B$ |   |   |   |   |   |   | $B$ |\n| $B$ |   |   |   |   |   |   | $B$ |\n| $B$ |   |   |   |   |   |   | $B$ |\n| $B$ |   |   |   |   |   |   | $B$ |\n| $B$ |   |   |   |   |   |   |   |\n\nTo see that there cannot be 15 bishops, observe that we have highlighted 15 right-down diagonals in the square above. Each diagonal can accommodate at most one bishop. Furthermore, the lower-left corner and the upper-right corner constitute diagonals of size 1 which cannot be both occupied. This gives the bound of 14 bishops.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a$ and $b$ are two positive numbers not greater than $1$ prove that\n$$\n\\frac{a+b}{1+a b} \\leq \\frac{1}{1+a}+\\frac{1}{1+b}\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds if and only if a = b = 1.",
    "solution": "Solution:\nIf we multiply both sides of the inequality with $(1+a b)(1+a)(1+b)$ the required inequality becomes equivalent to\n$$\n\\begin{gathered}\n(a+b)(1+a)(1+b) \\leq (1+a)(1+a b)+(1+b)(1+a b) \\Leftrightarrow \\\\\n(a+b)(1+a+b+a b) \\leq 1+a b+a+a^{2} b+1+a b+b+a b^{2} \\Leftrightarrow \\\\\na+a^{2}+a b+a^{2} b+b+a b+b^{2}+a b^{2} \\leq 1+a b+a+a^{2} b+1+a b+b+a b^{2} \\Leftrightarrow \\\\\na^{2}+b^{2} \\leq 2 .\n\\end{gathered}\n$$\nThe last inequality is true and the equality holds if and only if $a=b=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $k$ and $n$ be positive integers such that $k < 2^{n+1} - 1$. Prove that there is a sum of exactly $n$ powers of $2$ that is divisible by $k$. (Example: if $k = 9$ and $n = 4$, then $2 + 4 + 16 + 32$ is divisible by $9$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWrite $k$ in base $2$ and suppose there are $r$ $1$'s. Since $2^{n+1} - 1$ is the smallest number with at least $n+1$ $1$'s, we conclude $r \\leq n$. Also $k$ clearly divides a sum of $r$ powers of $2$, since $k$ is such a sum. Now we claim that if $k$ divides a sum of $i$ powers of $2$, it also divides a sum of $i+1$ such powers. To see this, suppose that $2^{l}$ is one of our $i$ powers with $l > 0$; then rewrite this as $2^{l-1} + 2^{l-1}$ and keep the remaining $i-1$ powers the same. This gives us the needed sum of $2^{i+1}$ powers. We can always do this unless all $i$ of our powers are equal to $2^{0}$; in this case we have $k \\mid i$, but then $k \\mid 2i$, so we can multiply everything by $2$ and then split one of the resulting $2^{1}$'s as described previously.\n\nNow by induction, it follows that $k$ divides a sum of $i$ powers of $2$ for all $i \\geq r$. In particular, since $n \\geq r$, we have what we need.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAre there natural numbers $a$ and $b$ that make the equation\n$$\n2 a^{2} + 1 = 4 b^{2}\n$$\ntrue? (The natural numbers are the counting numbers, $1, 2, 3, \\ldots$ )",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Suppose that $a$ and $b$ are any natural numbers. Then $2 a^{2}$ is even, and so is $4 b^{2}$. But an even number plus $1$ will be odd, not even, so the equation can never be made to work, i.e. there are no natural numbers that satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhich is larger, $A=20^{19^{20}}$ or $B=19^{20^{19}}$? (Here, $a^{b^{c}}$ means $a^{\\left(b^{c}\right)}$ and not $\\left(a^{b}\right)^{c}$.)",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nThe answer is that $A$ is larger. First,\n$$\n\\begin{aligned}\n\\left(\\frac{19}{20}\\right)^{20} & = 0.95^{20} = 0.9025^{10} > 0.9^{10} \\\\\n& = 0.81^{5} = 0.8^{5} = 0.32768 > 0.05 = \\frac{1}{20}\n\\end{aligned}\n$$\nwhich implies that $19^{20} > 20^{19}$. Thus, $A = 20^{19^{20}} > 20^{20^{19}} > 19^{20^{19}} = B$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine\n$$\nA = 1 + \\frac{1}{2 + \\frac{1}{3 + \\frac{1}{4 + \\ddots_{+} \\frac{1}{2006 + \\frac{1}{2007}}}}} \\text{ and } B = 1 + \\frac{1}{2 + \\frac{1}{3 + \\frac{1}{4 + \\ddots + \\frac{1}{2005 + \\frac{1}{2006}}}}}\n$$\nWhich of the two numbers is greater, $A$ or $B$? Explain your answer!",
    "options": [],
    "answer": "B > A",
    "solution": "Solution:\n\nWe will determine the sign of $A - B$. If that number happens to be positive then $A > B$, otherwise $A < B$. For each $n$ such that $1 \\leq n \\leq 2005$, let us define\n$$\nA_{n} = n + \\frac{1}{n+1 + \\frac{1}{n+2 + \\frac{1}{n+3 + \\ddots' + \\frac{1}{2006 + \\frac{1}{2007}}}}} \\quad B_{n} = n + \\frac{1}{n+1 + \\frac{1}{n+2 + \\frac{1}{n+3 + \\cdots + \\frac{1}{2005 + \\frac{1}{2006}}}}}\n$$\nThen $A_{1} = A$, $B_{1} = B$ and we always have that $A_{i} - B_{i}$ and $A_{i+1} - B_{i+1}$ have different signs. Indeed, since $A_{i} = i + \\frac{1}{A_{i+1}}$ and $B_{i} = i + \\frac{1}{B_{i+1}}$ we have $A_{i} - B_{i} = \\frac{B_{i+1} - A_{i+1}}{A_{i+1} B_{i+1}}$. Therefore if $A_{i} - B_{i} > 0$ then $B_{i+1} - A_{i+1} > 0$ and vice versa. Thus $A_{1} - B_{1}$, $A_{3} - B_{3}$, $\\ldots$, $A_{2005} - B_{2005}$ have the same signs. From\n$$\nA_{2005} - B_{2005} = \\left(2005 + \\frac{1}{2006 + \\frac{1}{2007}}\\right) - \\left(2005 + \\frac{1}{2006}\\right) < 0\n$$\nwe get $A < B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the number of ways to fill a $3 \\times 3$ grid with $0$'s and $1$'s such that each row and column has an even sum.",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\n| $a_{1}$ | $a_{2}$ | $a_{3}$ |\n| :--- | :--- | :--- |\n| $b_{1}$ | $b_{2}$ | $b_{3}$ |\n| $c_{1}$ | $c_{2}$ | $c_{3}$ |\n\nObserve that the condition is equivalent to saying that every entry in the table has the same parity as the sum of the entries in its row (resp. column).\n\nWe claim that upon filling $a_{1}$, $a_{2}$, $b_{1}$, $b_{2}$ there is exactly one way to fill the table. Indeed, we require that\n\n- $a_{3}$ has the same parity as $a_{1} + a_{2}$,\n- $b_{3}$ has the same parity as $b_{1} + b_{2}$,\n- $c_{1}$ has the same parity as $a_{1} + b_{1}$,\n- $c_{2}$ has the same parity as $a_{2} + b_{2}$.\n\nAll that's left to do is check that there is exactly one way to fill in the entry $c_{3}$. We require $c_{3}$ to have the same parity as $a_{3} + b_{3}$. But the parity of this is the same as that of $a_{1} + a_{2} + b_{1} + b_{2}$. Similarly, $c_{1} + c_{2}$ has the same parity as $a_{1} + b_{1} + a_{2} + b_{2}$. Hence there is exactly one way to fill in $c_{3}$.\n\nThus the claim is proved, and the number of ways to fill the entire table is merely the number of ways to fill the $2 \\times 2$ upper-left square, which is $2^{4} = 16$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCDE$ be a convex pentagon with perimeter $1$. Prove that triangle $ACE$ has perimeter less than $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy the triangle inequality, we have $AC < AB + BC$ and $CE < CD + DE$, so $AC + CE + EA < AB + BC + CD + DE + EA$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine a \"word\" to be a string of at most ten letters taken from the English alphabet. (The letters do not have to be distinct.) Prove that the number of \"words\" is divisible by $27$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe total number of $9$- and $10$-letter words is divisible by $27$, because they can be partitioned into groups of $27$ with each group containing a $9$-letter word and the twenty-six $10$-letter words formed by adding a letter at the end of it.\n\nFor the same reason, the number of $7$- and $8$-letter words is divisible by $27$; likewise for $5$- and $6$-, $3$- and $4$-, $1$- and $2$-letter words. Therefore the total number of words is divisible by $27$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$ and $y$ be positive integers. Can $x^{2}+2y$ and $y^{2}+2x$ both be squares?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Because of the symmetry between $x$ and $y$, we can assume that $x \\geq y$. Then\n$$\nx^{2} < x^{2} + 2y \\leq x^{2} + 2x < x^{2} + 2x + 1 = (x+1)^{2}.\n$$\nThus $x^{2} + 2y$, which lies between two consecutive squares, cannot be a square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $g_{1}(x), g_{2}(x), \\ldots, g_{5}(x)$ be polynomials with integer coefficients. Suppose that their product $f(x) = g_{1}(x) g_{2}(x) \\cdots g_{5}(x)$ satisfies $f(1999) = 2000$. Prove that for some $i \\in \\{1,2,3,4,5\\}$, the sum of the coefficients of $g_{i}(x)$ is odd.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFor some $i$, $g_{i}(1999)$ is odd. Indeed, if this were false, each $g_{i}(1999)$ would be divisible by $2$, so their product, $f(1999) = 2000$, would be divisible by $2^{5}$, which is not the case.\n\nNow write $g_{i}(x) = c_{n} x^{n} + c_{n-1} x^{n-1} + \\cdots + c_{0}$. Then\n$$\ng_{i}(1999) - (c_{n} + \\cdots + c_{0}) = c_{n}(1999^{n} - 1) + c_{n-1}(1999^{n-1} - 1) + \\cdots + c_{0}(1 - 1),\n$$\nwhich is even since $1999^{k} - 1$ is even for each $k$. Since $g_{i}(1999)$ is odd, $c_{n} + \\cdots + c_{0}$ is also odd, as desired.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 45892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDo there exist 100 consecutive positive integers such that their sum is a prime number?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nNo. Let $n, n+1, \\ldots, n+99$ be any 100 consecutive positive integers. Then\n$$\nn + (n+1) + (n+2) + \\cdots + (n+99) = 100n + (1+2+\\cdots+99).\n$$\nHowever,\n$$\n1+2+\\cdots+99 = (1+99) + (2+98) + (3+97) + \\cdots + (49+51) + 50 = 49 \\cdot 100 + 50 = 50(2 \\cdot 49 + 1) = 50 \\cdot 99.\n$$\nThus\n$$\nn + (n+1) + \\cdots + (n+99) = 100n + 50 \\cdot 99 = 50(2n + 99)\n$$\nand this is not prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x$ and $y$ are integers such that\n$$\nx^{2} y^{2} = x^{2} + y^{2},\n$$\nprove that $x = y = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe move all the terms to the left side of the equation and add $1$.\n$$\n\\begin{array}{r}\nx^{2} y^{2} - x^{2} - y^{2} + 1 = 1 \\\\\n\\left(x^{2} - 1\\right)\\left(y^{2} - 1\\right) = 1\n\\end{array}\n$$\nSince the factors are integers, they must be both $1$ or both $-1$. If both are $1$, we get $x^{2} = 2$ which is impossible for integer $x$. If both are $-1$, we get $x^{2} = 0$ and $y^{2} = 0$, so both $x$ and $y$ are zero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEach vertex of a cube is labeled with an integer. Prove that there exist four coplanar vertices the sum of whose numbers is even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLabel the vertices of the cube as shown. The quadruples of vertices\n$$\n\\begin{aligned}\n& a \\text{ and } b \\text{ and } c \\text{ and } d \\\\\n& a \\text{ and } b \\text{ and } e \\text{ and } f \\\\\n& c \\text{ and } d \\text{ and } e \\text{ and } f\n\\end{aligned}\n$$\nare all coplanar. Assume that all three quadruples have an odd sum; then the sum of all twelve terms is odd. But these twelve terms add to\n$$\n\\begin{aligned}\n& (a+b+c+d)+(a+b+e+f)+(c+d+e+f) \\\\\n& =2(a+b+c+d+e+f)\n\\end{aligned}\n$$\n![](attached_image_1.png)\nan even number. Contradiction!",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$ and $y$ be integers such that $\\frac{3x + 4y}{5}$ is an integer. Prove that $\\frac{4x - 3y}{5}$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe basic strategy is to combine the facts that $x$, $y$, and $\\frac{3x + 4y}{5}$ are all integers. Here is one solution:\n$$\n2(x) + 1(y) - 2\\left(\\frac{3x + 4y}{5}\\right) = \\frac{5(2x + y) - 2(3x + 4y)}{5} = \\frac{10x + 5y - 6x - 8y}{5} = \\frac{4x - 3y}{5}\n$$\nSince the left side is clearly an integer, so is the right side.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p$, $q$, and $r$ be distinct primes. Prove that $p+q+r+p q r$ is composite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNote that at most one of $p$, $q$, and $r$ can equal $2$. If $p=2$, then $q$ and $r$ are both odd, but $p q r$ is even. Therefore their sum is even. Similarly, if $q$ or $r$ is $2$, then $p+q+r+p q r$ is even. Finally, if $p$, $q$, and $r$ are all odd, then so is $p q r$, and $p+q+r+p q r$ is even. Thus $p+q+r+p q r$ is always divisible by $2$, and since it is obviously greater than $2$, it must be composite.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine whether there exists a natural number having exactly $10$ divisors (including itself and $1$), each ending in a different digit.",
    "options": [],
    "answer": "no",
    "solution": "Solution:\n\nThe answer is no.\nSuppose that such a number $n$ exists. Since $n$ has a divisor ending in $5$, $n$ is divisible by $5$. Then for each divisor $d$ of $n$ not divisible by $5$, $5d$ also divides $n$. This contradicts the assumption that $n$ has eight divisors ending in $1,2,3,4,6,7,8$, or $9$ and only two ending in $5$ or $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSix consecutive prime numbers have sum $p$. Given that $p$ is also prime, determine all possible values of $p$.",
    "options": [],
    "answer": "41",
    "solution": "Solution:\n\nWe consider two cases.\n- If $2$ is one of the six prime numbers, then the only possible sum is $p = 2 + 3 + 5 + 7 + 11 + 13 = 41$, which is indeed a prime.\n- If $2$ is not one of the six prime numbers, then all six primes are odd, so their sum must be even. Moreover, $p > 2$. Therefore $p$ is not prime.\n\nSo, the only possible value of $p$ is $p = 41$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the number of multiples of $3$ which have six digits, none of which is greater than $5$.",
    "options": [],
    "answer": "12960",
    "solution": "Solution:\n\nThe first digit can be any number from $1$ to $5$, making $5$ possibilities. Each of the succeeding digits, from the ten-thousands digit to the tens digit, can be any of the six digits from $0$ to $5$.\n\nFinally, we claim that there are exactly two possibilities for the last digit. Given the first five digits, if we append the digits $0$, $1$, and $2$ in turn, we get three consecutive integers, exactly one of which is a multiple of $3$. The same happens when we add the digits $3$, $4$, and $5$.\n\nThus the total number of multiples of $3$ is\n$$\n5 \\cdot 6 \\cdot 6 \\cdot 6 \\cdot 6 \\cdot 2 = 12960.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine whether it is possible to tile a standard $8 \\times 8$ chessboard with $15$ $L$-tiles and $1$ $T$-tile of the shapes below:\n\n![](attached_image_1.png)\n\n(Each tile covers four squares of the chessboard. The tiles can be flipped and rotated at will.)",
    "options": [],
    "answer": "Impossible",
    "solution": "Solution:\nLet the squares of the chessboard be colored black and white in the usual way. Note that the $L$-tile always covers two squares of each color, while the $T$-tile covers either $1$ black and $3$ white or $3$ black and $1$ white squares. Since the chessboard has equal numbers of black and white squares, it is impossible to cover it with these tiles.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $A B C$, let $D$ be the midpoint of side $B C$. Let $E$ and $F$ be the feet of the perpendiculars to $A D$ from $B$ and $C$, respectively. Prove that $B E = C F$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe have $\\angle D F C = \\pi / 2 = \\angle D E B$. Also, $\\angle C D F = \\angle B D E$ since they are vertical angles. (It seems possible that $E$ and $F$ could lie on the same side of $D$, so that $\\angle C D F$ and $\\angle B D E$ would be supplementary rather than equal; however, if they were supplementary and unequal, then one of them, say $\\angle B D E$, would be $> \\pi / 2$, so the sum of the angles of $\\triangle B D E$ would be $> \\pi$, a contradiction.) These equal angles imply $\\triangle D F C \\sim \\triangle D E B$. But $C D = B C / 2 = B D$, so in fact $\\triangle D F C \\cong \\triangle D E B$, giving $C F = B E$.\nSolution:\n\nLet $H$ be the foot of the perpendicular from $A$ to $B C$. Then, using the $b h / 2$ formula and the fact that $D$ is the midpoint of $B C$, we get\n$$\n\\frac{A D \\cdot B E}{2} = \\operatorname{Area}(\\triangle A B D) = \\frac{B D \\cdot A H}{2} = \\frac{C D \\cdot A H}{2} = \\operatorname{Area}(\\triangle A C D) = \\frac{A D \\cdot C F}{2} .\n$$\nMultiplying by $2 / A D$ now gives $B E = C F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven $n$ positive real numbers $a_{1}, a_{2}, \\ldots, a_{n}$ such that $a_{1} a_{2} \\cdots a_{n}=1$, prove that\n$$\n\\left(1+a_{1}\\right)\\left(1+a_{2}\\right) \\cdots\\left(1+a_{n}\\right) \\geq 2^{n}\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "At least 2^n; equality holds if and only if all the numbers are 1.",
    "solution": "Solution:\n\nBy the inequality $a+b \\geq 2 \\sqrt{a b}$ which holds for positive numbers $a, b$ (and equality is if and only if $a=b$), we see that $1+a_{1} \\geq 2 \\sqrt{a_{1}}$, $1+a_{2} \\geq 2 \\sqrt{a_{2}}$, $\\ldots$, $1+a_{n} \\geq 2 \\sqrt{a_{n}}$. Multiplying these inequalities we get\n$$\n\\left(1+a_{1}\\right) \\cdots \\left(1+a_{n}\\right) \\geq 2^{n} \\sqrt{a_{1} \\cdots a_{n}} = 2^{n}.\n$$\nThe equality holds if and only if $a_{1}=a_{2}=\\cdots=a_{n}=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeveral weights are given, each of which is not heavier than $1 \\mathrm{lb}$. It is known that they cannot be divided into two groups such that the weight of each group is greater than $1 \\mathrm{lb}$. Find the maximum possible total weight of these weights.",
    "options": [],
    "answer": "3 lb",
    "solution": "Solution:\n\nSuppose you start putting weights into two pans of a scale and stop when you cannot add another weight to either pan without it exceeding $1 \\mathrm{lb}$. At this point, you have put at most $2 \\mathrm{lb}$ of weight on the scale, and there can only be one weight left, otherwise you could add a weight to each pan and both would exceed $1 \\mathrm{lb}$. This remaining weight weighs at most $1 \\mathrm{lb}$, so the weights add up to at most $3 \\mathrm{lb}$ total. It is possible to have exactly $3 \\mathrm{lb}$ with three $1 \\mathrm{lb}$ weights, so this is the largest possible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a quadrilateral $ABCD$, show that the midpoints of its four edges form the vertices of a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nLet $M, N, P, Q$ be the midpoints of $AB, BC, CD$, and $DA$, respectively. Then $MN$ is the midline of $\\triangle ABC$ opposite $AC$, so it is parallel to $AC$ and of length $\\frac{1}{2} AC$. Similarly, $PQ$ is the midline of $\\triangle ACD$, so $PQ \\parallel AC$ and $PQ = \\frac{1}{2} AC$. Thus, the opposite sides $MN$ and $PQ$ are equal and parallel, and similarly, $NP$ and $QM$ are equal and parallel. Thus, $MNPQ$ is a parallelogram.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all integer solutions to $x y = 2003(x + y)$",
    "options": [],
    "answer": "All integer solutions (x, y) are (4014012, 2004), (2004, 4014012), (4006, 4006), (2002, -4010006), (-4010006, 2002), (0, 0).",
    "solution": "Solution:\nWe get\n$$\n\\begin{gathered}\n2003^2 - 2003x - 2003y - x y = 2003^2 \\\\\n(2003 - x)(2003 - y) = 2003^2\n\\end{gathered}\n$$\nSo $(2003 - x)$ is a divisor of $2003^2$. As $2003$ is a prime number, we conclude that $2003 - x$ is $-2003^2$, $-2003$, $-1$, $1$, $2003$, or $2003^2$, and $x$. Then for $x, y$ we obtain\n$$\n\\begin{array}{ll}\nx = 2003^2 + 2003 = 4014012, & y = 2004, \\\\\nx = 2003 + 2003 = 4006, & y = 4006, \\\\\nx = 1 + 2003 = 2004, & y = 4014012, \\\\\nx = -1 + 2003 = 2002, & y = -401006, \\\\\nx = -2003 + 2003 = 0, & y = 0, \\\\\nx = -2003^2 + 2002 = -401006, & y = 2002.\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $k$ is an integer, prove that the number $k^{2}+k+1$ is not divisible by $2006$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe number $k^{2}+k+1 = k(k+1)+1$ is odd, because $k(k+1)$ is even. Hence it can't be divisible by $2006$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose you have an unlimited number pennies, nickels, dimes, and quarters. Determine the number of ways to make 30 cents using these coins.",
    "options": [],
    "answer": "17",
    "solution": "Solution:\n\nWe use cases to organize our work, based first on the number of quarters and then the number of dimes. First note that the number of quarters must be $0$ or $1$, since $2$ quarters would be too much. This gives $2$ cases:\n\n- $1$ quarter: There are $2$ possibilities: a quarter and a nickel or a quarter and $5$ pennies.\n\n- $0$ quarters: If we don't use quarters, we can use at most $2$ dimes, so we can make $3$ subcases based on the number of dimes:\n\n  - $2$ dimes: We need to make $10$ cents more using nickels or pennies. We could use $0$, $1$, or $2$ nickels, so there are $3$ possibilities.\n\n  - $1$ dime: We need to make $20$ cents more using nickels or pennies. We could use $0$, $1$, $2$, $3$, or $4$ nickels, so there are $5$ possibilities.\n\n  - $0$ dimes: We need to make $30$ cents more using nickels and pennies. We could use $0$, $1$, $2$, $3$, $4$, $5$, or $6$ nickels, so there are $7$ possibilities.\n\nPutting this together, we get a total of $2+3+5+7=17$ possibilities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all positive prime numbers $p$ such that $p+2$ and $p+4$ are prime as well.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nFor $p=3$, $p+2=5$, $p+4=7$ and these are obviously prime.\n\nFor $p>3$, we know that $p$ is not divisible by $3$. The remainder of $p$ when divided by $3$ can be either $1$ or $2$. If it is $1$, then $p+2$ is divisible by $3$; if it is $2$, then $p+4$ is divisible by $3$. Hence $p=3$ is the only solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$ be an odd positive integer other than $1$. Prove that one can find positive integers $y$ and $z$ such that\n$$\nx^{2} + y^{2} = z^{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet\n$$\ny = \\frac{x^{2} - 1}{2} \\quad \\text{and} \\quad z = \\frac{x^{2} + 1}{2}.\n$$\nBecause $x$ is odd, $x^{2} - 1$ and $x^{2} + 1$ are both even and therefore $y$ and $z$ are integers. Moreover, because $x$ is more than $1$, $x^{2} - 1$ and $x^{2} + 1$ are more than $0$ and thus $y$ and $z$ are positive. Finally, the desired equation $x^{2} + y^{2} = z^{2}$ is equivalent to\n$$\n\\begin{aligned}\nx^{2} + \\left(\\frac{x^{2} - 1}{2}\\right)^{2} & = \\left(\\frac{x^{2} + 1}{2}\\right)^{2} \\\\\nx^{2} + \\frac{\\left(x^{2} - 1\\right)^{2}}{4} & = \\frac{\\left(x^{2} + 1\\right)^{2}}{4} \\\\\n4x^{2} + \\left(x^{2} - 1\\right)^{2} & = \\left(x^{2} + 1\\right)^{2} \\\\\n4x^{2} + x^{4} - 2x^{2} + 1 & = x^{4} + 2x^{2} + 1,\n\\end{aligned}\n$$\nwhich is true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTen different points are marked on a circle. Two players $A$ and $B$ play the following game. $A$ moves first and the players alternate their moves. In each of the moves a player connects two of the points with a straight line segment. A player whose segment crosses a segment previously drawn will lose the game. Which player has a winning strategy and what is the strategy.",
    "options": [],
    "answer": "Player A. Any maximal noncrossing drawing on ten points has exactly seventeen segments (a triangulation), so there are seventeen safe moves. Since A makes all odd-numbered moves, A makes the seventeenth safe move and leaves B with only crossing moves, so A wins.",
    "solution": "Solution:\n\nNotice that a player can draw a line segment if and only if the $10$-gon is not partitioned into triangles. Since there is a total of $17$ segments for any partition of $10$-gon into triangles, the first player will win the game no matter how he plays.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$ and $b$ be integers such that\n$$\n|a+b| > |1 + a b|.\n$$\nProve that $a b = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNotice that replacing both $a$ and $b$ by their negatives does not change either side of the given inequality. Therefore, we may assume that $a + b \\geq 0$. We now have $a + b > |1 + a b|$, so\n$$\n\\begin{array}{rll}\na + b > 1 + a b & \\text{ and } & a + b > -1 - a b \\\\\na b - a - b + 1 < 0 & \\text{ and } & a b + a + b + 1 > 0 \\\\\n(a - 1)(b - 1) < 0 & \\text{ and } & (a + 1)(b + 1) > 0.\n\\end{array}\n$$\nIf $a$ and $b$ are nonzero, then $a - 1$ and $a + 1$ are both nonnegative or both nonpositive, and similarly for $b - 1$ and $b + 1$. So $(a - 1)(b - 1)$ and $(a + 1)(b + 1)$ are both nonnegative or both nonpositive, which contradicts the information above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA fair coin is flipped nine times. Which is more likely, having exactly four heads or having exactly five heads?",
    "options": [],
    "answer": "Both are equally likely; each has probability 126/512.",
    "solution": "Solution:\nBoth outcomes are equally likely! Notice that for every sequence with four heads (such as HTTHHTHTT) there is a corresponding sequence with exactly five heads formed by reversing all the coin flips (in this case $T H H T H T H H$). Thus the number of sequences of nine flips with exactly four heads is equal to the number of sequences with exactly five heads; since every sequence is equally likely this completes the proof.\n\nIn fact, both probabilities are equal to\n$$\n\\frac{\\binom{9}{4}}{2^{9}} = \\frac{\\binom{9}{5}}{2^{9}} = \\frac{126}{512}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo players play a game with pennies, which are circles of radius $1$, on an $m \\times n$ rectangular table. Each player takes turns putting a penny on the table so that it touches no other penny. The first player who is unable to do so loses. The table starts with no pennies on it. Assuming that there is an infinite supply of pennies, for what values of $m$ and $n$ does the first player have a winning strategy?",
    "options": [],
    "answer": "m >= 2 and n >= 2",
    "solution": "Solution:\n\nIf $m<2$ or $n<2$, then it is not possible to fit even one penny on the table, so the first player loses immediately. On the other hand, if $m \\geq 2$ and $n \\geq 2$, we will show that the first player can win. Let $O$ denote the center of the table, and let the first player place his first penny so that it is centered at $O$. The resulting configuration is centrally symmetric (i.e. it remains the same under a rotation of $180^{\\circ}$ about $O$). Now, whenever the second player places a penny with some center $A$, the first player should respond with a penny centered at $B$, the $180^{\\circ}$ rotation of $A$ about $O$. It is evident that the first player always leaves a centrally symmetric configuration; we must show that this move is actually legal.\n\nSince placing the penny with center $A$ was legal, the symmetry of the previous configuration means that, before this penny was placed, a move centered at $B$ was also legal. So, after the second player's move at $A$, the first player can still legally move at $B$ unless the penny centered at $B$ intersects the penny centered at $A$. If $P$ is an intersection point, then $AP \\leq 1$ and $PB \\leq 1$, so $AB \\leq 2$. But $O$ is the midpoint of $AB$, so $AO = AB / 2 \\leq 1$. This means that $A$ lies inside the penny placed with center $O$; this is a contradiction. Hence, the move with center $B$ is, in fact, legal.\n\nThus, the first player can always legally move so as to preserve central symmetry. This move's availability means that the first player never loses. But since only a finite number of pennies can be placed without overlap, the game eventually ends, and it is the second player who loses.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIs there a positive integer $k$ such that\n$$\n(\\cdots((4 \\underbrace{!}_{k}) !) ! \\cdots) !>(\\cdots((3 \\underbrace{!}_{k+1}) !) ! \\cdots) ! ?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Since $3 != 6$, we have\n$$\n(\\cdots((3 \\underbrace{!}_{k+1}) !) ! \\cdots) !=(\\cdots((6 \\underbrace{!}_{k}) !) ! \\cdots) !>(\\cdots((4 \\underbrace{!}_{k}) !) ! \\cdots) !\n$$\nwhere the last step follows by using the obvious lemma that if $x > y$ then $x ! > y !$ (for positive integers $x$ and $y$) $k$ times.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven are $n+1$ real linear equations in $n$ variables (of the form $a_{1} x_{1}+a_{2} x_{2}+\\cdots+a_{n} x_{n}=a$). Prove that each $=$ sign can be replaced with either $\\leq$ or $\\geq$ so that the resulting $n+1$ inequalities have the following property: for every choice of real numbers $x_{1}, x_{2}, \\ldots, x_{n}$, at least one inequality is true.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nHenceforth, we assume that every equation has some $a_{i} \\neq 0$, since otherwise it is of the form $0=a$, and then we can change it to either $0 \\leq a$ or $0 \\geq a$ to form an inequality which is always true. We use induction on $n$.\n\nIf $n=1$, we have the two equations $a x=b$, $c x=d$. We can divide out by $a$ and $c$ to obtain the two equations $x=b'$, $x=d'$ equivalent to our original equations. Without loss of generality we can assume $b' \\leq d'$; then if we form the inequalities equivalent to $x \\geq b'$, $x \\leq d'$, at least one of these always holds.\n\nSuppose $n>1$ and that the statement has been proven for $n$ equations in $n-1$ variables. Given $n+1$ equations in $n$ variables, look at the last equation. By assumption, some coefficient - say the coefficient of $x_{n}$ - is nonzero, so we can solve this equation for $x_{n}$ in terms of the other variables. Plugging this result into the first $n$ equations, we obtain $n$ linear equations in $x_{1}, \\ldots, x_{n-1}$. By the induction hypothesis, we can replace each of these equalities with a $\\leq$ or $\\geq$ sign so that, for any choice of $x_{1}, \\ldots, x_{n-1}$, some one of these $n$ inequalities is true.\n\nWe thus assign either the $\\leq$ or the $\\geq$ sign for the first $n$ original equations, and we claim we can choose the sign in our $n+1$th equation so that this result still holds. Suppose, otherwise, that, either way we choose the sign, some choice of the variables causes all of the inequalities to be violated. If the $n+1$th equation was $a_{1} x_{1}+\\cdots+a_{n} x_{n}=a$, then we can choose $x_{1}', \\ldots, x_{n}'$ to violate the first $n$ inequalities so that $a_{1} x_{1}'+\\cdots+a_{n} x_{n}'=a'<a$, and we can also choose $x_{1}'', \\ldots, x_{n}''$ to violate the first $n$ inequalities so that $a_{1} x_{1}''+\\cdots+a_{n} x_{n}''=a''>a$. Now, consider what happens when we set\n$$\nx_{i}=\\frac{\\left[a''-a\\right] x_{i}'+\\left[a-a'\\right] x_{i}''}{a''-a'}.\n$$\nThe $n+1$th equation now holds (as an equality), since\n$$\n\\sum_{i=1}^{n} a_{i}\\frac{\\left[a''-a\\right] x_{i}'+\\left[a-a'\\right] x_{i}''}{a''-a'}=\\frac{\\left[a''-a\\right] \\sum_{i=1}^{n} a_{i} x_{i}'+\\left[a-a'\\right] \\sum_{i=1}^{n} a_{i} x_{i}''}{a''-a'}=\\frac{\\left[a''-a\\right] a'+\\left[a-a'\\right] a''}{a''-a'}=a.\n$$\nWe claim that each of the first $n$ inequalities is still violated. Suppose, for example, that the inequality $c_{1} x_{1}+\\cdots+c_{n} x_{n} \\leq c$ is violated, so that $c_{1} x_{1}'+\\cdots+c_{n} x_{n}'>c$ and $c_{1} x_{1}''+\\cdots+c_{n} x_{n}''>c$. We can multiply the first of these inequalities by $a''-a$ and the second by $a-a'$ (preserving their signs, since these quantities are positive), add them, and divide by $a''-a'$ to obtain\n$$\n\\sum_{i=1}^{n} c_{i}\\frac{\\left[a''-a\\right] x_{i}'+\\left[a-a'\\right] x_{i}''}{a''-a'}>c,\n$$\nas claimed. The $\\geq$ case is, of course, analogous. Thus, we have found values of $x_{i}$ for which the $n+1$th equality holds but none of the first $n$ inequalities hold. But this contradicts the way the first $n$ signs were chosen. So our assumption was wrong, and it was indeed possible to choose the sign of the $n+1$th inequality so that there would always be at least one true inequality, completing the induction step and the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn the sequence\n$$\n77492836181624186886128 \\ldots,\n$$\nall of the digits except the first two are obtained by writing down the products of pairs of consecutive digits. Prove that infinitely many 6 s appear in the sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $868$ appears in the sequence and $8 \\cdot 6 = 6 \\cdot 8 = 48$, $4848$ appears later in the sequence. Then $4 \\cdot 8 = 32$, so $3232$ appears somewhere later. And so on:\n$$\n868 \\rightarrow 4848 \\rightarrow 3232 \\rightarrow 666 \\rightarrow 3636 \\rightarrow 1818 \\rightarrow 888 \\rightarrow 6464 \\rightarrow 2424 \\rightarrow 888 \\ldots\n$$\nIt is clear that the three last sequences repeat indefinitely. Therefore $6464$ appears infinitely many times, and in particular $6$ appears infinitely many times.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn an $6 \\times 6$ chessboard, we randomly place counters on three different squares. What is the probability that no two counters are in the same row or column?",
    "options": [],
    "answer": "40/119",
    "solution": "Solution:\nThe number of ways to pick three squares is $\\binom{36}{3}$.\nWe now count the number of ways to pick three squares with no two in the same row or column. One can select three distinct rows, in $\\binom{6}{3}$ ways, that we will place the counters in. Afterwards, there are $6$ ways to pick the column for the first row, $5$ to pick the column in the second row, and $4$ for the final row.\nSo the answer is\n$$\n\\frac{\\binom{6}{3} \\cdot 6 \\cdot 5 \\cdot 4}{\\binom{36}{3}} = \\frac{20 \\cdot 6 \\cdot 5 \\cdot 4}{36 \\cdot 35 \\cdot 34 / 6} = \\frac{40}{119}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTen fair coins are flipped. Given that there are at least nine heads, what is the probability that all the coins show heads?",
    "options": [],
    "answer": "1/11",
    "solution": "Solution:\nThe answer is $\\frac{1}{11}$. Among the $2^{10} = 1024$ sequences of heads and tails, note that\n- There is only $1$ which is all heads, and\n- There are $10$ sequences which have exactly one tails.\nSo there are $11$ possible sequences with at least nine heads, only one of which is all heads. Hence the answer is $\\frac{1}{11}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $n$ is an integer, prove that the number\n$$\n1+n+n^{2}+n^{3}+n^{4}\n$$\nis not divisible by $4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe will prove that $1+n+n^{2}+n^{3}+n^{4}$ is an odd number and the required claim will follow immediately since no odd number is divisible by $4$.\n\nThere are two possibilities for $n$ — it can be even or odd.\n\nIn the case when $n$ is even, then $n+n^{2}+n^{3}+n^{4}$ is even as a sum of even numbers. Thus $1+n+n^{2}+n^{3}+n^{4}$ is an odd number.\n\nOn the other hand, if $n$ is odd then, all five numbers $1$, $n$, $n^{2}$, $n^{3}$, and $n^{4}$ are odd and hence their sum must be also odd.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all nonzero real numbers $x$ such that\n$$\nx^{2} + \\frac{36}{x^{2}} = 13.\n$$",
    "options": [],
    "answer": "-3, -2, 2, 3",
    "solution": "Solution:\nMultiplying through by $x^{2}$ and moving all terms to the left gives\n$$\nx^{4} - 13 x^{2} + 36 = 0.\n$$\nWe can factor this as\n$$\n\\left(x^{2} - 4\\right)\\left(x^{2} - 9\\right) = (x - 2)(x + 2)(x - 3)(x + 3) = 0.\n$$\nThus, the solutions are $x = \\pm 2$ and $x = \\pm 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn Predictable Island, everyone is either a liar (who always lies) or a truth-teller (who always tells the truth). You encounter Aerith who says, \"Bob and I are both liars.\" What are they actually?",
    "options": [],
    "answer": "Aerith is a liar and Bob is a truth-teller.",
    "solution": "Solution:\n\nIf Aerith was a truth-teller, she would be lying, because it wouldn't be the case that they're both liars. Thus, she must be a liar. Therefore, the statement must be false, which means Bob has to be a truth-teller.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA toy slot machine accepts two kinds of coins: red and green. When a coin is inserted, the machine returns 5 coins of the other color. Laura starts with one green coin. Can it happen that after a while, she has the same number of coins of each color?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that each time Laura uses the machine, the total number of coins she has increases by four, an even number. She begins with one coin, an odd number, so she will always have an odd number of coins and can never have the same number of each color.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind a positive integer satisfying the following three clues:\n- When multiplied by $2$, it yields a perfect square.\n- When multiplied by $3$, it yields a perfect cube.\n- When multiplied by $5$, it yields a perfect fifth power.",
    "options": [],
    "answer": "2^15 * 3^20 * 5^24",
    "solution": "Solution:\n\nIt is reasonable to try a number of the form $2^{a} 3^{b} 5^{c}$, where $a$, $b$, and $c$ are nonnegative integers. We want\n$$\n\\begin{array}{ll}\n2^{a+1} 3^{b} 5^{c} \\text{ is a square, i.e. } & 2|a+1,\\ 2| b, \\quad \\text{ and } 2 \\mid c \\\\\n2^{a} 3^{b+1} 5^{c} \\text{ is a cube, i.e. } & 3|a, \\quad 3| b+1, \\text{ and } 3 \\mid c \\\\\n2^{a} 3^{b} 5^{c+1} \\text{ is a fifth power, i.e. } 5|a, \\quad 5| b, \\quad \\text{ and } 5 \\mid c+1 .\n\\end{array}\n$$\nWe can now search for $a$, $b$, and $c$ separately. We find that $n_{0}=2^{15} 3^{20} 5^{24}$ is a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSix distinct numbers are chosen from the list $1, 2, \\ldots, 10$. Prove that their product is divisible by a perfect square greater than 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf all the odd numbers are chosen, then in particular $9$ is chosen and the product is divisible by $9$.\nIf not all the odd numbers are chosen, then at most $4$ odds and thus at least $2$ evens are chosen. Therefore the product is divisible by $4$.\nSolution:\n\nAt most one of the numbers is $1$, and each of the other numbers has at least one prime factor. Therefore the product of the six chosen numbers consists of at least five prime factors. But there are only four prime factors in the numbers from $1$ to $10$: $2, 3, 5$, and $7$. So some prime appears twice, making a square divisor greater than $1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the sequence\n$$\n1, 4, 7, 10, 13, 16, 19, \\ldots\n$$\neach term is 3 less than the next term. Find the 1000th term of the sequence.",
    "options": [],
    "answer": "2998",
    "solution": "Solution:\n\nTo get from the first term to the 1000th term, we must add three 999 times. That amounts to adding $3 \\cdot 999$ which is $2997$. So the 1000th term is $1 + 2997 = 2998$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCalculate, with proof, the last digit of\n\n$$\n3^{3^{3^{3^{3}}}}\n$$",
    "options": [],
    "answer": "7",
    "solution": "Solution:\nWhen $3$ is raised to the successive powers $1, 2, 3, 4, \\ldots$, the units digits are $3, 9, 7, 1, \\ldots$ From then on, since the digit $1$ has been reached, the units digits will repeat in this cycle of four elements. So it is necessary to find the remainder when\n$$\nn_{4} = 3^{3^{3^{3}}}\n$$\nis divided by $4$.\nWhen powers of $3$ are divided by $4$, the remainders are $3, 1, \\ldots$ Here the number $1$ appears after two steps, and the remainders therefore repeat in a two-element cycle. So it is necessary to find the remainder when\n$$\nn_{3} = 3^{3^{3}}\n$$\nis divided by $2$.\nBut $n_{3}$ is clearly odd, so $n_{4}$ has a remainder of $3$ when divided by $4$ and the original number $n_{5}$ has a last digit of $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $n$ such that $n(n+1)$ is a perfect square.",
    "options": [],
    "answer": "no positive integers",
    "solution": "Solution:\nSince $n$ and $n+1$ are coprime numbers, if $n(n+1)$ is a perfect square, then each of $n$ and $n+1$ has to be a perfect square itself. However, that is impossible since if $n = x^{2}$ and $n+1 = y^{2}$ we would have $1 = y^{2} - x^{2} = (y - x)(y + x)$ and $1$ can't be expressed as a product of two different integers. Hence there are no such integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFour friends, One, Two, Five, and Ten are located on one side of the dark tunnel, and have only one flashlight. It takes one minute for person One to walk through the tunnel, two minutes for Two, five for Five, and ten for Ten. The tunnel is narrow and at most two people can walk at the same time with the flashlight. Whenever two people walk together they walk at the speed of the slower one. Show that all four friends can go from one side of the tunnel to the other one in 17 minutes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFriends One and Two should walk to the other side. It will take them $2$ minutes. Then, One returns - this will last additional $1$ minute. Then, Ten and Five go to the other side - this will take $10$ minutes, but Two should return the flashlight ($2$ additional minutes), and together with One, go through the tunnel for the last time - additional $2$ minutes. It remains to notice that the time spent is exactly $17$ minutes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe rectangle $M N P Q$ is inside the rectangle $A B C D$. The portion of the rectangle $A B C D$ outside of $M N P Q$ is colored in green. Using just a straightedge construct a line that divides the green figure in two parts of equal areas.",
    "options": [],
    "answer": "Draw the line through the centers of the two rectangles, where each center is found as the intersection point of its diagonals.",
    "solution": "Solution:\n\nThe line passing through the centers of $A B C D$ and $M N P Q$ divides each of the rectangles in two pieces of equal areas. Hence that line will divide the green part into equal areas. The line can be easily constructed using only a straightedge because we can construct the centers of the given rectangles as intersections of their diagonals.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA cube $3 \\times 3 \\times 3$ is made of cheese and consists of 27 small cubical cheese pieces arranged in the $3 \\times 3 \\times 3$ pattern. A mouse is eating the cheese in such a way that it starts at one of the corners and eats smaller pieces one by one. After he finishes one piece, he moves to the adjacent piece (pieces are adjacent if they share a face). Is it possible that the last piece mouse has eaten is the central one?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nColor the pieces of cheese alternatively in red and green such that corners are green and any two adjacent cubes are of different colors. We easily see that the mouse is moving always from the cube of one color to the cube of the other color. There are 14 green and 13 red cubes, the central cube being red. Since mouse has started from green piece, it will finish at the green piece (after 27 moves), hence it can't finish at the central cube.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven 8 oranges on the table, 7 of them have exactly the same weight and the 8th is a little bit lighter. You are given a balance that can measure oranges against each other and you are allowed to use the balance at most twice! How can you determine which one of the oranges is lighter than the others? Explain your answer!",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst we put 3 oranges on the left and 3 on the right-hand side of the balance. In the case that the balance shows equal weights, one of the remaining two oranges is lighter and in the second measurement we can easily see which one. However, if one side (say left) is lighter, then we know that the orange we are looking for is one of the three that appear on the left side. Let us denote these oranges by $A$, $B$, $C$. In the second measurement we can measure $A$ against $B$. If one of them is lighter than the other, we are done (as the lighter orange is the one we are looking for). If the weights of $A$ and $B$ are the same, then $C$ is the lighter orange.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn a given street, there are $n$ houses numbered from $1$ to $n$. Let $a_{i}$ ($1 \\leq i \\leq n$) be the number of people living in the house numbered $i$, and let $b_{i}$ ($i \\geq 1$) be the number of houses on the street in which at least $i$ people live. Prove that\n$$\na_{1}+a_{2}+\\cdots+a_{n}=b_{1}+b_{2}+b_{3}+\\cdots\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet us number the people in each house from $1$ up to the total number of people living there. Then for each $k \\geq 1$, the label $k$ is assigned as often as there is a house with $k$ or more people; thus there are $b_{k}$ people labeled $k$. The quantity\n$$\nb_{1}+b_{2}+b_{3}+\\cdots\n$$\nthus represents the total number of people labeled with some positive integer; this is the total number of people in all the houses, $a_{1}+\\cdots+a_{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSolve\n\n$$\n2 \\sqrt{1+x \\sqrt{1+(x+1) \\sqrt{1+(x+2) \\sqrt{1+(x+3)(x+5)}}}}=x\n$$",
    "options": [],
    "answer": "no real solutions",
    "solution": "Solution:\n\nAs the left hand side is nonnegative, we see that any solution will have $x \\geq 0$. For such $x$ we have\n$$\n\\sqrt{1+(x+3)(x+5)} = \\sqrt{x^2+8x+16} = \\sqrt{(x+4)^2} = |x+4| = x+4.\n$$\nProceeding similarly we get\n$$\n\\begin{aligned}\n& 2 \\sqrt{1+x \\sqrt{1+(x+1) \\sqrt{1+(x+2) \\sqrt{1+(x+3)(x+5)}}}} \\\\\n= & 2 \\sqrt{1+x \\sqrt{1+(x+1) \\sqrt{1+(x+2)(x+4)}}} \\\\\n= & 2 \\sqrt{1+x \\sqrt{1+(x+1)(x+3)}} \\\\\n= & 2 \\sqrt{1+x(x+2)} \\\\\n= & 2(x+1)\n\\end{aligned}\n$$\nSolving $2(x+1)=x$ gives $x=-2$, which is negative. Therefore the equation has no solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA scalene triangle has side lengths which are all prime numbers. What is the smallest possible perimeter it could have?",
    "options": [],
    "answer": "15",
    "solution": "Solution:\nThe answer is $3 + 5 + 7 = 15$. This is achievable since the three sides $\\{3, 5, 7\\}$ form the sides of a triangle.\n\nTo show it is best possible, note that $\\{3, 5, 7\\}$ are the three smallest odd primes; so any smaller triangle would need to have a side of length $2$. However, the two longer lengths would then be different odd primes; this would cause them to differ by at least $2$, which is impossible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou are given an $m \\times n$ chocolate bar divided into $1 \\times 1$ squares. You can break a piece of chocolate by splitting it into two pieces along a straight line that does not cut through any of the $1 \\times 1$ squares. What is the minimum number of times you have to break the bar in order to separate all the $1 \\times 1$ squares?",
    "options": [],
    "answer": "mn - 1",
    "solution": "Solution:\n\nWe note that the number of separate pieces of chocolate increases by $1$ at each cut. We begin with $1$ piece and end with $m n$ pieces, so we must make $m n - 1$ cuts. Thus $m n - 1$ is the minimum (and also the maximum) number of cuts necessary to separate all the $1 \\times 1$ squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a$ and $b$ are positive integers prove that\n$$\na+b \\leq 1+a b.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $a$ and $b$ are positive integers we know that $a \\geq 1$ and $b \\geq 1$. This implies that $(a-1)(b-1) \\geq 0$ hence $a b - a - b + 1 \\geq 0$ which is equivalent to the given inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe five-digit number $9A65B$ is divisible by $36$, where $A$ and $B$ are digits. Find all possible values of $A$ and $B$.",
    "options": [],
    "answer": "A=5, B=2 or A=1, B=6",
    "solution": "Solution:\n\nA number is divisible by $36$ if and only if it is divisible by $9$ and $4$.\n\nFor it to be divisible by $4$, the last two digits must make a two-digit number divisible by $4$, and the only possibilities for this are $B = 2$ or $B = 6$.\n\nFor it to be divisible by $9$, the sum of the digits must be a multiple of $9$. Right now the sum is $9 + 6 + 5 = 20$, and the next multiple of $9$ is $27$, so we need $A + B = 7$.\n\nThus, the two solutions are $A = 5$, $B = 2$ or $A = 1$, $B = 6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers $1$ to $10$ are written on a blackboard in a row. Determine whether it is possible to put nine $+$ and $-$ signs between them (one between each pair of adjacent numbers) to make the resulting sum $0$.",
    "options": [],
    "answer": "Not possible",
    "solution": "Solution:\n\nIt is not possible. Note that the sum of the numbers is\n$$\n1+2+3+\\cdots+9+10=55\n$$\nan odd number. Thus, if we put all $+$ signs, the sum is odd. Every time we change a $+$ to a $-$, we change an $n$ term to a $-n$ term for some $n$, which subtracts $2n$ from the sum. This changes the sum by an even number, so the sum will still be odd. Thus, no matter what, the sum is odd and therefore cannot be $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all pairs $(m, n)$ of natural numbers such that $200 m + 6 n = 2006$.",
    "options": [],
    "answer": "(1, 301), (4, 201), (7, 101), (10, 1)",
    "solution": "Solution:\nFirst we divide both sides of the equation by $2$ and get: $100 m + 3 n = 1003$.\n\nSince $m$ and $n$ are natural numbers we immediately get that $m \\leq 10$.\n\nSince $3 n$ is divisible by $3$ and $1003$ gives remainder $1$ upon division by $3$, we conclude that $100 m$ must also give the remainder $1$ upon division by $3$.\n\nSince $100 m = 99 m + m$ and $99 m$ is divisible by $3$ we see that $m$ must give remainder $1$ when divided by $3$.\n\nThus $m$ has to be one of the numbers $1, 4, 7, 10$.\n\nCorresponding $n$'s are $301, 201, 101$ and $1$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIs the number $\\left|2^{3000}-3^{2006}\\right|$ bigger or smaller than $\\frac{1}{2}$?",
    "options": [],
    "answer": "greater than 1/2",
    "solution": "Solution:\n\nNotice that $\\left|2^{3000}-3^{2006}\\right|$ is a non-negative integer, so it is either $0$ or bigger than $1/2$. However, this number is not $0$ since $2^{3000} \\neq 3^{2006}$, hence $\\left|2^{3000}-3^{2006}\\right| > \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArrange the following number from smallest to largest: $2^{1000}, 3^{750}, 5^{500}$.",
    "options": [],
    "answer": "2^1000 < 5^500 < 3^750",
    "solution": "Solution:\n\nSince all the numbers are positive, taking the $250$th root of each number will not change their ordering. The resulting numbers are $2^{4} = 16$, $3^{3} = 27$, and $5^{2} = 25$. These have the ordering $2^{4} < 5^{2} < 3^{3}$, so the ordering of the original numbers is $2^{1000} < 5^{500} < 3^{750}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor an arrangement of the digits $0$ through $9$ around a circle, a number is called a neighbor sum if it is the sum of some two adjacent digits in the arrangement. For example, the arrangement\n![](attached_image_1.png)\nhas five neighbor sums: $4, 7, 8, 11$, and $14$. What is the minimal possible number of neighbor sums, given that each digit must be used just once?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nThe answer is $3$. It can be achieved using the arrangement\n![](attached_image_2.png)\nin which the only neighbor sums are $5, 9$, and $10$.\n\nTo prove that no smaller number is possible, first consider the digit $0$. It is adjacent to two different numbers between $1$ and $9$ inclusive, yielding two neighbor sums in the range $1, \\ldots, 9$. Now consider the digit $9$. It is adjacent to two numbers and at least one of them is not $0$. Thus $9$ has a neighbor between $1$ and $8$ inclusive, and thus there is a neighbor sum in the range $10, \\ldots, 17$. Since these ranges do not overlap, there are altogether at least three different neighbor sums in the range $1, \\ldots, 17$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA building has seven rooms numbered $1$ through $7$, all on one floor, and any number of doors connecting these rooms. These doors may be one-way, admitting motion in only one of the two directions, or two-way. In addition, there is a two-way door between room $1$ and the outside, and a treasure in room $7$. Your object is to choose the arrangement of the rooms and the locations of the doors in such a way that\n\na. it is possible to enter room $1$, reach the treasure, and make it back outside,\n\nb. the minimum number of steps required to do this (each step consisting of walking through a door) is as large as possible.",
    "options": [],
    "answer": "14",
    "solution": "Solution:\nOn the way to the treasure, no room need be entered twice; otherwise the path could be shortened by skipping the loop. Thus, the minimal path to the treasure, if it exists, is at most $7$ steps long. Similarly, the minimal path from the treasure to the outside is at most $7$ steps long, so the total number of steps cannot exceed $14$. The arrangement of the rooms in a line, with $1$ and $7$ at opposite ends and two-way doors between all adjacent rooms, shows that $14$ is attainable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo thousand and eleven positive integers are chosen, all different and less than or equal to $4020$. Prove that two of them have no common factors except $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSplit the numbers from $1$ to $4020$ into the $2010$ pairs $\\{1,2\\}, \\{3,4\\}, \\ldots, \\{4019,4020\\}$. Since $2011$ numbers are chosen but there are only $2010$ pairs, two numbers have to lie in the same pair. These numbers cannot have any common factors except $1$, because any divisor of both numbers would have to divide their difference, which is $1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the units digit of $17^{2021}$.",
    "options": [],
    "answer": "7",
    "solution": "Solution:\nThe units digits of powers of $17$ cycle: $7, 9, 3, 1, 7, 9, 3, 1, \\ldots$, so the units digit of $17^{n}$ is $1$ whenever $n$ is a multiple of $4$. Since $2020$ is a multiple of $4$, $17^{2020}$ has units digit $1$, so $17^{2021}$ has units digit $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind, with proof, all ways to write $1$ as a sum of three fractions, each with numerator $1$ and positive integer denominator. (The order of the fractions is irrelevant, so for instance $\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{4}$ is the same as $\\frac{1}{4}+\\frac{1}{4}+\\frac{1}{2}$.)",
    "options": [],
    "answer": "1 = 1/3 + 1/3 + 1/3 = 1/2 + 1/4 + 1/4 = 1/2 + 1/3 + 1/6",
    "solution": "Solution:\nThere are three solutions:\n$$\n1 = \\frac{1}{3} + \\frac{1}{3} + \\frac{1}{3} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6}\n$$\nNow we must prove that these are the only solutions. If the fraction $1/2$ appears in the expression, the remaining fractions must add to $1/2$, so one of them is greater than or equal to $1/4$. If this fraction is $1/3$, we get the solution $1/2 + 1/3 + 1/6$, and if this fraction is $1/4$, we get the solution $1/2 + 1/4 + 1/4$. Thus, if the fraction $1/2$ is used, we cannot get any new solutions.\nIf the fraction $1/2$ does NOT appear in the expression, then all three fractions are at most $1/3$. Then their sum will certainly be less than $1$ unless they are all equal to $1/3$. Thus in this case, we only get the third solution, $1/3 + 1/3 + 1/3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEach of Alice, Bob, and Carol is either a consistent truth-teller or a consistent liar. Alice states: \"At least one of Bob or Carol is a truth-teller.\" Bob states: \"Alice and Carol are both truth-tellers.\" Carol states: \"If Alice is a truth-teller, so too is Bob.\" Must they all be truth-tellers?",
    "options": [],
    "answer": "Yes, they must all be truth-tellers.",
    "solution": "Solution:\n\nYes. If Carol were a liar, Alice would have to be a truth-teller while Bob would have to be a liar. However, Bob would then be telling the truth, a contradiction.\n\nThus Carol is telling the truth. Alice's statement is then true as well, and thus Bob's statement is also true. Hence, all logicians must be telling the truth.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA triangle, two of whose sides are $3$ and $4$, is inscribed in a circle. Find the minimal possible radius of the circle.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nSince the circle has a chord of length $4$, its diameter is at least $4$ and so its radius is at least $2$. To achieve equality, choose a right triangle with hypotenuse $4$ and one leg $3$ (the other leg will, by the Pythagorean theorem, have length $\\sqrt{7}$). Then the midpoint of the hypotenuse is the center of a circle of radius $2$ passing through all three vertices.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are 25 people at a party and every pair of them is either friends or strangers. Prove that there are two people at the party who have the same number of friends.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume that no two people have the same number of friends. Because the number of friends a person can have ranges from $0$ to $24$, we can label the people with their numbers of friends and every label will be used once. Now consider the people labeled $0$ and $24$. They cannot be friends, because the person labeled $0$ has no friends; but they cannot be strangers, because the person labeled $24$ is friends with everybody else. We have a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying $f(x)+f(y) \\neq 0$ and\n$$\n\\frac{f(x)-f(x-y)}{f(x)+f(x+y)}+\\frac{f(x)-f(x+y)}{f(x)+f(x-y)}=0\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All functions of the form f(x) = c for all real x, where c is a nonzero constant.",
    "solution": "Solution:\nAny function of the form $f(x)=c$, where $c$ is a nonzero constant, clearly satisfies the requirements; we will show these are the only solutions. Bringing the original equation to a common denominator yields $2 f(x)^2 - f(x-y)^2 - f(x+y)^2 = 0$ for all $x$, $y$. Define $g(x)=f(x)^2$; clearly $g(x) \\geq 0$ for all $x$. Our equation states that $2 g(x) = g(x-y) + g(x+y)$. We will show this forces $g$ to be constant.\n\nIf $g$ is not constant, there exist some $a, b$ with $g(a) < g(b)$. We claim that $g(n a - (n-1) b) = n g(a) - (n-1) g(b)$ for all nonnegative integers $n$. This is shown by induction: the base cases $n=0,1$ are clear; if $n>1$ then our equation (with $x=(n-1)a-(n-2)b$, $y=a-b$) yields\n$$\n2 g((n-1)a-(n-2)b) = g(n a-(n-1) b) + g((n-2)a-(n-3)b)\n$$\n$\\Rightarrow 2(n-1) g(a) - 2(n-2) g(b) = g(n a-(n-1) b) + (n-2) g(a) - (n-3) g(b)$ (by induction hypothesis)\n$$\n\\Rightarrow g(n a-(n-1) b) = n g(a) - (n-1) g(b)\n$$\nand the induction step is complete. However, if $n > g(b)/(g(b)-g(a))$ then $g(n a-(n-1) b) = n g(a) - (n-1) g(b) < 0$ and we have a contradiction. This shows that $g$ must be constant.\n\nSo $g(x) = d$ for some constant $d \\geq 0$; then $f(x) = \\pm \\sqrt{d}$ for all $x$. If both values occur, then the $f(x)+f(y) \\neq 0$ constraint is violated; hence we must have $f(x) = \\sqrt{d}$ for all $x$ or $f(x) = -\\sqrt{d}$ for all $x$ (and $d \\neq 0$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine, with proof, the value of\n$$\n1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-\\cdots+97^{2}-98^{2}+99^{2} .\n$$",
    "options": [],
    "answer": "4950",
    "solution": "Solution:\nObserve that $3^{2}-2^{2}=(3-2)(3+2)=3+2$, $5^{2}-4^{2}=(5-4)(5+4)=5+4$, and so on. Thus this sum, call it $S$, is actually equal to\n$$\nS=1+(2+3)+(4+5)+\\cdots+(97+98)+99 .\n$$\nWe can also write it in reverse as\n$$\nS=99+98+97+96+95+\\cdots+3+2+1 .\n$$\nAdding these two, we get\n$$\n2 S=\\underbrace{100+100+\\cdots+100}_{99 \\text{ times}}\n$$\nThus, $S=\\frac{1}{2} \\cdot 100 \\cdot 99=4950$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that\n$$\n1 \\cdot 1! + 2 \\cdot 2! + \\cdots + n \\cdot n! = (n+1)! - 1\n$$\nfor all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n$$\n\\begin{aligned}\n& 1 \\cdot 1! + 2 \\cdot 2! + \\cdots + n \\cdot n! \\\\\n& = (2-1) \\cdot 1! + (3-1) \\cdot 2! + \\cdots + [(n+1)-n] \\cdot n! \\\\\n& = (2 \\cdot 1! - 1!) + (3 \\cdot 2! - 2!) + \\cdots + [(n+1) \\cdot n! - n!] \\\\\n& = (2! - 1!) + (3! - 2!) + \\cdots + [(n+1)! - n!]\n\\end{aligned}\n$$\nWhen we add together the factorials in the last row, all terms cancel except for the $-1!$ at the beginning and the $(n+1)!$ at the end, so the sum is $(n+1)! - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are 10 bags full of coins. All coins look the same and all weigh 10 grams, except the coins from one bag that are fake and all weigh 9 grams. Given a scale, how could you tell which bag has the wrong coins in just one measurement? Explain your answer!",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTake one coin from the first bag, 2 from the second, 3 from the third, ..., 10 from the 10th and place them on the scale. The total mass shown by the scale should be $$(1+2+\\cdots+10) \\cdot 10 - x$$ grams, where $x$ is the number of fake coins (each coin weighs 10 grams, except for the fake ones that are one gram lighter). Since the scale will show $550 - x$ grams, we can see how many coins are fake and that is enough to deduce from which bag those coins were taken.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhich is larger,\n$$\n3^{3^{3^{3}}} \\text{ or } 2^{2^{2^{2^{2}}}} ?\n$$",
    "options": [],
    "answer": "3^{3^{3^{3}}}",
    "solution": "Solution:\nStarting to evaluate, the first expression becomes\n$$\nA = 3^{3^{3^{3}}} = 3^{3^{27}}.\n$$\nThe second expression is\n$$\nB = 2^{2^{2^{2^{2}}}} = 2^{2^{2^{4}}} = 2^{2^{16}}.\n$$\nIt is now obvious that\n$$\nA = 3^{3^{27}} > 3^{3^{16}} > 3^{2^{16}} > 2^{2^{16}} = B\n$$\nso the first expression is the larger one.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle and suppose $AB = 3$, $BC = 4$, $CA = 5$. What is the distance from $B$ to line $AC$?",
    "options": [],
    "answer": "12/5",
    "solution": "Solution:\nFirst note that since $3^{2} + 4^{2} = 5^{2}$, we have $\\angle B = 90^{\\circ}$ by the (converse to the) Pythagorean theorem. We calculate the area of the triangle in two ways. Viewing $AB$ as the base and $BC$ as the height, we get that the area is $\\frac{1}{2} \\cdot 3 \\cdot 4 = 6$. But viewing $AC$ as the base and $h$ (the distance to be determined) as the height, we get an equation:\n\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n\\frac{1}{2} \\cdot 5 \\cdot h & = 6 \\\\\n\\frac{5}{2} \\cdot h & = 6 \\\\\n\\frac{2}{5} \\cdot \\frac{5}{2} \\cdot h & = 6 \\cdot \\frac{2}{5} \\\\\nh & = \\frac{6 \\cdot 2}{5} = \\frac{12}{5} .\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $6 \\times 6$ square is covered by nonoverlapping dominos ($2 \\times 1$ rectangles, placed horizontally or vertically). Prove that there must be a horizontal line or a vertical line that passes through the interior of the big square, but which does not cut the interior of any domino.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider all the grid lines of the big square. If some such line is intersected by $d$ dominos, we claim $d$ is even. Proof: Looking at the portion of the board on one side of the line, we find that the number of squares there is divisible by $6$ and so is even; on the other hand, each of the $d$ dominos covers exactly $1$ square there and any other domino covers either $0$ or $2$, so the number of squares has the same parity as $d$; hence, $d$ is even.\n\nNow if each of the $5$ horizontal and $5$ vertical lines were intersected by some domino, the claim implies that each would intersect at least $2$ dominos. Thus we would have $20$ intersections, and since a domino cannot cross more than one line, this gives $20$ distinct dominos. This is a contradiction since we only use $18$ dominos to cover the square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA house has several rooms. There are also several doors, each of which connects either one room to another or a room to the outside. Suppose that every room has an even number of doors leaving it. Prove that the number of outside entrance doors is even as well.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEvery door has two \"sides,\" one toward one room and one toward either another room or the outside. Clearly the total number of sides, being twice the number of doors, is even. However, for each room, the number of sides pointing to it is even. Since even subtracted from even gives even, the number of sides pointing to the outside is also even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJanuary 3, 1911 was an odd date as its abbreviated representation, $1 / 3 / 1911$, can be written using only odd digits (note all four digits are written for the year). To the nearest month, how many months will have elapsed between the most recent odd date and the next odd date (today is $3 / 3 / 2001$, an even date).",
    "options": [],
    "answer": "13333",
    "solution": "Solution:\n\nThe most recent odd date was $11 / 19 / 1999$ (November has 30 days, but the assumption that it has 31 days does not change the answer), and the next odd date will be $1 / 1 / 3111$. From $11 / 19 / 1999$ to $1 / 1 / 2000$ is about 1 month. From 2000 to 3111 is 1111 years, or $12 \\cdot 1111 = 13332$ months, so the total number of months is $13333$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a square. We randomly select a point $P$ inside the square, uniformly and at random. What is the probability that $\\angle BPC > 90^{\\circ}$?",
    "options": [],
    "answer": "π/8",
    "solution": "Solution:\nLet $\\omega$ be the semicircle whose diameter is $\\overline{BC}$ and whose interior lies inside $ABCD$. Let $M$ denote the center of $\\omega$. We use the result that\n- $\\angle BPC < 90^{\\circ}$ if $P$ is inside $\\omega$,\n- $\\angle BPC = 90^{\\circ}$ if $P$ is on $\\omega$, and\n- $\\angle BPC > 90^{\\circ}$ if $P$ is outside $\\omega$.\n\nIn order to prove this, we first show the result when $P$ is on $\\omega$ (this result is sometimes called Thales' Theorem.)\n\n![](attached_image_1.png)\n\nCompute\n$$\n\\begin{aligned}\n180^{\\circ} & = 360^{\\circ} - (\\angle PMB + \\angle PMC) & \\text{ as } & \\angle PMB + \\angle PMC = 180^{\\circ} \\\\\n& = \\left(180^{\\circ} - \\angle PMB\\right) + \\left(180^{\\circ} - \\angle PMC\\right) & & \\\\\n& = 2 \\angle BPM + \\left(180^{\\circ} - \\angle PMC\\right) & & MC = MP \\text{ in } \\triangle BMP \\\\\n& = 2 \\angle BPM + 2 \\angle CPM & & \\\\\n& = 2 \\angle BPC & &\n\\end{aligned}\n$$\n\nas claimed.\nIn the case that $P$ is outside $\\omega$, we may intersect segment $MP$ with $\\omega$ to obtain a point $K$. Then $90^{\\circ} = \\angle BKC > \\angle BPC$ (since, say, $\\angle BKM > \\angle BPM$ and $\\angle CKM > \\angle CPM$).\n\n![](attached_image_2.png)\n\nThe case where $P$ is inside $\\omega$ is dealt in the same way.\nThus, we see that we seek the probability that $P$ lies inside $\\omega$. The area of $\\omega$ inside $ABCD$ is simply $\\frac{1}{2} \\pi \\left(\\frac{1}{2}\\right)^2 = \\frac{1}{8} \\pi$ (choosing our units so that $AB = 1$), and the area of $ABCD$ is $1$; hence the answer is $\\frac{1}{8} \\pi$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo concentric circles have radii $r$ and $R > r$. Three new circles are drawn so that they are each tangent to the big two circles and tangent to the other two new circles. Find $\\frac{R}{r}$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nThe centers of the three new circles form a triangle. The diameter of the new circles is $R - r$, so the side length of the triangle is $R - r$. Call the center of the concentric circles $O$, two vertices of the triangle $A$ and $B$, and $AB$'s midpoint $D$. $OA$ is the average of $R$ and $r$, namely $\\frac{R + r}{2}$. Using the law of sines on triangle $DAO$, we get $\\frac{\\sin(30)}{AD} = \\frac{\\sin(90)}{AO} \\Rightarrow R = 3r$, so $\\frac{R}{r} = 3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n12 points are placed around the circumference of a circle. How many ways are there to draw 6 non-intersecting chords joining these points in pairs?",
    "options": [],
    "answer": "132",
    "solution": "Solution:\n\n$C$ (number of chords) $= C(6) = 132$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many distinct sets of 8 positive odd integers sum to 20?",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nThis is the same as the number of ways 8 nonnegative even integers sum to 12 (we subtract 1 from each integer in the above sum). All 11 possibilities are (leaving out 0s): $12$, $10+2$, $8+4$, $8+2+2$, $6+6$, $6+4+2$, $6+2+2+2+2$, $4+4+4$, $4+4+2+2$, $4+2+2+2+2$, $2+2+2+2+2+2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the number of real zeros of $x^{3}-x^{2}-x+2$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nLet $f(x) = x^{3} - x^{2} - x + 2$, so $f'(x) = 3x^{2} - 2x - 1$.\n\nThe slope is zero when $3x^{2} - 2x - 1 = 0$, where $x = -\\frac{1}{3}$ and $x = 1$.\n\nNow $f\\left(\\frac{1}{3}\\right) > 0$ and $f(1) > 0$, so there are no zeros between $x = -\\frac{1}{3}$ and $x = 1$.\n\nSince $\\lim_{x \\rightarrow +\\infty} f(x) > 0$, there are no zeros for $x > 1$.\n\nSince $\\lim_{x \\rightarrow -\\infty} f(x) < 0$, there is one zero for $x < -\\frac{1}{3}$, for a total of 1 zero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the exact value of $1+\\frac{1}{1+\\frac{2}{1+\\frac{1}{1+\\frac{2}{1+\\ldots}}}}$.",
    "options": [],
    "answer": "sqrt(2)",
    "solution": "Solution:\nLet $x$ be what we are trying to find.\n\n$x - 1 = \\frac{1}{1 + \\frac{2}{1 + \\frac{1}{1 + \\frac{2}{1 + \\ldots}}}}$\n\n$\\Rightarrow \\frac{1}{x - 1} - 1 = \\frac{2}{1 + \\frac{1}{1 + \\frac{2}{1 + \\cdots}}}$\n\n$\\Rightarrow \\frac{2}{\\frac{1}{x - 1} - 1} = x$\n\n$\\Rightarrow x^2 - 2 = 0$\n\nso $x = \\sqrt{2}$ since $x > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA beaver walks from $(0,0)$ to $(4,4)$ in the plane, walking one unit in the positive $x$ direction or one unit in the positive $y$ direction at each step. Moreover, he never goes to a point $(x, y)$ with $y>x$. How many different paths can he walk?",
    "options": [],
    "answer": "14",
    "solution": "Solution:\n\n$C(4)=14$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAfter walking so much that his feet get really tired, the beaver staggers so that, at each step, his coordinates change by either $(+1,+1)$ or $(+1,-1)$. Now he walks from $(0,0)$ to $(8,0)$ without ever going below the $x$-axis. How many such paths are there?",
    "options": [],
    "answer": "14",
    "solution": "Solution: $C(4)=14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFrank and Joe are playing ping pong. For each game, there is a $30\\%$ chance that Frank wins and a $70\\%$ chance Joe wins. During a match, they play games until someone wins a total of 21 games. What is the expected value of number of games played per match?",
    "options": [],
    "answer": "30",
    "solution": "Solution:\n\nThe expected value of the ratio of Frank's to Joe's score is $3:7$, so Frank is expected to win $9$ games for each of Frank's $21$. Thus the expected number of games in a match is $30$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 45968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest prime factor of $-x^{10}-x^{8}-x^{6}-x^{4}-x^{2}-1$, where $x=2i$, $i=\\sqrt{-1}$.",
    "options": [],
    "answer": "13",
    "solution": "Solution:\n13.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCalculate $\\sum_{n=1}^{2001} n^{3}$.",
    "options": [],
    "answer": "4012013006001",
    "solution": "Solution:\n\n$\\sum_{n=1}^{2001} n^{3} = \\left(\\sum_{n=1}^{2001} n\\right)^{2} = \\left(\\frac{2001 \\cdot 2002}{2}\\right)^{2} = 4012013006001$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that if two medians in a triangle are equal in length, then the triangle is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet equal medians $AD$ and $BE$ in triangle $ABC$ meet at $F$. It is well known that\n$$\nAF : FD = BF : FE = 2 : 1.\n$$\nTriangle $ABC$ is isosceles if we can show $AC = BC$ or equivalently, that $AE = BD$. But triangles $AFE$ and $BFD$ are congruent with vertical angles plus sides that are $1/3$ and $2/3$ of the equal median length.\n\n![](attached_image_1.png)\nSolution:\n\nLet medians $AM = BN$ in $\\triangle ABC$. Extend each median to $AM_1$ and $BN_1$ so that $M$ and $N$ are the midpoints of $AM_1$ and $BN_1$, respectively. By the property of bisecting diagonals, $ABM_1C$ and $ABC N_1$ are parallelograms. Hence $CM_1$ and $CN_1$ are each parallel and equal to $AB$. We conclude that $C$ lies on $N_1M_1$, $C$ is the midpoint of $N_1M_1$, and $AM_1 = BN_1$ as they are twice the lengths of the original medians $AM$ and $BN$. Summarizing, $ABM_1N_1$ is a trapezoid with equal diagonals.\n\nIt is easy to see that such a trapezoid is isosceles. One way to see this is to draw a line through $A$ parallel to diagonal $BN_1$, until it intersects line $N_1M_1$ in point $L$. Thus, $ABN_1L$ is a parallelogram, so $\\angle ALN_1 = \\angle ABN_1$. On the other hand, $\\triangle AM_1L$ is isosceles since $AL = BN_1 = AM_1$; hence, $\\angle ALN_1 = \\angle AM_1N_1$. Finally, $AB \\parallel N_1M_1$ implies $\\angle AM_1N_1 = \\angle BAM_1$. We conclude that $\\angle BAM_1 = \\angle ABN_1$, and $\\triangle ABN_1$ and $\\triangle BAM_1$ are congruent by two equal sides and angles between these sides. Therefore, $BM_1 = AN_1$ and our trapezoid is isosceles. Hence $\\angle AN_1C = \\angle BM_1C$.\n\nFinally, $\\triangle ACN_1$ and $\\triangle BCM_1$ are congruent by $AN_1 = BM_1$, $CN_1 = CM_1$ and $\\angle AN_1C = \\angle BM_1C$. We conclude that $AC = BC$ and our original $\\triangle ABC$ is also isosceles.\nSolution:\n\nAs a variation of the above solution, note that $NM$ is the midsegment of $\\triangle ABC$, and as such it is parallel to $AB$. Thus $ABMN$ is a trapezoid with equal diagonals, which by a similar argument as in Solution 1 is isosceles. Therefore, $\\angle BAC = \\angle ABC$ and $AC = BC$.\nSolution:\n\nA well-known formula for a parallelogram $ABM_1C$ says: $2(AB^2 + AC^2) = AM_1^2 + BC^2$ (it can be easily proved with vectors for example). From here one derives a formula for the median $AM$ of a triangle $\\triangle ABC$:\n$$\nAM^2 = \\frac{1}{2}(AB^2 + AC^2) - \\frac{1}{4} BC^2.\n$$\nSimilarly, the other median $BN$ in $\\triangle ABC$ satisfies:\n$$\nBN^2 = \\frac{1}{2}(AB^2 + BC^2) - \\frac{1}{4} AC^2.\n$$\nSince $AM = BN$, easy algebraic cancellations lead to $AC^2 = BC^2$, i.e. $AC = BC$ and our triangle is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many divisors does $2013^{13}$ have? (As usual, we count $1$ and $2013^{13}$ itself as divisors, but not negative integers.)",
    "options": [],
    "answer": "2744",
    "solution": "Solution:\n\nThe prime factorization of $2013$ is $3 \\cdot 11 \\cdot 61$, so\n$$\n2013^{13} = 3^{13} \\cdot 11^{13} \\cdot 61^{13}.\n$$\nA divisor of this number is found by choosing $0$ to $13$ of the factors $3$ (there are $14$ possible choices), $0$ to $13$ of the factors $11$ ($14$ choices), and $0$ to $13$ of the factors $61$ ($14$ choices). So the total number of divisors is $14 \\cdot 14 \\cdot 14 = 2744$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe set of positive integers is partitioned into finitely many subsets. Show that some subset $S$ has the following property: for every positive integer $n$, $S$ contains infinitely many multiples of $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the subsets be $S_{1}, S_{2}, \\ldots, S_{k}$. Suppose the statement is false and seek a contradiction. Then for each $S_{i}$ there exists some $n_{i}$ such that $S_{i}$ contains only finitely many multiples of $n_{i}$. Let $n = n_{1} n_{2} \\cdots n_{k}$; then every multiple of $n$ is a multiple of each $n_{i}$ and so each $S_{i}$ can contain only finitely many multiples of $n$. But this means that the $k$ sets together contain only finitely many multiples of $n$, and since they partition the positive integers (which contain infinitely many multiples of $n$), we have our contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all prime numbers $p$ such that $p^{2}+8$ is prime number, as well.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nFor $p=3$ we have $p^{2}+8=17$, which is prime.\n\nIf $p \\neq 3$ then $p$ is not divisible by $3$. The remainder of $p$ when divided by $3$ is either $1$ or $2$. This means that $p=3k+1$ for some integer $k$, or $p=3l+2$ for some integer $l$.\n\nIn the first case we get $p^{2}=(3k+1)^{2}=9k^{2}+6k+1$ and in the second, $p^{2}=(3l+2)^{2}=9l^{2}+12l+4$.\n\nIn both cases $p^{2}$ gives a remainder $1$ upon division by $3$. Hence $p^{2}+8$ is divisible by $3$ for all prime numbers different than $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKen is the best sugar cube retailer in the nation. Trevor, who loves sugar, is coming over to make an order. Ken knows Trevor cannot afford more than $127$ sugar cubes, but might ask for any number of cubes less than or equal to that. Ken prepares seven cups of cubes, with which he can satisfy any order Trevor might make. How many cubes are in the cup with the most sugar?",
    "options": [],
    "answer": "64",
    "solution": "Solution:\n\nThe only way to fill seven cups to satisfy the above condition is to use a binary scheme, so the cups must contain $1, 2, 4, 8, 16, 32$, and $64$ cubes of sugar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine, with proof, whether or not there exist positive integers $a$, $b$, and $c$ such that\n$$\na b + b c = a c \\quad \\text{and} \\quad a b c = 10! .\n$$",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Note that $10!$ has exactly one prime factor of $7$. Therefore, exactly one of $a$, $b$, and $c$ is divisible by $7$. If $7$ divides $b$ (we write this as $7 \\mid b$), then $7 \\mid a b + b c$ but $7 \\nmid a c$, so the equation cannot hold. Likewise, if $7 \\mid a$, then $7 \\mid a b$, $7 \\mid a c$ but $7 \\nmid b c$, and if $7 \\mid c$, then $7 \\mid b c$, $7 \\mid a c$ but $7 \\nmid a b$. So there are no solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA man has three pets: a mouse, a cat, and a dog. If the man leaves the cat and the dog alone, then the dog would kill the cat. If the man leaves the cat and the mouse alone, the cat would eat the mouse. One day the man decided to take his animals to the other side of the river. However, he has a small boat in which he can fit only one of the animals at a time. Show that it is possible for the man to take all of his animals to the other side of the river safely.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $A$ be the side of the river where the trip starts, and $B$ the other side. In the first trip the man should take the cat to the side $B$. He would leave the mouse and the dog together, but that is not dangerous. Then he will return back for a dog and take it to the side $B$. However when he brings the dog, then he returns the cat from $B$ to $A$. Then he takes the mouse from $A$ to $B$ and comes back to the side $A$ to finally take the cat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor which positive integers $n$ is $n^{4}+4$ equal to a prime number?",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nFor $n=1$ we get $1^{4}+4=5$, which works.\n\nFor all other values of $n$, the key idea is that\n$$\nn^{4}+4=n^{4}+4 n^{2}+4-4 n^{2}=(n^{2}+2)^{2}-(2 n)^{2}=(n^{2}+2 n+2)(n^{2}-2 n+2)\n$$\nwhich is the product of two integers greater than 1, and hence cannot be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $P$ be the point inside the square $ABCD$ such that $\\triangle ABP$ is equilateral. Calculate the angle $\\angle CPD$. Explain your answer!",
    "options": [],
    "answer": "150°",
    "solution": "Solution:\n\nThe triangle $DAP$ is isosceles because $AD = AP$, hence $\\angle ADP = \\angle APD = \\frac{180^\\circ - \\angle DAP}{2} = 75^\\circ$. Hence $\\angle PDC = 15^\\circ$. Similarly, $\\angle DCP = 15^\\circ$ and hence $\\angle CPD = 180^\\circ - 2 \\cdot 15^\\circ = 150^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven three points $A$, $B$, $C$ known to lie on a circle, prove that one can reconstruct the original circle with a straightedge and compass.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nHere is a suitable procedure: First draw circles centered at $A$ and $B$ with radius $AB$ and join their two points of intersection to create the perpendicular bisector of $AB$. This line contains all points that have the same distance from $A$ and $B$, so the center of a circle through $A$ and $B$ lies on it. Then, similarly construct the perpendicular bisector of $AC$. Because $A$, $B$, and $C$ are known to lie on a circle, the two perpendicular bisectors are not parallel (or else no point could have the same distance from $A$, $B$, and $C$), so they must meet at a point $O$. Draw a circle centered at $O$ with radius $OA$.\n\nSince $O$ lies on both perpendicular bisectors, this circle passes through $B$ and $C$. Finally, any other circle through the same three points would have to have its center on both perpendicular bisectors, and hence at $O$. Its radius must equal $OA$, implying that it coincides with the constructed circle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 45980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA car is moving at a constant speed. Every 15 minutes it makes a $90^{\\circ}$ turn to either left or right. If the car has started the trip at some point $A$, prove that it can return to $A$ only after an integer number of hours.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt is important to notice that an hour consists of 60 minutes and $60 / 15 = 4$. That's why it is common to say that 15 minutes is a \"quarter of an hour\". We may assume that the car is traveling in a square grid where the size of each square is the distance that the car travels in 15 minutes. Assume further that we know what is \"north\", \"east\", \"west\", and \"south\". Let $n$, $e$, $w$, and $s$ denote respectively the number of north, east, west, and south moves. If the car has returned to $A$, then $n = s$ and $w = e$. However, since the car at every vertex changes the direction from the set $\\{\\text{north}, \\text{south}\\}$ to the set $\\{\\text{east}, \\text{west}\\}$, or vice versa, we see that $n = w$. Thus $n = s = w = e$ and the total number of moves is $n + s + w + e = 4n$ which is divisible by 4. Thus the total time of travel is $4 \\cdot 15' n = n$ hours.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice picks an odd integer $n$ and writes the fraction\n$$\n\\frac{2 n+2}{3 n+2}\n$$\nShow that this fraction is already in lowest terms. (For example, if $n=5$ this is the fraction $\\frac{12}{17}$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $A=2 n+2$ and $B=3 n+2$. Now notice that\n$$\n3 A-2 B=3(2 n+2)-2(3 n+2)=2.\n$$\nSo if some integer $d \\geq 1$ divides both $A$ and $B$, it also divides $3 A-2 B=2$. Hence $d$ must be $1$ or $2$.\n\nBut since $n$ was odd, the number $3 n+2$ is odd, and so we can't have $d=2$. Thus the only common divisor of $A$ and $B$ is $1$, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the positive real numbers $a$ and $b$ satisfying $9 a^{2} + 16 b^{2} = 25$ such that $a \\cdot b$ is maximal. What is the maximum of $a \\cdot b$? Explain your answer!",
    "options": [],
    "answer": "Maximum ab = 25/24, attained at a = 5/(3√2) and b = 5/(4√2) (with 3a = 4b).",
    "solution": "Solution:\n\nApplying the inequality $x^{2} + y^{2} \\geq 2 x y$ on $x = 3a$ and $y = 4b$ we get\n$$\n25 = (3a)^{2} + (4b)^{2} \\geq 2 \\cdot 3a \\cdot 4b = 24ab.\n$$\nHence $ab \\leq \\frac{25}{24}$.\n\nThe equality is attained for $x = y$ or equivalently for $3a = 4b$. In that case\n$$\n25 = (3a)^{2} + (4b)^{2} = 2 \\cdot 9a^{2}\n$$\nhence $a = \\frac{5}{3\\sqrt{2}}$. Now we have\n$$\nb = \\frac{3}{4}a = \\frac{5}{4\\sqrt{2}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $x$ is a positive real number, find the smallest possible value of $2x + \\frac{18}{x}$.",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nThe answer is $12$. This is achieved when $x = 3$; to see it is optimal, note that\n$$\n2x + \\frac{18}{x} \\geq 12 \\Longleftrightarrow x^2 - 6x + 9 \\geq 0,\n$$\nwhich is obviously true since the left-hand side is $(x-3)^2 \\geq 0$.\n\nAlternatively, those who know the so-called AM-GM inequality may apply it directly.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo players, Cat and Mouse, play the following game on a $4 \\times 4$ checkerboard. Each player places a checker on a cell of the board (Cat goes first). Then, the two players take turns moving their checkers to an adjacent square, either vertically or horizontally (Cat again goes first). If, after either player's move, the two checkers occupy the same square, Cat wins. Otherwise, if each checker has made 2013 moves without this happening, Mouse wins. Determine, with proof, which player has a winning strategy.",
    "options": [],
    "answer": "Mouse has a winning strategy.",
    "solution": "Solution:\n\nMouse has a winning strategy. Color the cells of the checkerboard black and white alternately, in standard checkerboard fashion, so that adjacent cells are opposite colors. Whichever color Cat places his checker on, Mouse places his checker on a different cell of the same color. Then each time Cat moves, the checkers will be on opposite colors, and each time Mouse moves, the checkers will be on the same color. So Cat can never land on Mouse; it remains for Mouse to avoid landing on Cat. Since each cell has at least two neighbors, this is easy to do.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $a, b, c$ are positive integers such that\n$$\n\\begin{aligned}\n& b=a^{2}-a \\\\\n& c=b^{2}-b \\\\\n& a=c^{2}-c\n\\end{aligned}\n$$\nProve that $a=b=c=2$.",
    "options": [],
    "answer": "a = b = c = 2",
    "solution": "Solution:\nIf $a=1$, we get $b=0$ which is impossible. So it is enough to show that $a$ cannot be greater than $2$. If $a>2$, we have\n$$\nb=a^{2}-a=a(a-1)>a(2-1)=a.\n$$\nSo $b>a$; in particular $b>2$, so applying the same logic to the second equation we get $c>b$. Lastly, we have $c>2$ so applying the same logic to the third equation we get $a>c$. We have now proved $a>c>b>a$ which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequence\n$$\n5, 9, 49, 2209, \\ldots\n$$\nis defined by $a_{1} = 5$ and $a_{n} = a_{1} a_{2} \\cdots a_{n-1} + 4$ for $n > 1$. Prove that $a_{n}$ is a perfect square for $n \\geq 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThis is clear for $n = 2$. We use the relation\n$$\n\\begin{gathered}\na_{n-1} = a_{1} a_{2} \\cdots a_{n-2} + 4 \\\\\na_{1} a_{2} \\cdots a_{n-2} = a_{n-1} - 4\n\\end{gathered}\n$$\nfor $n \\geq 3$ to transform $a_{n}$:\n$$\n\\begin{aligned}\na_{n} & = a_{1} a_{2} \\cdots a_{n-2} a_{n-1} + 4 \\\\\n& = \\left(a_{n-1} - 4\\right) a_{n-1} + 4 \\\\\n& = a_{n-1}^{2} - 4 a_{n-1} + 4 \\\\\n& = \\left(a_{n-1} - 2\\right)^{2}.\n\\end{aligned}\n$$\nThis is clearly the square of an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBob has five airplane tickets with prices $\\$ 100, \\$ 120, \\$ 140, \\$ 160$, and $\\$ 180$. Bob gives an offer to Aerith: she can distribute his tickets among two bags, after which, without looking inside, Bob will randomly choose a bag and a ticket from it for Aerith to keep. What strategy should Aerith use to maximize the expected value of her ticket?",
    "options": [],
    "answer": "Put the most expensive ticket alone in one bag and all the remaining tickets in the other bag.",
    "solution": "Solution:\n\nLet the first bag be the one with less tickets. Let $a < b$ and $S, T$ be the numbers and prices of tickets in the respective bags. We have $a + b = 5$ and $S + T = \\$ 700$. The expected values of random tickets for each bag is then $S / a$ and $T / b$, the average of which is\n$$\n\\frac{S / a + T / b}{2} = \\frac{b S + a T}{2 a b} = \\frac{(b - a) S + a \\cdot \\$ 700}{2 a b}.\n$$\nFor fixed $a < b$, this is maximized when $S$ is as large as possible. Thus if $(a, b) = (1, 4)$, the first bag then should contain $\\$ 180$, whereas for $(a, b) = (2, 3)$, it should contain $\\$ 180 + \\$ 160 = \\$ 340$.\n\nUsing the first strategy, she has an expected value of $\\frac{\\$ 180 / 1 + \\$(700 - 180) / 4}{2} = \\$ 155$, and using the second, she has an expected value of $\\frac{\\$ 340 / 2 + \\$(700 - 340) / 3}{2} = \\$ 145$, so the best strategy is the first strategy: put the $\\$ 180$ ticket in one bag and the rest in another.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe country of Squareland is shaped like a square and is divided into $64$ congruent square cities. We want to divide Squareland into states and assign to each state a capital city so that the following rules are satisfied:\n\na. Every city lies entirely within one state.\n\nb. Given any two states, the numbers of cities in them differ by at most $1$.\n\nc. Any city in a state shares at least one corner with the state's capital.\n\nWhat is the smallest possible number of states?",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nIn the diagram below, no city shares a corner with any two of the cities marked X. Therefore the nine X's are\n![](attached_image_1.png)\nin nine different states. The diagram at right shows that nine states are also sufficient ( $*$ denotes capital).\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for $n \\geq 1$, the last $n+2$ digits of $11^{10^{n}}$ are $6000 \\ldots 0001$, with $n$ zeros between the 6 and the final 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe statement is that $11^{10^{n}} = k \\cdot 10^{n+2} + 6 \\cdot 10^{n+1} + 1$ for some positive integer $k$; we prove it by induction on $n$.\n\nFor the base case $n=1$, we expand $(10+1)^{10}$ by the binomial theorem and obtain\n$$\n\\left(\\text{ terms divisible by } 10^{3}\\right) + \\binom{10}{2} 10^{2} + \\binom{10}{1} 10^{1} + 1 = \\left(\\text{ terms divisible by } 10^{3}\\right) + 4500 + 100 + 1\n$$\nwhose last three digits are indeed 601.\n\nNow, for $n \\geq 2$, if the statement holds for $n-1$, then to prove it for $n$, we continue the process, using the binomial theorem twice:\n$$\n\\begin{gathered}\n11^{10^{n}} = \\left(11^{10^{n-1}}\\right)^{10} = \\left(k \\cdot 10^{n+1} + \\left[6 \\cdot 10^{n} + 1\\right]\\right)^{10} \\\\\n= \\left(\\text{ terms divisible by } 10^{2n+2}\\right) + \\binom{10}{1}\\left(k \\cdot 10^{n+1}\\right)\\left(6 \\cdot 10^{n} + 1\\right)^{9} + \\left(6 \\cdot 10^{n} + 1\\right)^{10}\n\\end{gathered}\n$$\nand all the terms except the last are divisible by $10^{n+2}$. Therefore it suffices to find the last $n+2$ digits of $\\left(6 \\cdot 10^{n} + 1\\right)^{10}$, which equals\n$$\n\\left(\\text{ terms divisible by } 10^{2n}\\right) + \\binom{10}{1} 6 \\cdot 10^{n} + 1 = \\left(\\text{ terms divisible by } 10^{2n}\\right) + 6 \\cdot 10^{n+1} + 1\n$$\nSince $2n \\geq n+2$, we have what we need.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that $k$ and $n$ are integers with $0 \\leq k \\leq n-3$. Prove that\n$$\n\\binom{n}{k},\\ \\binom{n}{k+1},\\ \\binom{n}{k+2},\\ \\binom{n}{k+3}\n$$\ncannot form an arithmetic progression.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose otherwise. The common difference $d$ satisfies $\\binom{n}{k+1} - \\binom{n}{k} = d = \\binom{n}{k+2} - \\binom{n}{k+1}$, giving $2\\binom{n}{k+1} = \\binom{n}{k} + \\binom{n}{k+2}$. Expanding the binomial coefficients in terms of factorials gives\n$$\n2 \\frac{n!}{(k+1)! (n-k-1)!} = \\frac{n!}{k! (n-k)!} + \\frac{n!}{(k+2)! (n-k-2)!}\n$$\nor, multiplying through by $(k+2)! (n-k)! / n!$,\n$$\n2(k+2)(n-k) = (k+2)(k+1) + (n-k)(n-k-1).\n$$\nA completely analogous analysis gives $2\\binom{n}{k+2} = \\binom{n}{k+1} + \\binom{n}{k+3}$ and\n$$\n2(k+3)(n-k-1) = (k+2)(k+3) + (n-k-2)(n-k-1).\n$$\nSubtracting (1) from (2) gives\n$$\n2(k+3)(n-k-1) - 2(k+2)(n-k) = 2(k+2) - 2(n-k-1).\n$$\nIf we divide by 2, this becomes $-n + 2k + 3 = (k+3)(n-k-1) - (k+2)(n-k) = n - 2k - 3$, which yields $n = 2k + 3$. Plugging this back into (1) gives us $2(k+2)(k+3) = (k+2)(k+1) + (k+3)(k+2)$; since $k \\geq 0$, we can divide by $k+2$ to get $2k + 6 = 2k + 4$, which yields $6 = 4$. This is nonsense, so our assumption - that the numbers form an arithmetic progression - must be false.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A B C D E$ be a convex pentagon. If $\\alpha=\\angle D A C$, $\\beta=\\angle E B D$, $\\gamma=\\angle A C E$, $\\delta=\\angle B D A$, and $\\epsilon=\\angle B E C$, as shown in the picture, calculate the sum $\\alpha+\\beta+\\gamma+\\delta+\\epsilon$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "180°",
    "solution": "Solution:\n\nLet $M, N, P, Q, R$ denote the intersections of the lines $A C$ and $B E$, $A C$ and $B D$, $C E$ and $B D$, $D A$ and $C E$, $E B$ and $A E$, respectively. Since the sum of the internal angles of a triangle is $180^{\\circ}$, from $\\triangle A M R$ we get $\\alpha=180^{\\circ}-\\angle A M R-\\angle M R A$. We also know that $\\angle A M R=\\angle B M N$ and we can denote these angles by $\\angle M$. Analogously $\\angle A R M=\\angle E R Q=\\angle R$. With similar equations for $\\beta, \\gamma, \\delta$, and $\\epsilon$ we get that\n\n$$\n\\alpha+\\beta+\\gamma+\\delta+\\epsilon=5 \\cdot 180^{\\circ}-2(\\angle M+\\angle N+\\angle P+\\angle Q+\\angle R).\n$$\n\nSince $\\angle M, \\angle N, \\angle P, \\angle Q$, and $\\angle R$ are exterior angles of the pentagon $M N P Q R$ their sum has to be $360^{\\circ}$ implying that\n\n$$\n\\alpha+\\beta+\\gamma+\\delta+\\epsilon=180^{\\circ}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA point $P$ lies inside a regular hexagon $A B C D E F$. The distances from $P$ to the sides $A B, B C, C D, D E, E F$, and $F A$ are respectively $1,2,5,7,6$, and $x$. Find $x$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nOpposite sides of a regular hexagon are parallel. The sum of the distances from $P$ to $A B$ and to $D E$ is simply the distance between $A B$ and $D E$, which must therefore equal $1+7=8$. Because of the symmetry of the hexagon, the distance between any pair of opposite sides must be $8$. Applying this to sides $C D$ and $F A$, we see that the distance from $P$ to $C D$ is $5$ so the distance from $P$ to $F A$ must be $8-5=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a convex quadrilateral. Let $I_{A}$ be the incenter of $BCD$ and define $I_{B}, I_{C}, I_{D}$ similarly. Show that $AC \\perp I_{B}I_{D}$ if and only if $I_{A}I_{C} \\perp BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe show that both are true if and only if $AB + CD = BC + DA$.\n\nLet's first consider the triangle $ABD$. We know the incircle of $ABD$ is centered at $I_{C}$. Let $X, Y, T_{A}$ denote the points where the incircle is tangent to $AB$, $AD$, and $BD$, respectively. We have that $AX = AY$, $BX = BT_{A}$, and $DT_{A} = DY$ since each of these are segments tangent to the incircle from a common point. Thus,\n$$\nAB - AD = (AX + XB) - (AY + YD) = BX - CY = BT_{A} - DT_{A}.\n$$\nWe now include the triangle $BCD$. Let $T_{C}$ denote the point at which the incircle of $BCD$, centered at $I_{A}$, is tangent to $BD$. Replicating the argument for $ABD$, we see that $BT_{C} - DT_{C} = BC - CD$. Now, $I_{A}I_{C} \\perp BD$ iff $T_{A} = T_{C}$, since the altitudes from $I_{A}$ and $I_{C}$ to diagonal $BD$ must meet at a point. Hence, $I_{A}I_{C} \\perp BD$ is equivalent to\n$$\nBA - DA = BT_{A} - DT_{A} = BT_{C} - DT_{C} = BC - CD,\n$$\nwhich in turn is equivalent to $AB + CD = AD + BC$.\n\nWe use a symmetric argument, but this time considering triangles $ABC$ and $ABD$, to show that a necessary and sufficient condition for $I_{B}I_{D} \\perp AC$ is also $AB + CD = AD + BC$. Since $AC \\perp I_{B}I_{D}$ and $I_{A}I_{C} \\perp BD$ are both equivalent to $AB + CD = AD + BC$, we have $AC \\perp I_{B}I_{D} \\Longleftrightarrow I_{A}I_{C} \\perp BD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA $3 \\times 3 \\times 3$ cube is made out of 27 subcubes. On every face shared by two subcubes, there is a door allowing you to move from one cube to the other. Is it possible to visit every subcube exactly once if\n(a) You may start and end wherever you like\n(b) You must start at the center subcube?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "Solution:\n(a) It is possible. Here is one of many possible routes.\n![](attached_image_1.png)\nLevel 1\nLevel 2\nLevel 3\n\n(b) It is impossible. Color the subcubes black and white alternately as shown:\n![](attached_image_2.png)\nLevel 1\n![](attached_image_3.png)\nLevel 2\n![](attached_image_4.png)\nLevel 3\nEvery door connects a black subcube to a white subcube. Since the central subcube is white, the route must begin\n$$\n\\text{White} \\rightarrow \\text{Black} \\rightarrow \\text{White} \\rightarrow \\text{Black} \\rightarrow \\cdots.\n$$\nExamining the first 27 subcubes visited, we see that 14 are white and 13 are black, a contradiction since the actual cube has 14 black and 13 white subcubes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $k$ be a rational number greater than $1$ (correction by Fengning Ding). Prove that there exist positive integers $a, b, c$ satisfying the equations\n$$\n\\begin{aligned}\na^{2}+b^{2} & =c^{2} \\\\\n\\frac{a+c}{b} & =k .\n\\end{aligned}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $k = x / y$, where $x$ and $y$ are positive integers. We find that $x > y$. It suffices to note that\n$$\na = x^{2} - y^{2}, \\quad b = 2 x y, \\quad c = x^{2} + y^{2}\n$$\nare positive integers satisfying both of the given equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 45996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that there exists a polynomial $f(x, y, z)$ with the following property: the numbers $|x|, |y|$, and $|z|$ are the sides of a triangle if and only if $f(x, y, z) > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n$$\nf(x, y, z) = (x + y + z)(-x + y + z)(x - y + z)(x + y - z) \\quad \\left[= x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} - x^{4} - y^{4} - z^{4}\\right]\n$$\nIt is easily seen that the transformation $x \\mapsto -x$, and symmetrically $y \\mapsto -y$ and $z \\mapsto -z$, do not change $f$, so it is enough to prove the following statement: If $x, y$, and $z$ are nonnegative reals, then $x, y$, and $z$ are the sides of a triangle if and only if $f(x, y, z) > 0$.\n\nMoreover, changing the order of $x, y$, and $z$ does not change $f$, so we may assume that $x \\leq y \\leq z$. Now three of the factors of $f(x, y, z)$, namely $x + y + z$, $-x + y + z$, and $x - y + z$, are clearly nonnegative.\n\nIf $x, y$, and $z$ are the sides of a triangle, the familiar triangle inequality $x + y \\geq z$ implies that the fourth factor $x + y - z$ is positive. Also, a side of a triangle cannot be zero, from which we get $x + y + z > 0$, $-x + y + z > 0$, $x - y + z > 0$, and hence $f(x, y, z) > 0$.\n\nConversely, if $f(x, y, z) > 0$, then the four factors must be positive, so $x, y$, and $z$ are positive and the triangle inequality $x + y > z$ holds. To construct the triangle, we may draw two circles of radii $x$ and $y$ whose centers $Y$, $X$ are a distance $z$ apart. Because each circle passes both inside and outside the other, the circles intersect at two points. Let $Z$ be one. Then $X Y Z$ is the desired triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 45997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are three prisoners in a prison. A warden has 2 red and 3 green hats and he has decided to play the following game: He puts the prisoners in a row one behind the other and on the head of each prisoner he puts a hat. The first prisoner in the row can't see any of the hats, the second prisoner can see only the hat on the head of the first one, and the third prisoner can see the hats of the first two prisoners. If some of the prisoners tells the color of his own hat, he is free; but if he is wrong, the warden will kill him. If a prisoner remains silent for sufficiently long, he is returned to his cell. Of course, each of them would like to be rescued from the prison, but if he isn't sure about the color of his hat, he won't guess.\n\nAfter noticing that second and third prisoner are silent for a long time, first prisoner (the one who doesn't see any hat) has concluded the color of his hat and told that to the warden. What is the color of the hat of the first prisoner? Explain your answer! (All prisoners know that there are 2 red and 3 green hats in total and all of them are good at mathematics.)",
    "options": [],
    "answer": "green",
    "solution": "Solution:\n\nIf the first two prisoners had red hats, the third one won't be silent (he would conclude that his hat is green). Hence, at least one of the first two prisoners has a green hat, and everybody knows that (because the third prisoner is silent). Thus if the first prisoner had red hat, the second one would conclude that he must be the one with the green hat. However, the second prisoner was silent and the hat at the head of the first prisoner must be green.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 45998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all prime numbers $p$ such that $p^{2} + 2007p - 1$ is prime as well.",
    "options": [],
    "answer": "p = 3",
    "solution": "Solution:\n\nIf $p = 3$, then $p^{2} + 2007p - 1 = 6029$ which is a prime.\n\nFor $p \\neq 3$, we know that $p = 3k \\pm 1$ for some integer $k$. Hence\n$$\np^{2} + 2007p - 1 = (3k \\pm 1)^{2} + 2007(3k \\pm 1) - 1 = 9k^{2} \\pm 6k + 1 + 2007 \\cdot 3k \\pm 2007 - 1 = 9k^{2} \\pm 6k + 2007 \\cdot 3k \\pm 2007.\n$$\nThis expression is divisible by $3$ and can't be prime (unless it equals $3$, but for $p > 3$ it is much larger than $3$). Thus the only such prime number is $p = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 45999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the value of the infinite continued fraction\n$$\n1 + \\frac{1}{2 + \\frac{1}{2 + \\frac{1}{2 + \\frac{1}{\\ldots}}}}\n$$",
    "options": [],
    "answer": "sqrt(2)",
    "solution": "Solution:\nLet the value of the fraction be $x$. Then we get\n$$\nx = 1 + \\frac{1}{x + 1} \\Longrightarrow x(x + 1) = (x + 1) + 1 \\Longrightarrow x^{2} = 2\n$$\nThus, since $x$ is positive, we must have $x = \\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe wish to distribute $12$ indistinguishable stones among $4$ distinguishable boxes $B_{1}$, $B_{2}$, $B_{3}$, $B_{4}$. (It is permitted some boxes are empty.)\n\na. Over all ways to distribute the stones, what fraction of them have the property that the number of stones in every box is even?\n\nb. Over all ways to distribute the stones, what fraction of them have the property that the number of stones in every box is odd?",
    "options": [],
    "answer": "a. 12/65; b. 1/13",
    "solution": "Solution:\n\nWe will invoke the following famous theorem, colloquially called stars and bars: the number of ways to distribute $a$ indistinguishable objects among $b$ boxes is $\\binom{a+b-1}{b-1}$. One can see this by considering a string of $a$ objects (\"stars\") and $b-1$ dividers (\"bars\"), which give a distribution into boxes.\n\nThe total number of ways to distribute the stones overall is equal to $\\binom{12+3}{3} = 455$, by stars-and-bars argument.\n\nFor a., we imagine distributing the stones in pairs; we are distributing $6$ doublestones. So the number of ways is $\\binom{6+3}{3} = 84$. Thus the fraction is $\\frac{84}{455} = \\frac{12}{65}$.\n\nFor b., we also distribute the stones in pairs, but we need to place one stone in each box first: thus, we are distributing $4$ double-stones (with the eight remaining stones). The number of ways to do this is $\\binom{4+3}{3} = 35$. Thus the fraction is $\\frac{35}{455} = \\frac{1}{13}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $M$ be the midpoint of the side $AC$ of triangle $ABC$. If $N$ is the point on the side $AB$, $O$ intersection of the lines $BM$ and $CN$, and if the areas of triangles $BON$ and $COM$ are equal, prove that $N$ is the midpoint of $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince the areas of $\\triangle BON$ and $\\triangle COM$ are equal we see that the areas of triangles $\\triangle BCN$ and $\\triangle CBM$ are also equal. Since these two triangles share the side, they must have the corresponding altitudes equal. Hence the length of perpendiculars from $M$ and $N$ to $BC$ are equal, implying that $NM \\parallel BC$. Thus $MN$ is the midsegment of $\\triangle ABC$ and consequently $N$ is the midpoint of $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $m, n$ be positive integers. Suppose that a given rectangle can be tiled (without overlaps) by a combination of horizontal $1 \\times m$ strips and vertical $n \\times 1$ strips. Show that it can be tiled using just one of the two types.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf the rectangle is $a \\times b$ ($a$ the horizontal dimension), it is clear that $a$ and $b$ are integers since the rectangle can be divided into rectangles of integer side lengths; we want to show that either $a$ is divisible by $m$ or $b$ is divisible by $n$.\n\nLet $\\zeta = \\cos \\frac{2\\pi}{m} + i \\sin \\frac{2\\pi}{m}$, $\\xi = \\cos \\frac{2\\pi}{n} + i \\sin \\frac{2\\pi}{n}$ be complex $m$th and $n$th roots of unity, respectively. Divide the rectangle into $ab$ unit squares, and write the number $\\zeta^{x} \\xi^{y}$ in the square with coordinates $(x, y)$.\n\nFor each vertical strip, the sum of the numbers written in it is zero: if its lowermost square has coordinates $(x, y)$, then the sum of these numbers is\n$$\n\\zeta^{x} \\xi^{y} + \\zeta^{x} \\xi^{y+1} + \\cdots + \\zeta^{x} \\xi^{y+n-1} = \\zeta^{x} \\xi^{y} (1 + \\xi + \\xi^{2} + \\cdots + \\xi^{n-1}) = \\zeta^{x} \\xi^{y} \\cdot \\frac{\\xi^{n} - 1}{\\xi - 1} = 0\n$$\nsince $\\xi^{n} = 1$.\n\nA similar calculation shows that the sum of the numbers in any horizontal strip is zero.\n\nSince the rectangle is tiled by these strips, the sum of all the numbers in the rectangle is zero. But this sum is also equal to\n$$\n\\left(\\zeta + \\zeta^{2} + \\zeta^{3} + \\cdots + \\zeta^{a}\\right) \\left(\\xi + \\xi^{2} + \\xi^{3} + \\cdots + \\xi^{b}\\right) = \\zeta \\xi \\cdot \\frac{\\zeta^{a} - 1}{\\zeta - 1} \\cdot \\frac{\\xi^{b} - 1}{\\xi - 1}\n$$\nFor this to be $0$, we must have $\\zeta^{a} = 1$ or $\\xi^{b} = 1$, which implies $a$ divisible by $m$ or $b$ divisible by $n$, respectively.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn how many different ways one can place 3 rooks on the cells of $6 \\times 2006$ chessboard such that they don't attack each other?",
    "options": [],
    "answer": "20 * 2006 * 2005 * 2004",
    "solution": "Solution:\n\nIn order not to attack each other, the rooks cannot share a row or a column. Since there are 6 columns in total, we have 20 possibilities for choosing 3 columns in which we will place our rooks. After choosing the columns, we will place our rooks one by one in these columns. The first rook can be placed in 2006 different ways. The second one, however, can't be placed in the row that contains the first rook. Hence there are 2005 possibilities for the second rook. Similarly, the third rook can be placed in 2004 different ways. In total, there are $20 \\cdot 2006 \\cdot 2005 \\cdot 2004$ ways to place 3 rooks in $6 \\times 2006$ chessboard so that they don't attack each other.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine whether there exists a number that begins with $2$ having the property that, when the $2$ is moved to the end, the number is\n(a) doubled;\n(b) tripled.",
    "options": [],
    "answer": "a) No such number exists. b) 285714.",
    "solution": "Solution.\nThere is no such number. Suppose that it existed; represent it by $2 \\cdots ab$ where $a$ and $b$ are the last two digits of the number. Let us write the digit-by-digit multiplication:\n\n![](attached_image_1.png)\n\nFrom the last column it is clear that there are only two choices for $b$: $1$ and $6$. We now look at the next-to-last column. If $b$ is $1$, there is no way that the multiplication of $a$ by $2$, without an added carry, can give the odd number $1$. On the other hand, if $b$ is $6$, there is no way that the multiplication of $b$ by $2$, with the added carry of $1$ resulting from $6 \\times 2 = 12$, can give the even number $6$.\n\nThe answer is yes; such a number is $285714$, which satisfies $285714 \\times 3 = 857142$. There is no need to say more; we have rigorously proved that such a number exists. However, we will mention two of many methods for finding this number. The first is the end-digit analysis used in part (a):\n$$\n\\begin{gathered}\n\\cdots x \\times 3 = \\cdots 2 \\text{ gives } x = 4 \\text{ (no other digit works) } \\\\\n\\cdots x4 \\times 3 = \\cdots 42 \\text{ gives } x = 1, \\text{ etc. }\n\\end{gathered}\n$$\nThis process continues until we reach a number that begins with $2$.\n\nThe second method is an algebraic one. Suppose that the number is $x$ and has $n$ digits. Then moving a $2$ from the beginning to the end of $x$ can be achieved by first multiplying $x$ by $10$ (thus adding a $0$ at the end), adding $2$ (thus changing this $0$ to a $2$), and subtracting $2 \\cdot 10^n$ (deleting the $2$ at the beginning). We have the condition\n$$\n10x + 2 - 2 \\cdot 10^n = 3x,\n$$\nwhich we can solve for $x$:\n$$\n\\begin{aligned}\n10x - 3x & = 2 \\cdot 10^n - 2 \\\\\n7x & = 2 \\cdot 10^n - 2 \\\\\nx & = \\frac{2 \\cdot 10^n - 2}{7}.\n\\end{aligned}\n$$\nThus for $x$ to exist, $2 \\cdot 10^n - 2$ must be divisible by $7$. As is easily verified, this does not hold for $n = 1, 2, 3, 4,$ or $5$, but it works for $n = 6$, yielding\n$$\nx = \\frac{2000000 - 2}{7} = 285714.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that any triangle can be dissected into five isosceles triangles. (The pieces must not intersect except at their boundaries, and they must cover the given triangle.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe present one of many possible solutions.\n\nFirst, draw the height to the longest side (or one of the longest sides), as shown in Figure 1. This height will always lie inside the triangle and will divide it into two right triangles. We will cut one of these right triangles into two isosceles pieces and the other into three.\n\nFor the former construction, we can simply join the midpoint of the hypotenuse to the opposite vertex, as shown in Figure 2. The two resulting triangles are isosceles by the familiar theorem that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices.\n\nWe now turn to the problem of cutting a right triangle $XYZ$ into three isosceles triangles. If the given triangle is not isosceles, the perpendicular bisector of the hypotenuse intersects the longer leg, labeled $YZ$ in Figure 3, at a point $W$. Drawing $XW$ cuts the triangle into an isosceles triangle $XWY$ and a right triangle $XWZ$; the latter can be chopped up into two isosceles triangles by the preceding technique. If instead the given right triangle is isosceles, it can easily be divided into three right isosceles triangles by the method shown in Figure 4.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n![](attached_image_3.png)\nFigure 3\n![](attached_image_4.png)\nFigure 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn a $5 \\times 5$ chessboard, a king moves according to the following rules:\n- It can move one square at a time, horizontally, vertically, or diagonally. (These are the usual moves of the king in chess.)\n- It can move in each of the eight allowable directions at most three times in its entire route.\nThe king can start at any square. Determine\n(a) whether the king can visit every square;\n(b) whether the king can visit every square except the center.",
    "options": [],
    "answer": "(a) No; (b) Yes.",
    "solution": "Solution:\n\na. The answer is no. To visit all 25 squares, the king must make his maximum of 24 moves and thus must move in each of the eight allowable directions exactly 3 times. Three of these directions (or 9 moves) take the king from a row to the next higher row; three directions (or 9 moves) go to the next lower row, and the remaining two directions and six moves keep the king in the same row. Because the numbers of upward and downward moves are equal, the king must start and end in the same row. Similarly, he must start and end in the same column and thus in the same square. Thus the 25th square visited will repeat the first and at least one square will be missed.\n\nb. The answer is yes. A solution is shown. Trickily, all known solutions are completely asymmetric.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a, b, c, d$ are positive real numbers such that $\\frac{5 a+b}{5 c+d}=\\frac{6 a+b}{6 c+d}$ and $\\frac{7 a+b}{7 c+d}=9$, calculate $\\frac{9 a+b}{9 c+d}$.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\nLet $\\frac{5 a+b}{5 c+d}=\\frac{6 a+b}{6 c+d}=k$. Then $5 a+b=k(5 c+d)$ and $6 a+b=k(6 c+d)$. Subtracting these two equations gives $a=k c$. Now we can easily get that $b=k d$. From $\\frac{7 a+b}{7 c+d}=\\frac{7 k c+k d}{7 c+d}=k=9$ we get $\\frac{9 a+b}{9 c+d}=9$ as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine whether it is possible to write the numbers $1, 2, \\ldots, 24$ on the edges of the $3 \\times 3$ grid of squares shown, one number to each edge, such that the sum of the six numbers on every path of minimal length from the upper left corner to the lower right corner is the same.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Yes, it is possible.",
    "solution": "Solution:\n\nIt is possible, and two especially simple solutions are shown. To check that they work without going through every path, one can use the following strategy: First notice that in each of the nine unit squares, the sum of the left and bottom edges equals the sum of the top and right edges. Now, given any path that includes a rightward move (R) followed by a downward move (D), we can replace RD by DR without changing the sum of the edges. Now a minimal path consists of three R's and three D's, and each $D$ can move past each $R$ at most once; so after at most nine steps the path will be converted to DDDRRR. This proves that all paths have the same sum.\n\n![](attached_image_2.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA $2012 \\times 2012$ table is to be filled with integers in such a way that each of the 4026 rows, columns, and main diagonals has a different sum. What is the smallest number of distinct values that must be used in the table?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nAnswer: 3.\nIf at most two numbers are used, say $x$ and $y$, the sum of every row and column is completely determined by the number of $y$'s it has, which ranges from 0 to 2012. Thus there are only 2013 possible sums, not enough for the 4026 rows, columns, and diagonals.\n\nOn the other hand, if $n$ is a large integer (say, 10000), then it is possible to fill the table using 0, 1, and $n$. Here is a $6 \\times 6$ example that generalizes readily:\n\n| 1 | 0 | 0 | 0 | 0 | 0 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 1 | 1 | 0 | 0 | 0 | $n$ |\n| 1 | 1 | 1 | 0 | $n$ | $n$ |\n| 1 | 1 | 0 | $n$ | $n$ | $n$ |\n| 1 | 0 | 0 | 0 | $n$ | $n$ |\n| 0 | 0 | 0 | 0 | 0 | $n$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe integers from $1$ to $16$ are arranged in a $4 \\times 4$ array so that each row, column and diagonal adds up to the same number.\n\na. Prove that this number is $34$.\n\nb. Prove that the four corners also add up to $34$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Add up all the numbers in the square in two ways. On the one hand, it consists of four rows, each adding to the common sum $S$, so the entire square adds to $4S$. But the numbers in the square are also the integers from $1$ to $16$, whose sum is\n$$\n\\frac{16 \\cdot 17}{2} = 136\n$$\nHowever, the OCR text says $152$; let's check: $1 + 2 + \\dots + 16 = \\frac{16 \\cdot 17}{2} = 136$.\n\nBut the solution text says $152$; this is an OCR error. The correct sum is $136$.\n\nSo,\n$$\n\\text{Sum of all numbers in square} = 4S = 136\n$$\nwhence $S = 34$.\n\nb. Denote the cells of the table as shown below.\n\n| $a$ | $b$ | $c$ | $d$ |\n| :---: | :---: | :---: | :---: |\n| $e$ | $f$ | $g$ | $h$ |\n| $i$ | $j$ | $k$ | $m$ |\n| $n$ | $p$ | $q$ | $r$ |\n\nAdding the equations\n$$\n\\begin{aligned}\na + b + c + d &= 34 \\\\\nn + p + q + r &= 34 \\\\\na + f + k + r &= 34 \\\\\nn + j + g + d &= 34 \\\\\n-b - f - j - p &= -34 \\\\\n-c - g - k - q &= -34\n\\end{aligned}\n$$\nyields $2a + 2d + 2n + 2r = 136$, or $a + d + n + r = 68$.\n\nBut since the correct sum for each is $34$, the sum $a + d + n + r = 34$.\n\nTherefore, the four corners also add up to $34$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve or disprove the following assertion: given 3 noncollinear points in the plane, the disk with the smallest radius which contains all three points is the circle passing through all three points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe assertion is false. Consider the points $A = (-\\sqrt{3}, 0)$, $B = (\\sqrt{3}, 0)$, $C = (0, 1)$. They are contained in the circle with diameter $AB$, which has radius $\\sqrt{3}$, but the circumcircle of the triangle $ABC$ has radius $2$ (centered at $O = (0, -1)$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA line $l$ and two points $A$ and $B$ are given in a plane in such a way that $A$ belongs to $l$ but $B$ doesn't. Construct the circle $k$ that passes through $B$ and touches $l$ at the point $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $m$ be the bisector of the segment $A B$. The center of the circle $k$ has to belong to $m$. Similarly, since the circle has to be tangent to $l$ at the point $A$ its center has to be located on the line $a$ perpendicular to $l$ that passes through $A$.\n\nThe lines $m$ and $a$ are easy to construct and their intersection is the center $O$ of the circle. Now we have the center and the point $B$ hence the circle is determined.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that no integer greater than $2008$ can be equal to the sum of squares of its digits.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $n = \\overline{a_k a_{k-1} \\ldots a_0}$ be an integer equal to the sum of squares of its own digits. Then\n$$\nn = 10^k a_k + 10^{k-1} a_{k-1} + \\cdots + a_0.\n$$\nOn the other hand,\n$$\na_0^2 + a_1^2 + \\cdots + a_k^2 \\leq 9^2 \\cdot (k+1) = 81 \\cdot (k+1) < 10^k \\quad \\text{for } k \\geq 4.\n$$\n(The last inequality is easy to prove by induction because it holds for $k=4$, and if it holds for some $k$, then\n$$\n81 \\cdot (k+2) = 81 \\cdot (k+1) + 81 \\leq 10^k + 81 < 10^k + 10^k < 10^{k+1}.\n$$\n)\n\nTherefore, for $k \\geq 4$, the sum of squares of the digits is less than $10^k$, so $n$ cannot have more than $4$ digits. Thus, $n \\leq 9999$.\n\nNow, for $n \\geq 2009$, $n$ must be a $4$-digit number. The maximum sum of squares of $4$ digits is $4 \\times 81 = 324$, which is much less than $2009$. Therefore, no integer greater than $2008$ can be equal to the sum of squares of its digits.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that there are infinitely many prime numbers whose last digit is not $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume there are only finitely many such primes $p_{1}, \\ldots, p_{k}$. Consider the number\n$$\nN = 10 p_{1} p_{2} \\ldots p_{k} - 1\n$$\nSince $N$ has last digit $9$, there must be a prime $p$ dividing $N$ which does not have last digit $1$ (otherwise $N$ must have last digit $1$). But by construction, $p$ cannot divide $N$, because it leaves a remainder of $p-1$ when divided by $p$. This is a contradiction, so our assumption was wrong and there must be infinitely many such primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo bikers, Bill and Sal, simultaneously set off from one end of a straight road. Neither biker moves at a constant rate, but each continues biking until he reaches one end of the road, at which he instantaneously turns around. When they meet at the opposite end from where they started, Bill has traveled the length of the road eleven times and Sal seven times. Find the number of times the bikers passed each other moving in opposite directions.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nDefine a pass $(\\mathrm{P})$ to be an time when Bill and Sal pass one another moving in opposite directions and a turn (T) to be a time when one of the bikers turns around. If both bikers turn around simultaneously, we may alter their speeds slightly, causing one turn to happen before the other, without affecting the number of passes. We distinguish three states:\nA. The bikers are moving in the same direction.\nB. The bikers are moving toward each other.\nC. The bikers are moving away from one another.\nWe notice that the bikers can only change state by turning or passing. Moreover, there are only three possible state changes:\n- From state A, one biker reaches the end and turns around, creating state B.\n- From state B, the bikers pass one another, creating state C.\n- From state $\\mathrm{C}$, one biker reaches the end and turns around, creating state A.\nSince there are no other possibilities and we begin at state A, the states must follow the sequence ABCABC $\\cdots \\mathrm{ABCA}$ and the turns and passes TPTTPT ... TPT. We are given that Bill makes ten turns and Sal six, or 16 in all. Consequently there are eight passes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a certain kingdom, the only coin values are $3$ and $5$. Determine all possible amounts of money you can have using only these coins.",
    "options": [],
    "answer": "All positive integers except 1, 2, 4, and 7.",
    "solution": "Solution:\n\nThe amounts of money that you can have are all positive integers except $1, 2, 4$, and $7$. We can see that all these are impossible. To show that all other amounts are possible, it is clear that $3$ and $5$ are possible. Also, $6, 8, 9$, and $10$ are all possible, since\n$$\n\\begin{aligned}\n6 & = 3 + 3, \\\\\n8 & = 3 + 5, \\\\\n9 & = 3 + 3 + 3, \\\\\n10 & = 5 + 5.\n\\end{aligned}\n$$\nNow, any higher amount of money has a remainder of $0, 1$, or $2$ when divided by $3$. If it has a remainder of $0$, it is $9$ plus a multiple of $3$; if it has a remainder of $1$, it is $10$ plus a multiple of $3$; and if it has a remainder of $2$, it is $8$ plus a multiple of $3$. No matter what, it is one of $8, 9$, or $10$ plus a multiple of $3$, so we can just add more $3$ coins to get from one of $8, 9$, or $10$ to our number. Thus, all higher amounts of money can be made.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCircles $k_{1}$ and $k_{2}$ intersect at points $A$ and $B$. Line $l$ is the common tangent to these circles and it touches $k_{1}$ at $C$ and $k_{2}$ at $D$ such that $B$ belongs to the interior of the triangle $ACD$. Prove that $\\angle CAD + \\angle CBD = 180^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nHere we use the fact that the angle between the tangent and the chord is equal to the peripheral angle corresponding to the chord (this fact is easy to verify and can be found in any book on geometry). We have $\\angle CAB = \\angle DCB$ and $\\angle DAB = \\angle DCB$. Hence $\\angle CAD = \\angle CAB + \\angle DAB = \\angle DCB + \\angle DCB = 180^{\\circ} - \\angle DBC$. The statement now follows directly.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA castle has infinitely many rooms labeled $1,2,3, \\ldots$, which are divided into several halls. Suppose room $n$ is on the same hall as rooms $3 n+1$ and $n+10$ for every $n$. Determine the maximum possible number of different halls in the castle.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nThere are at most three different halls in the castle. Because rooms $n$ and $n+10$ are on the same hall, any two rooms with the same units digit must be on the same hall.\n\nNow, repeatedly using the rule that rooms $n$ and $3 n+1$ are on the same hall, we find that room $1$ is on the same hall as rooms $4$, $13$, $40$, and $121$, so all rooms with units digit $0,1,3$, or $4$ are on the same hall. There need not be any other rooms on this hall, since whenever $n$ has units digit in this set, so does $3 n+1$, and so does $(n-1) / 3$, if it is a whole number.\n\nSimilarly, room $2$ is on the same hall as rooms $7$ and $22$, so all rooms with units digit $2$ or $7$ are on the same hall, and there need not be any other rooms on this hall. Finally, rooms $5$, $16$, $49$, and $148$ are on the same hall, so all rooms with units digit $5,6,8$ or $9$ are on the same hall. Thus, we see that there are at most three different halls in the castle.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n![](attached_image_1.png)\nAs shown in the diagram above, the vertices of a regular decagon are colored alternately black and white. We would like to draw colored line segments between the points in such a way that\n\na. Every black point is connected to every white point by a line segment.\n\nb. No two line segments of the same color intersect, except at their endpoints.\n\nDetermine, with proof, the minimum number of colors needed.",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\nNote that the five main diagonals of the decagon must all be different colors since they all intersect at the decagon's center. Thus, at least $5$ colors are needed. To see that $5$ colors are also sufficient, we can simply assign each black point a color and use that color to connect it with all the white points.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe sum of the digits of all counting numbers less than $13$ is\n$$\n1+2+3+4+5+6+7+8+9+1+0+1+1+1+2=51\n$$\nFind the sum of the digits of all counting numbers less than $1000$.",
    "options": [],
    "answer": "13500",
    "solution": "Solution:\nThe counting numbers less than $1000$ are simply the three-digit numbers, as long as we allow numbers to start with $0$ (except the number $000$, but including this does not affect the digit sum).\nEach of the ten digits appears as a hundreds digit $100$ times, because it can be paired with any of the $10$ tens digits and the $10$ units digits. So the sum of the $1000$ hundreds digits is\n$$\n100(0+1+2+3+4+5+6+7+8+9)=100 \\cdot 45=4500\n$$\nSimilarly, every digit appears as a tens digit $100$ times and as a units digit $100$ times, so the sums of the tens digits and of the units digits are also $4500$. Consequently, the sum of all the digits is $3 \\cdot 4500=13500$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x, y, z$ be nonzero real numbers such that the equations\n$$\n\\begin{aligned}\n& x+\\frac{1}{y}=y+\\frac{1}{x} \\\\\n& y+\\frac{1}{z}=z+\\frac{1}{y} \\\\\n& z+\\frac{1}{x}=x+\\frac{1}{z}\n\\end{aligned}\n$$\nall hold. Show that two of the three variables must be equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe equation $x+\\frac{1}{y}=y+\\frac{1}{x}$ rearranges as\n$$\n0=x^{2}+\\left(\\frac{1}{y}-y\\right) x-1=\\left(x+\\frac{1}{y}\\right)(x-y)\n$$\nso either $x=y$ or $x=-1 / y$.\nBy applying similar logic, we conclude that $y=z$ or $y=-1 / z$.\nIf we assume $x \\neq y, y \\neq z$, we thus see that $x=z$ follows. (Note the third equation was never used.)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs usual, let $n!$ denote the product of the integers from $1$ to $n$ inclusive. Determine the largest integer $m$ such that $m!$ divides $100! + 99! + 98!$.",
    "options": [],
    "answer": "98",
    "solution": "Solution:\n\nThe answer is $m = 98$. Set\n$$\nN = 98! + 99! + 100! = 98! (1 + 99 + 99 \\cdot 100)\n$$\nHence $N$ is divisible by $98!$. But\n$$\n\\frac{N}{98!} = 1 + 99 \\cdot 101\n$$\nis not divisible by $99$. Hence $N$ is not divisible by $99!$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA gadget has four dials in a row, each of which can be turned to point to one of three numbers: $0$ (left), $1$ (up) or $2$ (right). Initially the dials are in the respective positions $2,0,1,0$, so that the gadget reads \"2010.\" You may perform the following operation: choose two adjacent dials pointing at different numbers, and turn them to point to the third number. For example, taking the first two dials, you could change \"2010\" to \"1110.\" Is it possible to perform a sequence of such operations so that the gadget reads \"2011\"?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. We notice that initially the sum of the numbers on the dials is $3$. We claim that after each operation, the sum of the numbers on the dials remains a multiple of $3$. To see this, consider the three possible types of moves:\n(a) Changing a $0$ and a $2$ to two $1$'s does not change the digit sum.\n(b) Changing a $0$ and a $1$ to two $2$'s increases the sum by $3$.\n(c) Changing a $1$ and a $2$ to two $0$'s decreases the sum by $3$.\nThus, the sum always goes up or down by multiples of $3$, and thus we cannot reach the position $2011$ in which the sum is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven an $n \\times n$ matrix whose entries $a_{i j}$ satisfy $a_{i j} = \\frac{1}{i + j - 1}$, $n$ numbers are chosen from the matrix no two of which are from the same row or the same column. Prove that the sum of these $n$ numbers is at least $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose that $a_{i j}$ and $a_{k l}$ are among the chosen numbers and suppose that $i < k$ and $j < l$. It is straightforward to show that $a_{i j} + a_{k l} \\geqslant a_{i l} + a_{k j}$. Hence, whenever $a_{i j}$ and $a_{k l}$ with $i < k$ and $j < l$ are among chosen numbers, we can lower the sum by replacing these two numbers with $a_{i l}$ and $a_{k j}$. Hence the smallest possible sum is when we choose $a_{1 n}, a_{2, n-1}, \\ldots, a_{n 1}$—and in that case the sum is $1$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 46025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA group of mathematicians is lost in a forest. The forest has a shape of an infinite strip that is 1 mile wide. Prove that they can choose a path that will guarantee them a way out and that is at most $2 \\sqrt{2}$ miles long.\n\nRemark. The mathematicians have no device for orientation and no maps. All they know is that the forest is a region between two parallel lines 1 mile apart from each other. They can't see the end of the forest, unless they are at the edge. However, they can precisely follow any path they design, e.g. they can move along straight line, circle, etc.",
    "options": [],
    "answer": "2 sqrt(2) miles",
    "solution": "Solution:\n\nStarting from their initial point $A$, the mathematicians should first move $\\sqrt{2}$ miles in any direction. If they didn't get to the exit, they have arrived at point $B$. Then they should turn by $90^{\\circ}$ and walk for another $\\sqrt{2}$ miles to the point $C$. We claim that they did reach the exit. If not, then the triangle $ABC$ is entirely in the forest. The triangle is rectangular, isosceles and its altitude is 1 mile. Thus the triangle can't be placed in the infinite strip of width 1 unless the vertices are on the edges. Contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$ and $b$ be positive real numbers. Prove that\n$$\n\\sqrt{a^{2}-a b+b^{2}} \\geq \\frac{a+b}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSquaring both sides (which is OK since both sides are positive), it's equivalent to show that $4\\left(a^{2}-a b+b^{2}\\right) \\geq (a+b)^{2}$. But their difference is\n$$\n4\\left(a^{2}-a b+b^{2}\\right)-(a+b)^{2}=3 a^{2}-6 a b+3 b^{2}=3(a-b)^{2} \\geq 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a certain two-player game, you start with a rectangular $m \\times n$ grid of squares. On each turn, a player either makes a horizontal cut and takes away the portion of the rectangle above the cut, or makes a vertical cut and takes away the portion to the right. Whichever player takes the last square (in the bottom left corner) loses. If both players play perfectly, determine for which values of $m$ and $n$ the first player will win.",
    "options": [],
    "answer": "First player wins if and only if m ≠ n; second player wins when m = n.",
    "solution": "Solution:\n\nThe first player wins whenever the grid is not a square (i.e. $m \\neq n$). The strategy is to always leave the other player with a square grid (where $m = n$), since then they will eventually be left with a $1 \\times 1$ square and will lose. This works since starting from a square, it is impossible to get to a smaller square, but starting from a non-square, it is always possible to get to a square. Thus, the first player wins whenever the starting grid is not a square, and the second player wins whenever the starting grid is a square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWith two properly chosen weights and a balance scale, it is possible to determine the weight of an unknown object known to weigh an integer number of pounds from $1$ to $n$. Find the largest possible value of $n$.\n\nRemark. The balance scale tells whether the weights placed on each side are equal and, if not, which side is heavier. It may be used an unlimited number of times.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nLet $a$ and $b$ be the known weights. The balance scale allows one to compare the unknown weight with four known weights: $a$, $b$, $a+b$, and $a-b$ (the last of these is gotten by balancing $x+b$ on one side with $a$ on the other). After the comparisons are done, there are at most $4$ values that $x$ can be known to equal and at most $5$ gaps ($3$ between and $2$ on the ends) for $x$ to lie in. Thus at most $9$ values of $x$ can be distinguished.\n\nTo show $n=9$ is achievable, take the weights to be $a=6\\ \\mathrm{lb}$ and $b=2\\ \\mathrm{lb}$. Then the nine possibilities can be distinguished by\n$$\n1 < b, \\quad 2 = b, \\quad b < 3 < a-b, \\quad 4 = a-b, \\quad a-b < 5 < a, \\quad 6 = a, \\quad a < 7 < a+b, \\quad 8 = a+b, \\quad 9 > a+b\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an isosceles triangle with $AB = AC$, and let $P$ be a point moving along the side $BC$. Drop the heights $PX$, $PY$ from $P$ to the sides $AB$ and $AC$. Prove that the sum $PX + PY$ remains constant as $P$ moves.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe can cut triangle $ABC$ into triangles $ABP$ and $APC$, so\n$$\n\\text{Area } ABC = \\text{Area } ABP + \\text{Area } APC = \\frac{AB \\cdot PX}{2} + \\frac{AC \\cdot PY}{2}.\n$$\nBut sides $AB$ and $AC$ have the same length, so this can be rewritten as\n$$\n\\frac{AB \\cdot PX}{2} + \\frac{AB \\cdot PY}{2} = \\frac{AB}{2}(PX + PY)\n$$\nSince $AB$ and the area of $\\triangle ABC$ are constant, so is $PX + PY$.\n\n\nSolution 2:\n\nAs shown in the right-hand picture above, reflect $A$ and $X$ across $BC$ to form the rhombus $ABA' C$. Then $PX'$ and $PY$ are perpendiculars to parallel sides $AC$ and $A'B$, so they are collinear, and $PX + PY = PX' + PY = X'Y$. This is a constant because parallel lines are always the same distance apart.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice picks four numbers from the set $\\{1,2,3,4,5,6\\}$, tells Bob their product and asks him to guess their sum. Bob realizes he cannot even determine for sure whether the sum is odd or even. What is the product of the numbers Alice chose?",
    "options": [],
    "answer": "60",
    "solution": "Solution:\n\nLet $P$ be said product. Evidently there are two distinct sets of numbers $\\{x_{1}, x_{2}, x_{3}, x_{4}\\}$ and $\\{y_{1}, y_{2}, y_{3}, y_{4}\\}$ such that $x_{1} x_{2} x_{3} x_{4} = y_{1} y_{2} y_{3} y_{4} = P$, but $x_{1} + x_{2} + x_{3} + x_{4}$ and $y_{1} + y_{2} + y_{3} + y_{4}$ have different parity.\n\nInstead of considering the four numbers Alice picks, we consider instead the two numbers Alice does not pick; $\\{x_{5}, x_{6}\\}$ and $\\{y_{5}, y_{6}\\}$. They have the same properties we described above, since\n\n$$\nx_{5} x_{6} = \\frac{6!}{x_{1} x_{2} x_{3} x_{4}} = \\frac{720}{P} = \\frac{6!}{y_{1} y_{2} y_{3} y_{4}} = y_{5} y_{6}\n$$\nand analogously the sums also have different parities.\n\nThus we are looking for pairs of distinct numbers in $\\{1,2,3,4,5,6\\}$ which have the same product but different sums. We can record the entire multiplication table, as below.\n\n| $\\times$ | 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 |   | 2 | 3 | 4 | 5 | 6 |\n| 2 |   |   | 6 | 8 | 10 | 12 |\n| 3 |   |   |   | 12 | 15 | 18 |\n| 4 |   |   |   |   | 20 | 24 |\n| 5 |   |   |   |   |   | 30 |\n| 6 |   |   |   |   |   |   |\n\nThe numbers with duplicate entries are\n$$\n\\begin{aligned}\n6 & = 6 \\cdot 1 = 2 \\cdot 3 \\\\\n12 & = 2 \\cdot 6 = 3 \\cdot 4\n\\end{aligned}\n$$\n\nOf these, 12 has the desired property but 6 does not. Hence, Alice chose one of the quadruples $\\{1,3,4,5\\}$ or $\\{1,2,5,6\\}$; the product is then $P = 1 \\cdot 3 \\cdot 4 \\cdot 5 = 1 \\cdot 2 \\cdot 5 \\cdot 6 = 60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA finite number of points are drawn in the plane. Prove that one can select two of them, $A$ and $B$, such that:\n\na. $A$ and $B$ are not the same point.\n\nb. No drawn point, other than $A$ itself, is closer to $A$ than $B$ is.\n\nc. No drawn point, other than $B$ itself, is closer to $B$ than $A$ is.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider all the distances $X Y$ between two different drawn points. Since a finite number of points are drawn, there are only finitely many distances, and one of them, say $d$, is minimal. Let $A$ and $B$ be two of the drawn points such that $A B = d$. Now, by definition, $A$ and $B$ are not the same point.\n\nTo verify part (b), let $C$ be any drawn point other than $A$ and $B$. Then $A C \\geq d = A B$, so $C$ is not closer to $A$ than $B$ is.\n\nThe verification of part (c) is similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways are there to color the five vertices of a regular 17-gon either red or blue, such that no two adjacent vertices of the polygon have the same color?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nThe answer is zero! Call the polygon $A_{1} A_{2} \\ldots A_{17}$. Suppose for contradiction such a coloring did exist.\nIf we color $A_{1}$ red, then $A_{2}$ must be blue. From here we find $A_{3}$ must be red, then $A_{4}$ must be blue; thus $A_{5}$ must be red, $A_{6}$ must be blue. Proceeding in this manner, we eventually find that $A_{15}$ is red, $A_{16}$ is blue, and then $A_{17}$ is red. But $A_{1}$ and $A_{17}$ are adjacent and both red, impossible.\n\nThe exact same argument holds if we started by coloring $A_{1}$ blue. Therefore, there are no colorings at all with the desired property.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p$ be a prime number. Find all possible values of the remainder when $p^{2}-1$ is divided by $12$.",
    "options": [],
    "answer": "3, 8, 0",
    "solution": "Solution:\nThe answers are $3$, $8$, and $0$.\n\nIt is clear that $p=2$ gives $3$, $p=3$ gives $8$, and $p=5$ gives $0$. We claim that all primes $p \\geq 5$ give the remainder $0$ as well, i.e. that $p^{2}-1$ is divisible by $12$ for these $p$.\n\nWe factor:\n$$\np^{2}-1=(p+1)(p-1)\n$$\nSince $p \\neq 2$, $p$ is odd and so $p+1$ and $p-1$ are both even. This gives two factors of $2$ in $p^{2}-1$. Moreover, one of the three consecutive integers $p-1$, $p$, $p+1$ is divisible by $3$, and since $p \\neq 3$, it is not $p$. So either $p-1$ or $p+1$ has a factor of $3$, and so does $p^{2}-1$. Thus $p^{2}-1$ is divisible by $2 \\cdot 2 \\cdot 3=12$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAt a market, a buyer and a seller each have four exotic coins. You are allowed to label each of the eight coins with any positive integer value in cents. The labeling is called $n$-efficient if for any integer $k$, $1 \\leq k \\leq n$, it is possible for the buyer and the seller to give each other some of their coins in such a way that, as a net result, the buyer has paid the seller $k$ cents. Find the greatest positive integer $n$ such that an $n$-efficient labeling of the coins exists.",
    "options": [],
    "answer": "240",
    "solution": "Solution:\n\nThe answer is $240$.\n\nTo see that $n > 240$ is impossible, note that there are $2^{8} = 256$ ways for the transaction to happen, since each coin either changes hands or does not change hands. However, the $2^{4} = 16$ ways in which the buyer keeps all four of his coins clearly cannot allow the buyer to pay the seller a positive amount. Therefore there are at most $240$ different amounts which the buyer can pay the seller.\n\nTo make these amounts the consecutive integers $1, 2, 3, \\ldots, 240$, we assign values as follows:\n\nBuyer: $16, 32, 64, 128$\n\nSeller: $1, 2, 4, 8$.\n\nThen the buyer can give the seller any multiple $16x$ of $16$ cents up to $240$ by writing $x$ in binary as a sum of distinct powers of $2$, and the seller can correspondingly give the buyer any amount from $0$ to $15$ cents in change.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo regular polygons are said to be matching if the double of the interior angle of one of them equals the triple of the exterior angle of the other. Find all pairs of matching polygons.",
    "options": [],
    "answer": "(3,9), (4,6), (5,5), (8,4)",
    "solution": "Solution:\n\nThe answers, expressed in terms of numbers of sides, are $(3,9)$, $(4,6)$, $(5,5)$, and $(8,4)$. They are easily verified to be matching.\n\nTo prove that there are no others, first note that if the first polygon has at most five sides, the second one necessarily has one of the numbers of sides necessary to make one of the first three matching pairs. The second polygon cannot be a triangle (since then the first polygon would have an interior angle of $180^{\\circ}$), and if it is a square or pentagon, we recover the last two matching pairs. So if there are any matching pairs other than the four listed, both polygons must have at least six sides. But then the interior angle of the first is at least $120^{\\circ}$ and the exterior of the second is at most $60^{\\circ}$, which is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOscar draws a triangle $ABC$ on a sheet of paper. He finds that the side lengths of $ABC$ are all powers of $2$ (i.e. among $1, 2, 4, 8, \\ldots$). Prove that Oscar's triangle is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider a longest side of the triangle, $2^{a}$. We claim that another side must have this length too. Otherwise, suppose for contradiction they are $2^{b}$ and $2^{c}$ where $b, c < a$. Then\n$$\n2^{b} + 2^{c} < 2^{a-1} + 2^{a-1} = 2^{a}\n$$\nwhich contradicts the triangle inequality.\nHence there must be a second side of length $2^{a}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be an isosceles right triangle with $\\angle C = 90^{\\circ}$. Let $X$ be a point on the line segment $AC$, and let $Y$ and $Z$ be, respectively, the feet of the perpendiculars from $A$ and $C$ to the line $BX$. Prove that\n$$\nBZ = YZ + AY\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDraw the perpendicular $CW$ from $C$ to line $AY$. We get $\\triangle CBZ \\cong \\triangle CAW$ by AAS since:\n- $\\angle BCZ = \\angle ACW$ (this comes from the right angles $BCA$ and $ZCW$).\n- $\\angle BZC = \\angle AWC$ (both $90^{\\circ}$)\n- $BC = AC$.\n\n![](attached_image_1.png)\n\nQuadrilateral $CZYW$ must be a square, since it already has four right angles and we just proved that $CZ = CW$. Hence\n$$\nBZ = AW = YW + AY = YZ + AY.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDo there exist four consecutive positive integers whose product is a perfect square?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. If $x \\geq 1$ is an integer,\n$$\n\\begin{aligned}\n& x(x+1)(x+2)(x+3) \\\\\n& = [x(x+3)] \\cdot [(x+1)(x+2)] \\\\\n& = \\left[x^2 + 3x\\right] \\cdot \\left[x^2 + 3x + 2\\right] \\\\\n& = \\left[\\left(x^2 + 3x + 1\\right) - 1\\right] \\cdot \\left[\\left(x^2 + 3x + 1\\right) + 1\\right] \\\\\n& = \\left(x^2 + 3x + 1\\right)^2 - 1.\n\\end{aligned}\n$$\nTherefore the product of four consecutive positive integers is always one less than a square, and therefore cannot be a square since two positive squares cannot differ by $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe circle $\\omega$ passes through the vertices $A$ and $B$ of a unit square $A B C D$. It intersects $A D$ and $A C$ at $K$ and $M$ respectively. Find the length of the projection of $K M$ onto $A C$.",
    "options": [],
    "answer": "sqrt(2)/2",
    "solution": "Solution:\nLet $T$ be the point of intersection of $\\omega$ with $B C$. Then, as $\\angle A B T$ is a right angle, $A T$ is a diameter, and $\\angle A M T$ is also a right angle. Therefore the projections of $K M$ and $K T$ on $A C$ coincide. But the length of the projection of $K T$ is $\\frac{\\sqrt{2}}{2}$ because the length of $K T$ is one, and the angle between $K T$ and $A C$ is $45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $d$ be a fixed positive integer. Prove that there exists a unique polynomial $S(n)$ such that for every integer $n \\geq 0$, $S(n) = \\sum_{k=0}^{n} k^{d} = 0^{d} + 1^{d} + \\cdots + n^{d}$. Also prove that $S(n)$ can be expressed in the form\n$$\nc_{0} + c_{1}(2n+1) + c_{2}(2n+2)^{2} + \\cdots + c_{d+1}(2n+1)^{d+1}\n$$\nwhere the $c_{i}$ are rational numbers such that $c_{i} c_{i+1} = 0$ for $i = 0, 1, \\ldots, d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe shall prove by strong induction on $d \\geq 0$ the existence of a polynomial $P_{d}(x)$ of degree $d+1$ with rational coefficients such that\n- the exponents of $x$ which occur with nonzero coefficients are all of opposite parity from $d$;\n- $P_{d}(2n+1) - P_{d}(2n-1) = n^{d}$ (defined to be $1$ when $n = d = 0$);\n- $P_{d}(1) = 1/2$ if $d = 0$, and $P_{d}(1) = 0$ otherwise.\n\nThen, $S(n) = P_{d}(2n+1)$ will meet our requirements (we know $S(n) = 0^{d} + \\cdots + n^{d}$ by using the second property of $P_{d}$ to form a telescoping sum).\n\nThe base cases are $P_{0}(x) = x/2$, $P_{1}(x) = (x^{2} - 1)/8$. If $d > 1$, consider what happens when $(2n+1)^{d+1} - (2n-1)^{d+1}$ is expanded by the binomial theorem. The result is a polynomial in $n$, in which the powers having the same parity as $d+1$ cancel, leaving $c_{d} n^{d} + c_{d-2} n^{d-2} + \\cdots + c_{2\\lfloor d/2 \\rfloor} n^{2\\lfloor d/2 \\rfloor}$ where, using the binomial theorem, $c_{k} = 2^{k+1} \\binom{d+1}{k}$.\n\nNow first suppose $d$ is odd. By the induction hypothesis, for any odd $k < d$, there exists a rational polynomial $P_{k}$ meeting the conditions above (and every exponent of $x$ is even). Then\n$$\nP_{d}(x) = \\frac{x^{d+1} - c_{d-2} P_{d-2}(x) - c_{d-4} P_{d-4}(x) - \\cdots - c_{1} P_{1}(x) - 1}{c_{d}}\n$$\nmeets all the conditions.\n\nIf, instead, $d$ is even, then for any even $k < d$, there exists an odd rational polynomial $P_{k}$ meeting these conditions, and\n$$\nP_{d}(x) = \\frac{x^{d+1} - c_{d-2} P_{d-2}(x) - c_{d-4} P_{d-4}(x) - \\cdots - c_{0} P_{0}(x)}{c_{d}}\n$$\nmeets our requirements—the first two are clear, and, for the third, we use our earlier computation that $c_{0} = 2$ to find\n$$\n\\frac{1^{d+1} - c_{0} P_{0}(1)}{c_{d}} = \\frac{1 - 1}{c_{d}} = 0\n$$\nas claimed. Thus $P_{d}$ exists as needed.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P$ be a polynomial with integer coefficients. Let $S$ be the set of integers $n$ for which $P(n) / n$ is an integer. Show that $S$ contains either finitely many integers, or all but finitely many integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $c$ be the constant coefficient of $P$, so that $P(x)$ is of the form $Q(x) x + c$. If $n \\mid P(n)$, we then have $n \\mid (Q(n) n + c)$, or $n \\mid c$. $S$ is thus the set of divisors of $c$. If it is not finite, $c$ must then be $0$, and $S$ is the set of all nonzero integers, as desired.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that for each $n \\geq 17$ one can cut a square into $n$ smaller squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOne can cut a square into $n$ smaller squares for all $n \\geq 6$. Here is how.\nFor $n=4$\n![](attached_image_1.png)\n\nFor $n=6$\n![](attached_image_2.png)\n\nFor $n=8$\n![](attached_image_3.png)\n\nSuppose we know how to cut a square into $n$ smaller squares. If we take one of the square \"pieces\" and cut it onto four smaller squares (as in the picture above for $n=4$). Now we have cut our original square into $n+3$ pieces. Since we know how to cut a square into 4 pieces, we can cut it into $7,10, \\ldots, 3k+1, \\ldots$ pieces. Similarly, starting with 6 pieces we can get $6,9, \\ldots, 3k, \\ldots$ pieces, and from 8 pieces $8,11, \\ldots, 3k+2, \\ldots$ pieces. Hence we can get any number of pieces bigger or equal to 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn an infinite chessboard, two squares are said to touch if they share at least one vertex and they are not the same square. Suppose that the squares are colored black and white such that\n- there is at least one square of each color;\n- each black square touches exactly $m$ black squares;\n- each white square touches exactly $n$ white squares\nwhere $m$ and $n$ are integers. Must $m$ and $n$ be equal?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. There are many tilings to demonstrate this; one of the simplest is to divide the board into horizontal stripes and color every third stripe black. In this tiling, $m=2$ and $n=5$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest number $n$ having the following properties:\n(a) No two digits of $n$ are equal.\n(b) The number formed by reversing the digits of $n$ is divisible by 8.",
    "options": [],
    "answer": "8697543210",
    "solution": "Solution:\nBy condition (a), the number $n$ cannot have more than 10 digits. Write $m$ for the number formed by reversing the digits of $n$.\n\nThe first digit of $n$ is the last digit of $m$, and as such must be even, and thus at most 8. Assume that the first digit is 8.\n\nThen the second digit of $n$ is the tens digit of $m$, and must be chosen to make the last two digits of $m$ divisible by 4. As 98 and 78 are not divisible by 4, but 68 is, we conclude that the second digit is at most 6. Assume that the second digit is 6.\n\nThen the third digit must be chosen so that the last three digits of $m$ are divisible by 8, but since 968 is divisible by 8 we derive no information from this.\n\nAppending the remaining digits, selecting the greatest possibility at each turn, we conclude that $n$ is at most 8697543210. As this number indeed satisfies the conditions, it is the answer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that\n$$\nA=\\sqrt{4-2 \\sqrt{3}}-\\frac{\\sqrt{3}+1}{\\sqrt{3}-1}\n$$\nis an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n$\\sqrt{4-2 \\sqrt{3}} = \\sqrt{3^{2} - 2 \\sqrt{3} + 1^{2}} = \\sqrt{(\\sqrt{3}-1)^{2}} = |\\sqrt{3}-1| = \\sqrt{3}-1$.\n\nIf we multiply both numerator and denominator of $\\frac{\\sqrt{3}+1}{\\sqrt{3}-1}$ by $\\sqrt{3}+1$ we get:\n$$\n\\frac{\\sqrt{3}+1}{\\sqrt{3}-1} = \\frac{(\\sqrt{3}+1)^{2}}{\\sqrt{3}^{2}-1^{2}} = \\frac{4+2 \\sqrt{3}}{2} = 2+\\sqrt{3}\n$$\nThus $A = (\\sqrt{3}-1) - (2+\\sqrt{3}) = -1-2 = -3$, which is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an equilateral triangle, and let $P$ be a point on minor $\\operatorname{arc} BC$ of the circumcircle of $ABC$. Prove that $PA = PB + PC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nExtend line $PC$ through $C$ to point $D$ such that $CD = BP$. Note that $\\angle ACD = \\pi - \\angle PCA = \\angle ABP$ (since quadrilateral $ABPC$ is cyclic), and $AC = AB$ since $\\triangle ABC$ is equilateral. Consequently, $\\triangle ACD \\cong \\triangle ABP$. In particular, we have $\\angle PDA = \\angle CDA = \\angle BPA = \\angle BCA$ (again by cyclicity) $= \\pi/3$. But also $\\angle APD = \\angle APC = \\angle ABC$ (cyclicity) $= \\pi/3$. We conclude that triangle $APD$ is equilateral. So, $PA = PD = PC + CD = PC + BP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKaren has seven envelopes and seven letters of congratulations to various HMMT coaches. If she places the letters in the envelopes at random with each possible configuration having an equal probability, what is the probability that exactly six of the letters are in the correct envelopes?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\n$0$, since if six letters are in their correct envelopes the seventh is as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate $\\sum_{i=1}^{\\infty} \\frac{(i+1)(i+2)(i+3)}{(-2)^{i}}$.",
    "options": [],
    "answer": "96",
    "solution": "Solution:\nThis is the power series of $\\frac{6}{(1+x)^{4}}$ expanded about $x=0$ and evaluated at $x=-\\frac{1}{2}$, so the solution is $96$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA man is standing on a platform and sees his train move such that after $t$ seconds it is $2 t^{2} + d_{0}$ feet from his original position, where $d_{0}$ is some number. Call the smallest (constant) speed at which the man have to run so that he catches the train $v$. In terms of $n$, find the $n$th smallest value of $d_{0}$ that makes $v$ a perfect square.",
    "options": [],
    "answer": "4^{n-1}",
    "solution": "Solution:\nThe train's distance from the man's original position is $t^{2} + d_{0}$, and the man's distance from his original position if he runs at speed $v$ is $v t$ at time $t$. We need to find where $t^{2} + d_{0} = v t$ has a solution. Note that this is a quadratic equation with discriminant $D = \\sqrt{v^{2} - 4 d_{0}}$, so it has solutions for real $D$, i.e. where $v \\geq \\sqrt{4 d_{0}}$, so $4 d_{0}$ must be a perfect square. This happens when $4 d_{0}$ is an even power of $2$: the smallest value is $2^{0}$, the second smallest is $2^{2}$, the third smallest is $2^{4}$, and in general the $n$th smallest is $2^{2(n-1)}$, or $4^{n-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice, Bob, and Charlie each pick a 2-digit number at random. What is the probability that all of their numbers' tens' digits are different from each others' tens' digits and all of their numbers' ones digits are different from each others' ones' digits?",
    "options": [],
    "answer": "112/225",
    "solution": "Solution:\n\n$\\frac{9}{10} \\cdot \\frac{8}{10} \\cdot \\frac{8}{9} \\cdot \\frac{7}{9} = \\frac{112}{225}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSquare $ABCD$ has side length $1$. A dilation is performed about point $A$, creating square $AB' C' D'$. If $BC' = 29$, determine the area of triangle $BDC'$.",
    "options": [],
    "answer": "420",
    "solution": "Solution:\n\n$29^{2} - 2 \\cdot \\frac{1}{2}(29)\\left(\\frac{29}{2}\\right) - \\frac{1}{2} = 420$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the remainder when $100!$ is divided by $101$?",
    "options": [],
    "answer": "100",
    "solution": "Solution:\n\nWilson's theorem says that for $p$ a prime, $(p-1)! \\equiv -1 \\pmod{p}$, so the remainder is $100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle with center at $O$ has radius $1$. Points $P$ and $Q$ outside the circle are placed such that $PQ$ passes through $O$. Tangent lines to the circle through $P$ hit the circle at $P_{1}$ and $P_{2}$, and tangent lines to the circle through $Q$ hit the circle at $Q_{1}$ and $Q_{2}$. If $\\angle P_{1} P P_{2} = 45^{\\circ}$ and $\\angle Q_{1} Q Q_{2} = 30^{\\circ}$, find the minimum possible length of arc $P_{2} Q_{2}$.",
    "options": [],
    "answer": "π/12",
    "solution": "Solution:\n$$(45-30)^{\\circ} = \\frac{\\pi}{12}.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMona has $12$ match sticks of length $1$, and she has to use them to make regular polygons, with each match being a side or a fraction of a side of a polygon, and no two matches overlapping or crossing each other. What is the smallest total area of the polygons Mona can make?",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "$4 \\frac{\\sqrt{3}}{4} = \\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many different combinations of 4 marbles can be made from 5 indistinguishable red marbles, 4 indistinguishable blue marbles, and 2 indistinguishable black marbles?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\n$5 + 4 + 3 = 12$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCount the number of sequences $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ of integers such that $a_{i} \\leq 1$ for all $i$ and all partial sums $\\left(a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \\ldots, a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\\right)$ are non-negative.",
    "options": [],
    "answer": "132",
    "solution": "Solution:\n\n$C($ length $+1 ) = C(6) = 132$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be an integer greater than $12$. Points $P_{1}, P_{2}, \\ldots, P_{n}, Q$ in the plane are distinct. Prove that for some $i$, at least $n / 6 - 1$ of the distances\n$$\nP_{1} P_{i}, P_{2} P_{i}, \\ldots, P_{i-1} P_{i}, P_{i+1} P_{i}, \\ldots, P_{n} P_{i}\n$$\nare less than $P_{i} Q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCut the plane into six $60^{\\circ}$ \"pizza slices\" with vertex $Q$. Rotating if necessary, we may assume that none of the $P_{j}$ lie on the cuts. By the pigeonhole principle, one slice contains at least $n / 6$ of the $P_{j}$. Let $P_{i}$ be a point in this slice farthest from $Q$. It remains to show that all other points $P_{j}$ in this slice satisfy $P_{j} P_{i} < P_{i} Q$.\n\nThe average of the angles of $\\Delta P_{j} P_{i} Q$ is $180^{\\circ} / 3 = 60^{\\circ}$, so $\\angle P_{j} Q P_{i}$, which is less than $60^{\\circ}$, is less than one of the other angles. Smaller angles of a triangle are opposite shorter sides, so $P_{j} P_{i}$ is less than one of $P_{j} Q$ and $P_{i} Q$. By choice of $i$, $P_{j} Q \\leq P_{i} Q$, so in any case, $P_{j} P_{i} < P_{i} Q$. (Alternatively, one could use the Law of Cosines to show that the side of a triangle opposite an angle smaller than $60^{\\circ}$ is not the longest.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe integers from $1$ to $13$ are arranged around several rings such that every number appears once and every ring contains at least one two-digit number. Prove that there exist three one-digit numbers adjacent to one another on one ring.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFor any $n \\in S = 1, 2, \\ldots, 13$, define $f(n)$ to be the number immediately clockwise of $n$ on the same ring, where $f(n) = n$ if $n$ lies on a one-element ring. Notice that $f$ is a bijective function, since every number is immediately clockwise of exactly one number. Notice that there are four numbers $n \\in S$ such that $n$ has two digits, four numbers $n$ such that $f(n)$ has two digits, and four numbers $n$ such that $f(f(n))$ has two digits. This leaves at least $13 - 4 - 4 - 4 = 1$ number $n$ such that $n$, $f(n)$, and $f(f(n))$ are all one-digit numbers. No two of them can be equal, or else the ring containing them would have no two-digit numbers, so $n$, $f(n)$, and $f(f(n))$ are the desired adjacent one-digit numbers.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle. A line is drawn not passing through any vertex of $ABC$. Prove that some side of $ABC$ is not cut by the line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nConsider the two sides of the line $\\ell$. By the pigeonhole principle on the three vertices of $ABC$, two of these vertices, say $A$ and $B$, lie on the same side of $\\ell$. Then segment $AB$ does not intersect $\\ell$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers $1, 8, 4, 0$ are the first four terms of the infinite sequence. Every subsequent term of the sequence is obtained as the last digit of the sum of previous four terms. Therefore the fifth term of the sequence is $3$, because $1+8+4+0=13$; the sixth term is $5$ because $8+4+0+3=15$, and so on.\n\na. Will $2, 0, 0, 7$ ever appear as a subsequence?\n\nb. Will $1, 8, 4, 0$ appear again as a subsequence?\n\nExplain your answer!",
    "options": [],
    "answer": "a: yes; b: yes",
    "solution": "Solution:\n\na. Yes, very soon, in fact the next four terms (from $7$ to $10$th) are $2, 0, 0, 7$.\n\nb. We will prove that $1, 8, 4, 0$ will be a subsequence again. Assume the contrary. Since there are only finitely many combinations of four digits (precisely $10^4$), and the sequence is infinite, some combination of four digits (say $(a, b, c, d)$) has to reappear. Assume that $(x_n, x_{n+1}, x_{n+2}, x_{n+3}) = (a, b, c, d)$ is the first occurrence of $(a, b, c, d)$ and that $(x_m, x_{m+1}, x_{m+2}, x_{m+3}) = (a, b, c, d)$ is the second. Clearly $m > n$. However, $x_{n-1}$ and $x_{m-1}$ are uniquely determined and they have to be the same numbers. Thus $x_{n-1} = x_{m-1}$, $x_{n-2} = x_{m-2}$, and so on. This means that $x_1 = x_{m-n+1}$, $x_2 = x_{m-n+2}$, $x_3 = x_{m-n+3}$, and $x_4 = x_{m-n+4}$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTen cups lie upside down in a line. It is known that pennies lie under two of the cups which are consecutive in the line. Choosing several of the cups, you may ask for the total number of coins under them. Is it possible to determine the positions of the pennies by asking two such questions, without knowing the answer to the first question before making the second?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nThe answer is yes; here is one strategy that works. On the first turn choose the cups $1, 4, 5, 6, 7$, and on the second turn choose $1, 6, 7, 8, 9$. Depending on the locations of the coins, the answers will be:\n\n| Coins | $1,4,5,6,7$ | $1,6,7,8,9$ |\n| :---: | :---: | :---: |\n| 1,2 | 1 | 1 |\n| 2,3 | 0 | 0 |\n| 3,4 | 1 | 0 |\n| 4,5 | 2 | 0 |\n| 5,6 | 2 | 1 |\n| 6,7 | 2 | 2 |\n| 7,8 | 1 | 2 |\n| 8,9 | 0 | 2 |\n| 9,10 | 0 | 1 |\n\nSince each possible location for the coins gives a different pair of answers, the answers are enough to reconstruct the location of the coins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine the digitlength of a positive integer to be the total number of letters used in spelling its digits. For example, since \"two zero one one\" has a total of 13 letters, the digitlength of 2011 is 13. We begin at any positive integer and repeatedly take the digitlength. Show that after some number of steps, we must arrive at the number 4.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe first claim that if a number has at least two digits, then the digitlength is less than the number itself. To see this, note that any digit has at most five letters; thus a 2-digit number, which is at least $10$, has digitlength at most $5 \\cdot 2 = 10$; a 3-digit number, which is at least $100$, has digitlength at most $5 \\cdot 3 = 15$; and so on. The highest possible digitlength increases at each step by $5$, while the lowest possible number having a given number of digits increases by more than $5$. Thus the digitlength will always be strictly less than the number unless the number and the digitlength are both $10$, which does not happen since the digitlength of $10$ is $7$.\n\nWhen we repeatedly take the digitlength, the numbers keep decreasing as long as they have at least two digits. Since a sequence of positive integers cannot decrease forever, we must eventually hit a 1-digit number. Then, we quickly reach $4$ as shown by the following diagram:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven three squares of dimensions $2 \\times 2$, $3 \\times 3$, and $6 \\times 6$, choose two of them and cut each into 2 figures, such that it is possible to make another square from the obtained 5 figures.",
    "options": [],
    "answer": "A 7×7 square (e.g., cut the 3×3 into 3×2 and 3×1 and the 2×2 into two 2×1 pieces, then assemble with the 6×6)",
    "solution": "Solution:\nThe solution is shown in the picture below\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor any positive integer $n \\geq 3$, show that you can write $1$ as a sum of $n$ fractions with numerator $1$ and different denominators.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFor $n=3$, we can write\n$$\n1 = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6}.\n$$\nFor $n=4$, we can write\n$$\n1 = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{9} + \\frac{1}{18}.\n$$\nWe claim that we can continue this pattern by breaking the smallest fraction into two smaller fractions, such that the smallest fraction always has even denominator. We can show this inductively. Assume this is true for some $n$, where the smallest fraction is $\\frac{1}{2m}$. Then we can break the $\\frac{1}{2m}$ into\n$$\n\\frac{1}{2m} = \\frac{1}{3m} + \\frac{1}{6m}\n$$\nwhich creates a sum of $n+1$ distinct fractions with numerator $1$, and the largest denominator is still even. This completes our induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the maximum number of squares on an $8 \\times 8$ chessboard on which pieces may be placed so that no two of these squares touch horizontally, vertically, or diagonally?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\nThe answer is $16$.\n\nDivide the chessboard into sixteen $2 \\times 2$ regions (see figure). Notice that no two pieces can lie in the same region. So there are at most $16$ pieces. By placing a piece in the lower left corner of each region, we see that $16$ is achievable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA position of the hands of a (12-hour, analog) clock is called valid if it occurs in the course of a day. For example, the position with both hands on the 12 is valid; the position with both hands on the 6 is not. A position of the hands is called bivalid if it is valid and, in addition, the position formed by interchanging the hour and minute hands is valid. Find the number of bivalid positions.",
    "options": [],
    "answer": "143",
    "solution": "Solution:\nLet $h$ and $m$ denote the respective angles of the hour and minute hands, measured clockwise in degrees from 12 o'clock $(0 \\leq h, m < 360)$. Since the minute hand moves twelve times as fast as the hour hand, we have\n$$\nm = 12h - 360a\n$$\nwith $a$ an integer, for any valid time. Conversely, it is clear that any ordered pair $(h, m)$ satisfying (1) represents the valid time of $h / 30$ hours after 12 o'clock.\n\nThe condition that a time be valid after switching the hour and minute hands is, of course,\n$$\nh = 12m - 360b,\n$$\nwith $b$ an integer. Thus the problem of finding bivalid times is reduced to finding pairs $(h, m)$ satisfying (1) and (2). Note that if $h$ is known, $m$ is uniquely determined from (1) and the condition $0 \\leq m < 360$. Note also that the truth of (2) is not affected by increasing or decreasing $m$ by 360, as this simply increases or decreases $b$ by 12, giving a new integer value $b'$. Consequently we substitute $12h$ for $m$ in (2), getting\n$$\n\\begin{gathered}\nh = 144h - 360b' \\\\\n143h = 360b' \\\\\nh = \\frac{360b'}{143}.\n\\end{gathered}\n$$\nThis equation has 143 solutions in the range $0 \\leq h < 360$, each of which gives rise to one solution of (1) and (2). We conclude that there are 143 bivalid positions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA king is placed in the left bottom corner of the $6 \\times 6$ chessboard. At each step it can either move one square up, or one square to the right, or diagonally - one up and one to the right. How many ways are there for the king to reach the top right corner of the board?",
    "options": [],
    "answer": "1683",
    "solution": "Solution:\nWe shall make a $6 \\times 6$ table. In each cell of the table we will write a number of ways in which the king can reach that cell. We will fill it out gradually starting with a row of ones at the bottom and a column of ones at the left. To fill out the rest we use the following rule: the number in each cell is equal to the sum of the numbers immediately below, to the left, and diagonally (to the left and below). The result is:\n\n| 1 | 11 | 61 | 231 | 681 | 1683 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | 9 | 41 | 129 | 321 | 681 |\n| 1 | 7 | 25 | 63 | 129 | 231 |\n| 1 | 5 | 13 | 25 | 41 | 61 |\n| 1 | 3 | 5 | 7 | 9 | 11 |\n| 1 | 1 | 1 | 1 | 1 | 1 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice and Bob play the following game on the whiteboard. First, Alice writes an odd number in binary on the board. Then, beginning with Bob, the players take turns modifying the number in one of two ways: subtracting $1$ from it (preserving the binary notation), or erasing its last digit. When the whiteboard is blank, the last player to have played wins. Which player has a winning strategy?",
    "options": [],
    "answer": "Bob",
    "solution": "Solution:\n\nBob can win using the following strategy: Play so as to leave an empty whiteboard or a number with an odd number of terminal zeros (preceded by a $1$). We claim that Bob can fulfill this requirement at every move.\n\nOn his first move, or indeed any move in which Alice leaves an odd number on the board, it is easy to see how the requirement can be fulfilled. Unless the number on the board is $1$ (in which case Bob wins instantly), it ends in a $1$ followed by either an even or an odd number of zeros; in these two cases Bob should subtract one and erase the $1$ respectively.\n\nIf Alice leaves an even number on the board, it must be by erasing a final $0$ from a number left by Bob. Bob can then erase another zero, and the number of terminal zeros will remain odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn $\\triangle A B C$, let $O$ be the circumcenter, $D$ the foot of the altitude from $A$ to $B C$, and $E$ the foot of the altitude from $B$ to $A C$. Show that $D E \\perp C O$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy AA, $\\triangle A C D \\sim \\triangle B C E$, as $\\angle C$ is shared and both are right triangles. Thus $A C / C D = B C / C E$. It follows by SAS that\n$$\n\\triangle A B C \\sim \\triangle D E C \\Longrightarrow \\angle C D E = \\angle B A C.\n$$\nSince $O$ is the circumcenter of $\\triangle A B C$, $\\angle B O C = 2 \\angle B A C$, and since $\\triangle B O C$ is isosceles,\n$$\n\\angle O C B = \\angle O B C = 90^\\circ - \\angle B A C.\n$$\nThus, $\\angle C D E + \\angle O C B = 90^\\circ$, which implies that $D E \\perp O C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$, $y$, and $z$ be real numbers.\n\na. Prove that\n$$\nx + y > |x - y|\n$$\nif and only if $x$ and $y$ are both positive.\n\nb. Find an inequality involving the variables $x$, $y$, and $z$, using only the operations of addition, subtraction, multiplication and absolute value, that is true if and only if $x$, $y$, and $z$ are all positive.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that $|x-y|$ is the greater of $x-y$ and $-(x-y)$. So the above assertion is equivalent to\n$$\n\\begin{aligned}\n& x+y > x-y \\quad \\text{ and } \\quad x+y > -x+y \\\\\n& 2y > 0 \\quad \\text{ and } \\quad 2x > 0 \\\\\n& y > 0 \\quad \\text{ and } \\quad x > 0.\n\\end{aligned}\n$$\n\n\nSolution 1. The quantity\n$$\nx + |x|\n$$\nis $2x$, a positive number, if $x > 0$ and is $0$ otherwise. Therefore the inequality\n$$\n(x + |x|)(y + |y|)(z + |z|) > 0\n$$\nholds if and only if $x$, $y$, and $z$ are all positive.\n\n\nSolution 2. By part (a), the condition that $x$, $y$, and $z$ are all positive is equivalent to\n$$\nx + y - |x - y| > 0 \\quad \\text{ and } \\quad z > 0.\n$$\nBy part (a) again, this is equivalent to\n$$\nx + y - |x - y| + z > |x + y - |x - y| - z|.\n$$\nThis inequality involves no multiplication.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all pairs of integers $(x, y)$ for which $x^{2} + x y = y^{2}$.",
    "options": [],
    "answer": "(0, 0)",
    "solution": "Solution:\n\nThe only such pair is $(0, 0)$. If $x = 0$, we easily get $y = 0$ which satisfies the equation. Otherwise, dividing through by $x^{2}$ we get\n$$\n1 + \\frac{y}{x} = \\left(\\frac{y}{x}\\right)^{2}.\n$$\nThis implies $\\left(\\frac{y}{x}\\right)^{2} - \\frac{y}{x} - 1 = 0$ and the quadratic formula gives us $\\frac{y}{x} = \\frac{1 \\pm \\sqrt{5}}{2}$. Hence $\\frac{2y}{x} = 1 \\pm \\sqrt{5}$ or equivalently $\\frac{2y - x}{x} = \\pm \\sqrt{5}$. Since $\\sqrt{5}$ is irrational, this is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways are there to list the numbers $1$ to $10$ in some order such that every number is either greater or smaller than all the numbers before it?",
    "options": [],
    "answer": "512",
    "solution": "Solution:\n\nThe number of lists is $2^{9} = 512$. Imagine building a list from the end. First, note that the last number must be $1$ or $10$, since otherwise it would be less than some numbers before it and greater than others. Thus, there are two choices for the last number.\n\nSimilarly, after choosing the last number, the second to last number must be either the biggest number remaining or the smallest number remaining, since otherwise it would be less than some numbers before it and greater than some other numbers before it. Thus, we have two choices for the second to last number. Similarly, there are two choices for the 3rd number from the end once the last two are chosen, and so on until we reach the second number in the list. For the first number in the list, there is only one choice since there is only one number remaining. Putting this together, we've had $2$ choices for each of the last $9$ spots and $1$ choice for the first, giving $2^{9} = 512$ possible lists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn the sides of a convex quadrilateral $A B C D$, construct squares externally. Prove that the quadrilateral with vertices at the centers of the squares has equal and perpendicular diagonals.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst, a lemma: If squares constructed on $X Y$ and $Y Z$ of triangle $X Y Z$ have respective centers $U, V$, then a rotation through angle $\\pi / 2$ about the midpoint $M$ of $Z X$ carries $U$ into $V$.\n\nThe proof is as follows: Rotating quadrilateral $M X U Y$ by an angle $\\pi$ about $M$ gives us a quadrilateral $M Z W T$ (we know that $Z$ is the image of $X$ because $M$ is the midpoint of $Z X$). Now, a rotation by angle $\\pi / 2$ about $V$ carries $Z$ into $Y$. We will show that this same rotation carries $W$ into $U$. For this it suffices to check that triangles $V Z W$, $V Y U$ are congruent (and equally oriented). Since $V Z = V Y$, $Z W = X U = Y U$, and (using directed angles if necessary)\n$$\n\\angle W Z V = \\pi / 2 + \\angle T Z Y = 3 \\pi / 2 - \\angle Z Y X = \\angle U Y V\n$$\nthe congruence holds. So, as claimed, a $\\pi / 2$ rotation about $V$ takes $W$ into $U$, so that $\\triangle U V W$ is an isosceles right triangle, and $M$ is the midpoint of its hypotenuse. The lemma then follows.\n\nIn our original situation now, let the squares on $A B$, $B C$, $C D$, $D A$ have respective centers $E, F, G, H$. By the lemma, a rotation of angle $\\pi / 2$ about the midpoint of $A C$ maps $E$ into $F$. This rotation also maps $G$ into $H$. Hence it takes the segment $E G$ into $F H$, so these segments are equal and perpendicular.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $C$ be a circle with center at the origin $O$ of a system of rectangular coordinates, and let $M O N$ be the quarter circle of $C$ in the first quadrant. Let $P Q$ be an arc of $C$ of fixed length that lies in the arc $M N$. Let $K$ and $L$ be the feet of the perpendiculars from $P$ and $Q$ to $O N$, and let $V$ and $W$ be the feet of the perpendiculars from $P$ and $Q$ to $O M$, respectively. Let $A$ be the area of trapezoid $P K L Q$ and $B$ the area of trapezoid $P V W Q$. Prove that $A+B$ does not depend on where $\\operatorname{arc} P Q$ is chosen.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDraw $O P$, $O U$, and $O Q$, and note that $A+B$ is the area of rectangle $P V W U$, plus the area of rectangle $U K L Q$, plus twice the area of the triangle $P U Q$. But the area of triangle $P O U$ is half the area of rectangle $P V W U$, and the area of triangle $U O Q$ is half the area of rectangle $U K L Q$, so putting this together $A+B$ is twice the area of triangle $P O Q$, which depends only on the fixed length of arc $P Q$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A_{1}, B_{1}, C_{1}$ be points in the interior of sides $BC, CA, AB$, respectively, of equilateral triangle $ABC$. Prove that if the radii of the inscribed circles of $\\triangle C_{1}AB_{1}$, $\\triangle B_{1}CA_{1}$, $\\triangle A_{1}BC_{1}$, $\\triangle A_{1}B_{1}C_{1}$ are equal, then $A_{1}, B_{1}, C_{1}$ are the midpoints of the sides of $\\triangle ABC$ on which they lie.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst, suppose that $BA_{1} > CB_{1}$. We claim this forces $CB_{1} > AC_{1}$. If we rotate triangle $A_{1}CB_{1}$ by $2\\pi/3$ radians about the center of $\\triangle ABC$, we get $\\triangle A_{2}AB_{2}$ ($A_{2} \\in CA$, $B_{2} \\in AB$) whose incircle coincides with that of $\\triangle B_{1}AC_{1}$ (since they have the same radius and are both tangent to $AB$ and $CA$). But by our assumption, $AB_{1} > AA_{2}$; now if $AC_{1} \\geq CB_{1} = AB_{2}$ then segment $C_{1}B_{1}$ lies outside triangle $A_{2}AB_{2}$ and cannot be tangent to its incircle, a contradiction. Hence $BA_{1} > CB_{1}$ does indeed imply $CB_{1} > AC_{1}$ and, likewise, $AC_{1} > BA_{1}$; combining yields $BA_{1} > CB_{1} > AC_{1} > BA_{1}$, impossible. We conclude that our supposition was wrong, so $BA_{1} \\leq CB_{1}$; likewise $CB_{1} \\leq AC_{1} \\leq BA_{1}$ and we have equality throughout. This implies (by rotational symmetry) that $\\triangle A_{1}B_{1}C_{1}$ is equilateral, and that $AB_{1} + AC_{1} = AB_{1} + CB_{1} = AC$.\n\nIf the incenter of $\\triangle B_{1}AC_{1}$ is $I$, then $\\angle B_{1}IC_{1} = (\\pi + \\angle B_{1}AC_{1})/2 = 2\\pi/3$ and $I$ must lie on the arc of a circle passing through $B_{1}, C_{1}$; the unique point on this arc which is at maximal distance from $B_{1}C_{1}$—i.e., the position of $I$ which gives the largest radius for the incircle—is the midpoint of the arc, and $I$ is located there iff $\\triangle B_{1}AC_{1}$ is equilateral. However, looking at triangle $B_{1}A_{1}C_{1}$ which actually is equilateral, we see that the maximum possible radius is attained there; since it has the same radius as $\\triangle B_{1}AC_{1}$, this latter is also equilateral, and $B_{1}A = AC_{1}$. By our symmetry this yields $B_{1}A = AC_{1} = C_{1}B = BA_{1} = A_{1}C = CB_{1}$ as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine all integers $n$ for which $n^{2}+15$ is the square of an integer.",
    "options": [],
    "answer": "-7, -1, 1, 7",
    "solution": "Solution:\n\nWe may limit the search to nonnegative integers $n$, since $n^{2}+15 = (-n)^{2}+15$. Suppose there is a nonnegative integer $m$ such that\n$$\n\\begin{aligned}\nn^{2}+15 & = m^{2} \\\\\n15 & = m^{2} - n^{2} \\\\\n& = (m+n)(m-n) .\n\\end{aligned}\n$$\nNote that the factors $m+n$ and $m-n$ are not both negative, since their sum $2m$ is nonnegative, so they are both positive. Now the only ways to factor $15$ into two positive integer factors are $15 \\cdot 1$ and $5 \\cdot 3$. Also $m+n \\geq m-n$ since $n \\geq 0$.\n\nIf $m+n=15$ and $m-n=1$, then\n$$\nn = \\frac{(m+n)-(m-n)}{2} = \\frac{15-1}{2} = 7\n$$\nand $m=8$.\n\nIf $m+n=5$ and $m-n=3$, then\n$$\nn = \\frac{(m+n)-(m-n)}{2} = \\frac{5-3}{2} = 1\n$$\nand $m=4$.\n\nBoth of these solutions indeed work in the original problem. Finally, allowing for negative $n$, we compute the answers to be $-7, -1, 1$, and $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBart has 17 and 19 dollar bills only.\n\na. Prove that these bills are fake.\n\nb. Prove that there exists $m>0$ such that for each $n \\geq m$ Bart can give to Lisa exactly $n$ dollars using his bills.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Since there are no real 17 or 19 dollar bills, these bills must be fake.\n\nb. We can have $m = 17 \\cdot 19$. For each $n > 17 \\cdot 19$, we consider the set $S = \\{n, n-17, n-2 \\cdot 17, \\ldots, n-18 \\cdot 17\\}$. If none of these numbers is divisible by 19, two of them will give the same residue upon division by 19. Indeed, there are 19 numbers and without 0 there are only 18 residues. Let $n-17a$ and $n-17b$ give the same residue mod 19. Assuming $a < b$ we get that $n-17a-(n-17b) = 17(b-a)$ is divisible by 19 which is impossible.\n\nHence one of the numbers from $S$ is divisible by 19, say $n-k \\cdot 17$. Then $n-k \\cdot 17 = l \\cdot 19$ and $n = k \\cdot 17 + l \\cdot 19$ which solves the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$$\n\\frac{a}{b} \\quad \\text{ and } \\quad \\frac{c}{d}\n$$\nare called approximately equal if $a, b, c, d$ are positive integers and\n$$\n\\frac{a}{b}-\\frac{c}{d}=\\frac{1}{b d}\n$$\nProve that given two approximately equal fractions, we can multiply the four numerators and denominators by the same positive integer and then add or subtract 1 from each of them so that the resulting fractions are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nGiven the approximately equal fractions\n$$\n\\frac{a}{b} \\text{ and } \\frac{c}{d}\n$$\nwe multiply the four terms by $a+b+c+d$ to get\n$$\n\\frac{a^{2}+a b+a c+a d}{a b+b^{2}+b c+b d} \\quad \\text{ and } \\quad \\frac{a c+b c+c^{2}+c d}{a d+b d+c d+d^{2}}\n$$\nWe then adjust each member by 1 by changing the $a d$ terms to $b c$ and vice versa. The resulting fractions can be factorized:\n$$\n\\begin{aligned}\n& \\frac{a^{2}+a b+a c+b c}{a b+b^{2}+a d+b d} \\quad \\text{ and } \\quad \\frac{a c+a d+c^{2}+c d}{b c+b d+c d+d^{2}} \\\\\n& \\frac{(a+c)(a+b)}{(b+d)(a+b)} \\quad \\text{ and } \\quad \\frac{(a+c)(c+d)}{(b+d)(c+d)} \\text{. }\n\\end{aligned}\n$$\nThe last pair of fractions are visibly equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A_{1}, B_{1}, C_{1}$ be the points on the sides $BC, CA, AB$ (respectively) of the triangle $ABC$. Prove that the three circles circumscribed about the triangles $\\triangle AB_{1}C_{1}$, $\\triangle BC_{1}A_{1}$, and $\\triangle CA_{1}B_{1}$ intersect at one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDenote by $\\alpha$, $\\beta$, and $\\gamma$ the angles of $\\triangle ABC$. Assume that the circumcircles of $\\triangle AB_{1}C_{1}$ and $BC_{1}A_{1}$ intersect at the point $M$. Assume that $M$ is in the interior of $\\triangle ABC$ (the other cases are similar).\n\nBy the properties of the inscribed quadrilaterals $AB_{1}MC_{1}$ and $BA_{1}MC_{1}$ we get: $\\angle C_{1}MB_{1} = 180^{\\circ} - \\angle C_{1}AB_{1} = 180^{\\circ} - \\alpha$ and $\\angle C_{1}MA_{1} = 180^{\\circ} - \\angle C_{1}BA_{1} = 180^{\\circ} - \\beta$.\n\nHence $\\angle B_{1}MA_{1} = 360^{\\circ} - (\\angle C_{1}MB_{1} + \\angle C_{1}MA_{1}) = 360^{\\circ} - [360^{\\circ} - (\\alpha + \\beta)] = \\alpha + \\beta$.\n\nTherefore $\\angle B_{1}MA_{1} + \\angle B_{1}CA_{1} = \\angle B_{1}MA_{1} + \\gamma = \\alpha + \\beta + \\gamma = 180^{\\circ}$.\n\nThus the points $B_{1}, M, A_{1}$, and $C$ belong to a circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSome soldiers are standing in a line in the east-west direction, each of them facing north. Their officer commands, \"Right face!\" They should now all be facing east, but, as they are at the very beginning of their military career, some of them get the order wrong and turn to the west. Every soldier who is then facing his neighbor immediately concludes that he has made a mistake and performs a $180^{\\circ}$ turn within 1 second. The process continues, so that any two soldiers who are now facing each other perform $180^{\\circ}$ turns within 1 second. Prove that after some time, the soldiers will stop moving.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nGive the soldiers ID numbers from $1$ upwards from west to east. At any moment, define the confusion index to be the sum of the ID numbers of the soldiers who are facing west. Note that whenever a pair of soldiers turn around, soldier $n+1$ turns from west to east (decreasing the confusion index by $n+1$) while soldier $n$ turns from east to west (increasing the confusion index by $n$). So overall, the confusion index decreases by $1$ for every pair of turns. Since the confusion index is a nonnegative integer, it cannot keep decreasing forever.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe number $2011$ is written on a blackboard. It is permitted to transform the numbers on it by two types of moves:\n\na) Given a number $n$, we can erase $n$ and write two nonnegative integers $a$ and $b$ such that $a+b=n$.\n\nb) Given two numbers $a$ and $b$, we can erase them and write their difference $a-b$, assuming this is positive or $0$.\n\nIs it possible that after a sequence of such moves, there is only one number on the blackboard and it is $0$?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no.\n\nWe claim that the sum of the numbers on the blackboard is always odd. Indeed, the initial sum, $2011$, is odd, and when a move of type (a) is performed, the sum does not change. When a move of type (b) is performed, the sum decreases by\n$$\n(a+b)-(a-b)=2b,\n$$\nan even number. Since odd $-$ even $=$ odd, the sum will remain odd. Thus the desired final state, in which the sum ($0$) is even, is not achievable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $a$, $b$, and $c$ are positive real numbers with $2a + 4b + 8c = 16$, what is the largest possible value of $abc$?",
    "options": [],
    "answer": "64/27",
    "solution": "Solution:\n\nBy AM-GM,\n$$\n\\frac{16}{3} = \\frac{2a + 4b + 8c}{3} \\geq \\sqrt[3]{(2a) \\cdot (4b) \\cdot (8c)} = 4 \\sqrt[3]{abc}.\n$$\nRearranging, we get\n$$\n\\sqrt[3]{abc} \\leq \\frac{4}{3} \\Longleftrightarrow abc \\leq \\frac{64}{27}\n$$\nThis maximum is indeed attainable if we set $2a = 4b = 8c$, since that is the equality case of AM-GM, which means $2a = 4b = 8c = \\frac{16}{3}$, so $a = \\frac{8}{3}$, $b = \\frac{4}{3}$, $c = \\frac{2}{3}$. We conclude that the maximum is indeed $\\frac{64}{27}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine whether the number\n$$\n\\frac{1}{2 \\sqrt{1}+1 \\sqrt{2}}+\\frac{1}{3 \\sqrt{2}+2 \\sqrt{3}}+\\frac{1}{4 \\sqrt{3}+3 \\sqrt{4}}+\\cdots+\\frac{1}{100 \\sqrt{99}+99 \\sqrt{100}}\n$$\nis rational or irrational. Explain your answer.",
    "options": [],
    "answer": "rational",
    "solution": "Solution:\nNotice that\n$$\n\\begin{aligned}\n\\frac{1}{(n+1) \\sqrt{n}+n \\sqrt{n+1}} & =\\frac{1}{\\sqrt{n(n+1)}} \\cdot \\frac{1}{\\sqrt{n+1}+\\sqrt{n}} \\\\\n& =\\frac{1}{\\sqrt{n(n+1)}} \\cdot \\frac{\\sqrt{n+1}-\\sqrt{n}}{(\\sqrt{n+1}+\\sqrt{n})(\\sqrt{n+1}-\\sqrt{n})} \\\\\n& =\\frac{1}{\\sqrt{n}}-\\frac{1}{\\sqrt{n+1}} .\n\\end{aligned}\n$$\nNow we see that the given sum is equal to $1-\\frac{1}{\\sqrt{100}}=\\frac{9}{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo points $A$ and $C$ are marked on a circle; assume the tangents to the circle at $A$ and $C$ meet at $P$. Let $B$ be another point on the circle, and suppose $P B$ meets the circle again at $D$. Show that $A B \\cdot C D = B C \\cdot D A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $P A$ is tangent to the circle and $\\angle A B D$ is the inscribed angle opposite $A D$, $\\angle P A D = \\angle A B P$. Since $\\angle A P B$ is shared, it follows from AA similarity that $\\triangle P A D \\sim \\triangle P B A$. Thus we get\n$$\n\\frac{A B}{A D} = \\frac{P B}{P A}\n$$\nSimilarly, $\\triangle P C D \\sim \\triangle P B C$, so\n$$\n\\frac{B C}{C D} = \\frac{P B}{P C}\n$$\nSince $P A$ and $P C$ are tangents from the same point to the same circle, they are the same length. Thus,\n$$\n\\frac{A B}{A D} = \\frac{B C}{C D} \\Longrightarrow A B \\cdot C D = B C \\cdot D A\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $ABC$ and $A'B'C'$ are two triangles such that $\\angle A = \\angle A'$, $AB = A'B'$, and $BC = B'C'$. Suppose also that $\\angle C = 90^{\\circ}$. Prove that triangles $ABC$ and $A'B'C'$ are congruent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConstruct a point $D$ on ray $AC$ such that $AD = A'C'$. Then $\\triangle ABD \\cong \\triangle A'B'C'$ by SAS. If $D = C$, then we are done, so assume that $D \\neq C$. We have $BD = B'C'$ and also $BC = B'C'$, so right triangle $BCD$ has hypotenuse $BD$ equal to leg $BC$, which is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA point $M$ and a circle $k$ are given in the plane. If $ABCD$ is an arbitrary square inscribed in $k$, prove that the sum $MA^{4} + MB^{4} + MC^{4} + MD^{4}$ is independent of the positioning of the square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy scaling, assume $k$ has radius $1$. Position the circle in the coordinate plane with its center $O$ at the origin, and points $A$, $B$, $C$, $D$ at $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$, respectively. Let $M$ have coordinates $(x, y)$. Then, using the Pythagorean theorem to expand $MA^{2}$, $MB^{2}$, $MC^{2}$, $MD^{2}$, we get\n$$\n\\begin{aligned}\nMA^{4} & + MB^{4} + MC^{4} + MD^{4} = \\left[(x-1)^{2} + y^{2}\\right]^{2} + \\left[x^{2} + (y-1)^{2}\\right]^{2} + \\left[(x+1)^{2} + y^{2}\\right]^{2} + \\left[x^{2} + (y+1)^{2}\\right]^{2} \\\\\n& = \\left(x^{2} - 2x + 1 + y^{2}\\right)^{2} + \\left(x^{2} + y^{2} - 2y + 1\\right)^{2} + \\left(x^{2} + 2x + 1 + y^{2}\\right)^{2} + \\left(x^{2} + y^{2} + 2y + 1\\right)^{2}\n\\end{aligned}\n$$\nIf we set $s = x^{2} + y^{2} + 1 = OM^{2} + 1$, this becomes\n$$\n\\begin{gathered}\n(s - 2x)^{2} + (s - 2y)^{2} + (s + 2x)^{2} + (s + 2y)^{2} \\\\\n= \\left(s^{2} - 4sx + 4x^{2}\\right) + \\left(s^{2} - 4sy + 4y^{2}\\right) + \\left(s^{2} + 4sx + 4x^{2}\\right) + \\left(s^{2} + 4sy + 4y^{2}\\right) \\\\\n= 4s^{2} + 8(x^{2} + y^{2}) = 4 OM^{4} + 16 OM^{2} + 4.\n\\end{gathered}\n$$\nThus, our sum depends only on the circle's radius and the distance $OM$, which is what we need.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSix children are invited to a birthday party, and each pair of them are either mutual friends or mutual strangers. Prove that there are either three of them that are all friends or three of them that are all strangers to one another.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBegin by letting $A$ be any person at the party, and note that $A$ must be either friends or strangers with at least three of the others, for otherwise there would only be at most $2+2=4$ other people at the party. Because of the symmetry between friends and strangers, we can assume that $A$ has three friends, call them $B, C$, and $D$. Now if any two of these three are friends, then they together with $A$ are the desired triple. Otherwise, $B, C$, and $D$ are all mutual strangers, so they themselves furnish the desired triple.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x$ and $y$ are two positive numbers less than $1$, prove that\n$$\n\\frac{1}{1-x^{2}} + \\frac{1}{1-y^{2}} \\geq \\frac{2}{1-x y}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst we use the inequality $a + b \\geq 2 \\sqrt{a b}$ and get\n$$\n\\frac{1}{1-x^{2}} + \\frac{1}{1-y^{2}} \\geq \\frac{2}{\\sqrt{(1-x^{2})(1-y^{2})}}\n$$\nNow we notice that\n$$\n(1-x^{2})(1-y^{2}) = 1 + x^{2} y^{2} - x^{2} - y^{2} \\leq 1 + x^{2} y^{2} - 2 x y = (1-x y)^{2}\n$$\nwhich implies that\n$$\n\\frac{2}{\\sqrt{(1-x^{2})(1-y^{2})}} \\geq \\frac{2}{1-x y}\n$$\nand this completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $M$ be an interior point of a parallelogram $ABCD$. Prove that $MA + MB + MC + MD$ is strictly less than the length of the perimeter of $ABCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenote by $X$ and $Y$ the points of intersection of the segments $AB$ and $CD$ with the line through $M$ parallel to $BC$. Similarly, let $U$ and $V$ denote the points of intersection of the segments $AD$ and $BC$ with the line through $M$ parallel to $AB$.\n\nThen $MA < AU + UM = XM + UM$, $MB < MX + XB = MX + MV$, $MC < MV + VC = MV + MY$, and $MD < MY + YD = MY + MU$.\n\nHence\n$$\nMA + MB + MC + MD < 2(XM + YM) + 2(UM + VM) = 2AD + 2AB = AB + BC + CD + DA.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn isolated island has the shape of a circle. Initially there are $9$ flowers on the circumference of the island: $5$ of the flowers are red and the other $4$ are yellow. During the summer $9$ new flowers grow on the circumference of the island according to the following rule: between $2$ old flowers of the same color a new red flower will grow, between $2$ old flowers of different colors, a new yellow flower will grow. During the winter, the old flowers die, and the new survive. The same phenomenon repeats every year.\nIs it possible (for some configuration of initial $9$ flowers) to get all red flowers after finitely many years?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is \"no\". Assume that we got all red flowers in the year $n$ for the first time. Then in the year $n-1$ all the flowers were yellow. We will prove that this is impossible.\n\nLet's change the weird story into the one with the flowers labeled by $1$ (instead of red) and $-1$ (instead of yellow). What really happens is that between two flowers $a$ and $b$, the new flower will grow and will be labeled by $ab$. Notice that the initial product of all numbers is $1$, and at the end of each winter the product of the numbers is $1$ again, so it will never be equal to $-1$; hence it is impossible to get the configuration where all the flowers are yellow. This is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $x$ for which $\\tan (x / 2)$ is defined and greater than $\\sin (x)$.",
    "options": [],
    "answer": "x ∈ (kπ − π/2, kπ) for all integers k",
    "solution": "Solution:\nWe claim that the answer is $x \\in \\left(k \\pi - \\frac{\\pi}{2},\\ k \\pi\\right)$ for $k \\in \\mathbb{Z}$.\n\nThe statement is never true for $x / 2$ a multiple of $\\frac{\\pi}{2}$ because $\\tan (x / 2)$ is either undefined or equal to $0 = \\sin x$. Thus $x / 2$ is not a multiple of $\\frac{\\pi}{2}$, so $\\cos^2(x / 2) > 0$, and\n\n$$\n\\begin{aligned}\n\\tan (x / 2) &\\stackrel{?}{>} \\sin (x) \\\\\n\\cos^2(x / 2) \\tan (x / 2) &\\stackrel{?}{>} \\cos^2(x / 2) \\sin (x) \\\\\n2 \\cos (x / 2) \\sin (x / 2) &\\stackrel{?}{>} 2 \\cos^2(x / 2) \\sin (x) \\\\\n\\sin (x) &\\stackrel{?}{>} (1 + \\cos (x)) \\sin (x) \\\\\n0 &\\stackrel{?}{>} \\cos (x) \\sin (x) \\\\\n0 &\\stackrel{?}{>} \\sin (2x),\n\\end{aligned}\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo congruent line segments $AB$ and $CD$ intersect at a point $E$. The perpendicular bisectors of $AC$ and $BD$ intersect at a point $F$ in the interior of $\\angle AEC$. Prove that $EF$ bisects $\\angle AEC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $F$ is on the perpendicular bisectors of $AC$ and $BD$, $FA = FC$ and $FB = FD$. Also $AB = CD$ is given. Thus $\\triangle FAB \\cong \\triangle FCD$ by SSS. Since corresponding heights of congruent triangles are equal, $F$ is equidistant from $AB$ and $CD$, implying that $F$ is on the bisector of $\\angle AEC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a triangle $ABC$ such that $\\angle B = 90^{\\circ}$, denote by $k$ the circle with center on $BC$ that is tangent to $AC$. Denote by $T$ a point of tangency of $k$ and the tangent from $A$ to $k$ (different from $AC$). If $B'$ is the midpoint of $AC$ and $M$ the intersection of $BB'$ and $AT$, prove that $MB = MT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $O$ be the center of $k$. The quadrilateral $ABTO$ can be inscribed in a circle, hence $\\angle CBB' = \\angle B'CB$ and $\\angle TAO = \\angle OAC$. Thus $\\angle MTB = \\angle ATB = \\angle AOB = \\angle ACO + \\angle OAC = \\angle B'BC + \\angle TAO = \\angle B'BC + \\angle TBO = \\angle TBM$, which immediately implies the given statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe set $S$ is a finite subset of $[0,1]$ with the following property: for all $x \\in S$, there exist $a, b \\in S \\cup \\{0,1\\}$ with $a, b \\neq x$ such that $x = (a + b)/2$. Prove that all the numbers in $S$ are rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $x_{1}, x_{2}, \\ldots, x_{n}$ be the elements of $S$. They satisfy a system of $n$ linear equations of the form $x_{i} = (x_{a} + x_{b}) / 2$, where some of the $x_{a}, x_{b}$ may be replaced by $0$ or $1$. If the system is not redundant (its solution is unique), then it can be solved by the usual elimination method of solving systems of linear equations. Since this method computes the values of $x_{i}$ using only the existing coefficients $(0,1)$ and addition, subtraction, multiplication, and division, it follows that each $x_{i}$ must be rational. So it suffices to prove that the system has a unique solution.\n\nSuppose that there are two solutions, $(x_{1}, \\ldots, x_{n})$ (the distinct elements of $S$) and $(y_{1}, \\ldots, y_{n})$, and assume that $x_{k} > y_{k}$ for some particular $k$ (the $x_{k} < y_{k}$ case is analogous). Note that any linear equation satisfied by $(x_{1}, \\ldots, x_{n})$ and $(y_{1}, \\ldots, y_{n})$ is also satisfied by $(t x_{1} + (1-t) y_{1}, \\ldots, t x_{n} + (1-t) y_{n})$ for any real number $t$ (as is seen by multiplying the equation with the $x$'s by $t$ and the equation with the $y$'s by $1-t$, and adding). Thus the numbers $t x_{i} + (1-t) y_{i}$ form another solution of our system, so that each such number is an average of two quantities—another of the numbers, $1$, or $0$. If we choose $t > (1 - y_{k}) / (x_{k} - y_{k})$, then $t x_{k} + (1-t) y_{k} > 1$. It follows that the largest value of $t x_{i} + (1-t) y_{i}$ over all $i$ is also greater than $1$. If we can choose $t$ so that the largest value is unique (no other $i$ yields the same value of $t x_{i} + (1-t) y_{i}$), then this value is the average of two numbers which are strictly smaller than it, an impossibility.\n\nThus, what remains is to show that we can choose $t > (1 - y_{k}) / (x_{k} - y_{k})$ so that all the values of $t x_{i} + (1-t) y_{i}$ are different. There clearly are infinitely many $t$ large enough for the first condition. For the second condition, if $t x_{i} + (1-t) y_{i} = t x_{j} + (1-t) y_{j}$ ($i \\neq j$) then $t(x_{i} - y_{i} - x_{j} + y_{j}) = y_{j} - y_{i}$. If the coefficient of $t$ on the left side is $0$ then $y_{j} - y_{i} = 0$ forcing $x_{i} - x_{j} = 0$, an impossibility; otherwise, we can divide by this coefficient to see that the value of $t$ is uniquely determined by $i$ and $j$. Since there are a finite number of choices for $i$ and $j$, there are only a finite number of values of $t$ for which some equality $t x_{i} + (1-t) y_{i} = t x_{j} + (1-t) y_{j}$ holds, and we simply choose some $t$ which is not any of these. This meets our requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A_{1} A_{2} \\cdots A_{2 n}$ be a convex $2 n$-gon. Prove that there is an $i$ ($1 \\leq i \\leq n$) and a pair of parallel lines, each intersecting the $2 n$-gon only once, one at $A_{i}$ and one at $A_{n+i}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider the \"highest\" and \"lowest\" points of the $2 n$-gon, with respect to any chosen orientation. Unless a side of the $2 n$-gon is perfectly horizontal, these points will be unique and thus a horizontal line through them will not meet the $2 n$-gon again. These two points will also be vertices of the $2 n$-gon; call them $A_{i}$ and $A_{j}$ respectively. Say that the imbalance in this situation is $i-j \\bmod 2 n$, which (being nonzero) we can think of as an integer from $1$ to $2 n-1$ inclusive. We would like to prove that the $2 n$-gon can be rotated such that the imbalance is $n$.\n\nConsider smoothly rotating the $2 n$-gon counterclockwise (assume that the vertices are labeled $A_{1}, \\ldots, A_{2 n}$ in counterclockwise order). Note that the imbalance does not change unless one of the sides becomes momentarily horizontal. If the top side becomes horizontal, it goes down by $1$; if the bottom side becomes horizontal, it goes up by $1$; and if both sides momentarily become horizontal, it does not change at all. But after rotating by $180^{\\circ}$, the top and bottom vertices are switched: if the imbalance was between $0$ and $n$, it is now between $n$ and $2 n$ and vice versa. Since the imbalance can never be $0$ and never jumps by more than $1$, it must be $n$ at some point.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf the sum of digits in a decimal representation of a natural number $n$ is equal to 2006, prove that $n$ can't be a perfect square of an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe remainder of $n$ upon division by $3$ is equal to the sum of its digits, i.e. $2006$. Hence number $n$ has a remainder $2$ upon division by $3$ and no square can have that remainder.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $P(x)$ be a nonzero polynomial with real coefficients such that\n$$\nP(x) = P(0) + P(1) x + P(2) x^{2}\n$$\nholds for all $x$. What are the roots of $P(x)$?",
    "options": [],
    "answer": "(1 + sqrt(5))/2 and (1 - sqrt(5))/2",
    "solution": "Solution.\n\nLet $c = P(0)$. Selecting $x = 1$ in the given, we have that $P(0) + P(1) + P(2) = P(1)$, so $P(2) = -c$. Selecting $x = 2$, we find that\n$$\n-c = c + 2 P(1) + 4(-c) \\Longrightarrow P(1) = c .\n$$\nTherefore, $P(x) = c + c x - c x^{2}$. As $c \\neq 0$, we solve $1 + x - x^{2} = 0$ to get the answers $x = \\frac{1}{2}(1 \\pm \\sqrt{5})$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a, b, c, d, e$, and $f$ be decimal digits such that the six-digit number $\\overline{a b c d e f}$ is divisible by $7$. Prove that the six-digit number $\\overline{b c d e f a}$ is divisible by $7$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe have\n$$\n\\begin{aligned}\n\\overline{b c d e f a} & = \\overline{a b c d e f a} - \\overline{a 000000} \\\\\n& = \\overline{a b c d e f 0} - \\overline{a 000000} + a \\\\\n& = 10 \\cdot \\overline{a b c d e f} - 1000000 \\cdot a + a \\\\\n& = 10 \\cdot \\overline{a b c d e f} - 999999 \\cdot a \\\\\n& = 10 \\cdot \\overline{a b c d e f} - 7 \\cdot 142857 \\cdot a.\n\\end{aligned}\n$$\nThe first term is a multiple of $7$ because $\\overline{a b c d e f}$ is, and the second clearly has $7$ as a factor. So their difference $\\overline{b c d e f a}$ is divisible by $7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$p$ is a prime number such that the period of its decimal reciprocal is $200$. That is,\n\n$$\n\\frac{1}{p}=0 . X X X X \\ldots\n$$\n\nfor some block of $200$ digits $X$, but\n\n$$\n\\frac{1}{p} \\neq 0 . Y Y Y Y \\ldots\n$$\n\nfor all blocks $Y$ with less than $200$ digits. Find the $101$st digit, counting from the left, of $X$.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nLet $X$ be a block of $n$ digits and let $a=0 . X \\ldots$ Then $10^{n} a=X . X \\ldots$. Subtracting the previous two equalities gives us $\\left(10^{n}-1\\right) a=X$, i.e. $a=\\frac{X}{10^{n}-1}$.\n\nThen the condition that $a=\\frac{1}{p}$ reduces to $\\frac{1}{p}=\\frac{X}{10^{n}-1}$ or $p X=10^{n}-1$. For a given $p$ and $n$, such an $X$ exists if and only if $p$ divides $10^{n}-1$. Thus $p$ divides $10^{200}-1$ but not $10^{n}-1, 1 \\leq n \\leq 199$. Note that $10^{200}-1$ can be factored in this way:\n\n$$\n\\begin{aligned}\n10^{200}-1 & =\\left(10^{100}\\right)^{2}-1 \\\\\n& =\\left(10^{100}-1\\right)\\left(10^{100}+1\\right) .\n\\end{aligned}\n$$\n\nSince $p$ is prime and does not divide $10^{100}-1$, it must divide $10^{100}+1$, so that $10^{100}+1=k p$ for an integer $k$ and $X=\\frac{\\left(10^{100}-1\\right)\\left(10^{100}+1\\right)}{p}=\\left(10^{100}-1\\right) k$.\n\nIf $p=2,3,5$, or $7$, the fraction $\\frac{1}{p}$ either terminates or repeats less than $200$ digits. Therefore $p>10$ and $k<\\frac{10^{100}}{p}<10^{99}$. Now let us calculate the $101$st digit of $X=10^{100} k-k$, i.e. the digit representing multiples of $10^{99}$. Since $10^{100} k$ is divisible by $10^{100}$, its $10^{99}$s digit and all later digits are $0$. Since $k<10^{99}$, $k$ does not contribute a digit to the $10^{99}$s place, but it generates a borrow to this place, changing it into a $9$. Thus the $101$st digit of $X$ is a $9$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA building has the plan of a $5 \\times 5$ grid of rooms, each of which has a door in each of its four walls: thus there are 20 doors leading to the outside. The doors are to be opened and closed so that every room has exactly 3 open doors leading from it. Determine the minimum and maximum number of doors to the outside that may be left open.",
    "options": [],
    "answer": "minimum 5, maximum 19",
    "solution": "Solution:\nIf we consider the building as possessing 26 \"spaces,\" namely the outside and the 25 rooms, then each door belongs to two spaces; hence the total of the numbers of open doors to each space must be even. Each room has three open doors, an odd number; hence the number of open doors to the outside must also be odd.\n\nAlso, each of the four corner rooms clearly demands an outside door to be opened. So there are at least 4 open outside doors; coupling this with the parity observation yields at least 5 and at most 19 open outside doors. The diagrams show that both of these extremes are achievable.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n>3$ be an integer which is not divisible by $3$. Two players $A$ and $B$ play the following game with an $n \\times n$ chocolate table. First, player $A$ has to choose and remove one piece of the chocolate, without breaking other pieces. After his move, player $B$ tries to partition the remaining chocolate into $3 \\times 1$ (and $1 \\times 3$) rectangles. If $B$ manages to do so, then he/she is the winner. Otherwise the winner is $A$. Determine which player has a winning strategy and describe the strategy.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe player $A$ has a winning strategy. Imagine that the chocolate is painted in $3$ colors, $R$ (red), $G$ (green), and $B$ (blue) as shown in the table\n\n$$\n\\begin{array}{cccccccc}\nR & G & B & R & G & B & R & \\cdots \\\\\nB & R & G & B & R & G & B & \\cdots \\\\\nG & B & R & G & B & R & G & \\cdots \\\\\nR & G & B & R & G & B & R & \\cdots \\\\\nB & R & G & B & R & G & B & \\cdots \\\\\nG & B & R & G & B & R & G & \\cdots \\\\\nR & G & B & R & G & B & R & \\cdots \\\\\n& \\vdots & & & \\vdots & & & \\vdots\n\\end{array}\n$$\n\nNotice that every $1 \\times 3$ rectangle will contain exactly one red, one green, and one blue piece. Moreover, the original table had equal numbers of green and blue pieces, and one red piece extra. If $A$ takes out any green or blue piece, player $B$ won't be able to fulfill the requirement, and $A$ is the winner.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for any positive integer $n$, we have\n$$\n\\prod_{k=1}^{n} \\operatorname{lcm}\\left(1,2, \\ldots,\\left\\lfloor\\frac{n}{k}\\right\\rfloor\\right)=n !\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe show that the exponents of $p$ coincide for any prime $p$. Actually, we'll prove the stronger claim that for any prime power $q$, the number of terms on each side divisible by $q$ is exactly the same. By the fundamental theorem of arithmetic, that will imply the desired equality.\n\nClearly, the number of terms on the right-hand side which are divisible by $q$ is $\\lfloor n / q\\rfloor$. As for the left-hand side, the number of lcm's on the right which are divisible by $q$ is given by the number of $k$ for which $\\left\\lfloor\\frac{n}{k}\\right\\rfloor \\geq q$, which is exactly $k=1, \\ldots,\\lfloor n / q\\rfloor$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn this fragment of a computer keyboard, the keys are congruent squares touching along their edges, and each letter refers to the point at the center of the corresponding key. Prove that triangles $Q A Z$ and $E S Z$ have the same area.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet us use measuring units in which the side length of each key is $1$. We express the area of quadrilateral $Q A Z E$ in two ways:\n\na. By dividing into triangles $Q A Z$ and $Q Z E$. Since $\\triangle Q Z E$ has base $Q E = 2$ and height $2$, we get\n$$\n\\text{Area } Q A Z E = \\text{Area } Q A Z + \\frac{1}{2} \\cdot 2 \\cdot 2 = \\operatorname{Area} Q A Z + 2\n$$\n\nb. By dividing to triangles $E S Z$, $Q W A$, $W A S$, $W E S$, and $A S Z$. The four latter triangles all have base $1$, height $1$, and area $1/2$, so\n$$\n\\text{Area } Q A Z E = \\text{Area } E S Z + \\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} = \\operatorname{Area} E S Z + 2\n$$\n\nSince quadrilateral $Q A Z E$ must have the same area in both computations, we deduce that $\\text{Area } Q A Z = \\text{Area } E S Z$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle is inscribed in a sector that is one sixth of a circle of radius $6$. (That is, the circle is tangent to both segments and the arc forming the sector.) Find, with proof, the radius of the small circle.\n![](attached_image_1.png)",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nBecause the circles are tangent, we can draw the line $AT$, which passes through the center $O$ of the other circle. Note that triangles $ADO$ and $AEO$ are symmetric (this is HL congruence: $AO$ is shared, radii $OD$ and $OE$ are equal, and angles $ADO$ and $AEO$ are right). Therefore, since $\\angle BAC$ is $60^{\\circ}$, angles $BAT$ and $TAC$ are each $30^{\\circ}$. Now $ADO$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ right triangle. If we let $r$ be the radius of the small circle, then $OD = r$ and $OA = 2r$, but $OT = r$ so $AT = 3r$. But $AT$ is the radius of the large circle, so $3r = 6$ and $r = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe Cannibal Club of California (CCC) had 30 members yesterday morning - but that was before their festive annual dinner! After the dinner, it turned out that among any six members of the club, there was a pair one of whom ate the other. Prove that at least six members of the CCC are now nested inside one another.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe must assume that nobody was eaten by more than one person for the problem statement to make sense. To each cannibal we assign a numerical \"depth\" as follows: the depth of cannibal $C$ is the largest integer $n$ such that there exist cannibals $C_{1}, C_{2}, \\ldots, C_{n}=C$ such that $C_{i}$ ate $C_{i+1}$ for $i=1,2, \\ldots, n-1$. (A cannibal who was eaten by no one has depth 1. Also note that depth is definable: there is an upper bound on the value of $n$, since any chain of length greater than 30 would contain some cannibal twice, an impossibility; hence there is a maximum value of $n$ for which chains exist.) Note that no cannibal ate another of the same depth, since the inner cannibal always has higher depth.\n\nNow, if we can find a cannibal of depth $\\geq 6$ we are done, so assume that the only depths which occur are $1,2,3,4,5$. By the pigeonhole principle, some depth was assumed by at least 6 cannibals; from the given, one of these six ate another. But we know this is impossible, so our assumption was wrong and some higher depth does occur.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of ordered pairs $(a, b)$ of positive integers such that $a$ and $b$ both divide $20^{19}$, but $a b$ does not.",
    "options": [],
    "answer": "444600",
    "solution": "Solution:\nWrite $N = 20^{19} = 2^{38} \\cdot 5^{19}$.\nThe number of divisors of $N$ is $39 \\cdot 20 = 780$.\n\nThe number of pairs $(a, b)$ with $a b$ dividing $N$ is equal to the number of nonnegative integer solutions to $x + y \\leq 38$ times the number of nonnegative integer solutions to $x + y \\leq 19$; by a stars-and-bars argument this equals\n$$\n\\binom{40}{2} \\binom{21}{2} = 163800.\n$$\nTherefore the answer is $780^2 - 163800 = 444600$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine, with proof, whether the following statement is true or false: Out of any six natural numbers, one can find either three which are pairwise relatively prime or three whose greatest common divisor is greater than $1$.",
    "options": [],
    "answer": "false",
    "solution": "Solution:\nThe statement is false. To avoid having three pairwise relatively prime numbers, we should give each pair of numbers a prime factor in common; to avoid having three numbers with a nontrivial gcd, we should not use any factor to cover more than two numbers. Utilizing the first fifteen primes for the $\\binom{6}{2}$ pairs, we arrive at the numbers:\n$$\n\\begin{gathered}\n2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\\\\n2 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 23 \\\\\n3 \\cdot 13 \\cdot 29 \\cdot 31 \\cdot 37 \\\\\n5 \\cdot 17 \\cdot 29 \\cdot 41 \\cdot 43 \\\\\n7 \\cdot 19 \\cdot 31 \\cdot 41 \\cdot 47 \\\\\n11 \\cdot 23 \\cdot 37 \\cdot 43 \\cdot 47\n\\end{gathered}\n$$\nIn fact, no pair of these numbers are relatively prime, while no three have a common factor.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine all triples $(x, y, n)$ of integers such that $x^{2} + 2 y^{2} = 2^{n}$.",
    "options": [],
    "answer": "All solutions are (±2^r, 0, 2r) and (0, ±2^r, 2r+1) for any nonnegative integer r.",
    "solution": "Solution:\n\nIt is easy to check that $\\left( \\pm 2^{r}, 0, 2r \\right)$ and $\\left(0, \\pm 2^{r}, 2r+1 \\right)$ satisfy this equation for any nonnegative integer $r$. We will show that these are all the solutions by an infinite descent method.\n\nSo suppose we have some solution $\\left(x_{0}, y_{0}, n_{0}\\right)$. If $x_{0}$ is odd, then $2^{n_{0}}$ is odd, which forces $n_{0}=0$ and then $x_{0}^{2} + 2 y_{0}^{2} = 1$, so $y_{0} = 0$ (or else $2 y_{0}^{2} > 1$ already) and then $x_{0} = \\pm 1$.\n\nOn the other hand, if $x_{0}$ is even, we can let $x_{0} = 2 x_{0}'$ and then $4 x_{0}'^{2} + 2 y_{0}^{2} = 2^{n_{0}} \\Rightarrow y_{0}^{2} + 2 x_{0}'^{2} = 2^{n_{0}-1}$, so $\\left(x_{1}, y_{1}, n_{1}\\right) = \\left(y_{0}, x_{0}/2, n_{0}-1\\right)$ is another solution to our equation, where $n_{0}$ has been replaced by $n_{0}-1$. Now if $x_{1}$ is even, we can repeat this construction to get another new solution $\\left(x_{2}, y_{2}, n_{2}\\right)$ with $n_{0}-1$ replaced by $n_{0}-2$, and so on. These integers $n$ cannot go on decreasing forever, since there does not exist an integral solution where $n < 0$. Thus, eventually our process terminates, which means we get to a solution $\\left(x_{k}, y_{k}, n_{k}\\right)$ with $x_{k}$ odd. By the above, this is possible only if $x_{k} = \\pm 1, y_{k} = 0, n_{k} = 0$.\n\nOn the other hand, the above construction can be performed in reverse: we have $x_{i} = 2 y_{i+1}, y_{i} = x_{i+1}, n_{i} = n_{i+1} + 1$ for each value of $i \\geq 0$. Now we claim that $\\left(x_{i}, y_{i}, n_{i}\\right) = \\left( \\pm 2^{(k-i)/2}, 0, k-i \\right)$ when $k-i$ is even, and $\\left(0, \\pm 2^{(k-i-1)/2}, k-i \\right)$ when $k-i$ is odd. The proof is by downward induction: the base case $i = k$ is certainly true; given that the statement holds for some $i > 0$, it is simple algebra to check that it holds for $i-1$ by applying our reverse construction. Thus, the claim is true for each $i \\geq 0$. In particular, $\\left(x_{0}, y_{0}, n_{0}\\right) = \\left( \\pm 2^{k/2}, 0, k \\right)$ or $\\left(0, \\pm 2^{(k-1)/2}, k \\right)$, which fits the form above. So, every solution is of this form.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle. A circle is tangent to segments $BC$, $CA$, $AB$ at points $D$, $E$, $F$, respectively. Given that the measures of $\\angle CAB$, $\\angle ABC$, $\\angle BCA$ form an arithmetic progression in some order, prove that the measures of $\\angle FDE$, $\\angle DEF$, $\\angle EFD$ also form an arithmetic progression in some order.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe main observation is that the three angles of a triangle form an arithmetic progression if and only if one of the angles is $60^{\\circ}$. Indeed, suppose that a triangle has angles $x \\leq y \\leq z$ in arithmetic progression. Then $y = \\frac{x + y + z}{3} = 60^{\\circ}$. Conversely, if a triangle has angles $x$, $60^{\\circ}$, $y$, where $x \\geq y$, we have $x + y = 120^{\\circ}$, so $x - 60^{\\circ} = 60^{\\circ} - y$.\n\n![](attached_image_1.png)\n\nSuppose without loss of generality that $\\angle CAB = 60^{\\circ}$. We will now prove that $\\angle FDE = 60^{\\circ}$ as well. Let $I$ be the center of the inscribed circle. Since $\\angle IFA = \\angle IEA = 90^{\\circ}$ we have that\n$$\n\\angle EIF = 360^{\\circ} - 2 \\cdot 90^{\\circ} - \\angle BAC = 120^{\\circ}.\n$$\nBut then\n$$\n\\angle EDF = \\frac{1}{2} \\angle EIF = 60^{\\circ}\n$$\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Prove that for all real numbers $x$ and $y$,\n$$\nx^{2}-2 y^{2}=-\\left[(x+2 y)^{2}-2(x+y)^{2}\\right] .\n$$\n\nb. How many positive integer solutions does the equation $x^{2}-2 y^{2}=1$ have?\n\nc. How many positive integer solutions does the equation $x^{2}-2 y^{2}=5$ have?",
    "options": [],
    "answer": "a) The identity holds. b) Infinitely many positive integer solutions. c) No positive integer solutions.",
    "solution": "Solution:\n\na. This is simply an algebra calculation:\n$$\n\\begin{aligned}\n-\\left[(x+2 y)^{2}-2(x+y)^{2}\\right] & =-\\left[x^{2}+4 x y+4 y^{2}-2\\left(x^{2}+2 x y+y^{2}\\right)\\right] \\\\\n& =-\\left[x^{2}+4 x y+4 y^{2}-2 x^{2}-4 x y-2 y^{2}\\right] \\\\\n& =-\\left[-x^{2}+2 y^{2}\\right] \\\\\n& =x^{2}-2 y^{2} .\n\\end{aligned}\n$$\n\nb. There are infinitely many. Begin at the solution $x_{0}=3, y_{0}=2$ and apply the transformation\n$$\nx_{n+1}=x_{n}+2 y_{n}, \\quad y_{n+1}=x_{n}+y_{n} .\n$$\nBy part (a), the value of $x_{n}^{2}-2 y_{n}^{2}$ is alternately 1 and $-1$ (to be precise, $(-1)^{n}$). Also, all the $x_{n}$ are positive so $y_{0}<y_{1}<y_{2}<\\cdots$ and we never repeat a pair $(x, y)$. Thus we get infinitely many solutions to $x^{2}-2 y^{2}=1$. (With some more work one can prove that the $\\left(x_{2 n}, y_{2 n}\\right)$ are all positive integer solutions.)\n\nc. There are no solutions. Look at the remainders of the terms mod $5$. The term $x^{2}$ must be $0,1$, or $4$; the term $2 y^{2}$ must be $0,2$, or $3$. The only way these can equal each other is if $5 \\mid x$ and $5 \\mid y$. But then $25 \\mid x^{2}-2 y^{2}=5$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequence of numbers $1, 2, 3, \\ldots, 100$ is written on the blackboard. Between each two consecutive numbers a square box is drawn. Player $A$ starts the game and the players $A$ and $B$ alternate the moves. In each turn a player chooses an empty box and places \"+\" or \".\" sign in it. After all the boxes are filled the expression on the blackboard is evaluated and if the result is an odd number the winner is $A$. Otherwise the winner is $B$. Determine which of the players has a winning strategy and what the strategy is.",
    "options": [],
    "answer": "Player A. First place plus between one and two to isolate one as a separate summand. Then partition the remaining boxes into pairs around each odd integer and, whenever Player B plays in a box of a pair, immediately play in the other box of that same pair with a multiplication sign. This guarantees at least one multiplication adjacent to every odd integer other than one, making all those contributions even, so the total sum is odd.",
    "solution": "Solution:\n\n$A$ has the winning strategy. She should first place the sign \"+\" between the numbers $1$ and $2$. After that she should group the square boxes into pairs: each pair consisting of two boxes adjacent to the same odd number. Then the player $A$ should make sure that there is at least one sign \".\" in each pair of boxes. Since $\\cdot$ has the priority over $+$, after all products of numbers are calculated, $1$ will be the only odd summand in the whole expression, hence the sum is odd. $A$ can achieve this goal by placing $\\cdot$ sign in the box of the pair where $B$ has previously put his sign.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are infinitely many bowls arranged on the number line, one at each integer. Initially each bowl has one fruit in it. In a move, one may take any fruit and move it to an adjacent bowl (bowls may hold more than one fruit, or no fruits at all).\nIs it possible that after 999 moves, every bowl still has exactly one fruit remaining?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no.\nAfter 999 moves, only finitely many bowls were touched (i.e. received or lost any fruit), so we can consider a large interval $[L, M]$ such that all touched bowls are in this interval.\n\nNow, consider the sum $S$ of the indices of all fruits inside this interval. By hypothesis, $S$ is the same before and after the moves. However, every move changes the parity of $S$, so after 999 moves the value of $S$ should have opposite parity, contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA triangular number is one of the numbers $1, 3, 6, 10, 15, \\ldots$ of the form $T_{n} = 1 + 2 + \\cdots + n$ or, equivalently, $T_{n} = \\left(n^{2} + n\\right) / 2$.\nFind, with proof, all ways of writing $2015$ as the difference of two triangular numbers.",
    "options": [],
    "answer": "T_2015 − T_2014, T_1008 − T_1006, T_405 − T_400, T_206 − T_196, T_161 − T_148, T_90 − T_64, T_80 − T_49, T_63 − T_1",
    "solution": "Solution:\nSince\n$$\nT_{m} - T_{n} = \\frac{m^{2} + m}{2} - \\frac{n^{2} + n}{2} = \\frac{m^{2} - n^{2} + m - n}{2} = \\frac{(m + n + 1)(m - n)}{2},\n$$\nany expression of $2015$ as the difference of two triangular numbers yields a factorization $2 \\cdot 2015 = 4030 = (m + n + 1)(m - n)$. Conversely, assigning values $a = m + n + 1$, $b = m - n$ to the factors (necessarily both positive) yields a unique solution\n$$\nm = \\frac{a + b - 1}{2} \\quad n = \\frac{a - b - 1}{2}\n$$\nfor $m$ and $n$, which are positive integers if the following two conditions are satisfied:\n- $a$ and $b$ have unlike parity (that is, one is even and one is odd). Since $4030$ has exactly one prime factor of $2$, this is no restriction.\n- $a \\geq b + 2$. In particular, out of every pair of factors, we must put the larger factor for $a$.\n\nThe primes in the factorization $4030 = 2 \\cdot 5 \\cdot 13 \\cdot 31$ may be apportioned to two factors in $16$ ways, or $8$ ways if the order of the factors is insignificant. So we get eight solutions:\n\n| $a$ | $b$ | Expression |\n| :---: | :---: | :---: |\n| $2 \\cdot 5 \\cdot 13 \\cdot 31$ | $1$ | $T_{2015} - T_{2014}$ |\n| $5 \\cdot 13 \\cdot 31$ | $2$ | $T_{1008} - T_{1006}$ |\n| $2 \\cdot 13 \\cdot 31$ | $5$ | $T_{405} - T_{400}$ |\n| $13 \\cdot 31$ | $2 \\cdot 5$ | $T_{206} - T_{196}$ |\n| $2 \\cdot 5 \\cdot 31$ | $13$ | $T_{161} - T_{148}$ |\n| $5 \\cdot 31$ | $2 \\cdot 13$ | $T_{90} - T_{64}$ |\n| $2 \\cdot 5 \\cdot 13$ | $31$ | $T_{80} - T_{49}$ |\n| $5 \\cdot 13$ | $2 \\cdot 31$ | $T_{63} - T_{1}$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a square in the coordinate plane such that $A$ is on the $x$-axis and $C$ is on the $y$-axis. Prove that one of the vertices $B$ and $D$ lies on the line $y = x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume without loss of generality that the vertices $A, B, C, D$ are labeled in counterclockwise order. Let $A$ and $C$ have the coordinates $(a, 0)$ and $(0, c)$ respectively. The center $M$ of the square is the midpoint of $AC$ and therefore has the coordinates\n$$\n\\left(\\frac{a}{2}, \\frac{c}{2}\\right).\n$$\n\nTo get from $M$ to $C$, we can move upward $c/2$ units and then leftward $a/2$ units (of course, these are signed distances, indicating movement in the opposite directions if they are negative). Since $B$ is the $90^{\\circ}$ clockwise rotation of $C$ around $M$, we can get from $M$ to $B$ by moving rightward $c/2$ units and then upward $a/2$ units. This takes us from $M$ to the point\n$$\n\\left(\\frac{a}{2} + \\frac{c}{2}, \\frac{c}{2} + \\frac{a}{2}\\right)\n$$\nwhich clearly lies on the line $y = x$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA natural number $n$ is chosen between two consecutive square numbers. The smaller square is obtained by subtracting $k$ from $n$, and the larger one is obtained by adding $\\ell$ to $n$. Prove that the number $n-k \\ell$ is the square of an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $n-k = x^{2}$ and $n+\\ell = (x+1)^{2}$. Then $k = n - x^{2}$ and $\\ell = (x+1)^{2} - n$. We express everything in terms of $n$ and $x$:\n$$\n\\begin{aligned}\nn - k \\ell & = n - \\left(n - x^{2}\\right)\\left((x+1)^{2} - n\\right) \\\\\n& = n + \\left(x^{2} - n\\right)\\left(x^{2} + 2x + 1 - n\\right) \\\\\n& = n + \\left(x^{2} - n\\right)\\left(\\left(x^{2} - n\\right) + 2x + 1\\right) \\\\\n& = n + \\left(x^{2} - n\\right)^{2} + 2x\\left(x^{2} - n\\right) + x^{2} - n \\\\\n& = \\left(x^{2} - n\\right)^{2} + 2x\\left(x^{2} - n\\right) + x^{2} \\\\\n& = \\left(x^{2} - n + x\\right)^{2}.\n\\end{aligned}\n$$\nThis is the square of an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x \\geq 4$ is a real number prove that\n$$\n\\sqrt{x}-\\sqrt{x-1} \\geq \\frac{1}{x}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNotice that\n$$\n\\sqrt{x}-\\sqrt{x-1}=\\frac{(\\sqrt{x}-\\sqrt{x-1}) \\cdot (\\sqrt{x}+\\sqrt{x-1})}{\\sqrt{x}+\\sqrt{x-1}}=\\frac{1}{\\sqrt{x}+\\sqrt{x-1}}\n$$\nNow the required inequality is equivalent to $x \\geq \\sqrt{x}+\\sqrt{x-1}$ or after dividing both sides by $\\sqrt{x}$:\n$$\n\\sqrt{x} \\geq 1+\\sqrt{1-\\frac{1}{x}}\n$$\nThe last inequality holds for $x \\geq 4$ because $1+\\sqrt{1-\\frac{1}{x}} \\leq 2 \\leq \\sqrt{x}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46117,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAerith has written down two letters on a piece of paper. Bob will pick a positive integer and she'll try to factor Bob's positive integer into two others, such that when spelled in English, one contains her first letter and the other her second.\nFor example, if Aerith had chosen letters \"v\" and \"w\" and Bob chose \"ten\", she could spell it out as \"five\" times \"two,\" but would fail if Bob chose \"twenty\".\nShow that there is a unique pair of letters Aerith can choose that allows her to succeed no matter what number Bob picks.",
    "options": [],
    "answer": "o and e",
    "solution": "Solution:\n\nWe can narrow down her possible choices quickly:\n- If Bob chose \"one,\" Aerith would have to have to say \"one times one\", so her two letters must both be in the word \"one\".\n- If Bob chose \"three,\" she would have to say \"one times three\", so one of her letters must be in \"three\" and thus must be \"e\".\n- If Bob chose \"four,\" she could not have said \"two times two\" as neither word contains \"e\", so she must have said \"one times four\". Thus her second letter must be \"o\".\nThus, Aerith has letters \"o\" and \"e\".\n\nWe now show that all positive integers can be represented this way. We note that any number containing either one of \"o\" or \"e\" works, because multiplying itself by \"one\" will supply the other letter. As such,\n- any number 1000 or higher works because it must contain \"o\" (\"thousand\", \"million\", \"billion\", ...), and\n- any number from 100 to 999 works, because it must contain \"hundred\",\n- 10 through 19 work, because all of \"ten\", \"eleven\", \"twelve\", and \"thirteen\" through \"nineteen\" contain \"e\",\n- any other number ending in digits other than 0 or 6 works, because all of \"one\", \"two\" \"three\", \"four\" \"five\", \"seven\", \"eight\" and \"nine\" also contain either \"e\" or \"o\",\n- all numbers ending in 6 can be written as \"two\" times a number ending in \"three\", \"eight\", \"thirteen\" or \"eighteen\", all of which contain \"e\",\n- 30, 50, and 60 can be factored as \"two\" times \"fifteen\", \"twentyfive\", or \"twentyfive\", respectively, and\n- the remaining multiples of 10 , i.e., \"twenty\", \"forty\", \"seventy\", \"eighty\" and \"ninety\", all contain either \"e\" or \"o\".\nThis covers all natural numbers, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be an integer greater than $2$. Positive real numbers $x$ and $y$ satisfy $x^{n} = x + 1$ and $y^{n+1} = y^{3} + 1$. Prove that $x < y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt is clear from the given that $x^{n} > 1$ and $y^{n+1} > 1$; therefore $x > 1$, $y > 1$. From this we get\n$$\n0 < (y - 1)(y^{2} - 1) = y^{3} - y^{2} - y + 1 \\Rightarrow y^{2} + y < y^{3} + 1 = y^{n+1}\n$$\nand, dividing by $y$, we obtain $y + 1 < y^{n}$. Thus $y^{n} - y > 1 = x^{n} - x$. Now if $y \\leq x$ then we also have $0 < y^{n-1} - 1 \\leq x^{n-1} - 1$ and so\n$$\ny^{n} - y = y(y^{n-1} - 1) \\leq x(x^{n-1} - 1) = x^{n} - x,\n$$\nwhich is a contradiction; hence we must have $y > x$ as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNeville Nevermiss and Benjamin Baskethound are two players on the Simpson School basketball team. During Season I, Neville made a higher percentage of his attempted baskets than Ben. The same happened in Season II. Prove or disprove: When the statistics of the two seasons are combined, Neville necessarily made a higher percentage of his attempted baskets than Ben.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIt does not follow that Neville made a higher percentage over the two seasons. Consider the following (deliberately extreme) situation. In Season I:\n- Neville attempts $80$ baskets and makes $1$ of them;\n- Ben attempts $20$ baskets and misses them all.\n\nThen Neville obviously has a higher success rate ($1.25\\%$ versus $0\\%$). In Season II:\n- Neville attempts $20$ baskets and makes them all;\n- Ben attempts $80$ baskets and misses $1$.\n\nThen Neville again has the higher success rate ($100\\%$ versus $98.75\\%$). But overall, both players have attempted $100$ baskets, and Ben has made $79\\%$ to Neville's $21\\%$!",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ which satisfy\n$$\nf(x+y)=f(x-y)+4 x y\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x^2 + c for some real constant c",
    "solution": "Solution:\nThe answer is the functions $f(x)=x^{2}+c$, where $c$ is a constant. It is easy to check that all such functions work, since\n$$\n(x+y)^{2}+c=(x-y)^{2}+c+4 x y\n$$\nis trivially true.\n\nNow, we prove these are the only functions. Put $x=y=\\frac{1}{2} a$ to obtain\n$$\nf(a)=f(0)+a^{2}\n$$\nfor all real numbers $a$, which implies the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven the hypotenuse and the difference of the two legs of a right triangle, show how to reconstruct the triangle with ruler and compass.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nHere is one construction. Let $AB = d$ be the segment representing the difference of the two legs. Extend $AB$ to $C$ and raise a perpendicular $BD$ to $AC$ at $B$. Bisect angle $DBC$ to make ray $BE$ with $\\angle EBC = 45^{\\circ}$. Now set the compass to the length $c$ of the hypotenuse and draw a circle $k$ of radius $c$ centered at $A$. Because the hypotenuse of a right triangle is longer than the difference of the legs, $c > AB$, and thus $k$ will intersect ray $BE$ at a point $X$. Drop a perpendicular $XY$ from $X$ to $AC$. Then $\\triangle AXY$ is a right triangle with hypotenuse $AX$, and the difference $AY - XY$ of the legs equals $AY - BY = AB$ since $BXY$ is an isosceles right triangle.\n\nIf there were any other right triangle satisfying the same specifications, we could put it in the position $AX'Y'$ with the smallest angle at $A$, the longer leg $AY'$ on ray $AC$, and $X'$ above $Y'$. Then $X'$ would lie on $k$ because $AX' = c$; also, since $AX' - X'Y' = AB = AY' - BY'$, $BX'Y'$ is an isosceles right triangle, and so $X'$ lies on $\\overrightarrow{BE}$. However, since $B$ is inside $k$, $k$ and $\\overrightarrow{BE}$ can only intersect once. Thus $X' = X$ and $Y' = Y$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(n)$ be the number of digits of a positive integer $n$ (in base 10). Prove that\n\n$$\nf\\left(2^{n}\\right)+f\\left(5^{n}\\right)=n+1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $f\\left(2^{n}\\right)=x$. Then since the smallest number with $x$ digits is $10^{x-1}$ and the largest is $10^{x}-1$, we have\n$$\n10^{x-1} \\leq 2^{n}<10^{x}\n$$\nHowever, a power of $2$ (other than $1$) cannot also be a power of $10$, so the inequality is strict:\n$$\n10^{x-1}<2^{n}<10^{x}.\n$$\nSimilarly, if $f\\left(5^{n}\\right)=y$, then\n$$\n10^{y-1}<5^{n}<10^{y}.\n$$\nMultiplying (1) by [2],\n$$\n\\begin{gathered}\n10^{x-1} \\cdot 10^{y-1}<2^{n} \\cdot 5^{n}<10^{x} \\cdot 10^{y} \\\\\n10^{x+y-2}<10^{n}<10^{x+y} \\\\\nx+y-2<n<x+y .\n\\end{gathered}\n$$\nSince the only integer between $x+y-2$ and $x+y$ is $x+y-1$,\n$$\nn=x+y-1\n$$\nso $x+y=n+1$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $k$ be a positive integer. Prove that there is a positive integer $N$ with the following properties:\n\na. $N$ has $k$ digits, none of which is $0$.\n\nb. No matter how the digits of $N$ are rearranged, the resulting number is not divisible by $13$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $N_{1}$ be the number consisting of $k$ ones. If $N_{1}$ is not divisible by $13$, it is the number $N$ we seek, since no matter how its digits are rearranged, it is still the same number.\n\nIf $N_{1}$ is divisible by $13$, consider $N_{1} + 1$, that is to say the number consisting of $k-1$ ones and one $2$. When the digits of this number are rearranged, the result has the form\n$$\n11 \\cdots 121 \\cdots 11 = 11 \\cdots 111 \\cdots 11 + 10 \\cdots 00 = N_{1} + 10^{k}\n$$\nwhere $k \\geq 0$ is an integer. This is not divisible by $13$ since $N_{1}$ is divisible by $13$ and $10^{k}$ is not.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine an $n$-staircase to be the union of all squares of an $n \\times n$ grid lying on or below its main diagonal. How many ways are there to divide a 10-staircase into 10 rectangles, each having a side of length 1? (Reflections are not included.)",
    "options": [],
    "answer": "256",
    "solution": "Solution:\n\nA 10-staircase has 10 \"upper right corners\" $P$, each of which must be the upper right corner of some rectangle, and no two of which can belong to the same rectangle. It also has a single lower left corner $Q$ which must belong to the same rectangle as one of the ten points $P$. Since this rectangle has one side of length 1, it must be a $10 \\times 1$ rectangle placed either vertically or horizontally along the long side of the staircase. The remainder of the figure is then a 9-staircase to be filled with 9 rectangles.\n\nWe can then repeat the argument to find that one of the long sides of the 9-staircase must be filled by a $9 \\times 1$ rectangle, leaving an 8-staircase. This continues until we reach the 1-staircase, a single square, which can be filled in only one way.\n\n![](attached_image_1.png)\n\nThe placement of the $10 \\times 1$ rectangle is irrelevant because of the symmetry of the shape. But the $9 \\times 1$ through $2 \\times 1$ rectangles each involve a choice between two alternatives, so the number of tilings is $2^{8} = 256$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCheryl chooses a word in this problem and tells its first letter to Aerith and its last letter to Bob. The following conversation ensues over a series of emails:\nAerith: \"I don't know her word, do you?\"\nBob: \"No, in fact, I don't know if we can ever figure out what her word is without having more information.\"\nAerith: \"Then I do know what it is!\"\nBob: \"Now I also do.\"\nWhat is Cheryl's word?",
    "options": [],
    "answer": "then",
    "solution": "Solution:\nWe will process their conversation message by message.\n\"I don't know her word, do you?\"\nAerith would know Cheryl's word if the first letter was unique, so Aerith does not have any of $b$ (Bob), $k$ (know), $m$ (more), $p$ (problem), $s$ (series), $y$ (you).\n\n\"No, in fact, I don't know if we can ever figure out what her word is without having more information.\"\nIf they can't ever figure out what Cheryl's word is, even if they just outright told each other their letters, it would be because Cheryl's word is not identifiable from these letters alone. The unidentifiable words are: can/conversation, Cheryl's/chooses, emails/ensues, fact/first, in/information, is/its, tells/this, what/without.\nBob's statement is then equivalent to the claim \"The word could be in this list but does not have to be.\" The only last letters which satisfy this property are $n$ and $t$. While some words in the list end in $s$, $s$ does not satisfy the property because no words not in the list end in $s$. (Series was the only one but it was eliminated by Aerith's statement.)\n\n\"Then I do know what it is!\"\nIn order for Aerith to make this claim, there must be only one possible word beginning with her letter that ends with $n$ or $t$. Her letter must then be one of $d$ (don't), $l$ (last), $o$ (out), $t$ (then).\n\n\"Now I also do.\"\nFor Bob to know Cheryl's word, the last letter must be unique among the remaining options. Inspecting we see that the word must then be \"then\".",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A$ and $B$ be two points on the plane with $AB = 7$. What is the set of points $P$ such that $PA^2 = PB^2 - 7$?",
    "options": [],
    "answer": "The line perpendicular to AB through the point on AB that is 3 units from A and 4 units from B (in coordinates with A at the origin and B at seven on the horizontal axis, this is the line x = 3).",
    "solution": "Solution:\n\nIf we let $K$ be the point on $AB$ with $AK = 4$, $BK = 3$, then the answer is the line through $K$ perpendicular to $AB$. To see this, set $A = (0, 0)$ and $B = (7, 0)$. Then the points $P = (x, y)$ are exactly those satisfying\n$$\n(x - 0)^2 + (y - 0)^2 = (x - 7)^2 + (y - 0)^2 - 7\n$$\nwhich rearranges to $9 = -14x + 42$, i.e., $x = 3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArt and Ben play a game while sharing an $m \\times n$ chocolate bar. They take turns breaking the bar into two rectangular pieces along one of the lines and eating the smaller piece. (If the two pieces are equal, they can choose which piece to eat.) Whoever is left with the last $1 \\times 1$ square of chocolate loses. If Art moves first, describe all pairs $(m, n)$ for which Ben has a winning strategy.",
    "options": [],
    "answer": "All pairs with (m+1)/(n+1) equal to an integer power of two; equivalently, (a−1, 2^k a−1) or (2^k a−1, a−1) for integers a ≥ 2 and k ≥ 0.",
    "solution": "Solution:\n\nThe answer is all pairs $(m, n)$ such that the ratio\n$$\nR=\\frac{m+1}{n+1}\n$$\nis an integer power of two, that is, the pairs $\\left(a-1,2^{k} a-1\\right)$ and $\\left(2^{k} a-1, a-1\\right)$ for $a \\geq 2$ and $k \\geq 0$. We will show that\n\na. If $R$ is a power of two, then it will not remain so after one move;\n\nb. If $R$ is not a power of two, then some move will make it one.\n\nTo prove (a), simply note that at each move, one of the dimensions is decreased at most by half, so either $m+1$ or $n+1$ is diminished by a factor strictly between 1 and 2. Thus the new value of $R$ is sandwiched between two powers of two: $R / 2$ and $R$ in the first case, $R$ and $2 R$ in the second.\n\nAs for (b), assume $m \\geq n$ (the other case is symmetric) so $2^{k}<R<2^{k+1}$ for some $k \\geq 0$. We have\n$$\n2^{k}(n+1)<m+1<2^{k+1}(n+1) \\text{.}\n$$\nThese inequalities imply that one can move to a bar that is $\\left[2^{k}(n+1)-1\\right] \\times(n+1)$, with $R$-value $2^{k}$.\n\nIf $R$ is initially a power of two, Ben wins by making it that way at each move; otherwise Art wins by the same strategy (in either case ending at the $1 \\times 1$ square, where $R=1$ ).",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn each cell of $n \\times n$ table there is an arrow pointing in one of eight principal directions (i.e. one of the following: $\\rightarrow, \\leftarrow, \\uparrow, \\downarrow, \\nearrow, \\searrow, \\nwarrow, \\swarrow$ ), in such a way that the arrows in any two adjacent cells form an angle of no more than 45 degrees, and the arrows in any two cells adjacent diagonally (i.e. sharing a vertex) form an angle of no more than 90 degrees. We start at some cell and follow the arrows (e.g. if the arrow in our cell is $\\uparrow$ we move up by one, if it is $\\searrow$ we move diagonally down and to the right, etc.). Prove that we will eventually escape from the table (i.e. will reach a cell on the boundary of the table where the arrow will point \"out of the table\").",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose that there exists such a table of arrows and a cell in it so that we can follow the arrows without ever escaping the table. Then, since the number of cells in the table is finite, we must at some point return to a cell we have already visited, and then continue in a loop. Let's rotate all the arrows in our original $n \\times n$ table by 45 degrees, so that the arrows in the cells along our loop point into the loop after the rotation. Note that because the adjacent arrows of the loop are at no more than 90 degrees, the rotated arrows will indeed point to the cells inside of the loop. The new table of (rotated) arrows still has the property that any two adjacent cells form an angle of no more than 45 degrees, and the arrows in any two cells adjacent diagonally form an angle of no more than 90 degrees. If we now start at some point on the loop and follow the (rotated) arrows, then since all the arrows on the boundary of the loop point inward, we will never escape from inside of the loop, and so there is going to be a yet smaller loop inside of our first loop. We can now repeat the process, obtaining smaller and smaller loops along the way. But the number of cells inside the loop (counting those on the loop itself) is an integer, and so it can not decrease indefinitely. This contradiction shows that one will always escape from the table.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a square table $n \\times n$, two players $A$ and $B$ are playing the following game: At the beginning all cells of the table are empty, and the players alternate playing with coins. Player $A$ has the first move, and in each of the moves a player will put a coin on some of the cells that doesn't contain a coin and is not adjacent to any of the cells that already contains a coin. The player who makes the last move is the winner. Which player has a winning strategy and what is the strategy?\n\nRemark. The cells are adjacent if they share an edge.",
    "options": [],
    "answer": "If the side length is even, the second player wins by always placing a coin on the square centrally symmetric to the opponent’s last move. If the side length is odd, the first player wins by first placing a coin on the center square and then always responding on the centrally symmetric square to the opponent’s last move.",
    "solution": "Solution:\n\nIf $n$ is even, $B$ has the winning strategy - he just needs to put the coin in the centrally symmetric cell to the cell where $A$ has put the coin in his previous move. Following this strategy, $B$ will always have a free cell to put his coin, and thus he can never lose. Since the game has to end, $A$ is the loser.\n\nIf $n$ is odd, $A$ has a winning strategy: first he needs to put a coin in the central cell, and after that he puts the coin in the centrally symmetric cell to the cell where $B$ has put his coin.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA circle is inscribed in $\\triangle ABC$ that touches side $BC$ at $D$, side $AC$ at $E$, and side $AB$ at $F$. Show that $\\triangle DEF$ must be acute.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $AE$ and $AF$ are tangents from the same point to the same circle, we must have $AE = AF$. Thus, $\\triangle AEF$ is isosceles, so\n$$\n\\angle AEF = \\angle AFE = \\frac{180^\\circ - \\angle A}{2} = 90^\\circ - \\frac{1}{2} \\angle A.\n$$\nNow, since $AE$ is a tangent to the circle, $\\angle EDF = \\angle AEF = 90^\\circ - \\frac{1}{2} \\angle A$. Thus, $\\angle EDF < 90^\\circ$. Similarly, the other two angles must also be less than $90^\\circ$, so the triangle is acute.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of triangulations of a general convex $7$-gon into $5$ triangles by $4$ diagonals that do not intersect in their interiors.",
    "options": [],
    "answer": "42",
    "solution": "Solution:\nDefine the Catalan numbers by $C(n) = \\frac{1}{n+1} \\binom{2n}{n}$. The current solution is the $C($ number of triangles $) = C(5) = 42$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMr. Fat moves around on the lattice points according to the following rules: From point $(x, y)$ he may move to any of the points $(y, x)$, $(3x, -2y)$, $(-2x, 3y)$, $(x+1, y+4)$ and $(x-1, y-4)$. Show that if he starts at $(0,1)$ he can never get to $(0,0)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nObserve that for each of Mr. Fat's moves, the value of $x + y \\pmod{5}$ is invariant. Therefore, Mr. Fat can never reach $(0,0)$ from $(0,1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIs there an integer $x$ such that\n$$\n2010 + 2009 x + 2008 x^{2} + 2007 x^{3} + \\cdots + 2 x^{2008} + x^{2009} = 0?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Solution:\nThe answer is no.\n\nIt is clear that if $x$ is positive, the left side is positive, and if $x = 0$, the left side is $2010$. If $x = -1$, the left side is\n$$\n(2010 - 2009) + (2008 - 2007) + \\cdots + (2 - 1) = 1 + 1 + \\cdots + 1,\n$$\nlikewise a positive number.\n\nIf $x \\leq -2$, we claim that the left side is negative. Pair the terms again and factor:\n$$\n(2010 + 2009 x) + x^{2}(2008 + 2007 x) + \\cdots + x^{2006}(4 + 3 x) + x^{2008}(2 + x).\n$$\nEach of the binomials in parentheses has the form $(a + 1) + a x$, with $a \\geq 1$, and its value is at most\n$$\n(a + 1) + a(-2) = 1 - a \\leq 0.\n$$\nMoreover, only the last binomial, for $a = 1$, is capable of equaling $0$; the others are strictly negative. The coefficients $1, x^{2}, x^{4}, \\ldots$ of these binomials are of course positive, yielding a negative sum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind at least one non-zero polynomial $P(x, y, z)$ such that $P(a, b, c)=0$ for every three real numbers that satisfy $\\sqrt[3]{a}+\\sqrt[3]{b}=\\sqrt[3]{c}$.",
    "options": [],
    "answer": "27xyz - (z - x - y)^3",
    "solution": "Solution:\nCube both sides of the condition $\\sqrt[3]{x}+\\sqrt[3]{y}=\\sqrt[3]{z}$:\n$$\n\\begin{gathered}\nx+3 \\sqrt[3]{x} \\sqrt[3]{x} \\sqrt[3]{y}+3 \\sqrt[3]{x} \\sqrt[3]{y} \\sqrt[3]{y}+y=z \\\\\n3 \\sqrt[3]{x} \\sqrt[3]{x} \\sqrt[3]{y}+\\sqrt[3]{x} \\sqrt[3]{y} \\sqrt[3]{y}=z-x-y \\\\\n3 \\sqrt[3]{x} \\sqrt[3]{y}(\\sqrt[3]{x}+\\sqrt[3]{y})=z-x-y\n\\end{gathered}\n$$\nBut since $\\sqrt[3]{x}+\\sqrt[3]{y}=\\sqrt[3]{z}$\n$$\n\\begin{gathered}\n3 \\sqrt[3]{x} \\sqrt[3]{y} \\sqrt[3]{z}=z-x-y \\\\\n27 x y z=(z-x-y)^{3}\n\\end{gathered}\n$$\nHence $P(x, y, z)=27 x y z-(z-x-y)^{3}$ is one such polynomial.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are $m$ friends with $n$ cupcakes each weighing 1 ounce. They wish to split the cupcakes equally by dividing each cupcake into some number of parts, and allocating some parts to each person.\n\na) Assume $m=3$ and $n=5$. Show that they may divide the cupcakes with all pieces being larger than $\\frac{1}{3}$ ounces.\n\nb) Assume $m=5$ and $n=3$. Show that they may divide the cupcakes with all pieces being larger than $\\frac{1}{5}$ ounces.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nEach person needs $\\frac{5}{3} = \\frac{20}{12}$ of a cupcake. If they cut four of the cupcakes into $\\frac{5}{12}$ and $\\frac{7}{12}$ and the last cupcake in half, then one person can take the four $\\frac{5}{12}$ pieces, giving them\n$$\n4 \\cdot \\frac{5}{12} = \\frac{20}{12}\n$$\nand the other two can each take two $\\frac{7}{12}$ pieces and one $\\frac{1}{2} = \\frac{6}{12}$ piece, giving them as well\n$$\n\\frac{7}{12} + \\frac{7}{12} + \\frac{6}{12} = \\frac{20}{12}\n$$\nAll pieces are larger than $\\frac{1}{3}$ ounces.\n\nb.\nEach person needs $\\frac{3}{5} = \\frac{12}{20}$ of a cupcake. We can divide two of the cupcakes into $\\frac{6}{20}$, $\\frac{7}{20}$ and $\\frac{7}{20}$ pieces and the last cupcake into four $\\frac{1}{4} = \\frac{5}{20}$ pieces. Then four of the people can get one $\\frac{7}{20}$ piece and one $\\frac{5}{20}$ piece and the fifth person can get the two $\\frac{6}{20}$ pieces.\n\nAll pieces are larger than $\\frac{1}{5}$ ounces.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many roots does $\\arctan x = x^{2} - 1.6$ have, where the arctan function is defined in the range $-\\frac{\\pi}{2} < \\arctan x < \\frac{\\pi}{2}$?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\n2 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf two fair dice are tossed, what is the probability that their sum is divisible by $5$?",
    "options": [],
    "answer": "1/4",
    "solution": "Solution:\n\n$\\frac{1}{4}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCount the number of permutations $a_{1} a_{2} \\ldots a_{7}$ of $1234567$ with longest decreasing subsequence of length at most two (i.e. there does not exist $i<j<k$ such that $a_{i}>a_{j}>a_{k}$ ).",
    "options": [],
    "answer": "429",
    "solution": "Solution:\n\n$C(7) = 429$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA line of soldiers 1 mile long is jogging. The drill sergeant, in a car, moving at twice their speed, repeatedly drives from the back of the line to the front of the line and back again. When each soldier has marched 15 miles, how much mileage has been added to the car, to the nearest mile?",
    "options": [],
    "answer": "30",
    "solution": "Solution:\n30 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all the values of $m$ for which the zeros of $2x^{2} - m x - 8$ differ by $m-1$.",
    "options": [],
    "answer": "6, -10/3",
    "solution": "Solution:\n6, $-\\frac{10}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest integer that divides $m^{5}-5 m^{3}+4 m$ for all $m \\geq 5$.",
    "options": [],
    "answer": "120",
    "solution": "Solution: $.5!=120$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCount the number of sequences $1 \\leq a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{5}$ of integers with $a_{i} \\leq i$ for all $i$.",
    "options": [],
    "answer": "42",
    "solution": "Solution:\n\n$C$ (number of terms) $= C(5) = 42$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlex and Bob have $30$ matches. Alex picks up somewhere between one and six matches (inclusive), then Bob picks up somewhere between one and six matches, and so on. The player who picks up the last match wins. How many matches should Alex pick up at the beginning to guarantee that he will be able to win?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\n$2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe cafeteria in a certain laboratory is open from noon until 2 in the afternoon every Monday for lunch. Two professors eat 15 minute lunches sometime between noon and 2. What is the probability that they are in the cafeteria simultaneously on any given Monday?",
    "options": [],
    "answer": "15/64",
    "solution": "Solution:\n\n$\\frac{15}{64}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the remainder when $2^{2001}$ is divided by $2^{7}-1$?",
    "options": [],
    "answer": "64",
    "solution": "Solution:\n$2^{2001 \\bmod 7} = 2^{6} = 64$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are 1000 cities in the country of Euleria, and some pairs of cities are linked by dirt roads. It is possible to get from any city to any other city by traveling along these roads. Prove that the government of Euleria may pave some of the roads so that every city will have an odd number of paved roads leading out of it.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCall a city \"even\" or \"odd\" according to whether the number of paved roads coming out of it is even or odd. Note the following.\n\nLemma: No matter which roads are paved, there will be an even (possibly zero) number of even cities.\n\nTo see why this is true, let $d_{i}$ be the number of paved roads leading out of city $i$. The sum\n$$\nd_{1}+d_{2}+\\cdots+d_{1000}\n$$\ncounts each paved road exactly twice, and hence will be an even number (this is known as the \"Handshake Lemma\"). If there were an odd number of even cities, then there would be an odd number of odd cities (since there are 1000 cities and 1000 is even). But then the sum above would be odd, a contradiction.\n\nUsing this lemma, we can create an algorithm which will eventually pave the roads so that all cities are odd.\n\n1. Start by paving all the roads. If each city is odd, we are done.\n\n2. Otherwise, find an even city $x$. By the lemma, there will be at least one other even city, $y$. Consider the path joining $x$ and $y$. Change the \"state\" of all the roads in this path (in other words, if a road is paved, unpave it; if it is dirt, pave it). This procedure changes the parity of $x$ and $y$ (i.e., changes them from even to odd), but does not alter the parity of any other city in Euleria, because if $z$ is a city on the path from $x$ to $y$, both the road going into $z$ (from the $x$-direction) and the road leaving it (heading towards $y$) will have changed.\n\n3. Step 2 thus reduces the number of even cities by 2. Repeat this step as much as needed until the number of even cities is zero.\nSolution:\n\nWe will think of Euleria as a connected graph (a network of vertices joined by edges), where each city is a vertex and each dirt road joining two cities is an edge. Number the vertices $1,2, \\ldots, 1000$. For each $i=1,2, \\ldots, 500$, consider any path from vertex $2i-1$ to vertex $2i$ (we know such a path exists, since the graph is connected), and place a mark on each edge used in the path. Then pave all the roads (edges) that have an odd number of marks. To show that this scheme meets the requirements, it is enough to prove that, for each vertex $v$, the edges incident to it have an odd total number of marks, since the number of such edges with an odd number of marks will then be odd. But consider all the occurrences of $v$ in any of our 500 paths. Notice that $v$ is an endpoint of one such path, which therefore contributes one mark on an edge incident to $v$; any other occurrence of $v$ is internal to a path, which therefore contributes two marks, one on the edge leading into $v$ and another on the edge leading out of it. It follows that the total number of marks on edges incident to $v$ is odd. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $\\triangle ABC$ be a triangle such that $\\angle A = 90^\\circ$. Determine whether it is possible to partition $\\triangle ABC$ into 2006 smaller triangles in such a way that\n\n$1^\\circ$ Each triangle in the partition is similar to $\\triangle ABC$;\n\n$2^\\circ$ No two triangles in the partition have the same area.\n\nExplain your answer!",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe required partition is always possible. We consider 2 cases:\n\n(a) The triangle is not isosceles - First we construct the perpendicular from the vertex of the right angle. The triangle is divided into two similar, but non-congruent triangles. Now we divide the smaller triangle, and keep going until the total number of triangles becomes 2006.\n\n(b) The triangle is isosceles - Repeat the procedure from the previous problem until we get 2001 triangles. Then we divide one of the smallest triangles into 6 smaller and noncongruent triangles as shown in the second picture.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle, and let $M$ and $N$ be the respective midpoints of $AB$ and $AC$. Suppose that\n$$\n\\frac{CM}{AC} = \\frac{\\sqrt{3}}{2}\n$$\nProve that\n$$\n\\frac{BN}{AB} = \\frac{\\sqrt{3}}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $L$ be the midpoint of $BC$. Let $BC = 2a$, $AC = 2b$, and $AB = 2c$. Applying the parallelogram law to $CLMN$ gives\n$$\nCM^{2} + c^{2} = a^{2} + b^{2} + a^{2} + b^{2}\n$$\nor\n$$\nCM^{2} = 2a^{2} + 2b^{2} - c^{2}.\n$$\nSquaring both sides of the given equation and substituting yields\n$$\n\\frac{2a^{2} + 2b^{2} - c^{2}}{4b^{2}} = \\frac{3}{4}\n$$\nwhich simplifies to\n$$\nb^{2} + c^{2} = 2a^{2}\n$$\nThis equation is equivalent to $CM / AC = \\sqrt{3} / 2$. Because it is symmetric in $b$ and $c$, we conclude that it is also equivalent to $BN / AB = \\sqrt{3} / 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. Prove that there exist distinct positive integers $x$, $y$, $z$ such that\n$$\nx^{n-1} + y^n = z^{n+1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOne solution is\n$$\nx = 2^{n^2} 3^{n+1}, \\quad y = 2^{n^2 - n} 3^n, \\quad z = 2^{n^2 - 2n + 2} 3^{n-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a triangle $A B C$, a circle $k$ is tangent to the lines $A B$ and $A C$ at $B$ and $P$. Let $H$ be the foot of perpendicular from the center $O$ of $k$ to $B C$, and let $T$ be the intersection point of $O H$ and $B P$. Prove that $A T$ bisects the segment $B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $X$ be the intersection of $A T$ and $B C$, $S$ the intersection of $B C$ and $A O$, and $Y$ the intersection of $B P$ and $A O$. Since $B Y \\perp S O$ and $O H \\perp B S$, $T$ is the orthocenter of $\\triangle B O S$ and $S T \\perp B O$, hence $S T \\parallel A B$. Thus\n$$\nS T : A B = T X : X A \\text{.}\n$$\nHowever since $A B = A P$ and the triangles $S Y T$ and $A Y P$ are similar we get\n$$\nS T : A P = T Y : Y P .\n$$\nFrom (1) and (2) we now get $X Y \\parallel A P$ and since $Y$ is the midpoint of $B P$, $X Y$ is the middle line of $\\triangle C B P$ and $B X = X C$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a convex quadrilateral such that $\\angle ABD = \\angle ACD$. Prove that $ABCD$ can be inscribed in a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThere are many ways to structure the proof. The following method seems to have minimal logical difficulties.\nBecause points $A$, $B$, and $C$ are not collinear, we can draw the circumscribed circle $\\omega$ of $\\triangle ABC$. The arc $AC$ of $\\omega$, not containing $B$, is intercepted by inscribed angle $ABC$ and thus has measure $2 \\angle ABC$. On this arc we may find a point $E$ such that $AE$ has the smaller measure $2 \\angle ABD$. Then angles $ABD$ and $ABE$ have the same measure and orientation, so $E$ is on $BD$; also, angles $ACD$ and $ACE$ have the same measure and orientation, so $E$ is on $CD$. Since lines $BD$ and $CD$ have only one point in common, $D=E$ and thus $D$ lies on the circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a parallelogram. Suppose that the circumcenter of $\\triangle ABC$ lies on diagonal $BD$. Prove that $ABCD$ is either a rectangle or a rhombus (or both).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTo get a conclusion of the appropriate type (a rectangle OR a rhombus), we must divide up the problem into two cases. Here is one way of accomplishing this:\n\nCase 1. The circumcenter $O$ of $ABC$ is the center of $ABCD$, the common midpoint of diagonals $AC$ and $BD$. Then since radii $OA$ and $OB$ are equal, we get $AC = 2OA = 2OB = BD$. Thus $ABCD$ is a parallelogram whose diagonals are congruent, i.e. a rectangle.\n\nCase 2. The circumcenter $O$ of $ABC$ does not coincide with the midpoint $M$ of $AC$ and $BD$. Then since $O$ is on the perpendicular bisector of $AC$, we have $OM \\perp AC$. But $O$ and $M$ are both on line $BD$, so $BD \\perp AC$. Thus $ABCD$ is a parallelogram whose diagonals are perpendicular, i.e. a rhombus.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a positive integer one-full if it satisfies the following criteria:\n\na. Every digit is either $0$, $1$, or $2$.\n\nb. Out of every two consecutive digits, at least one of them is a $1$.\n\nFor $n \\geq 2$, calculate the number of $n$-digit one-full numbers. (Numbers beginning with the digit $0$ are not allowed.)",
    "options": [],
    "answer": "2^n",
    "solution": "Solution:\n\nThe answer is $2^{n}$. It is clear that there are four $2$-digit one-full numbers: $10$, $11$, $12$, and $21$.\n\nTo prove that the required number is $2^{n}$ for all $n \\geq 3$, it is enough to show that there are twice as many $n$-digit one-full numbers as $(n-1)$-digit one-full numbers for every $n$.\n\nGiven an $(n-1)$-digit one-full number, we construct two different $n$-digit one-full numbers via the following rules:\n\n- If the first digit is $1$, we replace it with $11$ and then with $21$.\n- If the first digit is $2$, we replace it with $12$ and then with $10$.\n\nSince $2$'s and $0$'s have the same function in the middle of a number, the two numbers that we get using this process are both one-full. Moreover, since any $n$-digit one-full number must begin with one of the four $2$-digit one-full numbers, we can get all $n$-digit one-full numbers in this way. Thus there are twice as many $n$-digit one-full numbers as $(n-1)$-digit ones.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $\\angle ABC = 90^{\\circ}$. Points $D$ and $E$ on $AC$ and $BC$ respectively satisfy $BD \\perp AC$ and $DE \\perp BC$. The circumcircle of $\\triangle CDE$ intersects $AE$ at two points, $E$ and $F$. Prove that $BF \\perp AE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy Power of a Point,\n$$\nAF \\cdot AE = AD \\cdot AC\n$$\nbecause $\\triangle ABC$ is right,\n$$\nAD \\cdot AC = AB^2\n$$\nCombining,\n$$\nAF \\cdot AE = AB^2.\n$$\nOn the other hand, if $F'$ is the foot of the altitude from $B$ to $AE$, then\n$$\nAF' \\cdot AE = AB^2\n$$\nConsequently $AF' = AF$ and $F' = F$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all pairs $(a, b)$ of positive integers such that\n$$\n1 + 5^{a} = 6^{b}.\n$$",
    "options": [],
    "answer": "(1, 1)",
    "solution": "Solution:\nThe only solution is $(1, 1)$.\n\nIt is clear that if $b = 1$ then $a = 1$, and that $(1, 1)$ is a solution. Consequently, assume $b > 1$. Then $6^{b}$ is divisible by $4$. On the other hand, since $5^{a} \\equiv 1^{a} = 1 \\pmod{4}$ for all $a$, the left side is $2 \\pmod{4}$. Thus there are no solutions for $b > 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $F$ be a point, $d$ be a line, and let $\\mathcal{P}$ be the parabola with focus $F$ and directrix $d$, i.e. the set of points which are equidistant from $F$ and $d$. Show that there is a line $\\ell$ such that for any point $A$ on $\\mathcal{P}$, the circle with diameter $A F$ is tangent to $\\ell$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $\\ell$ be the line parallel to $d$ equidistant from $F$ and $d$,\n$V$ be the vertex of $\\mathcal{P}$, i.e., the foot of the altitude from $F$ to $\\ell$,\n$X$ be the foot of the altitude from $A$ to $d$,\n$O$ be the midpoint of $\\overline{F A}$,\n$B$ be the foot of the altitude from $A$ to $\\ell$, and\n$M$ be the midpoint of $\\overline{B V}$, i.e., the foot of the altitude from $O$ to $\\ell$.\n\n![](attached_image_1.png)\n\nWe have $A B + X B = A X$. By definition of $\\mathcal{P}$, $A X = A F$, and by definition of $\\ell$, $X B = F V$, so $A B + F V = A F$. Since $O M$ is the average of $F V$ and $A B$, this means that $O M = A F / 2$. Thus the circle with diameter $A F$ has center $O$ and radius $O M$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven ten points in the plane, show that it is always possible to cover all of them with non-overlapping unit circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose we place the circles in a hexagonal tiling so they are all tangent to each other. Then each circle has area $\\pi$, and each circle is inscribed in a hexagon. In each hexagon, the distance from the center to each side is $1$. Thus, each of the six equilateral triangles making up the hexagon has height $1$ and thus side length $2 / \\sqrt{3}$. Therefore, the triangle's area is\n$$\n\\frac{1}{2} \\cdot 1 \\cdot \\frac{2}{\\sqrt{3}} = \\frac{1}{\\sqrt{3}}\n$$\nSince the hexagon is made of six triangles, it has area $6 \\cdot 1 / \\sqrt{3} = 2 \\sqrt{3}$. Thus, the portion of the plane covered by circles is the area of each circle divided by the area of the hexagon it is inscribed in, or\n$$\n\\frac{\\pi}{2 \\sqrt{3}} \\approx 0.907\n$$\nSo, if we use unit circles in a hexagonal tiling with a randomly chosen center, each point has a $90.7\\%$ chance of being covered by a circle, so the expected number of our $10$ points that are covered is $9.07 > 9$. Since every tiling covers a whole number of points and the expected number covered is more than $9$, there must be some tiling that covers at least $10$ points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo villages, $A$ and $B$, lie on opposite sides of a straight river in the positions shown. There will be a market in village $B$, and residents of village $A$ wish to attend. The people of village $A$ would like to build a bridge across and perpendicular to the river so that the total route walked by residents of $A$ to the bridge, across the bridge, and onward to $B$ is as short as possible. How can they find the exact location of the bridge, and how long will the route be?",
    "options": [],
    "answer": "Place the bridge where the line from village A to the point obtained by translating village B perpendicular to the river by the river’s width meets the bank; the minimal total route length is 6 km.",
    "solution": "Solution:\n\nFor convenience, we label the point $X$, the foot of the perpendicular from $A$ to the river. We also label $Y$, the point where the perpendicular $B Y$ to the river through $B$ meets the parallel $A Y$ to the river through $A$. The key construction is to translate $B$ southward $1 \\mathrm{~km}$ to point $E$. Let $A C D B$ be any route. Then segments $D C$ and $B E$ are congruent and parallel,\n\n![](attached_image_1.png)\n\nmaking a parallelogram $D C E B$. The total route is $A C + C D + D B = A C + C E + E B$. But by the triangle inequality, $A C + C E \\geq A E$ so the total route cannot be shorter than $A E + E B = 5 + 1 = 6 \\mathrm{~km}$. (continued)\n\nThe residents of $A$ can ensure a 6-kilometer route by putting the start $C$ of the bridge at the point where $A E$ crosses the lower bank of the river. In this situation, since $A$, $C$, and $E$ lie on a line, the triangle inequality is an equality, so the total route is $A C + C D + D B = A C + C E + E B = A E + E B = 5 + 1 = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers $1, 2, \\ldots, 50$ are written on a blackboard. We may erase two numbers $a$ and $b$, and replace both with $a + b + 2 a b$; we repeat this operation until only one number remains. Prove that the value of this last number does not depend on how the operations were performed.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose at some point the numbers on the board are $x_{1}, \\ldots, x_{k}$. We claim that the quantity $\\left(2 x_{1}+1\\right)\\left(2 x_{2}+1\\right) \\ldots\\left(2 x_{k}+1\\right)$ does not change. Indeed, this follows from the identity $1+2(a+b+2 a b)=(1+2 a)(1+2 b)$.\n\nThus, if there is exactly one number $M$ on the board, it is given exactly by $2 M+1=3 \\cdot 5 \\cdots 101$, and in particular does not depend on the choice of operations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with incenter $I$. A line through $I$ parallel to $BC$ intersects sides $AB$ and $AC$ at $D$ and $E$ respectively. Prove that the perimeter of $\\triangle ADE$ is equal to $AB + AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBecause $DE \\parallel BC$, $\\angle BID = \\angle IBC$ which is the same as $\\angle DBI$ since $I$ is on the bisector of $\\angle ABC$. Thus $\\triangle BDI$ is isosceles, implying $BD = DI$. Similarly $CE = EI$. Thus the perimeter of $\\triangle ADE$ is\n$$\nAD + AE + DE = AD + AE + DI + EI = AD + AE + BD + CE = AB + AC\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWally has a very unusual combination lock number. It has five digits, all different, and is divisible by $111$. If he removes the middle digit and replaces it at the end, the result is a larger number that is still divisible by $111$. If he removes the digit that is now in the middle and replaces it at the end, the result is a still larger number that is still divisible by $111$. What is Wally's combination lock number? Explain your answer!",
    "options": [],
    "answer": "74259",
    "solution": "Solution:\n\nThe solution is $74259$. The numbers $74259$, $74592$, and $74925$ are all divisible by $111$. Denote the original number by $\\overline{abcde}$ (the line prevents confusion with $a \\cdot b \\cdot c \\cdot d \\cdot e$). Then we have\n$$\n\\begin{aligned}\n& 111 \\mid \\overline{abcde} \\\\\n& 111 \\mid \\overline{abdec}\n\\end{aligned}\n$$\nSubtracting,\n$$\n\\begin{gathered}\n111 \\mid \\overline{abdec} - \\overline{abcde} \\\\\n111 \\mid \\overline{dec} - \\overline{cde}\n\\end{gathered}\n$$\nSince the number $\\overline{dec} - \\overline{cde} = 90d + 9e - 99c$ is divisible by $9$ we get $33 \\mid \\overline{dec} - \\overline{cde}$. Since it is given that $\\overline{abdec} > \\overline{abcde}$, $\\overline{cde} \\leq \\overline{dec} - 333$.\nHowever, we could have done the above logic with $d, e, c$ instead of $c, d, e$ and gotten $\\overline{dec} \\leq \\overline{ecd} - 333$. Consequently $\\overline{cde} \\leq \\overline{ecd} - 666$. Since obviously $\\overline{ecd} \\leq 999$, $\\overline{cde}$ is one of the multiples of $37$ up to $333$:\n$$\n000\\quad 037\\quad 074\\quad 111\\quad 148\\quad 185\\quad 222\\quad 259\\quad 296\\quad 333\n$$\nWe can immediately eliminate $000, 111, 222$, and $333$, since we know that the digits are all different. We can also eliminate $074, 185$, and $296$, since $\\overline{ecd} > \\overline{dec}$. The three remaining choices are all of the form $111 \\cdot k + 37$. So\n$$\n\\begin{aligned}\n111 \\mid \\overline{abcde} & \\\\\n& = \\overline{ab000} + \\overline{cde} \\\\\n& = 999 \\cdot \\overline{ab} + \\overline{ab} + 111 \\cdot k + 37 \\\\\n111 \\mid \\overline{ab} + 37 &\n\\end{aligned}\n$$\nwhence $\\overline{ab} = 74$. This leaves only three possibilities for Wally's combination lock number: $74037$, $74148$, and $74259$, of which only the last has all unlike digits.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $z = \\cos \\dfrac{2\\pi}{n} + i \\sin \\dfrac{2\\pi}{n}$ where $n$ is a positive odd integer. Prove that\n$$\n\\frac{1}{1+z} + \\frac{1}{1+z^{2}} + \\frac{1}{1+z^{3}} + \\cdots + \\frac{1}{1+z^{n}} = \\frac{n}{2}\n$$",
    "options": [],
    "answer": "n/2",
    "solution": "Solution:\n\nConsider the polynomial $P(x) = x^{n} + (x-1)^{n}$. We claim that for each integer $i$, $x = 1/(1+z^{i})$ is a root. Indeed, we have $(-z^{i})^{n} = (-1)^{n} z^{i n} = -1$ (since $z^{n} = 1$ and $n$ is odd) and so\n$$\nx^{n} = \\frac{1}{(1+z^{i})^{n}} ; \\quad (x-1)^{n} = \\left(\\frac{-z^{i}}{1+z^{i}}\\right)^{n} = \\frac{-1}{(1+z^{i})^{n}} \\Rightarrow x^{n} + (x-1)^{n} = 0 .\n$$\nMoreover, since the values of $z^{i}$ are distinct for $i = 1, 2, \\ldots, n$, we have found $n$ different roots of this polynomial which has degree $n$, so by the Fundamental Theorem of Algebra, they are all the roots and we can write\n$$\nP(x) = c\\left(x - \\frac{1}{1+z}\\right)\\left(x - \\frac{1}{1+z^{2}}\\right) \\cdots \\left(x - \\frac{1}{1+z^{n}}\\right) \\text{ for some constant } c \\text{. }\n$$\nIf we expand the expression on the right side, the coefficient of $x^{n}$ is $c$ and the coefficient of $x^{n-1}$ is $-c\\left[1/(1+z) + 1/(1+z^{2}) + \\cdots + 1/(1+z^{n})\\right]$. On the other hand, on the left side, the coefficient of $x^{n}$ is $2$ and the coefficient of $x^{n-1}$ is $-n$ (found by expanding $(x-1)^{n}$ using the binomial theorem). Equating coefficients, we get $c = 2$ and thus $-2\\left[1/(1+z) + 1/(1+z^{2}) + \\cdots + 1/(1+z^{n})\\right] = -n$ which yields what we need.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind $\\prod_{n=2}^{\\infty}\\left(1-\\frac{1}{n^{2}}\\right)$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n$\\prod_{n=2}^{\\infty}\\left(1-\\frac{1}{n^{2}}\\right) = \\prod_{n=2}^{\\infty} \\frac{n^{2}-1}{n^{2}} = \\prod_{n=2}^{\\infty} \\frac{(n-1)(n+1)}{n \\cdot n}$\n\n$= \\frac{1 \\cdot 3}{2 \\cdot 2} \\cdot \\frac{2 \\cdot 4}{3 \\cdot 3} \\cdot \\frac{3 \\cdot 5}{4 \\cdot 4} \\cdot \\frac{4 \\cdot 6}{5 \\cdot 5} \\cdot \\frac{5 \\cdot 7}{6 \\cdot 6} \\cdots$\n\n$= \\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdots}{2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 4 \\cdot 4 \\cdot 5 \\cdot 5 \\cdots}$\n\n$= \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A O B$ be a $60$-degree angle. For any point $P$ in the interior of $\\angle A O B$, let $A'$ and $B'$ be the feet of the perpendiculars from $P$ to $A O$ and $B O$ respectively. Denote by $r$ and $s$ the distances $O P$ and $A' B'$. Find all possible pairs of real numbers $(r, s)$.",
    "options": [],
    "answer": "All pairs (r, r*sqrt(3)/2) with r > 0.",
    "solution": "Solution:\n\nExtend $A' P$ to meet $O B$ at $Z$. Notice that, because $\\angle O A' P$ and $\\angle O B' P$ are both right, the circle with diameter $O P$ passes through $O, P, A'$, and $B'$. Thus $\\angle B' O P = \\angle B' A' P$ since both intercept the same arc on this circle, and $\\triangle Z O P \\sim \\triangle Z A' B'$ by AA. We get\n$$\n\\frac{s}{r} = \\frac{B' A'}{O P} = \\frac{Z A'}{Z O} = \\frac{\\sqrt{3}}{2}\n$$\nbecause $Z O A'$ is a $30$-$60$-$90$ triangle. Since $r$ can obviously take on any value, the possibilities for $(r, s)$ are $\\left(r, \\frac{r \\sqrt{3}}{2}\\right)$ for\n\n![](attached_image_1.png)\n\nevery positive real $r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that there exists an infinite sequence of $a_{1}, a_{2}, \\ldots$ positive integers such that the following condition holds: $\\operatorname{gcd}\\left(a_{m}, a_{n}\\right)=1$ if and only if $|m-n|=1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEnumerate the primes $p_{1}, q_{1}, p_{2}, q_{2}, \\ldots$ and define\n$$\na_{n}=p_{n} q_{n} \\cdot \\begin{cases}\\prod_{k=1}^{n-2} p_{k} & n \\text{ even } \\\\ \\prod_{k=1}^{n-2} q_{k} & n \\text{ odd. }\\end{cases}\n$$\nThis works by construction. The idea is that you just take every pair $i<j$ you want to not be relatively prime (meaning $|i-j| \\geq 2$) and throw in a prime. You can't do this by using a different prime for every pair (since each $a_{i}$ must be finite) and you can't use the same prime for a fixed $i$, so you do the next best thing and alternate using even and odd and you're done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle, and let the bisector of $\\angle BAC$ meet $BC$ at $D$. Let $O, O_{1}, O_{2}$ be the circumcenters of triangles $ABC$, $ABD$, and $ADC$, respectively. Prove that $OO_{1} = OO_{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe use the standard fact that the circumcenter of a triangle is the intersection of the perpendicular bisectors of the three sides. Note that $OO_{1} \\perp AB$ since both $O$ and $O_{1}$ lie on the perpendicular bisector of $AB$. Similarly, $OO_{2} \\perp AC$ and $O_{1}O_{2} \\perp AD$. Hence the (counterclockwise) angle of rotation between lines $OO_{1}$ and $O_{1}O_{2}$ is the same as that between the perpendicular lines $AB$ and $AD$; thus $\\angle BAD = \\angle OO_{1}O_{2}$, and similarly $\\angle DAC = \\angle O_{1}O_{2}O$. We are given that $\\angle BAD = \\angle DAC$; thus $\\angle OO_{1}O_{2} = \\angle O_{1}O_{2}O$, that is, $\\triangle OO_{1}O_{2}$ is isosceles with $OO_{1} = OO_{2}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose you have three children and 40 pieces of candy. How many ways are there to distribute the candy such that each child gets more than one but fewer than 20 pieces?",
    "options": [],
    "answer": "171",
    "solution": "Solution:\nWe can use the \"stars and bars\" method, since this is the equivalent of giving 40 pieces of candy to three children, such that each child gets at least two pieces. This is the same as giving $40-6=34$ pieces to the three children with no restrictions (since we can pretend six pieces are already given out, two to each child). This is the same as arranging the 36 pieces of candy and two \"dividers\" (one between the first child's candy and the second child's candy and one between the second child's candy and the third child's candy) in a row. Thus, the number of ways to do this is\n$$\n\\left(\\begin{array}{c}\n36 \\\\\n2\n\\end{array}\\right)=\\frac{36 \\cdot 35}{2}=630\n$$\nHowever, this counts cases where one child gets at least 20 pieces of candy, which we don't want, so we need to subtract these cases. Only one child could get at least 20 pieces, or we would run out of candy, so there are three ways to choose the child who gets at least 20 pieces. Then, we have 20 pieces of candy to distribute among three children so that two of them (the ones who didn't already get 20) get at least 2 pieces each, which is the equivalent of distributing $20-4=16$ pieces among 3 children with no restrictions (since we can think of four of the pieces as being already given out). This can be done in $\\left(\\begin{array}{c}18 \\\\ 2\\end{array}\\right)$ ways, since it is the same as lining up 16 pieces of candy and two \"dividers.\" Putting this together, we get\n$$\n\\left(\\begin{array}{c}\n36 \\\\\n2\n\\end{array}\\right)-3\\left(\\begin{array}{c}\n18 \\\\\n2\n\\end{array}\\right)=630-3 \\cdot 153=171\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $X$, $Y$, and $Z$ be the points on the sides $BC$, $CA$, and $AB$ of the triangle $ABC$, such that $\\triangle XYZ \\sim \\triangle ABC$ ($\\angle X = \\angle A$, $\\angle Y = \\angle B$). Prove that the orthocenter of $\\triangle XYZ$ coincides with the circumcenter of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $x$, $y$, and $z$ be the points passing through $X$, $Y$, and $Z$, parallel to $YZ$, $ZX$, and $XY$. Let $P$, $Q$, and $R$ be the intersection point of the lines $y$ and $z$; $z$ and $x$; $x$ and $y$, respectively. Then we have $\\triangle PQR \\sim \\triangle XYZ$. The points $X$, $Y$, and $Z$ are respectively the midpoints of $RQ$, $QP$, and $PR$. Let $M$ be the orthocenter of $\\triangle XYZ$. Obviously, $M$ is the circumcenter of $\\triangle PQR$ and $\\angle ZMY = 180^{\\circ} - \\angle X = 180^{\\circ} - \\angle P = 180^{\\circ} - \\angle A$. Hence the points $P$, $A$, $Z$, $M$, and $Y$ belong to a circle. In a similar way we prove that the points $Z$, $B$, $R$, $X$ and $M$ belong to a circle. Then $\\angle PMA = \\angle PZA = \\angle BZR = \\angle BMR$. Since $MR = MP$ and $\\angle PAM = \\angle BRM = 90^{\\circ}$, we conclude that $MA = MB$. Analogously we conclude that $MA = MC$ implying that $M$ is the circumcenter of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA positive integer is written in each cell of an $8 \\times 8$ table so that each entry is the arithmetic mean of some two of its neighbors. Find the maximum number of distinct integers that may appear in the table.",
    "options": [],
    "answer": "58",
    "solution": "Solution:\n\nFirst consider the minimum number $m$ in the table. If it appears in some cell $A$, two neighboring cells $B, C$ must also contain $m$ because there is no other way for $m$ to be the arithmetic mean of two numbers in the table. Since $B$ cannot neighbor $C$, it must have another neighbor $D$ in addition to $A$ that contains $m$.\n\nSo the minimum number in the table appears at least four times. Likewise, the maximum number appears at least four times. We can now find 8 cells that contain at most 2 distinct numbers. The remaining 56 cells of course contain at most 56 distinct numbers. So there are at most 58 distinct numbers in the table. The diagram below shows that this bound can be achieved.\n\n| 1 | 1 | 21 | 22 | 37 | 38 | 58 | 58 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | 1 | 20 | 23 | 36 | 39 | 58 | 58 |\n| 3 | 2 | 19 | 24 | 35 | 40 | 57 | 56 |\n| 4 | 5 | 18 | 25 | 34 | 41 | 54 | 55 |\n| 7 | 6 | 17 | 26 | 33 | 42 | 53 | 52 |\n| 8 | 9 | 16 | 27 | 32 | 43 | 50 | 51 |\n| 11 | 10 | 15 | 28 | 31 | 44 | 49 | 48 |\n| 12 | 13 | 14 | 29 | 30 | 45 | 46 | 47 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $E$ be an ellipse that is not a circle. For which $n \\geq 3$ is it possible to inscribe a regular $n$-gon in $E$? (For each $n$, either show how to construct such an $n$-gon or prove that none exists.)",
    "options": [],
    "answer": "n = 3 and n = 4",
    "solution": "Solution:\nWe claim $n=3,4$ are the only solutions. To see that $3$ is possible, let $P$ be the endpoint of one of the axes, and draw two lines at angles of $\\pi / 6$ to that axis through $P$. By symmetry, these lines intersect the ellipse at points $Q, R$ equidistant from $P$, so $PQR$ is an equilateral triangle.\n\nTo see that $4$ is possible, draw the four lines through the center of $E$ at angles of $\\pi / 4$ to its axes; by symmetry, they intersect the ellipse at points which are equidistant from (and subtend equal angles at) its center and therefore form a square.\n\nNow, the vertices of any regular polygon lie on a common circle. By Bezout's theorem, this circle can only intersect $E$ in $2^{2}$ points (since both are curves of degree $2$), so only $4$ vertices of our polygon can lie on the ellipse, and we cannot have $n>4$. (Alternatively, one can see this using the fact that $5$ distinct points determine a unique conic section - that conic cannot be both $E$ and a circle.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a \\leq b \\leq c \\leq d$ be real numbers such that\n\n$$\na+b+c+d=0 \\quad \\text{ and } \\quad \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}=0\n$$\n\nProve that $a+d=0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet us rearrange the first equation to read\n$$\na+b=-(c+d)\n$$\nand the second to give\n$$\n\\begin{aligned}\n\\frac{1}{a}+\\frac{1}{b} & =-\\frac{1}{c}-\\frac{1}{d} \\\\\n\\frac{a+b}{a b} & =-\\frac{c+d}{c d}\n\\end{aligned}\n$$\nIf $a+b \\neq 0$, then we can divide (1) by (2) to yield $a b=c d$. So we conclude that\n$$\na+b=0 \\quad \\text{ or } \\quad a b=c d\n$$\nIn exactly the same manner, we derive that\n$$\na+c=0 \\quad \\text{ or } \\quad a c=b d\n$$\nand\n$$\na+d=0 \\quad \\text{ or } \\quad a d=b c \\quad \\text{ (this equation will not be needed). }\n$$\nFor the sake of contradiction, we are assuming that $a+d \\neq 0$. Suppose that $a+c=0$ holds; since $a+b+c+d=0$, we get $b+d=0$, whereas we know $a \\leq b$ and $c \\leq d$. We conclude that $a=b$ and $c=d$, whence $a+d=a+c=0$. Thus $a+c \\neq 0$, and for similar reasons $a+b \\neq 0$.\nMultiplying together the equations $a b=c d$ and $a c=b d$ and canceling the nonzero (why?) factor $b c$ gives $a^{2}=d^{2}$. So either $a=-d$, which is what we assumed impossible, or $a=d$, which yields the absurdity $a=b=c=d=0$.\nSolution:\n\nConsider the polynomial\n$$\nf(x)=(x-a)(x-b)(x-c)(x-d)\n$$\nwith roots $a, b, c, d$. The cubic coefficient of $f$ is $-(a+b+c+d)=0$; the linear coefficient is\n$$\n-(a b c+a b d+a c d+b c d)=-a b c d\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}\\right)=0 .\n$$\nConsequently $f$ has the form $f(x)=x^{4}+p x^{2}+q$, and there are complex numbers $u$ and $v$ such that\n$$\nf(x)=\\left(x^{2}\\right)^{2}+p x^{2}+q=\\left(x^{2}-u\\right)\\left(x^{2}-v\\right)=(x+\\sqrt{u})(x-\\sqrt{u})(x-\\sqrt{v})(x+\\sqrt{v}) .\n$$\nSince $f$ has only real roots, $u$ and $v$ must be real and nonnegative. WLOG $u \\leq v$; then $-\\sqrt{v} \\leq-\\sqrt{u} \\leq \\sqrt{u} \\leq \\sqrt{v}$ and we can identify the two factorizations of $f$ :\n$$\na=-\\sqrt{v}, \\quad b=-\\sqrt{u}, \\quad c=\\sqrt{u}, \\quad d=\\sqrt{v} .\n$$\nHence $a+d=0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a rectangle. Let $E$ be the foot of perpendicular from $A$ to $BD$. Let $F$ be an arbitrary point of the diagonal $BD$ between $D$ and $E$. Let $G$ be the intersection of the line $CF$ with the perpendicular from $B$ to $AF$. Let $H$ be the intersection of the line $BC$ with the perpendicular from $G$ to $BD$. Prove that $\\varangle EGB = \\varangle EHB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $X$ be the intersection of lines $BG$ and $AE$. Since $BX \\perp AF$ and $AX \\perp BF$, $X$ is the orthocenter of $\\triangle ABF$; hence $FX \\perp AB \\Rightarrow FX \\parallel BC \\parallel AD$. It follows:\n$$\n\\frac{EF}{ED} = \\frac{FX}{DA} = \\frac{FX}{BC} = \\frac{GF}{GC}\n$$\n(the first equality follows from $\\triangle EFX \\sim \\triangle EDA$, and the last from $\\triangle GFX \\sim \\triangle GCB$). We conclude that $GE \\parallel CD \\perp BH$. Since $BE \\perp GH$ it follows that $E$ is the orthocenter of $\\triangle GHB$ and $\\varangle EGB = \\varangle EHB$ follows easily (if we denote by $H'$ and $G'$ the intersections of $GE$ and $HE$, the triangles $GG'B$ and $HH'B$ are similar).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all polynomials $f$ that satisfy the equation\n$$\n\\frac{f(9 x)}{f(3 x)} = \\frac{243 x - 729}{x - 81}\n$$\nfor infinitely many values of $x$.",
    "options": [],
    "answer": "f(x) = a x^2 (x - 27)(x - 81)(x - 243) for real constants a",
    "solution": "Solution:\nWe have\n$$\n(x-81) f(9 x) = (243 x - 729) f(3 x)\n$$\nfor infinitely many values of $x$. Since both sides of this equation are polynomials, they must then be equal for all $x$.\n\nPlugging in $x=3$, we get $f(27)=0$. Plugging in $x=9$ then gives $f(81)=0$. Plugging in $x=27$ then gives $f(243)=0$. Thus we have that $f(x) = g(x) \\cdot (x-27)(x-81)(x-243)$ for some polynomial $g(x)$.\n\nPlugging this into the original equation, we get that for infinitely many $x$,\n$$\n243 \\frac{x-3}{x-81} = \\frac{f(9 x)}{f(3 x)} = \\frac{g(9 x)}{g(3 x)} \\cdot \\frac{(9 x-27)(9 x-81)(9 x-243)}{(3 x-27)(3 x-81)(3 x-243)} = \\frac{g(9 x)}{g(3 x)} \\cdot 27 \\frac{x-3}{x-81},\n$$\nso $\\frac{g(9 x)}{g(3 x)} = \\frac{243}{27} = 9$ and thus $f$ works if and only if $g(9 x) = 9 g(3 x)$ for all $x$. Thus, the roots of $g(9 x)$ are those of $g(3 x)$, so they must all be zero, so $g$ must be of the form $a x^{n}$. Plugging in $x=1$ then gives $9^{n} = 9 \\cdot 3^{n}$, so $n=2$. Indeed, $g(x) = x^{2}$ works.\n\nThe solutions are then\n$$\nf(x) = a x^{2} (x-27)(x-81)(x-243)\n$$\nfor real constants $a$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$, $y$, and $z$ be real numbers such that $x y z = 1$. Prove that\n$$\nx^{2} + y^{2} + z^{2} \\geq \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nReplacing the $1$ in the numerators of the fractions on the right by $x y z$, it suffices to prove that\n$$\nx^{2} + y^{2} + z^{2} \\geq y z + z x + x y\n$$\nwhich is true because\n$$\nx^{2} + y^{2} + z^{2} - y z - z x - x y = \\frac{(x - y)^{2} + (y - z)^{2} + (z - x)^{2}}{2} \\geq 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, $c$, $x$ be real numbers such that\n$$\na x^{2}-b x-c = b x^{2}-c x-a = c x^{2}-a x-b.\n$$\nProve that $a = b = c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $u = a - b$, $v = b - c$, and $w = c - a$. Then $u + v + w = 0$, and we would like to prove that $u = v = w = 0$. We have\n$$\n\\begin{aligned}\na x^{2} - b x - c &= b x^{2} - c x - a \\\\\n(a - b) x^{2} + (-b + c) x + (-c + a) &= 0 \\\\\nu x^{2} - v x - w &= 0.\n\\end{aligned}\n$$\n\nIn a similar manner, we get\n$$\nv x^{2} - w x - u = 0\n$$\nand\n$$\nw x^{2} - u x - v = 0.\n$$\n\nTaking $v$ times (1) minus $u$ times (2) eliminates the $x^{2}$ term; we get\n$$\n\\left(v^{2} - w u\\right) x = u^{2} - v w.\n$$\n\nLetting $A = u^{2} - v w$, $B = v^{2} - w u$, $C = w^{2} - u v$, we now have $B x = A$, and analogously $C x = B$ and $A x = C$. We see that if any of $A$, $B$, and $C$ are zero, then they all are. So we have two cases:\n\nCase 1. $A = B = C = 0$, that is, $u^{2} = v w$, $v^{2} = w u$, and $w^{2} = u v$. Clearly if one of $u$, $v$, and $w$ is $0$, then they all are and we are done. Otherwise, dividing the first equation by the second leads to $v^{3} = w^{3}$ and $v = w$. Similarly $u = v = w$. But since $u + v + w = 0$, the only common value that $u$, $v$, and $w$ can have is $0$ and we are done.\n\nCase 2. $A$, $B$, and $C$ are all nonzero. Multiplying the three equations $B x = A$, $C x = B$, and $A x = C$ together, we derive that $x^{3} = 1$ so $x = 1$. Plugging $x = 1$ into the original equations gives\n$$\na - b - c = b - c - a = c - a - b,\n$$\nso $a = b = c$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA large integer is divisible by all the integers between $1$ and $30$ inclusive, except for two consecutive integers. Determine those two consecutive integers.",
    "options": [],
    "answer": "16 and 17",
    "solution": "Solution:\nThe two numbers must be $16$ and $17$.\n\nFirst, note that if a number $n$ is not a power of a prime, then it has two smaller factors $a$ and $b$ that are relatively prime and multiply to $n$. Thus, $n$ is the least common multiple of $a$ and $b$. Then, any multiple of both $a$ and $b$ must also be a multiple of $n$, so any multiple of all the numbers less than $n$ would have to be a multiple of $n$ (since two of the numbers less than $n$ are $a$ and $b$). Thus, for there to be some number that is a multiple of all the numbers less than $n$ but not $n$ itself, $n$ must be a prime power. Therefore, both consecutive numbers are prime powers.\n\nNote also that if $n$ is at most $15$, then $2n \\leq 30$. Thus, if we chose one of the consecutive numbers to be $n \\leq 15$, the big number would have to be a multiple of $2n$, and therefore also of $n$. Thus, this cannot happen, so both numbers must be more than $15$. The only pair of consecutive numbers from $16$ to $30$ that are both prime powers is $16$ and $17$, so this must be the answer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that there exist infinitely many triples of positive integers $x, y, z$ which satisfy $x^{999} + y^{1000} = z^{1001}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe will choose $x$, $y$, and $z$ to be powers of $2$ such that the terms $x^{999}$ and $y^{1000}$ are equal and $z^{1001}$ is their sum. Writing $x = 2^{a}$, $y = 2^{b}$, $z = 2^{c}$, we have the conditions\n$$\n999a = 1000b = 1001c - 1.\n$$\nThe first equation suggests that we try $a = 1000d$, $b = 999d$, which yields the equation\n$$\n999000d = 1001c - 1.\n$$\nSince $999000 = 1001 \\cdot 998 + 2$, we can rewrite this as\n$$\n2d = 1001(c - 998d) - 1.\n$$\nIt is now clear that we can plug any odd positive integer $2k + 1$ for $c - 998d$ and get an integer for $d$, and hence for $c$. The results are as follows:\n$$\n\\begin{aligned}\na & = 1000(1001k + 500) \\\\\nb & = 999(1001k + 500) \\\\\nc & = 999000k + 499001.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all integers $x$ such that the product $x(x+1)(x+2)$ is the square of an integer.",
    "options": [],
    "answer": "x = -2, -1, 0",
    "solution": "Solution:\nThe answers are $x=0$, $x=-1$, and $x=-2$, for all of which the product is $0=0^{2}$.\n\nAssume that there is another solution. We can immediately rule out the case $x<-2$, as then the product is negative. So $x$ is a positive integer. Note that the greatest common divisor of $x$ and $x+1$ is $1$, as two consecutive integers cannot be multiples of the same prime. Likewise, $\\gcd(x+2, x+1)=1$ and so $\\gcd(x(x+2), x+1)=1$. Now $x(x+2)$ and $x+1$ are relatively prime positive integers whose product is a square, so they must both be squares (any prime dividing either of them must do so to an even power). But $x(x+2)=x^{2}+2x=(x+1)^{2}-1$, so now $x(x+2)$ and $(x+1)^{2}$ are two consecutive positive integers which are both squares, which is impossible (the difference between any two consecutive squares $a^{2}$ and $(a+1)^{2}$ is $2a+1 \\geq 3$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoes there exist a convex polygon that can be partitioned into non-convex quadrilaterals?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Assume that, on the contrary, it is possible to partition a polygon $P$ into non-convex quadrilaterals. Let $n$ be the number of quadrilaterals. Denote by $S$ the total sum of all internal angles of all the quadrilaterals. Since the sum of internal angles of each quadrilateral is $360^{\\circ}$, we have $S = 360^{\\circ} n$. However, each of the non-convex angles has to be in the interior of $P$, hence the sum of angles around the vertex of that angle has to be $360^{\\circ}$. This immediately gives $360^{\\circ} n$ as the sum of angles around such vertices. Since those are not the only vertices (at least the vertices of $P$ will contribute to the sum $S$), we have that $S > 360^{\\circ} n$ and this is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle. The incircle, centered at $I$, touches side $BC$ at $D$. Let $E$ be the reflection of $D$ through $I$, and let $F$ be the reflection of $D$ through the midpoint of $BC$. Prove that $A$, $E$, and $F$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the tangent line to the incircle at $E$ (which is of course parallel to $BC$) meet $AB$ and $AC$ at $X$ and $Y$, respectively.\n\n![](attached_image_1.png)\n\nNote that\n$$\n\\angle IXE = \\frac{\\angle BXE}{2} = \\frac{180^\\circ - \\angle DBX}{2} = 90^\\circ - \\angle DBI = BID.\n$$\nTherefore the right triangles $IXE$ and $BID$ are similar. We get\n$$\n\\frac{XE}{EI} = \\frac{ID}{BD}\n$$\nthat is, $XE \\cdot BD = r^2$ where $r$ is the inradius. Likewise, $EY \\cdot DC = r^2$. So\n$$\n\\frac{XE}{EY} = \\frac{DC}{BD} = \\frac{BF}{FC}\n$$\nNow $XE / BF = XY / BC = AX / AB$, so $\\triangle AXE \\sim \\triangle ABF$ and $A$, $E$, and $F$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$, $y$, $z$, and $u$ be real numbers satisfying the equation\n$$\n\\frac{x-y}{x+y}+\\frac{y-z}{y+z}+\\frac{z-u}{z+u}+\\frac{u-x}{u+x}=0\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $x$, $y$, and $z$ are rational (i.e. each is the quotient of two integers) and distinct. Prove that $u$ is rational as well.\n\nLet us begin by combining the fractions in pairs.\n$$\n\\begin{aligned}\n0 & =\\left(\\frac{x-y}{x+y}+\\frac{y-z}{y+z}\\right)+\\left(\\frac{z-u}{z+u}+\\frac{u-x}{u+x}\\right) \\\\\n& =\\frac{(x-y)(y+z)+(x+y)(y-z)}{(x+y)(y+z)}+\\frac{(z-u)(u+x)+(z+u)(u-x)}{(z+u)(u+x)}\n\\end{aligned}\n$$\nAfter expanding the numerators and collecting like terms, we get\n$$\n\\begin{aligned}\n0 & =\\frac{2 x y-2 y z}{(x+y)(y+z)}+\\frac{2 z u-2 u x}{(z+u)(u+x)} \\\\\n& =2(x-z)\\left[\\frac{y}{(x+y)(y+z)}-\\frac{u}{(z+u)(u+x)}\\right] .\n\\end{aligned}\n$$\nBecause $x \\neq z$, we can divide out $2(x-z)$ to get\n$$\n\\begin{aligned}\n0 & =\\frac{y}{(x+y)(y+z)}-\\frac{u}{(x+u)(u+z)} \\\\\n& =\\frac{y(x+u)(u+z)-u(x+y)(y+z)}{(x+y)(y+z)(x+u)(u+z)} .\n\\end{aligned}\n$$\nWe now expand the numerator and get, after canceling some terms,\n$$\n\\begin{aligned}\n0 & =u^{2} y-u y^{2}+x y z-u x z \\\\\n& =u y(u-y)+x z(y-u) .\n\\end{aligned}\n$$\nIf $u=y$, then we are done since $y$ is rational. Assuming $u \\neq y$, we divide through by $(u-y)$ to get $0=u y-x z$. Solving for $u$, we get $u=x z / y$ which is rational given that $x$, $y$, and $z$ are.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEight friends, Aerith, Bob, Chebyshev, Descartes, Euler, Fermat, Gauss, and Hilbert, bought tickets for adjacent seats at the opera. However when they arrived they mixed up their seats:\n- Bob sat in his assigned seat,\n- Chebyshev sat two seats to the right of Gauss' assigned seat,\n- Descartes sat one seat to the left of Fermat's assigned seat,\n- Euler sat four seats to the left of Hilbert's assigned seat,\n- Fermat sat five seats to the right of Descartes' assigned seat,\n- Gauss sat one to the right of Euler's assigned seat,\n- Hilbert sat three seats to the left of Aerith's assigned seat.\nIn whose seat did Aerith sit?",
    "options": [],
    "answer": "Chebyshev",
    "solution": "Solution:\n\nNumber the seats $1$ through $8$ and let $a, \\ldots, h$ be the seat assignments. Let $A$ be the seat occupied by Aerith. As each seat is assigned to exactly one person we must have $a+\\cdots+h=1+\\cdots+8$. As each seat is occupied by exactly one person we must have\n$$\n\\begin{aligned}\n1+\\cdots+8 & =A+b+(g+2)+(f-1)+(h-4)+(d+5)+(e+1)+(a-3) \\\\\n& =A+(a+b+d+e+f+g+h) .\n\\end{aligned}\n$$\nThus $A=c$, so Aerith occupies Chebyshev's seat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that for each $n \\geq 1$, there is a number $N$ having $n$ digits, each of which is either $1$ or $2$, such that $N$ is divisible by $2^{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe proceed by induction. For $n=1$, the number $N=2$ works. Suppose $N$ works for a given $n$. Consider the two $(n+1)$-digit numbers\n$$\nN_{1}=10^{n}+N \\quad \\text{ and } \\quad N_{2}=2 \\cdot 10^{n}+N .\n$$\nformed by attaching a single $1$ or $2$ to the left-hand side of $N$. Note that $N_{1}$ and $N_{2}$ are both divisible by $2^{n}$ (since $N$ and $10^{n}$ are) and their difference, $N_{2}-N_{1}=10^{n}=2^{n} \\cdot 5^{n}$, is divisible by $2^{n}$ but not by $2^{n+1}$. Therefore, one of $N_{1}$ and $N_{2}$ is divisible by $2^{n+1}$, which completes the induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAerith and Bob take turns picking a nonnegative integer, each time changing exactly one digit from the other's last number. The first person to pick a number that (s)he picked before loses. If Aerith goes first, and both play optimally, who wins?\n\n(Note: There are no leading zeroes, except in the number $0$ itself. For instance, if one person picks $2020$, the other could respond by picking $0020=20$, however the reverse does not hold.)",
    "options": [],
    "answer": "Bob wins",
    "solution": "Solution:\n\nBob wins. One winning strategy for Bob is: each time Aerith picks an even number, add one, and each time Aerith picks an odd number, subtract one. This only changes the last digit, since there are no carryovers in either case.\n\nBob would only get into a situation where he repeated an even number if Aerith had repeated the succeeding number twice, and similarly he would only need to repeat an odd number if Aerith had picked the preceding number twice. In either case, Aerith would have already lost.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P$ be a $2023$-sided polygon. All but one side has length $1$. What is the maximum possible area of $P$?",
    "options": [],
    "answer": "1011/2 * cot(pi/4044)",
    "solution": "Solution:\nFirst, we claim $P$ must be convex to maximize its area. If not, let $A$ and $B$ be consecutive vertices on the perimeter of its convex hull that aren't consecutive vertices of $P$. Reflecting the path between $A$ and $B$ over line $AB$ must increase the area of $P$ as the new shape strictly contains $P$.\n\nThus we assume $P$ is convex. Let $\\ell$ be the line containing the side with length not equal to $1$. Let $P'$ be the reflection of $P$ over $\\ell$. By convexity, $P$ and $P'$ do not overlap, so the union of $P$ and $P'$ is a polygon, specifically an equilateral $4044$-gon with sides of length $1$.\n\nThe area of an equilateral polygon is maximized when it is regular, so this union has maximum area that of a regular $4044$-gon, which is $1011 \\cot \\dfrac{\\pi}{4044}$. Thus, the answer is half of this, i.e.\n$$\n\\frac{1011}{2} \\cdot \\cot \\frac{\\pi}{4044}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a triangle $ABC$, let $D$ be the point of the ray $BA$ such that $BD = BA + AC$. If $K$ and $M$ are points on the sides $BA$ and $BC$, respectively, such that the triangles $BDM$ and $BCK$ have the same areas, prove that $\\angle BKM = \\frac{1}{2} \\angle BAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince the area of the triangle $BDM$ is equal to $BD \\cdot BM \\cdot \\sin \\angle DBM$ and a similar formula holds for the area of $\\triangle BCK$, we immediately have that $BD \\cdot BM = BC \\cdot BK$. This implies that $\\frac{BD}{BK} = \\frac{BC}{BM}$, hence $KM \\parallel CD$. Thus $\\angle BKM = \\angle BDC = \\angle ADC$. The last angle is equal to $\\frac{1}{2} \\angle BAC$ since $\\triangle DAC$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn acute angle $\\angle ABC$ and interior ray $BD$ are given, as shown. Laura is given an infinite ruler which consists of two parallel rays joined on one end by a segment perpendicular to both of them. One may place the infinite ruler onto the diagram so that one of the infinite edges (marked with an arrow) passes through any two selected points in the diagram, or so that any edge of the ruler coincides with (i.e. exactly overlaps with) a portion of a segment, ray or line already in the diagram. Once the ruler is placed, one may draw any edge of the ruler \"onto the diagram\", in the usual fashion. One may also plot points where any two straight objects intersect.\n\n![](attached_image_1.png)\n\nUsing only an infinite ruler, describe how to construct points $P$ on ray $BA$ and $Q$ on ray $BC$ such that ray $BD$ intersects ray $PQ$ at a point $R$ with $PR = 2(QR)$. Then prove that your construction works.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPlace the infinite ruler so that one infinite edge aligns with ray $BD$ and the other infinite edge intersects ray $BC$, then trace its outline. Do the same thing twice on the other side of ray $BD$, so that the final edge intersects ray $BA$. These points of intersection give the desired segment $PQ$.\n\n![](attached_image_2.png)\n\nWe outline the proof, which is mainly self-evident. The construction produces a set of four equally spaced rays, which therefore cut $PQ$ into three congruent segments. It follows that $PR = 2(QR)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x)$ be a quadratic polynomial. Prove that there exist quadratic polynomials $g(x)$ and $h(x)$ such that\n$$\nf(x) f(x+1)=g(h(x)) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWrite $f(x)=a x^{2}+b x+c$, $a \\neq 0$. It is a familiar fact that the graph of a quadratic function always has an axis of symmetry, specifically the line $x=-\\frac{b}{2 a}$. By substituting\n$$\nu=x+\\frac{b}{2 a}+\\frac{1}{2}$$\nwe can make $f(x)=a \\nu^{2}-a \\nu+d$ symmetric about the line $\\nu=1 / 2$. Then $f(x+1)=a \\nu^{2}+a \\nu+d$ is symmetric about the line $\\nu=-1 / 2$, and their product is symmetric about the line $\\nu=0$ :\n$$\n\\begin{aligned}\nf(x) f(x+1) & =\\left(a \\nu^{2}-a \\nu+d\\right)\\left(a \\nu^{2}+a \\nu+d\\right) \\\\\n& =a^{2} \\nu^{4}+2 a d \\nu^{2}-a^{2} \\nu^{2}+d^{2} \\\\\n& =g\\left(\\nu^{2}\\right)\n\\end{aligned}\n$$\nwhere $g(v)=a^{2} v^{2}+\\left(2 a d-a^{2}\\right) v+d^{2}$. But now\n$$\n\\nu^{2}=\\left(x+\\frac{b}{2 a}+\\frac{1}{2}\\right)^{2}\n$$\nis a quadratic function $h(x)$ of $x$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate the sum\n$$\n\\frac{1}{1+\\tan 1^{\\circ}}+\\frac{1}{1+\\tan 2^{\\circ}}+\\frac{1}{1+\\tan 3^{\\circ}}+\\cdots+\\frac{1}{1+\\tan 89^{\\circ}}\n$$\n(The tangent $(\\tan)$ of an angle $\\alpha$ is the ratio $BC/AC$ in a right triangle $ABC$ with $\\angle C=90^{\\circ}$ and $\\angle A=\\alpha$, and its value does not depend on the triangle used.)",
    "options": [],
    "answer": "89/2",
    "solution": "Solution:\nBy examining a triangle with angles $x$, $90-x$, and $90$, it is not hard to see the trigonometric identity\n$$\n\\tan \\left(90^{\\circ}-x\\right)=\\frac{1}{\\tan x} .\n$$\nLet us pair up the terms\n$$\n\\frac{1}{1+\\tan x} \\text{ and } \\frac{1}{1+\\tan \\left(90^{\\circ}-x\\right)}\n$$\nfor $x=1,2,3, \\ldots, 44$. The sum of such a pair of terms is\n$$\n\\begin{aligned}\n\\frac{1}{1+\\tan x}+\\frac{1}{1+\\tan \\left(90^{\\circ}-x\\right)} & =\\frac{1}{1+\\tan x}+\\frac{1}{1+\\frac{1}{\\tan x}} \\\\\n& =\\frac{1}{1+\\tan x}+\\frac{\\tan x}{\\tan x+1} \\\\\n& =\\frac{1+\\tan x}{1+\\tan x}=1\n\\end{aligned}\n$$\nThere is one more term in the sum, the middle term:\n$$\n\\frac{1}{1+\\tan 45^{\\circ}}=\\frac{1}{1+1}=\\frac{1}{2} .\n$$\nTherefore the sum is $44(1)+1/2=89/2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the minimal natural number $n$ with the following property: It is possible to tile the plane with squares whose side lengths belong to the set $\\{1,2, \\ldots, n\\}$ so that no two squares with the same side length touch along a segment of an edge.\n\nRemark. Squares with the same side length can touch at a vertex, however. Not all of the side lengths from $1$ to $n$ need to be used in the tiling.\n\n\nFigure 1\n\nFigure 2\n\nFigure 3\n\nFigure 4",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nThe answer is $n=5$. The desired tiling is shown in Figure 1. It is formed by translation from the L-shaped region in bold borders. Since none of the five squares in this region border on squares of like side length, neither does any square in the infinite tiling.\n\nTo show that $n \\leq 4$ does not work, it is necessary to plow through many arrangements of the tiles until reaching a contradiction. We present one method of structuring the argument. Some square in the tiling must have minimal size. Its four sides must be covered by squares larger than itself. If one side is covered by squares that protrude on both ends (pictorially, $\\square$ or $\\square$ ), then it becomes impossible to cover the opposite side; consequently any minimal square must be covered by four squares in the pinwheel arrangement $\\Vdash$.\n\nAssume first that the smallest square is the $1 \\times 1$ and it is surrounded by a $2 \\times 2$ and a $3 \\times 3$ in the manner of A, B, C in Figure 2. Now the upper left corner of $\\mathrm{D}$ must be filled by a $4 \\times 4$ (a $1 \\times 1$ would lack the necessary pinwheel layout) and likewise there is a $3 \\times 3$ at $\\mathrm{E}$. Now the corner $\\mathrm{F}$ cannot be filled with a $4 \\times 4$ square without wrecking the pinwheel at $\\mathrm{A}$, so it must be a $1 \\times 1$. The space above $\\mathrm{F}$ is now calling for either a $3 \\times 3$ or a $4 \\times 4$, either of which disrupts the pinwheel at $\\mathrm{A}$. Thus the arrangement $\\mathrm{ABC}$ at Figure 2 is impossible.\n\nSo if a $2 \\times 2$ borders on a $1 \\times 1$, the resulting cavity must be filled by a $4 \\times 4$ as in $\\mathrm{ABC}$ of Figure 3. The $3 \\times 3$ at $\\mathrm{D}$ is clear. The left side of the A-pinwheel must have a $3 \\times 3$ ($E$) since a $4 \\times 4$ would leave untilable space below $B$. The remaining spot on the pinwheel is necessarily occupied by a $2 \\times 2$ ($F$). Now $AFE$ of Figure 3 is the same configuration as $ABC$ of Figure 2. Hence this case is also impossible. So a $2 \\times 2$ cannot touch a $1 \\times 1$.\n\nThus any $1 \\times 1$ must be covered alternately by $3 \\times 3$'s and $4 \\times 4$'s as at $ABC$ of Figure 4. The $2 \\times 2$ at $D$ follows immediately, and since a $1 \\times 1$ cannot touch a $2 \\times 2$, we must use a $4 \\times 4$ at $\\mathrm{E}$ and a $2 \\times 2$ at $\\mathrm{F}$. Now the space around $\\mathrm{F}$ must be covered by a $3 \\times 3$ and a $4 \\times 4$ which is impossible.\n\nWe have left for last the case where there are no $1 \\times 1$'s. Since a tiling using only $3 \\times 3$'s and $4 \\times 4$'s is clearly impossible, the minimal square must be $2 \\times 2$, covered alternately by $3 \\times 3$'s and $4 \\times 4$'s as at $\\mathrm{D}$ and $\\mathrm{CBE}$ of Figure 4 (ignore square $A$). We derive the $2 \\times 2$ at $\\mathrm{F}$ and the resulting contradiction in the same manner as the preceding case.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\n\n![](attached_image_4.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c$ be positive real numbers satisfying $a b c=1$. Prove that\n$$\na(a-1)+b(b-1)+c(c-1) \\geq 0 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAt least two of $a, b, c$ are either not less than $1$ or not greater than $1$. Assume that $a$ and $b$ are on the same side of $1$. Next, transform the inequality as follows:\n$$\n\\begin{aligned}\na(a-1)+b(b-1)+c(c-1) &\\stackrel{?}{\\geq} 0 \\\\\na(a-1)+b(b-1)+c^{2}\\left(1-\\frac{1}{c}\\right) &\\stackrel{?}{\\geq} 0 \\\\\na(a-1)+b(b-1)+c^{2}(1-a b) &\\stackrel{?}{\\geq} 0 \\\\\na(a-1)+b(b-1)-c^{2}(a-1)-c^{2}(a b-a) &\\stackrel{?}{\\geq} 0 \\\\\n\\left(a-c^{2}\\right)(a-1)+\\left(b-c^{2} a\\right)(b-1) &\\stackrel{?}{\\geq} 0 \\\\\n\\left(a-\\frac{1}{a^{2} b^{2}}\\right)(a-1)+\\left(b-\\frac{1}{a^{2} b}\\right)(b-1) &\\stackrel{?}{\\geq} 0\n\\end{aligned}\n$$\nUsing the hypotheses concerning $a$ and $b$, it is not hard to see that the four factors in parentheses are all nonnegative or all nonpositive, and therefore the left side is nonnegative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $23 \\times 23$ square is divided into smaller squares of dimensions $1 \\times 1$, $2 \\times 2$, and $3 \\times 3$. What is the minimum possible number of $1 \\times 1$ squares?",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nColor the rows of the square black and white alternately, so the top and bottom rows are black. Then each $2 \\times 2$ tile covers two cells of each color, and each $3 \\times 3$ tile covers six of one color and three of the other. In particular, if only $2 \\times 2$ and $3 \\times 3$ tiles are used, the difference between the number of black and white cells covered is divisible by $3$. But the entire board has $23$ more black than white cells (if the bottom row were removed, the colors would be equally represented). So there must\n![](attached_image_1.png)\nbe at least one $1 \\times 1$ tile.\n\nTo construct the required tiling with only one $1 \\times 1$ tile, first use $3 \\times 3$ tiles to build a $9 \\times 12$ rectangle and $2 \\times 2$ tiles to build a $2 \\times 12$ rectangle. Join these two rectangles to form an $11 \\times 12$ rectangle. Then use four copies of this $11 \\times 12$ rectangle, together with the $1 \\times 1$ square, to build a $23 \\times 23$ square as shown in the diagram.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDenote by $f(n)$ the integer obtained by reversing the digits of a positive integer $n$. Find the greatest integer that is certain to divide $n^{4}-f(n)^{4}$ regardless of the choice of $n$.",
    "options": [],
    "answer": "99",
    "solution": "Solution:\n\nThe answer is $99$.\n\nLet $x = \\overline{d_{n-1} \\ldots d_{2} d_{1} d_{0}}$, i.e. $x = 10^{n-1} d_{n-1} + \\cdots + 10^{2} d_{2} + 10 d_{1} + d_{0}$. Then $y = f(x) = 10^{n-1} d_{0} + 10^{n-2} d_{1} + \\cdots + 10 d_{n-2} + d_{n-1}$.\n\nLet us show separately that $9 \\mid x^{4} - y^{4}$ and that $11 \\mid x^{4} - y^{4}$.\n\nSince $10 \\equiv 1 \\pmod{9}$, $x \\equiv y \\equiv d_{n-1} + d_{n-2} + \\cdots + d_{1} + d_{0} \\pmod{9}$, so $9 \\mid x - y$ and therefore also $9 \\mid (x - y)(x^{3} + x^{2} y + x y^{2} + y^{3}) = x^{4} - y^{4}$.\n\nSince $10 \\equiv -1 \\pmod{11}$, $x \\equiv d_{n-1} - d_{n-2} + \\cdots \\mp d_{1} \\pm d_{0} \\pmod{11}$ and $y \\equiv d_{0} + d_{1} + \\cdots \\mp d_{n-2} \\pm d_{n-1} \\pmod{11}$, i.e. $x \\equiv \\pm y$. In either case $x^{4} \\equiv y^{4} \\pmod{11}$, so $11 \\mid x^{4} - y^{4}$.\n\nSince $9$ and $11$ are relatively prime, their product $99$ must divide $x^{4} - y^{4}$.\n\nTo see that there is no larger integer that always divides $n^{4} - f(n)^{4}$, let $n = 10$, so that $f(n) = 01 = 1$, $n^{4} - f(n)^{4} = 9999 = 3^{2} \\cdot 11 \\cdot 101$, and let $n = 21$, so that $f(n) = 12$ and $n^{4} - f(n)^{4} = 173745 = 3^{5} \\cdot 5 \\cdot 11 \\cdot 13$. The greatest common divisor of these two values of $n^{4} - f(n)^{4}$ is $3^{2} \\cdot 11 = 99$, so there is no larger integer certain to divide $n^{4} - f(n)^{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn a distant planet, there are $2014$ cities, some pairs of which are connected by two-way roads. It turns out that the population of each city is the average of the populations of the cities to which it is connected by a single road, and moreover that it is possible to travel from every city to every other city by a sequence of roads.\nProve that all cities have the same population.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider the city $C_{\\max}$ with the maximal population $M$ (breaking ties arbitrarily). Then $M$ is the average of the populations of the neighboring cities, say $p_{1}, p_{2}, \\ldots, p_{n}$, meaning that\n$$\n\\frac{p_{1}+p_{2}+\\cdots+p_{n}}{n}=M\n$$\nBut $p_{1}, p_{2}, \\ldots, p_{n} \\leq M$, and hence $p_{1}+p_{2}+\\cdots+p_{n} \\leq n M$. So this can only occur if $p_{1}=p_{2}=\\cdots=p_{n}=M$. Hence all neighbors of $C_{\\max}$ have population $M$.\n\nProceeding in the same fashion, we find that all neighbors of neighbors of $C_{\\max}$ also must have population $M$, and so on. Because the network of cities is connected, this implies that all cities must have population $M$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all pairs of integers $x, y$ such that $2 x^{2} - 6 x y + 3 y^{2} = -1$.",
    "options": [],
    "answer": "For any nonnegative integer n, let A_n = (2 + √3)^n + (2 − √3)^n and B_n = ((2 + √3)^n − (2 − √3)^n)/√3. Then all integer solutions are given by x = ± A_n/2 and y = x ± B_n/2, with the two signs chosen independently.",
    "solution": "Solution:\n\nWe first characterize all solutions to the equation $a^{2} - 3 b^{2} = 1$ in nonnegative integers. This is a Pell equation, and the method used is standard. Suppose we have $(a + b \\sqrt{3})(a - b \\sqrt{3}) = a^{2} - 3 b^{2} = 1$ with $a > 1$; then we can set $a' = 2a - 3b$, $b' = 2b - a$. Then $(a + b \\sqrt{3}) = (2 + \\sqrt{3})(a' + b' \\sqrt{3})$ and $a'^{2} - 3 b^{2} = (4a^{2} - 12ab + 9b^{2}) - 3(4b^{2} - 4ab + a^{2}) = a^{2} - 3b^{2} = 1$. Moreover, we claim $a'$ and $b'$ are also nonnegative, with $a' < a$. To see this, note that $9b^{2} = 3(a^{2} - 1) < 4a^{2} \\Rightarrow 3b < 2a \\Rightarrow a' = 2a - 3b > 0$; $3b^{2} = a^{2} - 1 > 0 \\Rightarrow b > 0 \\Rightarrow a^{2} = 3b^{2} + 1 \\leq 4b^{2} \\Rightarrow b' = 2b - a \\geq 0$; and $b > 0 \\Rightarrow 9b^{2} > 3b^{2} + 1 = a^{2} \\Rightarrow a - a' = 3b - a > 0$.\n\nThus, a solution $(a, b)$ in nonnegative integers with $a > 1$ yields a new solution $(a', b')$ such that $a' < a$ and $(a + b \\sqrt{3}) = (2 + \\sqrt{3})(a' + b' \\sqrt{3})$. If $a' > 1$, we can repeat this process to obtain nonnegative integers $a'', b''$ such that $a'' < a'$ and $(a' + b' \\sqrt{3}) = (2 + \\sqrt{3})(a'' + b'' \\sqrt{3})$. Continuing in this manner, we obtain a sequence of solutions with $a > a' > a'' > \\cdots$. But this sequence of nonnegative integers cannot decrease forever, so eventually we get to some $n$ for which $a^{(n)} = 1$. The equation then implies $b^{(n)} = 0$, so $a + b \\sqrt{3} = (2 + \\sqrt{3})^{n}(a^{(n)} + b^{(n)} \\sqrt{3}) = (2 + \\sqrt{3})^{n}$. Then, it follows that $(2 - \\sqrt{3})^{n} = a - b \\sqrt{3}$. (This can be proven by comparing the binomial expansions of $(2 + \\sqrt{3})^{n}$ and $(2 - \\sqrt{3})^{n}$, or by a straightforward induction on $n$; note that the equation $a + b \\sqrt{3} = (2 + \\sqrt{3})^{n}$ uniquely determines the integers $a$ and $b$.) Thus, we conclude that\n$$\na = \\frac{(2 + \\sqrt{3})^{n} + (2 - \\sqrt{3})^{n}}{2}, \\quad b = \\frac{(2 + \\sqrt{3})^{n} - (2 - \\sqrt{3})^{n}}{2 \\sqrt{3}}.\n$$\nIf we had wanted all integer solutions to the equation $a^{2} - 3 b^{2} = 1$, rather than all nonnegative integer solutions, we simply note that the equation is invariant under switching signs of $a$ and/or $b$, so the set of all solutions is simply the set of solutions given above with possible sign changes thrown in.\n\nNow we consider the given equation, $2 x^{2} - 6 x y + 3 y^{2} = -1$. This can be rewritten as $x^{2} - 3(y - x)^{2} = 1$. Thus, by letting $a = x$, $b = y - x$, we have an equivalence between our given equation and the Pell equation, so we can express our solutions to the former in terms of those of the latter: $x = a$ and $y = a + b$ gives\n$$\nx = \\pm \\frac{(2 + \\sqrt{3})^{n} + (2 - \\sqrt{3})^{n}}{2}, \\quad y = x \\pm \\frac{(2 + \\sqrt{3})^{n} - (2 - \\sqrt{3})^{n}}{2 \\sqrt{3}}\n$$\nwhere $n$ is any nonnegative integer. It can be checked by computation that any such pair is a valid, integral solution, and the foregoing argument shows that all solutions are of this form.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $n$ is a natural number such that $2n+1$ and $3n+1$ are perfect squares, prove that $5n+3$ can't be a prime number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose that $5n+3$ is a prime number. Let $x$ and $y$ be natural numbers such that $x^2 = 2n+1$ and $y^2 = 3n+1$. Then\n$$\n5n+3 = 4(2n+1) - (3n+1) = 4x^2 - y^2 = (2x - y)(2x + y).\n$$\nSince $5n+3$ is a prime number and $x$ and $y$ are natural numbers, we must have $2x - y = 1$, implying that $y = 2x - 1$.\n\nSince $n = y^2 - x^2$, we have that\n$$\nx^2 = 2n + 1 = 2(y^2 - x^2) + 1 = 2((2x - 1)^2 - x^2) + 1 = 6x^2 - 8x + 3.\n$$\nThus,\n$$\n5x^2 - 8x + 3 = 0\n$$\nand the solutions are $x = 1$ or $x = 3/5$. Since $x^2 = 2n+1$ for a natural number $n$, neither of these $x$ would work and we have a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $b$ and $c$ are real numbers such that the equation\n$$\nx^{2}+b x+c=0\n$$\nhas two different solutions $x_{1}, x_{2}$. Suppose that\n\na. The (positive) difference between $x_{1}$ and $x_{2}$ is $1$;\n\nb. The (positive) difference between $b$ and $c$ is also $1$.\n\nFind all possible values of $b$ and $c$.",
    "options": [],
    "answer": "(b, c) = (-1, 0), (5, 6), (1, 0), (3, 2)",
    "solution": "Solution:\n\nGiven the two solutions $x_{1}$ and $x_{2}$, we know that the equation factors:\n$$\nx^{2}+b x+c=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right)=0.\n$$\nFrom this we derive that\n$$\nb=-x_{1}-x_{2} \\text{ and } c=x_{1} x_{2}.\n$$\nIn condition (a), we switch the labels on $x_{1}$ and $x_{2}$, if necessary, to assume that $x_{2}=x_{1}+1$. However, for condition (b) we still have the two cases $c-b=1, c-b=-1$.\nWe compute\n$$\n\\begin{aligned}\nc-b & =x_{1} x_{2}+x_{1}+x_{2} \\\\\n& =x_{1}\\left(x_{1}+1\\right)+x_{1}+\\left(x_{1}+1\\right) \\\\\n& =x_{1}^{2}+3 x_{1}+1\n\\end{aligned}\n$$\nSetting $c-b=1$ gives $x_{1}^{2}+3 x_{1}=0$, which has two solutions:\n(a) $x_{1}=0$, giving $x_{2}=1, b=-1, c=0$;\n(b) $x_{1}=-3$, giving $x_{2}=-2, b=5, c=6$.\nLikewise, setting $c-b=-1$ gives $x_{1}^{2}+3 x_{1}+2=0$, which also has two solutions:\n(a) $x_{1}=-1$, giving $x_{2}=0, b=1, c=0$;\n(b) $x_{1}=-2$, giving $x_{2}=-1, b=3, c=2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe have a row of boxes that is infinite in one direction, as shown.\n![](attached_image_1.png)\nDetermine if it is possible to fill each box with a positive integer such that the number in every box (except the leftmost one) is greater than the average of the numbers in the two neighboring boxes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe answer is no.\nDenote the numbers in the boxes by $a_{0}, a_{1}, a_{2}, \\ldots$ Then, if the conditions are satisfied, we have for all $n \\geq 1$\n$$\n\\begin{aligned}\na_{n} & >\\frac{a_{n-1}+a_{n+1}}{2} \\\\\n2 a_{n} & >a_{n-1}+a_{n+1} \\\\\na_{n}-a_{n-1} & >a_{n+1}-a_{n} .\n\\end{aligned}\n$$\nThis says that the differences $a_{n}-a_{n-1}$ form a strictly decreasing sequence. Since the differences are all integers, they must eventually become negative.\nAt this point the sequence $a_{n}$ is itself strictly decreasing, so by the same token, eventually the $a_{n}$'s become negative, contradicting the condition that they are all positive.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the triangle $A B C$ the angle $B$ is not a right angle, and $A B : B C = k$. Let $M$ be the midpoint of $A C$. The lines symmetric to $B M$ with respect to $A B$ and $B C$ intersect $A C$ at $D$ and $E$. Find $B D : B E$.",
    "options": [],
    "answer": "k^2",
    "solution": "Solution:\n\nAs $B C$ is the angle bisector in the triangle $M B E$, we have $\\frac{C E}{B E} = \\frac{C M}{B M}$ (by a well-known property of the angle bisector). Similarly, $\\frac{A D}{B D} = \\frac{A M}{B M}$. Draw a line $B M'$ symmetric to $B M$ with respect to the angle bisector of $A B C$ (point $M'$ is on the line $A C$). $B M'$ bisects the angle $D B E$. Using the same property of the angle bisector, we get $\\frac{E M'}{B E} = \\frac{D M'}{B D}$. Subtracting from this $\\frac{C E}{B E} = \\frac{C M}{B M}$ we get $\\frac{C M'}{B E} = \\frac{A M'}{B D}$ or $\\frac{B D}{B E} = \\frac{A M'}{C M'}$.\n\nNow it remains only to find the ratio in which $M'$ divides $A C$. To do that, note that $M B C$ and $M B A$ have equal areas: $\\frac{1}{2} B M \\cdot B C \\cdot \\sin \\angle M B C = \\frac{1}{2} B M \\cdot B A \\cdot \\sin \\angle M B A$. Therefore $\\frac{\\sin \\angle M B C}{\\sin \\angle M B A} = \\frac{A B}{B C} = k$. Hence\n\n$$\n\\begin{aligned}\n\\frac{A M'}{C M'} = \\frac{S_{A B M'}}{S_{B C M'}} & = \\frac{\\frac{1}{2} B M' \\cdot B A \\cdot \\sin \\angle M' B A}{\\frac{1}{2} B M' \\cdot B C \\cdot \\sin \\angle M' B C} \\\\\n& = k \\cdot \\frac{\\sin \\angle M' B A}{\\sin \\angle M' B C} = \\frac{\\sin \\angle M B C}{\\sin \\angle M B A} = k \\cdot k = k^{2}\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $P(x)$ be a polynomial with integer coefficients. Let $a_{0}=0$ and for $i \\geq 0$ define $a_{i+1}=P\\left(a_{i}\\right)$. Prove that $\\operatorname{gcd}\\left(a_{m}, a_{n}\\right)=a_{\\gcd(m, n)}$ for any integers $m, n \\geq 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf $Q(x), R(x)$ are any two polynomials with integer coefficients, it is not hard to see that their composition $Q(R(x))$ is also such a polynomial. In particular, for any nonnegative integer $n$, $P^{n}(x)$, the function obtained by iterating $P$ $n$ times, is an integer polynomial. Note that $a_{n}=P^{n}(0)$.\n\nNow, we prove the result by induction on the larger one of $m, n$. The base case is $m, n=1$, where we want $\\operatorname{gcd}\\left(a_{1}, a_{1}\\right)=a_{\\operatorname{gcd}(1,1)}$, and this is obvious. Now suppose one of $n, m$ is greater than $1$. For definiteness take $n \\geq m$. If $n=m$, the result is again obvious, so assume $n>m$. Since $a_{n-m}=P^{n-m}(0)$, $P^{n-m}$ is a polynomial whose constant term is $a_{n-m}$. Then $a_{n}=P^{n}(0)=P^{n-m}\\left(P^{m}(0)\\right)=P^{n-m}\\left(a_{m}\\right)$ which equals $a_{n-m}$ (the constant term) plus a sum of terms divisible by $a_{m}$. It follows that $\\operatorname{gcd}\\left(a_{n}, a_{m}\\right)=\\operatorname{gcd}\\left(a_{n-m}, a_{m}\\right)$. But by the induction hypothesis (since $m$ and $n-m$ are less than $n$), we have $\\operatorname{gcd}\\left(a_{n-m}, a_{m}\\right)=a_{\\operatorname{gcd}(n-m, m)}=a_{\\operatorname{gcd}(n, m)}$; combining this with the previous equality completes the induction step and the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $X$, $Y$, and $Z$ be points on one side of a line $AB$ such that\n$$\n\\triangle XAB \\sim \\triangle BYA \\sim \\triangle ABZ\n$$\nProve that $\\triangle XYZ$ is similar to all these triangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe will prove that $\\triangle XYZ \\sim \\triangle XAB$. Because $\\angle AXB = \\angle YXZ$ if and only if $\\angle AXY = \\angle BXZ$ and\n$$\n\\frac{ZX}{BX} = \\frac{YX}{AX} \\quad \\text{if and only if} \\quad \\frac{ZX}{YX} = \\frac{BX}{AX}\n$$\nthis is the same as proving $\\triangle XAY \\sim \\triangle XBZ$. Note that\n$$\n\\angle YAX = \\angle BAX - \\angle BAY = \\angle ZBA - \\angle XBA = \\angle ZBX.\n$$\nAlso note that\n$$\n\\frac{XA}{YA} = \\frac{XA}{AB} \\cdot \\frac{AB}{YA} = \\frac{AB}{BZ} \\cdot \\frac{BX}{AB} = \\frac{BX}{BZ}\n$$\nTherefore $\\triangle XAY \\sim \\triangle XBZ$ by SAS.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA Mystic Four Calculator has a four-digit display and four buttons. The calculator works as follows: Pressing button 1 replaces the number in the display with $1$; Pressing button 2 divides the number in the display by $2$; Pressing button 3 subtracts $3$ from the number in the display; Pressing button 4 multiplies the number in the display by $4$.\n\nInitially the display shows $0$. Any operation yielding a negative, fractional, or five-digit answer is ignored.\n\na. Can $2007$ appear in the display?\n\nb. Can $2008$ appear in the display?",
    "options": [],
    "answer": "a: No; b: Yes",
    "solution": "Solution:\n\na. No. Notice that if the number on the display is not divisible by $3$, then none of the operations can have as a result a number divisible by $3$. At the start, $1$ is the only button producing a result, so we are required to press it at some point. After that the number will never be divisible by $3$ again, and $2007$ is divisible by $3$.\n\nb. Yes. For instance, press $1$ so that the display shows $1$, press $4$ six times so that the display shows $4096$, and press $3$ $3696$ times so that the display shows $2008$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a quadrilateral whose diagonals are perpendicular and intersect at $P$. Let $h_{1}, h_{2}, h_{3}, h_{4}$ be the lengths of the altitudes from $P$ to $AB, BC, CD, DA$. Show that\n$$\n\\frac{1}{h_{1}^{2}}+\\frac{1}{h_{3}^{2}}=\\frac{1}{h_{2}^{2}}+\\frac{1}{h_{4}^{2}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe area of triangle $ABP$ is equal to $h_{1} \\cdot AB / 2$ and also to $AP \\cdot BP / 2$. Hence\n$$\n\\frac{1}{h_{1}^{2}}=\\frac{AB^{2}}{AP^{2} \\cdot BP^{2}}=\\frac{AP^{2}+BP^{2}}{AP^{2} \\cdot BP^{2}}=\\frac{1}{BP^{2}}+\\frac{1}{AP^{2}}\n$$\nApplying the same transformation to all the terms of (1) yields\n$$\n\\left(\\frac{1}{AP^{2}}+\\frac{1}{BP^{2}}\\right)+\\left(\\frac{1}{CP^{2}}+\\frac{1}{DP^{2}}\\right)=\\left(\\frac{1}{BP^{2}}+\\frac{1}{CP^{2}}\\right)+\\left(\\frac{1}{DP^{2}}+\\frac{1}{AP^{2}}\\right)\n$$\na triviality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $ABC$, $\\angle A = 50^\\circ$, $\\angle B = 60^\\circ$, and $\\angle C = 70^\\circ$. A ray of light bounces from point $D$ on $BC$ to $E$ on $CA$ to $F$ on $AB$ and then back to $D$. Find the angles of $\\triangle DEF$.\n\n(Light always takes the shortest path between points, meaning it bounces off an edge at equal angles.)",
    "options": [],
    "answer": "∠EDF = 80°, ∠DEF = 60°, ∠DFE = 40°",
    "solution": "Solution:\n\nLet $\\angle BDF = \\angle CDE = x$, $\\angle CED = \\angle AEF = y$, and $\\angle AFE = \\angle BFD = z$.\n\n![](attached_image_1.png)\n\nSince $\\triangle AEF$, $\\triangle BDF$, and $\\triangle CDE$ all have angle sum $180^\\circ$, we get\n$$\n\\begin{aligned}\n& x + y + 70^\\circ = 180^\\circ, \\\\\n& x + z + 60^\\circ = 180^\\circ, \\\\\n& y + z + 50^\\circ = 180^\\circ.\n\\end{aligned}\n$$\n\nAdding the three equations, subtracting $180^\\circ$ from both sides, and dividing by 2 gives\n$$\nx + y + z = 180^\\circ.\n$$\n\nFrom this, we get\n$$\nx = 50^\\circ, \\quad y = 60^\\circ, \\quad z = 70^\\circ.\n$$\n\n(Thus, the triangles $\\triangle AEF$, $\\triangle BDF$, and $\\triangle CDE$ are in fact all congruent to $\\triangle ABC$!) Now we get\n$$\n\\begin{aligned}\n& \\angle EDF = 180^\\circ - 2 \\cdot 50^\\circ = 80^\\circ, \\\\\n& \\angle DEF = 180^\\circ - 2 \\cdot 60^\\circ = 60^\\circ, \\\\\n& \\angle DFE = 180^\\circ - 2 \\cdot 70^\\circ = 40^\\circ.\n\\end{aligned}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence $a_{1}, a_{2}, \\ldots$ of positive integers satisfies\n$$\na_{n+1}=a_{n}^{3}+103\n$$\nfor every positive integer $n$. Prove that the sequence contains at most one perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt's easy to check that no two consecutive terms can be perfect squares, since the only squares which differ by $103$ are $51^{2}$ and $52^{2}$.\n\nNow, note that squares are $0$, $1$, or $4$ mod $8$. After a perfect square appears, the next term must be $-1$ or $0 \\bmod 8$, and thereafter all terms are $-1$, $-2$ modulo $8$, so no more squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers whose sum is equal to $1$. If\n$$\n\\begin{aligned}\nS= & \\frac{a_{1}^{2}}{2 a_{1}}+\\frac{a_{1} a_{2}}{a_{1}+a_{2}}+\\frac{a_{1} a_{3}}{a_{1}+a_{3}}+\\cdots+\\frac{a_{1} a_{n}}{a_{1}+a_{n}} \\\\\n& +\\frac{a_{2} a_{1}}{a_{2}+a_{1}}+\\frac{a_{2}^{2}}{2 a_{2}}+\\frac{a_{2} a_{3}}{a_{2}+a_{3}}+\\cdots+\\frac{a_{2} a_{n}}{a_{2}+a_{n}} \\\\\n& \\vdots \\\\\n& +\\frac{a_{n} a_{1}}{a_{n}+a_{1}}+\\frac{a_{n} a_{2}}{a_{n}+a_{2}}+\\frac{a_{n} a_{3}}{a_{n}+a_{3}}+\\cdots+\\frac{a_{n}^{2}}{2 a_{n}}\n\\end{aligned}\n$$\nprove that $S \\leq \\frac{n}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIt is easy to show that $\\left(\\frac{a+b}{2}\\right)^{2} \\geq a b$. Now applying this inequality to $a_{1}, a_{2}$ we get $\\frac{a_{1} a_{2}}{a_{1}+a_{2}} \\leq \\frac{\\left[\\left(a_{1}+a_{2}\\right) / 2\\right]^{2}}{a_{1}+a_{2}} = \\frac{a_{1}+a_{2}}{4}$. We get similar inequalities for all terms of the given sum. Now, using that $a_{1}+a_{2}+a_{3}+\\cdots+a_{n}=1$ we get:\n$$\n\\frac{a_{1}^{2}}{2 a_{1}}+\\frac{a_{1} a_{2}}{a_{1}+a_{2}}+\\frac{a_{1} a_{3}}{a_{1}+a_{3}}+\\cdots+\\frac{a_{1} a_{n}}{a_{1}+a_{n}} \\leq \\frac{a_{1}+a_{1}}{4}+\\frac{a_{1}+a_{2}}{4}+\\frac{a_{1}+a_{3}}{4}+\\cdots+\\frac{a_{1}+a_{n}}{4}=\\frac{n a_{1}+1}{4}.\n$$\nSimilarly\n$$\n\\frac{a_{2} a_{1}}{a_{2}+a_{1}}+\\frac{a_{2}^{2}}{2 a_{2}}+\\frac{a_{2} a_{3}}{a_{2}+a_{3}}+\\cdots+\\frac{a_{2} a_{n}}{a_{2}+a_{n}} \\leq \\frac{n a_{2}+1}{4}, \\text{ etc. }\n$$\nAdding these inequalities and using again that $a_{1}+\\cdots+a_{n}=1$ we get the desired inequality. Equality holds if and only if $a_{1}=a_{2}=\\cdots=a_{n}=1 / n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the remainder when $10^{2^{0}} + 10^{2^{1}} + \\cdots + 10^{2^{2021}}$ is divided by $\\underbrace{44 \\cdots 44}_{44}$.",
    "options": [],
    "answer": "2020202020202020202020202020202020202020110",
    "solution": "Solution:\nLet $N = 10^{2^{0}} + 10^{2^{1}} + \\cdots + 10^{2^{2021}}$. By the Chinese Remainder Theorem (CRT), it suffices to find $N \\bmod 4$ and $N \\bmod M = \\underbrace{11 \\cdots 11}_{44}$.\n\nModulo $4$, the remainder is $2$.\n\nModulo $M = \\frac{10^{44} - 1}{9}$, $10^{44} \\equiv 1$, so $10^{2^{i}} \\bmod M$ is determined by $2^{i} \\bmod 44$.\n\nMod $44$, $2^{0}$ is $1$ and $2^{1}$ is $2$. The remaining powers of two are all divisible by $4$. Furthermore, the period of $2 \\bmod 11$ is $10$, so the remaining $2020$ powers of $2$ are in fact divided equally among $10$ nonzero remainders that are multiples of $4$. Therefore,\n$$\nN \\bmod M = 10^{1} + 10^{2} + \\frac{2020}{10} \\cdot \\sum_{i=1}^{10} 10^{4i} = 2020202020202020202020202020202020202020110\n$$\nwhich is $2 \\bmod 4$, and is therefore the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCertain cities are connected by roads connecting pairs of them. The roads intersect only at the cities. A subset of the roads is called important if destroying those roads would make it so that there are two cities such that it is impossible to go from the first to the second. A subset $S$ of the roads is called strategic if it is important and no proper subset of $S$ is important. Let $S$ and $T$ be distinct strategic sets of roads. Let $U$ be the set of roads that are either in $S$ or $T$, but not both. Prove that $U$ is important.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $G$ be the graph having the cities as its vertices and the roads as its edges. (This just restates the problem in graph-theoretic language, which we use from now on.) We will look at the components of the graph $G$ and its subgraphs, where a component of a graph is defined to be a maximal connected subset: two vertices are in the same component if and only if it is possible to get from one to the other by following the edges.\n\nThe graph $G-S$ has more than one component, since it is disconnected. On the other hand, adding one edge of $S$ to $G-S$ can only connect two of these components; if $G-S$ had more than two components, adding this edge would still leave a disconnected graph and $S$ would not be strategic. So $G-S$ has exactly two components; call them $A, B$. Since these components are not connected to each other in $G-S$, every edge of $G$ connecting $A$ to $B$ must lie in $S$. Conversely, every edge of $S$ must connect $A$ to $B$, since, if it were entirely within $A$ or $B$, it could be added to $G-S$ while leaving the graph disconnected, and $S$ would not be strategic. Thus, $S$ consists exactly of those edges of $G$ which connect $A$ to $B$. Similarly, $G-T$ has two components $C, D$, and $T$ consists precisely of the edges of $G$ which connect $C$ to $D$.\n\nNow, the vertices of $G$ can be represented as a disjoint union of the four sets of vertices $A \\cap C, A \\cap D, B \\cap C$, and $B \\cap D$. From our characterizations of $S$ and $T$, we see that $U$ consists of precisely those edges of $G$ which link $A$ to $B$ or $C$ to $D$, but not both. It follows that, when we remove $U$ from $G$, all the remaining edges either link neither pair - and therefore lie entirely within one of our four sets of vertices - or both pairs, in which case they connect $A \\cap C$ to $B \\cap D$ or $A \\cap D$ to $B \\cap C$. Thus, $G-U$ contains no edges connecting the set of vertices $(A \\cap C) \\cup (B \\cap D)$ with the vertices $(A \\cap D) \\cup (B \\cap C)$. Furthermore, neither set of vertices is empty: we know that all the edges in $U$ connect these two sets, and $U$ is nonempty because $S \\neq T$. So we can choose one vertex from each set, and these two vertices are not connected by any path in $G-U$; thus, $U$ is important.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequences $\\{x_{n}\\}$ and $\\{y_{n}\\}$ are defined by $x_{0}=2$, $y_{0}=1$ and, for $n \\geq 0$,\n$$\nx_{n+1}=x_{n}^{2}+y_{n}^{2} \\quad \\text{ and } \\quad y_{n+1}=2 x_{n} y_{n}\n$$\nFind and prove an explicit formula for $x_{n}$ in terms of $n$.",
    "options": [],
    "answer": "x_n = (3^{2^n} + 1) / 2",
    "solution": "Solution:\n\nIf we add the formulas for $x_{n+1}$ and $y_{n+1}$ together, we get\n$$\nx_{n+1}+y_{n+1}=x_{n}^{2}+y_{n}^{2}+2 x_{n} y_{n}=(x_{n}+y_{n})^{2}.\n$$\nThus, increasing $n$ by 1 raises $x_{n}+y_{n}$ to the power 2. Since $x_{0}+y_{0}=3$, we get\n$$\nx_{n}+y_{n}=(\\cdots(3^{2})^{2} \\cdots)^{2}[n \\text{ squarings}]=3^{2^{n}}.\n$$\nSimilarly, we can subtract the formulas for $x_{n+1}$ and $y_{n+1}$ to get\n$$\nx_{n+1}-y_{n+1}=x_{n}^{2}+y_{n}^{2}-2 x_{n} y_{n}=(x_{n}-y_{n})^{2}.\n$$\nHowever, $x_{0}-y_{0}=1$. Since $1^{2}=1$, we get\n$$\nx_{n}-y_{n}=1\n$$\nfor all $n$.\nFinally, we add together equations (1) and (2), getting\n$$\n2 x_{n}=3^{2^{n}}+1\n$$\nwhich yields an explicit formula\n$$\nx_{n}=\\frac{3^{2^{n}}+1}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that $S$ is a set of 2001 positive integers, and $n$ different subsets of $S$ are chosen so that their sums are pairwise relatively prime. Find the maximum possible value of $n$. (Here the \"sum\" of a finite set of numbers means the sum of its elements; the empty set has sum 0.)",
    "options": [],
    "answer": "2^{2000}+1",
    "solution": "Solution:\nThe answer is $2^{2000}+1$. To see that we cannot do better than this, note that at least half of the $2^{2001}$ possible subsets of $S$ have even sums. Indeed, if all elements of $S$ are even, then all subsets have even sums; on the other hand, if there exists some odd $s \\in S$, we can divide the subsets of $S$ into pairs of the form $\\{T, T \\cup\\{s\\}\\}$ for each subset $T$ not containing $s$. Since the sum of $T$ and that of $T \\cup\\{s\\}$ are of opposite parity, each pair contains exactly 1 subset with an even sum. So, in this case, half the subsets of $S$ have even sums. The upshot is that, in either case, there are at most $2^{2000}$ subsets of $S$ with odd sums. Since our chosen subsets can include at most one subset whose sum is even (because no two sums can have a common factor of 2), we cannot choose more than $2^{2000}+1$ subsets altogether.\n\nNow, we must construct an example to show that we can have $n=2^{2000}+1$. To do this, let $k=\\left(2^{2000}\\right)!$, and let $S=\\left\\{k, 2k, 4k, 8k, \\ldots, 2^{1999}k, 1\\right\\}$. We consider the $2^{2000}$ subsets containing the element $1$, plus the one subset $\\{k\\}$. It is evident that $k$, the sum of the last subset, is relatively prime to the sum of any subset containing $1$, since this latter sum is of the form $ak+1$ for some $a$. So now we just need to prove that any two distinct subsets containing $1$ have relatively prime sums. Well, any such set consists of several distinct powers of $2$, multiplied by $k$, plus $1$. The sum of these powers of $2$ is some number $a$, $0 \\leq a < 2^{2000}$. Thus the subset's sum is $ak+1$. However, it follows from the uniqueness of binary representation that, for each possible value of $a$, there is only one subset whose sum is $ak+1$. Consequently, if we choose another, different subset (also containing $1$), its sum is $bk+1$ for some $b$, $0 \\leq b < 2^{2000}$ with $a \\neq b$. Now suppose $ak+1$ and $bk+1$ are not relatively prime; then they have some common prime factor $p$. So $p \\mid ak+1$ and $p \\mid bk+1$, hence $p \\mid (ak+1)-(bk+1) = (a-b)k$. Then, $p \\mid a-b$ or $p \\mid k$. But $a-b$ is nonzero and has absolute value $<2^{2000}$, so $a-b$ is one of the factors in the product $1 \\cdot 2 \\cdot 3 \\cdots 2^{2000} = k$, and we get $a-b \\mid k$. Thus, we are guaranteed that $p$ divides $k$. But then $p$ cannot divide $ak+1$, so we have a contradiction. We conclude that our subset sums are, in fact, pairwise relatively prime, completing the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe Moria Indestructible Phone Co. has hired you to test the hardiness of their newest smartphone model, the Mithril II. Your assignment is to determine the lowest floor of the Burj Khalifa tower (the world's tallest building, with 163 floors) from which the phone must be dropped to break it. You can ride the elevator to any floor, drop the phone to the ground, and then test whether it is intact. You may assume that if the phone breaks at a given floor, it consistently breaks at that floor and all higher floors. But the company has given you only two Mithril II's to test, and once one of them breaks, it remains broken.\nWhat is the minimum number of drops needed to determine the minimum floor of breaking, or else to conclude that the phone will withstand dropping from any of the floors?",
    "options": [],
    "answer": "18",
    "solution": "Solution:\n\nHere is a strategy requiring at most 18 drops. Drop the first phone from the 18th floor. If it breaks, drop the second phone from floors $1, 2, \\ldots, 17$ in that order to determine the minimum breaking floor. Otherwise, drop the first phone from the $18+17=35$th floor. If it breaks, use the 16 remaining drops to test the second phone on floors 19 through 34. Otherwise, drop the first phone from the $18+17+16$th floor, and so on. If the first phone makes it to the $18+17+\\cdots+5=161$st floor, then there are 4 drops left, more than enough to test it on the last two floors.\n\nSuppose that there existed a strategy requiring at most 17 drops. There are 164 possible \"strengths\" of the Mithril II (the breaking floor could be $1, 2, \\ldots, 163$ or undefined). When the strategy is applied to one of these strengths, the result is a sequence of intact (I) or broken (B) outcomes containing at most two B's and having length at most 17. In fact, we can arrange for the sequence to have length exactly 17 by declaring that, if at a given point the strategy stops and declares the phone strength, the outcomes of any unused drops are arbitrarily designated I. Then each phone strength has a \"signature\" of 17 letters, each B or I, with at most two B's. But the number of possible signatures is only\n$$\n\\binom{17}{0} + \\binom{17}{1} + \\binom{17}{2} = 1 + 17 + 136 = 154\n$$\nThis shows that there are some two phone strengths that have the same signature and therefore cannot be distinguished by the claimed strategy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn convex hexagon $A B C D E F$, $\\angle A = \\angle B$, $\\angle C = \\angle D$, and $\\angle E = \\angle F$. Prove that the perpendicular bisectors of $\\overline{A B}$, $\\overline{C D}$, and $\\overline{E F}$ pass through a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLines $A F$, $B C$, $D E$ determine a triangle $\\Delta$ whose angle bisectors are the lines in question. Hence the lines concur at the incenter of $\\Delta$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A$ be the number of 4-tuples $(x, y, z, t)$ of positive integers smaller than $2006^{2006}$ such that\n$$\nx^{3}+y^{2}=z^{3}+t^{2}+1,\n$$\nand let $B$ the number of 4-tuples $(x, y, z, t)$ of positive integers smaller than $2006^{2006}$ such that\n$$\nx^{3}+y^{2}=z^{3}+t^{2}.\n$$\nProve that $B>A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFor each natural number $k$, denote by $l_{k}$ the number of pairs $(x, y)$ such that $x^{3}+y^{2}=k$. Then $B=l_{2}^{2}+l_{3}^{2}+\\cdots+l_{r}^{2}$ where $r$ is the maximal $k$ for which $l_{k} \\neq 0$. Similarly, $A=l_{2} l_{3}+l_{3} l_{4}+\\cdots+l_{r-1} l_{r}$. Now\n$$\nB=\\frac{l_{2}^{2}}{2}+\\frac{1}{2}\\left(l_{2}^{2}+l_{3}^{2}\\right)+\\frac{1}{2}\\left(l_{3}^{2}+l_{4}^{2}\\right)+\\cdots+\\frac{1}{2}\\left(l_{r-1}^{2}+l_{r}^{2}\\right)+\\frac{l_{r}^{2}}{2}>l_{2} l_{3}+l_{3} l_{4}+\\cdots+l_{r-1} l_{r}\n$$\nWe have used that $a^{2}+b^{2} \\geq 2 a b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that every positive real number $y$ satisfies\n$$\n2 y \\geq 3-\\frac{1}{y^{2}}\n$$\nWhen does equality occur?",
    "options": [],
    "answer": "y=1",
    "solution": "Solution:\nSince $y$ is positive, we can multiply both sides by $y$ and rearrange to get the equivalent inequality\n$$\n2 y^{3}-3 y^{2}+1 \\geq 0\n$$\nThis factors as\n$$\n(2 y+1)(y-1)^{2} \\geq 0.\n$$\nThe first factor is positive since $y>0$, and the second is nonnegative since it is a real number squared. Thus, the product must be nonnegative, and will be $0$ iff $(y-1)^{2}=0$, which happens iff $y=1$.\n\nAlternate solution: By AM-GM, we get\n$$\n\\frac{\\frac{1}{y^{2}}+y+y}{3} \\geq \\sqrt[3]{\\frac{1}{y^{2}} \\cdot y \\cdot y}=1\n$$\nMultiplying by $3$ and rearranging gives the desired inequality, and equality occurs iff $\\frac{1}{y^{2}}=y$, which is true iff $y=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a circle $k$, let $AB$ be its diameter. An arbitrary line $l$ intersects the circle $k$ at the points $P$ and $Q$. If $A_{1}$ and $B_{1}$ are the feet of perpendiculars from $A$ and $B$ to $PQ$, prove that $A_{1}P = B_{1}Q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $A_{2}$ and $B_{2}$ be the intersections of the lines $AA_{1}$ and $BB_{1}$ with the circle $k$. Furthermore, let $s$ be the line through the center of $k$ perpendicular to $l$. Then $AB_{2}BA_{2}$ is a rectangle (because $AA_{2} \\parallel BB_{2}$ and $\\angle AB_{2}B = 90^{\\circ}$ since $AB$ is a diameter) and $s$ is an axis of symmetry of that rectangle. $A_{1}B_{1}B_{2}A$ is also a rectangle and $s$ is also an axis of symmetry for that rectangle. Since $k$ is also symmetric with respect to $s$, we conclude that $P$ and $Q$ are symmetric with respect to $s$. Thus $A_{1}Q = B_{1}P$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $0<\\alpha<1$. Prove that there exists a real number $x$, $0<x<1$, such that $\\alpha^{n}<\\{n x\\}$ for every positive integer $n$. (Here $\\{n x\\}$ is the fractional part of $n x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt suffices to find some real (noninteger) $x$ with this last property; we then replace $x$ by $\\{x\\}$ to satisfy the $0<x<1$ condition (note that the other condition is still satisfied, since $\\{n x\\}=\\{n\\{x\\}\\}$). We claim that $x=\\sqrt{m^{2}-1}$ for a sufficiently large integer $m$ meets our requirements.\n\nNote that, for all $n$ sufficiently large, $1 / 2 n^{2}>\\alpha^{n}$. (Proof: Choose integer $s>0$ large enough so that $1 / s<\\alpha^{-1 / 4}-1$; then, for $t \\geq s$, $1 / t \\leq 1 / s<\\alpha^{-1 / 4}-1 \\Rightarrow ([t+1] / t)^{2}=(1+1 / t)^{2}<\\alpha^{-1 / 2}$ and, writing $n^{2} / s^{2}=\\left(\\frac{n}{n-1}\\right)^{2}\\left(\\frac{n-1}{n-2}\\right)^{2} \\cdots\\left(\\frac{s+1}{s}\\right)^{2}$ for $n>s$, we see $n^{2} / s^{2}<\\alpha^{-(n-s) / 2} \\Rightarrow 2 n^{2} \\alpha^{n}<2 s^{2} \\alpha^{(n+s) / 2}$ which is less than $1$ when $n$ is large enough.) We claim that, if $n \\geq m$, then $\\{n x\\}>1 / 2 n^{2}$. Indeed, letting $k=\\lfloor n x\\rfloor=\\left\\lfloor n \\sqrt{m^{2}-1}\\right\\rfloor$, we have\n$$\n\\left(n \\sqrt{m^{2}-1}-k\\right)\\left(n \\sqrt{m^{2}-1}+k\\right)=n^{2}\\left(m^{2}-1\\right)-k^{2} \\geq 1\n$$\n(since it is an integer and, from the first factorization, is positive)\n$$\n\\Rightarrow \\{n x\\}=n \\sqrt{m^{2}-1}-k \\geq 1 /\\left(n \\sqrt{m^{2}-1}+k\\right)>1 / 2 n m \\geq 1 / 2 n^{2}\n$$\nThese results show that, if $m$ is large enough, then $\\{n x\\}>1 / 2 n^{2}>\\alpha^{n}$ for all $n \\geq m$.\n\nNow let $s$ be chosen so that $\\alpha^{s}<1 / 3$; we claim that, if $m>\\max (1,2 s / 3(1-\\alpha))$, then $\\{n x\\}>\\alpha^{n}$ for all $n<m$. Indeed, we have $m-\\frac{2}{3 m}<\\sqrt{m^{2}-1}$ (this is verified by squaring both sides), so $n m-1<n\\left(m-\\frac{2}{3 m}\\right)<n \\sqrt{m^{2}-1}<n m$ and hence $\\{n x\\}>n\\left(m-\\frac{2}{3 m}\\right)-(n m-1)=1-\\frac{2 n}{3 m}$. When $s \\leq n<m$, we have $1-\\frac{2 n}{3 m}>1 / 3>\\alpha^{n}$ and our condition is met; when $n<s$, $1-\\frac{2 s}{3 m}>\\alpha>\\alpha^{n}$ and we are again done.\n\nThus, by choosing $x=\\sqrt{m^{2}-1}$ for $m$ sufficiently large, we can ensure $\\{n x\\}>\\alpha^{n}$ for both all $n \\geq m$ and $n<m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe tangents at $A$ and $B$ to the circumcircle of an acute triangle $ABC$ intersect at $T$. Point $D$ lies on line $BC$ such that $DA = DC$. Prove that $TD \\parallel AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUsing the properties of tangent lines and chords, we have\n$$\n\\angle TAB = \\angle ACB = \\angle ABT = \\alpha\n$$\nMoreover, $\\triangle DAC$ is isosceles so $\\angle DAC = \\alpha$ as well. Now triangles $TAB$ and $DAC$ are similar, so\n$$\n\\frac{AT}{AB} = \\frac{AD}{AC} \\quad \\text{and} \\quad \\angle TAD = \\angle BAC\n$$\nimplying that triangles $TAD$ and $BAC$ are similar. Therefore\n$$\n\\angle ADT = \\angle ACB = \\alpha = \\angle DAC,\n$$\nso $TD \\parallel AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor a positive integer $n$, let $f(n)$ be the number of divisors of $n$ which are perfect squares, and let $g(n)$ be the number of divisors of $n$ which are perfect cubes. Determine whether there exists an integer $n$ such that\n$$\n\\frac{f(n)}{g(n)}=2012\n$$",
    "options": [],
    "answer": "Yes; for example n = 2^{4022} * 3^{2680} * 5^{1786} * 7^{1190} * 11^{792} * 13^{528} * 17^{352} * 19^{234} * 23^{156} * 29^{104} * 31^{68} * 37^{44} * 41^{28} * 43^{18} * 47^{12} * 53^{8} * 59^{4} * 61^{2}.",
    "solution": "Solution:\n\nIf $n$ has the prime factorization\n$$\nn=p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{k}^{e_{k}}\n$$\nthen the divisors of $n$ which are perfect squares are the numbers of the form\n$$\np_{1}^{2 a_{1}} p_{2}^{2 a_{2}} \\cdots p_{k}^{2 a_{k}}\n$$\nwhere each $a_{i}$ is an integer such that $0 \\leq a_{i} \\leq e_{i} / 2$. There are $\\left\\lfloor e_{i} / 2\\right\\rfloor+1$ such integers, and we conclude that\n$$\nf(n)=\\left(\\left\\lfloor\\frac{e_{1}}{2}\\right\\rfloor+1\\right)\\left(\\left\\lfloor\\frac{e_{2}}{2}\\right\\rfloor+1\\right) \\cdots\\left(\\left\\lfloor\\frac{e_{k}}{2}\\right\\rfloor+1\\right) .\n$$\nBy the same argument we find that\n$$\ng(n)=\\left(\\left\\lfloor\\frac{e_{1}}{3}\\right\\rfloor+1\\right)\\left(\\left\\lfloor\\frac{e_{2}}{3}\\right\\rfloor+1\\right) \\cdots\\left(\\left\\lfloor\\frac{e_{k}}{3}\\right\\rfloor+1\\right),\n$$\nand thus the problem reduces to writing 2012 as a product of factors of the form\n$$\n\\frac{\\lfloor x / 2\\rfloor+1}{\\lfloor x / 3\\rfloor+1}\n$$\nfor various positive integers $x$. Taking $x=2(2012-1)$ gives a fraction whose numerator is 2012, and this procedure can be applied iteratively:\n$$\n2012=\\frac{2012}{1341} \\cdot \\frac{1341}{894} \\cdot \\frac{894}{596} \\cdot \\frac{596}{397} \\cdot \\frac{397}{265} \\cdot \\frac{265}{177} \\cdot \\frac{177}{118} \\cdot \\frac{118}{79} \\cdot \\frac{79}{53} \\cdot \\frac{53}{35} \\cdot \\frac{35}{23} \\cdot \\frac{23}{15} \\cdot \\frac{15}{10} \\cdot \\frac{10}{7} \\cdot \\frac{7}{5} \\cdot \\frac{5}{3} \\cdot \\frac{3}{2} \\cdot \\frac{2}{1}=\\frac{f(n)}{g(n)}\n$$\nwhere $n=2^{4022} \\cdot 3^{2680} \\cdot 5^{1786} \\cdot 7^{1190} \\cdot 11^{792} \\cdot 13^{528} \\cdot 17^{352} \\cdot 19^{234} \\cdot 23^{156} \\cdot 29^{104} \\cdot 31^{68} \\cdot 37^{44} \\cdot 41^{28} \\cdot 43^{18} \\cdot 47^{12} \\cdot 53^{8} \\cdot 59^{4} \\cdot 61^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that the product of 6 consecutive positive integers is never equal to $n^{5}$ for some positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIt is easy to show that at least one of 6 consecutive numbers is relatively prime to each of the others (3 of these numbers are odd, exactly one of these odd numbers is divisible by 3, and for each prime $p > 3$ at most one is divisible by $p$). Assume that $x-2, x-1, x, x+1, x+2, x+3$, $x \\geq 3$ are given numbers. The number relatively prime to the others has to be a perfect 5th power. Let $F$ be the product of the other 5 numbers. We have:\n\n$$\nP(x) = (x-2)(x-1)x(x+1)(x+2) = x^{5} - 5x^{3} + 4x \\leq F \\leq x^{5} + 5x^{4} + 5x^{3} - 5x^{2} - 6x = (x-1)x(x+1)(x+2)(x+3) = Q(x)\n$$\n\nSince for $x \\geq 3$ we have $(x+1)^{5} > Q(x) \\geq F \\geq P(x) > (x-1)^{5}$, we conclude that $F = x^{5}$. However, this is a contradiction since $x$ is relatively prime to $x-1$ and $x+1$ and at least one of these numbers is a factor of $F$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$n$ boxes initially contain $1, 2, \\ldots, n$ marbles respectively ($n \\geq 1$). Charlotte first adds a marble to each box. Then she adds a marble to each box in which the number of marbles is divisible by $2$, then a marble to each box in which the number of marbles is divisible by $3$, and so on, until she adds a marble to each box in which the number of marbles is divisible by $n$. For which values of $n$ does the procedure end with exactly $n+1$ marbles in every box?",
    "options": [],
    "answer": "All n such that n+1 is prime",
    "solution": "Solution:\n\nThe answer is: all $n$ such that $n+1$ is prime.\n\nLemma 1. If box $A$ is to the left of box $B$ (we can arrange the boxes so that they initially contain $1, \\ldots, n$ marbles respectively from left to right), then at no stage of the process can box $A$ have more marbles than box $B$.\n\nProof. At the outset box $A$ certainly has fewer marbles than box $B$. Each subsequent stage adds at most one marble to each box, with one of two outcomes: (1) $A$ still has fewer marbles than $B$; (2) $A$ and $B$ have the same number of marbles, at which point the rules force $A$ and $B$ to have the same number of marbles throughout the process. This proves the lemma.\n\nIn view of this lemma, for all the boxes to end up with $n+1$ marbles, it is necessary and sufficient for the first box and the last box to end up with $n+1$ marbles. The first box will certainly have $n+1$ marbles because Charlotte adds a marble to it on every turn. The last box, which begins with $n$ marbles, will have $n+1$ after the first turn. If $n+1$ is prime, then since it is not divisible by any of the numbers $2, 3, 4, \\ldots, n$, the box will not pick up any more marbles and thus will end up with $n+1$. Then the intervening boxes, being sandwiched between two boxes with $n+1$ marbles, will also have $n+1$ marbles.\n\nIf $n+1$ is composite, then it is divisible by one of the numbers $2, 3, 4, \\ldots, n$. Hence the box will pick up another marble and will finish with at least $n+2$ marbles.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are seven green amoeba and three blue amoeba in a dish. Every minute, each amoeba splits into two identical copies; then, we randomly remove half the amoeba (thus there are always $10$ amoeba remaining). This process continues until all amoeba are the same color. What is the probability that this color is green?",
    "options": [],
    "answer": "70%",
    "solution": "Solution:\n\nThe answer is $70\\%$. Rather than considering the given colors, imagine the amoeba are colored with the following ten colors:\n- Forest green\n- Hunter green\n- Jungle green\n- Myrtle green\n- Mint green\n- Tea green\n- Aqua green\n- Sky blue\n- Navy blue\n- Midnight blue\n\nNow, we simply wait until one of these particular shades becomes the only remaining shade. By symmetry, each shade has a $10\\%$ chance of being the sole survivor. Thus, the chance it is a \"green\" shade is $70\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $f(1) = 1$ and $f(1) + f(2) + \\cdots + f(n) = n^{2} f(n)$ for every integer $n \\geq 2$, evaluate $f(2008)$.",
    "options": [],
    "answer": "2/(2009*2008)",
    "solution": "Solution:\n$n^{2} f(n) - f(n) = f(1) + f(2) + \\cdots + f(n-1) = (n-1)^{2} f(n-1)$\nhence $f(n) = \\frac{(n-1)^{2}}{n^{2} - 1} f(n-1) = \\frac{n-1}{n+1} f(n-1)$.\n\nThus\n$$\n\\begin{aligned}\nf(2008) &= \\frac{2007}{2009} f(2007) \\\\\n&= \\frac{2007}{2009} \\cdot \\frac{2006}{2008} f(2006) \\\\\n&= \\frac{2007}{2009} \\cdot \\frac{2006}{2008} \\cdot \\frac{2005}{2007} f(2005) \\\\\n&= \\cdots \\\\\n&= \\frac{2007!}{2009 \\cdot 2008 \\cdots 4 \\cdot 3} f(1) \\\\\n&= \\frac{2}{2009 \\cdot 2008}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sum of the squares of five real numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ equals $1$. Prove that the least of the numbers $(a_{i}-a_{j})^{2}$, where $i, j=1,2,3,4,5$ and $i \\neq j$, does not exceed $1/10$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume w.l.o.g. that $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4} \\leq a_{5}$. If $m$ is the least value of $|a_{i}-a_{j}|$, $i \\neq j$, then $a_{i+1}-a_{i} \\geq m$ for $i=1,2, \\ldots, 4$, and consequently $a_{i}-a_{j} \\geq (i-j) m$ for any $i, j \\in \\{1, \\ldots, 5\\}$, $i>j$. Then it follows that\n$$\n\\sum_{i>j}(a_{i}-a_{j})^{2} \\geq m^{2} \\sum_{i>j}(i-j)^{2} = 50 m^{2}\n$$\nOn the other hand, by the condition of the problem,\n$$\n\\sum_{i>j}(a_{i}-a_{j})^{2} = 5 \\sum_{i=1}^{5} a_{i}^{2} - (a_{1}+\\cdots+a_{5})^{2} \\leq 5\n$$\nTherefore $50 m^{2} \\leq 5$; i.e., $m^{2} \\leq \\frac{1}{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that each nonnegative integer can be represented in the form $a^{2}+b^{2}-c^{2}$, where $a, b, c$ are positive integers with $a<b<c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe note that\n$$\n0=3^{2}+4^{2}-5^{2}, \\quad 2=5^{2}+11^{2}-12^{2},\n$$\nand for $n>1$, we can write $2 n$ as\n$$\n2 n=(3 n)^{2}+(4 n-1)^{2}-(5 n-1)^{2},\n$$\nand $3 n<4 n-1<5 n-1$. For odd numbers greater than $7$, we can use\n$$\n2 n+3=(3 n+2)^{2}+(4 n)^{2}-(5 n+1)^{2},\n$$\nand for the first four odd numbers we have\n$$\n\\begin{gathered}\n1=4^{2}+7^{2}-8^{2}, \\quad 3=4^{2}+6^{2}-7^{2} \\\\\n5=4^{2}+5^{2}-6^{2}, \\quad 7=6^{2}+14^{2}-15^{2}\n\\end{gathered}\n$$\nThis covers all cases, so the statement holds for all nonnegative integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $2 \\times n$ grid has a light bulb in each square. Each bulb has a switch that flips the state of its corresponding bulb as well as all (horizontally or vertically) adjacent bulbs. Determine whether it is always possible, regardless of the initial state of the bulbs, to turn all the bulbs off if\n\na) $n=2011$\n\nb) $n=2012$.",
    "options": [],
    "answer": "a) No; b) No",
    "solution": "Solution:\n\nThe answer to both questions is no. Call a lamp \"hot\" if it is in one of the squares marked $\\times$ in the following two patterns:\n\nFor $2 \\times 2011$ :\n\n| $\\times$ |   | $\\times$ |   | $\\times$ | $\\cdots$ | $\\times$ |   | $\\times$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\times$ |   | $\\times$ |   | $\\times$ | $\\cdots$ | $\\times$ |   | $\\times$ |\n\nFor $2 \\times 2012$ :\n\n|   | $\\times$ | $\\times$ | $\\times$ |   | $\\times$ | $\\times$ | $\\times$ | $\\cdots$ |   | $\\times$ | $\\times$ | $\\times$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|   |   | $\\times$ |   |   |   | $\\times$ |   | $\\cdots$ |   |   | $\\times$ |   |\n\nIt is not hard to see that any flip of a switch affects precisely zero, two, or four hot bulbs, which keeps the parity of the number of lit hot bulbs the same. Thus if we begin with only one lit hot bulb, the number of lit hot bulbs will always be odd and we can never turn all the bulbs off.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $2011$ light bulbs are arranged in a row. Each bulb has a button under it. Pressing the button will change the state of the bulb above it (on to off or vice versa) and will also change the two neighboring bulbs, or the single neighboring bulb in the case of one of the two end buttons. Is it always possible, regardless of the initial state of the bulbs, to turn them all off by pressing some buttons?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe answer is yes.\nLet us number the bulbs $1$ to $2011$ from left to right. Given any initial state of the bulbs, let us begin by following this algorithm: As long as at least one bulb other than bulb $1$ is on, let $n$ be the number of the rightmost lit bulb and push the button for bulb $n-1$. This will turn bulb $n$ off and move the location of the rightmost lit bulb to the left. We may continue until either (1) all bulbs are off or (2) only bulb $1$ is on. In the former case, we are done; in the latter case we then push the buttons marked $\\times$ :\n\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $\\cdots$ | 2007 | 2008 | 2009 | 2010 | 2011 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\times$ |  | $\\times$ | $\\times$ |  | $\\times$ | $\\times$ |  |  | $\\times$ | $\\times$ |  | $\\times$ | $\\times$ |\n\nIt is evident that every bulb will then change state twice, except the first, which will change state once. Thus all the bulbs will be turned off.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA product of five primes is of the form $ABC,ABC$, where $A$, $B$, and $C$ represent digits. If one of the primes is $491$, find the product $ABC,ABC$.",
    "options": [],
    "answer": "982,982",
    "solution": "Solution:\n$491 \\cdot 1001 \\cdot 2 = 982,982$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $\\left(a+\\frac{1}{a}\\right)^{2}=3$, find $\\left(a+\\frac{1}{a}\\right)^{3}$ in terms of $a$.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n0 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSolve $x = \\sqrt{x - \\frac{1}{x}} + \\sqrt{1 - \\frac{1}{x}}$ for $x$.",
    "options": [],
    "answer": "(1+sqrt(5))/2",
    "solution": "Solution:\n\n$\\frac{1 + \\sqrt{5}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhen a single number is added to each member of the sequence $20, 50, 100$, the sequence becomes expressible as $x, a x, a^{2} x$. Find $a$.",
    "options": [],
    "answer": "5/3",
    "solution": "Solution:\n$\\frac{5}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThrough a point in the interior of a triangle $ABC$, three lines are drawn, one parallel to each side. These lines divide the sides of the triangle into three regions each. Let $a$, $b$, and $c$ be the lengths of the sides opposite $\\angle A$, $\\angle B$, and $\\angle C$, respectively, and let $a'$, $b'$, and $c'$ be the lengths of the middle regions of the sides opposite $\\angle A$, $\\angle B$, and $\\angle C$, respectively. Find the numerical value of $\\dfrac{a'}{a} + \\dfrac{b'}{b} + \\dfrac{c'}{c}$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\n$1$ .",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA stacking of circles in the plane consists of a base, or some number of unit circles centered on the $x$-axis in a row without overlap or gaps, and circles above the $x$-axis that must be tangent to two circles below them (so that if the ends of the base were secured and gravity were applied from below, then nothing would move). How many stackings of circles in the plane have 4 circles in the base?",
    "options": [],
    "answer": "14",
    "solution": "Solution:\n$C(4)=14$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDraw a rectangle. Connect the midpoints of the opposite sides to get 4 congruent rectangles. Connect the midpoints of the lower right rectangle for a total of 7 rectangles. Repeat this process infinitely. Let $n$ be the minimum number of colors we can assign to the rectangles so that no two rectangles sharing an edge have the same color and $m$ be the minimum number of colors we can assign to the rectangles so that no two rectangles sharing a corner have the same color. Find the ordered pair $(n, m)$.",
    "options": [],
    "answer": "(3, 4)",
    "solution": "Solution:\n\n$(3,4)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor the sequence of numbers $n_{1}, n_{2}, n_{3}, \\ldots$, the relation $n_{i}=2 n_{i-1}+a$ holds for all $i>1$. If $n_{2}=5$ and $n_{8}=257$, what is $n_{5}$?",
    "options": [],
    "answer": "33",
    "solution": "Solution:\n\n33 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the smallest positive integer $x$ for which $x^{2}+x+41$ is not a prime?",
    "options": [],
    "answer": "40",
    "solution": "Solution:\n40",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $\\frac{1}{9}$ of $60$ is $5$, what is $\\frac{1}{20}$ of $80$?",
    "options": [],
    "answer": "4",
    "solution": "Solution: In base 15, 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $D$ be a dodecahedron which can be inscribed in a sphere with radius $R$. Let $I$ be an icosahedron which can also be inscribed in a sphere of radius $R$. Which has the greater volume, and why?\n\nNote: A regular polyhedron is a geometric solid, all of whose faces are congruent regular polygons, in which the same number of polygons meet at each vertex. A regular dodecahedron is a polyhedron with 12 faces which are regular pentagons and a regular icosahedron is a polyhedron with 20 faces which are equilateral triangles. A polyhedron is inscribed in a sphere if all of its vertices lie on the surface of the sphere.\n\nThe illustration below shows a dodecahedron and an icosahedron, not necessarily to scale.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "The dodecahedron has the greater volume.",
    "solution": "Solution:\n\n$D$ has the greater volume.\n\nThe key idea that we will use is a beautiful relationship between $I$ and $D$; they are dual polyhedra. To understand this, note that $D$ has 12 faces, 30 edges, and 20 vertices. (The 12 pentagonal faces produce 60 edges and 60 vertices, but these are, respectively, double- and triple-counted. Hence there are $60/2 = 30$ edges and $60 / 3=20$ vertices). Likewise, $I$ has 20 faces, 30 edges, and 12 vertices. Notice that $I$ has as many vertices as $D$ has faces, and vice-versa.\n\nConsequently, if we join each center of the faces of an icosahedron to the centers of adjacent faces, we will get a dodecahedron, and vice-versa. (For the other platonic solids, the cube and octahedron are duals of each other, and the tetrahedron is its own dual).\n\nWe will use duality to prove an important lemma.\n\nLemma: If a dodecahedron and an icosahedron can be inscribed in the same sphere, then they circumscribe the same sphere as well.\n\nTo see why this is true, imagine $I$ and $D$ both sitting inside a sphere of radius $R$ with center $O$, which circumscribes both polyhedra. Thus all the vertices of both polyhedra lie on the surface of this sphere. Because the polyhedra are dual, we can place the polyhedra so that for every vertex $A$ of $I$, the line $A O$ is perpendicular to and passes through the center of a face of $D$. Likewise, for every vertex $B$ of $D$, the line $B O$ is perpendicular to and passes through the center of a face of $I$.\n\nThe illustration below is an imperfect depiction of a portion of this situation. The left picture is a \"top view,\" as seen by an observer looking directly down at the center of a face of $D$. Point $A$ is a vertex of $I$, and $B$ is a vertex of $D$.\n\n![](attached_image_2.png)\nTop View\n\n![](attached_image_3.png)\nSide View\n\nThe picture on the right is a \"side view\" (not-to-scale), showing the projections of $A$ and $B$ down to the center $O$ of the sphere. Let $C O$ be the perpendicular projection of $A O$ onto $B O$. Note that point $C$ is the center of a face of the icosahedron $I$ (in the Top View, you cannot see $C$ because it is directly \"underneath\" $B$ ), and $h=C O$ is the radius of the inscribed sphere of $I$. Also $r=A C$ is the radius of the circumscribed circle about a face of $I$.\n\nLet $E O$ be the perpendicular projection of $B O$ on to $A O$. Clearly $E O=C O=h$ (since $A O= B O=R$ ), but by duality, $E$ must be a center of a face of the dodecahedron $D$. Consequently, the radius of the inscribed sphere of $D$ must also equal $h$.\n\nNow it is a simple matter to compute and compare the two volumes. Notice that $D$ consists of 12 pyramids with regular pentagonal bases (each with circumradius $r$ ) and height $h$. Likewise, $I$ consists of 20 pyramids with equilateral triangular bases (each with circumradius $r$ ) and height $h$. Recall that the volume of a pyramid with base area $B$ and height $h$ is $B h / 3$. Now we can easily compare the two volumes.\n\n$$\n\\frac{\\operatorname{vol}(D)}{\\operatorname{vol}(I)}=\\frac{\\frac{h}{3} \\times 12 \\times \\text{ area of pentagon }}{\\frac{h}{3} \\times 20 \\times \\text{ area of triangle }}\n$$\n$$\n\\begin{aligned}\n& =\\frac{3 \\times \\text{ area of pentagon }}{5 \\times \\text{ area of triangle }} \\\\\n& =\\frac{3 \\times 5 \\times \\frac{r^{2}}{2} \\sin 72^{\\circ}}{5 \\times 3 \\times \\frac{r^{2}}{2} \\sin 120^{\\circ}} \\\\\n& =\\frac{\\sin 72^{\\circ}}{\\sin 60^{\\circ}} \\\\\n& >1 .\n\\end{aligned}\n$$\n\nWe are using the fact that a regular $n$-gon that can be inscribed in a circle of radius $r$ can be dissected into $n$ \"pie slices\" that are isosceles triangles with sides of length $r$ and vertex angle $(360 / n)^{\\circ}$. Thus the area of the regular $n$-gon is $n \\times\\left(r^{2} / 2\\right) \\times \\sin (360 / n)^{\\circ}$.\nSolution:\n\n(sketch) Instead of showing that the inscribed spheres have equal radii, we can use duality and the \"top view\" used in the previous solution to show that the radius of the circumscribed circle of each face of $D$ is equal to the radius of the circumscribed circle of each face of $I$. In other words, the two circles depicted below have the same radii. This fact is attributed to the ancient Greek mathematician Apollonius.\n\n![](attached_image_4.png)\n\nOnce this is known, it is easy to deduce that the two polyhedra have identical inscribed sphere radii, and the rest of the solution proceeds as before.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSquares $A B D E$, $B C F G$ and $C A H I$ are drawn exterior to a triangle $A B C$. Parallelograms $D B G X$, $F C I Y$ and $H A E Z$ are completed. Prove that $\\angle A Y B + \\angle B Z C + \\angle C X A = 90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $\\rho$ be the $90^{\\circ}$ rotation about the center of square $A B D E$, counterclockwise (orienting $\\triangle A B C$ to have its vertices in counterclockwise order). Note that segments $C A$ and $Z E$ are congruent and perpendicular (thanks to square $C A H I$ and parallelogram $H A E Z$), so $\\rho(C) = Z$. Likewise, segments $B C$ and $D X$ are congruent and perpendicular, implying $\\rho(X) = C$. Now $\\rho(X C) = C Z$ which implies $\\angle Z C X = 90^{\\circ}$. Likewise $\\angle X A Y = \\angle Y B Z = 90^{\\circ}$. With three of the angles of the reentrant hexagon $X A Y B Z C$ known, the sum of the other three is readily computed:\n\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n& \\angle A Y B + \\angle B Z C + \\angle C X A \\\\\n& = \\left(180^{\\circ} - \\angle B A Y - Y B A\\right) + \\left(180^{\\circ} - \\angle C B Z - \\angle Z C B\\right) + \\left(180^{\\circ} - \\angle A C X - \\angle X A C\\right) \\\\\n& = 540^{\\circ} - (\\angle B A Y + \\angle X A C) - (\\angle C B Z + \\angle Y B A) - (\\angle A C X + \\angle Z C B) \\\\\n& = 540^{\\circ} - (\\angle X A Y + \\angle B A C) - (\\angle Y B Z + \\angle C B A) - (\\angle Z C X + \\angle A C B) \\\\\n& = 540^{\\circ} - 3 \\cdot 90^{\\circ} - (\\angle B A C + \\angle C B A + \\angle A C B) \\\\\n& = 540^{\\circ} - 270^{\\circ} - 180^{\\circ} = 90^{\\circ}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $\\angle ACB = 90^{\\circ}$. The inscribed circle of $\\triangle ABC$ touches sides $AC$ and $BC$ at $D$ and $E$, respectively. On the circumscribed circle of $\\triangle ABC$, the midpoints of minor arcs $AC$ and $BC$ are respectively $P$ and $Q$. Prove that $D$, $E$, $P$, and $Q$ are all collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $M$ be the midpoint of $AC$, let $O$ be the circumcenter of $\\triangle ABC$, and let $F$ be the point where the incircle touches $AB$.\n\nNote that $\\triangle CDE$ is a right isosceles triangle and therefore $\\angle CDE = 45^{\\circ}$. Also, $\\angle PMD$ is right since $OM \\perp AC$ and $OM$ passes through $P$. Let us prove that $\\triangle PMD$ is right isosceles:\n$$\nPM = OP - OM = \\frac{AB - BC}{2} = \\frac{AF + FB - BE - EC}{2} = \\frac{AD - DC}{2} = \\frac{AC - 2DC}{2} = MC - DC = MD.\n$$\nTherefore $\\angle PDM = 45^{\\circ}$ and\n$$\n\\angle PDM + \\angle MDE = \\angle PDM + 180^{\\circ} - \\angle CDE = 45^{\\circ} + 180^{\\circ} - 45^{\\circ} = 180^{\\circ}.\n$$\nThis proves that $P$ lies on line $DE$. Similarly, we can prove that $Q$ lies on line $DE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe have four bowls labeled $A, B, C, D$ in a row, and we have four indistinguishable marbles which we can initially distribute among the bowls any way we like. A move consists of transferring one marble from a bowl to one of the adjacent bowls in the row. Is it possible to perform a succession of moves in which every distribution of the marbles appears exactly once?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Call a position of the marbles \"even\" if bowls $A$ and $C$ have an even total number of marbles and \"odd\" otherwise. It is easy to see that any move changes an even position to an odd position and vice versa. However, note that, for any nonnegative integer $k$, the number of ways to distribute $k$ marbles among two bowls is $k+1$ (as the first bowl can hold any number of marbles from 0 to $k$). Therefore it is easy to count the number of even and odd positions:\n\n| Marbles in $A$ and $C$ | Marbles in $B$ and $D$ | Results |\n| :--- | :--- | :--- |\n| 4, in 5 ways | 0, in 1 way | $5 \\cdot 1=5$ even positions |\n| 3, in 4 ways | 1, in 2 ways | $4 \\cdot 2=8$ odd positions |\n| 2, in 3 ways | 2, in 3 ways | $3 \\cdot 3=9$ even positions |\n| 1, in 2 ways | 3, in 4 ways | $2 \\cdot 4=8$ odd positions |\n| 0, in 1 way | 4, in 5 ways | $1 \\cdot 5=5$ even positions |\n\nThe totals are 19 even positions and 16 odd. It is not possible to visit all 19 even positions without the 18 odd positions separating them containing some duplicates.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\ominus$ be an operation on the set of real numbers such that\n$$\n(x \\ominus y)+(y \\ominus z)+(z \\ominus x)=0\n$$\nfor all real $x$, $y$, and $z$. Prove that there is a function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$\nx \\ominus y=f(x)-f(y)\n$$\nfor all real $x$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst we plug in $x=y=z=0$ to get\n$$\n(0 \\ominus 0)+(0 \\ominus 0)+(0 \\ominus 0)=0\n$$\nthat is, $0 \\ominus 0=0$. Then we plug $y=z=0$, keeping $x$ undetermined, into the original equation to get\n$$\n(x \\ominus 0)+0+(0 \\ominus x)+0=0 .\n$$\nSo $0 \\ominus x=-(x \\ominus 0)$. Define $g(x)=x \\ominus 0$. Plugging $z=0$ into the original equation gives\n$$\n\\begin{aligned}\n(x \\ominus y)+(y \\ominus 0)+(0 \\ominus x) & =0 \\\\\n(x \\ominus y)+(y \\ominus 0)-(x \\ominus 0) & =0 \\\\\n(x \\ominus y) & =(x \\ominus 0)-(y \\ominus 0) \\\\\n& =g(x)-g(y)\n\\end{aligned}\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p$ and $q$ be positive real numbers with $p+q<1$. Teams $A$ and $B$ play a series of games. For each game, $A$ wins with probability $p$, $B$ wins with probability $q$, and they tie with probability $1-p-q$. The series ends when one team has won two more games than the other, that team being declared the winner of the series. What is the probability that $A$ wins the series?",
    "options": [],
    "answer": "p^2/(p^2+q^2)",
    "solution": "Solution:\n\nWe can break the series into rounds, where each round is a series of ties followed by one team winning a game. Thus, a round that $A$ wins has the form $A$, $tA$, $ttA$, $tttA$, $\\ldots$, where $t$ represents a tied game and $A$ a game $A$ wins. Thus, the probability $A$ wins a round is the geometric series\n$$\np' = p + p(1-p-q) + p(1-p-q)^2 + \\cdots = \\frac{p}{p+q}\n$$\nSimilarly, the probability that $B$ wins a round is $\\frac{q}{p+q}$. Now, the series must last an even number of rounds, since one team must be ahead by $2$. If it lasts $2n$ rounds, then for the first $2n-2$ rounds, the teams must alternate winning. The probability that each team wins one round from an adjacent pair is $p'q' + q'p' = 2p'q'$. This must happen $n-1$ times, and then $A$ must win the last two rounds, with probability $(p')^2$. Thus, the probability $A$ wins the series is another geometric series\n$$\n\\sum_{n=1}^{\\infty} (2p'q')^{n-1} (p')^2 = \\frac{(p')^2}{1-2p'q'} = \\frac{p^2}{(p+q)^2 - 2pq} = \\frac{p^2}{p^2 + q^2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $M_{n}$ be the number of integers $N$ such that\n(a) $0 \\leq N < 10^{n}$;\n(b) $N$ is divisible by $4$;\n(c) The sum of the digits of $N$ is also divisible by $4$.\nProve that $M_{n} \\neq 10^{n} / 16$ for all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $10^{n} / 16$ is not an integer for $n=1,2,3$, we may assume that $n \\geq 4$. Let\n$$\n\\tilde{M}_{n} = M_{n} - \\frac{10^{n}}{16}\n$$\nWe note that numbers whose hundreds digit is between $2$ and $9$ inclusive make no total contribution to $\\tilde{M}_{n}$ for $n \\geq 3$. This is because, given the remaining digits, divisibility by $4$ (which depends only on the tens and units digits) occurs one-fourth of the time, and exactly two of the eight choices for the hundreds digit brings the total of the digits to a multiple of $4$. Similarly, numbers with hundreds digit $0$ or $1$ but thousands digit at least $2$ make no contribution to $\\tilde{M}_{n}$. (We require the hundreds digit to be less than $2$ to avoid double counting.) Continuing in this way, we find that only the $25 \\cdot 2^{n}$ numbers with all but the last two digits equal to $0$ or $1$ can make $\\tilde{M}_{n}$ nonzero. Let $L_{n}$ be the number of such numbers satisfying conditions (a), (b), and (c). We thus have\n$$\n\\tilde{M}_{n} = L_{n} - \\frac{25 \\cdot 2^{n}}{16}\n$$\nNow let $f(i, k)$ be the number of binary strings of length $k$ whose sum is congruent to $i \\bmod 4$. Since there are respectively $9, 4, 6, 6$ multiples of $4$ less than $100$ with digit sums $\\equiv 0, 1, 2, 3 \\bmod 4$, we have\n$$\nL_{n} = 9 f(0, n-2) + 4 f(3, n-2) + 6 f(2, n-2) + 6 f(1, n-2).\n$$\nIf we let $\\tilde{f}(i, k) = f(i, k) - 2^{k} / 4$, we can subtract $25 \\cdot 2^{n} / 16$ from both sides of the above equation to get\n$$\n\\tilde{M}_{n} = 9 \\tilde{f}(0, n-2) + 4 \\tilde{f}(3, n-2) + 6 \\tilde{f}(2, n-2) + 6 \\tilde{f}(1, n-2)\n$$\nOn the other hand, dividing a binary $k$-string with sum $i$ into a $k-1$-string and a single bit and noting that the sum of the $k-1$-string is either $i$ or $i-1$, we get the recurrence\n$$\nf(i, k) = f(i, k-1) + f(i-1, k-1)\n$$\nwhich is equivalent to\n$$\n\\tilde{f}(i, k) = \\tilde{f}(i, k-1) + \\tilde{f}(i-1, k-1).\n$$\nTogether with the boundary conditions\n$$\n\\begin{array}{ll}\nf(0,0)=1, & f(1,0)=f(2,0)=f(3,0)=0, \\\\\n\\tilde{f}(0,0)=\\frac{3}{4}, & \\tilde{f}(1,0)=\\tilde{f}(2,0)=\\tilde{f}(3,0)=-\\frac{1}{4},\n\\end{array}\n$$\nthis enables us to calculate values of $\\tilde{f}(i, k)$ and therefore of $\\tilde{M}_{n}$. It is trivial to prove by induction that for $k \\geq 1$,\n$$\n\\tilde{f}(i, k) = -\\tilde{f}(i+2, k),\n$$\nfrom which it follows after a short computation using (4) that\n$$\n\\tilde{f}(i, k+4) = -4 \\tilde{f}(i, k), \\quad k \\geq 1\n$$\nNow we substitute this equation into (3), yielding\n$$\n\\tilde{M}_{n+4} = -4 \\tilde{M}_{n}, \\quad n \\geq 3\n$$\nThis implies that $\\tilde{M}_{n+4}$ is nonzero if $\\tilde{M}_{n}$ is. We have already noted that $\\tilde{M}_{3}$ is nonzero because it is not an integer. Computing $\\tilde{M}_{4}=2$, $\\tilde{M}_{5}=-1$, and $\\tilde{M}_{6}=-6$ by hand using (5), (4), and (3) completes the proof that $\\tilde{M}_{n}$ is nonzero for all $n \\geq 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose you have only an unmarked straightedge (no compass), and you are given a line segment $A B$ with midpoint $O$ and a point $P$ not on line $A B$.\n\na. Construct a line through $P$ parallel to $A B$.\n\nb. If you are also given the circle with center $O$ and radius $O A$ and $P$ does not lie on the circle, construct a line through $P$ perpendicular to $A B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nExtend line $A P$ to some point $Q$ on the opposite side of $P$ from $A$. Let $R$ be the intersection of lines $Q O$ and $P B$, and let $S$ be the intersection of lines $A R$ and $Q B$, as shown below.\n\n![](attached_image_1.png)\n\nBy Ceva's theorem,\n$$\n\\frac{Q P}{P A} \\cdot \\frac{A O}{O B} \\cdot \\frac{B S}{S Q} = 1\n$$\nIt follows that\n$$\n\\frac{Q P}{P A} = \\frac{Q S}{Q B}\n$$\nwhich means $\\triangle Q P S \\sim \\triangle Q A B$, therefore $\\angle Q P S = \\angle Q A B$ and so $P S \\parallel A B$.\n\nb.\nLet $P A$ and $P B$ meet the circle again at points $Q$ and $R$, respectively, and let $S$ be the intersection of $A R$ and $B Q$.\n\n![](attached_image_2.png)\n\nThen $\\angle A Q B = \\angle A R B = 90^\\circ$ since both angles are inscribed in a semicircle, so $S$ is the orthocenter of $\\triangle A B P$, which means $P S$ is the desired perpendicular.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDefine a function $f$ on the real numbers by\n$$\nf(x)= \\begin{cases}2 x & \\text{ if } x<1 / 2 \\\\ 2 x-1 & \\text{ if } x \\geq 1 / 2\\end{cases}\n$$\nDetermine all values $x$ satisfying $f(f(f(f(f(x)))))=x$.",
    "options": [],
    "answer": "0, 1/31, 2/31, ..., 30/31, 1",
    "solution": "Solution:\nThe answer is the 32 values $0, \\frac{1}{31}, \\frac{2}{31}, \\ldots, \\frac{30}{31}, 1$.\n\nIf $x<0$, then $f(x)=2 x<x$ so the sequence $x, f(x), f(f(x)), \\ldots$ is strictly decreasing and cannot return to $x$.\n\nIf $x>1$, similarly $f(x)=2 x-1>x$ so the sequence $x, f(x), f(f(x)), \\ldots$ is strictly increasing and cannot return to $x$.\n\nIf $x=1$, then $f(x)=1$ and we have $1$ as a solution.\n\nFinally, we assume that $0 \\leq x<1$, so $0 \\leq f(x)<1$ as well. For simplicity let $x_{0}=x$ and $x_{n+1}=f\\left(x_{n}\\right)$ so the equation we are trying to solve is $x_{5}=x_{0}$. Note that $2 x_{n}-x_{n+1}$ is an integer (either $0$ or $1$) for each $n$, so\n$$\n32 x_{0}-x_{5}=16\\left(2 x_{0}-x_{1}\\right)+8\\left(2 x_{1}-x_{2}\\right)+4\\left(2 x_{2}-x_{3}\\right)+2\\left(2 x_{3}-x_{4}\\right)+\\left(2 x_{4}-x_{5}\\right)\n$$\nmust be an integer as well. If we assume $x_{5}=x_{0}$ we deduce that $31 x_{0}$ is an integer. Conversely, if $31 x_{0}$ is an integer, then $x_{5}-x_{0}$ is an integer and this integer must be $0$ because $0 \\leq x_{0}, x_{5}<1$. Thus the solutions in this range are exactly the multiples of $\\frac{1}{31}$: $0, \\frac{1}{31}, \\frac{2}{31}, \\ldots, \\frac{30}{31}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that the functional equations\n$$\n\\begin{aligned}\nf(x+y) &= f(x) + f(y) \\\\\n\\text{and}\\quad f(x+y+xy) &= f(x) + f(y) + f(xy) \\quad (x, y \\in \\mathbb{R})\n\\end{aligned}\n$$\nare equivalent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet us assume that $f(x+y) = f(x) + f(y)$ for all reals. In this case we trivially apply the equation to get $f(x+y+xy) = f(x+y) + f(xy) = f(x) + f(y) + f(xy)$. Hence the equivalence is proved in the first direction.\n\nNow let us assume that $f(x+y+xy) = f(x) + f(y) + f(xy)$ for all reals. Plugging in $x = y = 0$ we get $f(0) = 0$. Plugging in $y = -1$ we get $f(x) = -f(-x)$. Plugging in $y = 1$ we get $f(2x+1) = 2f(x) + f(1)$ and hence $f(2(u+v+uv)+1) = 2f(u+v+uv) + f(1) = 2f(uv) + 2f(u) + 2f(v) + f(1)$ for all real $u$ and $v$. On the other hand, plugging in $x = u$ and $y = 2v+1$ we get $f(2(u+v+uv)+1) = f(u + (2v+1) + u(2v+1)) = f(u) + 2f(v) + f(1) + f(2uv + u)$. Hence it follows that $2f(uv) + 2f(u) + 2f(v) + f(1) = f(u) + 2f(v) + f(1) + f(2uv + u)$, i.e.,\n$$\nf(2uv + u) = 2f(uv) + f(u).\n$$\nPlugging in $v = -1/2$ we get $0 = 2f(-u/2) + f(u) = -2f(u/2) + f(u)$. Hence, $f(u) = 2f(u/2)$ and consequently $f(2x) = 2f(x)$ for all reals. Now (1) reduces to $f(2uv + u) = f(2uv) + f(u)$. Plugging in $u = y$ and $x = 2uv$, we obtain $f(x) + f(y) = f(x + y)$ for all nonzero reals $x$ and $y$. Since $f(0) = 0$, it trivially holds that $f(x + y) = f(x) + f(y)$ when one of $x$ and $y$ is $0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle is tangent to the continuations of sides $CA$ and $CB$ of the triangle $ABC$, and is also tangent to the side $AB$ at point $P$. Prove that the radius of the circle tangent to $AP$, $CP$ and the circle circumscribed around $ABC$ is equal to the radius of the circle inscribed in $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $K$ and $M$ be the tangency points of the circle with $AP$ and $CP$ respectively, $L$ the point where it touches the circumscribed circle of $ABC$, $T$ the middle of the arc $AB$ of the circumscribed circle, $I$ the center of the inscribed circle of $ABC$.\n\nThe tangent to the circle $ABC$ at $T$ is parallel to the line $AB$. So the similarity transformation (\"stretching\") centered at $K$ taking the circle $ABC$ to the circle $KLM$ takes $AB$ to this tangent, and hence $T$ to $L$. So $K$, $L$ and $T$ are collinear. We now prove that points $K$, $M$ and $I$ are collinear. Let $M'$ be the point of intersection of the line $KI$ with the circle $KLM$. We want to show $M = M'$.\n\nFirst, let's note that the quadrilateral $LCIM'$ can be inscribed into a circle. In fact, $\\angle LCT$ is equal to the half-sum of the $\\operatorname{arcs} LA$ and $AT$, which is equal to $\\frac{1}{2}(\\breve{LA} + TB) = \\angle LKA$. But $LKA$ is the angle formed by the tangent $AK$ and the chord $LK$ of the circle $LKM$, and is therefore equal to $LM'K$, subtended by this chord. Hence $\\angle LCT = \\angle LM'K$ and the quadrilateral $LCMI$ can be inscribed in a circle. Therefore $\\angle LM'C = \\angle LIC$.\n\nFurther, $\\angle AKT = \\frac{1}{2}(\\breve{AT} + \\breve{BL}) = \\frac{1}{2}(\\check{TB} + \\check{BL}) = \\angle LAT$. Hence the triangles $TAK$ and $TLA$ are similar, and so $TK \\cdot TL = TA^2$. Also $\\angle AIT = \\angle CAI + \\angle ACI = \\angle CAI + \\angle TAB = \\angle BAI + \\angle TAB = \\angle TAI$. Therefore $TAI$ is an equilateral triangle, and so $TI^2 = TA^2 = TK \\cdot TL$, which implies that the triangles $TKI$ and $TIL$ are similar, and so $\\angle LIC = \\angle LKI$. Hence $\\angle LM'C = \\angle LIC = \\angle LKM'$, so the angle subtended by the chord $LM'$ is equal to the angle between this chord and $M'C$, so $M'C$ is a tangent, and so $M' = M$.\n\nIt is time to use the definition of the point $P$. Draw a tangent to the inscribed circle of $ABC$ parallel to $AB$. Let it be tangent to this circle at point $F$. The circle tangent to the continuations of $AC$, $BC$ and to $AB$ at $P$ can be obtained from the inscribed circle of $ABC$ by a similarity transformation centered at $C$. This transformation would take $F$ to $P$. Hence $C$, $F$ and $P$ are collinear. Let $V$ be the midpoint of $PE$, where $E$ is the tangency point of $AB$ with the inscribed circle of $ABC$. As $IV \\parallel FP$ ($IV$ connects the midpoints of the sides of $FEP$), $KIV$ is similar to $KMP$, and so $VI = VK$. Hence $IE^2 = VI^2 - VE^2 = VK^2 - VE^2 = (VK + VE)(VK - VE) = EK \\cdot PK$.\n\nNow let $D$ be the center of the circle $KLM$. As $\\angle KDP = \\angle IKE$, the triangles $KDP$ and $EKI$ are similar. Hence $DK \\cdot IE = EK \\cdot PK$, and so $IE^2 = EK \\cdot PK = DK \\cdot IE$, and so $IE = DK$, as wanted.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the greatest real number $a$ such that the inequality\n$$\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \\geq a\\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\\right)\n$$\nholds for every five real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$.",
    "options": [],
    "answer": "2/√3",
    "solution": "Solution:\n\nNote that\n$$\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} = \\left(x_{1}^{2}+\\frac{x_{2}^{2}}{3}\\right) + \\left(\\frac{2 x_{2}^{2}}{3}+\\frac{x_{3}^{2}}{2}\\right) + \\left(\\frac{x_{3}^{2}}{2}+\\frac{2 x_{4}^{2}}{3}\\right) + \\left(\\frac{x_{4}^{2}}{3}+x_{5}^{2}\\right) .\n$$\n\nNow applying the inequality $a^{2}+b^{2} \\geq 2 a b$ we get\n$$\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \\geq \\frac{2}{\\sqrt{3}}\\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\\right) \\text{.}\n$$\n\nThis proves that $a \\geq \\frac{2}{\\sqrt{3}}$. In order to prove $a \\leq \\frac{2}{\\sqrt{3}}$ it is enough to notice that for $(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (1, \\sqrt{3}, 2, \\sqrt{3}, 1)$ we have\n$$\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} = \\frac{2}{\\sqrt{3}}\\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\\right) \\text{.}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, $c$ be positive real numbers with $a + b + c = 1$. Prove that\n$$\na^{4} + b^{4} + c^{4} \\geq a b c .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $a + b + c = 1$, we can multiply the right side by $a + b + c$ to get the equivalent inequality\n$$\na^{4} + b^{4} + c^{4} \\geq a^{2} b c + a b^{2} c + a b c^{2} .\n$$\nBy AM-GM,\n$$\n\\frac{2 a^{4} + b^{4} + c^{4}}{4} = \\frac{a^{4} + a^{4} + b^{4} + c^{4}}{4} \\geq \\sqrt[4]{a^{4} a^{4} b^{4} c^{4}} = a^{2} b c .\n$$\nSimilarly,\n$$\n\\begin{aligned}\n& \\frac{a^{4} + 2 b^{4} + c^{4}}{4} \\geq a b^{2} c \\\\\n& \\frac{a^{4} + b^{4} + 2 c^{4}}{4} \\geq a b c^{2}\n\\end{aligned}\n$$\nAdding these three inequalities proves the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n > 1$ be an odd integer. Prove that every integer $l$ satisfying $1 \\leq l \\leq n$ can be represented as a sum or difference of two integers each of which is less than $n$ and relatively prime to $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe will use the following lemma (it is known as the Chinese Remainder Theorem).\n\nLemma. Let $m_{1}, m_{2}, \\ldots, m_{k}$ be different relatively prime numbers. If $q_{1}, q_{2}, \\ldots, q_{k}$ are arbitrary non-negative integers then there exists a natural number $x$ less than $m_{1} m_{2} \\ldots m_{k}$ such that\n$$\n\\begin{aligned}\nx & \\equiv q_{1}\\left(\\bmod m_{1}\\right) \\\\\nx & \\equiv q_{2}\\left(\\bmod m_{2}\\right) \\\\\n& \\vdots \\\\\nx & \\equiv q_{k}\\left(\\bmod m_{k}\\right) .\n\\end{aligned}\n$$\nProof of the Lemma. We can assume that $q_{i} < m_{i}$ for $1 \\leq i \\leq k$. We will prove this by induction. For $k = 2$ we consider the numbers $x_{1} = q_{1}, x_{2} = m_{1} + q_{1}, x_{3} = 2 m_{1} + q_{1}, \\ldots, x_{m_{2}} = (m_{2} - 1) m_{1} + q_{1}$. If two of the numbers $x_{i}$ and $x_{j}$ give the same remainder upon division by $m_{2}$ we have that $x_{j} - x_{i}$ is divisible by $m_{2}$ which is impossible since that difference is equal to $(j - i) m_{1}$. Thus all $x_{i}$'s give different remainders modulo $m_{2}$ and one of them has to give a remainder $q_{2}$. This finishes the proof for the case $k = 2$.\n\nAssume now that the statement holds for $k$ and we want to prove it for $k + 1$. Applying the inductional hypothesis $q_{1}, \\ldots, q_{k}$ we find a number $x'$ such that $x' \\equiv q_{i}\\left(\\bmod m_{i}\\right)$ for $1 \\leq i \\leq k$. Now by the case $k = 2$ we get that there is an $x$ such that $x \\equiv x'\\left(\\bmod m_{1} \\cdots m_{k}\\right)$ and $x \\equiv q_{k+1}\\left(\\bmod m_{k+1}\\right)$. Such $x$ satisfies the required condition.\n\nLet $n = p_{1}^{\\alpha_{1}} \\ldots p_{k}^{\\alpha_{k}}$, where $p_{1}, \\ldots, p_{k}$ are prime numbers and $\\alpha_{1}, \\ldots, \\alpha_{k}$ positive integers. Assume that $l$ is given and that $l \\equiv t_{i}\\left(\\bmod p_{i}\\right)$. Since $p_{i} > 2$ we can choose $s_{i}$ such that $s_{i} \\not\\equiv 0\\left(\\bmod p_{i}\\right)$ and $s_{i} \\not\\equiv t_{i}\\left(\\bmod p_{i}\\right)$. By the Chinese Remainder Theorem there exists an integer $s < p_{1} \\ldots p_{k}$ such that $s \\equiv s_{i}\\left(\\bmod p_{i}\\right)$ for every $i = 1, \\ldots, k$. If $s < l$ then choose $a = s$ and $b = l - s$. If $s > l$ then we can choose $a = s$ and $b = s - l$. It is easy to verify that such $a$ and $b$ satisfy the conditions of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEach of three schools contain $n$ students. Each student has at least $n+1$ friends among students of the other two schools. Prove that there are three students, all from different schools who are friends to each other. (Friendship is symmetric: If $A$ is a friend to $B$, then $B$ is a friend to $A$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose to the contrary that there doesn't exist three students from different schools who know each other. Denote schools by $A$, $B$ and $C$. Consider the student (or one of the students if there are more of them) who has the maximal number of friends in one of the schools. Let $a$ be the name of that student, suppose he is from the school $A$ and suppose that he has $k$ friends denoted by $b_{1}, b_{2}, \\ldots, b_{k}$ from the school $B$. Then he must have at least $n+1-k$ friends in the school $C$, say $c_{1}, c_{2}, \\ldots, c_{n+1-k}$. Then student $c_{1}$ doesn't know any of $b_{1}, b_{2}, \\ldots, b_{k}$ (otherwise, if he knows $b_{i}$, then $a, b_{i}$ and $c_{1}$ would be the desired triple), so he can have at most $n-k$ friends in the school $B$. Since $c_{1}$ has at least $n+1$ friends in the schools $B$ and $A$, he must have at least $n+1-(n-k)=k+1$ friends in $A$. However, this is impossible since we assumed that $k$ is the maximal number of friends a student can have in one of the schools.\n\nThe obtained contradiction proves the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven five vertices of a regular heptagon, construct the two remaining vertices using straightedge alone.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $A, B, C, D$, and $E$ be the known vertices and $F$ and $G$ the unknown vertices. The arrangement of $A, B, C, D$, and $E$ depends on the relative positions of $F$ and $G$ as shown in the diagram.\n![](attached_image_1.png)\nBy connecting the intersections of $B C$ with $D E$ and $B D$ with $C E$ one finds $l$, the line of symmetry through $G$. Let $A B$ and $A C$ meet $l$ in $H$ and $I$ respectively. By symmetry about $l, H E$ and $I D$ both pass through $F$. Now perform the same construction on $B, C, D, E, F$ to get $G$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA number bracelet in base $m$ is made by choosing two nonnegative integers less than $m$ (not both $0$) and continuing in a clockwise loop, each succeeding number being the mod $m$ sum of its two predecessors. The figure is closed up as soon as it starts to repeat. The figure to the right shows two number bracelets in base $10$, starting with the pairs $(1,3)$, and $(2,2)$, respectively. Prove that the lengths of all number bracelets in a given base are divisors of the length of the number bracelet beginning with $(0,1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe elements of a number bracelet $N$ will be denoted $N_{0}, N_{1}, N_{2}, \\ldots$; $N_{0}$ and $N_{1}$ being the starting numbers. Let $F$ be the bracelet with starting numbers $F_{0}=0, F_{1}=1$. All congruences are modulo $m$ unless otherwise noted.\n\nSince there are only $m^{2}$ possible pairs $(N_{k}, N_{k+1})$, there must be repetition, so that\n$$\n(N_{k}, N_{k+1}) = (N_{p+k}, N_{p+k+1})\n$$\nfor some $p \\geq 1$ and $k$. If this is true for one $k$, it must be true for the next $k$, since $N_{k+1} = N_{p+k+1}$ by hypothesis and $N_{k+2} \\equiv N_{k} + N_{k+1} = N_{p+k} + N_{p+k+1} \\equiv N_{p+k+2}$, and so it is true for all $k$ bigger than that $k$. It must also be true for the previous $k$, since $N_{k} = N_{p+k}$ by hypothesis and $N_{k-1} \\equiv N_{k+1} - N_{k} = N_{p+k+1} - N_{p+k} \\equiv N_{p+k-1}$, and so it is true for all $k$ smaller than that $k$. Therefore it is true for all $k$.\n\nLet $P_{N}$, the period of $N$, be the smallest $p$ such that this is true for some $k$ (and therefore for all $k$).\n\nIt is clear that if $0 \\leq i, j < P(N), i \\neq j$, then $(N_{i}, N_{i+1}) \\neq (N_{j}, N_{j+1})$. Therefore, among the terms $N_{0}$ through $N_{P_{N}}$, there is no repetition even of pairs of adjacent terms. But because of the property of $P_{N}$, $(N_{0}, N_{1}) = (N_{P_{N}}, N_{P_{N}+1})$ and the repetition will continue. Thus the number bracelet $N$ consists of a single loop of $P_{N}$ elements.\n\nWe will prove by induction that for all $n \\geq 1$,\n$$\nN_{n} \\equiv N_{0} F_{n-1} + N_{1} F_{n}.\n$$\nThe case $n=1$ is trivial, and $n=2$ follows directly from the definition of $N$. If this is true for $n=k$ and $n=k-1$, $N_{k+1} \\equiv N_{k} + N_{k-1} \\equiv N_{0} F_{k-1} + N_{1} F_{k} + N_{0} F_{k-2} + N_{1} F_{k-1} = N_{0} F_{k} + N_{1} F_{k+1}$ so it is also true for $n=k+1$.\n\nIt is clear from the circularity of the number bracelet that $N_{k+R} = N_{k}$ for all $k \\geq 1$ if and only if $P_{N} \\mid R$. Since $N_{k+P_{F}} = N_{0} F_{P_{F}+k-1} + N_{1} F_{P_{F}+k} = N_{0} F_{k-1} + N_{1} F_{k} = N_{k}$ for all $k \\geq 1$, $P_{N} \\mid P_{F}$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 46254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvery two members of a certain society are either friends or enemies. Suppose that there are $n$ members of the society, that there are exactly $q$ pairs of friends, and that in every set of three persons there are two who are enemies to each other. Prove that there is at least one member of the society among whose enemies there are at most $q \\cdot\\left(1-\\frac{4 q}{n^{2}}\\right)$ pairs of friends.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenote by $S$ the set of all members of the society, by $A$ the set of all pairs of friends, and by $N$ the set of all pairs of enemies. For every $x \\in S$, denote by $f(x)$ the number of friends of $x$ and by $F(x)$ the number of pairs of friends among enemies of $x$. It is easy to prove:\n$$\n\\begin{gathered}\nq = |A| = \\frac{1}{2} \\sum_{x \\in S} f(x) ; \\\\\n\\sum_{\\{a, b\\} \\in A}(f(a)+f(b)) = \\sum_{x \\in S} f^{2}(x) .\n\\end{gathered}\n$$\nIf $a$ and $b$ are friends, then the number of their common enemies is equal to $(n-2)-(f(a)-1)-(f(b)-1)=n-f(a)-f(b)$. Thus\n$$\n\\frac{1}{n} \\sum_{x \\in S} F(x) = \\frac{1}{n} \\sum_{\\{a, b\\} \\in A}(n-f(a)-f(b)) = q - \\frac{1}{n} \\sum_{x \\in S} f^{2}(x)\n$$\nUsing the inequality between arithmetic and quadratic mean on the last expression, we get\n$$\n\\frac{1}{n} \\sum_{x \\in S} F(x) \\leq q - \\frac{4 q^{2}}{n^{2}}\n$$\nand the statement of the problem follows immediately.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a_{1}, a_{2}, \\ldots, a_{n}$ be distinct integers. Prove that there do not exist two nonconstant integer-coefficient polynomials $p$ and $q$ such that\n$$\n\\left(x-a_{1}\\right)\\left(x-a_{2}\\right) \\cdots\\left(x-a_{n}\\right)-1=p(x) q(x)\n$$\nfor all $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume for the sake of contradiction that $p$ and $q$ exist. If we substitute $x=a_{i}$ for $i=1, \\ldots, n$, the left side of (1) becomes $-1$. Since $p\\left(a_{i}\\right)$ and $q\\left(a_{i}\\right)$ are both integers, we either have\n$$\np\\left(a_{i}\\right)=1, \\quad q\\left(a_{i}\\right)=-1\n$$\nor\n$$\np\\left(a_{i}\\right)=-1, \\quad q\\left(a_{i}\\right)=1\n$$\nIn either case,\n$$\n(p+q)\\left(a_{i}\\right)=0\n$$\nThus $p+q$ is a polynomial with $n$ distinct roots $a_{1}, a_{2}, \\ldots, a_{n}$. Such a polynomial must have degree at least $n$—unless it is the zero polynomial.\n\nCase 1. $p+q$ has degree $\\geq n$. Then one of $p$ and $q$ has degree $\\geq n$, and the other, of course, has degree $\\geq 1$. It follows that $pq$ has degree $\\geq n+1$, a contradiction since the left side of (1) has degree $n$.\n\nCase 2. $p(x)+q(x)=0$ for all $x$. Substituting $q=-p$ into (1),\n$$\n\\left(x-a_{1}\\right)\\left(x-a_{2}\\right) \\cdots\\left(x-a_{n}\\right)-1=-p(x)^{2}.\n$$\nNow substitute an integer for $x$ that is so huge that each of the factors $\\left(x-a_{i}\\right)$ is greater than $1$. This makes the left side strictly positive. Since the right side is negative or zero, we have a contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n \\geq 1$ be an integer. How many ways can the rectangle having vertices $(0,0)$, $(n, 0)$, $(n, 1)$, $(0,1)$ be dissected into $2n$ triangles, all vertices of which have integer coordinates?\n\nThe triangles are considered as positioned on the coordinate plane; in particular, tilings related by rotation and reflection are considered distinct.",
    "options": [],
    "answer": "binomial(2n, n)",
    "solution": "Solution:\nSince the vertices of the tiles must lie within the rectangle and cannot be collinear, all tiles have the form\n$$\n(a, 0)(b, 0)(c, 1) \\text{ or } \\quad (a, 1)(b, 1)(c, 0) \\quad (a < b).\n$$\nSuch a triangle has area $(b-a)/2$; for $2n$ such triangles to tile a rectangle of area $n$, it is necessary that they all have area $1/2$, i.e. be of the form\n$$\n(a, 0)(a+1, 0)(c, 1) \\quad \\text{ or } \\quad (a, 1)(a+1, 1)(c, 0).\n$$\nLet $T_{a, b}$, for $a, b \\geq 0$, be the trapezoid (or rectangle, triangle, degenerate segment) having vertices $(0,0)$, $(a, 0)$, $(b, 1)$, $(0,1)$, and let $t_{a, b}$ be the number of ways to tile it with triangles of type (11). If $a$ and $b$ are positive, the right-hand edge can be covered in only two ways:\n- Laying the tile $(a-1, 0)(a, 0)(b, 1)$ and tiling the resulting $T_{a-1, b}$;\n- Laying the tile $(a, 0)(b-1, 1)(b, 1)$ and tiling the resulting $T_{a, b-1}$.\nThus in this case $t_{a, b} = t_{a-1, b} + t_{a, b-1}$. This, together with the initial conditions $t_{a, 0} = t_{0, b} = 1$, is the recursion of Pascal's triangle; it shows that $t_{a, b} = \\binom{a+b}{a}$. Thus the desired answer is $t_{n, n} = \\binom{2n}{n}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle such that $\\angle A = 90^{\\circ}$ and $\\angle B < \\angle C$. The tangent at $A$ to its circumcircle $\\omega$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ across $BC$, $X$ the foot of the perpendicular from $A$ to $BE$, and $Y$ the midpoint of $AX$. Let the line $BY$ meet $\\omega$ again at $Z$. Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $M$ be the point of intersection of $AE$ and $BC$, and let $N$ be the point on $\\omega$ diametrically opposite $A$. Since $\\angle B < \\angle C$, points $N$ and $B$ are on the same side of $AE$.\n\nFurthermore, $\\angle NAE = \\angle BAX = 90^{\\circ} - \\angle ABE$; hence the triangles $NAE$ and $BAX$ are similar. Consequently, $\\triangle BAY$ and $\\triangle NAM$ are also similar, since $M$ is the midpoint of $AE$. Thus $\\angle ANZ = \\angle ABZ = \\angle ABY = \\angle ANM$, implying that $N, M, Z$ are collinear.\n\nNow we have $\\angle ZMD = 90^{\\circ} - \\angle ZMA = \\angle EAZ = \\angle ZED$ (the last equality because $ED$ is tangent to $\\omega$); hence $ZMED$ is a cyclic quadrilateral.\n\n![](attached_image_1.png)\n\nIt follows that $\\angle ZDM = \\angle ZEA = \\angle ZAD$, which is enough to conclude that $MD$ is tangent to the circumcircle of $AZD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that\n$$\ntan \\left(\\frac{3 \\pi}{11}\\right)+4 \\sin \\left(\\frac{2 \\pi}{11}\\right)=\\sqrt{11}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe work with complex roots of unity. Let $\\zeta=\\cos \\frac{\\pi}{11}+i \\sin \\frac{\\pi}{11}$; then we know that $\\zeta^{n}=\\cos \\frac{n \\pi}{11}+i \\sin \\frac{n \\pi}{11}$ and, in particular, $\\zeta^{11}=-1$. Now, the given expression is clearly positive (since each term is positive), so we need only check that its square is $11$. To do that, observe that $\\sin \\frac{2 \\pi}{11}=\\frac{\\zeta^{2}+\\zeta^{9}}{2i}$, $\\sin \\frac{3 \\pi}{11}=\\frac{\\zeta^{3}+\\zeta^{8}}{2i}$, and $\\cos \\frac{3 \\pi}{11}=\\frac{\\zeta^{3}-\\zeta^{8}}{2}$; hence our objective is to show that\n$$\n\\left(2 \\frac{\\zeta^{2}+\\zeta^{9}}{i}+\\frac{\\zeta^{3}+\\zeta^{8}}{(\\zeta^{3}-\\zeta^{8}) i}\\right)^{2}=11\n$$\nor, equivalently, that $\\left[2(\\zeta^{2}+\\zeta^{9})(\\zeta^{3}-\\zeta^{8})+(\\zeta^{3}+\\zeta^{8})\\right]^{2}=-11(\\zeta^{3}-\\zeta^{8})^{2}$. The expression inside brackets on the left multiplies out to $2 \\zeta^{5}+2 \\zeta^{12}-2 \\zeta^{10}-2 \\zeta^{17}+\\zeta^{3}+\\zeta^{8}=-2 \\zeta^{10}+\\zeta^{8}+2 \\zeta^{6}+2 \\zeta^{5}+\\zeta^{3}-2 \\zeta$; when we square this and collect terms (remembering again that $\\zeta^{11}=-1$), we obtain\n$$\n\\begin{gathered}\n4 \\zeta^{10}-4 \\zeta^{9}+4 \\zeta^{8}-4 \\zeta^{7}-7 \\zeta^{6}+7 \\zeta^{5}+4 \\zeta^{4}-4 \\zeta^{3}+4 \\zeta^{2}-4 \\zeta-18 \\\\\n=4\\left(\\zeta^{10}-\\zeta^{9}+\\zeta^{8}-\\cdots+1\\right)+11\\left(\\zeta^{5}-2-\\zeta^{6}\\right) .\n\\end{gathered}\n$$\nSince the first expression in parentheses is $\\frac{\\zeta^{11}+1}{\\zeta+1}=0$ and the second is $-\\left(\\zeta^{3}-\\zeta^{8}\\right)^{2}$, we have what we wanted.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that any prime which is the difference of two cubes is also the sum of a square and three times a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose that\n$$\np = a^{3} - b^{3}\n$$\nwhere $p$ is prime and $a$ and $b$ are natural numbers. The right side can be factored:\n$$\np = (a - b)\\left(a^{2} + a b + b^{2}\\right)\n$$\nSince $p$ is prime, one of the factors must be $\\pm 1$. We note that $a^{2} + a b + b^{2} \\geq 1 + 1 + 1 = 3$, so $a - b = 1$ and\n$$\np = a^{2} + a b + b^{2} = (b + 1)^{2} + b(b + 1) + b^{2} = 3 b^{2} + 3 b + 1.\n$$\nTrying to express $3 b^{2} + 3 b + 1$ directly as $f(b)^{2} + 3 g(b)^{2}$, where $f(b)$ and $g(b)$ are integer polynomials, leads to difficulties (the coefficient of $b$ can never be made odd). However, we note that $b$ is either even or odd. If $b = 2x$, then\n$$\np = 12 x^{2} + 6 x + 1 = (3x + 1)^{2} + 3 x^{2}\n$$\nwhile if $b = 2x - 1$, then\n$$\np = 12 x^{2} - 6 x + 1 = (3x - 1)^{2} + 3 x^{2}.\n$$\nThus $p = c^{2} + 3 d^{2}$ where $c$ and $d$ are integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that there exists an infinite set $S$ of positive integers with the property that if we take any finite subset $T$ of $S$, the sum of the elements of $T$ is not a perfect $k$th power for any $k \\geq 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$$\nS = \\left\\{2, 2^{2} \\cdot 3, 2^{2} \\cdot 3^{2} \\cdot 5, 2^{2} \\cdot 3^{2} \\cdot 5^{2} \\cdot 7, 2^{2} \\cdot 3^{2} \\cdot 5^{2} \\cdot 7^{2} \\cdot 11, \\ldots \\right\\}.\n$$\nLet $T \\subseteq S$ be a finite nonempty set, and $m$ its smallest element. Let $p$ be the prime factor which appears exactly once in $m$. Then $p$ divides every element of $T$, and $p^{2}$ divides every element of $T$ except $m$. From this it follows that the sum of the elements of $T$ has exactly one factor of $p$; thus, it cannot be a $k$th power.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor $x$ a positive real number with finitely many decimal places, denote by $r(x)$ the number formed by reversing the digits and decimal point of $x$. For instance, $r(98.6)=6.89$ and $r(740)=0.047$.\n\na. Prove that for all positive real numbers $x$ and $y$ with finitely many decimal places,\n$$\nr(x y) \\leq 10 r(x) r(y) .\n$$\n\nb. Determine whether there exist values of $x$ and $y$, each having at least 2015 nonzero digits, such that equality holds.",
    "options": [],
    "answer": "Yes. For example, let x be the integer whose decimal representation is two thousand fifteen consecutive ones, and let y be the integer whose decimal representation consists of two thousand fifteen ones, each pair of consecutive ones separated by two thousand fourteen zeros. For these, equality holds.",
    "solution": "Solution:\n\na. Write\n$$\nx=\\sum_{i=i_{0}}^{i_{1}} x_{i} 10^{i} \\quad \\text{ and } \\quad y=\\sum_{j=j_{0}}^{j_{1}} y_{j} 10^{j}\n$$\nwhere the $x_{i}$ and $y_{j}$ are digits, and the limits $i_{0}, i_{1}, j_{0}, j_{1}$ are integers, possibly negative. Then\n$$\nr(x)=\\sum_{i=i_{0}}^{i_{1}} x_{i} 10^{-1-i} \\quad \\text{ and } \\quad r(y)=\\sum_{j=j_{0}}^{j_{1}} y_{j} 10^{-1-j}\n$$\nand we may multiply:\n$$\nx y=\\sum_{i, j} x_{i} y_{j} 10^{i+j} \\text{ and } \\quad r(x) r(y)=\\sum_{i, j} x_{i} y_{j} 10^{-2-i-j} .\n$$\nLet (initially) $z_{k}=\\sum_{i+j=k} x_{i} y_{j}$, a finite sum since $x$ has only finitely many nonzero digits. Then\n$$\nA:=\\sum_{k} z_{k} 10^{k}=x y, \\quad \\text{ and } \\quad B:=\\sum_{k} z_{k} 10^{-1-k}=10 r(x) r(y)\n$$\nIf all $z_{k}<10$, then the $z_{k}$ are the digits of $A$ and we simply get $B=r(A)$. However, in general, there will be carrying involved in the multiplication of $x$ by $y$. Consider the following \"carry transform\": find the least $k$ such that $z_{k} \\geq 10$, decrease $z_{k}$ by 10, and increase $z_{k+1}$ by 1. This does not change the sum $A$ at all, and it decreases the sum $B$ by $99 \\cdot 10^{-2-k}$. We can only apply the carry transform finitely many times since, once $z_{k}$ is brought below 10, we move on to $z_{k+1}$, and continue increasing $k$ until $10^{k}>A$, at which point $z_{k}$ must remain zero. Now $A$ still equals $x y$, and the $z_{k}$ are its digits; and $B$ has been decreasing so\n$$\nr(x y)=r(A)=B \\leq 10 r(x) r(y) .\n$$\n\nb. The answer is yes; the choice\n$$\nx=\\overbrace{111 \\cdots 1}^{2015}, \\quad y=\\overbrace{1 \\overbrace{000 \\cdots 0}^{2014} 1 \\overbrace{000 \\cdots 0}^{2014} 1 \\cdots 1}^{20151^{1 ' s}}\n$$\nis readily seen to yield\n$$\nr(x y)=10 r(x) r(y)=0 . \\overbrace{11111 \\cdots 1}^{2015^{2}} .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\{a_{1}, a_{2}, a_{3}, \\ldots\\}$ be a sequence of real numbers such that for each $n \\geq 1$,\n$$\na_{n+2}=a_{n+1}+a_{n}\n$$\nProve that for all $n \\geq 2$, the quantity\n$$\n\\left|a_{n}^{2}-a_{n-1} a_{n+1}\\right|\n$$\ndoes not depend on $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt suffices to prove that increasing $n$ to $n+1$ does not change the value, i.e. that for $n \\geq 2$,\n$$\n\\left|a_{n}^{2}-a_{n-1} a_{n+1}\\right|=\\left|a_{n+1}^{2}-a_{n} a_{n+2}\\right| .\n$$\nWe will prove more specifically that\n$$\na_{n}^{2}-a_{n-1} a_{n+1}=-\\left(a_{n+1}^{2}-a_{n} a_{n+2}\\right)\n$$\nby transforming:\n$$\n\\begin{aligned}\na_{n}^{2}-a_{n-1} a_{n+1} & \\stackrel{?}{=}-\\left(a_{n+1}^{2}-a_{n} a_{n+2}\\right) \\\\\na_{n}^{2}-a_{n-1} a_{n+1} & \\stackrel{?}{=}-a_{n+1}^{2}+a_{n} a_{n+2} \\\\\na_{n+1}^{2}-a_{n-1} a_{n+1} & \\stackrel{?}{=} a_{n} a_{n+2}-a_{n}^{2} \\\\\na_{n+1}\\left(a_{n+1}-a_{n-1}\\right) & \\stackrel{?}{=} a_{n}\\left(a_{n+2}-a_{n}\\right) .\n\\end{aligned}\n$$\nUsing the given relation $a_{n+2}=a_{n+1}+a_{n}$, we see that the right side equals $a_{n} \\cdot a_{n+1}$. Replacing $n$ by $n-1$ in the given relation gives $a_{n+1}=a_{n}+a_{n-1}$, so the left side equals $a_{n+1} \\cdot a_{n}$ and thus the equality is true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Let $a$, $b$, and $n$ be positive integers such that $a b = n^{2} + 1$. Prove that\n$$\n|a-b| \\geq \\sqrt{4 n - 3}\n$$\n\nb. Prove that there are infinitely many such triples $(a, b, n)$ such that equality occurs.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nWe may assume that $a \\geq b$ (thus eliminating the absolute value sign). Write $t = a - b$, so $a = b + t$. Then the equality $n^{2} + 1 = a b$ can be rewritten as $n^{2} + 1 = (b + t) b$ or $b^{2} + t b - (n^{2} + 1) = 0$.\n\nReplacing $b$ by $x$, we consider the quadratic equation $x^{2} + t x - (n^{2} + 1) = 0$. It has integer coefficients and the integer root $x = b$. Hence the discriminant\n$$\nD = t^{2} + 4(n^{2} + 1)\n$$\nis a perfect square. Clearly $D > 4 n^{2} = (2 n)^{2}$. So\n$$\n\\begin{aligned}\nD & \\geq (2 n + 1)^{2} \\\\\nt^{2} + 4 n^{2} + 4 & \\geq 4 n^{2} + 4 n + 1 \\\\\nt^{2} & \\geq 4 n - 3 \\\\\nt & \\geq \\sqrt{4 n - 3} .\n\\end{aligned}\n$$\n\nb.\nThe construction $a = k^{2} + 1$, $b = (k + 1)^{2} + 1$, $n = k^{2} + k + 1$ (for any integer $k \\geq 0$) is easily verified to work.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoes there exist a function $f$ from the positive integers to itself, such that for any positive integers $a$ and $b$, we have $\\operatorname{gcd}(a, b)=1$ if and only if $\\operatorname{gcd}(f(a), f(b))>1$ holds?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Assume that $f$ satisfies the hypothesis. Let $k$ denote the number of distinct primes dividing $f(1)$. For every integer $e$, the number $f\\left(2^{e}\\right)$ shares some prime factor with $f(1)$. So among $f(2), f(4), \\ldots, f\\left(2^{k+1}\\right)$ two of them have the same shared prime factor, which is impossible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA set $S$ of irrational real numbers has the property that among any subset of five numbers in $S$, one can find two with irrational sum. How large can $|S|$ be?",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nThe answer is $|S| \\leq 8$. An example is $S=\\{n \\pm \\sqrt{2} \\mid n=1,2,3,4\\}$ (and any of its subsets).\n\nIn general, construct a graph with vertex set $S$ in which we join two numbers with rational sum. We claim this graph is bipartite; indeed if $a_{1}+a_{2}, a_{2}+a_{3}, \\ldots, a_{n}+a_{1}$ are all rational for some odd $n$, solving the resulting system of equations gives $a_{1}$, $\\ldots$, $a_{n}$ all rational numbers.\n\nAccordingly we may 2-color $S$. If $|S| \\geq 9$, then we may find a set of five numbers with rational sum, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $a, b, c$ are rational numbers such that\n$$\n\\begin{aligned}\n& \\left(a^{2}+1\\right)^{3}=b+1 \\\\\n& \\left(b^{2}+1\\right)^{3}=c+1 \\\\\n& \\left(c^{2}+1\\right)^{3}=a+1\n\\end{aligned}\n$$\nProve that $a=b=c=0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe have that $b=\\left(a^{2}+1\\right)^{3}-1$, $c=\\left(b^{2}+1\\right)^{3}-1$, and $a=\\left(c^{2}+1\\right)^{3}-1$. By direct substitution we derive that $a$ satisfies the following polynomial equation of degree 216:\n$$\n\\left(\\left(\\left(\\left(\\left(a^{2}+1\\right)^{3}-1\\right)^{2}+1\\right)^{3}-1\\right)^{2}+1\\right)^{3}-(a+1)=0\n$$\nWe observe that the polynomial can be rewritten as\n$$\na^{216}+c_{215} a^{215}+\\cdots+c_{2} a^{2}-a=0\n$$\nfor some integers $c_{2}, \\ldots, c_{215}$. Hence by the Rational Root Theorem, if $a \\neq 0$ then it follows that $a= \\pm 1$. So $a \\in\\{-1,0,1\\}$. Similarly, $b, c \\in\\{-1,0,1\\}$ as well.\n\nBut if $a= \\pm 1$, then we have $b=(1+1)^{3}-1=7$, which is impossible. Hence only $a=0$ can occur. Thus $a=b=c=0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeveral positive integers are written on the blackboard. You can erase any two numbers and write their greatest common divisor (GCD) and least common multiple (LCM) instead. Prove that eventually the numbers will stop changing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst of all, since\n$$\na \\cdot b = \\operatorname{gcd}(a, b) \\cdot \\operatorname{lcm}(a, b)\n$$\nfor all positive integers $a$ and $b$, the product of the numbers on the board does not change.\n\nSecond, we claim that at each step in which the numbers change, their sum increases. Suppose we pick two numbers $a$ and $b$ ($a \\leq b$) and replace them with $g = \\operatorname{gcd}(a, b)$ and $\\ell = \\operatorname{lcm}(a, b)$. If $g = a$, then $\\ell = b$ and the numbers do not change, so $g < a$ and therefore $g < b$. The change in the sum of the numbers is\n$$\n(g + \\ell) - (a + b) = \\frac{g^2 + g \\ell - g a - g b}{g} = \\frac{g^2 + a b - g a - g b}{g} = \\frac{(a - g)(b - g)}{g} > 0.\n$$\nSince each number on the blackboard cannot exceed the product of the numbers that were originally there, the sum cannot increase unboundedly, and thus the numbers must stop changing.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe positive integers from $1$ to $100$ are written, in some order, in a $10 \\times 10$ square. In each row, the five smallest numbers are crossed out. In each column, the five largest numbers (including those that have already been crossed out) are circled. Prove that at least $25$ numbers will be circled but not crossed out.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe generalize to the following:\nLet $m \\geq p$, $n \\geq q$ be integers. Each square of an $m$-row, $n$-column grid is filled with a different positive integer.\nThen at least $p q$ numbers are among the $p$ largest numbers in their columns and the $q$ largest numbers in their rows.\nWe prove this latter statement by induction on $m+n$. The base case, when $m=1$ or $n=1$, is trivial. Given an $m \\times n$ grid with $m \\geq 2$ and $n \\geq 2$, let $L$ be the largest number that is one of the largest in its row or its column, but not both. Assume without loss of generality that $L$ is among the largest in its column and not among the largest in its row, so the $q$ largest numbers in $L$'s row are all larger than $L$ and all among the largest in their columns. Temporarily remove this row from the matrix. By the induction hypothesis, at least $(p-1) q$ numbers are now among the $q$ largest numbers in their rows and the $p-1$ largest numbers in their columns. These are also among the $p$ largest numbers in their columns in the original matrix. Adding the $q$ largest numbers in the deleted row, we find $(p-1) q+q=p q$ numbers satisfying the original condition. The problem follows by taking $m=n=10$ and $p=q=5$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIt's a week before Thanksgiving, and a family is trying to find their turkey. There are 5 boxes in a row, and the turkey is hiding in one of the 5 boxes.\n\nEvery day, the family is allowed to check one box to try to find the turkey, and every night, the turkey moves to a box right next to the box it was in. For example, from box 3 it could move to box 2 or 4, and from box 5 it must move to box 4. Determine a strategy for the family to catch the turkey before Thanksgiving.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOne strategy is as follows:\n\nFirst assume the turkey starts in an even numbered box (box 2 or 4). On day 1, check box 2. If the turkey is not there, he must have been in box 4 to begin with, so tomorrow he will be in box 3 or box 5. On day 2, check box 3. If the turkey is not there, then he must currently be in box 5 (since he started in box 4), so tomorrow he will be in box 4. Thus, on day 3, check box 4.\n\nIf he is not there, then the assumption that he began in an even numbered box must have been wrong. Thus, he started in an odd numbered box, which means on day 4 he will be in an even numbered box. Thus, we can repeat the same strategy of checking boxes 2, 3, and 4 on days 4, 5, and 6, and we are guaranteed to find the turkey in at most 6 days.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x_{1}, x_{2}, \\ldots, x_{1000}, y_{1}, y_{2}, \\ldots, y_{1000}$ be 2000 different real numbers, and form the $1000 \\times 1000$ matrix whose $(i, j)$-entry is $x_{i}+y_{j}$. If the product of the numbers in each row is 1, show that the product of the numbers in each column is -1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe given says that $\\left(x_{i}+y_{1}\\right)\\left(x_{i}+y_{2}\\right) \\cdots\\left(x_{i}+y_{1000}\\right)=1$ for each $i=1,2, \\ldots, 1000$. So, if we let $P(x)$ be the polynomial $\\left(x+y_{1}\\right)\\left(x+y_{2}\\right) \\cdots\\left(x+y_{1000}\\right)-1$, the numbers $x_{i}$ are all roots of $P$. These numbers are all distinct, and there are 1000 of them. But $P$, being of degree 1000, can only have 1000 roots, so these are all the roots of $P$ and the polynomial factors as $P(x)=c\\left(x-x_{1}\\right)\\left(x-x_{2}\\right) \\cdots\\left(x-x_{1000}\\right)$ for some constant $c$. Since the leading coefficient of $P$ is 1, we conclude that $c=1$. Thus,\n$$\n\\left(x+y_{1}\\right)\\left(x+y_{2}\\right) \\cdots\\left(x+y_{1000}\\right)-1=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right) \\cdots\\left(x-x_{1000}\\right)\n$$\nis a polynomial identity, valid for all $x$.\nNow choose any $j$ ($1 \\leq j \\leq 1000$); we wish to show that the product of the numbers in the $j$th column of the matrix is $-1$. Letting $x=-y_{j}$ in the above equation, we get\n$$\n\\begin{aligned}\n\\left(-y_{j}+y_{1}\\right)\\left(-y_{j}+y_{2}\\right) \\cdots\\left(-y_{j}+y_{1000}\\right)-1 & =\\left(-y_{j}-x_{1}\\right)\\left(-y_{j}-x_{2}\\right) \\cdots\\left(-y_{j}-x_{1000}\\right) \\\\\n& =(-1)^{1000}\\left(x_{1}+y_{j}\\right)\\left(x_{2}+y_{j}\\right) \\cdots\\left(x_{1000}+y_{j}\\right)\n\\end{aligned}\n$$\nHowever, the product on the left-hand side is $0$, since the $j$th factor is $-y_{j}+y_{j}=0$; also, $(-1)^{1000}=1$. Thus, we obtain $-1=\\left(x_{1}+y_{j}\\right)\\left(x_{2}+y_{j}\\right) \\cdots\\left(x_{1000}+y_{j}\\right)$, which is what we wanted to prove.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 46271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOne marks 16 points on a circle. What is the maximum number of acute triangles with vertices in these points?",
    "options": [],
    "answer": "168",
    "solution": "Solution:\n\nConsider the set of all angles $M_{1} M_{2} M_{3}$, where $M_{1}$, $M_{2}$ and $M_{3}$ is an arbitrary triple of selected points. There are $\\frac{16 \\cdot 15 \\cdot 14}{2} = 1680$ different angles in this set. Suppose $n$ of them are not acute. We shall prove $n \\geq 392$.\n\nFor each integer $m$ between 1 and 7, take a chord with endpoints among the selected points such that there are exactly $m$ selected points to one side of the chord (not including the endpoints). We will call such a chord an $m$-chord. Each $m$-chord subtends not less than $m$ nonacute angles with vertices among the marked points. For each $m \\leq 6$ there are exactly $16$ $m$-chords, and for $m=7$ there are exactly $8$ of them. So the total number of nonacute angles is at least $16(1+2+\\ldots+6) + 8 \\times 7 = 392$.\n\nThere are $\\frac{16 \\cdot 15 \\cdot 14}{6} = 560$ triangles with vertices among the marked points. Each nonacute angle will \"spoil\" exactly one triangle, so the number of acute triangles is not greater than $560 - 392 = 168$.\n\nIt is only left to construct an example with exactly 168 acute triangles. Mark eight consecutive vertices of a right 16-gon $V_{1}, \\ldots, V_{8}$. Draw a line through the center of the 16-gon not parallel to $V_{1} V_{8}$, such that all $V$'s lie on the same side of that line (and not on the line). Reflecting the $V$'s with respect to that line we get points $V_{1}' , \\ldots, V_{8}'$. We claim that the set $V_{1}, \\ldots, V_{8}, V_{1}', \\ldots, V_{8}'$ is as wanted. Indeed, there are no diametrically opposite points in this set (otherwise we would get $V_{1} V_{1}' = 0$ or $V_{8} V_{8}' = 0$), and so for $m=7$ each $m$-chord subtends exactly 8 nonacute angles. Moreover, for each $m \\leq 6$ each $m$-chord subtends exactly $m$ nonacute angles, and so $n=392$, and the number of acute triangles is 168.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe tower function of twos, $T(n)$, is defined by $T(1)=2$ and $T(n+1)=2^{T(n)}$ for $n \\geq 1$. Prove that $T(n)-T(n-1)$ is divisible by $n!$ for $n \\geq 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $U(n) = T(n) - T(n-1)$. Note that\n$$\nU(n) = T(n) - T(n-1) = 2^{T(n-1)} - 2^{T(n-2)} = 2^{T(n-2)}\\left(2^{T(n-1)-T(n-2)} - 1\\right) = T(n-1)\\left(2^{U(n-1)} - 1\\right)\n$$\n\nWe first prove two lemmas, the second a refinement of the first.\n\nLemma 1. $U(n-1) \\mid U(n)$.\n\nProof. By induction on $n$. It is clear when $n=2$ and $n=3$. Suppose that $U(n-2) \\mid U(n-1)$, $n \\geq 3$. Then\n$$\n\\frac{U(n)}{U(n-1)} = \\frac{T(n-1)}{T(n-2)} \\cdot \\frac{2^{U(n-1)}-1}{2^{U(n-2)}-1}.\n$$\nThe first fraction is clearly an integer. Since $a \\mid b$ implies $2^a-1 \\mid 2^b-1$, the second fraction is an integer as well.\n\nLemma 2. If $p$ is a prime and $U(k)$ is divisible by $p^r$, $r \\geq 1$, then $U(k+1)$ is divisible by $p^{r+1}$.\n\nProof. If $p=2$, the $r$ factors of $2$ in $U(k)$ must be contained in $T(k-1)$. Since $T(k) > T(k-1)$, $T(k)$ has at least $r+1$ factors of $2$, and so does $U(k+1)$.\n\nIf $p>2$, we use induction on $n$. Let $n_0$ be the smallest value such that $p \\mid U(n_0)$. Let $r_0$ be the number of factors of $p$ in $U(n_0)$. Since $T(n_0-1)$ is a power of $2$, the $r_0$ factors are all contained in $2^{U(n_0-1)}-1$, i.e.\n$$\n2^{U(n_0-1)} \\equiv 1 \\pmod{p^{r_0}}.\n$$\nBy the preceding lemma, $U(n_0)$ is divisible by $U(n_0-1)$; however, by definition of $n_0$, $U(n_0)$ is divisible by $p$ while $U(n_0-1)$ is not. Consequently\n$$\nU(n_0) = U(n_0-1) \\cdot p \\cdot J\n$$\nwhere $J$ is an integer. We find that\n$$\n\\begin{aligned}\n2^{U(n_0-1)} &\\equiv 1 \\pmod{p^{r_0}} \\\\\n2^{U(n_0-1) \\cdot J} &\\equiv 1 \\pmod{p^{r_0}} \\\\\n2^{U(n_0)} = 2^{U(n_0-1) \\cdot J \\cdot p} &\\equiv 1 \\pmod{p^{r_0+1}}\n\\end{aligned}\n$$\nwhere we have used the fact that $a \\equiv b \\pmod{p^r}$ implies $a^p \\equiv b^p \\pmod{p^{r+1}}$. We conclude that $p^{r_0+1} \\mid 2^{U(n_0)}-1 \\mid U(n_0+1)$. This completes the base case. The induction step is similar, except that $n_0$ and $r_0$ are replaced by $n$ and $r$, and the analogous equation follows from the induction hypothesis rather than the definition of $n_0$.\n\nWe prove the following statement by strong induction for $n \\geq 2$:\n\nLemma 3. $U(n)$ is divisible by all prime powers less than $3 \\cdot 2^{n-2} - 1$.\n\nProof. We can verify the statement by hand for $n=2$ and $n=3$. Thus we suppose that $U(k) = T(k) - T(k-1)$ is divisible by $k!$ for $k < n$, where $n \\geq 4$. We prove the second assertion first. Let $p^r$ be a prime power less than $3 \\cdot 2^{n-2} - 1$. If $r > 1$, we note that\n$$\np^{r-1} \\leq \\frac{p^r}{2} \\leq \\frac{3 \\cdot 2^{n-2} - 2}{2} = 3 \\cdot 2^{n-3} - 1\n$$\nfurthermore, at least one of these inequalities is strict, since the first becomes equality only if $p=2$, and $2^n - 1 < 3 \\cdot 2^{n-2} < 2^n$ when $n \\geq 4$. Thus by the induction hypothesis, $p^{r-1}$ divides $U_{n-1}$, and we are done by Lemma 2. Now we assume that $r=1$. We may assume that $p \\geq 5$ since $U_3$ is already divisible by $2$ and $3$. Then $p-1$ is composite, and we can resolve it into prime powers\n$$\np-1 = q_1^{a_1} q_2^{a_2} \\cdots q_s^{a_s}\n$$\nwhere each factor $q_i^{a_i}$ is at most\n$$\n\\frac{p-1}{2} < \\frac{3 \\cdot 2^{n-2} - 2}{2} = 3 \\cdot 2^{n-3} - 1.\n$$\nWe infer that each $q_i^{a_i}$ divides $U_{n-1}$, and therefore $p-1$ divides $U_{n-1}$. We now have\n$$\np \\mid 2^{p-1} - 1 \\mid 2^{U_{n-1}} - 1 \\mid U_n\n$$\nas desired.\n\nWe are now ready to solve the problem. Assume that $n \\geq 4$; the cases $n=2$ and $n=3$ are easily checked by hand. Let $p$ be a prime dividing $n!$, that is, $p \\leq n$. The number of factors of $p$ in $n!$ is\n$$\nR = \\sum_{i=1}^{\\infty} \\left\\lfloor \\frac{n}{p^i} \\right\\rfloor < \\sum_{i=1}^{\\infty} \\frac{n}{p^i} = n \\sum_{i=1}^{\\infty} \\frac{1}{p^i} = n \\cdot \\frac{1/p}{1-1/p} = \\frac{n}{p-1}\n$$\nSo\n$$\nR \\leq \\frac{n-1}{p-1}\n$$\nWe observe that $p \\mid U(p)$: by hand calculation if $p=2$, and because $p < 3 \\cdot 2^{p-2} - 1$ if $p > 3$. By Lemma 2, applied $n-p$ times,\n$$\np^{n-p+1} \\mid U(n).\n$$\nSince $n-p+1 \\geq \\frac{n-1}{p-1}$ if $n \\geq p$, we have $p^R \\mid U(n)$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn convex hexagon $A X B Y C Z$, sides $A X$, $B Y$ and $C Z$ are parallel to diagonals $B C$, $X C$ and $X Y$, respectively. Prove that $\\triangle A B C$ and $\\triangle X Y Z$ have the same area.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $[\\mathcal{P}]$ denote the area of a polygon $\\mathcal{P}$.\n\nThe important claim is that if $\\overline{K L} \\parallel \\overline{M N}$, then $[K L M] = [K L N]$. This is a simple consequence of the formula $A = \\frac{1}{2} b h$.\n\nThen, we find that\n$$\n\\begin{aligned}\n& [A B C] = [X B C] \\quad (\\text{since } \\overline{A X} \\parallel \\overline{B C}) \\\\\n& = [X Y C] \\quad (\\text{since } \\overline{B Y} \\parallel \\overline{X C}) \\\\\n& = [X Y Z] \\quad (\\text{since } \\overline{C Z} \\parallel \\overline{X Y})\n\\end{aligned}\n$$\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWeighted coins numbered $2, 3, 4, \\ldots, 2015$ are tossed. The coin numbered $i$ comes up heads with probability $1/(2i^2)$. What is the probability that an odd number of coins come up heads?",
    "options": [],
    "answer": "1007/4030",
    "solution": "Solution:\n\nLet $P_n$ be the probability that an odd number out of the coins whose numbers are at most $n$ ($1 \\leq n \\leq 2015$) come up heads. For $2 \\leq n \\leq 2015$, there are two ways for this to happen: coin $n$ is tails and an odd number of the preceding coins are heads, or coin $n$ is heads and an even number of the preceding coins are heads. So we have a recursion\n$$\n\\begin{aligned}\nP_n & = \\left(1 - \\frac{1}{2n^2}\\right) P_{n-1} + \\frac{1}{2n^2} (1 - P_{n-1}) \\\\\n& = \\frac{2n^2 P_{n-1} - 2P_{n-1} + 1}{2n^2} \\\\\n& = \\frac{2(n^2 - 1) P_{n-1} + 1}{2n^2}.\n\\end{aligned}\n$$\nIf we subtract $1/2$ from each side, we have the factorization\n$$\nP_n - \\frac{1}{2} = \\frac{n^2 - 1}{n^2} \\left(P_{n-1} - \\frac{1}{2}\\right)\n$$\nNow, using the initial value $P_1 = 0$, we have\n$$\n\\begin{aligned}\nP_{2015} - \\frac{1}{2} & = -\\frac{1}{2} \\cdot \\prod_{n=2}^{2015} \\frac{n^2 - 1}{n^2} \\\\\n& = -\\frac{1}{2} \\cdot \\prod_{n=2}^{2015} \\frac{(n-1)(n+1)}{n \\cdot n} \\\\\n& = -\\frac{1}{2} \\cdot \\frac{(1 \\cdot 2 \\cdots 2014)(3 \\cdot 4 \\cdots 2016)}{(2 \\cdot 3 \\cdots 2015)(2 \\cdot 3 \\cdots 2015)} \\\\\n& = -\\frac{1}{2} \\cdot \\frac{1 \\cdot 2016}{2015 \\cdot 2} \\\\\n& = -\\frac{504}{2015}\n\\end{aligned}\n$$\nand thus\n$$\nP_{2015} = \\frac{1}{2} - \\frac{504}{2015} = \\frac{1007}{4030}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $0 < a_{0} \\leq a_{1} \\leq \\cdots \\leq a_{n}$. If $z$ is a complex number such that $a_{0} z^{n} + a_{1} z^{n-1} + \\cdots + a_{n} = 0$ prove that $|z| \\geq 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume that $|z| < 1$. If $a_{0} z^{n} + a_{1} z^{n-1} + \\cdots + a_{n} = 0$ then $a_{0} z^{n+1} + a_{1} z^{n} + \\cdots + a_{n} z = 0$ and subtracting these two equations leads to $a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \\cdots + (a_{n} - a_{n-1}) z - a_{n} = 0$, or equivalently $a_{n} = a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \\cdots + (a_{n} - a_{n-1}) z$ hence\n\n$$\n\\begin{aligned}\n|a_{n}| & = \\left| a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \\cdots + (a_{n} - a_{n-1}) z \\right| \\\\\n& \\leq a_{0} |z|^{n+1} + (a_{1} - a_{0}) |z|^{n} + \\cdots + (a_{n} - a_{n-1}) |z| \\\\\n& < a_{0} + (a_{1} - a_{0}) + \\cdots + (a_{n} - a_{n-1}) = a_{n},\n\\end{aligned}\n$$\n\nwhich is impossible. Thus $|z| \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer which also divides $2^{n}-1$. Show that $n=1$.",
    "options": [],
    "answer": "n = 1",
    "solution": "Solution:\n\nAssume not and let $p$ be the smallest prime divisor of $n$. We have $p \\mid 2^{n}-1$, and also $p \\mid 2^{p-1}-1$ by Fermat's little theorem. By using the classical fact that $\\gcd\\left(2^{x}-1,2^{y}-1\\right)=2^{\\gcd(x, y)}-1$, we conclude $p$ divides $2^{\\gcd(p-1, n)}-1$.\n\nBut since $p$ is the smallest divisor of $n$, it follows $\\gcd(p-1, n)=1$. So $p$ divides $2^{1}-1=1$ which is absurd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a square. Consider four circles $k_{1}, k_{2}, k_{3}, k_{4}$ which pass respectively through $A$ and $B$, $B$ and $C$, $C$ and $D$, $D$ and $A$, and whose centers are outside the square. Circles $k_{4}$ and $k_{1}$, $k_{1}$ and $k_{2}$, $k_{2}$ and $k_{3}$, $k_{3}$ and $k_{4}$ intersect respectively at $L$, $M$, $N$, $P$ inside the square. Prove that quadrilateral $LMNP$ can be inscribed in a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe use a standard theorem that states that a convex quadrilateral $WXYZ$ can be inscribed in a circle if and only if $\\angle W + \\angle Y = 180$. We have\n$$\n\\begin{aligned}\n\\angle PLM + \\angle MNP & = (360 - \\angle MLA - \\angle ALP) + (360 - \\angle PNC - \\angle CNM) \\\\\n& = (180 - \\angle MLA) + (180 - \\angle ALP) + (180 - \\angle PNC) + (180 - \\angle CNM) \\\\\n& = \\angle ABM + \\angle PDA + \\angle CDP + \\angle MBC \\\\\n& = (\\angle ABM + \\angle MBC) + (\\angle PDA + \\angle CDP) \\\\\n& = 90 + 90 = 180,\n\\end{aligned}\n$$\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn apple is in the shape of a ball of radius $31~\\mathrm{mm}$. A worm gets into the apple and digs a tunnel of total length $61~\\mathrm{mm}$, and then leaves the apple. (The tunnel need not be a straight line.) Prove that one can cut the apple with a straight slice through the center so that one of the two halves is not rotten.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $A$ and $B$ be the endpoints of the tunnel, and let $C$ be the point diametrically opposite to $A$. We claim that the plane $\\pi$ which perpendicularly bisects $BC$ gives the desired cut. Indeed, this plane contains the center of the apple since it is equidistant from $B$ and $C$; now assume for the sake of contradiction that the tunnel enters both halves of the apple. Then it must cross $\\pi$ at some point $R$. Reflect that portion of the tunnel lying between $R$ and $B$ across $\\pi$, thus obtaining a tunnel from $R$ to $C$. Note that reflection preserves the length of this segment of tunnel. Thus we obtain a tunnel from $A$ to $R$ to $C$, which has the same length as the original $(61~\\mathrm{mm})$. On the other hand, since $A$ and $C$ are diametrically opposite, by the triangle inequality the tunnel has length $\\geq AC = 62~\\mathrm{mm}$. This is a contradiction. Thus our tunnel cannot intersect both halves, which is what we want.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that there are infinitely many primes $p$ with the following property: there exists a positive integer $k$ such that $2^{k}-3$ is divisible by $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose that there were only finitely many such primes $p_{1}, p_{2}, p_{3}, \\ldots, p_{n}$. Consider the number\n$$\nN=2^{\\left(p_{1}-1\\right)\\left(p_{2}-1\\right) \\cdots\\left(p_{n}-1\\right)+2}-3 .\n$$\nClearly, none of the $p_{i}$'s is $2$. Using Fermat's little theorem, we can prove that $N$ is not divisible by any of the $p_{i}$:\n$$\nN \\equiv 2^{\\left(p_{1}-1\\right) \\cdots\\left(p_{i}-1\\right) \\cdots\\left(p_{n}-1\\right)} \\cdot 4-3 \\equiv 1 \\cdot 4-3 \\equiv 1 \\quad (\\bmod\\ p_{i})\n$$\nHowever, $N>1$ since the product $\\left(p_{1}-1\\right) \\cdots\\left(p_{n}-1\\right)$ is at least $1$. Therefore $N$ is divisible by a prime other than the $p_{i}$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, $c$ be positive real numbers. Assume that\n$$\n\\frac{a^{19}}{b^{19}} + \\frac{b^{19}}{c^{19}} + \\frac{c^{19}}{a^{19}} \\leq \\frac{a^{19}}{c^{19}} + \\frac{b^{19}}{a^{19}} + \\frac{c^{19}}{b^{19}}\n$$\nProve that\n$$\n\\frac{a^{20}}{b^{20}} + \\frac{b^{20}}{c^{20}} + \\frac{c^{20}}{a^{20}} \\leq \\frac{a^{20}}{c^{20}} + \\frac{b^{20}}{a^{20}} + \\frac{c^{20}}{b^{20}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIf we multiply the first equation by $(a b c)^{19}$ then it can be rewritten as\n$$\n\\left(a^{19} - b^{19}\\right)\\left(b^{19} - c^{19}\\right)\\left(c^{19} - a^{19}\\right) \\leq 0.\n$$\nSimilarly, the desired equation is equivalent to\n$$\n\\left(a^{20} - b^{20}\\right)\\left(b^{20} - c^{20}\\right)\\left(c^{20} - a^{20}\\right) \\leq 0.\n$$\nThese are evidently equivalent since $a^{19} - b^{19}$ and $a^{20} - b^{20}$ have the same sign, etc.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a triangle $ABC$, we draw three circles with respective diameters $AB$, $BC$, and $CA$. Prove that there exists a point that is inside all three circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe claim that the center $I$ of the triangle's inscribed circle is such a point. To see this, note that\n$$\n\\angle BIC = 180 - \\angle ICB + \\angle CBI = 180 - \\frac{\\angle C + \\angle B}{2} = 180 - \\frac{180 - A}{2} = 90 + \\frac{A}{2} > 90.\n$$\nThus $\\triangle BIC$ is obtuse and if we drop a perpendicular $BD$ from $B$ to $CI$, then $D$ lies on the extension of ray $CI$. The circle with diameter $BC$ passes through $D$ since $\\angle BDC = 90$, and thus $I$, an interior point of chord $CD$, lies inside the circle. By symmetry, we can conclude that $I$ lies inside the other two circles as well.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that $n$ divides $\\varphi\\left(a^{n}-1\\right)$ for any integers $a$ and $n$, where $\\varphi$ is Euler's totient function.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $N = a^{n} - 1$. Then $\\gcd(a, N) = 1$ and the order of $a \\pmod{N}$ is exactly equal to $n$. But $a^{\\varphi(N)} \\equiv 1 \\pmod{N}$ too. Thus $n$ divides $\\varphi(N)$.",
    "topic": "Number Theory",
    "subtopic": "Number-Theoretic Functions"
  },
  {
    "id": 46283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPentagon $A B C D E$ is cyclic, i.e., inscribed in a circle. Diagonals $A C$ and $B D$ meet at $P$, and diagonals $A D$ and $C E$ meet at $Q$. Triangles $A B P$, $A E Q$, $C D P$, $C D Q$, and $A P Q$ have equal areas. Prove that the pentagon is regular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA society has $100$ members and every two members are either friends or enemies. Prove that there are two persons from the society that have an even number of common enemies.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume that for each two members the set of common enemies has an odd number of elements. Then the set of their common friends has also an odd number of elements. Fix a person $a$. Let $F=\\{f_{1}, \\ldots, f_{k}\\}$ be the set of friends of the member $a$, and $E=\\{e_{1}, \\ldots, e_{99-k}\\}$ the set of its enemies. First we will show that $k$ is even. Assume that $f_{1}$ has $m_{1}$ friends inside $F$, $f_{2}$ has $m_{2}$ friends inside $F$, etc. Then $m_{1}+m_{2}+\\cdots+m_{k}$ is equal to twice the number of all friendships inside $F$. However, each of the numbers $m_{1}, \\ldots, m_{k}$ must be odd because $m_{i}$ represents the number of common friends for $a$ and $f_{i}$. Since the sum of $k$ odd numbers ended up being even, we conclude that $k$ is even.\n\nThus, every member has an even number of friends (the argument in the previous paragraph works for any member of the society, not just $a$). Now we will calculate the parity of the number of friendships between members of $E$ and members of $F$. We will do that in two different ways and get a contradiction.\n\nFor every $i$, $1 \\leq i \\leq 99-k$, $e_{i}$ has to have an odd number of friends among $F$, because $e_{i}$ and $a$ have an odd number of common friends. The number of elements of $E$ is odd, hence the total number of friendships between $E$ and $F$ has to be odd.\n\nOn the other hand, $b_{i}$ ($1 \\leq i \\leq k$) has an odd number of friends inside $F$, $a$ is also a friend of $b_{i}$, and the total number of friends of $b_{i}$ has to be even. Thus, $b_{i}$ has an even number of friends inside $E$. Hence the number of friendships between $F$ and $E$ is even. A contradiction!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoes there exist an integer such that its cube is equal to $3 n^{2} + 3 n + 7$, where $n$ is an integer?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose that there exist integers $n$ and $m$ such that $m^{3} = 3 n^{2} + 3 n + 7$. Then from $m^{3} \\equiv 1 \\pmod{3}$ it follows that $m = 3k + 1$ for some $k \\in \\mathbb{Z}$. Substituting into the initial equation we obtain $3k\\left(3k^{2} + 3k + 1\\right) = n^{2} + n + 2$. It is easy to check that $n^{2} + n + 2$ cannot be divisible by $3$, and so this equality cannot be true. Therefore our equation has no solutions in integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all pairs $(n, k)$ of integers such that $0 < k < n$ and\n$$\n\\binom{n}{k-1} + \\binom{n}{k+1} = 2 \\binom{n}{k}\n$$",
    "options": [],
    "answer": "All integer pairs given by n = c^2 − 2 and k = (c^2 + c − 2)/2 for any integer c with c ≤ −3 or c ≥ 3.",
    "solution": "Solution:\nIn the factorial form,\n$$\n\\frac{n!}{(k-1)!(n-k+1)!} + \\frac{n!}{(k+1)!(n-k-1)!} = \\frac{2 \\cdot n!}{k!(n-k)!}\n$$\nwe multiply through by $(k+1)!(n-k+1)!$ to clear the fractions and then divide through by $n!$:\n$$\nk(k+1) + (n-k)(n-k+1) = 2(k+1)(n-k+1)\n$$\nTo decrease the number of terms, we let $k+1 = a$ and $n-k+1 = b$:\n$$\n\\begin{gathered}\n(a-1)a + (b-1)b = 2ab \\\\\na^2 - 2ab + b^2 = a + b \\\\\n(a-b)^2 = a + b\n\\end{gathered}\n$$\nIf we let $a-b = c$, then $a + b = c^2$ and we get\n$$\na = \\frac{c^2 + c}{2} \\quad \\text{and} \\quad b = \\frac{c^2 - c}{2}\n$$\nHere any integer value of $c$ will yield nonnegative integer values of $a$ and $b$; however, the condition $0 < k < n$ requires that $a = k+1$ and $b = n-k+1$ are each at least $2$. Hence the values $c = -2, -1, 0, 1, 2$ are excluded, while every $c \\leq -3$ and every $c \\geq 3$ will yield permissible values for\n$$\nk = a - 1 = \\frac{c^2 + c - 2}{2}\n$$\nand\n$$\nn = a + b - 2 = c^2 - 2\n$$\nwhich satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that, for positive integers $n$ and $m$,\n$$\n\\operatorname{gcd}\\left(2^{m}-1,2^{n}-1\\right)=2^{\\operatorname{gcd}(m, n)}-1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOur main method of attacking the gcd is the following fact:\n$$\n\\operatorname{gcd}(m, n)=\\operatorname{gcd}(m+k n, n)\n$$\nfor positive integers $n$ and $k$. (Proof: If $d \\mid m$ and $d \\mid n$, then $d \\mid m+k n$; if $d \\mid m+k n$ and $d \\mid n$, then $d \\mid m$. Thus the pairs on both sides have the same set of common divisors.)\nSuppose that (1) does not hold and $m+n$ is as small as possible (here we are using the \"well-ordering principle\"). Note that (1) clearly holds when $m=n$, so we may assume that $m>n$. Let $m-n=r$ and note that $\\operatorname{gcd}(m, n)=\\operatorname{gcd}(r, n)$ Also,\n$$\n\\begin{aligned}\n\\operatorname{gcd}\\left(2^{m}-1,2^{n}-1\\right) & =\\operatorname{gcd}\\left(2^{m}-2^{r}+2^{r}-1,2^{n}-1\\right) \\\\\n& =\\operatorname{gcd}\\left(2^{r}\\left(2^{n}-1\\right)+2^{r}-1,2^{n}-1\\right) \\\\\n& =\\operatorname{gcd}\\left(2^{r}-1,2^{n}-1\\right)\n\\end{aligned}\n$$\nTherefore we may replace $m$ by $r$ without affecting the claim that (1) is false. But now $r+n<m+n$, contradicting the minimality of $m+n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, a_{3}, \\ldots$ be an infinite sequence of positive real numbers such that for all $n \\geq 1$,\n$$\na_{n} \\leq a_{2 n}+a_{2 n+1}\n$$\nProve that there exists an $N \\geq 1$ such that\n$$\n\\sum_{n=1}^{N} a_{n}>1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet us prove by induction that for any positive integer $k$, there is an $N$ for which\n$$\n\\sum_{n=1}^{N} a_{n} \\geq k a_{1}\n$$\nThe base case, $k=1$ and $N=1$, is trivial. Given that (1) is true for a given $k$, we have\n$$\nk a_{1} \\leq \\sum_{n=1}^{N} a_{n} \\leq \\sum_{n=1}^{N}\\left(a_{2 n}+a_{2 n+1}\\right)=\\sum_{\\substack{2 \\leq n \\leq 2 N \\\\ n \\text{ even }}} a_{n}+\\sum_{\\substack{3 \\leq n \\leq 2 N+1 \\\\ n \\text{ odd }}} a_{n}=\\sum_{n=2}^{2 N+1} a_{n}\n$$\nAdding $a_{1}$, we get\n$$\n\\sum_{n=1}^{2 N+1} a_{n} \\geq(k+1) a_{1}\n$$\nas desired.\n\nThe problem now follows by taking $k$ large enough so that $k>1 / a_{1}$, so $k a_{1}>1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWithout a calculator, find a factor $85^{9}-21^{9}+6^{9}$ that is between 2000 and 3000.",
    "options": [],
    "answer": "2240",
    "solution": "Solution:\nWe know that $85^{9}-21^{9}$ has $85-21=64$ as a factor, and $6^{9}$ also has $64$ as a factor, so the sum is divisible by $64$.\n\nSimilarly, $-21^{9}+6^{9}$ is divisible by $-21+6=-15$, which means it is divisible by $5$. Since $85^{9}$ is also divisible by $5$, the whole sum is divisible by $5$.\n\nFinally, $85^{9}+6^{9}$ is divisible by $85+6=91$, so it is divisible by $7$. Since $21^{9}$ is also divisible by $7$, the sum is divisible by $7$.\n\nSince the sum is divisible by $64$, $5$, and $7$, it is also divisible by $64 \\cdot 5 \\cdot 7 = 2240$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with incenter $I$ and circumcenter $O$. Let the circumradius be $R$. What is the least upper bound of all possible values of $IO$?",
    "options": [],
    "answer": "R",
    "solution": "Solution:\n\n$I$ always lies inside the convex hull of $ABC$, which in turn always lies in the circumcircle of $ABC$, so $IO < R$. On the other hand, if we first draw the circle $\\Omega$ of radius $R$ about $O$ and then pick $A$, $B$, and $C$ very close together on it, we can force the convex hull of $ABC$ to lie outside the circle of radius $R - \\epsilon$ about $O$ for any $\\epsilon$. Thus the answer is $R$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A B C D E F$ be a convex hexagon such that the quadrilaterals $A B D E$ and $A C D F$ are parallelograms. Prove that $B C E F$ is also a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIt is well known that a quadrilateral is a parallelogram if and only if the two diagonals bisect each other, that is, have a common midpoint. Since $A B D E$ is a parallelogram, the midpoints of $A D$ and $B E$ coincide; since $A C D F$ is a parallelogram, the midpoints of $A D$ and $C F$ coincide. Consequently the midpoints of $B E$ and $C F$ coincide and $B C E F$ is a parallelogram.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPentagon $A B C D E$ is cyclic, i.e., inscribed in a circle. Diagonals $A C$ and $B D$ meet at $P$, and diagonals $A D$ and $C E$ meet at $Q$. Triangles $A B P$, $A E Q$, $C D P$, $C D Q$, and $A P Q$ have equal areas. Prove that the pentagon is regular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAdding the area of triangle $B C P$ to those of $\\triangle A B P$, $\\triangle C D P$, we see that $\\triangle A B C$, $\\triangle D B C$ have equal areas, so $A D$, $B C$ are parallel. Then $A B C D$ is a cyclic trapezoid, so it is isosceles and $A B = C D$. Similarly, $A C D E$ is a trapezoid and $C D = E A$. Now construct parallelogram $A B C R$; we see that $R$ lies on ray $A D$ and $\\triangle A R P$ has the same area as $\\triangle A B P$ (because they have equal base $A P$ and, by symmetry, equal altitudes). Since these properties uniquely determine $R$, we conclude that $R = Q$, so lines $A B$, $C Q = C E$ are parallel. Then $A B C E$ is a trapezoid, so it is isosceles and $E A = B C$. Similarly, $D E A P$ is shown to be a parallelogram, so $A B D E$ is a trapezoid and $A B = D E$. We have now shown that $D E = A B = C D = E A = B C$, so all sides of the pentagon are equal. Since it is cyclic, all sides subtend arcs of equal measure $\\theta$; then every angle subtends an arc of measure $3 \\theta$, so each angle is $3 \\theta / 2$ and the pentagon is regular.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n > 3$ be a positive integer. Define an integer $k$ to be snug if $1 \\leq k < n$ and\n$$\n\\gcd(k, n) = \\gcd(k+1, n).\n$$\nProve that the product of all snug integers is congruent to $1$ modulo $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $k$ be a snug integer. Note that any factor that divides $n$, $k$, and $k+1$ must also divide $(k+1)-k=1$, so\n$$\n\\gcd(k, n) = \\gcd(k+1, n) = 1.\n$$\nIn particular, $k$ has a multiplicative inverse $h \\bmod n$ (we can choose $h$ such that $0 < h < n$). We claim that $h$ is also snug. Clearly $\\gcd(h, n) = 1$; note that\n$$\n(h+1) \\cdot k = h k + k \\equiv k + 1 \\pmod{n};\n$$\nsince $k$ and $k+1$ are invertible $\\bmod n$, so is $h+1$.\nThus we can pair up snug residues $\\bmod n$ into pairs with product $1$, unless there is a snug $k$ that is its own multiplicative inverse. We claim that there is no such $k$ except possibly $k=1$. Indeed, if $k^2 \\equiv 1 \\pmod{n}$, then $n$ divides\n$$\nk^2 - 1 = (k+1)(k-1).\n$$\nSince $k$ is snug, $n$ is relatively prime to $k+1$ and hence divides $k-1$, implying that $k=1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that every positive real number $x$ satisfies\n$$\n\\sqrt{x^{2}-x+\\frac{1}{2}} \\geq \\frac{1}{x+\\frac{1}{x}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nBy the QM-AM inequality,\n$$\n\\sqrt{x^{2}-x+\\frac{1}{2}} = \\sqrt{\\frac{x^{2} + (1-x)^{2}}{2}} \\geq \\frac{x + (1-x)}{2} = \\frac{1}{2}.\n$$\nNow, by AM-GM (since we know $x$ is positive),\n$$\n\\frac{x + \\frac{1}{x}}{2} \\geq \\sqrt{x \\cdot \\frac{1}{x}} \\Longrightarrow \\frac{1}{2} \\geq \\frac{1}{x + \\frac{1}{x}}\n$$\nPutting this together gives\n$$\n\\sqrt{x^{2}-x+\\frac{1}{2}} \\geq \\frac{1}{2} \\geq \\frac{1}{x + \\frac{1}{x}}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe point $K$ lies in the interior of the unit circle. Four lines are drawn through $K$ such that each two adjacent lines form an angle of $45^{\\circ}$, as shown in the picture. In such a way the circle is divided into 8 regions. Four of these regions are colored such that no two colored regions share more than one point. Find the maximal and minimal possible value for the total area of colored regions.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Minimal = π/2, Maximal = π/2",
    "solution": "Solution:\n\nWe will show that the total area of colored regions is equal to $\\frac{1}{2}$ the area of the circle no matter how the lines are drawn.\n\nFirst we will inscribe our circle in a square whose sides are parallel to 2 of these lines. If we color the regions of the square in the same way as the regions of the circle are colored we can easily prove that the total area of colored regions is equal to the area of non-colored regions for the square. Hence it remains to prove the equality of the areas of colored and uncolored regions that are outside the circle. Now we will look at the pictures below. For each picture it is obvious that it has the same area of colored parts as the previous picture (as some colored parts are just moved from one location to the other).\n\n![](attached_image_2.png)\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a bag are $n$ fair, six-sided dice whose faces are colored white and red in such a way that the total numbers of white and red sides are equal. Let $p$ be the probability that the same color comes up twice when taking one die randomly out of the bag and throwing it twice. Let $q$ be the probability that the same color comes up twice when taking two dice randomly out of the bag and throwing them at the same time. Prove that\n$$\np+(n-1) q=\\frac{n}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider the following procedure: Remove one die randomly from the bag, roll it, replace it in the bag, remove another die randomly from the bag, and roll it. It is clear that each roll is an independent and random choice of one of the $6 n$ sides of all the dice; hence the probability of getting the same color twice is $1 / 2$. On the other hand, we can decompose the probability as follows:\n- With probability $1 / n$, we will pick the same die twice. Then the probability that the same color comes up is $p$.\n- With probability $(n-1) / n$, we will not pick the same die twice. Then the two dice we roll form a random pair of distinct dice, and the probability that they will come up the same color is $q$.\n\nAdding up the probabilities, we conclude that\n$$\n\\frac{1}{n} \\cdot p+\\frac{n-1}{n} \\cdot q=\\frac{1}{2}\n$$\nthat is,\n$$\np+(n-1) q=\\frac{n}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAerith has an unlimited number of coins of values $5$, $m-1$, and $m$, and $71$ is the greatest value that cannot be made with a combination of these coins. Find all possible values of $m$.",
    "options": [],
    "answer": "20 and 38",
    "solution": "Solution:\n\nBy the Chicken McNugget Theorem, the largest coin value that can't be achieved with $5$ and $m-1$ is\n$$\n5(m-1)-5-(m-1)=4m-9 \\geq 71,\n$$\nso $4m \\geq 80$ and thus $m \\geq 20$.\n\nSince $71$ is not possible, there cannot be a number less than $71$ which is $1 \\pmod{5}$ and divisible by $m$ or $m-1$, since then you could add some number of $5$'s to it to get $71$. In particular, $m$ and $m-1$ cannot be $1 \\pmod{5}$, so $m$ can't be $1$ or $2 \\pmod{5}$.\n\nSince all numbers above $71$ are achievable, $76$ has to be achievable, and without using any $5$'s. If $m \\geq 36$, the only way this can happen is if it's either $2m$ or $2(m-1)$, so $m=38$ or $m=39$. If $m=39$, then there's no way to get $72$. However, if $m=38$, then we can get:\n$$\n\\begin{aligned}\n&72=37+7\\cdot 5, \\\\\n&73=38+7\\cdot 5, \\\\\n&74=2\\cdot 37, \\\\\n&75=37+38, \\\\\n&76=2\\cdot 38.\n\\end{aligned}\n$$\nWe also can't get $71$, so $m=38$ is a solution.\n\nIf $m \\leq 37$, then $2m$ also can't be $1 \\pmod{5}$, so $m$ can't be $3 \\pmod{5}$. Also, $2(m-1)$ can't be $1 \\pmod{5}$, so $m$ can't be $4 \\pmod{5}$. Thus $m$ has to be a multiple of $5$. But then $m-1 \\equiv 4 \\pmod{5}$, so the only way we could get $76$ is if it's $4(m-1)$, which gives $m=20$. This works since $71$ can't be achieved but\n$$\n\\begin{aligned}\n&72=3\\cdot 19+3\\cdot 5, \\\\\n&73=2\\cdot 19+20+3\\cdot 5, \\\\\n&74=19+2\\cdot 20+3\\cdot 5, \\\\\n&75=15\\cdot 5, \\\\\n&76=2\\cdot 38.\n\\end{aligned}\n$$\nThus, the two solutions are $m=20$ and $m=38$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest prime $p > 100$ for which there exists an integer $a > 1$ such that $p$ divides $\\frac{a^{89} - 1}{a - 1}$.",
    "options": [],
    "answer": "179",
    "solution": "Solution:\nThe answer is $p = 179$. To see this works, take $a = 4$; by Fermat's little theorem, $4^{89} - 1 = 2^{178} - 1$ is divisible by $179$.\n\nNow suppose $a^{89} \\equiv 1 \\pmod{p}$. We consider two cases:\n- If $a \\equiv 1 \\pmod{p}$, then\n$$\n0 \\equiv 1 + a + \\cdots + a^{88} \\equiv 89 \\pmod{p}\n$$\nwhich forces $p = 89$.\n- Otherwise, since $89$ is prime, it follows $a$ has order $89$ modulo $p$. So $89 \\mid p - 1$. The smallest prime which obeys this is $p = 179$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAcute triangle $A B C$ is made of solid metal, and it is on top of a wooden table. Points $P$ on $A B$ and $Q$ on $A C$ are such that the perpendicular to $A B$ through $P$ intersects the perpendicular to $A C$ through $Q$ inside the triangle. Nails are hammered into the table at $P$ and $Q$. (The nails do not go through the triangle but are at its edges, bounding the triangle.) Prove that there is a unique position $R$ on $B C$ such that, if a nail is hammered into the table at $R$, the triangle will no longer be able to move (within the plane of the table).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst we prove existence. Let $O$ be the intersection of the perpendiculars at $P$ and $Q$ to $A B$ and $A C$ respectively. Insert a nail at $R$, the foot of the perpendicular from $O$ to $B C$. Observe that $O P \\cdot A B, O Q \\cdot A C, O R \\cdot B C$ equal twice the respective areas of triangles $A O B, A O C, B O C$; since $O$ is inside the triangle, their sum equals $2[A B C]$. Now if the triangle can be moved at all, then it can be moved a very small amount, to a position $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ such that segments $O P, O Q, O R$ still intersect the respective sides $A^{\\prime} B^{\\prime}, A^{\\prime} C^{\\prime}, B^{\\prime} C^{\\prime}$ at some points $P^{\\prime}, Q^{\\prime}, R^{\\prime}$. (We know that the intersection points lie within these segments because the nails cannot go inside $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$.) The distance from $O$ to $A^{\\prime} B^{\\prime}$ is then at most $O P^{\\prime}$, so twice the area of triangle $O A^{\\prime} B^{\\prime}$ is $\\leq O P^{\\prime} \\cdot A^{\\prime} B^{\\prime} \\leq O P \\cdot A^{\\prime} B^{\\prime}=O P \\cdot A B$. Equality can only occur in two cases: either $P^{\\prime}=P$ and $O P$ is still perpendicular to $A^{\\prime} B^{\\prime}$, or $P^{\\prime}=O$, i.e. $O$ lies on $A^{\\prime} B^{\\prime}$. We get analogous results for triangles $O A^{\\prime} C^{\\prime}, O B^{\\prime} C^{\\prime}$; summing these gives\n$$\n2[A B C]=2\\left[A^{\\prime} B^{\\prime} C^{\\prime}\\right] \\leq O P \\cdot A B+O Q \\cdot A C+O R \\cdot B C=2[A B C]\n$$\nand we have equality throughout. We showed that only two equality cases can occur for each of the three sides; thus, two sides, say $A^{\\prime} B^{\\prime}, B^{\\prime} C^{\\prime}$, satisfy the same case. Thus either $O$ lies on both $A^{\\prime} B^{\\prime}$ and $B^{\\prime} C^{\\prime} \\Rightarrow O=B^{\\prime}$ which will not occur assuming the triangle was moved a small enough distance, or $P^{\\prime}=P, R^{\\prime}=R$, and $O P, O R$ are perpendicular to $A^{\\prime} B^{\\prime}, B^{\\prime} C^{\\prime}$, but this, given $\\triangle A^{\\prime} B^{\\prime} C^{\\prime} \\cong \\triangle A B C$, can only occur when in fact $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}=\\triangle A B C$, contradicting the assumption that the triangle was moved at all. Thus either way we get a contradiction, so the triangle cannot in fact be budged.\n\nNow we prove uniqueness. Given a point $R$ on $B C$, it is a fact (known as the \"Pivot Theorem\" see e.g. solution to BAMO $\\# 2$ for proof) that the circumcircles of triangles $A P Q, B P R, C Q R$ meet in a common point $O$. Suppose that $\\angle O P A$ is acute. Then, using cyclic quadrilaterals, we quickly find that angles $O R B, O Q C$ are acute (in fact, equal to $\\angle O P A$ ). It follows that we can rotate the triangle about $O$ in the direction from $A$ to $B$ for some distance without running into the nails at $P, Q, R$. The case where $\\angle O P A$ is obtuse is similar, so the only way that the triangle can be immobile is if $\\angle O P A=\\pi / 2$ and (again by cyclic quadrilaterals) $\\angle O R B, \\angle O Q C$ are also right angles, which means the position of $O$ must be that identified in the previous paragraph.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A_{0}, A_{1}, \\ldots, A_{n}$ be points in a plane such that\n(i) $A_{0}A_{1} \\leq \\frac{1}{2}A_{1}A_{2} \\leq \\cdots \\leq \\frac{1}{2^{n-1}}A_{n-1}A_{n}$ and\n(ii) $0 < \\measuredangle A_{0}A_{1}A_{2} < \\measuredangle A_{1}A_{2}A_{3} < \\cdots < \\measuredangle A_{n-2}A_{n-1}A_{n} < 180^{\\circ}$,\nwhere all these angles have the same orientation. Prove that the segments $A_{k}A_{k+1}, A_{m}A_{m+1}$ do not intersect for each $k$ and $n$ such that $0 \\leq k \\leq m-2 < n-2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose that $A_{k}A_{k+1} \\cap A_{m}A_{m+1} \\neq \\emptyset$ for some $k, m > k+1$. Without loss of generality we may suppose that $k=0, m=n-1$ and that no two segments $A_{k}A_{k+1}$ and $A_{m}A_{m+1}$ intersect for $0 \\leq k < m-1 < n-1$ except for $k=0$, $m=n-1$. Also, shortening $A_{0}A_{1}$, we may suppose that $A_{0} \\in A_{n-1}A_{n}$. Finally, we may reduce the problem to the case that $A_{0} \\ldots A_{n-1}$ is convex: Otherwise, the segment $A_{n-1}A_{n}$ can be prolonged so that it intersects some $A_{k}A_{k+1}$, $0 < k < n-2$.\n\nIf $n=3$, then $A_{1}A_{2} \\geq 2A_{0}A_{1}$ implies $A_{0}A_{2} > A_{0}A_{1}$, hence $\\angle A_{0}A_{1}A_{2} > \\angle A_{1}A_{2}A_{3}$, a contradiction.\n\nLet $n=4$. From $A_{3}A_{2} > A_{1}A_{2}$ we conclude that $\\angle A_{3}A_{1}A_{2} > \\angle A_{1}A_{3}A_{2}$. Using the inequality $\\angle A_{0}A_{3}A_{2} > \\angle A_{0}A_{1}A_{2}$ we obtain that $\\angle A_{0}A_{3}A_{1} > \\angle A_{0}A_{1}A_{3}$ implying $A_{0}A_{1} > A_{0}A_{3}$. Now we have $A_{2}A_{3} < A_{3}A_{0} + A_{0}A_{1} + A_{1}A_{2} < 2A_{0}A_{1} + A_{1}A_{2} \\leq 2A_{1}A_{2} \\leq A_{2}A_{3}$, which is not possible.\n\nNow suppose $n \\geq 5$. If $\\alpha_{i}$ is the exterior angle at $A_{i}$, then $\\alpha_{1} > \\cdots > \\alpha_{n-1}$; hence $\\alpha_{n-1} < \\frac{360^{\\circ}}{n-1} \\leq 90^{\\circ}$. Consequently $\\angle A_{n-2}A_{n-1}A_{0} \\geq 90^{\\circ}$ and $A_{0}A_{n-2} > A_{n-1}A_{n-2}$. On the other hand, $A_{0}A_{n-2} < A_{0}A_{1} + A_{1}A_{2} + \\cdots + A_{n-3}A_{n-2} < \\left(\\frac{1}{2^{n-2}} + \\frac{1}{2^{n-3}} + \\cdots + \\frac{1}{2}\\right)A_{n-1}A_{n-2} < A_{n-1}A_{n-2}$, which contradicts the previous relation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor integers $n \\geq 1$, prove that the product\n$$\n3 \\cdot 12 \\cdot 21 \\cdot 30 \\cdots (9n-6)\n$$\nis divisible by $n!$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $p$ be a prime number. We will prove that the number $A$ of factors of $p$ in $3 \\cdot 12 \\cdots (9n-6)$ is greater than or equal to the number $B$ of factors of $p$ in $n!$.\n\nWe first explain the widely known method for computing $B$. Out of the numbers from $1$ to $n$, exactly $\\lfloor n / p \\rfloor$ of them are multiples of $p$; they will contribute $\\lfloor n / p \\rfloor$ \"first\" factors of $p$ to the product $n!$. In addition, $\\left\\lfloor n / p^{2} \\right\\rfloor$ of these numbers are also divisible by $p^{2}$, giving $\\left\\lfloor n / p^{2} \\right\\rfloor$ \"second\" factors of $p$. This continues, and we get\n$$\nB = \\left\\lfloor \\frac{n}{p} \\right\\rfloor + \\left\\lfloor \\frac{n}{p^{2}} \\right\\rfloor + \\left\\lfloor \\frac{n}{p^{3}} \\right\\rfloor + \\cdots,\n$$\nwhere the sum continues until eventually all of its terms become $0$ due to a lack of terms divisible by very high powers of $p$.\n\nNow we estimate $A$. If $p = 3$, then $A = n$ since every term of the arithmetic sequence $3, 12, 21, \\ldots$ is divisible by $3$ but not $9$. In this case it is clear that\n$$\nB = \\left\\lfloor \\frac{n}{3} \\right\\rfloor + \\left\\lfloor \\frac{n}{3^{2}} \\right\\rfloor + \\left\\lfloor \\frac{n}{3^{3}} \\right\\rfloor + \\cdots \\leq \\frac{n}{3} + \\frac{n}{3^{2}} + \\frac{n}{3^{3}} + \\cdots = \\frac{n}{2} < n.\n$$\n\nNow assume that $p \\neq 3$. Divide the arithmetic progression $3, 12, 21, \\ldots, (9n-6)$ into $\\lfloor n / p \\rfloor$ blocks of length $p$, discarding any terms that remain; because the common difference $9$ is relatively prime to $p$, each block will have one representative of each congruence class $\\bmod\\ p$, and in particular exactly one multiple of $p$. Thus the product $3 \\cdot 12 \\cdot 21 \\cdot 30 \\cdots (9n-6)$ will have at least $\\lfloor n / p \\rfloor$ \"first\" factors of $p$. By the same argument, using blocks of length $p^{2}$, there are at least $\\left\\lfloor n / p^{2} \\right\\rfloor$ \"second\" factors of $p$, and so on, so\n$$\nA \\geq \\left\\lfloor \\frac{n}{p} \\right\\rfloor + \\left\\lfloor \\frac{n}{p^{2}} \\right\\rfloor + \\left\\lfloor \\frac{n}{p^{3}} \\right\\rfloor + \\cdots = B\n$$\nas desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers from the table\n\n| $a_{11}$ | $a_{12}$ | $\\cdots$ | $a_{1 n}$ |\n| :---: | :---: | :---: | :---: |\n| $a_{21}$ | $a_{22}$ | $\\cdots$ | $a_{2 n}$ |\n| $\\vdots$ | $\\vdots$ |  | $\\vdots$ |\n| $a_{n 1}$ | $a_{n 2}$ | $\\cdots$ | $a_{n n}$ |\n\nsatisfy the inequality\n\n$$\n\\sum_{i=1}^{n}\\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\\cdots+a_{i n} x_{n}\\right| \\leq M\n$$\n\nfor every choice $x_{i}= \\pm 1$. Prove that $\\left|a_{11}\\right|+\\left|a_{22}\\right|+\\cdots+\\left|a_{n n}\\right| \\leq M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe will sum the given inequality over all possible choices for $\\left(x_{1}, \\ldots, x_{n}\\right)$. Since the number of such choices is $2^{n}$ we obtain\n\n$$\n\\sum_{i=1}^{n} \\sum_{\\left(x_{1}, \\ldots, x_{n}\\right)}\\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\\cdots+a_{i n} x_{n}\\right| \\leq 2^{n} M\n$$\n\nNow we will consider the expressions $\\sum_{\\left(x_{1}, \\ldots, x_{n}\\right)}\\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\\cdots+a_{i n} x_{n}\\right|$. We will group summands in pairs - members of each pair will have all terms $x_{j}$ the same, except for the $i$-th. One member will have $x_{i}=1$ and the other $x_{i}=-1$. To each of those terms we will apply the inequality $\\left|a_{i i}+B\\right|+\\left|a_{i i}-B\\right| \\geq\\left|2 a_{i i}\\right|$. More precisely,\n\n$$\n\\begin{aligned}\n& \\sum_{\\left(x_{1}, \\ldots, x_{n}\\right)}\\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\\cdots+a_{i n} x_{n}\\right| \\\\\n= & \\sum_{\\left(x_{1}, \\ldots, x_{i-1}, x_{i+1}, \\ldots, x_{n}\\right)}\\left(\\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\\cdots+a_{i i}+\\cdots+a_{i n} x_{n}\\right|+\\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\\cdots-a_{i i}+\\cdots+a_{i n} x_{n}\\right|\\right) \\\\\n\\geq & \\sum_{\\left(x_{1}, \\ldots, x_{i-1}, x_{i+1}, \\ldots, x_{n}\\right)}\\left|2 a_{i i}\\right|=2^{n-1}\\left|a_{i i}\\right| .\n\\end{aligned}\n$$\n\nSumming over $i$ from $1$ to $n$, we get:\n\n$$\n\\sum_{i=1}^{n} 2^{n-1}\\left|a_{i i}\\right| \\leq 2^{n} M\n$$\n\nwhich implies that $\\left|a_{11}\\right|+\\left|a_{22}\\right|+\\cdots+\\left|a_{n n}\\right| \\leq M$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S > 0$. If $a, b, c, x, y, z$ are positive real numbers such that $a + x = b + y = c + z = S$, prove that\n$$\na y + b z + c x < S^{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDenote $T = S / 2$. One of the triples $(a, b, c)$ and $(x, y, z)$ has the property that at least two of its members are greater than or equal to $T$. Assume that $(a, b, c)$ is the one, and choose $\\alpha = a - T$, $\\beta = b - T$, and $\\gamma = c - T$. We then have $x = T - \\alpha$, $y = T - \\beta$, and $z = T - \\gamma$. Now the required inequality is equivalent to\n$$\n(T + \\alpha)(T - \\beta) + (T + \\beta)(T - \\gamma) + (T + \\gamma)(T - \\alpha) < 4 T^{2}\n$$\nAfter simplifying we get that what we need to prove is\n$$\n-(\\alpha \\beta + \\beta \\gamma + \\gamma \\alpha) < T^{2}\n$$\nWe also know that at most one of the numbers $\\alpha, \\beta, \\gamma$ is negative. If all are positive, there is nothing to prove. Assume that $\\gamma < 0$. Now (1) can be rewritten as $-\\alpha \\beta - \\gamma(\\alpha + \\beta) < T^{2}$. Since $-\\gamma < T$ we have that $-\\alpha \\beta - \\gamma(\\alpha + \\beta) < -\\alpha \\beta + T(\\alpha + \\beta)$ and the last term is less than $T$ since $(T - \\alpha)(T - \\beta) > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that $f(f(x)+y+1)=x+f(y)+1$ for every two integers $x$ and $y$.",
    "options": [],
    "answer": "f(n) = n for all integers n; and f(n) = -n - 2 for all integers n",
    "solution": "Solution:\n\nSubstituting $x=0$ we get $f(f(0)+y+1)=f(y)+1$.\n\nNow we have $x+f(y)+1=f(f(x)+y+1)=f[(f(x)+1)+y]=f[f(f(0)+x+1)+(y-1)+1]=f(y-1)+(f(0)+x+1)+1$ from where we conclude that\n\n$$\nf(y)=f(y-1)+f(0)+1 .\n$$\n\nUsing induction we obtain $f(n)=f(0)+n(f(0)+1)$ for all integers $n$.\n\nIf we set $y=0$ in (1) we get $f(-1)=-1$ and substituting $y=-1$ in the original equation yields $f(f(x))=x$.\n\nIf we apply $f$ to both sides of $f(n)=f(0)+n(f(0)+1)$ we get $n=f(0)+[f(0)+n(f(0)+1)](f(0)+1)$ which is equivalent to $n=f(0)(f(0)+2)+n(f(0)+1)^2$ and this can be satisfied for all integers $n$ if and only if $f(0)=0$ or $f(0)=-2$.\n\nIn the first case we get $f(n)=n$ and in the second $f(n)=-n-2$.\n\nIt is easy to verify that both functions satisfy the required conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that\n$$\n-\\frac{1}{2} \\leq \\frac{(x+y)(1-x y)}{\\left(1+x^{2}\\right)\\left(1+y^{2}\\right)} \\leq \\frac{1}{2}\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $x$ and $y$ can be replaced by their negatives, it suffices to prove the right-hand inequality. On cross-multiplication, this is\n$$\n\\begin{aligned}\n0 & \\stackrel{?}{\\leq} \\left(x^{2}+1\\right)\\left(y^{2}+1\\right)-2(x+y)(1-x y) \\\\\n& = x^{2} y^{2}-2 x^{2} y + x^{2} - 2 x y^{2} + 2 x + y^{2} + 2 y + 1 \\\\\n& = (x y - x - y - 1)^{2}\n\\end{aligned}\n$$\nwhich is always non-negative.\nSolution:\nLet $x=\\tan \\alpha$ and $y=\\tan \\beta$. Then\n$$\n\\begin{aligned}\n\\frac{(x+y)(1-x y)}{\\left(1+x^{2}\\right)\\left(1+y^{2}\\right)} & = \\frac{(\\tan \\alpha+\\tan \\beta)(1-\\tan \\alpha \\tan \\beta)}{\\sec^{2} \\alpha \\sec^{2} \\beta} \\\\\n& = \\cos \\alpha \\cos \\beta (\\tan \\alpha + \\tan \\beta) \\cdot \\cos \\alpha \\cos \\beta (1 - \\tan \\alpha \\tan \\beta) \\\\\n& = (\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta)(\\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta) \\\\\n& = \\sin (\\alpha + \\beta) \\cos (\\alpha + \\beta) \\\\\n& = \\frac{1}{2} \\sin (2\\alpha + 2\\beta),\n\\end{aligned}\n$$\nwhich lies between $-1/2$ and $1/2$ inclusive.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor positive integers $n$, define\n$$\na_{n} = 3^{n} + 6^{n} - 2^{n}.\n$$\nFind, with proof, all primes that do not divide any of the numbers $a_{1}, a_{2}, a_{3}, \\ldots$.",
    "options": [],
    "answer": "2 and 3",
    "solution": "Solution:\n\nAnswer: 2 and 3. It is clear that, for $n \\geq 1$, the terms $6^{n}$ and $2^{n}$ are even while $3^{n}$ is odd, so $2 \\nmid a_{n}$. Similarly, $3^{n}$ and $6^{n}$ are divisible by $3$ but $2^{n}$ is not, so $3 \\nmid a_{n}$.\n\nLet $p \\geq 5$ be a prime. We claim that $a_{p-2} \\equiv 0 \\pmod{p}$. We have\n$$\n\\begin{aligned}\n6 a_{p-2} &= 6\\left(3^{p-2} + 6^{p-2} - 2^{p-2}\\right) \\\\\n&\\equiv 2 \\cdot 3^{p-1} + 6^{p-1} - 3 \\cdot 2^{p-1} \\\\\n&\\equiv 2 + 1 - 3 \\quad (\\text{by Fermat's little theorem}) \\\\\n&\\equiv 0 \\pmod{p}\n\\end{aligned}\n$$\nSince $p$ does not divide $6$, $p$ must divide $a_{p-2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that there exist pairwise distinct positive integers $a_{0}, a_{1}, a_{2}, \\ldots, a_{1000}$ such that\n$$\na_{0}! = a_{1}!\\, a_{2}!\\, \\ldots\\, a_{1000}!\n$$\nHere $n! = 1 \\times 2 \\times \\cdots \\times n$ as usual.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe proceed by induction on $n \\geq 2$. First, we can have $a_{1}=3$, $a_{2}=5$ and $a_{0}=6$. Now, given a working tuple $\\left(a_{0}, a_{1}, \\ldots, a_{n}\\right)$, note that the tuple\n$$\n\\left(a_{0}!, a_{1}, \\ldots, a_{n},\\left(a_{0}-1\\right)!\\right)\n$$\nis a working tuple of length $n+1$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine, with proof, whether there is a function $f(x, y)$ of two positive integers, taking positive integer values, such that\n- For each fixed $x$, $f(x, y)$ is a polynomial function of $y$;\n- For each fixed $y$, $f(x, y)$ is a polynomial function of $x$;\n- However, $f(x, y)$ does not equal any polynomial function of $x$ and $y$.",
    "options": [],
    "answer": "Yes. For example, f(x, y) = 1 + (x−1)(y−1) + (x−1)(y−1)(x−2)(y−2) + (x−1)(y−1)(x−2)(y−2)(x−3)(y−3) + ⋯, which truncates for fixed x or y, is separately polynomial but not a polynomial in two variables.",
    "solution": "Solution:\nThe answer is yes. Consider the following expression:\n$$\nf(x, y) = 1 + (x-1)(y-1) + (x-1)(y-1)(x-2)(y-2) + (x-1)(y-1)(x-2)(y-2)(x-3)(y-3) + \\cdots.\n$$\nHere, although the sum appears to be infinite, if we fix a value $y = y_{0}$, all but the first $y_{0}$ terms contain the factor $(y - y_{0})$ and therefore equal $0$. Therefore $f(x, y_{0})$ is defined and indeed is a polynomial in $x$. (The initial term $1$ is merely to ensure that $f(x, y)$ is always positive.) Symmetrically, when $x$ is fixed, $f(x, y)$ becomes a polynomial in $y$.\n\nIt remains to prove that $f$ is not a polynomial in $x$ and $y$. If so, we can expand $f$ as a finite sum of terms $c x^{a} y^{b}$; let $a_{0}$ be the largest exponent $a$ occurring. Then, for every $y_{0}$, $f(x, y_{0})$ is a polynomial in $x$ of degree at most $a_{0}$. However, we see that $f(x, y_{0})$ is a polynomial in $x$ of degree $y_{0} - 1$. Taking $y_{0} = a_{0} + 2$ yields a contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $S_{1}, S_{2}, S_{3}, \\ldots$ are sets of integers such that no integer is contained in more than one $S_{n}$; every $S_{n}$ has exactly two elements; and the sum of the elements of $S_{n}$ is $n$. Prove that there exist infinitely many values of $n$ with the following property: one of the elements of $S_{n}$ is greater than $13 n / 7$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider any set $S$ representable as the union of $n$ disjoint pairs of integers such that, if any negative number $-m$ is in $S$, then the other element of its pair (henceforth called its \"partner\") is $\\geq 13 m / 6$. Any such $S$ will be called \"plausible\" of order $n$. We will work with plausible sets (and their partner decompositions) in general; later, we will show how these are relevant to the original problem.\nLet $S$ be plausible of order $n$. Then it consists of $n$ disjoint pairs, each of which has a nonnegative sum. Hence, the sum of the elements of $S$ is a nonnegative integer. Now, if we consider the set of all possible values for the sum of $S$, this set consists of nonnegative integers, and it is nonempty (since at least one plausible set exists). Hence, it has a minimal element. So we can meaningfully attempt to characterize a plausible set $S$ (of order $n$) with minimal sum.\nThis set consists of some negative numbers $-m_{1},-m_{2}, \\ldots,-m_{k}$, their (positive) partners, and $2(n-k)$ other nonnegative numbers, all distinct. Clearly, for any fixed choice of the $m_{i}$ and their partners, the sum is minimized by choosing the extra numbers to be the $2(n-k)$ smallest nonnegative integers not equal to the partner of any $-m_{i}$, so we may assume the set to be of this form. Now define $f(m)=\\lceil 13 m / 6\\rceil$ for any integer $m$. We claim it is no loss to assume the partner of $-m_{i}$ is $f\\left(m_{i}\\right)$ for each $i$. Indeed, if not, then this partner is $p>f\\left(m_{i}\\right)$. If $f\\left(m_{i}\\right) \\notin S$, then replacing $p$ by $f\\left(m_{i}\\right)$ gives us a new plausible set with smaller sum than before, contradicting minimality. So $f\\left(m_{i}\\right) \\in S$. But then $f\\left(m_{i}\\right)$ has some partner $q$. We claim we can switch partnerships, matching $-m_{i}$ with $f\\left(m_{i}\\right)$ and $p$ with $q$, so that our partnership decomposition remains plausible. Indeed, if $q \\geq 0$, this is clear. If $q<0$ then $q=$ some $-m_{j} \\Rightarrow f\\left(m_{i}\\right) \\geq f\\left(m_{j}\\right)$ since $f\\left(m_{i}\\right)$ is the partner of $-m_{j}$, but $p>f\\left(m_{i}\\right) \\geq f\\left(m_{j}\\right)$, so we are again safe. Moreover, in either case, neither of the pairs involved is initially of the form $-m_{l}, f\\left(m_{l}\\right)$ (in the case $q=-m_{j}$, we cannot have $f\\left(m_{i}\\right)=f\\left(m_{j}\\right)$ because $f$ is injective), and after the switch, one of them is of this form, so the total number of such pairs strictly increases. So we can perform this switch as many times as necessary, and eventually, we have each $-m_{i}$ partnered with $f\\left(m_{i}\\right)$.\nNext, we claim that, in the context of the partnering result obtained in the previous paragraph, $-j \\in S$ implies $-i \\in S$ for $0<i<j$. Indeed, suppose a counterexample exists. If $f(i) \\notin S$, then we can replace $-j$ and its partner by $-i$ and $f(i)$ to obtain a new plausible set $S$ whose sum has been changed by $\\leq-i+f(i)+j-f(j)=\\lceil 7 i / 6\\rceil-\\lceil 7 j / 6\\rceil<0$, contradicting minimality. So we do have $f(i) \\in S$. It cannot be partnered with a negative number, since this number would have to be $-i$ (by injectivity of $f)$ but $-i \\notin S$; so it is partnered with some positive number. Then, changing this number to $-i$ would again give us a new plausible set with strictly smaller sum, contradiction. This proves the claim.\nBut this claim shows that the negative elements of $S$ must be exactly $-1,-2, \\ldots,-k$, and now we are in a position to estimate the sum of $S$. $S$ consists of these negative elements, their partners $f(1), \\ldots, f(k)$, and $2(n-k)$ other, distinct \"leftover\" nonnegatives which are as small as possible. Let $a$ be the largest leftover, so $0, \\ldots, a$ are all in $S$. We claim $b=\\lfloor 6 a / 13\\rfloor \\leq k$; if not, then $-b \\notin S$ but $f(b) \\leq a$ is in $S$, so it is a leftover (because if its partner is $-c$, injectivity implies $c=b$, impossible), and replacing its nonnegative partner by $-b$ gives a new plausible set with smaller sum, contradiction. So $k \\geq b$, and it follows that fully $b$ of the numbers $0, \\ldots, a$ are partners of negative numbers. Hence, the number of leftovers can be counted in two ways: $a+1-b=2(n-k)$. But $b \\geq 6 a / 13-1 \\Rightarrow 2(n-k) \\leq a-6 a / 13+2=7 a / 13+2 \\Rightarrow a \\geq 26(n-k-1) / 7$. Now, the elements of $S$ consist of $-k, \\ldots,-1,0, \\ldots, a$, and the partners of $-k, \\ldots,-(b+1)$. So the sum of these numbers is\n$$\n\\frac{a(a+1)}{2}-\\frac{k(k+1)}{2}+\\sum_{i=b+1}^{k}\\lceil 13 i / 6\\rceil \\geq \\frac{a(a+1)}{2}-\\frac{k(k+1)}{2}+\\frac{13}{6}\\left(\\frac{k(k+1)}{2}-\\frac{b(b+1)}{2}\\right) ;\n$$\nusing the (rather crude) estimates $0 \\leq a, b, k \\leq 2 n$ and $b \\leq 6 a / 13$ to simplify our calculations, this is\n$$\n\\begin{gathered}\n\\geq \\frac{a^{2}}{2}-\\frac{k^{2}}{2}+\\frac{13 k^{2}}{12}-\\frac{13 b^{2}}{12}-4 n \\geq \\frac{a^{2}}{2}+\\frac{7 k^{2}}{12}-\\frac{13(6 a / 13)^{2}}{12}-4 n=\\frac{7 a^{2}}{26}+\\frac{7 k^{2}}{12}-4 n \\\\\n\\geq \\frac{26(n-k-1)^{2}}{7}+\\frac{7 k^{2}}{12}-4 n \\geq \\frac{26(n-k)^{2}}{7}-8 n+\\frac{7 k^{2}}{12}-4 n=\\frac{361 k^{2}}{84}-\\frac{52 n k}{7}+\\frac{26 n^{2}}{7}-12 n\n\\end{gathered}\n$$\n$$\n=\\frac{361}{84}\\left(k-\\frac{312 n}{361}\\right)^{2}+\\frac{1274}{2527} n^{2}-12 n \\geq \\alpha n^{2}-12 n\n$$\nwhere $\\alpha=1274 / 2527>1 / 2$. So we now know that any plausible set of order $n$ has sum $\\geq \\alpha n^{2}-12 n$.\nNow let us return to the original problem statement. Suppose that the statement is false; we seek a contradiction. Let $i_{1}<i_{2}<\\cdots<i_{c}$ be the (finitely many) values of $i$ for which $S_{i}$ has an element $>13 i / 7$, and let $d=i_{1}+i_{2}+\\cdots+i_{c}$. Now suppose $i$ is not equal to any of $i_{1}, \\ldots, i_{c}$, and suppose $-m \\in S_{i}(m>0)$. Then the other element of $S_{i}$ is $i+m$, and we have $i+m \\leq$ $13 i / 7 \\Rightarrow m \\leq 6 i / 7 \\Rightarrow i+m \\geq 13 m / 6$. Consequently, if we take any $n>i_{c}-c$, we see that the set $S=\\left(S_{1} \\cup S_{2} \\cup \\cdots \\cup S_{n+c}\\right)-\\left(S_{i_{1}} \\cup S_{i_{2}} \\cup \\cdots \\cup S_{i_{c}}\\right)$ is plausible of order $n$. So the sum of its elements is $\\geq \\alpha n^{2}-12 n$. However, this sum should also equal\n$$\n(1+2+\\cdots+(n+c))-\\left(i_{1}+i_{2}+\\cdots+i_{c}\\right)=(n+c)(n+c+1) / 2-d\n$$\nThis is a quadratic function of $n$ with leading coefficient $1 / 2$, so it will actually be less than $\\alpha n^{2}-12 n$ when $n$ is sufficiently large. This is our contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for every positive integer $n$, there is an integer $x$ such that $x^{2}-17$ is divisible by $2^{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe prove this by induction on $n$.\n\nIf $n=1,2$, or $3$, then $x=1$ works.\n\nSuppose that $x^{2}-17$ is divisible by $2^{n}$ and $n \\geq 3$. We seek to find $y$ such that $y^{2}-17$ is divisible by $2^{n+1}$. Let\n$$\nx^{2}-17=k \\cdot 2^{n} \\text{.}\n$$\nIf $k$ is even, we are done since $x^{2}-17$ is divisible by $2^{n+1}$. If $k$ is odd, $k=2 m+1$, we have\n$$\n\\begin{aligned}\n\\left(x+2^{n-1}\\right)^{2}-17 & =x^{2}+2 \\cdot x \\cdot 2^{n-1}+2^{2(n-1)}-17 \\\\\n& =x^{2}-17+x \\cdot 2^{n}+2^{2 n-2} \\\\\n& =(2 m+1) \\cdot 2^{n}+x \\cdot 2^{n}+2^{2 n-2} \\\\\n& =2^{n}(x+1)+2^{n+1}\\left(m+2^{n-3}\\right)\n\\end{aligned}\n$$\nSince $x$ is obviously odd and $n \\geq 3$, this is a multiple of $2^{n+1}$, and $y=x+2^{n-1}$ works.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a polynomial $P(x)$ with integer coefficients, assume that for every positive integer $n$ we have $P(n) > n$. Consider the sequence\n$$\nx_{1} = 1,\\quad x_{2} = P(x_{1}),\\quad \\ldots,\\quad x_{n} = P(x_{n-1}),\\quad \\ldots\n$$\nIf for every positive integer $N$ there exists a member of the sequence divisible by $N$, prove that $P(x) = x + 1$.",
    "options": [],
    "answer": "P(x) = x + 1",
    "solution": "Solution:\n\nAssume the contrary. The polynomial $Q(x) = P(x) - x$ is non-decreasing for all $x$ greater than some $M$, as otherwise it wouldn't be positive, and it has to be bigger than $1$ for each $n$.\n\nIf there are infinitely many integers $n$ such that $P(n) = n + 1$ then we would have $P(x) = x + 1$ (nonzero polynomial $P(x) - x - 1$ can have only finitely many zeroes). Therefore there exists an index $k \\in \\mathbb{N}$ such that $x_{k} > M$ and $P(x_{l}) - x_{l} \\geq 2$ for $l \\geq k$.\n\nAssume that $d = x_{l+1} - x_{l} \\geq 2$. Then $d$ is a divisor of $P(x_{l+1}) - P(x_{l})$, i.e., of $x_{l+2} - x_{l+1}$. Since, if $a \\equiv b \\pmod{d}$, then $P(a) \\equiv P(b)$, by induction all the numbers $x_{l}, x_{l+1}, \\ldots$ have the same remainder modulo $d$. If that remainder is not zero then no term of the sequence could be divisible by $d^{m}$ for sufficiently large $m$, which would contradict the assumptions.\n\nHence $x_{l}$ is divisible by $x_{l+1} - x_{l}$ for all $l \\geq k$. Let $x_{l} = c_{l} (P(x_{l}) - x_{l})$. Then $P(x_{l}) = \\frac{c_{l} + 1}{c_{l}} x_{l}$ holds for infinitely many pairs of integers $(x_{l}, c_{l})$ such that $x_{l} \\to \\infty$. Because $\\frac{c_{l} + 1}{c_{l}} x_{l} = \\left(1 + \\frac{1}{c_{l}}\\right) x_{l} \\leq 2 x_{l}$, the degree of $P$ cannot exceed $1$ so that either $P(x) = 2x$ or $P(x) = x$. Neither of these polynomials satisfy the given condition, and this is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that $\\sin 10^{\\circ}$ is irrational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOne can show the triple-angle identity\n$$\n\\sin (3 \\theta) = 3 \\sin \\theta - 4 \\sin^{3} \\theta.\n$$\nThus letting $x = 2 \\sin (10^{\\circ})$ we derive\n$$\nx^{3} - 3x + 1 = 0.\n$$\nThis polynomial has no rational roots (by, say, Rational Root Theorem), hence $x$ is irrational, whence $\\sin 10^{\\circ}$ is irrational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a trapezoid with $AB \\parallel CD$. Let $M$ and $N$ be the respective midpoints of $AB$ and $CD$. Diagonals $AC$ and $BD$ meet at $P$, and lines $AD$ and $BC$ meet at $Q$. Prove that $M$, $N$, $P$, and $Q$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that $\\triangle QAB \\sim \\triangle QDC$ (by corresponding angles). Since $QM$ and $QN$ are corresponding medians, we have $\\angle AQ M = \\angle DQ N$ and hence $Q$, $M$, and $N$ are collinear.\n\n![](attached_image_1.png)\n\nSimilarly, note that $\\triangle PAB \\sim \\triangle PCD$. Since $PM$ and $PN$ are corresponding medians, we have $\\angle AP M = \\angle CP N$ and hence $P$, $M$, and $N$ are collinear. Since $P$ and $Q$ lie on line $MN$, all four of these points are collinear.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAerith and Bob take turns picking a nonnegative integer, each time subtracting a (positive) divisor from the other's last number. The first person to pick $0$ loses. For example, if Aerith reached $2020$ on some turn, Bob could pick $2020-20=2000$, as $20$ is a divisor of $2020$.\nContinuing this example (with Aerith now picking a divisor of $2000$), if both of them play optimally, who wins?",
    "options": [],
    "answer": "Aerith",
    "solution": "Solution:\n\nIf $1998 = 2000 - 2$ were a losing position, Aerith would pick it. Otherwise, she can choose $1999 = 2000 - 1$. And since this is prime, Bob must pick either $1999 - 1999 = 0$ or $1999 - 1 = 1998$. In any case, Aerith wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo dice are loaded so that the numbers $1$ through $6$ come up with various (possibly different) probabilities on each die. Is it possible that, when both dice are rolled, each of the possible totals $2$ through $12$ has an equal probability of occurring?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Suppose that each of the totals from $2$ to $12$ has an equal probability, which must be $1/11$ since the sum of all probabilities is $1$. Let $a$ and $b$ be the probabilities of a $1$ and a $6$, respectively, on the first die, and let $c$ and $d$ be the corresponding probabilities on the second die.\n\nSince $1/11$ is the probability of rolling a total of $2$, $a c = 1/11$ so $c = 1/(11a)$; since $1/11$ is the probability of rolling $12$, $b d = 1/11$ so $d = 1/(11b)$. Since the probability of rolling a $7$ through the combination $1+6$ or $6+1$ is at most $1/11$,\n$$\n\\begin{aligned}\n\\frac{1}{11} &\\geq a d + b c \\\\\n\\frac{1}{11} &\\geq \\frac{a}{11b} + \\frac{b}{11a} \\\\\n1 &\\geq \\frac{a}{b} + \\frac{b}{a}\n\\end{aligned}\n$$\nSince $a/b$ and $b/a$ are reciprocals, one of them is at least $1$, so this inequality cannot hold.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be an acute triangle with orthocenter $H$, and let $M$ and $N$ denote the midpoints of $AB$ and $AC$. Rays $MH$ and $NH$ intersect the circumcircle of $ABC$ again at points $X$ and $Y$. Prove that the four points $M, N, X, Y$ lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $\\Omega$ be the circumcircle of $\\triangle ABC$. Let $P$ and $Q$ be the reflections of $H$ across $M$ and $N$.\n![](attached_image_1.png)\nSince $\\angle APB = \\angle AHB = 180^\\circ - \\angle C$, point $P$ lies on $\\Omega$; thus so does $Q$. Moreover, $MN \\parallel PQ$, so\n$$\n\\angle XMN = \\angle XPQ = \\angle XYQ\n$$\nand we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\omega$ be a circle with diameter $A B$. A circle $\\gamma$, whose center $C$ lies on $\\omega$, is tangent to $A B$ at $D$ and cuts $\\omega$ at $E$ and $F$. Prove that triangles $C E F$ and $D E F$ have the same area.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet segments $C D$ and $E F$ intersect at $M$. Extend $C D$ to meet $\\gamma$ at $G$ and $\\omega$ at $H$, noting that $G C = C D = D H$. By Power of a Point,\n$$\n\\begin{gathered}\nM G \\cdot M D = M E \\cdot M F = M C \\cdot M H \\\\\n(C G + M C) \\cdot M D = M C \\cdot (M D + D H) \\\\\nC G \\cdot M D = M C \\cdot D H \\\\\nM D = M C .\n\\end{gathered}\n$$\nSo $M$ is the midpoint of $C D$. Drop the perpendiculars $C X$ and $D Y$ to $E F$. Triangles $M C X$ and $M D Y$ are congruent by AAS, so the heights $C X, D Y$ are equal and the triangles $C E F$ and $D E F$ have the same area.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe Fibonacci sequence $F_{1}, F_{2}, F_{3}, \\ldots$ is defined by $F_{1}=F_{2}=1$ and $F_{n+2}=F_{n+1}+F_{n}$. Find the least positive integer $t$ such that for all $n>0$, $F_{n}=F_{n+t}$.",
    "options": [],
    "answer": "60",
    "solution": "Solution:\n\n60 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSome people like to write with larger pencils than others. Ed, for instance, likes to write with the longest pencils he can find. However, the halls of MIT are of limited height $L$ and width $L$. What is the longest pencil Ed can bring through the halls so that he can negotiate a square turn?",
    "options": [],
    "answer": "3L",
    "solution": "Solution:\n\n$3L$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all ordered pairs $(m, n)$ of integers such that $231 m^{2} = 130 n^{2}$.",
    "options": [],
    "answer": "(0, 0)",
    "solution": "Solution:\nThe unique solution is $(0, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of the infinite series\n$$\n\\frac{1}{3^{2}-1^{2}}\\left(\\frac{1}{1^{2}}-\\frac{1}{3^{2}}\\right)+\\frac{1}{5^{2}-3^{2}}\\left(\\frac{1}{3^{2}}-\\frac{1}{5^{2}}\\right)+\\frac{1}{7^{2}-5^{2}}\\left(\\frac{1}{5^{2}}-\\frac{1}{7^{2}}\\right)+\\cdots\n$$",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n$1-\\frac{1}{3}+\\frac{1}{3}-\\frac{1}{5}+\\cdots=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many multiples of $7$ between $10^{6}$ and $10^{9}$ are perfect squares?",
    "options": [],
    "answer": "4375",
    "solution": "Solution:\n\n$\\left[\\sqrt{\\frac{10^{9}}{7^{2}}}\\right] - \\left[\\sqrt{\\frac{10^{6}}{7^{2}}}\\right] = 4517 - 142 = 4375$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA triangle has sides of length $888$, $925$, and $x > 0$. Find the value of $x$ that minimizes the area of the circle circumscribed about the triangle.",
    "options": [],
    "answer": "259",
    "solution": "Solution: 259 .",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x = 2001^{1002} - 2001^{-1002}$ and $y = 2001^{1002} + 2001^{-1002}$. Find $x^2 - y^2$.",
    "options": [],
    "answer": "-4",
    "solution": "Solution:\n\nWe are given $x = 2001^{1002} - 2001^{-1002}$ and $y = 2001^{1002} + 2001^{-1002}$. We are to find $x^2 - y^2$.\n\nRecall that $x^2 - y^2 = (x - y)(x + y)$.\n\nCompute $x + y$ and $x - y$:\n\n$x + y = (2001^{1002} - 2001^{-1002}) + (2001^{1002} + 2001^{-1002}) = 2 \\times 2001^{1002}$\n\n$x - y = (2001^{1002} - 2001^{-1002}) - (2001^{1002} + 2001^{-1002}) = -2 \\times 2001^{-1002}$\n\nTherefore,\n\n$x^2 - y^2 = (x + y)(x - y) = [2 \\times 2001^{1002}] \\times [-2 \\times 2001^{-1002}] = -4 \\times 2001^{1002} \\times 2001^{-1002}$\n\n$2001^{1002} \\times 2001^{-1002} = 2001^{1002 - 1002} = 2001^0 = 1$\n\nSo,\n\n$x^2 - y^2 = -4 \\times 1 = -4$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $(x, y)$ be a point in the cartesian plane, $x, y > 0$. Find a formula in terms of $x$ and $y$ for the minimal area of a right triangle with hypotenuse passing through $(x, y)$ and legs contained in the $x$ and $y$ axes.",
    "options": [],
    "answer": "2xy",
    "solution": "Solution:\n$2 x y$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrevor and Edward play a game in which they take turns adding or removing beans from a pile. On each turn, a player must either add or remove the largest perfect square number of beans that is in the heap. The player who empties the pile wins. For example, if Trevor goes first with a pile of 5 beans, he can either add 4 to make the total 9, or remove 4 to make the total 1, and either way Edward wins by removing all the beans. There is no limit to how large the pile can grow; it just starts with some finite number of beans in it, say fewer than 1000.\n\nBefore the game begins, Edward dispatches a spy to find out how many beans will be in the opening pile, call this $n$, then \"graciously\" offers to let Trevor go first. Knowing that the first player is more likely to win, but not knowing $n$, Trevor logically but unwisely accepts, and Edward goes on to win the game. Find a number $n$ less than 1000 that would prompt this scenario, assuming both players are perfect logicians. A correct answer is worth the nearest integer to $\\log_{2}(n-4)$ points.",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nThe correct answers are 0 (worth imaginary points), 5 (worth 0 points), 20 (4 points), 29, 45 (5 points), 80 (6 points), 101, 116, 135, 145, 165, 173 (7 points), 236, 257 (8 points), 397, 404, 445, 477, 540, 565, 580, 629, 666 (9 points), $836, 845, 885, 909, 944$, 949, 954, 975 (10 points). This game is called Epstein's Put-or-Take-a-Square game. It is unknown whether or not these numbers (or the first player's win positions) have positive density.",
    "topic": "Discrete Mathematics",
    "subtopic": "Logic"
  },
  {
    "id": 46327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven five nonnegative real numbers with sum $1$, prove that it is possible to arrange them at the vertices of a regular pentagon such that no two numbers connected by a side of the pentagon have product exceeding $1/9$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLabel the numbers $a, b, c, d, e$ in increasing order. Place them around the pentagon in the order $e, a, d, c, b$. Then it is clear that the products of the numbers on the sides follow the inequalities\n$$\na d \\leq a e \\leq b e \\quad \\text{and} \\quad b c \\leq c d\n$$\nThus it suffices to prove that $b e \\leq 1/9$ and $c d \\leq 1/9$. Using the AM-GM inequality,\n$$\n1 = a + b + c + d + e \\geq 0 + 0 + c + d + d = c + 2d \\geq \\frac{3c}{2} + \\frac{3d}{2} \\geq 2 \\sqrt{\\frac{3c}{2} \\cdot \\frac{3d}{2}} = 3 \\sqrt{c d}\n$$\nso $c d \\leq 1/9$. Also,\n$$\n1 = a + b + c + d + e \\geq 0 + b + b + b + e = 3b + e \\geq 2 \\sqrt{3b \\cdot e}\n$$\nso $b e \\leq 1/12 < 1/9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe take a $6 \\times 6$ chessboard, which has six rows and columns, and indicate its squares by $(i, j)$ for $1 \\leq i, j \\leq 6$. The $k$th northeast diagonal consists of the six squares satisfying $i-j \\equiv k \\pmod{6}$ (and so there are six such diagonals); hence there are six such diagonals.\nDetermine if it is possible to fill the entire chessboard with the numbers $1,2, \\ldots, 36$ (each exactly once) such that each row, each column, and each of the six northeast diagonals has the same sum.",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. Assume for contradiction such a coloring existed; then each row, column and northeast diagonal would have sum exactly\n$$\nN = \\frac{1}{6}(1+2+\\cdots+36) = 111.\n$$\nNow consider the marked squares shown below.\n\n| $A$ | $B$ | $A$ | $B$ | $A$ | $B$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|      | $C$ |      | $C$ |      | $C$ |\n| $A$ | $B$ | $A$ | $B$ | $A$ | $B$ |\n|      | $C$ |      | $C$ |      | $C$ |\n| $A$ | $B$ | $A$ | $B$ | $A$ | $B$ |\n|      | $C$ |      | $C$ |      | $C$ |\n\nLet $\\mathcal{A}, \\mathcal{B}, \\mathcal{C}$ denote the sum of numbers of the squares labelled $A, B, C$, respectively. We deduce that\n$$\n\\begin{aligned}\n\\mathcal{A} + \\mathcal{B} &= 3N \\\\\n\\mathcal{B} + \\mathcal{C} &= 3N \\\\\n\\mathcal{A} + \\mathcal{C} &= 3N\n\\end{aligned}\n$$\nby considering three rows, three columns, and three northeast diagonals. Summing all these equations gives\n$$\n2(\\mathcal{A} + \\mathcal{B} + \\mathcal{C}) = 9N\n$$\nwhich is impossible, because the left-hand side is even while the right-hand side is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute triangle, and $P$ a point on the interior of side $BC$. Let $I$ be the incenter of triangle $ABC$, and denote by $D$ the foot of the altitude from $I$ to $BC$. Line $BI$ meets the internal angle bisector of $\\angle APC$ at $X$, while line $CI$ meets the internal angle bisector of $\\angle APB$ at $Y$. Show that the points $D, P, X, Y$ lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst, notice that $\\angle YPX$ is a right angle. Thus, we claim that in fact $D$ and $P$ lie on a circle with diameter $\\overline{XY}$; in light of this it suffices to prove that\n$$\nDX^{2} + DY^{2} = PX^{2} + PY^{2}\n$$\nLet $K$ and $L$ be the feet of the altitudes from $X$ and $Y$ to $\\overline{BC}$, respectively. By the Pythagorean Theorem we may write these as\n$$\n\\begin{aligned}\n& DX^{2} = DK^{2} + XK^{2} \\\\\n& DY^{2} = DL^{2} + YL^{2} \\\\\n& PX^{2} = PK^{2} + XK^{2} \\\\\n& PY^{2} = PL^{2} + YL^{2}\n\\end{aligned}\n$$\n![](attached_image_1.png)\nIn light of this it is sufficient to show that\n$$\nDK^{2} + DL^{2} = PK^{2} + PL^{2}.\n$$\nWithout loss of generality, assume $L, D, P, K$ lie in that order. Note that $X$ is the $B$-excenter of $\\triangle ABP$. From this configuration we derive that\n$$\n\\begin{aligned}\n& BK = \\frac{1}{2}(AB + AP + BP) \\\\\n& PK = \\frac{1}{2}(AB + AP - BP)\n\\end{aligned}\n$$\nFrom this we derive that\n$$\n\\begin{aligned}\nDK + PK & = BK + PK - BD \\\\\n& = \\frac{1}{2}(2AB + 2AP - (AB + BC - AC)) \\\\\n& = \\frac{1}{2}(2AP + AB + AC - BC)\n\\end{aligned}\n$$\nRepeating the same calculation with $\\triangle ACP$, we see this equals $DL + PL$ too. Thus, we have\n$$\nDK + PK = DL + PL.\n$$\nMultiplying both sides by $DP = DK - PK = PL - DL$, we obtain $DK^{2} - PK^{2} = PL^{2} - DL^{2}$, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, $c$ be pairwise distinct integers. Prove that\n$$\n\\frac{a^{3}+b^{3}+c^{3}}{3} \\geq a b c+\\sqrt{3(a b+b c+c a+1)} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $3 k^{2}-1=a b+b c+c a$, so we need $a^{3}+b^{3}+c^{3} \\geq 3(a b c+3 k)$. Now, we have\n$$\n(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\geq 2^{2}+1^{2}+1^{2}=6 .\n$$\nIn particular, we get\n$$\n(a+b+c)^{2}=\\frac{1}{2}\\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\\right]+3\\left(3 k^{2}-1\\right) \\geq 9 k^{2}\n$$\nThus $a+b+c \\geq 3 k$. Now, using the factorization of $a^{3}+b^{3}+c^{3}-3 a b c$ gives\n$$\na^{3}+b^{3}+c^{3}-3 a b c \\geq(3 k)\\left(\\frac{1}{2} \\cdot 6\\right)=9 k\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConvex quadrilateral $ABCD$ with $BC = CD$ is inscribed in circle $\\Omega$; the diagonals of $ABCD$ meet at $X$. Suppose $AD < AB$, the circumcircle of triangle $BCX$ intersects segment $AB$ at a point $Y \\neq B$, and ray $\\overrightarrow{CY}$ meets $\\Omega$ again at a point $Z \\neq C$. Prove that ray $\\overrightarrow{DY}$ bisects angle $ZDB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThis is mostly just angle chasing. In this case $Y$ and $Z$ lie between $A$ and $B$, on the respective segment/arc. We'll prove $Y$ is the incenter of $\\triangle ZDB$; it will follow that ray $\\overrightarrow{DY}$ indeed internally bisects $\\angle ZDB$. It suffices to prove the following two facts:\n- $BY$ is the internal angle bisector of $\\angle DBZ$. This is true in general; it doesn't require $CB = CD$. It's part of the spiral similarity configuration centered at $B$: $YX \\rightarrow ZA$ and $B: ZY \\rightarrow AX$, due to $YZ \\cap AX = C$ and $B = (CYX) \\cap (CZA)$. More explicitly, this follows from the angle chase\n$$\n\\angle DBA = \\angle XBY = \\angle XCY = \\angle ACZ = \\angle ABZ.\n$$\n- $ZY$ is the internal angle bisector of $\\angle BZD$, since $CB = CD$. Indeed (more explicitly), arcs $BC$ and $CD$ are equal, so $\\angle BZC = \\angle CZD$, i.e. $YZ$ bisects $\\angle BZD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf triangle $ABC$ has perimeter $2$, prove that not all its altitudes can exceed $1 / \\sqrt{3}$ in length.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $a, b, c$ be the side lengths of the triangle, where $a$ is the longest side, so $a \\geq 2 / 3$, and let $h_{a}$ be the length of the altitude to that side. The semiperimeter is $1$, so by Heron's formula we get\n$$\n\\frac{1}{2} a h_{a} = \\sqrt{(1-a)(1-b)(1-c)} \\Longrightarrow h_{a} \\leq 3 \\sqrt{(1-a)(1-b)(1-c)}\n$$\nBut by AM-GM,\n$$\n\\frac{(1-a)+(1-b)+(1-c)}{3} = \\frac{1}{3} \\geq \\sqrt[3]{(1-a)(1-b)(1-c)} \\Longrightarrow \\sqrt{(1-a)(1-b)(1-c)} \\leq \\frac{1}{3 \\sqrt{3}}\n$$\nThus, we get\n$$\nh_{a} \\leq 3 \\cdot \\frac{1}{3 \\sqrt{3}} = \\frac{1}{\\sqrt{3}}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine whether there exists a polynomial $f(x, y)$ of two variables, with real coefficients, with the following property: A positive integer $m$ is a triangular number if and only if there do not exist positive integers $x$ and $y$ such that $f(x, y)=m$.",
    "options": [],
    "answer": "Yes. For example, f(x, y) = ((x + y)(x + y − 1))/2 + y works.",
    "solution": "Solution:\nThe answer is yes.\n\nThe difference between the $n$th and $(n+1)$st triangular numbers is $n+1$. Therefore, a positive integer $m$ is not a triangular number if and only if it has the form\n$$\n\\frac{n^{2}+n}{2}+y\n$$\nwhere $1 \\leq y \\leq n$. Define $x = n - y + 1$; then the conditions $1 \\leq y$ and $y \\leq n$ are equivalent to $y \\geq 1$ and $x \\geq 1$. Conversely, $n = x + y - 1$, so the polynomial\n$$\nf(x, y) = \\frac{(x + y - 1)^{2} + (x + y - 1)}{2} + y = \\frac{(x + y)(x + y - 1)}{2} + y\n$$\nhits exactly the non-triangular numbers as $x$ and $y$ range over positive integers.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with circumcircle $\\Gamma$, whose incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$. We draw a circle tangent to segment $BC$ at $D$ and to minor $\\operatorname{arc} \\widehat{BC}$ of $\\Gamma$ at the point $A_{1}$. Define $B_{1}$ and $C_{1}$ in a similar way. Prove that lines $A_{1}D$, $B_{1}E$, $C_{1}F$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy the so-called \"shooting lemma\" (which is proved by taking homothety at $A_{1}$) we find that line $A_{1}D$ passes through the arc midpoint of $\\widehat{BAC}$ of $\\Gamma$; denote this arc midpoint by $X$. Define $Y$ and $Z$ similarly, so that $Y$ lies on line $B_{1}E$ and $Z$ lies on line $C_{1}F$.\n\nWe note the triangles $XYZ$ and $DEF$ are homothetic, since their corresponding sides are parallel: line $YZ$ and $EF$ are both known to be perpendicular to the internal $\\angle A$-bisector. Thus $DX$, $EY$, $FZ$ meet at a point—which thus is also the concurrency point of $A_{1}D$, $B_{1}E$, $C_{1}F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46335,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine whether there exist two distinct finite subsets $A$ and $B$ of the reals such that for every polynomial $f$ of degree $2012$ with real coefficients,\n$$\n\\sum_{x \\in A} f(x) = \\sum_{x \\in B} f(x)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe will prove by induction on $n \\geq 0$ that there exist distinct sets $A_n$ and $B_n$ satisfying the desired equality\n$$\n\\sum_{x \\in A_n} f(x) = \\sum_{x \\in B_n} f(x)\n$$\nfor all polynomials $f$ of degree at most $n$.\n\nThe base case is $n=0$; we take $A = \\{0\\}$ and $B = \\{1\\}$.\n\nNow we suppose $A_{n-1}$ and $B_{n-1}$, $n \\geq 1$, to be constructed and define\n$$\n\\begin{aligned}\n& A_n = A_{n-1} \\cup (B_{n-1} + t) \\\\\n& B_n = B_{n-1} \\cup (A_{n-1} + t)\n\\end{aligned}\n$$\nwhere the parameter $t$ of the translations is a real number large enough that the unions are actually disjoint unions.\n\nNow the condition becomes\n$$\n\\begin{aligned}\n\\sum_{x \\in A_{n-1}} f(x) + \\sum_{x \\in B_{n-1}} f(x+t) &= \\sum_{x \\in B_{n-1}} f(x) + \\sum_{x \\in A_{n-1}} f(x+t) \\\\\n\\sum_{x \\in A_{n-1}} [f(x) - f(x+t)] &= \\sum_{x \\in B_{n-1}} [f(x) - f(x+t)] .\n\\end{aligned}\n$$\nIf $f(x)$ has degree at most $n$, then the polynomial $g(x) = f(x) - f(x+t)$ has degree at most $n-1$ since the leading terms of $f(x)$ and $f(x+t)$ are identical. Thus this last condition holds by the induction hypothesis.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSix students taking a test sit in a row of seats with aisles only on the two sides of the row. If they finish the test at random times, what is the probability that some student will have to pass by another student to get to an aisle?",
    "options": [],
    "answer": "43/45",
    "solution": "Solution:\n\nThe probability $p$ that no student will have to pass by another student to get to an aisle is the probability that the first student to leave is one of the students on the end, the next student to leave is on one of the ends of the remaining students, etc.:\n\n$$\np = \\frac{2}{6} \\cdot \\frac{2}{5} \\cdot \\frac{2}{4} \\cdot \\frac{2}{3}\n$$\n\nso the desired probability is\n\n$$\n1 - p = \\frac{43}{45}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA bulldozer is touring Pascal's triangle. It starts at the top of the triangle, at $\\left(\\begin{array}{l}0 \\\\ 0\\end{array}\\right)=1$. Each move, it travels to an adjacent positive integer, but can never return to a spot it has already visited. Moreover, if it has visited two numbers $a>b$, it may not visit $a+b$ or $a-b$. Finally, the bulldozer is confined to the first 140 rows of Pascal's triangle.\nProve that the bulldozer may visit at least 2017 numbers. (By convention, the $n$th row contains the entries $\\left(\\begin{array}{c}n-1 \\\\ k\\end{array}\\right)$ for $k=0, \\ldots, n-1$, hence the $n$th row has $n$ entries.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe main idea is to visit odd numbers!\nWe claim inductively that the first $2^{n}$ rows of Pascal's triangle satisfy the following properties:\n- The $2^{n}$th row contains only odd numbers.\n- The first $2^{n}$ rows contain $3^{n}$ odd numbers.\n- When taken modulo 2, there is 120 degree symmetry\n- There is a path starting at any corner to any other corner through only odd numbers.\nIndeed this is clear for $n=1$. For the inductive step, let $T$ denote the shape of the first $2^{n}$ rows modulo 2. Note that row $2^{n}+1$ contains all even numbers except the endpoints $\\left(\\begin{array}{c}2^{n} \\\\ 0\\end{array}\\right)=\\left(\\begin{array}{c}2^{n} \\\\ 2^{n}\\end{array}\\right)=1$. Thus in fact we get two side-by-side copies of the triangle $T$, which meet on row $2^{n+1}$ and thus have all ones. (Between the two copies of $T$ we get an inverted triangle having all entries 0.) From these observations, we see that all statements in the inductive hypothesis hold.\nThus, we may visit odd numbers from rows 1 to 128. In doing so, we visit $3^{7}=2187$ odd numbers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that the polynomial $\\left(x^{2}+x\\right)^{2^{1000}}+1$ cannot be factored as the product of two nonconstant polynomials with integer coefficients.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume for contradiction this is not the case, and the polynomial can be written as\n$$\n\\left(x^{2}+x\\right)^{2^{1000}}+1=f(x) g(x)\n$$\nfor some nonconstant $f$ and $g$ with integer coefficients. Clearly we may assume $f$ and $g$ have leading coefficient one. Taking modulo 2 we obtain that\n$$\nf(x) g(x) \\equiv\\left(x^{2}+x\\right)^{2^{1000}}+1 \\equiv\\left(x^{2}+x+1\\right)^{2^{1000}} \\quad(\\bmod 2)\n$$\nwhere the last step follows by noticing that $(a+b)^{2} \\equiv a^{2}+b^{2}(\\bmod 2)$.\nAs $x^{2}+x+1$ is irreducible modulo 2, it follows that we must have\n$$\nf(x)=\\left(x^{2}+x+1\\right)^{\\alpha}+2 \\hat{f}(x)\n$$\nand\n$$\ng(x)=\\left(x^{2}+x+1\\right)^{\\beta}+2 \\hat{g}(x)\n$$\nfor some polynomials $\\hat{f}, \\hat{g}$ with integer coefficients, where $\\alpha+\\beta=1000$. By hypothesis, $\\alpha, \\beta \\geq 1$.\nLet $\\varepsilon$ be a root to $x^{2}+x+1$, namely $\\varepsilon=\\frac{1}{2}(-1+i \\sqrt{3})$. Substituting $\\varepsilon$ into the original relation, we derive\n$$\n2=(-1)^{2^{1000}}+1=4 \\hat{f}(\\varepsilon) \\hat{g}(\\varepsilon)\n$$\nand hence $\\frac{1}{2}=\\hat{f}(\\varepsilon) \\hat{g}(\\varepsilon)$. But we claim this is impossible. Indeed, the right-hand side is a polynomial in $\\varepsilon$ with integer coefficients. Repeatedly applying the identity $\\varepsilon^{2}=-(\\varepsilon+1)$, we find that the right-hand side must be of the form $a+b \\varepsilon$, where $a, b$ are integers. By equating real and imaginary parts this gives $a=\\frac{1}{2}$ and $b=0$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all perfect squares that can be written as the sum of two powers of $2$.",
    "options": [],
    "answer": "All such squares are 4^{k+1} or 9*4^{k} for k >= 0.",
    "solution": "Solution:\nWe claim that the squares which work are $4^{k+1}$ or $9 \\cdot 4^{k}$ for $k \\in \\mathbb{Z}_{\\geq 0}$. These work because $4^{k+1} = 2^{2k+1} + 2^{2k+1}$ and $9 \\cdot 4^{k} = 2^{2k+3} + 2^{2k}$.\n\nWrite $n^{2} = 2^{a} + 2^{b}$ for $n, a, b \\in \\mathbb{Z}_{\\geq 0}$ and $a \\geq b$. If $a = b$, $n^{2} = 2^{b+1}$, and since the perfect squares that are powers of two are the powers of $4$, we have $n = 4^{k+1}$ for some $k \\in \\mathbb{Z}_{\\geq 0}$.\n\nOtherwise, $a - b > 0$ so $2^{a-b}$ is even and $2^{a-b} + 1$ is odd. Therefore, $2^{b}$ and $2^{a-b} + 1$ are relatively prime, so since $n^{2} = 2^{a} + 2^{b} = 2^{b} (2^{a-b} + 1)$, $2^{b}$ and $2^{a-b} + 1$ are perfect squares, so $2^{b}$ is a power of $4$. Let $m^{2} = 2^{a-b} + 1$, so $(m-1)(m+1) = 2^{a-b}$. Then, $m-1$ and $m+1$ are powers of $2$ that differ by $2$. By inspection, we see that $m = 3$ so $2^{a-b} + 1 = 9$ and $n^{2}$ is $9$ times a power of $4$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine if it is possible to color each of the rational numbers either red or blue such that the following three conditions are all satisfied:\n(i) $x$ and $-x$ are opposite colors, for all rational $x \\neq 0$;\n(ii) $x$ and $1-x$ are opposite colors, for all rational $x \\neq 1 / 2$;\n(iii) $x$ and $1 / x$ are opposite colors, for all rational $x \\neq 0, \\pm 1$.",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\nThe answer is yes.\nWe will prove the following statement by induction: It is possible to color the rational numbers with denominator at most $n$ red and blue such that the conditions (i)-(iii) hold whenever the two rational numbers in question both have denominator at most $n$, and such that no number reverses color at any step. (The last condition ensures that, when the induction is complete, every number will have a well-defined color.)\n\nWe make two base cases. For $n=1$, color all the positive integers red and all the nonpositive integers blue. Then, for $n=2$, color the number $x+\\frac{1}{2}$ (for integer $x$) blue if $x \\geq 0$ and red if $x<0$. It is easy to see that conditions (i)-(iii) are satisfied so far.\n\nAssume that all rational numbers with denominator less than $n$ ($n \\geq 3$) have been colored. We must color the fractions $m / n$, where $m$ is an integer coprime to $n$. First, when $0<m<n$, color $m / n$ the opposite of $n / m$ (which has already been colored). Then color $1+m / n$ the same color as $m / n$ for each $m>0$ coprime to $n$. Finally, color $-m / n$ the opposite of $m / n$. Condition (i) is obviously satisfied. Condition (iii), that $m / n$ and $n / m$ are opposite colors, is relevant only if $|m|<n$; it is satisfied by definition if $m>0$, and otherwise it is enough to note that the row of numbers\n$$\n\\frac{m}{n}, \\quad \\frac{n}{m}, \\quad-\\frac{n}{m}, \\quad-\\frac{m}{n}\n$$\nhas been colored alternately red and blue, implying that the two end terms have opposite colors. It remains to verify condition (ii). If $x>1$ or $x<0$, this follows from (i) and the fact that $y$ and $y+1$ have been colored alike for $y>0$. So we are left with the case $0<x<1$, that is, proving that $m / n$ and $(n-m) / n$ are the same color when $0<m<n$. The six numbers\n$$\n\\frac{m}{n}, \\frac{n}{m},-\\frac{n-m}{m},-\\frac{m}{n-m}, \\frac{n}{n-m}, \\frac{n-m}{n}\n$$\nhave been colored alternately red and blue by (ii) and (iii), so the two end terms have opposite colors.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor which $n \\geq 3$ can an $n \\times n$ board with all four corners removed be tiled with \"L\" and \"J\" tetris pieces?",
    "options": [],
    "answer": "All n ≥ 3 with n ≡ 2 (mod 4).",
    "solution": "Solution:\n\nWe claim that $n$ works iff $n \\equiv 2 \\pmod{4}$.\n\nThe board has $n^{2}-4$ squares. If $n$ is odd, this is odd, so the board wouldn't be coverable with tetris tiles (which have 4 cells each).\n\nThus $n$ is even. Alternately color the rows of the board black and white. Each of the tetris pieces must cover three squares of one color and one of the other. Let $b$ be the number of blocks which cover three black squares and $w$ be the number of blocks which cover three white squares.\n\nBy vertical symmetry, there are an equal amount of black squares and white squares, so $b=w$. The total number of tiles is thus $b+w=2b$, so the total number of cells is $8b=n^{2}-4$. Thus, $8 \\mid n^{2}-4$, so $8 \\nmid n^{2}$, so $4 \\nmid n$. Thus, $n \\equiv 2 \\pmod{4}$, as desired.\n\nIt remains to show that all $n$ such that $n \\equiv 2 \\pmod{4}$ work. In fact, one only needs \"J\" tiles to construct such $n$; we present the following proof by induction:\n\n![](attached_image_1.png)\n\nOur base case is $n=2$, which loses its tiles once we remove the corners, so it is trivially tiled by 0 pieces. For $n \\equiv 2 \\pmod{4}$, we can follow the process illustrated in the illustration: The pink rectangles have length $n-6 \\equiv 0 \\pmod{4}$ and width 2, so we can nest $\\mathrm{J}$ blocks to form a rectangle of width 2 and length 4, which we stack to tile the pink regions. The red region is a board with length $n-4 \\equiv 2 \\pmod{4}$ and its corners removed. By induction, such a board can be tiled by our blocks, so we're done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAt a certain school, there are 6 subjects offered, and a student can take any combination of them. It is noticed that for any two subjects, there are fewer than 5 students taking both of them and fewer than 5 students taking neither. Determine the maximum possible number of students at the school.",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nLet us count in two ways the number of ordered pairs $\\left(s,\\left\\{c_{1}, c_{2}\\right\\}\\right)$, where $s$ is a student, $\\left\\{c_{1}, c_{2}\\right\\}$ is an unordered pair of two distinct courses, and $s$ is either taking both $c_{1}$ and $c_{2}$ or neither $c_{1}$ nor $c_{2}$.\n\nFirst, we consider any of the $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)=15$ pairs $\\left\\{c_{1}, c_{2}\\right\\}$ of distinct courses. By the given conditions, at most 4 students are taking neither $c_{1}$ nor $c_{2}$ and at most 4 take both, implying that $\\left\\{c_{1}, c_{2}\\right\\}$ gives rise to at most 8 pairs $\\left(s,\\left\\{c_{1}, c_{2}\\right\\}\\right)$, so the total number of such pairs is at most 120.\n\nSecond, let $N$ be the number of students. Consider any student $s$. If $k$ denotes the number of courses $s$ is taking, then the number of pairs of courses $\\left\\{c_{1}, c_{2}\\right\\}$ such that $s$ is taking both or neither of them is\n$$\n\\left(\\begin{array}{l}\nk \\\\\n2\n\\end{array}\\right)+\\left(\\begin{array}{c}\n6-k \\\\\n2\n\\end{array}\\right)\n$$\nPlugging in $k=0,1, \\ldots, 6$ shows that the minimum value of this expression is 6, attained at $k=3$. Thus there are at least $6 N$ pairs $\\left(s,\\left\\{c_{1}, c_{2}\\right\\}\\right)$.\n\nPutting this together, we see that $6 N \\leq 120$, so $N$ is at most 20. A total of 20 students is indeed possible, as can be seen by letting each student take a distinct triplet of courses from the $\\left(\\begin{array}{l}6 \\\\ 3\\end{array}\\right)=20$ possible triplets.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCircles $j$ and $k$, centered at $O$ and $P$ respectively, do not intersect. The two tangent rays from $O$ to $k$ meet $j$ at $A$ and $B$, respectively, and the two tangent rays from $P$ to $j$ meet $k$ at $C$ and $D$, respectively. Prove that $A, B, C$, and $D$ are the vertices of a rectangle.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWithout loss of generality, we may assume that $A, B, C$, and $D$ have the relative positions shown. We label the points of tangency $W, X, Y$, and $Z$. We also note that the entire construction is symmetric about the line of centers $OP$, which therefore perpendicularly bisects segments $AB$, $CD$, $WX$, and $YZ$ at their respective midpoints $K, L, M$, and $N$. Let $r_1$ and $r_2$ be the respective radii of $j$ and $k$. We will first prove that $AK = CL$. Since $\\triangle OAK \\sim \\triangle OYN \\sim \\triangle OPY$, we have\n$$\n\\frac{AK}{OA} = \\frac{PY}{OP}, \\quad \\text{so } AK = \\frac{OA \\cdot PY}{OP} = \\frac{r_1 r_2}{OP}\n$$\nSymmetrically, $CL = r_1 r_2 / OP$ so $AK = CL$. Now quadrilateral $AKLC$ has right angles at $K$ and $L$ and equal, parallel sides $AK = CL$, so it is a rectangle. Symmetrically, $BKLD$ is a rectangle so $ABDC$ is a rectangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $R$ be the region consisting of all points inside or on the boundary of a given circle of radius $1$. Find, with proof, all positive real numbers $d$ such that it is possible to color each point of $R$ red, green or blue such that any two points of the same color are separated by a distance less than $d$.",
    "options": [],
    "answer": "d >= sqrt(3)",
    "solution": "Solution:\n\nThe answer is all $d \\geq \\sqrt{3}$.\n\nIf $d \\geq \\sqrt{3}$, refer to the diagram at right. Color $120^{\\circ}$ sectors $OAB$, $OBC$, and $OCA$ red, green, and blue respectively, including their boundaries on the circle. For the boundaries between the sectors, color $OA$ red, $OB$ green, and $OC$ blue, making point $O$ red. Then it is clear that the red sector can be covered by a circle of diameter $AB = \\sqrt{3}$, so all points in it have a distance less than $\\sqrt{3}$ except $A$ and $B$, but $B$ is not even red. So all red points have a distance less than $\\sqrt{3}$, and a similar argument can be made for the green and the blue.\n\n![](attached_image_1.png)\n\nNow assume that $d < \\sqrt{3}$. Let $\\theta < 120^{\\circ}$ be the angle corresponding to a chord of length $d$. Let $n$ be a positive integer large enough so that\n$$\n\\frac{1}{3n+1} < \\frac{120^{\\circ} - \\theta}{120^{\\circ}}\n$$\nthat is,\n$$\n\\theta < 120^{\\circ} - \\frac{120^{\\circ}}{3n+1}\n$$\nDefine points $P_0, P_1, P_2, \\ldots$ recursively as follows: $P_0$ is any point on the circumference, and for $i \\geq 0$, $P_{i+1}$ is the counterclockwise rotation of $P_i$ by the angle\n$$\n\\alpha = 120^{\\circ} - \\frac{120^{\\circ}}{3n+1} = 360^{\\circ} \\cdot \\frac{n}{3n+1}.\n$$\nNote that\n$$\n\\theta < \\alpha < 360^{\\circ} - \\theta\n$$\nso for each $i \\geq 0$, $P_i P_{i+1} > d$ and $P_i$ and $P_{i+1}$ are different colors. Also note that\n$$\n\\theta < 2\\theta < 2\\alpha < 240^{\\circ} < 360^{\\circ} - \\theta,\n$$\nso $P_i$ and $P_{i+2}$ are different colors. Thus $P_i, P_{i+1}$, and $P_{i+2}$ are all different colors for each $i$. If we assume $P_0$ is red, $P_1$ is green, and $P_2$ is blue, we will get $P_3$ red, $P_4$ green, and so on until $P_{3n+1}$ is green. But the total angle of rotation from $P_0$ to $P_{3n+1}$ is\n$$\n(3n+1)\\alpha = (3n+1) \\cdot 360^{\\circ} \\cdot \\frac{n}{3n+1} = n \\cdot 360^{\\circ},\n$$\nso $P_0$ and $P_{3n+1}$ are the same point and we have a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAerith writes $50$ consecutive positive integers in a circle on a whiteboard. Each minute after, she simultaneously replaces each number $x$ with $2020a - x + 2020b$, where $a$ and $b$ were the numbers next to $x$. Can she choose her initial numbers such that she will never write down a negative number?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $R_{0}$ be the initial sum of every other number that Aerith wrote down, and let $S_{0}$ be the sum of the remaining numbers. Let $R_{t}$ and $S_{t}$ be the corresponding sums after $t$ minutes. Because $D = R_{0} + S_{0}$ is a sum of $25$ even and $25$ odd numbers, it is itself odd, so $R_{0} - S_{0} \\neq 0$.\n\nWe have\n$$\nR_{t} = 2020 S_{t-1} - R_{t-1} + 2020 S_{t-1} = 4040 S_{t-1} - R_{t-1}\n$$\nand similarly\n$$\nS_{t} = 4040 S_{t-1} - R_{t-1}.\n$$\nAdding and subtracting, we get\n$$\nR_{t} + S_{t} = 4039\\left(R_{t-1} + S_{t-1}\\right)\n$$\nand\n$$\nR_{t} - S_{t} = -4041\\left(R_{t-1} - S_{t-1}\\right)\n$$\nso $|R_{t} - S_{t}|$ grows exponentially faster than $R_{t} + S_{t}$. Thus, for some $T$, $|R_{T} - S_{T}| > R_{T} + S_{T}$. This could not be true if $R_{T}$ and $S_{T}$ were both nonnegative, so Aerith will eventually write down a negative number.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers $1, 2, \\ldots, 2014$ are arranged evenly around a circle in arbitrary order. We are permitted to swap two adjacent numbers, as long as they do not sum to $2015$. Prove that it is impossible to perform finitely many swaps so that each number ends up diametrically opposite from its starting point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume for contradiction it's possible. Each time a number moves one slot counterclockwise, assign a score of $+1$; each time a number moves one slot clockwise, assign a score of $-1$. Thus every move produces a total score of $1 + (-1) = 0$.\n\nLet $S(a)$ denote the total score accumulated by moves of $a$. Observe that if $a + b = 2015$, we see that in fact $S(a) = S(b)$. Thus\n$$\n0 = \\sum_{a=1}^{2014} S(a) = \\sum_{a=1}^{1007} S(a) + S(2015 - a) = 2 \\sum_{a=1}^{1007} S(a)\n$$\nHence\n$$\n0 = \\sum_{a=1}^{1007} S(a)\n$$\nOn the other hand, since $a$ is supposed to end up diametrically opposite its starting point, we are supposed to have $S(a) \\equiv 1007 \\pmod{2014}$ for every $a$. This gives\n$$\n0 \\equiv 1007 \\cdot 1007 \\equiv 1007 \\pmod{2014}\n$$\nwhich is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbb{Q} \\rightarrow \\mathbb{R}$ from rational to real numbers such that for all rational $p, q$,\n$$\nf(p+2q) - f(p-2q) = 2(f(p+q) - f(p-q)).\n$$",
    "options": [],
    "answer": "All quadratic polynomials: f(x) = ax^2 + bx + c for real a, b, c.",
    "solution": "Solution:\nLet $P(p, q)$ be the given condition. Expanding $P(p-q, q) + 2P(p, q) + P(p+q, q)$ gives\n$$\nf(p+3q) - f(p-3q) = 3(f(p+q) - f(p-q)).\n$$\nLet $F$ be the quadratic that equals $f$ at $-1, 0, 1$. Plugging in $p=0.5, q=0.5$, we get that it also intersects $f$ at $2$. Plugging in $p=1.5, q=0.5$ then gives that it also intersects $f$ at $3$, and continuing this induction gives that it is equal to $f$ for all positive integers. Continuing this reasoning in the other direction gives that it is equal to $f$ for all negative integers as well.\n\nWe now claim that $f$ is in fact equal to $F$. Let $r/s$ be a rational number. Consider the arithmetic sequence from $-r$ to $r$ with difference $r/s$. By similar reasoning, $f$ is a quadratic over this sequence. However, $F$ contains three of the terms: $-r, 0, r$. Therefore, $f$ is equal to $F$ over this sequence; in particular, $f(r/s) = F(r/s)$.\n\nTherefore, $f$ is a quadratic, and all quadratics work, so the answer is all quadratics.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVandal Evan cut a rectangular portrait of Professor Zvezda along a straight line. Then he cut one of the pieces along a straight line, and so on. After he had made 100 cuts, Professor Zvezda walked in and forced him to pay 2 cents for each triangular piece and 1 cent for each quadrilateral piece. Prove that Vandal Evan paid more than $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst note that the total number of sides increases by at most $4$ at each cut. This is because two new sides are created along the cut, and the two endpoints of the cut may optionally divide other sides into two parts. Therefore, since there are initially $4$ sides, at the end there are at most $404$ sides.\n\nNow note that the cost of a piece ($2$, $1$, or $0$ cents as specified in the problem) is at least five minus its number of sides. So the total of the costs of all $101$ pieces is at least $505$ minus the total number of sides, or at least $505 - 404 = 101$ cents as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a convex quadrilateral. Suppose that the area of $ABCD$ is equal to\n$$\n\\frac{AB+CD}{2} \\cdot \\frac{AD+BC}{2}\n$$\nProve that $ABCD$ is a rectangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIn the picture shown, we have replicated quadrilateral $ABCD$ four times, rotating by $180^{\\circ}$ successively about the midpoints of $BC$, $CF$, and $CH$. Because the angles of the quadrilateral sum to $360^{\\circ}$, the figure \"closes up\" so that the final quadrilateral has $CD$ as a side and would, if further rotated $180^{\\circ}$ about $CD$, coincide with $ABCD$.\nWe note that $AD$ and $EF$ are parallel and congruent, as are $FG$ and $DI$; thus triangles $ADI$\n![](attached_image_1.png)\nand $EFG$ are translations of one another, and $AI$ is parallel and congruent to $IG$. The same can be said for triangles $ABE$ and $IHG$. So\n$$\n\\text{4. Area } ABCD = \\text{Area } ABEFGHID = \\text{Area } ABEGHI = \\text{Area } AEGI.\n$$\nBecause $AEGI$ is a parallelogram, with base $AE$ and height at most $AI$, we can continue:\n$$\n\\text{Area } AEGI \\leq AE \\cdot AI \\leq (AB + BE)(AD + DI) = (AB + CD)(AD + BC) = 4 \\text{Area } ABCD.\n$$\nAll these inequalities must be equalities. So $A$, $B$, $E$ are collinear (i.e. $AB \\parallel CD$); $A$, $D$, $I$ are collinear (i.e. $AD \\parallel BC$); and angle $EAI$, which is also angle $BAD$, equals $90^{\\circ}$. From this we conclude that $ABCD$ is a rectangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven $a > b > c > 0$, prove that\n$$\na^{4} b + b^{4} c + c^{4} a > a b^{4} + b c^{4} + c a^{4}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nObserve that equality holds whenever $a = b$, $b = c$, or $a = c$. Knowing this, it is not difficult to factor the difference between the two sides:\n$$\n\\begin{aligned}\n& a^{4} b + b^{4} c + c^{4} a - a b^{4} - b c^{4} - c a^{4} \\\\\n& = (a^{4} b - a b^{4}) + (a c^{4} - b c^{4}) + (-a^{4} c + b^{4} c) \\\\\n& = (a - b)\\left[ a^{3} b + a^{2} b^{2} + a b^{3} + c^{4} - a^{3} c - a^{2} b c - a b^{2} c - b^{3} c \\right] \\\\\n& = (a - b)\\left[ (a^{3} - a^{3} c) + (a^{2} b^{2} - a^{2} b c) + (a b^{3} - a b^{2} c) + (-b^{3} c + c^{4}) \\right] \\\\\n& = (a - b)(b - c)\\left[ a^{3} + a^{2} b + a b^{2} - b^{2} c - b c^{2} - c^{3} \\right] \\\\\n& = (a - b)(b - c)\\left[ (a^{3} - c^{3}) + (a^{2} b - b c^{2}) + (a b^{2} - b^{2} c) \\right] \\\\\n& = (a - b)(b - c)(a - c)\\left[ a^{2} + a c + c^{2} + a b + b c + b^{2} \\right]\n\\end{aligned}\n$$\nThis last is clearly positive.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute triangle with circumcenter $O$, incenter $I$, orthocenter $H$. If $OI = HI$, what are the possible values of the angles of triangle $ABC$?",
    "options": [],
    "answer": "Some angle is 60 degrees.",
    "solution": "Solution:\n\nAnswer: this occurs if and only if some angle is $60$ degrees.\n\nOne direction is immediate; if $\\angle A = 60^{\\circ}$ then $BHOIC$ are cyclic since $\\angle BHC = \\angle BIC = \\angle BOC = 120^{\\circ}$.\n\nFor the other direction, note that we have an \"SSA congruence\" of triangles $AIH$ and $AIO$. Consequently, either $A$ lies on the circle $(OIH)$ or $\\triangle AIH \\cong \\triangle AIO$. In the latter case, $AH = AO$, but since it's known that $AH = 2AO \\cos A$, it follows that $\\cos A = \\frac{1}{2} \\Longrightarrow \\angle A = 60^{\\circ}$.\n\nNow it's impossible for $A, B, C \\in (OIH)$ since the points $A, B, C, O$ are not concyclic. Thus some angle must be $60^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $N$ be a positive integer such that $N$ is divisible by $81$ and the number formed by reversing the digits of $N$ is also divisible by $81$. Prove that the sum of the digits of $N$ is divisible by $81$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe begin with a lemma.\nLemma. For all $k \\geq 0$, $10^{k} \\equiv 1 + 9k \\pmod{81}$.\nProof. Binomial theorem:\n$$\n10^{k} = (1 + 9)^{k} = 1^{k} + \\binom{k}{1} 1^{k-1} 9 + \\text{terms divisible by } 9^{2}.\n$$\nWe now write $N$ in terms of its digits as\n$$\n\\begin{aligned}\nN &= a_{0} + 10 a_{1} + \\cdots + 10^{n} a_{n} \\\\\n&\\equiv (a_{0} + 9 \\cdot 0 \\cdot a_{0}) + (a_{1} + 9 \\cdot 1 \\cdot a_{1}) + \\cdots + (a_{n} + 9 \\cdot n \\cdot a_{n}) \\\\\n&\\equiv a_{0} + a_{1} + \\cdots + a_{n} + 9(0 a_{0} + 1 a_{1} + 2 a_{2} + \\cdots + n a_{n}) \\pmod{81}\n\\end{aligned}\n$$\nCorrespondingly, the number formed by reversing the digits of $N$ is\n$$\na_{n} + 10 a_{n-1} + \\cdots + 10^{n} a_{0} \\equiv a_{0} + a_{1} + \\cdots + a_{n} + 9(n a_{0} + (n-1) a_{1} + \\cdots + 0 a_{n}) \\pmod{81}.\n$$\nIf we add these two numbers, we get that $81$ divides\n$$\n2(a_{0} + \\cdots + a_{n}) + 9(n a_{0} + n a_{1} + \\cdots + n a_{n}) = (9n + 2)(a_{0} + \\cdots + a_{n}).\n$$\nSince $9n + 2$ is not divisible by $3$, the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$\nf(x y+z)=f(x) f(y)+f(z)\n$$\nfor all real numbers $x$, $y$, and $z$.",
    "options": [],
    "answer": "f(x) = 0 for all real x, or f(x) = x for all real x",
    "solution": "Solution:\nAnswer: $f(x)=0$ or $f(x)=x$. Both of these trivially satisfy the equation.\n\nWe plug in $x=y=z=0$ to get\n$$\n\\begin{array}{r}\nf(0)=f(0)^{2}+f(0) \\\\\n0=f(0)^{2} \\\\\n0=f(0)\n\\end{array}\n$$\nThen we plug in $x=y=1$, $z=0$ to get\n$$\n\\begin{aligned}\nf(1) & =f(1)^{2}+f(0) \\\\\n0 & =f(1)^{2}-f(1)\n\\end{aligned}\n$$\nso $f(1)=0$ or $f(1)=1$. If $f(1)=0$, we can immediately plug in $y=1$ and $z=0$ to conclude that\n$$\nf(x)=f(x) \\cdot 0+0=0\n$$\nfor all $x$.\n\nIf $f(1)=1$, then we can derive that $f$ has two simple properties: the addition property from plugging in $y=1$,\n$$\nf(x+z)=f(x)+f(z)\n$$\nand the multiplication property from plugging in $z=0$,\n$$\nf(x y)=f(x) f(y) .\n$$\nPlugging $z=-x$ into the addition property tells us that $f(-x)=-f(x)$, i.e. $f$ is an odd function. Also, the addition property lends itself to use for an induction, starting at $f(1)=1$, that proves $f(n)=n$ for all positive integers $n$.\n\nSuppose that $f$ is not the identity function. Then there is a number $x$ such that $f(x) \\neq x$. Changing $x$ to $-x$ if necessary, we can assume that $f(x)<x$. We can then double $x$ repeatedly, which will also double $f(x)$ by the addition property, until the difference $x-f(x)$ becomes at least $2$. Then there is an integer $n$ between $f(x)$ and $x$:\n$$\nf(x)<n<x \\text{.}\n$$\nWe note that $x-n$ is positive, while $f(x)-n=f(x-n)$ is negative. Thus we have a number $u>0$ such that $f(u)<0$. This is impossible by the multiplication property, since for $u>0$,\n$$\nf(u)=f(\\sqrt{u} \\cdot \\sqrt{u})=f(\\sqrt{u}) \\cdot f(\\sqrt{u}) \\geq 0 .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine whether there exists a $2011 \\times 2011$ matrix with the following properties:\n- Every cell is filled with an integer from $1$ to $4021$.\n- For every integer $i$ ($1 \\leq i \\leq 2011$), the $i$th row and the $i$th column together contain every integer from $1$ to $4021$.",
    "options": [],
    "answer": "no",
    "solution": "Solution:\n\nAnswer: no. Fix an integer $k$ from $1$ to $4021$. Let us say that an index $i$ \"hits\" a cell containing the number $k$ if the cell is in either the $i$th row or the $i$th column. The conditions stipulate that each index hits exactly one instance of $k$, so the total number of hits is $4021$. On the other hand, every cell not lying on the main diagonal is hit by exactly two indices (its row number and its column number), while those on the diagonal are hit only once. In particular, to create an odd total number of hits, $k$ must appear on the diagonal. This is a contradiction since there are $4021$ permissible values of $k$ and only $2011$ spots on the diagonal.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 46355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $c > 0$ be a positive real number. We define the sequence $(x_n)$ by $x_0 = 0$ and\n$$\nx_{n+1} = x_n^2 + c\n$$\nfor each $n \\geq 0$. For which values of $c$ is it true that $|x_n| < 2016$ for all $n$?",
    "options": [],
    "answer": "c ≤ 1/4",
    "solution": "Solution:\nThe answer is $c \\leq \\frac{1}{4}$.\n\nFirst, we show that $c \\leq \\frac{1}{4}$ all work. Clearly it suffices to prove the result when $c = \\frac{1}{4}$. In that case, the sequence is defined by $x_{n+1} = x_n^2 + \\frac{1}{4}$. We claim that $x_n \\leq \\frac{1}{2}$ for all $n$. Indeed, this follows by induction, since it is true for $n = 1$ and for the inductive step we have\n$$\nx_{n+1} = x_n^2 + \\frac{1}{4} \\leq \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}.\n$$\n\nNow we show all $c > \\frac{1}{4}$ fail. Assume on the contrary that $|x_n| < 2016$ for all $n$. Then since the real interval $[0, 2016]$ is compact, we must have that $x_n$ converges to some limit $x$. This limit must then satisfy\n$$\nx^2 = x + c.\n$$\nThus the discriminant $1 - 4c$ is nonnegative, so $c \\leq \\frac{1}{4}$ must hold.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A$ be a set of size $2023$. Find the maximum number of pairs of elements $x, y \\in A$ so that $x-y$ is a power of $e$.",
    "options": [],
    "answer": "11043",
    "solution": "Solution:\nLet $a_{n}$ be the maximum possible number of such pairs for a set of size $n$. Let $s_{2}(n)$ be the number of ones in $n$'s binary representation. Let $S(n)=\\sum_{k=0}^{n-1} s_{2}(k)$. We show that $a_{n}=S(n)$.\n\nFor the construction, we can take the binary representations of all numbers from $0$ to $n-1$, and interpret them as numbers \"base $e$\". Every $x$ corresponding to some integer $0 \\leq k < n$ then has $s_{2}(k)$ working values of $y$, corresponding to all ways to replace a $1$ with a $0$ in $k$'s binary representation.\n\nFor optimality, we use strong induction. The base case of $n=1$ holds as $a_{1}=0=s_{2}(0)$.\n\nNow assume $n>1$. If $A$ has no working pairs $x, y$, we are done. Otherwise, let $t$ be an integer so that there is at least one pair $x, y \\in A$ so that $x-y=e^{t}$.\n\nLet $G$ be the graph of such pairs in $A$. If $G$ is not connected, we can increase the number of edges of $G$ by shifting the vertices of one component of $G$ to create at least one edge to another component. Thus we can assume that all elements of $A$ are sums of powers of $e$. For an element $z \\in A$, let $c_{z}$ be the coefficient of $e^{t}$ in the representation of $A$ as a sum.\n\nLet $X$ be the set of $z$ so that $c_{z} \\geq c_{x}$ and let $Y$ be the set of $z$ so that $c_{z} \\leq c_{y}$. Note that $X \\sqcup Y = A$. By strong induction, there are at most $a_{|X|}$ working pairs in $X$, and at most $a_{|Y|}$ pairs in $Y$. By definition of $X$ and $Y$, any pair between them can only have one possible difference, namely $e^{t}$. Thus, there are at most $\\min (|X|,|Y|)$ pairs between them.\n\nThus, we have the recurrence $a_{n} \\leq \\max_{X+Y=n}\\left(a_{X}+a_{Y}\\right)+\\min (X, Y)$. It thus suffices to show that if $Y \\geq X$, $S(X+Y)-S(Y) \\geq S(X)+X$, which expands to\n$$\n\\sum_{k=Y}^{Y+X-1} s_{2}(k) \\geq \\sum_{k=0}^{X-1}\\left(1+s_{2}(k)\\right).\n$$\nAn exercise to the interested reader is to show this by strong induction on $X$.\n\nNow it remains to evaluate $S(2023)$. By linearity of expectation, $S(2048)$ is equal to $2048 \\cdot 11 / 2$. For every number from $2032=2048-16$ to $2048$, $7$ digits must be $1$ and the remaining four each have a half chance of being $1$, giving $S(2048)-S(2032)=16 \\cdot(7+4 / 2)$. Similarly $S(2032)-S(2024)=8 \\cdot(7+3 / 2)$, $S(2024)-S(2023)$ is just the number of ones in $2023=11111100111_{2}$ is $9$. Thus the answer is\n$$\n2048 \\cdot 11 / 2 - 16 \\cdot (7+4/2) - 8 \\cdot (7+3/2) - 9 = 11043.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEquilateral triangles $\\triangle ABD$, $\\triangle ACE$, and $\\triangle BCF$ are drawn outside $\\triangle ABC$ on each of its sides, with centers $G$, $H$, and $I$, respectively.\n\na) Prove that $\\triangle GHI$ is equilateral.\n\nb) Let $G'$, $H'$, and $I'$ be the reflections of $G$, $H$, and $I$ across $AB$, $BC$, and $CA$, respectively. Prove that $\\triangle GHI$ and $\\triangle G'H'I'$ have the same circumcenter.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\na) Let $O$ be the intersection of circles $ABD$ and $ACE$ (besides $A$). Since opposite angles in a cyclic quadrilateral sum to $180^\\circ$, $\\angle AOB = \\angle AOC = 120^\\circ$, so $\\angle BOC = 120^\\circ$ as well. Thus, quadrilateral $BOCF$ is also cyclic. Since $A$ and $O$ are the two intersection points of circles with centers $G$ and $H$, $AO \\perp GH$. Similarly, $BO \\perp GI$. Since $\\angle AOB = 120^\\circ$, $\\angle GHI = 60^\\circ$. Similarly, $\\angle HGI = \\angle HIG = 60^\\circ$, so $\\triangle GHI$ is equilateral.\n\nb) We can use similar logic to show that $\\triangle G'H'I'$ is equilateral as well (since $G'$, $H'$, and $I'$ are the centers of equilateral triangles drawn in the opposite direction on each side of $\\triangle ABC$).\n\nNext, we claim that\n$$\n\\triangle CHI' \\cong \\triangle CH'I \\sim \\triangle ABC.\n$$\nWe can see that $CH'$ is a $60^\\circ$ rotation of $CH$ about point $C$, and $CI$ is a $60^\\circ$ rotation of $CI'$ about $C$ in the same direction. Also, $\\triangle ACH$ and $\\triangle BCI$ are isosceles triangles with base angle $30^\\circ$. Thus,\n$$\n\\begin{array}{r}\nCH = CH' = AC / \\sqrt{3}, \\\\\nCI' = CI = BC / \\sqrt{3}, \\\\\n\\angle HCI' = \\angle ACB = \\angle H'CI.\n\\end{array}\n$$\nThe desired similarity and congruence follows by SAS, and we conclude that\n$$\nIH' = I'H = AB / \\sqrt{3} = GG'.\n$$\nBy similar logic,\n$$\nHH' = IG' = G'I.\n$$\nThus by SSS,\n$$\n\\triangle IHH' \\cong \\triangle HI'G \\cong \\triangle GG'I.\n$$\nThus, the vertices of $\\triangle G'H'I'$ are positioned symmetrically with respect to the sides of $\\triangle GHI$, and it follows that the two equilateral triangles have the same center.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn a certain island, there are knights, who always tell the truth, knaves, who always lie, and spies, who could do either. Suppose you meet three people, and you know one is a knight, one is a knave, and one is a spy, but you don't know which is which.\nFind a method to ask three yes/no questions, each to one of the three people, so you can determine for sure which is which. You may question the same person multiple times, and your questions can depend on answers to previous questions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOne strategy is as follows:\n1. Ask the first person, \"Is the second person more likely to tell the truth than the third person?\"\nIf the response is \"yes\", then if the first person is the knight, the second must be the spy and the third must be the knave. If the first person is the knave, the second must be the spy and the third the knight. Either way, if the first person is not the spy, the second person is the spy.\nIf the response is \"no\", then if the first person is the knight, the second must be the knave and the third the spy. If the first is the knave, the second must be the knight and the third the spy. Either way, the first person or the third is the spy.\nThus, at this point, you either know for certain that the second person is not the spy, or that the third person is not the spy.\n\n2. Ask the person whom you now know to be either the knight or the knave and not the spy (either the second or the third person) some question you know the answer to, like, \"Does $1+1$ equal 2?\" Then, depending on their response, you know whether they are a knight or a knave.\n\n3. Now ask this same person about the identity of one of the other two people. Since you know for sure whether this person is lying, their answer will tell you the identities of the other two people. For instance, if you know the second person to be a knave, you could ask, \"Is the first person a knight?\" If the second person says yes, the first person is the spy and the third the knight, and if the second person says no, the first person is the knight and the third the spy. No matter what, you can now figure out the identities of all three people.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ be a function such that $f(0)=2$ and for all integers $x$,\n$$\nf(x+1)+f(x-1)=f(x) f(1) .\n$$\nProve that for all integers $x$ and $y$,\n$$\nf(x+y)+f(x-y)=f(x) f(y) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet us first prove the result for nonnegative $y$ by strong induction on $y$. If $y=0$, (2) becomes\n$$\nf(x)+f(x)=f(x) f(0),\n$$\nwhich is true since $f(0)=2$, and if $y=1$, (2) is the same as (1). Let us assume (2) for $y=z-1$ and $y=z$ and try to prove it for $y=z+1$. We have\n$$\n\\begin{gathered}\nf(x+z-1)+f(x-z+1)=f(x) f(z-1) \\\\\nf(x+z)+f(x-z)=f(x) f(z)\n\\end{gathered}\n$$\nMultiplying (4) by $f(1)$ and subtracting (3) yields\n$$\n\\begin{gathered}\nf(x+z) f(1)-f(x+z-1)+f(x-z) f(1)-f(x-z+1)=f(x)[f(z) f(1)-f(z-1)] \\\\\nf(x+z+1)+f(x-z-1)=f(x) f(z+1)\n\\end{gathered}\n$$\nas desired.\nIt remains to prove (2) for $y<0$. If $y=-z$ and $z>0$, (2) becomes\n$$\nf(x-z)+f(x+z)=f(x) f(-z)\n$$\nwhich would follow from (2) for $y=z$ if $f(-z)=f(z)$. To prove this, let $x=0$ and $y=z$ in (2), getting\n$$\n\\begin{aligned}\nf(z)+f(-z) & =f(0) f(z) \\\\\nf(z)+f(-z) & =2 f(z) \\\\\nf(-z) & =f(z) .\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the interior of a triangle $ABC$ with area $1$, points $D$, $E$, and $F$ are chosen such that $D$ is the midpoint of $AE$, $E$ is the midpoint of $BF$, and $F$ is the midpoint of $CD$. Find the area of triangle $DEF$.",
    "options": [],
    "answer": "1/7",
    "solution": "Solution:\n\nLet $x$ be the area of $\\triangle DEF$. Comparing triangles $ABE$ and $DEF$, we find that base $AE$ is twice base $DE$ but, since $E$ bisects $BF$, the heights to these bases are equal. Thus $\\triangle ABE$ has area $2x$. Symmetrically, triangles $BCF$ and $CAD$ have area $2x$. Since these four triangles fill up $\\triangle ABC$, we have $1 = x + 2x + 2x + 2x = 7x$, so $x = 1/7$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor all positive integers $n$, prove that\n$$\n\\sum_{k=1}^{n} \\phi(k)\\left\\lfloor\\frac{n}{k}\\right\\rfloor=\\frac{n(n+1)}{2}\n$$\n(For a positive integer $n$, $\\phi(n)$ denotes the number of positive integers less than or equal to $n$ and relatively prime to $n$. For a real number $x$, $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider the fractions $a / b$, where $a$ and $b$ range over integers such that $1 \\leq a \\leq b \\leq n$. We will count these fractions in two ways:\n\na. By unreduced form. For each denominator $b$, there are $b$ possible numerators $a=1,2, \\ldots, b$, so the total number of fractions is\n$$\n1+2+\\cdots+n=\\frac{n(n+1)}{2}\n$$\n\nb. By reduced form. Suppose a fraction $a / b$ has been reduced to $c / d$. Given the denominator $d$, there are $\\phi(d)$ possible numerators $c$ such that $c \\leq d$ and $c / d$ is in lowest terms. To get from $c / d$ back to $a / b$, we must multiply numerator and denominator by a positive integer $k$ such that $k c \\leq k d \\leq n$. Since $k c \\leq k d$ always holds if $c \\leq d$, the choice of $k$ is limited only by the inequality $k d \\leq n$, which has $\\lfloor n / d\\rfloor$ solutions. Thus there are $\\phi(d)\\lfloor n / d\\rfloor$ fractions $a / b$ for a given choice of $d$, so the total number of fractions is\n$$\n\\sum_{d=1}^{n} \\phi(d)\\left\\lfloor\\frac{n}{d}\\right\\rfloor\n$$\nSince both methods of counting must yield the same answer, the identity follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCDEF$ be a hexagon circumscribing a circle $\\omega$. The sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$ touch $\\omega$ at $U$, $V$, $W$, $X$, $Y$, and $Z$ respectively; moreover, $U$, $W$, and $Y$ are the midpoints of sides $AB$, $CD$, and $EF$, respectively. Prove that $UX$, $VY$, and $WZ$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $U$ is the midpoint of $AB$, we have $ZA = AU = UB = BV$ and so (letting $O$ be the center of $\\omega$) $\\triangle OZA \\cong \\triangle OUA \\cong \\triangle OUB \\cong \\triangle OVB$. Thus arcs $ZU$ and $UV$ are equal, and so $UX$ is the bisector of $VXZ$. Similarly, $VY$ and $WZ$ are the bisectors of the other two angles of $\\triangle VXZ$, so these three lines are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that if $n$ is a positive integer for which $2+2 \\sqrt{1+12 n^{2}}$ is an integer, then it is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet the value of this expression be $2k$, say. Then we may conclude that\n$$\n12 n^{2} + 1 = (k-1)^{2} \\Longrightarrow 12 n^{2} = k(k-2).\n$$\nObviously $k$ is even and hence $\\gcd(k, k-2) = 2$. So we may write this factorization as $2x^{2}$ and $6y^{2}$ in some order.\nIf $k = 2x^{2}$ then $2k = 4x^{2}$ and we are done. But otherwise $k-2 = 2x^{2}$ and $k = 6y^{2}$ which gives $3y^{2} - 1 = x^{2}$, and this is a contradiction modulo $3$. This solves the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe inscribed circle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $D$, $E$, and $F$ respectively. Let $X$, $Y$, and $Z$ be the incenters of triangles $AEF$, $BFD$, and $CDE$, respectively. Prove that $DX$, $EY$, and $CZ$ meet at one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider the midpoint $M$ of arc $EF$ on the incircle of $\\triangle ABC$. Angles $AFM$ and $MFE$ are equal since they intercept equal arcs $FM$ and $ME$, and so $M$ is on the bisector of $\\angle AFE$. Similarly, $M$ is on the bisector of $\\angle FEA$, and therefore $M$ coincides with $X$. Moreover, angles $FDX$ and $XDE$ are equal since they intercept equal arcs $FX$ and $XE$, and so $DX$ is the angle bisector of $\\angle D$ in $\\triangle DEF$. Similarly, $EY$ and $FZ$ are the other two angle bisectors in $\\triangle DEF$. But the three angle bisectors in a triangle always meet!",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle with $\\angle A = 120^\\circ$. The bisector of $\\angle A$ meets side $BC$ at $D$. Prove that\n$$\n\\frac{1}{AD} = \\frac{1}{AB} + \\frac{1}{AC}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe area of $\\triangle ABC$ is the sum of the areas of triangles $ABD$ and $ADC$, so\n$$\n\\begin{aligned}\n\\frac{1}{2} AB \\cdot AC \\cdot \\sin 120^\\circ & = \\frac{1}{2} AB \\cdot AD \\cdot \\sin 60^\\circ + \\frac{1}{2} AD \\cdot AC \\cdot \\sin 60^\\circ \\\\\n\\frac{1}{2} AB \\cdot AC \\cdot \\frac{\\sqrt{3}}{2} & = \\frac{1}{2} AB \\cdot AD \\cdot \\frac{\\sqrt{3}}{2} + \\frac{1}{2} AD \\cdot AC \\cdot \\frac{\\sqrt{3}}{2} \\\\\nAB \\cdot AC & = AB \\cdot AD + AD \\cdot AC.\n\\end{aligned}\n$$\nDividing through by $AB \\cdot AC \\cdot AD$ gives the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind an $n$ such that $n! - (n-1)! + (n-2)! - (n-3)! + \\cdots \\pm 1!$ is prime. Be prepared to justify your answer for\n$$\n\\left\\{\\begin{array}{c}\nn, \\\\\n{\\left[\\frac{n+225}{10}\\right],} \\\\\nn \\leq 25\n\\end{array}\\right.\n$$\npoints, where $[N]$ is the greatest integer less than $N$.",
    "options": [],
    "answer": "n ∈ {3, 4, 5, 6, 7, 8, 10, 15, 19, 41, 59, 61, 105, 160} (these are exactly those ≤ 335 producing a prime value)",
    "solution": "Solution:\n\n$3, 4, 5, 6, 7, 8, 10, 15, 19, 41$ (26 points), $59, 61$ (28 points), $105$ (33 points), $160$ (38 points) are the only ones less than or equal to $335$. If anyone produces an answer larger than $335$, then we ask for justification to call their bluff. It is not known whether or not there are infinitely many such $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$, $y$, and $z$ be positive integers satisfying $x y = z^{2} + 1$. Prove that there are integers $a$, $b$, $c$, and $d$ such that $x = a^{2} + b^{2}$, $y = c^{2} + d^{2}$, and $z = a c + b d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe will use strong induction on $z$. Assume first that $z = 1$. Then $x y = 1^{2} + 1 = 2$, so $x$ and $y$ are $1$ and $2$ in some order. If $x = 2$ and $y = 1$, then there is a solution $a = b = c = 1$, $d = 0$; if $x = 1$ and $y = 2$, we may take $a = c = d = 1$, $b = 0$.\n\nNow assume that $z > 1$ and that, whenever $Z < z$, the equation $X Y = Z^{2} + 1$ implies that there are integers $A$, $B$, $C$, and $D$ such that $X = A^{2} + B^{2}$, $Y = C^{2} + D^{2}$, and $Z = A C + B D$. We note that neither $x$ nor $y$ can equal $z$, for then $x y$ and $z^{2}$ would be two multiples of $z$ although their difference is $1$. We note that $x$ and $y$ cannot be both greater or both less than $z$ since\n$$\n(z-1)^{2} < z^{2} + 1 < (z+1)^{2}.\n$$\nThus, because the symmetry of the problem allows us to switch $x$ and $y$, we can assume that $x > z > y$.\n\nLet $X = x + y - 2z$, $Y = y$, and $Z = z - y$. Multiplying out shows that the equation $X Y = Z^{2} + 1$ is satisfied. Also, $Z < z$, so we have integers $A$, $B$, $C$, and $D$ such that the following equations hold:\n$$\n\\begin{aligned}\n& X = A^{2} + B^{2} \\\\\n& Y = C^{2} + D^{2} \\\\\n& Z = A C + B D\n\\end{aligned}\n$$\nLet $a = A + C$, $b = B + D$, $c = C$, and $d = D$. Equation (4) already tells us that $y = c^{2} + d^{2}$. Since $Y + Z = y + z - y = z$, we can add equation (4) and (5) together to get\n$$\nz = Y + Z = A C + C^{2} + B D + D^{2} = (A + C) C + (B + D) D = a c + b d\n$$\nFinally, since $X + Y + 2Z = x + y - 2z + y + 2z - 2y = x$, we add (3), (4), and twice (5) to get\n$$\nx = X + 2Z + Y = A^{2} + 2 A C + C^{2} + B^{2} + 2 B D + D^{2} = (A + C)^{2} + (B + D)^{2} = a^{2} + b^{2}.\n$$\nThis completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA lattice point is a point in the coordinate plane both of whose coordinates are integers. In $\\triangle ABC$, all three vertices are lattice points and the area of the triangle is $1/2$. Prove that the orthocenter of $\\triangle ABC$ is also a lattice point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet us position our coordinate system so that $A$ is the origin. Let $B = (a, b)$ and $C = (c, d)$. Then the formula for the area of a triangle with given vertex coordinates gives $\\frac{1}{2} = \\frac{1}{2}|ad - bc|$, i.e. $ad - bc = \\pm 1$.\n\nLet $H = (x, y)$ be the orthocenter of $\\triangle ABC$. The condition $CH \\perp AB$ can be expressed in vector form as\n$$\n0 = \\overrightarrow{AB} \\cdot \\overrightarrow{CH} = \\langle a, b \\rangle \\cdot \\langle x - c, y - d \\rangle\n$$\nor\n$$\na x + b y = a c + b d.\n$$\nSimilarly, the condition $BH \\perp AC$ can be expressed as\n$$\nc x + d y = a c + b d.\n$$\nAdding $d$ times equation (2) to $-b$ times equation (3)$:$\n$$\n(a d - b c) x = d(a c + b d) - b(a c + b d).\n$$\nSince $ad - bc = \\pm 1$, it follows that $x$ is an integer. Similarly, we can prove that $y$ is an integer as well.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose there are 100 prisoners, each of whom is given a number between 1 and 100. There is also a room with 100 boxes, labeled 1 to 100, and 100 pieces of paper with the numbers 1 to 100 on them. Each piece of paper is randomly placed inside one of the 100 boxes.\n\nOne at a time, each prisoner is allowed to enter the room and open at most 50 boxes to see the numbers inside. If every prisoner opens the box with his own number inside it, they will all be released. They are not allowed to communicate at all during this process, but they can come up with a strategy beforehand. Show that there is a strategy that gives them at least a $30\\%$ chance of winning.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThey can use the following strategy: each prisoner opens the box labeled with his own number. He then looks at the number on the paper inside and opens the box labeled with that number next. He continues doing so until he finds his own number or he has opened 50 boxes. For instance, prisoner 2 might open box 2 and find the number 5, then open box 5 and find the number 17, then open box 17, and so on.\n\nThus, the prisoners will win iff the permutation of paper numbers in box numbers has no cycle of length more than 50. If the permutation does have such a cycle, it has at most one, and its length could be $51, 52, \\ldots, 99$, or $100$. If the cycle has length $n$, then there are $100! / n$ possible permutations, since there are\n$$\n100 \\cdot 99 \\cdots \\cdot (100-n+1)\n$$\nways to choose the numbers in the cycle (in order) and $(100-n)!$ ways to permute the remaining numbers, but then we divide by $n$ since it doesn't matter to the permutation which of the $n$ numbers is first when we write out the cycle. Thus, the probability there is a cycle of length $n > 50$ is $1/n$ for each $n$. There is at most one such cycle, so the total probability is\n$$\n\\begin{gathered}\n\\frac{1}{51} + \\frac{1}{52} + \\cdots + \\frac{1}{100} \\\\\n= \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{100}\\right) - \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{50}\\right) \\\\\n\\approx \\ln 100 - \\ln 50 = \\ln 2 \\approx 30\\%.\n\\end{gathered}",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $a, b, c, d$, and $e$ are objects that we can multiply together, but the multiplication doesn't necessarily satisfy the associative law, i.e. $(x y) z$ does not necessarily equal $x(y z)$. How many different ways are there to interpret the product $a b c d e$?",
    "options": [],
    "answer": "14",
    "solution": "Solution:\n$C$ (number of letters $-1$) $= C(4) = 14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine a sequence $a_{0}, a_{1}, a_{2}, \\ldots$ in the following way: $a_{0}=0$, and for $n \\geq 0$,\n$$\na_{n+1}=a_{n}+5^{a_{n}} .\n$$\nLet $k$ be any positive integer. Prove that the remainders when $a_{0}, a_{1}, \\ldots, a_{2^{k-1}}$ are divided by $2^{k}$ are all different.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe begin with a simple numerical lemma.\n\nLemma 1. For all $k \\geq 0, 5^{2^{k}}-1$ is divisible by $2^{k+2}$.\n\nProof. By induction. For $k=0$, the statement may be checked directly. To step from $k$ to $k+1$, we write\n$$\n5^{2^{k+1}}-1=\\left(5^{2^{k}}\\right)^{2}-1=\\left(5^{2^{k}}-1\\right)\\left(5^{2^{k}}+1\\right)\n$$\nand note that the first factor is divisible by $2^{k+2}$ by the induction hypothesis and the second factor is clearly divisible by 2, so the product is divisible by $2^{k+3}$.\n\nNow we turn to the main lemma of the proof.\n\nLemma 2. Fix $k \\geq 0$. For all $r \\geq 0$, the difference\n$$\na_{r+2^{k}}-a_{r}\n$$\nis divisible by $2^{k}$, not divisible by $2^{k+1}$, and independent of $r$ mod $2^{k+2}$.\n\nProof. By induction. For $k=0$,\n$$\na_{r+1}-a_{r}=5^{a_{r}}\n$$\nis clearly divisible by 1, not divisible by 2, and congruent to the constant value $1^{a_{r}}=1 \\bmod 4$. Now assume that the statement is true for $k$; we will prove it for $k+1$. We know that $a_{r+2^{k}}-a_{r}$ has the form $2^{k} \\cdot m$ for $m$ odd. The difference $a_{r+2^{k+1}}-a_{r+2^{k}}$ has the same value, $2^{k} \\cdot m$, to the modulus $2^{k+2}$. Therefore, $\\bmod 2^{k+2}$,\n$$\na_{r+2^{k+1}}-a_{r}=\\left(a_{r+2^{k+1}}-a_{r+2^{k}}\\right)+\\left(a_{r+2^{k}}-a_{r}\\right) \\equiv 2^{k} \\cdot m+2^{k} \\cdot m=2^{k+1} m\n$$\nso this difference is divisible by $2^{k+1}$ but not $2^{k+2}$.\n\nIt remains to prove that $a_{r+2^{k+1}}-a_{r}$ is constant $\\bmod 2^{k+3}$, i.e. that $2^{k+3}$ divides the difference between two values for consecutive values of $r$ :\n$$\n\\begin{aligned}\n\\left(a_{r+2^{k+1}+1}-a_{r+1}\\right)-\\left(a_{r+2^{k+1}}-a_{r}\\right) & =\\left(a_{r+2^{k+1}+1}-a_{r+2^{k+1}}\\right)-\\left(a_{r+1}-a_{r}\\right) \\\\\n& =5^{a_{r+2^{k+1}}}-5^{a_{r}} \\\\\n& =5^{a_{r}}\\left(5^{a_{r+2^{k+1}}-a_{r}}-1\\right) .\n\\end{aligned}\n$$\nWe have already proved that $a_{r+2^{k+1}}-a_{r}$ is divisible by $2^{k+1}$. By Lemma $1, 5^{2^{k+1}} \\equiv 1 \\bmod 2^{k+3}$, so $5^{a_{r+2^{k+1}}-a_{r}}$ is also 1 $\\bmod 2^{k+3}$, completing the proof.\n\nTo solve the problem, suppose that $r$ and $s$ are two nonnegative integers such that $0 \\leq r<s \\leq 2^{k}-1$ and $a_{r} \\equiv a_{s} \\bmod 2^{k}$. Let $s-r=2^{\\ell} \\cdot m$, where $m$ is odd. Mod $2^{\\ell+1}, a_{t+2^{\\ell}}-a_{t} \\equiv 2^{\\ell}$ for all $t \\geq 0$ and therefore\n$$\na_{s}-a_{r} \\equiv\\left(a_{r+2^{\\ell}}-a_{r}\\right)+\\left(a_{r+2 \\cdot 2^{\\ell}}-a_{r+2^{\\ell}}\\right)+\\cdots+\\left(a_{r+m \\cdot 2^{\\ell}}-a_{r+(m-1) 2^{\\ell}}\\right) \\equiv m \\cdot 2^{\\ell} \\not \\equiv 0,\n$$\nimplying that $\\ell+1 \\geq k+1$, i.e. $\\ell \\geq k$. Thus $s-r \\geq 2^{k}$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$ and $b$ be real numbers. Prove that the polynomial\n$$\nP(x) = x^{3} + (2a + 1)x^{2} + (2a^{2} + 2a - 3)x + b\n$$\ndoes not have three distinct rational roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume the contrary, and suppose that the polynomial has rational roots $u, v, w$. Then by Viète's formulas,\n$$\n\\begin{aligned}\nu + v + w & = -(2a + 1) \\\\\nu v + v w + w u & = 2a^{2} - 2a - 3\n\\end{aligned}\n$$\nSquaring the first equation and subtracting twice the second yields\n$$\n\\begin{aligned}\n(u + v + w)^{2} - 2uv - 2vw - 2wu & = (2a + 1)^{2} - (4a^{2} + 4a - 6) \\\\\nu^{2} + v^{2} + w^{2} & = 4a^{2} + 4a + 1 - 4a^{2} - 4a + 6 = 7.\n\\end{aligned}\n$$\nWe claim that there are no rational numbers $u, v, w$ satisfying $u^{2} + v^{2} + w^{2} = 7$. Assuming that there were, let $s$ be their least common denominator, so $u = p/s$, $v = q/s$, $w = r/s$ for integers $p, q, r, s$ with GCD 1. Then\n$$\n\\begin{aligned}\n& \\frac{p^{2} + q^{2} + r^{2}}{s^{2}} = 7 \\\\\n& p^{2} + q^{2} + r^{2} = 7s^{2}\n\\end{aligned}\n$$\nIf we work mod $8$, this equation assumes the symmetrical form\n$$\n\\begin{aligned}\np^{2} + q^{2} + r^{2} & \\equiv (-1)s^{2} \\\\\np^{2} + q^{2} + r^{2} + s^{2} & \\equiv 0\n\\end{aligned}\n$$\nUsing the fact that the squares $\\bmod\\ 8$ are $0, 1$, and $4$, it is not hard to conclude that $p, q, r$, and $s$ must all be even, contradicting the assumption that their GCD is $1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A X Y Z B$ be a convex pentagon inscribed in a semicircle with diameter $\\overline{A B}$, and let $K$ be the foot of the altitude from $Y$ to $\\overline{A B}$. Let $O$ denote the midpoint of $\\overline{A B}$ and $L$ the intersection of $\\overline{X Z}$ with $\\overline{Y O}$. Select a point $M$ on line $K L$ with $M A = M B$, and finally, let $I$ be the reflection of $O$ across $\\overline{X Z}$. Prove that if quadrilateral $X K O Z$ is cyclic then so is quadrilateral $Y O M I$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nExtend the semicircle to a circle $\\Gamma$. Let line $L K$ meet $\\Gamma$ again at two points $P$ and $Q$. Let $W$ be the point on ray $O M$ such that $O W \\cdot O M = O A \\cdot O B$. So points $P, Q, W, O$ are concyclic, say on $\\gamma$.\n\nNow, $L$ is the radical center of $\\gamma, \\Gamma$, and the circumcircles of $X K O Z$, because lines $X Z$ and $P Q$ are radical axes. So, line $Y O$ is the radical axis of $\\Gamma$ and $\\gamma$.\n\nLet $T$ denote the intersection of lines $X Z$ and $A B$. We have that $K O \\cdot K T = K A \\cdot K B = K P \\cdot K Q$, so point $T$ also lies on $\\gamma$. Also, according to $T A \\cdot T B = T K \\cdot T Z = T K \\cdot T O$, we deduce that $\\overline{T Y}$ is tangent to $\\Gamma$.\n\nFinally, let $S$ denote the midpoint of $\\overline{Y W}$. By a homothety of ratio $2$ at $W$, we have that the line passing through $S$ and the midpoint of $\\overline{W T}$ is perpendicular to $\\overline{Y O}$. Moreover, $S$ lies on the perpendicular bisector of line $\\overline{K O}$. Therefore, $S$ is the center of the circumcircle of quadrilateral $X K O Z$.\n\nFinally, the collinearity of $Y, S, W$ implies quadrilateral $Y O M I$ is concyclic, since one can readily show that $O S \\cdot O I = O W \\cdot O M = O Y^{2}$, hence $\\angle O I M = \\angle O W S = \\angle O W Y = \\angle O Y M$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn 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": "Solution:\n\nLet the bisector of $\\angle BAC$ intersect the circumcircle of $\\triangle ACE$ at $K'$. The arcs, and hence the chords, $K'C$ and $K'E$ are equal; since $CB = BE$ is given, we have $\\triangle K'BC \\cong \\triangle K'BE$ and so $K'$ is on the bisector of $\\angle CBE$. This shows that $K'$ is the excenter of $\\triangle ABC$ opposite $A$. By symmetry, we could have defined $K'$ as the intersection of the bisector of $\\angle BAC$ and the circumcircle of $\\triangle ABF$, and it would have been the same excenter.\n\nTo prove that $K' = K$, it remains to show that $F$, $K'$, and $E$ are collinear. Since $\\angle CK'E = \\angle BK'F = 120^\\circ$ (by cyclic quads $AC K' E$ and $AB K' F$), it suffices to show that $\\angle BK'C = 60^\\circ$. But this follows from the properties of the excenter:\n$$\n\\begin{aligned}\n\\angle BK'C & = 180 - \\angle CBK' - \\angle K'CB \\\\\n& = 180 - \\frac{1}{2} \\angle CBE - \\frac{1}{2} \\angle FCB \\\\\n& = 180 - \\frac{360 - \\angle ABC - \\angle BCA}{2} \\\\\n& = 180 - \\frac{180 + \\angle CAB}{2} \\\\\n& = 180 - \\frac{180 + 60}{2} = 60.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$, $b$, and $c$ be positive real numbers such that\n$$\na b c = 1\n$$\nProve that for every positive integer $n$,\n$$\na + b + c > \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUsing $a b c = 1$, we transform the given condition to\n$$\na^{n} + b^{n} + c^{n} > \\frac{1}{a^{n}} + \\frac{1}{b^{n}} + \\frac{1}{c^{n}} .\n$$\n$$\na + b + c > b c + c a + a b\n$$\nor\n$$\n- b c - c a - a b + a + b + c > 0 .\n$$\nWe then add $a b c - 1 (= 0)$ to the left side, getting\n$$\na b c - b c - c a - a b + a + b + c - 1 > 0\n$$\nwhich factors as\n$$\n(a - 1)(b - 1)(c - 1) > 0 .\n$$\nIn exactly the same way we transform the condition to be proved to\n$$\n\\left(a^{n} - 1\\right)\\left(b^{n} - 1\\right)\\left(c^{n} - 1\\right) > 0 .\n$$\nHowever, for any positive real $x$, the numbers $x - 1$ and $x^{n} - 1$ are both positive, both negative, or both zero. Consequently (1) and (2) are equivalent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWe are given a right isosceles triangle with legs of length $1$ inside which every point (including vertices, points on sides, and all points in the interior) is colored red, yellow, green, or blue. Prove that there are two points of the same color such that the distance between them is at least $2-\\sqrt{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nDefine points $D, E, F, G$ on the sides of the triangle as shown such that $AD = AF = GB = BE = 2 - \\sqrt{2}$. Then $CD = CE = \\sqrt{2} - 1$, and since $\\triangle CDE$ is isosceles and right, $DE = \\sqrt{2}(\\sqrt{2} - 1) = 2 - \\sqrt{2}$. Now segments $DE$ and $AF$ are equal and parallel, so $ADEF$ is a parallelogram and $EF = AD = 2 - \\sqrt{2}$. Similarly, $DG = 2 - \\sqrt{2}$.\n\nAssume that the conclusion is false, i.e. any two points at a distance of at least $2 - \\sqrt{2}$ are different colors. Since $\\triangle ABC$ has all side lengths greater than $2 - \\sqrt{2}$, all of its vertices are different colors. Without loss of generality, assume that $A$ is red, $B$ is yellow, and $C$ is green. Points $F$ and $G$, which are at distances of at least $2 - \\sqrt{2}$ from all three vertices, must both be blue. Point $E$, which is at a distance of at least $2 - \\sqrt{2}$ from $A, F$, and $B$, must be green. Similarly, $D$ is green. But $D$ and $E$ are a distance of $2 - \\sqrt{2}$ from each other, so we have a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDecide whether there exist positive integers $a, b, c$ such that $3(a b + b c + c a)$ divides $a^{2} + b^{2} + c^{2}$.",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no: such integers do not exist. In what follows, $\\nu_{p}(n)$ will denote the exponent of $p$ in the prime factorization of $n$.\n\nAssume without loss of generality that $a, b, c$ do not have some common divisor, and $a^{2} + b^{2} + c^{2} = 3k(a b + b c + c a)$. Write\n$$\n(3k + 2)\\left(a^{2} + b^{2} + c^{2}\\right) = 3k \\cdot (a + b + c)^{2}.\n$$\n\nSince $3k + 2 \\equiv 2 \\pmod{3}$, there is a prime $p \\equiv 2 \\pmod{3}$ with $\\nu_{p}(3k + 2)$ odd (in particular, $p \\mid 3k + 2$).\nWe first show that $p \\neq 2$. Let us assume on the contrary that $\\nu_{2}(3k + 2)$ is odd (in particular, $k$ is even). Remark that since $a, b, c$ are not all even, $\\nu_{2}\\left(a^{2} + b^{2} + c^{2}\\right) \\leq 1$. Furthermore,\n$$\n\\nu_{2}(a + b + c) = 0 \\Longleftrightarrow \\nu_{2}\\left(a^{2} + b^{2} + c^{2}\\right) = 0.\n$$\n\nNow we consider two cases.\n- Assume $\\nu_{2}(k) \\geq 2$. Then $\\nu_{2}(3k + 2) = 1$, $\\nu_{2}\\left(a^{2} + b^{2} + c^{2}\\right) \\leq 1$. Therefore\n$$\n\\nu_{2}(k) + 2\\nu_{2}(a + b + c) \\geq 2 \\geq \\nu_{2}(3k + 2) + \\nu_{2}\\left(a^{2} + b^{2} + c^{2}\\right)\n$$\nbut equality cannot occur since the relations\n$$\n\\nu_{2}(a + b + c) = 0 \\quad \\text{and} \\quad \\nu_{2}\\left(a^{2} + b^{2} + c^{2}\\right) = 1\n$$\ncannot hold simultaneously.\n\n- Assume $\\nu_{2}(k) = 1$. Then $\\nu_{2}(3k + 2) > 1$ and is odd. Now\n$$\n\\nu_{2}(3k + 2) + \\nu_{2}\\left(a^{2} + b^{2} + c^{2}\\right) = 1 + 2\\nu_{2}(a + b + c).\n$$\nThus $\\nu_{2}\\left(a^{2} + b^{2} + c^{2}\\right)$ must be even, so it is zero; consequently $\\nu_{2}(a + b + c) = 0$ as well and we obtain $1 < v_{2}(3k + 2) = 1$.\n\nNow for the interesting part. Remark $p \\mid a + b + c$ and $p \\mid a^{2} + b^{2} + c^{2}$. Without loss of generality $b \\not\\equiv 0 \\pmod{p}$, so that\n$$\na^{2} + b^{2} + (a + b)^{2} \\equiv 0 \\pmod{p} \\Longrightarrow a^{2} + a b + b^{2} \\equiv 0 \\pmod{p}\n$$\nThen if $x = a b^{-1}$, we get that $x^{2} + x + 1 \\equiv 0 \\pmod{p}$. But the left-hand side is the third cyclotomic polynomial, so either $p = 3$ or $3 \\mid p - 1$, but neither is the case.\n\nTherefore, such a triple $(a, b, c)$ does not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be a positive integer. Prove that there exist POSITIVE integers $a_{0}, \\ldots, a_{k}$ such that for all integers $x \\geq 0$,\n$$\nx^{k}=a_{0}\\left(\\begin{array}{l}\nx \\\\\nk\n\\end{array}\\right)+a_{1}\\left(\\begin{array}{c}\nx+1 \\\\\nk\n\\end{array}\\right)+\\cdots+a_{k-1}\\left(\\begin{array}{c}\nx+k-1 \\\\\nk\n\\end{array}\\right) \\text{.}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nGiven $k$ and $x$, let $A$ be the set of all sets of the form\n$$\n\\left\\{m_{1}+\\frac{1}{k}, m_{2}+\\frac{2}{k}, \\ldots, m_{k}+\\frac{k}{k}\\right\\}\n$$\nwhere the $m_{i}$ are integers, $1 \\leq m_{i} \\leq x$. Clearly, $A$ consists of $x^{k}$ sets, each of which has $k$ distinct elements. Now we count the elements of $A$ in another way.\n\nArrange the elements of a member of $A$ in increasing order:\n$$\nn_{1}+\\frac{p_{1}}{k}<n_{2}+\\frac{p_{2}}{k}<\\cdots<n_{k}+\\frac{p_{k}}{k}\n$$\nFor each permutation $p=\\left(p_{1}, \\ldots, p_{k}\\right)$ of the numbers $(1, \\ldots, k)$, let $A_{p} \\subseteq A$ be the subset consisting of sets whose sorted presentation (1) displays the given sequence $p$ of numerators. We claim that the size of $A_{p}$ is a binomial coefficient $\\left(\\begin{array}{c}x+r_{p} \\\\ k\\end{array}\\right)$, where $r_{p}$ depends on $p$ (hence on $k$) but not on $x$.\n\nFor $n_{1}, \\ldots, n_{k}\\left(1 \\leq n_{i} \\leq x\\right)$ to define an element of $A_{p}$, it is necessary and sufficient that the inequalities (1) hold. So we must have\n$$\n1 \\leq n_{1} \\leq n_{2} \\leq \\cdots \\leq n_{k} \\leq x\n$$\nMoreover, the strict inequality $n_{i}<n_{i+1}$ is required whenever $p_{i}>p_{i+1}$. To condense all these conditions, it is convenient to define nonnegative integers $e_{1}, e_{2}, \\ldots, e_{k}$ by\n$$\n\\begin{aligned}\ne_{1} & =0 \\\\\ne_{i+1} & = \\begin{cases}e_{i}+1 & \\text{ if } p_{i}<p_{i+1} \\\\\ne_{i} & \\text{ if } p_{i}>p_{i+1}\\end{cases}\n\\end{aligned}\n$$\nThen the conditions may be written as\n$$\n1 \\leq n_{1}+e_{1}<n_{2}+e_{2}<\\cdots<n_{k}+e_{k} \\leq x+e_{k} .\n$$\nWe observe that $\\left\\{\\left(n_{1}+e_{1}, \\ldots, n_{k}+e_{k}\\right)\\right\\}$ may be any $k$-element subset of $\\left\\{1,2, \\ldots, x+e_{k}\\right\\}$, written in increasing order. So, writing $r_{p}=e_{k}$,\n$$\n\\left|A_{p}\\right|=\\left(\\begin{array}{c}\nx+r_{p} \\\\\nk\n\\end{array}\\right)\n$$\nObserve that $0 \\leq r_{p} \\leq k-1$. Letting $a_{i}$ be the number of $p$ for which $r_{p}=i$, we obtain\n$$\nx^{k}=|A|=\\sum_{p}\\left|A_{p}\\right|=\\sum_{i=0}^{k-1} a_{i}\\left(\\begin{array}{c}\nx+i \\\\\nk\n\\end{array}\\right)\n$$\nHere the $a_{i}$ are clearly nonnegative integers. To prove that they are positive, it suffices to exhibit, for each $i$, a $p$ with $r_{p}=i$; the permutation $(n, n-1, \\ldots, i+1,1,2, \\ldots, i)$ is readily seen to work.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the hold of a pirate ship are ten treasure chests lying on pedestals in a circular arrangement. The captain would like to move each chest clockwise by one pedestal. However, the chests are so heavy that the captain and his assistant can only switch two chests at a time. What is the minimum number of switches needed to accomplish the task?",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nThe answer is $9$. It is easy to see that nine moves are sufficient; they can move a single chest counterclockwise one pedestal at a time, until after nine moves all of the other chests have been moved clockwise one pedestal.\n\nSuppose that, at each stage of the game, we draw an arrow from each pedestal to the pedestal on which the chest which it now holds originally sat. Then every pedestal has one incoming and one outgoing arrow, and the overall graph can be decomposed into some number of disconnected cycles. Let $c$ be the number of cycles. We note that at the beginning of the process, $c=10$ (every arrow is its own cycle) while at the end, $c=1$ (all the pedestals are connected in a ring).\n\nWe now claim that at each move, $c$ cannot decrease by more than $1$. Each move affects only two pedestals, which may initially belong to one or two cycles; the performance of the move may fuse two cycles into one or break up cycles into more cycles, but it certainly cannot replace two cycles by zero cycles. Therefore, $c$ decreases by at most $1$ at each move and hence cannot reach as low as $1$ before the ninth move.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that for sufficiently large primes $p$, there is an Eulerian circuit on the complete graph with $p$ vertices that does not contain any cycles of length at most $2023$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nTake a generator $g \\pmod{p}$ so that $g$ is not equivalent to anything of the form $-a / b$ for integers $a, b \\leq 2023$. For big enough primes, such a $g$ must exist as there are at most a constant number of such fractions.\n\nNow number the vertices with the residues $\\bmod\\ p$. For any residue $r \\pmod{p}$, consider the sequence of vertices $P_{r}$ formed by the multiples of $r$, starting from $r$ and ending at $p r \\equiv 0 \\pmod{p}$. We claim that the concatenation of $P_{1}, P_{g}, P_{g^{2}}, \\ldots$ works.\n\nClearly no $P_{r}$ contains a cycle. Thus, if a cycle were to exist, it would have to be formed by the end of one $P$ and the start of another. In other words, we would have $c g^{i+1} \\equiv (p-d) g^{i} \\pmod{p}$ for $c+d \\leq 2023$. However, this is impossible by choice of $g$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all primes $p$ such that there exist integers $a, b, c$, and $k$ satisfying the equations\n$$\n\\begin{aligned}\n& a^{2}+b^{2}+c^{2}=p \\\\\n& a^{4}+b^{4}+c^{4}=k p .\n\\end{aligned}\n$$",
    "options": [],
    "answer": "2 and 3",
    "solution": "Solution:\nThe answers are $2$ and $3$. It is clear that for them we can use the values $a = b = k = 1$ and $c = 0$ or $1$ respectively. Assume $p \\geq 5$. Subtracting twice the second equation from the square of the first, we find that $p$ divides\n$$\n2 a^{2} b^{2} + 2 b^{2} c^{2} + 2 c^{2} a^{2} - a^{4} - b^{4} - c^{4},\n$$\nan expression which factors as\n$$\n(a + b + c)(a + b - c)(a - b + c)(-a + b + c) .\n$$\nOne of the four factors must be divisible by $p$. By flipping the signs on $a, b$, and $c$, we can assume it is the first one. Using the inequality $x^{2} - x \\geq 0$ (for integers $x$, equality holding at $x = 0$ and $x = 1$), we have\n$$\np \\leq a + b + c \\leq a^{2} + b^{2} + c^{2} = p .\n$$\nThis equality condition can only hold if $a, b$, and $c$ are each equal to their squares, implying that they are at most $1$ and $p \\leq 1 + 1 + 1 = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle, and let $X, Y, Z$ be the excenters opposite $A, B, C$. The incircle of triangle $ABC$ touches $BC, CA, AB$ at points $D, E, F$. Finally, let $I$ and $O$ denote the incenter and circumcenter of triangle $ABC$.\nProve that lines $DX, EY, FZ, IO$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe fact that $DX, EY, FZ$ are concurrent follows from the fact that triangles $DEF$ and $XYZ$ are homothetic; indeed, note that $EF$ and $YZ$ are both perpendicular to the internal angle bisector of $\\angle BAC$.\n\nNow, to see that the concurrence point lies on $IO$, note that point $I$ is the orthocenter of triangle $XYZ$, and $O$ is the nine-point center of triangle $XYZ$. Thus line $IO$ is the Euler line of triangle $XYZ$ and thus passes through the circumcenter $S$ of triangle $XYZ$. But $I$ is the circumcenter of triangle $DEF$, hence line $SI$ passes through the concurrency point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute triangle with circumcenter $O$ and incenter $I$. Points $E, M$ lie on $AC$ and $F, N$ on $AB$ so that $BE \\perp AC$, $CF \\perp AB$, $\\angle ABM = \\angle CBM$ and $\\angle ACN = \\angle BCN$. Prove that $I$ lies on $EF$ if and only if $O$ lies on $MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $a = BC$, $b = CA$, $c = AB$. It is well-known (and follows from, say, Stewart's Theorem) that $AM = \\frac{bc}{a + c}$ and $AN = \\frac{bc}{a + b}$.\n\nNow, the distances from $O$ to $BC$, $AC$, $AB$ are $R \\cos \\alpha$, $R \\cos \\beta$, $R \\cos \\gamma$, respectively, where $\\alpha, \\beta, \\gamma$ are the angles of $\\triangle ABC$, and $R$ is the circumradius of $ABC$.\n\nSo, $O$ is on line $MN$ if and only if\n$$\n\\begin{aligned}\n{[ANM]} & = [ANO] + [AOM] \\\\\n\\Longleftrightarrow [ABC] \\cdot \\frac{AN}{AB} \\cdot \\frac{AM}{AC} & = [ANO] + [AOM] \\\\\n\\Longleftrightarrow \\left(\\frac{1}{2} a R \\cos \\alpha + \\frac{1}{2} b R \\cos \\beta + \\frac{1}{2} c R \\cos \\gamma\\right) \\cdot \\frac{b}{a + b} \\cdot \\frac{c}{a + c}\n\\end{aligned}\n$$\n(Here we use $[\\mathcal{P}]$ for the area of polygon $\\mathcal{P}$.) Next, recall that $AE = c \\cos \\alpha$, $AF = b \\cos \\alpha$. Thus $I$ is on $EF$ if and only if\n$$\n\\begin{aligned}\n{[AFE]} & = [AFI] + [AIE] \\\\\n\\Longleftrightarrow [ABC] \\cdot \\frac{AF}{AB} \\cdot \\frac{AE}{AC} & = \\frac{1}{2} r \\cdot c \\cos \\alpha + \\frac{1}{2} r \\cdot b \\cos \\alpha \\\\\n\\Longleftrightarrow \\frac{1}{2} r(a + b + c) \\cos^2 \\alpha & = \\frac{1}{2} r(b + c) \\cos \\alpha \\\\\n\\Longleftrightarrow (a + b + c) \\cos \\alpha & = b + c.\n\\end{aligned}\n$$\nBecause $AC = AE + EC$, we know $b = c \\cos \\alpha + a \\cos \\gamma$. Similarly, $c = b \\cos \\alpha + a \\cos \\beta$. Thus $I$ is on $EF$ if and only if\n$$\n\\begin{aligned}\n(a + b + c) \\cos \\alpha & = (c \\cos \\alpha + a \\cos \\gamma) + b(\\cos \\alpha + a \\cos \\beta) \\\\\n\\Longleftrightarrow \\cos \\alpha & = \\cos \\beta + \\cos \\gamma.\n\\end{aligned}\n$$\nThis implies the result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ and $k$ be positive integers with $n < \\sqrt{(k-1) 2^{k}}$. Prove that it is possible to color each element of the set $\\{1,2, \\ldots, n\\}$ red or green such that no $k$-term arithmetic progression is monochromatic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $A$ be the number of $k$-term arithmetic progressions in $\\{1,2, \\ldots, n\\}$. For any common difference $d$, an arithmetic progression of difference $d$ fits in $\\{1,2, \\ldots, n\\}$ iff its initial term $a$ satisfies\n$$\n1 \\leq a < a + (k-1)d \\leq n\n$$\nwhich is equivalent to\n$$\n1 \\leq a \\leq n - (k-1)d ;\n$$\nthis inequality has $n - (k-1)d$ solutions, as long as $n - (k-1)d \\geq 0$. Thus the total number of arithmetic progressions is\n$$\nA = \\sum_{d=1}^{\\left\\lfloor \\frac{n}{k-1} \\right\\rfloor} (n - (k-1)d)\n$$\nAs shown in the figure, the terms of this sum can be seen as the areas of nonoverlapping rectangles lying under the graph of $y = n - (k-1)x$. Their sum therefore does not exceed the area of the triangle enclosed by this line and the axes:\n$$\nA \\leq \\frac{1}{2} \\cdot \\frac{n}{k-1} \\cdot n = \\frac{n^{2}}{2(k-1)}\n$$\nNow consider any $k$-term arithmetic progression. The number of colorings in which it is monochromatic is\n$$\n2^{n} \\cdot \\frac{2}{2^{k}}\n$$\n![](attached_image_1.png)\nsince, of the $2^{k}$ ways that its terms might be colored, only 2 are monochromatic. Therefore the number of colorings that make no $k$-term arithmetic progression monochromatic is at least\n$$\n\\begin{aligned}\n& 2^{n} - A \\left(2^{n} \\cdot \\frac{2}{2^{k}}\\right) \\\\\n& \\geq 2^{n} - \\frac{n^{2}}{2(k-1)} \\cdot 2^{n} \\cdot \\frac{2}{2^{k}} \\\\\n& = 2^{n} \\left(1 - \\frac{n^{2}}{(k-1) 2^{k}}\\right)\n\\end{aligned}\n$$\nIf $n^{2} < (k-1) 2^{k}$ (equivalently, $n < \\sqrt{(k-1) 2^{k}}$) then this lower bound will be positive, implying that there is at least one coloring with the desired property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAerith and Bob are playing tag at Lake Round, a perfectly circular lake. Aerith tags Bob right next to the lake and dives in. Aerith can swim at a speed of $2 \\mathrm{mph}$, while Bob can't swim but runs at a speed of $9 \\mathrm{mph}$. Can Aerith leave the lake without getting tagged?",
    "options": [],
    "answer": "Yes, Aerith can escape.",
    "solution": "Solution:\n\nWe claim that Aerith can escape, even when Bob plays optimally and always runs towards Aerith's current location.\nLet a dash (d) be a unit of distance and a tick ($t$) be a unit of time such that the radius of the lake is $9$ dashes and such that one mph is equal to one dash per tick $(\\mathrm{d} / \\mathrm{t})$. Let $\\Omega$ be the perimeter of the lake, and let $\\omega$ be a concentric circle with radius $1.98$ dashes.$^{1}$\n\n![](attached_image_1.png)\n\nb)\n![](attached_image_2.png)\n\nc)\n![](attached_image_3.png)\n\nWe outline a four-step strategy for Aerith to escape the lake:\n\na) First, Aerith swims to the edge of $\\omega$.\n\nb) Then, Aerith begins swimming along $\\omega$. Her angular velocity is then\n$$\n\\left(2 \\frac{\\mathrm{d}}{\\mathrm{t}}\\right) /(1.98 \\mathrm{~d})>1 \\mathrm{t}^{-1}\n$$\nHowever, Bob's angular velocity is exactly $\\left(9 \\frac{d}{t}\\right) /(9 \\mathrm{~d})=1 \\mathrm{t}^{-1}$, which is slightly less, so Aerith outpaces Bob. Thus after some amount of time, Aerith will eventually be at the furthest point on $\\omega$ from Bob. At this point in time, let Aerith's current position be $A$ and let Bob's current position be $B$.\n\nc) Aerith now swims outward on line $AB$ until she reaches the point $A'$ that is $2 \\mathrm{~d}$ away from the center of the lake. This amount of time this takes is\n$$\n(2 \\mathrm{~d}-1.98 \\mathrm{~d}) /\\left(2 \\frac{\\mathrm{d}}{\\mathrm{t}}\\right)=0.01 \\mathrm{t}\n$$\nand in this time Bob travels $0.01t \\cdot 9 \\frac{\\mathrm{d}}{\\mathrm{t}}=0.09 \\mathrm{~d}$.\n\nd) Finally, Aerith swims away from Bob along the straight line through $A'$ perpendicular to line $AB$. Aerith's angular velocity along this journey is now less than $1 \\mathrm{t}^{-1}$ because, as we saw earlier, she has to be within a circle of radius $2$ in order to outpace Bob. Therefore, she remains ahead of Bob, so Bob will continue in his direction when chasing her. It takes her $\\left(\\sqrt{(9 \\mathrm{~d})^{2}-(2 \\mathrm{~d})^{2}}\\right) /\\left(2 \\frac{\\mathrm{d}}{\\mathrm{t}}\\right)=\\frac{\\sqrt{77}}{2} \\mathrm{t}$ to reach the edge, and in this time Bob can travel $\\frac{\\sqrt{77}}{2} \\mathrm{t} \\cdot 9 \\frac{\\mathrm{d}}{\\mathrm{t}}=\\frac{9 \\sqrt{77}}{2} \\mathrm{~d}$.\n\nThe total distance Bob can travel along the circle in the time it takes Aerith to reach the edge is thus $0.09 \\mathrm{~d}+\\frac{9 \\sqrt{77}}{2} \\mathrm{~d}<40 \\mathrm{~d}$. However, in order to arrive on time to tag Aerith, he would need to travel $9\\left(\\frac{3 \\pi}{2}-\\arcsin \\left(\\frac{2}{9}\\right)\\right)>40 \\mathrm{~d}$, so Aerith will indeed escape before Bob catches up.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle. The incircle, centered at $I$, touches side $BC$ at $D$. Let $M$ be the midpoint of the altitude from $A$ to $BC$. Lines $MI$ and $BC$ meet at $E$. Prove that $BD = CE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $E'$ be the point on side $BC$ such that $BD = E'C$. Then it is well known that $E'$ is the tangency point of the $A$-excircle $\\Omega$ of $\\triangle ABC$ with $BC$. Let $H$ be the foot of the altitude from $A$ to $BC$, and let $K$ be the point on the incircle $\\omega$ opposite $D$.\n\nCircles $\\omega$ and $\\Omega$ are homothetic about $A$ with $K$ and $E'$ corresponding points, so $A$, $K$, and $E'$ are collinear. Thus triangles $AHE'$ and $KDE'$ are homothetic; $M$ and $I$ are midpoints of corresponding sides, so they are collinear with $E'$. So $E' = E$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine if there exist positive integers $a, b, m, n$ such that $a \\neq b$, $m \\geq 2$, $n \\geq 2$, and\n$$\n\\underbrace{a^{a^{\\cdots a}}}_{m} = \\underbrace{b^{b^{\\cdots b}}}_{n}.\n$$",
    "options": [],
    "answer": "No; such integers do not exist.",
    "solution": "Solution:\n\nThere do not exist such positive integers. Assume for a contradiction that there do, however. We may assume $m > n$, so that $b > a$. Since we have equality between a power of $a$ and a power of $b$, $b$ is a rational power of $a$. We write $b = a^{x}$, where we know that $x \\geq 1$ is a rational number. Then we may rewrite the $b$ power tower (using $n \\geq 2$) as\n$$\n\\underbrace{b^{b^{\\cdots}}}_{n\\ b\\ \\text{'s}} = \\underbrace{a^{x a^{x b} \\cdots^{\\cdot b}}}_{n-2\\ b\\ \\text{'s}}\n$$\nOn the other hand, we know that this is equal to the power tower of $a$'s in the given equation, so removing the bottom $a$ gives\n$$\n\\underbrace{a^{a^{\\cdots a}}}_{m-1\\ a'\\mathrm{s}} = \\underbrace{x a^{x b^{\\cdots b}}}_{n-2\\ b\\ b'\\mathrm{s}}\n$$\nIn particular, this tells us that $x$ is a rational power of $a$, $x = a^{y}$ for some nonnegative rational number $y$. Substituting and again removing the bottom $a$ gives\n$$\n\\underbrace{a^{a^{\\cdots}}}_{m-2} = y + a^{y} \\underbrace{b^{b^{\\cdots}}}_{n-2}\n$$\nObserve that the right side is $y$ plus a rational power of $a$ (simply $a^{y}$ if $n=2$). Now consider the possibilities for this power $a^{p/q}$. If $p/q < 0$, then $y < 0$ and the right side of (2) is less than 1, an impossibility. So either $a^{p/q}$ is irrational (again impossible) or it is the $q$th root of a $q$th power, which is necessarily a positive integer. Then $y$ is also an integer, namely, the difference between the tower of $m-2$ $a$'s on the left and another power of $a$. Dividing both sides of (2) by $y$ indicates that $y$ is at least $1 - \\frac{1}{a} = \\frac{a-1}{a}$ times the tower of $m-2$ $a$'s, which is certainly larger than a power tower of $m-3$ $a$'s. But then $a^{y}$ is larger than a power tower of $m-2$ $a$'s, and clearly the right side of (2) is even larger than this, so we have a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that there are infinitely many positive integers $n$ for which $n^{2}+1$ has no repeated prime factors (that is, $n^{2}+1$ is squarefree).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy Fermat's Christmas theorem, the only primes which may divide $n^{2}+1$ other than $2$ are those which are $1 \\pmod{4}$, and moreover $2^{2} \\nmid n^{2}+1$ for any $n$.\n\nConsider primes $p \\equiv 1 \\pmod{4}$. Observe that for any $x$, we have\n$$\n\\#\\{n \\leq x \\mid n^{2}+1 \\equiv 0 \\pmod{p^{2}}\\} \\leq \\frac{2}{p^{2}} x + 2\n$$\nsince there are at most two solutions $\\bmod\\ p^{2}$ to $n^{2} \\equiv -1 \\pmod{p^{2}}$.\n\nSumming over all primes $p \\equiv 1 \\pmod{4}$ now implies the result, since\n$$\n\\sum_{p \\equiv 1} \\frac{2}{p^{2}} < 2\\left(\\frac{1}{4^{2}} + \\frac{1}{8^{2}} + \\ldots\\right) = \\frac{\\pi^{2}}{6} \\cdot \\frac{1}{8} < 1\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\sigma(n)$ denote the sum of the positive divisors of $n$. We say $n$ is perfect if $\\sigma(n)=2 n$. If $n$ is a positive integer such that\n$$\n\\frac{\\sigma(n)}{n}=\\frac{5}{3}\n$$\nshow that $5 n$ is an odd perfect number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNote that $\\sigma(n) / n$ is the sum of the reciprocals of the divisors of $n$. Since $3 \\sigma(n)=5 n$, $n$ must be divisible by $3$. If $n$ were even, then $n$ would be divisible by $6$, so we would get\n$$\n\\frac{\\sigma(n)}{n} \\geq 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=2>\\frac{5}{3}\n$$\na contradiction. Thus, $n$ is odd, so $\\sigma(n)$ is odd as well. This means $n$ is a perfect square, since $\\sigma(n)$ is the sum of the (odd) divisors of $n$, and $n$ has an odd number of divisors if and only if it is a square. Thus, since $n$ is divisible by $3$, it is also divisible by $9$. Now we claim that $n$ is not divisible by $5$. If so, it would also be divisible by $45$, so we would get\n$$\n\\frac{\\sigma(n)}{n} \\geq 1+\\frac{1}{3}+\\frac{1}{5}+\\frac{1}{9}+\\frac{1}{15}+\\frac{1}{45}=\\frac{78}{45}=\\frac{26}{15}>\\frac{5}{3} .\n$$\nThus, $n$ is not divisible by $5$, so $\\sigma(5 n)=\\sigma(5) \\sigma(n)=6 \\sigma(n)$. Thus we get\n$$\n\\frac{\\sigma(5 n)}{5 n}=\\frac{6}{5} \\cdot \\frac{\\sigma(n)}{n}=\\frac{6}{5} \\cdot \\frac{5}{3}=2 .\n$$\nThus, $5 n$ is a perfect number, and since $n$ is odd, $5 n$ is odd as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$ and $y$ be real numbers, and define a sequence $a_{0}, a_{1}, a_{2}, \\ldots$ by\n$$\na_{n}=\\sum_{k=0}^{n} x^{k} y^{n-k}\n$$\nSuppose that $a_{m}, a_{m+1}, a_{m+2}, a_{m+3}$ are integers for some $m \\geq 0$. Prove that $a_{n}$ is an integer for all $n \\geq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nBy cancellation of terms we see that\n$$\na_{n+1}-x a_{n}=y^{n+1} \\quad \\text{ and } \\quad a_{n+1}-y a_{n}=x^{n+1}\n$$\nIn particular, $a_{n+2}-x a_{n+1}=y^{n+2}=y \\cdot y^{n+1}=y\\left(a_{n+1}-x a_{n}\\right)$, which we can write as\n$$\na_{n+2}=(x+y) a_{n+1}-x y a_{n} .\n$$\nWe let $s=x+y$ and $t=x y$. Then the four given integral values of $a_{n}$ yield a pair of linear equations in $s$ and $t$ (or, to be precise, $s$ and $-t$ ):\n$$\n\\begin{aligned}\n& a_{m+2}=s a_{m+1}-t a_{m} \\\\\n& a_{m+3}=s a_{m+2}-t a_{m+1}\n\\end{aligned}\n$$\nIf the determinant $a_{m+1}^{2}-a_{m+2} a_{m}$ is nonzero, these equations have a unique solution. In fact,\n$$\n\\begin{aligned}\na_{m+1}^{2}-a_{m+2} a_{m} & =a_{m+1}\\left(a_{m+1}-x a_{m}\\right)-a_{m}\\left(a_{m+2}-x a_{m+1}\\right) \\\\\n& =a_{m+1} y^{m+1}-a_{m} y^{m+2}=y^{m+1}\\left(a_{m+1}-y a_{m}\\right)=x^{m+1} y^{m+1}=t^{m+1}\n\\end{aligned}\n$$\nSo we distinguish two cases.\n\nCase 1. $t=0$. Then without loss of generality $y=0$, so $a_{n}=x^{n}$. The conclusion follows from the following lemma:\n\nLemma 1. If $x$ is a real number and $n$ a nonnegative integer such that $x^{n}$ and $x^{n+1}$ are integers, then $x$ is an integer.\n\nProof. Note that $x=x^{n+1} / x^{n}$ is rational (if $x=0$, the conclusion is immediate). Write $x=p / q$ in lowest terms with $q>0$; then $x^{n}=p^{n} / q^{n}$ is also in lowest terms, hence $q=1$.\n\nCase 2. $t \\neq 0$. Then $t^{m+1}$ is an integer, and likewise $t^{m+2}=a_{m+2}^{2}-a_{m+3} a_{m+1}$ is an integer. So by Lemma 1 again, $t$ is an integer. Now $s$ is rational by Cramer's rule applied to (7). If we can prove $s$ is an integer, then since $a_{0}=1$ and $a_{1}=s$, we will be done by 6 .\n\nWrite $s=u / v$ in lowest terms and assume that $v>1$. Every $a_{n}$ is rational; write $a_{n}=u_{n} / v_{n}$ in lowest terms. We claim that $v_{n}=v^{n}$ for every $n \\geq 0$. The cases $n=0$ and $n=1$ are clear. We now induct, using (6):\n$$\n\\frac{u_{n+2}}{v_{n+2}}=\\frac{u}{v} \\cdot \\frac{u_{n+1}}{v^{n+1}}-t \\cdot \\frac{u_{n}}{v^{n}}=\\frac{u u_{n+1}-t v^{2} u_{n}}{v^{n+2}}\n$$\nSince $u$ and $u_{n+1}$ are coprime to $v$, the last fraction is reduced. So $v_{n+2}=v^{n+2}$ as desired. Hence $a_{n} \\notin \\mathbb{Z}$ for $n \\geq 1$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAre there positive integers $a$ and $b$ satisfying $a^{2}-23=b^{11}$?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. We may write the given equation as\n$$\na^{2}+45^{2}=b^{11}+2^{11}.\n$$\nTaking modulo $4$, we have $b^{11} \\equiv a^{2}+45^{2} \\equiv a^{2}+1 \\pmod{4}$ which forces $b \\equiv 1 \\pmod{4}$ and $a \\equiv 0 \\pmod{2}$. Thus $b+2 \\equiv 3 \\pmod{4}$.\n\nLet $\\nu_{p}$ be the usual $p$-adic notation. Let $p$ be an odd prime divisor of $b+2$ which is $3 \\pmod{4}$, and with $\\nu_{p}(b+2)$ being odd (in particular nonzero). As a consequence of Fermat's Christmas theorem, we have\n$$\n\\nu_{p}\\left(a^{2}+45^{2}\\right)= \\begin{cases}4 & p=3 \\text{ and } \\nu_{3}(a) \\geq 2 \\\\ 2 & p=3 \\text{ and } \\nu_{3}(a)=1 \\\\ 2 & p=5 \\text{ and } \\nu_{5}(a) \\geq 1 \\\\ 0 & \\text{ otherwise }\\end{cases}\n$$\nOn the other hand for any $p \\neq 11$ we have $\\nu_{p}\\left(b^{11}+2^{11}\\right)=\\nu_{p}(11)+\\nu_{p}(b+2)=\\nu_{p}(b+2)$, by the exponent lifting lemma (since we assumed $p \\mid b+2$). This is odd, and hence we get a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$ and $b$ be positive integers. Define a sequence $x_{0}, x_{1}, x_{2}, \\ldots$ by $x_{0}=0$, $x_{1}=1$, and $x_{n+2}=a x_{n+1}+b x_{n}$ for $n \\geq 0$. Prove that\n$$\n\\frac{x_{m+1} x_{m+2} \\cdots x_{m+n}}{x_{1} x_{2} \\cdots x_{n}}\n$$\nis an integer for all positive integers $m$ and $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $c_{m, n}$ be the quotient in the problem. Inspired by the similarity to $\\left(\\begin{array}{c}m+n \\\\ m\\end{array}\\right)$, we seek a relation between $c_{m, n}$, $c_{m-1, n}$, and $c_{m, n-1}$. First we prove the addition identity:\nLemma 1. For $m \\geq 1$ and $n \\geq 0$,\n$$\nx_{m+n}=x_{m} x_{n+1}+b x_{m-1} x_{n} .\n$$\nProof. Induction on $m$. The cases $m=1$ and $m=2$ are trivial. Adding $b$ times the equation for $m=m_{0}$ to $a$ times the equation for $m=m_{0}+1$ yields the equation for $m=m_{0}+2$.\n\nWe can now expand\n$$\n\\begin{aligned}\nc_{m, n} & =\\frac{x_{m+1} \\cdots x_{m+n-1} x_{m+n}}{x_{1} x_{2} \\cdots x_{n}} \\\\\n& =\\frac{x_{m+1} \\cdots x_{m+n-1}\\left(x_{m} x_{n+1}+b x_{m-1} x_{n}\\right)}{x_{1} x_{2} \\cdots x_{n}} \\\\\n& =x_{n+1} c_{m-1, n}+b x_{m-1} c_{m, n-1} .\n\\end{aligned}\n$$\nWe can now solve the problem by induction on $m+n$. The base cases $c_{m, 0}=c_{0, n}=1$ are integers, and the foregoing shows that if $c_{m-1, n}$ and $c_{m, n-1}$ are integers, so is $c_{m, n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the pattern shown below, row 1 (the bottom row) consists of two $1$'s, and row $n$ is formed by taking row $n-1$ and inserting between each adjacent pair of numbers $a$ and $b$ their sum $a+b$:\n\n| 1 | 5 | 4 | 7 | 3 | 8 | 5 | 7 | 2 | 7 | 5 | 8 | 3 | 7 | 4 | 5 | 1 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 1 |  | 4 |  | 3 |  | 5 |  | 2 |  | 5 |  | 3 |  | 4 |  | 1 |\n| 1 |  |  |  | 3 |  |  |  | 2 |  |  |  | 3 |  |  |  | 1 |\n| 1 |  |  |  |  |  |  |  | 2 |  |  |  |  |  |  |  | 1 |\n| 1 |  |  |  |  |  |  |  |  |  |  |  |  |  |  | 1 |  |\n| 1 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |\n\nIn row 2019 of the pattern, how many copies of 2019 will there be?",
    "options": [],
    "answer": "1344",
    "solution": "Solution:\n\nThe number of copies of $n$ in row $n$ is $\\phi(n)$, where $\\phi$ is the totient function, or the number of positive integers less than or equal to $n$ that are relatively prime to $n$. To see this, consider the following pattern of fractions, where we start with $\\frac{0}{1}$ and $\\frac{1}{1}$ and insert $\\frac{a+c}{b+d}$ between $\\frac{a}{b}$ and $\\frac{c}{d}$:\n\n![](attached_image_1.png)\n\nThe denominators of these fractions form the pattern of numbers in the problem, and the fractions are in increasing order. Row $n$ contains all the fractions in the $n$th Farey sequence exactly once, i.e. all the fractions with denominator at most $n$ (plus some other fractions). Thus, the number of copies of $n$ in row $n$ equals the number of fractions between 0 and 1 with denominator $n$ in simplest form. Each of these fractions must have numerator less than or equal to $n$ and relatively prime to $n$, so there are $\\phi(n)$ such fractions. Thus, there are $\\phi(n)$ copies of $n$ in row $n$.\n\nThe prime factorization of $2019$ is $3 \\cdot 673$, thus in row $2019$ the number of copies of $2019$ is\n$$\n\\phi(2019) = 2019\\left(1-\\frac{1}{3}\\right)\\left(1-\\frac{1}{673}\\right) = 2 \\cdot 672 = 1244\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that if positive real numbers $x, y, z$ have sum $1$, then\n$$\n\\frac{x}{x+yz} + \\frac{y}{y+zx} + \\frac{z}{z+xy} \\leq \\frac{2}{1-3xyz}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe idea is to use the identity\n$$\n\\frac{x}{x+yz} = \\frac{x}{x(x+y+z) + yz} = \\frac{x}{(x+y)(x+z)} = \\frac{x(y+z)}{(x+y)(y+z)(z+x)}\n$$\nSo the left-hand side is exactly equal to\n$$\n\\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)} = \\frac{2(xy+yz+zx)(x+y+z)}{(x+y)(y+z)(z+x)}.\n$$\nWe let $k = (xy+yz+zx)(x+y+z)$. On the other hand, we claim that $k \\leq 1/3$, which is sufficient. Indeed, to prove $k \\leq 1/3$ it suffices to prove that\n$$\nxy+yz+zx \\leq \\frac{(x+y+z)^2}{3}\n$$\nwhich is a direct consequence of the Cauchy-Schwarz inequality.\n\nOn the other hand, the left-hand side equals, exactly,\n$$\n\\frac{2k}{k-xyz} = 2 + \\frac{2xyz}{k-xyz}\n$$\nSo simply noting $k \\leq 1/3$ implies the desired conclusion, since this is a decreasing function of $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer which is divisible by $5$ and which can be written as the sum of two (not necessarily distinct) squares. Prove that $n$ can be written as the sum of two squares one of which is greater than or equal to four times the other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $n = a^2 + b^2$, where $a$ and $b$ are nonnegative integers. Suppose that each of the squares $a^2$ and $b^2$ is less than $4$ times the other, so\n$$\na < 2b \\text{ and } b < 2a.\n$$\nSince $a^2 + b^2$ is divisible by $5$, so is $a^2 + b^2 - 5b^2 = a^2 - 4b^2 = (a + 2b)(a - 2b)$. Thus $5$ divides either $a + 2b$ or $a - 2b$. If $5 \\mid a - 2b$, then $5 \\mid 2(a - 2b) + 5b = 2a + b$. Thus $5$ divides either $a + 2b$ or $2a + b$, and by switching the labels $a$ and $b$ we may assume that $5 \\mid a + 2b$.\n\nThen\n$$\n5 \\mid 4(a + 2b) - 5b = 4a + 3b\n$$\nand\n$$\n5 \\mid 3(a + 2b) - 10b = 3a - 4b.\n$$\nNote that\n$$\n\\begin{aligned}\n\\left(\\frac{4a + 3b}{5}\\right)^2 + \\left(\\frac{3a - 4b}{5}\\right)^2 &= \\frac{16a^2 + 24ab + 9b^2 + 9a^2 - 24ab + 16b^2}{25} \\\\\n&= \\frac{25a^2 + 25b^2}{25} = a^2 + b^2 = n.\n\\end{aligned}\n$$\nWe claim that\n$$\n\\left(\\frac{4a + 3b}{5}\\right)^2 \\geq 4\\left(\\frac{3a - 4b}{5}\\right)^2\n$$\nwhich is equivalent to\n$$\n4a + 3b \\geq 2|3a - 4b|\n$$\nwhich in turn is equivalent to the pair of inequalities\n$$\n4a + 3b \\geq 2(3a - 4b) \\text{ and } 4a + 3b \\geq -2(3a - 4b)\n$$\nSimplifying the first inequality yields $11b \\geq 2a$ which is true since $2b \\geq a$. Simplifying the second inequality yields $10a \\geq 5b$ which is true since $2a \\geq b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a > 2$ be given, and define a sequence $a_{0}, a_{1}, a_{2}, \\ldots$ by\n$$\na_{0} = 1, \\quad a_{1} = a, \\quad a_{n+1} = \\left(\\frac{a_{n}^{2}}{a_{n-1}^{2}} - 2\\right) \\cdot a_{n} .\n$$\nShow that for all integers $k \\geq 0$, we have\n$$\n\\sum_{i=0}^{k} \\frac{1}{a_{i}} < \\frac{a+2-\\sqrt{a^{2}-4}}{2} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWrite the recursion as\n$$\n\\frac{a_{n+1}}{a_{n}} = \\left(\\frac{a_{n}}{a_{n-1}}\\right)^{2} - 2\n$$\nLet $b_{n} = a_{n+1} / a_{n}$; then $b_{n} = b_{n-1}^{2} - 2$ and $b_{0} = a$. Let\n$$\nt = \\frac{a + \\sqrt{a^{2} - 4}}{2}\n$$\nThen $a = t + 1 / t$ and it is a simple induction to show that $b_{n} = t^{2^{n}} + t^{-2^{n}}$ for $n \\geq 0$. Then\n$$\na_{n} = a_{0} b_{0} b_{1} \\cdots b_{n-1} = \\prod_{i=0}^{n-1} \\left(t^{2^{i}} + t^{-2^{i}}\\right) = \\prod_{i=0}^{n-1} \\frac{t^{2^{i+1}} - t^{-2^{i+1}}}{t^{2^{i}} - t^{-2^{i}}} = \\frac{t^{2^{n}} - t^{-2^{n}}}{t - t^{-1}}\n$$\nSo\n$$\n\\begin{aligned}\n\\sum_{i=0}^{k} \\frac{1}{a_{i}} & = \\left(t - t^{-1}\\right) \\sum_{i=0}^{k} \\frac{1}{t^{2^{i}} - t^{-2^{i}}} \\\\\n& = \\left(t - t^{-1}\\right) \\sum_{i=0}^{k} \\frac{t^{2^{i}} + 1 - 1}{t^{2^{i+1}} - 1} \\\\\n& = \\left(t - t^{-1}\\right) \\sum_{i=0}^{k} \\left(\\frac{1}{t^{i} - 1} - \\frac{1}{t^{2^{i+1}} - 1}\\right) \\\\\n& = \\left(t - t^{-1}\\right)\\left(\\frac{1}{t - 1} - \\frac{1}{t^{2^{k+1}} - 1}\\right) \\\\\n& < \\left(t - t^{-1}\\right)\\left(\\frac{1}{t - 1}\\right) \\\\\n& = 1 + \\frac{1}{t} \\\\\n& = \\frac{a + 2 - \\sqrt{a^{2} - 4}}{2} .\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $N$ be a positive integer. Define a sequence $a_{n}$, $n \\geq 0$, by\n$$\na_{0}=0, \\quad a_{1}=1, \\quad a_{n+1}+a_{n-1}=a_{n}\\left(2-\\frac{1}{N}\\right) \\quad(n \\geq 1)\n$$\nProve that $a_{n}<\\sqrt{N+1}$ for all $n \\geq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLemma. For all $n \\geq 0$, $a_{n}$ and $a_{n+1}$ are related by a quadratic equation, namely\n$$\na_{n+1}^{2}+a_{n}^{2}-\\left(2-\\frac{1}{N}\\right) a_{n} a_{n+1}=1\n$$\nProof. By induction. The case $n=0$ is obvious. Suppose that for some $n \\geq 0$, (1) holds. Then\n$$\n\\begin{aligned}\na_{n+2}^{2}+a_{n+1}^{2}-\\left(2-\\frac{1}{N}\\right) a_{n+1} a_{n+2} & =a_{n+1}^{2}-\\left[\\left(2-\\frac{1}{N}\\right) a_{n+1}-a_{n+2}\\right] a_{n+2} \\\\\n& =a_{n+1}^{2}-a_{n} a_{n+2} \\\\\n& =a_{n+1}^{2}-\\left[\\left(2-\\frac{1}{N}\\right) a_{n+1}-a_{n}\\right] a_{n} \\\\\n& =a_{n+1}^{2}+a_{n}^{2}-\\left(2-\\frac{1}{N}\\right) a_{n} a_{n+1} \\\\\n& =1 .\n\\end{aligned}\n$$\nIf $a_{n}$ is treated as a constant, equation (1), the quadratic\n$$\na_{n+1}^{2}-\\left(2-\\frac{1}{N}\\right) a_{n} a_{n+1}+\\left(a_{n}^{2}-1\\right)=0\n$$\nmust have at least one real root $a_{n+1}$, so its discriminant is nonnegative:\n$$\n\\begin{aligned}\n& \\left(2-\\frac{1}{N}\\right)^{2} a_{n}^{2} \\geq 4 \\cdot 1 \\cdot\\left(a_{n}^{2}-1\\right) \\\\\n& \\left(\\left(2-\\frac{1}{N}\\right)^{2}-4\\right) a_{n}^{2} \\geq-4 \\\\\n& \\left(\\frac{-4 N+1}{N^{2}}\\right) a_{n}^{2} \\geq-4 \\\\\n& a_{n}^{2} \\leq \\frac{4 N^{2}}{4 N-1}\n\\end{aligned}\n$$\nThus it suffices to show that\n$$\n\\begin{aligned}\n\\frac{4 N^{2}}{4 N-1} & <N+1 \\\\\n4 N^{2} & <(N+1)(4 N-1)=4 N^{2}+3 N-1 \\\\\n1 & <3 N,\n\\end{aligned}\n$$\nwhich is true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n > 1$ be an integer. Three complex numbers have the property that their sum is $0$ and the sum of their $n$th powers is also $0$. Prove that two of the three numbers have the same absolute value.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nGiven that\n$$\na + b + c = 0 \\quad \\text{and} \\quad a^{n} + b^{n} + c^{n} = 0,\n$$\nlet\n$$\nt = ab + bc + ca \\quad \\text{and} \\quad u = abc.\n$$\nThen $a$, $b$, and $c$ are the roots of the polynomial\n$$\nf(z) = (z - a)(z - b)(z - c) = z^{3} + t z - u.\n$$\nIf $t = 0$, then $a$, $b$, and $c$ are the three cube roots of $u$ and thus all have the same absolute value; otherwise we can normalize, dividing $a$, $b$, and $c$ by a square root of $t$, so that $t = 1$. Now we have\n$$\nf(z) = (z - a)(z - b)(z - c) = z^{3} + z - u.\n$$\nDefine $p_{k} = a^{k} + b^{k} + c^{k}$ for nonnegative integers $k$ (here we make the convention that $0^{0} = 1$). It is not hard to compute that $p_{0} = 3$, $p_{1} = 0$, $p_{2} = -2$, and, for $k \\geq 0$,\n$$\np_{k+3} = u p_{k} - p_{k+1}.\n$$\nThis recursion allows us to think of each $p_{k}$ as a polynomial (with integer coefficients) in $u$. Our plan will be to prove that, for $k > 1$, these polynomials have only real roots. Then, given that $p_{n} = 0$, we can deduce that $f(z)$ is a real polynomial and therefore either has all real roots (which is impossible, since the sum of the 2nd powers of the roots is $-2$) or has a pair of complex conjugate roots which have the same absolute value.\nThe leading term of $p_{k}$ follows a pattern which is not difficult to verify by induction. It is:\n$$\n\\begin{array}{ll}\n3 u^{i} & \\text{if } k = 3i,\\ i \\geq 0 \\\\\n\\frac{k i}{2} u^{i-1} & \\text{if } k = 3i + 1,\\ i \\geq 1 \\\\\n-k u^{i} & \\text{if } k = 3i + 2,\\ i \\geq 0\n\\end{array}\n$$\nIn particular, for $k \\neq 1$, the leading coefficients of $p_{k}$ and $p_{k+3}$ have the same sign, and the degree of $p_{k+3}$ is one more than the degree of $p_{k}$. We will prove that the roots of $p_{k}$ alternate with the roots of $p_{k+3}$ on the number line, with no two coinciding, beginning and ending with a root of $p_{k+3}$. When $p_{k}$ is constant (that is, for $k = 0, 2$, or $4$) this statement is trivial, and we will use it as the base of an induction. For all other $k$, the induction hypothesis tells us there is exactly one root of $p_{k-3}$ between each pair of roots of $p_{k}$, and it suffices to prove the following statement:\n$$\n\\text{If } p_{k} = 0, \\text{ then } p_{k+3} \\text{ and } p_{k-3} \\text{ have opposite signs.}\n$$\nFor brevity we will prove this only when $x = p_{k-3}$ is positive and $u$ is also positive; the proof readily generalizes when one or both are negative. (By the induction hypothesis $p_{k-3} \\neq 0$, and if $u = 0$ then (1) gives $p_{k+3} = -p_{k+1} = p_{k-1} = -p_{k-3}$ for $k \\geq 4$.) Assume for contradiction that $y = p_{k+3} \\geq 0$. Using (1) repeatedly, we get\n$$\n\\begin{gathered}\np_{k+1} = u p_{k} - p_{k+3} = -y \\\\\np_{k-2} = u p_{k-3} - p_{k} = u x \\\\\np_{k-1} = u p_{k-2} - p_{k+1} = u^{2} x + y\n\\end{gathered}\n$$\nNote that $p_{k-1}$, $p_{k-2}$, $p_{k-3}$ are all positive. Then, using the reverse recursion\n$$\np_{i} = \\frac{p_{i+1} + p_{i+3}}{u}\n$$\nwe get that $p_{i} > 0$ for all $i < k$. Since $p_{2} = -2$, this is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, \\ldots, a_{n}$ be distinct integers. Prove that the polynomial\n$$\n\\left(x-a_{1}\\right)\\left(x-a_{2}\\right) \\ldots\\left(x-a_{n}\\right)-1\n$$\ncannot be written as the product of two nonconstant polynomials with integer coefficients (i.e. it is irreducible over the integers).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume there exist polynomials $f$ and $g$ satisfying\n$$\nf(x) g(x)=\\left(x-a_{1}\\right) \\ldots\\left(x-a_{n}\\right)-1 .\n$$\nLet $h(x)=f(x)+g(x)$. Now, for every $a_{i}$ we have $f\\left(a_{i}\\right) g\\left(a_{i}\\right)=-1$, so\n$$\nh\\left(a_{i}\\right)=f\\left(a_{i}\\right)+g\\left(a_{i}\\right)=0 .\n$$\nThus $h(x)$ has at least $n$ distinct roots, so it has degree at least $n$. But since $\\operatorname{deg} f+\\operatorname{deg} g=n$, this can only occur if one of $f$ and $g$ is a constant polynomial, which is what we wanted to prove.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a fixed triangle $\\triangle ABC$ and a point $P$, find the maximum value of\n$$\n\\frac{AB^{2} + BC^{2} + CA^{2}}{PA^{2} + PB^{2} + PC^{2}}\n$$",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nWe use the following lemma.\n\nLemma. Given $a, b, c, p \\in \\mathbb{R}$ such that not all of $a, b, c$ are equal,\n$$\n(a-b)^{2} + (b-c)^{2} + (c-a)^{2} \\leq 3\\left((p-a)^{2} + (p-b)^{2} + (p-c)^{2}\\right)\n$$\nwith equality if and only if $p = (a+b+c)/3$.\n\nProof. For fixed $a, b, c$, the right hand side is a quadratic in $p$, namely\n$$\n\\begin{aligned}\n3\\left((p-a)^{2} + (p-b)^{2} + (p-c)^{2}\\right) &= 3\\left((p^{2} - 2ap + a^{2}) + (p^{2} - 2bp + b^{2}) + (p^{2} - 2cp + c^{2})\\right) \\\\\n&= 9p^{2} - 6(a+b+c)p + 3(a^{2} + b^{2} + c^{2}) .\n\\end{aligned}\n$$\nand is thus minimized at $-\\frac{-6(a+b+c)}{2 \\cdot 9} = (a+b+c)/3$. By plugging $p = (a+b+c)/3$ into the original equation, we observe equality, as desired.\n\nUsing Cartesian coordinates, let $A = (a_{x}, a_{y})$, $B = (b_{x}, b_{y})$, $C = (c_{x}, c_{y})$, and $P = (p_{x}, p_{y})$. By the Pythagorean theorem,\n$$\n\\begin{gathered}\nAB^{2} + BC^{2} + CA^{2} = [(a_{x} - b_{x})^{2} + (a_{y} - b_{y})^{2}] + [(b_{x} - c_{x})^{2} + (b_{y} - c_{y})^{2}] \\\\\n+ [(c_{x} - a_{x})^{2} + (c_{y} - a_{y})^{2}] \\\\\n= [(a_{x} - b_{x})^{2} + (b_{x} - c_{x})^{2} + (c_{x} - a_{x})^{2}] \\\\\n+ [(a_{y} - b_{y})^{2} + (b_{y} - c_{y})^{2} + (c_{y} - a_{y})^{2}]\n\\end{gathered}\n$$\nso by the above,\n$$\n\\begin{aligned}\nAB^{2} + BC^{2} + CA^{2} \\leq 3[ & (p_{x} - a_{x})^{2} + (p_{x} - b_{x})^{2} + (p_{x} - c_{x})^{2} ] \\\\\n& + 3[ (p_{y} - a_{y})^{2} + (p_{y} - b_{y})^{2} + (p_{y} - c_{y})^{2} ] \\\\\n= & 3[ (p_{x} - a_{x})^{2} + (p_{y} - a_{y})^{2} ] + 3[ (p_{x} - b_{x})^{2} + (p_{y} - b_{y})^{2} ] \\\\\n& + 3[ (p_{x} - c_{x})^{2} + (p_{y} - c_{y})^{2} ] \\\\\n= & 3(PA^{2} + PB^{2} + PC^{2}) .\n\\end{aligned}\n$$\nThis gives\n$$\n\\frac{AB^{2} + BC^{2} + CA^{2}}{PA^{2} + PB^{2} + PC^{2}} \\leq 3,\n$$\nwith equality at\n$$\nP = \\left(\\frac{a_{x} + b_{x} + c_{x}}{3}, \\frac{a_{y} + b_{y} + c_{y}}{3}\\right),\n$$\nthe centroid of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe wish to place exactly $100$ dominoes (of size $2 \\times 1$ or $1 \\times 2$) without overlapping on a $20 \\times 20$ chessboard so that every $2 \\times 2$ square contains at least two uncovered unit squares which lie in the same row or column. In how many ways can this be done?",
    "options": [],
    "answer": "(20 choose 10)^2",
    "solution": "Solution:\n\nThe answer is $\\left(\\begin{array}{l}20 \\\\ 10\\end{array}\\right)^{2}$.\n\nGeneralizing the problem slightly, the answer is $\\left(\\begin{array}{c}m+n \\\\ n\\end{array}\\right)^{2}$ for a $2m \\times 2n$ rectangle. We provide a \"proof without words\" with the following bijection:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be a positive integer, and let $(a_{1}, a_{2}, \\ldots, a_{2k})$ and $(b_{1}, b_{2}, \\ldots, b_{2k})$ be two sequences of real numbers such that $\\frac{1}{2} \\leq a_{1} \\leq \\cdots \\leq a_{2k}$ and $\\frac{1}{2} \\leq b_{1} \\leq \\cdots \\leq b_{2k}$. Let $M$ and $m$ be the maximum and minimum respectively of\n$$\n(a_{1}+c_{1})(a_{2}+c_{2}) \\cdots (a_{2k}+c_{2k})\n$$\nas $(c_{1}, \\ldots, c_{2k})$ ranges through all possible permutations of $(b_{1}, \\ldots, b_{2k})$. Prove that\n$$\nM-m \\geq k(a_{k}-a_{k+1})(b_{k}-b_{k+1}) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $c_{i}=b_{\\sigma(i)}$ where $\\sigma$ is a permutation of the numbers $1$ through $2k$ that will vary. We begin with $\\sigma$ being the numbers $1$ through $2k$ written in order,\n$$\n12 \\cdots 2k,\n$$\nso $c_{i}=b_{i}$ for each $i$. We then switch $1$ with $2$, $1$ with $3$, $1$ with $4$, and so on, until $1$ arrives to the right of $2k$. Then we switch $2$ with $3$, $2$ with $4$, etc., until $2$ is just to the left of $1$. We continue in this way until the order of all the numbers $\\sigma(i)$ is reversed. Each of these switches changes the permutation $\\sigma$ and therefore the product (1). We claim that after all of them have been performed, the value of (1) is increased by at least $k(a_{k}-a_{k+1})(b_{k}-b_{k+1})$.\n\nLemma. No switch decreases the value of (1).\n\nProof. Each switch interchanges the values of two consecutive $c_{i}$'s, replacing two factors $(a_{i}+c_{i})(a_{i+1}+c_{i+1})$ of the product by $(a_{i}+c_{i+1})(a_{i+1}+c_{i})$ and increasing the value by a positive constant times\n$$\n(a_{i}+c_{i+1})(a_{i+1}+c_{i})-(a_{i}+c_{i})(a_{i+1}+c_{i+1})=(a_{i+1}-a_{i})(c_{i+1}-c_{i}) .\n$$\nThe factor $(a_{i+1}-a_{i})$ is obviously nonnegative, and the factor $(c_{i+1}-c_{i})$ is nonnegative also due to the way we performed the switches: a smaller number always moves to the right while a larger number moves to the left, so $c_{i}=b_{m}$ and $c_{i+1}=b_{n}$ with $m<n$.\n\nLemma. At least $k$ switches occur between $c_{k}$ and $c_{k+1}$ in which initially $\\sigma(k) \\leq k$ and $\\sigma(k+1)>k$ (and subsequently these inequalities are reversed).\n\nProof. Let $m(\\sigma)$ be the number of indices $i$ such that $i \\leq k$ and $\\sigma(i) \\leq k$. Initially, $m(\\sigma)=k$ and ultimately $m(\\sigma)=0$. Only switches of the type described alter $m(\\sigma)$, and only by $1$, so there must be at least $k$ of them (in fact there are exactly $k$).\n\nWe complete the proof by showing that each switch of the type described in the above lemma increases the value of (1) by at least $(a_{k}-a_{k+1})(b_{k}-b_{k+1})$. By the calculation in Lemma 7, the product of the two factors altered by the switch is increased by\n$$\n(a_{i+1}-a_{i})(c_{i+1}-c_{i}) \\geq (a_{k+1}-a_{k})(b_{k}-b_{k+1})\n$$\nthe remaining factors all have the form\n$$\na_{i}+c_{i} \\geq \\frac{1}{2}+\\frac{1}{2}=1\n$$\nso the increase in the product is at least $(a_{k}-a_{k+1})(b_{k}-b_{k+1})$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p$ be a prime number, and $f\\left(x_{1}, \\ldots, x_{n}\\right)$ be a polynomial with integer coefficients of total degree less than $n$. Prove that the number of ordered $n$-tuples $\\left(x_{1}, \\ldots, x_{n}\\right)$ with $0 \\leq x_{i}<p$ such that $f\\left(x_{1}, \\ldots, x_{n}\\right)$ is an integer multiple of $p$ is an integer multiple of $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy Fermat's little theorem, the expression\n$$\nf\\left(x_{1}, \\ldots, x_{n}\\right)^{p-1}\n$$\nequals $0 \\bmod p$ if $f\\left(x_{1}, \\ldots, x_{n}\\right)$ is $0 \\bmod p$ and $1 \\bmod p$ otherwise. So\n$$\nS=\\sum_{0 \\leq x_{1}, \\ldots, x_{n}<p} f\\left(x_{1}, \\ldots, x_{n}\\right)^{p-1} \\equiv p^{n}-z \\equiv -z\n$$\nwhere $z$ is the number of zeros of $f \\bmod p$. To prove $z \\equiv 0$, it suffices to prove that $S \\equiv 0$. Expanding $f^{p-1}$ as a polynomial, we see that $S$ is a sum of terms of the form\n$$\n\\sum_{0 \\leq x_{1}, \\ldots, x_{n}<p} c_{a_{1}, \\ldots, a_{n}} x_{1}^{a_{1}} \\cdots x_{n}^{a_{n}}=c_{a_{1}, \\ldots, a_{n}}\\left(\\sum_{0 \\leq x_{1}<p} x_{1}^{a_{1}}\\right) \\cdots\\left(\\sum_{0 \\leq x_{n}<p} x_{n}^{a_{n}}\\right)\n$$\nWe claim that one of the factors in parentheses is zero $\\bmod p$. We know $f$ has total degree at most $n-1$, so $f^{p-1}$ has total degree at most $(n-1)(p-1)$, and hence some $a_{i}$ is less than $p-1$. So it suffices to prove that\n$$\nT:=\\sum_{0 \\leq x<p} x^{a} \\equiv 0 \\quad \\bmod p\n$$\nfor $0 \\leq a<p-1$. For $a=0$ it is obvious (note that we are using the convention $0^{0}=1$ ). Otherwise, note that the equation $y^{a} \\equiv 1$ has at most $a$ roots $\\bmod p$, so there is a $y$ such that $y^{a} \\not \\equiv 1$. Now\n$$\n\\left(y^{a}-1\\right) T=\\sum_{0 \\leq x<p}(x y)^{a}-\\sum_{0 \\leq x<p} x^{a}\n$$\nwhich is $0 \\bmod p$ since the terms of the first sum are just a permutation of those of the second. Consequently $T \\equiv 0 \\bmod p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A B C D E$ be a convex pentagon circumscribed around a circle $\\omega$ such that $A B \\| C D$ and $B C \\| D E$. Locate points $X$ and $Y$ on rays $A B$ and $E D$, respectively, such that $B X = A B$ and $D Y = D E$. Prove that $X Y$ is tangent to $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFor convenience, let $Z$ be the intersection of sides $A B$ and $D E$ (so we have rhombus $Z B C D$), and let $P$ and $Q$ be the tangency points of $\\omega$ with $A B$ and $D E$, respectively. Then it is evident that a necessary condition for $X Y$ to touch $\\omega$ is\n$$\nP X + Q Y = X Y\n$$\nTo see that it is also sufficient, suppose $\\omega$ and $X Y$ are not tangent and consider the concentric circle $\\omega'$ tangent to $X Y$ at $T'$. Then $X P'$ and $Y Q'$ are either both greater or both less than $X T'$ and $Y T'$, respectively, and $X T' + Y T' = X Y$, so (1) is not satisfied.\n\nLet $x = B X$ and $y = D Y$; we will think of $x$ and $y$ as varying, keeping $X Y$ tangent to $\\omega$, while rhombus $Z B C D$ is fixed. Squaring both sides and using the Law of Cosines, we transform (1) to the form\n$$\n\\begin{aligned}\n(P X + Q Y)^2 & = Z X^2 + Z Y^2 - 2 Z X \\cdot Z Y \\cdot \\cos Z \\\\\n(B P + x + D Q + y)^2 & = (Z B + x)^2 + (Z D + y)^2 - 2(Z B + x)(Z D + y) \\cos Z\n\\end{aligned}\n$$\nIt is evident that when this equation is expanded, $x^2$ and $y^2$ will cancel, leaving an equation of the form\n$$\na x y + b x + c y + d = 0 .\n$$\nWe claim that $b$ and $c$ are both zero. If $x \\rightarrow \\infty$ (that is, $X$ runs out along ray $Z B$), then $Y$ tends toward $D$, i.e. $y \\rightarrow 0$. This can only happen if $b = 0$; symmetrically $c = 0$. So we have an equation of the form\n$$\na x y + d = 0 .\n$$\nWe note that $(x, y)$ satisfies this equation if and only if $(-x, -y)$ does. But $a(-x)(-y) + d = 0$ is exactly the equation obtained by applying the Law of Cosines to the condition that $A E$ is tangent to $\\omega$, as the reader can check.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all nonnegative integer solutions $(a, b, c, d)$ to the equation\n$$\n2^{a} 3^{b}-5^{c} 7^{d}=1.\n$$",
    "options": [],
    "answer": "(1,0,0,0), (3,0,0,1), (1,1,1,0), (2,2,1,1)",
    "solution": "Solution:\nThe answer is $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$ and $(2,2,1,1)$. The solution involves several cases. It's clear that $a \\geq 1$, otherwise the left-hand side is even. The remainder of the solution involves several cases.\n\n- First, suppose $b=0$.\n- If $c \\geq 1$, then modulo 5 we discover $2^{a} \\equiv 1 \\pmod{5}$ and hence $4 \\mid a$. But then modulo 3 this gives $-5^{c} 7^{d} \\equiv 0$, which is a contradiction.\n- Hence assume $c=0$. Then this becomes $2^{a}-7^{d}=1$. This implies $1+7^{d} \\equiv 2^{a} \\pmod{16}$, and hence $a \\leq 3$. Exhausting the possible values of $a=0,1,2,3$ we discover that $(3,0,0,1)$ and $(1,0,0,0)$ are solutions.\n\n- Henceforth suppose $b>0$. Taking modulo 3, we discover that $5^{c} \\equiv -1 \\pmod{3}$, so $c$ must be odd and in particular not equal to zero. Then, taking modulo 5 we find that\n$$\n1 \\equiv 2^{a} 3^{b} \\equiv 2^{a-b} \\pmod{5}\n$$\nThus, $a \\equiv b \\pmod{4}$. Now we again have several cases.\n\n- First, suppose $d=0$. Then $2^{a} 3^{b}=5^{c}+1$. Taking modulo 4, we see that $a=1$ is necessary, so $b \\equiv 1 \\pmod{4}$. Clearly we have a solution $(1,1,1,0)$ here. If $b \\geq 2$, however, then taking modulo 9 we obtain $5^{c} \\equiv -1 \\pmod{9}$, which occurs only if $c \\equiv 0 \\pmod{3}$. But then $5^{3}+1=126$ divides $5^{c}+1=2^{a} 3^{b}$, which is impossible.\n\n- Now suppose $d \\neq 0$ and $a, b$ are odd. Then $6 M^{2} \\equiv 1 \\pmod{7}$, where $M=2^{\\frac{a-1}{2}} 3^{\\frac{b-1}{2}}$ is an integer. Hence $M^{2} \\equiv -1 \\pmod{7}$, but this is not true for any integer $M$.\n\n- Finally, suppose $b, c, d \\neq 0$, and $a=2x, b=2y$ are even integers with $x \\equiv y \\pmod{2}$, and that $c$ is odd. Let $M=2^{x} 3^{y}$. We obtain $(M-1)(M+1)=5^{c} 7^{d}$. As $\\gcd(M-1, M+1) \\leq 2$, this can only occur in two situations.\n\n* In one case, $M-1=5^{c}$ and $M+1=7^{d}$. Then $5^{c}+1=2^{x} 3^{y}$. We have already discussed this equation; it is valid only when $x=y=1$ and $c=1$, which gives $(a, b, c, d)=(2,2,1,1)$.\n* In the other case, $M+1=5^{c}$ and $M-1=7^{d}$. Taking the first relation modulo 3, we obtain that $M \\equiv 2^{c}-1 \\equiv 1 \\pmod{3}$. Hence $y=0$, and $x$ is even. Now $2^{x}+1=5^{c}$ and $2^{x}-1=7^{d}$. But if $x$ is even then $3=2^{2}-1\\mid 2^{x}-1\\mid 7^{d}$, which is impossible. Hence there are no solutions here.\n\nIn summary, the only solutions are $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$ and $(2,2,1,1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a scalene triangle inscribed in circle $\\Gamma$. The internal bisector of $\\angle A$ meets $\\overline{BC}$ and circle $\\Gamma$ at points $D$ and $E$. The circle with diameter $\\overline{DE}$ meets $\\Gamma$ at a second point $F$. Prove that $\\frac{AB}{AC}=\\frac{FB}{FC}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $K$ be the midpoint of $\\operatorname{arc} \\widehat{BC}$ containing $A$. Let $M$ be the midpoint of $\\overline{BC}$.\n![](attached_image_1.png)\nSince $\\angle DME=\\angle DFE=90^\\circ$, the points $D, M, E, F$ are concyclic. Moreover, since $\\angle EFK=\\angle EFD=90^\\circ$, the points $F, D, K$ are collinear. Finally, since $\\angle KAD=\\angle KMD=90^\\circ$, the points $A, D, M, K$ are concyclic.\nTherefore,\n$$\n\\angle FAD=\\angle FAE=\\angle FKE=\\angle DKM=\\angle DAM.\n$$\nThis implies that $\\angle BAF=\\angle CAM$ and $\\angle CAF=\\angle BAM$. It follows that\n$$\n\\frac{FB}{FC}=\\frac{\\sin \\angle BAF}{\\sin \\angle CAF}=\\frac{\\sin \\angle CAM}{\\sin \\angle BAM}=\\frac{AB \\cdot \\frac{BM}{\\sin \\angle AMB}}{AC \\cdot \\frac{CM}{\\sin \\angle AMC}}=\\frac{AB}{AC}\n$$\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a_{1}=1$, $a_{2}=2$, and for $n \\geq 3$, let $a_{n}$ be the smallest positive integer such that $a_{n} \\neq a_{i}$ for $i<n$ and $\\operatorname{gcd}\\left(a_{n}, a_{n-1}\\right)>1$. Prove that every positive integer appears as some $a_{i}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOur solution will proceed in three steps.\n\nStep 1. We show that there is a prime $p$ such that infinitely many of the terms $a_{i}$ are divisible by $p$.\n\nProof. Suppose that such a prime $p$ does not exist. In particular, taking $p=2$, we find that there is an $N_{1}$ such that for all $i \\geq N_{1}$, $a_{i}$ is odd. Now let $p_{1}$ be the largest prime dividing any of the numbers $a_{1}, \\ldots, a_{N_{1}}$, and choose $N_{2}>N_{1}$ such that for $i \\geq N_{2}$, all prime factors of $a_{i}$ exceed $p_{1}$. The terms $a_{N_{2}}$ and $a_{N_{2}+1}$ have a common prime divisor $p_{2}>p_{1}$. Note that the term $2 p_{2}$ never appears in the sequence, since all the even terms have prime factors bounded by $p_{1}$. Therefore $p_{2}$ never appears in the sequence, since if $a_{n}=p_{2}$ for some $n$, we would have $a_{n+1}=2 p_{2}$. Thus $a_{N_{2}+1} \\geq 3 p_{2}$, which is impossible, since $p_{2}<a_{N_{2}+1}$ has not appeared at stage $N_{2}+1$ and shares a factor with $a_{N_{2}}$.\n\nStep 2. We show that every multiple of $p$ is in the sequence $\\left\\{a_{i}\\right\\}$.\n\nProof. Suppose that $p, 2 p, \\ldots,(k-1) p$ are in the sequence but $k p$ is not $(k \\geq 1)$. Choose $N$ such that all positive integers less than $k p$, if they appear in the sequence at all, are among the terms $a_{1}, \\ldots, a_{N}$. Then, using Step 1, choose $n \\geq N$ such that $p \\mid a_{n}$. What can $a_{n+1}$ be? By hypothesis $a_{n+1} \\geq k p$; we note that $k p$ is not relatively prime to $a_{n}$ and not equal to any of the terms $a_{1}, \\ldots, a_{n}$. So $a_{n+1}=k p$, a contradiction.\n\nStep 3. We show that every positive integer is in the sequence $\\left\\{a_{i}\\right\\}$.\n\nProof. Suppose that $1,2, \\ldots, k-1$ are in the sequence but $k$ is not $(k \\geq 3)$. Choose $N$ such that $1,2, \\ldots, k-1$ appear among the terms $a_{1}, a_{2}, \\ldots, a_{N}$. By Step 2, there is an $n \\geq N$ such that $a_{n}$ is divisible by $k p$, hence by $k$. Using an argument like that of the preceding step, we get $a_{n+1}=k$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDo there exist positive integers $a_{1}, \\ldots, a_{100}$ such that for each $k=1, \\ldots, 100$, the number $a_{1}+\\cdots+a_{k}$ has exactly $a_{k}$ divisors?",
    "options": [],
    "answer": "yes",
    "solution": "Solution:\n\nAnswer: yes.\n\nThe idea is to define the partial sums instead as follows. Let $d(n)$ denote the divisor function. Let $s_{N}$ be suitably large, then define by downwards recursion\n$$\ns_{n}=s_{n+1}-d\\left(s_{n+1}\\right)\n$$\nwith $N$ set such that $s_{0}=1$. Then $s_{k}=a_{1}+\\cdots+a_{k}$ works fine.\n\nAll we need is to ensure this sequence stays positive. But $x-d(x) \\geq x / 2$ for $x \\geq 8$, so it's enough to take $s_{N}=2^{103}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the game Pogémon, Pogé Balls contain Pichagus with a $50\\%$ chance and otherwise contain nothing. The new Pogémon universe started with one Pichagu and has existed for five hours. Every hour, every Pichagu at the time opened a new Pogé Ball.\n\na) What is the expected number of Pichagus created in the last hour?\n\nb) Every two currently existing Pichagus will now battle each other. What is the expected number of battles?",
    "options": [],
    "answer": "a) 81/32; b) 17091/512",
    "solution": "Solution:\n\na) Every hour, the expected number of Pichagus increases $50\\%$. Thus, the expected number of Pichagus after $4$ hours is $\\left(\\frac{3}{2}\\right)^{4}$. Each of these Pichagus has a $50\\%$ chance of creating a new Pichagu in the last hour, so the answer is $\\frac{1}{2}\\left(\\frac{3}{2}\\right)^{4}=\\frac{81}{32}=2.53125$.\n\nb) We show via induction that the answer for $n$ hours is $E(n):=\\left(\\frac{3}{2}\\right)^{2 n-1}-\\left(\\frac{3}{2}\\right)^{n-1}$. This is clearly true for $n=0$; no battles would happen at that point.\n\nNow we show that this is true for $n+1$ given that it's true for $n$. After one hour, there are two cases: either there is one Pichagu or there are two.\n\nIn the first case, the expected value is just $E(n)$ as there are $n$ hours left.\n\nOtherwise, say that the Pichagus are divided into two \"factions\": the Pichagus that descend from the original Pichagu and the ones that descend from the second one. Each faction has an expected value of $E(n)$ internal battles (battles between two members of the faction). Additionally, as in part (a), each faction has an expected value of $\\left(\\frac{3}{2}\\right)^{n}$ members, so there will be an expected value of $\\left(\\left(\\frac{3}{2}\\right)^{n}\\right)^{2}$ battles between factions. Thus in this case there will be an expected value of $2 E(n)+\\left(\\frac{3}{2}\\right)^{2 n}$ battles.\n\nCombining the cases we have\n$$\n\\begin{aligned}\nE(n+1) & =\\frac{1}{2} E(n)+\\frac{1}{2}\\left(2 E(n)+\\left(\\frac{3}{2}\\right)^{2 n}\\right) \\\\\n& =\\frac{1}{2}\\left(\\frac{3}{2}\\right)^{2 n}+\\frac{3}{2} E(n) \\\\\n& =\\frac{3}{4}\\left(\\frac{3}{2}\\right)^{2 n-1}+\\frac{3}{2}\\left(\\left(\\frac{3}{2}\\right)^{2 n-1}-\\left(\\frac{3}{2}\\right)^{n-1}\\right) \\\\\n& =\\left(\\frac{3}{2}\\right)^{2 n+1}-\\left(\\frac{3}{2}\\right)^{n},\n\\end{aligned}\n$$\nas desired. Plugging in $n=5$ gives $\\left(\\frac{3}{2}\\right)^{9}-\\left(\\frac{3}{2}\\right)^{4}=\\frac{3^{9}-2^{5} \\cdot 3^{4}}{2^{9}}=\\frac{17091}{512}=$ $33.380859375$ .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute scalene triangle with centroid $G$. The rays $BG$ and $CG$ meet the circumcircle of $ABC$ again at points $P$ and $Q$. Let $D$ denote the foot of the altitude from $A$ to $BC$. Suppose ray $GD$ meets the circumcircle of $ABC$ again at $E$. Show that the circumcircle of triangle $ADE$ lies on line $PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $M$ and $N$ be the midpoints of $CA$ and $AB$. By Pascal's theorem on $AABPQC$, we find that the tangent to the circumcircle at $A$, the line $MN$, and the line $PQ$ are concurrent at a single point $X$.\n\n![](attached_image_1.png)\n\nOur claim is that $X$ is the desired circumcenter.\n\nFirst note that $XA = XD$ simply because $MN$ is the perpendicular bisector of segment $AD$.\n\nTo proceed further we'll introduce point $F$, the intersection of ray $DG$ with the circumcircle. Let's assume for simplicity that $\\angle B > \\angle C$ (the other case is analogous). Owing to the negative homothety at $G$ mapping the nine-point circle to the circumcircle, we find that $ABCF$ is an isosceles trapezoid. Thus\n$$\n\\angle AED = \\angle AEF = \\angle B - \\angle C = \\angle ANM - \\angle XAB = \\angle AXM = \\frac{1}{2} \\angle AXD\n$$\nand this completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with incenter $I$. The incircle of $ABC$ is tangent to sides $BC$, $CA$, $AB$ at $D$, $E$, $F$. Let $H$ denote the orthocenter of triangle $BIC$, and let $P$ denote the midpoint of the altitude from $D$ to $EF$. Prove that $HP$ bisects $EF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWithout loss of generality, $AB \\leq AC$. Let $B_1$ and $C_1$ be the projections of $C$ and $B$ onto lines $BI$ and $CI$, respectively. Then $\\angle IEC = \\angle IB_1C = 90^{\\circ}$, so quadrilateral $IEB_1C$ is cyclic. But then\n$$\n\\angle B_1EC = \\angle B_1IC = 180^{\\circ} - \\angle BIC = 180^{\\circ} - \\angle FEC\n$$\nand so we find that $B_1$ lies on line $EF$. Similarly, $C_1$ lies on line $EF$.\n\nObserve that $I$ is the orthocenter of triangle $HBC$. Then it is well-known that $I$ is the incenter of triangle $DB_1C_1$. Moreover, $H$ is the intersection of the external angle bisectors of $\\angle DC_1B_1$ and $\\angle DB_1C_1$. So it is the center of the circle $\\Gamma$ tangent to $B_1C_1$ and the extensions of rays $DB_1$ and $DC_1$ past $B_1$ and $C_1$.\n\nLet $N$ be the foot of the altitude from $I$ to $\\overline{B_1C_1}$, noting that $N$ is in fact the midpoint of $\\overline{EF}$. Let $K$ be the tangency point of the $\\Omega$ to $\\overline{B_1C_1}$. Consider the homothety which takes the incircle of $\\triangle DB_1C_1$ to the circle $\\Gamma$; it sends $N$ to the point diametrically opposite $K$ on $\\Omega$. Call this point $L$; we see $D, N, L$ are collinear by the homothety mentioned. Moreover, $M, N, K$ are collinear. As $H$ is the midpoint of $\\overline{KL}$ and $P$ is the midpoint of $\\overline{DM}$, it follows that the points $H, N, P$ are collinear, which is what we wanted to prove.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute $1 \\cdot 2 + 2 \\cdot 3 + \\cdots + (n-1)n$.",
    "options": [],
    "answer": "(n-1)n(n+1)/3",
    "solution": "Solution:\nLet $S = 1 \\cdot 2 + 2 \\cdot 3 + \\cdots + (n-1)n$. We know $\\sum_{i=1}^{n} i = \\frac{n(n+1)}{2}$ and $\\sum_{i=1}^{n} i^{2} = \\frac{n(n+1)(2n+1)}{6}$. So\n$$\nS = 1(1+1) + 2(2+1) + \\cdots + (n-1)n = \\left(1^{2} + 2^{2} + \\cdots + (n-1)^{2}\\right) + (1 + 2 + \\cdots + (n-1))\n$$\n$$\n= \\frac{(n-1)n(2n-1)}{6} + \\frac{(n-1)n}{2} = \\frac{(n-1)n(n+1)}{3}\n$$\n\nWe can also arrive at the solution by realizing that\n$$\n\\sum_{i=1}^{n} i^{2} = \\sum_{i=1}^{n} i^{2} + \\sum_{i=2}^{n} i^{2} + \\sum_{i=3}^{n} i^{2} + \\cdots + \\sum_{i=n}^{n} i^{2} = n \\sum_{i=1}^{n} i^{2} - \\left(\\sum_{i=1}^{1} i^{2} + \\sum_{i=1}^{2} i^{2} + \\sum_{i=1}^{3} i^{2} + \\cdots + \\sum_{i=1}^{n-1} i^{2}\\right)\n$$\n$$\n= n \\frac{n(n+1)}{2} - \\left(\\frac{1 \\cdot 2}{2} + \\frac{2 \\cdot 3}{2} + \\cdots + \\frac{(n-1)n}{2}\\right) = n \\frac{n(n+1)}{2} - \\frac{1}{2} S = \\frac{n(n+1)(2n+1)}{6}\n$$\nso $S = \\frac{(n-1)n(n+1)}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $x$ satisfies $x^{3}+x^{2}+x+1=0$. What are all possible values of $x^{4}+2 x^{3}+2 x^{2}+2 x+1$?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n$x^{4}+2 x^{3}+2 x^{2}+2 x+1 = (x+1)\\left(x^{3}+x^{2}+x+1\\right) = 0$ is the only possible solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(n)$ be the minimal number of colors needed to properly color the directed edges of a tournament on $n$ vertices so that no two edges of the same color form a directed path of length $2$. Determine $f(n)$ for all $n \\ge 1$.",
    "options": [],
    "answer": "⌈log n⌉",
    "solution": "Let $f(n)$ be the answer to the problem. We claim that $f(n) = \\lceil \\log n \\rceil$ for all $n \\ge 1$. The proof is divided into showing that this quantity is both an upper and a lower bound.\n\n**Lower Bound.** We show that every tournament on $n$ vertices has directed-edge-chromatic number at least $\\lceil \\log n \\rceil$. We prove this by strong induction on $n$. As our base case, if $n = 1$, then clearly $f(1) \\ge 0 = \\lceil \\log 1 \\rceil$. Now, for $n \\ge 2$, assume by the strong inductive hypothesis that $f(k) \\ge \\lceil \\log k \\rceil$ for all $k < n$. Consider any proper directed-edge-coloring of an $n$-vertex directed graph (vertex set $V$) with $f(n) \\ge 1$ colors. Pick an arbitrary color, and let $U_1$ be the set of vertices $u$ for which there exists an edge $\\vec{uv}$ of that color, and let $U_2$ be the set of vertices $u$ for which there exists an edge of the form $\\vec{vu}$. Clearly since the coloring is proper $U_1 \\cap U_2 = \\emptyset$. Thus, there exists $i \\in \\{1, 2\\}$ such that $U_i$ has size at most $n/2$. Notice that $V \\setminus U_i$ has size at least $n/2$ and also has no edges of the originally chosen color. Thus, the graph induced by $V \\setminus U_i$ has a proper directed-edge-coloring of $f(n)-1$ colors. By the inductive hypothesis, this implies that\n$$\nf(n) - 1 \\ge f(|V \\setminus U_i|) \\ge \\lceil \\log |V \\setminus U_i| \\rceil \\ge \\lceil \\log(n/2) \\rceil = \\lceil \\log n \\rceil - 1.\n$$\nTherefore, $f(n) \\ge \\lceil \\log n \\rceil$.\n\n**Upper Bound.** We exhibit a proper directed-edge-coloring with $\\lceil \\log n \\rceil$ colors of a tournament with $n$ vertices. Consider the tournament with vertices $0, \\dots, n-1$ such that there is a directed edge from $i$ to $j$ if and only if $i > j$. Label our colors $0, \\dots, \\lceil \\log n \\rceil - 1$. Color the edge $\\vec{ij}$ the largest $k$ such that $i$ has a $1$ and $j$ has a $0$ in the $2^k$th place of their binary representations. Such a $k$ always exists and is at most $\\lceil \\log n \\rceil - 1$ because $i > j$ and $n-1$ requires at most $\\lceil \\log n \\rceil$ digits in its binary representation. Assume for sake of contradiction that this is not a proper directed-edge-coloring. Then, there would exist edge $\\overrightarrow{ij}$ and $\\overrightarrow{jk}$ of color $l$. Thus, $j$ would have both $0$ and $1$ in the $2^l$th place of its binary representation, contradiction. Hence, this exhibited coloring is proper so $f(n) \\le \\lceil \\log n \\rceil$.\n\nThus, the minimal directed-edge-chromatic number among tournaments on $n$ vertices is $\\lceil \\log n \\rceil$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe cost of 3 hamburgers, 5 milk shakes, and 1 order of fries at a certain fast food restaurant is $\\$ 23.50$. At the same restaurant, the cost of 5 hamburgers, 9 milk shakes, and 1 order of fries is $\\$ 39.50$. What is the cost of 2 hamburgers, 2 milk shakes, and 2 orders of fries at this restaurant?",
    "options": [],
    "answer": "$15",
    "solution": "Solution:\nLet $H = $ hamburger, $M = $ milk shake, and $F = $ order of fries. Then $3H + 5M + F = \\$ 23.50$. Multiplying the equation by 2 yields $6H + 10M + 2F = \\$ 47$. Also, it is given that $5H + 9M + F = \\$ 39.50$. Then subtracting the following equations\n$$\n\\begin{aligned}\n& 6H + 10M + 2F = \\$ 47.00 \\\\\n& 5H + 9M + F = \\$ 39.50\n\\end{aligned}\n$$\nyields $H + M + F = \\$ 7.50$. Multiplying the equation by 2 yields $2H + 2M + 2F = \\$ 15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate\n$$\nsin \\left(1998^{\\circ}+237^{\\circ}\\right) \\sin \\left(1998^{\\circ}-1653^{\\circ}\\right)\n$$",
    "options": [],
    "answer": "-1/4",
    "solution": "Solution:\n\n$\\sin \\left(1998^{\\circ}+237^{\\circ}\\right) \\sin \\left(1998^{\\circ}-1653^{\\circ}\\right) = \\sin \\left(2235^{\\circ}\\right) \\sin \\left(345^{\\circ}\\right) = \\sin \\left(75^{\\circ}\\right) \\sin \\left(-15^{\\circ}\\right) = -\\sin \\left(75^{\\circ}\\right) \\sin \\left(15^{\\circ}\\right) = -\\sin \\left(15^{\\circ}\\right) \\cos \\left(15^{\\circ}\\right) = -\\frac{\\sin \\left(30^{\\circ}\\right)}{2} = -\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S$ be the locus of all points $(x, y)$ in the first quadrant such that $\\frac{x}{t}+\\frac{y}{1-t}=1$ for some $t$ with $0<t<1$. Find the area of $S$.",
    "options": [],
    "answer": "1/6",
    "solution": "Solution:\n\nSolving for $t$ in the given equation, we get $t^{2}+(y-x-1) t+x=0$. Using the quadratic equation, we get\n$$\nt=\\frac{(x+1-y) \\pm \\sqrt{(y-x-1)^{2}-4 x}}{2}.\n$$\nFor all valid combinations of $(x, y)$, $t$ is positive and less than $1$ (this is easy to see by inspection). All valid combinations of $(x, y)$ are those that make $(y-x-1)^{2}-4 x \\geq 0$.\n\nSolving for $y$ in the equation $(y-x-1)^{2}-4 x=0$ yields\n$$\ny^{2}-(2 x+2) y+(x-1)^{2}=0 \\implies y=(x+1) \\pm 2 \\sqrt{x}.\n$$\nIn the original equation, it is given that $\\frac{x}{t}+\\frac{y}{1-t}=1$, and $0<t<1$. This implies that $x, y<1$. Then the only possible $y<1$ that satisfies $(y-x-1)^{2}-4 x=0$ is $y=x+1-2 \\sqrt{x}$.\n\nThen to satisfy the inequality $(y-x-1)^{2}-4 x \\geq 0$, we must have $y \\leq x+1-2 \\sqrt{x}$. Recall that this is when $0<y<1$. Hence we integrate in the interval $[0,1]$:\n$$\n\\int_{0}^{1} x+1-2 \\sqrt{x} \\, dx = \\frac{1}{6}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nG.H. Hardy once went to visit Srinivasa Ramanujan in the hospital, and he started the conversation with: \"I came here in taxi-cab number 1729. That number seems dull to me, which I hope isn't a bad omen.\" \"Nonsense,\" said Ramanujan. \"The number isn't dull at all. It's quite interesting. It's the smallest number that can be expressed as the sum of two cubes in two different ways.\" Ramanujan had immediately seen that $1729=12^{3}+1^{3}=10^{3}+9^{3}$. What is the smallest positive integer representable as the sum of the cubes of three positive integers in two different ways?",
    "options": [],
    "answer": "251",
    "solution": "Solution:\n\nLet this smallest positive integer be represented as $a^{3}+b^{3}+c^{3}=d^{3}+e^{3}+f^{3}$. By inspection, a solution is not possible with the first 4 cubes. We prove that it is impossible to write the same number as two different sums of the first 5 cubes. Because we necessarily need to use the 5th cube (otherwise, this proof would be for the first 4 cubes), we have $5^{3}+b^{3}+c^{3}=d^{3}+e^{3}+f^{3}$. Without loss of generality, suppose $d=5$. By inspection, there is no solution to $b^{3}+c^{3}=e^{3}+f^{3}$, such that $b, c, e, f \\leq 5$ and $b, c$ and $e, f$ are unique.\nThen none of $d, e, f$ are $5$. Then at least two must be $4$, otherwise the RHS would be too small. Without loss of generality, suppose $d=e=4$. Then $b^{3}+c^{3}=3+f^{3}$. By inspection, there are no possible solutions if $b, c, f \\leq 4$.\nThus if $a=5$, there are no solutions.\nSuppose that there is a solution within the first 6 cubes. Then $a=6$. By the same analysis as above, $d=e=5$, otherwise the RHS would be too small. Then $b^{3}+c^{3}=34+f^{3}$. By inspection, we see that a possible solution is $b=3, c=2, f=1$. Then the desired integer is $6^{3}+3^{3}+2^{3}=251$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the fourth annual Swirled Series, the Oakland Alphas are playing the San Francisco Gammas. The first game is played in San Francisco and succeeding games alternate in location. San Francisco has a $50\\%$ chance of winning their home games, while Oakland has a probability of $60\\%$ of winning at home. Normally, the series will stretch on forever until one team gets a three-game lead, in which case they are declared the winners. However, after each game in San Francisco there is a $50\\%$ chance of an earthquake, which will cause the series to end with the team that has won more games declared the winner. What is the probability that the Gammas will win?",
    "options": [],
    "answer": "34/73",
    "solution": "Solution:\n\nLet $F(x)$ be the probability that the Gammas will win the series if they are ahead by $x$ games and are about to play in San Francisco, and let $A(x)$ be the probability that the Gammas will win the series if they are ahead by $x$ games and are about to play in Oakland. Then we have\n$$\n\\begin{gathered}\nF(2)=\\frac{3}{4}+\\frac{A(1)}{4} \\\\\nA(1)=\\frac{6 F(0)}{10}+\\frac{4 F(2)}{10} \\\\\nF(0)=\\frac{1}{4}+\\frac{A(1)}{4}+\\frac{A(-1)}{4} \\\\\nA(-1)=\\frac{6 F(-2)}{10}+\\frac{4 F(0)}{10} \\\\\nF(-2)=\\frac{A(-1)}{4}\n\\end{gathered}\n$$\nPlugging $A(1)=\\frac{6 F(0)}{10}+\\frac{4 F(2)}{10}$ into $F(2)=\\frac{3}{4}+\\frac{A(1)}{4}$, we get\n$$\n\\begin{gathered}\nF(2)=\\frac{3}{4}+\\frac{1}{4}\\left(\\frac{6 F(0)}{10}+\\frac{4 F(2)}{10}\\right) \\\\\n\\frac{9 F(2)}{10}=\\frac{3}{4}+\\frac{6 F(0)}{40} \\Leftrightarrow F(2)=\\frac{5}{6}+\\frac{F(0)}{6}\n\\end{gathered}\n$$\nPlugging $A(-1)=\\frac{6 F(-2)}{10}+\\frac{4 F(0)}{10}$ into $F(-2)=\\frac{A(-1)}{4}$, we get\n$$\n\\frac{34 A(-1)}{40}=\\frac{4 F(0)}{10} \\Leftrightarrow F(-2)=\\frac{2 F(0)}{17}\n$$\nNow,\n$$\nF(0)=\\frac{1}{4}+\\frac{1}{4}\\left(\\frac{6 F(0)}{10}+\\frac{4 F(2)}{10}\\right)+\\frac{1}{4}\\left(\\frac{6 F(-2)}{10}+\\frac{4 F(0)}{10}\\right)\n$$\nThis simplifies to $F(0)=\\frac{1}{4}+\\frac{F(0)}{4}+\\frac{F(2)}{10}+\\frac{6 F(-2)}{40}$. Then, plugging our formulas in, we get\n$$\n\\begin{gathered}\nF(0)=\\frac{1}{4}+\\frac{F(0)}{4}+\\frac{1}{10}\\left(\\frac{5}{6}+\\frac{F(0)}{6}\\right)+\\frac{3 F(0)}{170} \\\\\n\\frac{73 F(0)}{102}=\\frac{1}{3} \\Leftrightarrow F(0)=\\frac{34}{73}\n\\end{gathered}\n$$\nSince $F(0)$ is the situation before the Series has started, the probability that the Gammas will win is $\\frac{34}{73}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBobbo starts swimming at 2 feet/s across a 100 foot wide river with a current of 5 feet/s. Bobbo doesn't know that there is a waterfall 175 feet from where he entered the river. He realizes his predicament midway across the river. What is the minimum speed that Bobbo must increase to make it to the other side of the river safely?",
    "options": [],
    "answer": "3 feet per second",
    "solution": "Solution:\n\nWhen Bobbo is midway across the river, he has travelled 50 feet. Going at a speed of 2 feet/s, this means that Bobbo has already been in the river for $\\frac{50 \\text{ feet}}{2 \\text{ feet} / \\mathrm{s}} = 25~\\mathrm{s}$. Then he has traveled $5 \\text{ feet} / \\mathrm{s} \\cdot 25~\\mathrm{s} = 125$ feet down the river. Then he has $175$ feet $- 125$ feet $= 50$ feet left to travel downstream before he hits the waterfall.\n\nBobbo travels at a rate of 5 feet/s downstream. Thus there are $\\frac{50~\\mathrm{feet}}{5 \\text{ feet} / \\mathrm{s}} = 10~\\mathrm{s}$ before he hits the waterfall. He still has to travel 50 feet horizontally across the river. Thus he must travel at a speed of $\\frac{50 \\text{ feet}}{10~\\mathrm{s}} = 5$ feet $/ \\mathrm{s}$. This is a 3 feet/s difference from Bobbo's original speed of 2 feet $/ \\mathrm{s}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cube with sides $1$ m in length is filled with water, and has a tiny hole through which the water drains into a cylinder of radius $1$ m. If the water level in the cube is falling at a rate of $1~\\mathrm{cm}/\\mathrm{s}$, at what rate is the water level in the cylinder rising?",
    "options": [],
    "answer": "1/pi cm/s",
    "solution": "Solution:\nThe magnitude of the change in volume per unit time of the two solids is the same. The change in volume per unit time of the cube is $1~\\mathrm{cm} \\cdot \\mathrm{m}^2 / \\mathrm{s}$. The change in volume per unit time of the cylinder is $\\pi \\cdot \\frac{d h}{d t} \\cdot m^2$, where $\\frac{d h}{d t}$ is the rate at which the water level in the cylinder is rising.\n\nSolving the equation $\\pi \\cdot \\frac{d h}{d t} \\cdot m^2 = 1~\\mathrm{cm} \\cdot m^2 / \\mathrm{s}$ yields $\\frac{1}{\\pi}~\\mathrm{cm}/\\mathrm{s}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many values of $x$, $-19 < x < 98$, satisfy\n$$\n\\cos^2 x + 2 \\sin^2 x = 1?\n$$",
    "options": [],
    "answer": "38",
    "solution": "Solution:\n\nAnswer: 38. For any $x$, $\\sin^2 x + \\cos^2 x = 1$. Subtracting this from the given equation gives $\\sin^2 x = 0$, or $\\sin x = 0$. Thus $x$ must be a multiple of $\\pi$, so $-19 < k\\pi < 98$ for some integer $k$, or approximately $-6.1 < k < 31.2$. There are 38 values of $k$ that satisfy this, so there are 38 values of $x$ that satisfy $\\cos^2 x + 2 \\sin^2 x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the area of the region bounded by the graphs of $y = x^{2}$, $y = x$, and $x = 2$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nThere are two regions to consider. First, there is the region bounded by $y = x^{2}$ and $y = x$, in the interval $[0, 1]$. In this interval, the values of $y = x$ are greater than the values of $y = x^{2}$, thus the area is calculated by\n$$\n\\int_{0}^{1} (x - x^{2}) \\, dx\n$$\nSecond, there is the region bounded by $y = x^{2}$ and $y = x$ and $x = 2$, in the interval $[1, 2]$. In this interval, the values of $y = x^{2}$ are greater than the values of $y = x$, thus the area is calculated by\n$$\n\\int_{1}^{2} (x^{2} - x) \\, dx\n$$\nThen the total area of the region bounded by the three graphs is\n$$\n\\int_{0}^{1} (x - x^{2}) \\, dx + \\int_{1}^{2} (x^{2} - x) \\, dx = 1\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 46424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of every even positive integer less than $233$ not divisible by $10$.",
    "options": [],
    "answer": "10812",
    "solution": "Solution:\nWe find the sum of all positive even integers less than $233$ and then subtract all the positive integers less than $233$ that are divisible by $10$.\n\n$2 + 4 + \\ldots + 232 = 2(1 + 2 + \\ldots + 116) = 116 \\cdot 117 = 13572$.\n\nThe sum of all positive integers less than $233$ that are divisible by $10$ is $10 + 20 + \\ldots + 230 = 10(1 + 2 + \\ldots + 23) = 2760$.\n\nThen our answer is $13572 - 2760 = 10812$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of the infinite series\n$$\n1 + 2\\left(\\frac{1}{1998}\\right) + 3\\left(\\frac{1}{1998}\\right)^{2} + 4\\left(\\frac{1}{1998}\\right)^{3} + \\ldots\n$$",
    "options": [],
    "answer": "(1998/1997)^2",
    "solution": "Solution:\nAnswer: $\\left(\\frac{1998}{1997}\\right)^{2}$ or $\\frac{3992004}{3988009}$. We can rewrite the sum as\n$\\left(1 + \\frac{1}{1998} + \\left(\\frac{1}{1998}\\right)^{2} + \\ldots\\right) + \\left(\\frac{1}{1998} + \\left(\\frac{1}{1998}\\right)^{2} + \\left(\\frac{1}{1998}\\right)^{3} + \\ldots\\right) + \\left(\\left(\\frac{1}{1998}\\right)^{2} + \\left(\\frac{1}{1998}\\right)^{3} + \\ldots\\right) + \\ldots$.\n\nEvaluating each of the infinite sums gives\n$\\frac{1}{1 - \\frac{1}{1998}} + \\frac{\\frac{1}{1998}}{1 - \\frac{1}{1998}} + \\frac{\\left(\\frac{1}{1998}\\right)^{2}}{1 - \\frac{1}{1998}} + \\ldots = \\frac{1998}{1997} \\cdot \\left(1 + \\frac{1}{1998} + \\left(\\frac{1}{1998}\\right)^{2} + \\ldots\\right) = \\frac{1998}{1997} \\cdot \\left(1 + \\frac{1}{1998} + \\left(\\frac{1}{1998}\\right)^{2} + \\ldots\\right)$,\nwhich is equal to $\\left(\\frac{1998}{1997}\\right)^{2}$, or $\\frac{3992004}{3988009}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the range of\n$$\nf(A) = \\frac{(\\sin A)\\left(3 \\cos^{2} A + \\cos^{4} A + 3 \\sin^{2} A + (\\sin^{2} A)(\\cos^{2} A)\\right)}{(\\tan A)(\\sec A - (\\sin A)(\\tan A))}\n$$\nif $A \\neq \\frac{n \\pi}{2}$.",
    "options": [],
    "answer": "(3,4)",
    "solution": "Solution:\nAnswer: $(3,4)$.\n\nWe factor the numerator and write the denominator in terms of fractions to get\n$$\n\\frac{(\\sin A)\\left(3+\\cos^{2} A\\right)\\left(\\sin^{2} A+\\cos^{2} A\\right)}{\\left(\\frac{\\sin A}{\\cos A}\\right)\\left(\\frac{1}{\\cos A}-\\frac{\\sin^{2} A}{\\cos A}\\right)} = \\frac{(\\sin A)\\left(3+\\cos^{2} A\\right)\\left(\\sin^{2} A+\\cos^{2} A\\right)}{\\frac{(\\sin A)\\left(1-\\sin^{2} A\\right)}{\\cos^{2} A}}.\n$$\nBecause $\\sin^{2} A + \\cos^{2} A = 1$, $1 - \\sin^{2} A = \\cos^{2} A$, so the expression is simply equal to $3 + \\cos^{2} A$.\n\nThe range of $\\cos^{2} A$ is $(0,1)$ (0 and 1 are not included because $A \\neq \\frac{n \\pi}{2}$), so the range of $3 + \\cos^{2} A$ is $(3,4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = 1 + \\frac{x}{2} + \\frac{x^{2}}{4} + \\frac{x^{3}}{8} + \\ldots$, for $-1 \\leq x \\leq 1$. Find $\\sqrt{e^{\\int_{0}^{1} f(x) d x}}$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nObserve that $f(x)$ is merely an infinite geometric series. Thus $f(x) = \\frac{1}{1 - \\frac{x}{2}} = \\frac{2}{2 - x}$.\n\nThen\n$$\n\\int_{0}^{1} \\frac{2}{2 - x} \\, dx = 2 \\int_{0}^{1} \\frac{1}{2 - x} \\, dx = 2 [ -\\ln|2 - x| ]_{0}^{1} = 2 ( -\\ln(1) + \\ln(2) ) = 2 \\ln 2.\n$$\n\nThen\n$$\n\\sqrt{e^{2 \\ln 2}} = \\sqrt{2^{2}} = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven that $r$ and $s$ are relatively prime positive integers such that $\\frac{r}{s}=\\frac{2(\\sqrt{2}+\\sqrt{10})}{5(\\sqrt{3+\\sqrt{5}})}$, find $r$ and $s$.",
    "options": [],
    "answer": "r = 4, s = 5",
    "solution": "Solution:\nSquaring both sides of the given equation yields\n$$\n\\frac{r^{2}}{s^{2}}=\\frac{4(12+4 \\sqrt{5})}{25(3+\\sqrt{5})}=\\frac{16(3+\\sqrt{5})}{25(3+\\sqrt{5})}=\\frac{16}{25}.\n$$\nBecause $r$ and $s$ are positive and relatively prime, then by inspection, $r=4$ and $s=5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many positive integers less than $1998$ are relatively prime to $1547$? (Two integers are relatively prime if they have no common factors besides $1$.)",
    "options": [],
    "answer": "1487",
    "solution": "Solution:\n\nAnswer: $1487$. The factorization of $1547$ is $7 \\cdot 13 \\cdot 17$, so we wish to find the number of positive integers less than $1998$ that are not divisible by $7$, $13$, or $17$. By the Principle of Inclusion-Exclusion, we first subtract the numbers that are divisible by one of $7$, $13$, and $17$, add back those that are divisible by two of $7$, $13$, and $17$, then subtract those divisible by three of them. That is,\n$$\n1997-\\left\\lfloor\\frac{1997}{7}\\right\\rfloor-\\left\\lfloor\\frac{1997}{13}\\right\\rfloor-\\left\\lfloor\\frac{1997}{17}\\right\\rfloor+\\left\\lfloor\\frac{1997}{7 \\cdot 13}\\right\\rfloor+\\left\\lfloor\\frac{1997}{7 \\cdot 17}\\right\\rfloor+\\left\\lfloor\\frac{1997}{13 \\cdot 17}\\right\\rfloor-\\left\\lfloor\\frac{1997}{7 \\cdot 13 \\cdot 17}\\right\\rfloor,\n$$\nor $1487$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA man named Juan has three rectangular solids, each having volume $128$. Two of the faces of one solid have areas $4$ and $32$. Two faces of another solid have areas $64$ and $16$. Finally, two faces of the last solid have areas $8$ and $32$. What is the minimum possible exposed surface area of the tallest tower Juan can construct by stacking his solids one on top of the other, face to face? (Assume that the base of the tower is not exposed).",
    "options": [],
    "answer": "688",
    "solution": "Solution:\nSuppose that $x$, $y$, $z$ are the sides of the following solids. Then Volume $= x y z = 128$.\n\nFor the first solid, without loss of generality (with respect to assigning lengths to $x$, $y$, $z$), $x y = 4$ and $y z = 32$. Then $x y^{2} z = 128$. Then $y = 1$. Solving the remaining equations yields $x = 4$ and $z = 32$. Then the first solid has dimensions $4 \\times 1 \\times 32$.\n\nFor the second solid, without loss of generality, $x y = 64$ and $y z = 16$. Then $x y^{2} z = 1024$. Then $y = 8$. Solving the remaining equations yields $x = 8$ and $z = 2$. Then the second solid has dimensions $8 \\times 8 \\times 2$.\n\nFor the third solid, without loss of generality, $x y = 8$ and $y z = 32$. Then $y = 2$. Solving the remaining equations yields $x = 4$ and $z = 16$. Then the third solid has dimensions $4 \\times 2 \\times 16$.\n\nTo obtain the tallest structure, Juan must stack the boxes such that the longest side of each solid is oriented vertically. Then for the first solid, the base must be $1 \\times 4$, so that the side of length $32$ can contribute to the height of the structure. Similarly, for the second solid, the base must be $8 \\times 2$, so that the side of length $8$ can contribute to the height. Finally, for the third solid, the base must be $4 \\times 2$.\n\nThus the structure is stacked, from bottom to top: second solid, third solid, first solid. This order is necessary, so that the base of each solid will fit entirely on the top of the solid directly beneath it.\n\nAll the side faces of the solids contribute to the surface area of the final solid. The side faces of the bottom solid have area $8 \\cdot (8 + 2 + 8 + 2) = 160$. The side faces of the middle solid have area $16 \\cdot (4 + 2 + 4 + 2) = 192$. The sides faces of the top solid have area $32 \\cdot (4 + 1 + 4 + 1) = 320$.\n\nFurthermore, the top faces of each of the solids are exposed. The top face of the bottom solid is partially obscured by the middle solid. Thus the total exposed area of the top face of the bottom solid is $8 \\cdot 2 - 4 \\cdot 2 = 8$. The top face of the middle solid is partially obscured by the top solid. Thus the total exposed area of the top face of the middle solid is $4 \\cdot 2 - 4 \\cdot 1 = 4$. The top face of the top solid is fully exposed. Thus the total exposed area of the top face of the top solid is $4 \\cdot 1 = 4$.\n\nThen the total surface area of the entire structure is $160 + 192 + 320 + 8 + 4 + 4 = 688$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many pairs of positive integers $(a, b)$ with $a \\leq b$ satisfy $\\frac{1}{a}+\\frac{1}{b}=\\frac{1}{6}$?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n$\\frac{1}{a}+\\frac{1}{b}=\\frac{1}{6} \\Rightarrow \\frac{a+b}{a b}=\\frac{1}{6} \\Rightarrow a b=6 a+6 b \\Rightarrow a b-6 a-6 b=0$.\nFactoring yields $(a-6)(b-6)=36$.\nBecause $a$ and $b$ are positive integers, only the factor pairs of $36$ are possible values of $a-6$ and $b-6$.\nThe possible pairs are:\n$$\n\\begin{aligned}\n& a-6=1,\\ b-6=36 \\\\\n& a-6=2,\\ b-6=18 \\\\\n& a-6=3,\\ b-6=12 \\\\\n& a-6=4,\\ b-6=9 \\\\\n& a-6=6,\\ b-6=6\n\\end{aligned}\n$$\nBecause $a \\leq b$, the symmetric cases, such as $a-6=12,\\ b-6=3$ are not applicable. Then there are $5$ possible pairs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEdward, the author of this test, had to escape from prison to work in the grading room today. He stopped to rest at a place $1,875$ feet from the prison and was spotted by a guard with a crossbow. The guard fired an arrow with an initial velocity of $100\\ \\mathrm{ft} / \\mathrm{s}$. At the same time, Edward started running away with an acceleration of $1\\ \\mathrm{ft} / \\mathrm{s}^{2}$. Assuming that air resistance causes the arrow to decelerate at $1\\ \\mathrm{ft} / \\mathrm{s}^{2}$ and that it does hit Edward, how fast was the arrow moving at the moment of impact (in $\\mathrm{ft} / \\mathrm{s}$)?",
    "options": [],
    "answer": "75 ft/s",
    "solution": "Solution:\n\nWe use the formula for distance, $d = \\frac{1}{2} a t^{2} + v t + d_{0}$. Then after $t$ seconds, Edward is at location $1875 + \\frac{1}{2}(1)(t^{2})$ from the prison. After $t$ seconds, the arrow is at location $\\frac{1}{2}(-1)(t^{2}) + 100 t$ from the prison. When the arrow hits Edward, both objects are at the same distance away from the tower. Hence\n$$\n1875 + \\frac{1}{2}(1)(t^{2}) = \\frac{1}{2}(-1)(t^{2}) + 100 t.\n$$\nSolving for $t$ yields\n$$\nt^{2} - 100 t + 1875 = 0 \\Rightarrow t = 25 \\text{ or } t = 75.\n$$\nThen it must be $t = 25$, because after the arrow hits Edward, he will stop running.\n\nAfter $25$ seconds, the arrow is moving at a velocity of $100 - 25(1) = 75\\ \\mathrm{ft} / \\mathrm{s}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA parabola is inscribed in equilateral triangle $ABC$ of side length $1$ in the sense that $AC$ and $BC$ are tangent to the parabola at $A$ and $B$, respectively. Find the area between $AB$ and the parabola.",
    "options": [],
    "answer": "sqrt(3)/6",
    "solution": "Solution:\nSuppose $A = (0, 0)$, $B = (1, 0)$, and $C = \\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$. Then the parabola in question goes through $(0, 0)$ and $(1, 0)$ and has tangents with slopes of $\\sqrt{3}$ and $-\\sqrt{3}$, respectively, at these points. Suppose the parabola has equation $y = a x^{2} + b x + c$. Then $\\frac{d y}{d x} = 2 a x + b$.\n\nAt point $(0, 0)$, $\\frac{d y}{d x} = b$. Also the slope at $(0, 0)$, as we determined earlier, is $\\sqrt{3}$. Hence $b = \\sqrt{3}$. Similarly, at point $(1, 0)$, $\\frac{d y}{d x} = 2 a + b$. The slope at $(1, 0)$, as we determined earlier, is $-\\sqrt{3}$. Then $a = -\\sqrt{3}$.\n\nSince the parabola goes through $(0, 0)$, $c = 0$. Hence the equation of the parabola is $y = -\\sqrt{3} x^{2} + \\sqrt{3} x$. The desired area is simply the area under the parabolic curve in the interval $[0, 1]$.\n\nHence\n$$\n\\int_{0}^{1} \\left( -\\sqrt{3} x^{2} + \\sqrt{3} x \\right) dx = \\frac{\\sqrt{3}}{6}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven that three roots of $f(x) = x^{4} + a x^{2} + b x + c$ are $2$, $-3$, and $5$, what is the value of $a + b + c$?",
    "options": [],
    "answer": "79",
    "solution": "Solution:\nBy definition, the coefficient of $x^{3}$ is negative the sum of the roots. In $f(x)$, the coefficient of $x^{3}$ is $0$. Thus the sum of the roots of $f(x)$ is $0$. Then the fourth root is $-4$. Then $f(x) = (x - 2)(x + 3)(x - 5)(x + 4)$. Notice that $f(1)$ is $1 + a + b + c$. Thus our answer is $f(1) - 1 = (1 - 2)(1 + 3)(1 - 5)(1 + 4) - 1 = 79$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46435,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe Houson Association of Mathematics Educators decides to hold a grand forum on mathematics education and invites a number of politicians from the United States to participate. Around lunch time the politicians decide to play a game. In this game, players can score 19 points for pegging the coordinator of the gathering with a spit ball, 9 points for downing an entire cup of the forum's interpretation of coffee, or 8 points for quoting more than three consecutive words from the speech Senator Bobbo delivered before lunch. What is the product of the two greatest scores that a player cannot score in this game?",
    "options": [],
    "answer": "1209",
    "solution": "Solution:\n\nAnswer: 1209. Attainable scores are positive integers that can be written in the form $8a + 9b + 19c$, where $a, b$, and $c$ are nonnegative integers. Consider attainable number of points modulo $8$.\n\nScores that are $0 \\pmod{8}$ can be obtained with $8a$ for positive $a$.\n\nScores that are $1 \\pmod{8}$ greater than or equal to $9$ can be obtained with $9 + 8a$ for nonnegative $a$.\n\nScores that are $2 \\pmod{8}$ greater than or equal to $18$ can be obtained with $9 \\cdot 2 + 8a$.\n\nScores that are $3 \\pmod{8}$ greater than or equal to $19$ can be obtained with $19 + 8a$.\n\nScores that are $4 \\pmod{8}$ greater than or equal to $19 + 9 = 28$ can be obtained with $19 + 9 + 8a$.\n\nScores that are $5 \\pmod{8}$ greater than or equal to $19 + 9 \\cdot 2 = 37$ can be obtained with $19 + 9 \\cdot 2 + 8a$.\n\nScores that are $6 \\pmod{8}$ greater than or equal to $19 \\cdot 2 = 38$ can be obtained with $19 \\cdot 2 + 8a$.\n\nScores that are $7 \\pmod{8}$ greater than or equal to $19 \\cdot 2 + 9 = 47$ can be obtained with $19 \\cdot 2 + 9 + 8a$.\n\nSo the largest two unachievable values are $39$ and $31$. Multiplying them gives $1209$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven any two positive real numbers $x$ and $y$, then $x \\diamond y$ is a positive real number defined in terms of $x$ and $y$ by some fixed rule. Suppose the operation $x \\diamond y$ satisfies the equations $(x \\cdot y) \\diamond y = x(y \\diamond y)$ and $(x \\diamond 1) \\diamond x = x \\diamond 1$ for all $x, y > 0$. Given that $1 \\diamond 1 = 1$, find $19 \\diamond 98$.",
    "options": [],
    "answer": "19",
    "solution": "Solution:\n\nNote first that $x \\diamond 1 = (x \\cdot 1) \\diamond 1 = x \\cdot (1 \\diamond 1) = x \\cdot 1 = x$.\n\nAlso, $x \\diamond x = (x \\diamond 1) \\diamond x = x \\diamond 1 = x$.\n\nNow, we have $(x \\cdot y) \\diamond y = x \\cdot (y \\diamond y) = x \\cdot y$.\n\nSo $19 \\diamond 98 = \\left(\\frac{19}{98} \\cdot 98\\right) \\diamond 98 = \\frac{19}{98} \\cdot (98 \\diamond 98) = \\frac{19}{98} \\cdot 98 = 19$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the set of solutions for $x$ in the inequality $\\frac{x+1}{x+2} > \\frac{3x+4}{2x+9}$ when $x \\neq -2$, $x \\neq \\frac{9}{2}$.",
    "options": [],
    "answer": "(-9/2, -2) ∪ ((1 - √5)/2, (1 + √5)/2)",
    "solution": "Solution:\nThere are 3 possible cases of $x$:\n\n1) $-\\frac{9}{2} < x$\n\n2) $\\frac{9}{2} \\leq x \\leq -2$\n\n3) $-2 < x$\n\nFor the cases (1) and (3), $x+2$ and $2x+9$ are both positive or negative, so the following operation can be carried out without changing the inequality sign:\n$$\n\\begin{aligned}\n\\frac{x+1}{x+2} &> \\frac{3x+4}{2x+9} \\\\\n\\Rightarrow 2x^{2} + 11x + 9 &> 3x^{2} + 10x + 8 \\\\\n\\Rightarrow 0 &> x^{2} - x - 1\n\\end{aligned}\n$$\nThe inequality holds for all $\\frac{1-\\sqrt{5}}{2} < x < \\frac{1+\\sqrt{5}}{2}$. The initial conditions were $-\\frac{9}{2} < x$ or $-2 < x$. The intersection of these three conditions occurs when $\\frac{1-\\sqrt{5}}{2} < x < \\frac{1+\\sqrt{5}}{2}$.\n\nCase (2) is $\\frac{9}{2} \\leq x \\leq -2$. For all $x$ satisfying these conditions, $x+2 < 0$ and $2x+9 > 0$. Then the following operations will change the direction of the inequality:\n$$\n\\begin{aligned}\n\\frac{x+1}{x+2} &> \\frac{3x+4}{2x+9} \\\\\n\\Rightarrow 2x^{2} + 11x + 9 &< 3x^{2} + 10x + 8 \\\\\n\\Rightarrow 0 &< x^{2} - x - 1\n\\end{aligned}\n$$\nThe inequality holds for all $x < \\frac{1-\\sqrt{5}}{2}$ and $\\frac{1+\\sqrt{5}}{2} < x$. The initial condition was $-\\frac{9}{2} \\leq x \\leq -2$. Hence the intersection of these conditions yields all $x$ such that $-\\frac{9}{2} \\leq x \\leq -2$.\n\nThen all possible cases of $x$ are $-\\frac{9}{2} \\leq x \\leq -2 \\cup \\frac{1-\\sqrt{5}}{2} < x < \\frac{1+\\sqrt{5}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the slopes of all lines passing through the origin and tangent to the curve $y^{2}=x^{3}+39 x-35$.",
    "options": [],
    "answer": "m = ±√51/2, ±√285/5",
    "solution": "Solution:\nAny line passing through the origin has equation $y = m x$, where $m$ is the slope of the line. If a line is tangent to the given curve, then at the point of tangency $(x, y)$, $\\frac{d y}{d x} = m$.\n\nFirst, we calculate $\\frac{d y}{d x}$ of the curve:\n\n$2 y d y = 3 x^{2} d x + 39 d x \\Rightarrow \\frac{d y}{d x} = \\frac{3 x^{2} + 39}{2 y}$.\n\nSubstituting $m x$ for $y$, we get the following system of equations:\n\n$$\n\\begin{aligned}\nm^{2} x^{2} & = x^{3} + 39 x - 35 \\\\\nm & = \\frac{3 x^{2} + 39}{2 m x}\n\\end{aligned}\n$$\n\nSolving for $x$ yields the equation $x^{3} - 39 x + 70 = 0 \\Rightarrow (x - 2)(x + 7)(x - 5) = 0 \\Rightarrow x = 2$ or $x = -7$ or $x = 5$.\n\nThese solutions indicate the $x$-coordinate of the points at which the desired lines are tangent to the curve. Solving for the slopes of these lines, we get $m = \\pm \\frac{\\sqrt{51}}{2}$ for $x = 2$, no real solutions for $x = -7$, and $m = \\pm \\frac{\\sqrt{285}}{5}$ for $x = 5$.\n\nThus $m = \\pm \\frac{\\sqrt{51}}{2}, \\pm \\frac{\\sqrt{285}}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate $\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot 2^{n-1}}$.",
    "options": [],
    "answer": "2 ln 2",
    "solution": "Solution:\nNote that if we take the integral of $f(x)$ in problem 4, we get the function $F(x) = x + \\frac{x^{2}}{2 \\cdot 2} + \\frac{x^{3}}{3 \\cdot 2^{2}} + \\ldots$. Evaluating this integral in the interval $[0,1]$, we get $1 + \\frac{1}{2 \\cdot 2} + \\frac{1}{3 \\cdot 2^{2}} + \\ldots$, which is the desired sum.\n\nHence $\\int_{0}^{1} \\frac{2}{2-x} d x = 2 \\ln 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBob's Rice ID number has six digits, each a number from $1$ to $9$, and any digit can be used any number of times. The ID number satisfies the following property: the first two digits is a number divisible by $2$, the first three digits is a number divisible by $3$, etc., so that the ID number itself is divisible by $6$. One ID number that satisfies this condition is $123252$. How many different possibilities are there for Bob's ID number?",
    "options": [],
    "answer": "324",
    "solution": "Solution:\n\nAnswer: $324$.\n\nWe will count the number of possibilities for each digit in Bob's ID number, then multiply them to find the total number of possibilities for Bob's ID number.\n\nThere are $3$ possibilities for the first digit given any last $5$ digits, because the entire number must be divisible by $3$, so the sum of the digits must be divisible by $3$.\n\nBecause the first two digits are a number divisible by $2$, the second digit must be $2, 4, 6$, or $8$, which is $4$ possibilities.\n\nBecause the first five digits are a number divisible by $5$, the fifth digit must be a $5$.\n\nNow, if the fourth digit is a $2$, then the last digit has two choices, $2, 8$, and the third digit has $5$ choices, $1, 3, 5, 7, 9$.\n\nIf the fourth digit is a $4$, then the last digit must be a $6$, and the third digit has $4$ choices, $2, 4, 6, 8$.\n\nIf the fourth digit is a $6$, then the last digit must be a $4$, and the third digit has $5$ choices, $1, 3, 5, 7, 9$.\n\nIf the fourth digit is an $8$, then the last digit has two choices, $2, 8$, and the third digit has $4$ choices, $2, 4, 6, 8$.\n\nSo there are a total of $3 \\cdot 4 (2 \\cdot 5 + 4 + 5 + 2 \\cdot 4) = 3 \\cdot 4 \\cdot 27 = 324$ possibilities for Bob's ID number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $f(x)$ is a rational function such that $3 f\\left(\\frac{1}{x}\\right)+\\frac{2 f(x)}{x}=x^{2}$ for $x \\neq 0$. Find $f(-2)$.",
    "options": [],
    "answer": "67/20",
    "solution": "Solution:\nLet $x=\\frac{-1}{2}$. Then\n$$\n\\begin{aligned}\n& 3 f(-2)+\\frac{2 f\\left(\\frac{-1}{2}\\right)}{\\frac{-1}{2}}=\\frac{1}{4} \\\\\n& \\Rightarrow 3 f(-2)-4 f\\left(\\frac{-1}{2}\\right)= \\frac{1}{4}\n\\end{aligned}\n$$\nLet $x=-2$. Then\n$$\n\\begin{aligned}\n& 3 f\\left(\\frac{-1}{2}\\right)+\\frac{2 f(-2)}{-2}=4 \\\\\n\\Rightarrow 3 f\\left(\\frac{-1}{2}\\right)-f(-2) & =4\n\\end{aligned}\n$$\nSolving this system of equations $\\{(1),(2)\\}$ for $f(-2)$ yields $f(-2)=\\frac{67}{20}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor any finite set $S$, let $f(S)$ be the sum of the elements of $S$ (if $S$ is empty then $f(S)=0$). Find the sum over all subsets $E$ of $S$ of $\\frac{f(E)}{f(S)}$ for $S=\\{1,2, \\ldots, 1999\\}$.",
    "options": [],
    "answer": "2^{1998}",
    "solution": "Solution:\n\nAn $n$ element set has $2^{n}$ subsets, so each element of $S$ appears in $2^{1998}$ subsets $E$, so our sum is $2^{1998} \\cdot \\frac{1+2+\\ldots+1999}{1+2+\\ldots+1999} = 2^{1998}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that among any 12 consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (The proper divisors of a positive integer $n$ are all positive integers other than 1 and $n$ which divide $n$. For example, the proper divisors of 14 are 2 and 7.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOne of the twelve numbers must be a multiple of twelve; call it $a = 12 n$. Among the proper divisors of $a$ are the integers $2 n$, $3 n$, $4 n$, $6 n$. These sum to $15 n > a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $C$ be a circle in the $x y$-plane with center on the $y$-axis and passing through $A=(0, a)$ and $B=(0, b)$ with $0<a<b$. Let $P$ be any other point on the circle, let $Q$ be the intersection of the line through $P$ and $A$ with the $x$-axis, and let $O=(0,0)$. Prove that $\\angle B Q P=\\angle B O P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe make use of the fact that an angle inscribed in a circle has measure equal to one-half of the arc subtended. Since the $x$- and $y$-axes meet in a right angle, the circle $C_{1}$ through $B$, $O$, and $Q$ has $Q B$ as a diameter. Also, $\\angle A P B$ is a right angle, since $A B$ is the diameter of $C$. But this means that $\\angle Q P B$ and $\\angle Q O B$ are both right angles, so that $P$, $O$, $Q$, $B$ all lie on circle $C_{1}$. Thus the two angles in question, $\\angle B Q P$ and $\\angle B O P$, are inscribed in $C_{1}$, subtend the same arc, and are therefore equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFinitely many cards are placed in two stacks, with more cards in the left stack than the right. Each card has one or more distinct names written on it, although different cards may share some names. For each name, we define a \"shuffle\" by moving every card that has this name written on it to the opposite stack. Prove that it is always possible to end up with more cards in the right stack by picking several distinct names, and doing in turn the shuffle corresponding to each name.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the number of cards be $c$ and let the number of distinct names be $n$. Each card contains a set of names; denote these sets by $S_{1}, S_{2}, \\ldots, S_{c}$ (some of these sets may share elements). Now let $E$ be a subset of the set of $n$ names, and denote by $D(E)$ the difference of the number of cards in the left stack and the number of cards in the right stack, after the cards are shuffled by the names in $E$. We are given that $D(\\emptyset)>0$; we must show that there exists a subset $E$ such that $D(E)<0$.\n\nConsider the sum of all $D(E)$ as $E$ ranges through all $2^{n}$ possible subsets of names. If we can show that this sum is equal to zero, we will be done, since $D(\\emptyset)>0$ is just one term in this sum, forcing at least one other term to be negative. For $i=1,2, \\ldots, c$, let $v_{i}=+1$ or $-1$ if the $i$th card is initially in the left or right stack, respectively. Then\n$$\nD(\\emptyset)=v_{1}+v_{2}+\\cdots+v_{c}=\\sum_{i=1}^{c} v_{i}\n$$\nNow consider what happens when we shuffle the cards corresponding to a subset $E$ of names. The $i$th card will move back and forth from one stack to the other a total of $|E \\cap S_{i}|$ times ($|A|$ means the number of elements in the set $A$). The only thing that matters is whether this value is even or odd. Thus we have\n$$\nD(E)=\\sum_{i=1}^{c}(-1)^{|E \\cap S_{i}|} v_{i}\n$$\nIt remains to sum this expression over all subsets $E$. Let us examine what happens just for one card; i.e., let us compute\n$$\n\\sum_{E}(-1)^{|E \\cap S_{i}|} v_{i}\n$$\nfor a fixed $i$ as $E$ ranges over all $2^{n}$ subsets of names. If we can show that this equals zero, then the entire sum will equal zero and we are done. Let $|S_{i}|=k$. Then $|E \\cap S_{i}|$ will range from $0$ to $k$ inclusive. For $0<r<k$, how many subsets $E$ are there such that $|E \\cap S_{i}|=r$? There are $\\binom{k}{r}$ subsets of $S_{i}$ with $r$ elements. Fix one of them, call it $T$. Then $E$ must contain all the elements of $T$, plus any subset of the names that are not contained in $S_{i}$. In other words, there are $2^{n-k}$ subsets containing $T$, and thus there are $\\binom{k}{r} 2^{n-k}$ subsets $E$ altogether satisfying $|E \\cap S_{i}|=r$. Therefore\n$$\n\\sum_{E}(-1)^{|E \\cap S_{i}|} v_{i}=v_{i} \\sum_{r=0}^{k}(-1)^{r}\\binom{k}{r} 2^{n-k}=2^{n-k} v_{i} \\sum_{r=0}^{k}(-1)^{r}\\binom{k}{r}\n$$\nBy the binomial theorem, we have\n$$\n2^{n-k} v_{i} \\sum_{r=0}^{k}(-1)^{r}\\binom{k}{r}=2^{n-k} v_{i}(1-1)^{k}=0\n$$\n\nWe will prove the statement by induction on $n$, the number of distinct names present. Call the left stack $L$ and the right stack $R$. For the case $n=1$, let $L$ have $x$ cards and $R$ have $y$ cards with $x>y$; all cards have the same single name written on them. We perform the shuffle corresponding to the single name present, and we are done (now we have $x$ in $R$ and $y$ in $L$).\n\nNext, assume that we have proven the statement for $n$ names, and consider the case with $n+1$ names. Call the first $n$ names $a_{1}, a_{2}, \\ldots, a_{n}$, and let the new $(n+1)$-st name be $a$. There are two cases:\n\nCase 1. The number of $L$ cards containing only name $a$ is less than or equal to the number of $R$ cards containing only name $a$. In this case, we need only look at the remaining cards (the ones that do not contain only $a$). We can ignore the name $a$, and use the inductive hypothesis to perform the required shuffles using some subset of the $n$ names $a_{1}, a_{2}, \\ldots, a_{n}$, and we will be done: there are now more of these remaining cards in stack $R$ than in stack $L$, and since there were at least as many \"only $a$\" cards in $R$ than in $L$, the final configuration has more cards in $R$ than in $L$. (Note that if the original configuration contains as many \"remaining\" cards in $R$ as in $L$, we don't have to perform any shuffles at all, given that the \"only $a$\" cards in $R$ are more than those in $L$.)\n\nCase 2: The number of $L$ cards containing only name $a$ is greater than the number of $R$ cards containing only $a$. In this case, we first perform the shuffle corresponding to the name $a$. This will switch the \"only $a$\" cards, so that there are more of them in $R$ than in $L$. Unfortunately, some of the remaining cards may also have moved around. But now there are two possibilities: if there are at least as many remaining cards in $R$ than in $L$, we are done. If not, then we can again employ the inductive hypothesis, ignoring the name $a$, and just applying the shuffles to some subset of the $n$ names $a_{1}, a_{2}, \\ldots, a_{n}$. The end result will be that there will be as many remaining cards in $R$ than in $L$, and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a cyclic quadrilateral (a quadrilateral which can be inscribed in a circle). Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE / EB = CF / FD$. Let $P$ be the point on the segment $EF$ such that $PE / PF = AB / CD$. Prove that the ratio between the areas of triangle $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThere are two cases to consider.\n\nFirst, assume that the lines $AD$ and $BC$ are not parallel and meet at $S$. Since $ABCD$ is cyclic, $\\triangle ASB$ and $\\triangle CSD$ are similar. Then, $AE / AB = CF / CD$ and $AE / CF = AB / CD = AS / CS$, so that $\\triangle ASE$ and $\\triangle CSF$ are also similar ($\\angle SAE = \\angle SCD$), and therefore, $\\angle DSE = \\angle CSF$.\n\n![](attached_image_1.png)\n\nBy similarity, we have\n$$\n\\frac{SE}{SF} = \\frac{SA}{SC} = \\frac{AB}{CD} = \\frac{PE}{PF}\n$$\nwhich means that $SP$ is the bisector of angle $S$ in $\\triangle FSE$. This implies that $\\angle ESP = \\angle FSP$ and hence $\\angle ASP = \\angle BSP$, so $SP$ is also the bisector of angle $S$ in $\\triangle ASB$. This means that $P$ is equidistant from the lines $AD$ and $BC$. Thus\n\n$$\n\\frac{[APD]}{[BPC]} = \\frac{AD}{BD}\n$$\n\nwhich is a constant (we use the notation $[ABC]$ for the area of $\\triangle ABC$).\n\nFor the second case, assume that $AD$ and $BC$ are parallel. Then $ABCD$ is an isosceles trapezoid with $AB = CD$, and we have $BE = DF$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. Then $ME = NF$ and $E$ and $F$ are equidistant from the line $MN$. Thus $P$, the midpoint of $EF$, lies on $MN$. Thus $P$ is equidistant from $AD$ and $BC$, and hence\n\n$$\n\\frac{[APD]}{[BPC]} = \\frac{AD}{BD}.\n$$\n\nwhich is a constant.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a cyclic quadrilateral (a quadrilateral which can be inscribed in a circle). Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE / EB = CF / FD$. Let $P$ be the point on the segment $EF$ such that $PE / PF = AB / CD$. Prove that the ratio between the areas of triangle $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThere are two cases to consider. First, assume that the lines $AD$ and $BC$ are not parallel and meet at $S$. Since $ABCD$ is cyclic, $\\triangle ASB$ and $\\triangle CSD$ are similar. Since $AE / AB = CF / CD$, then $AE / CF = AB / CD = AS / CS$, and $\\triangle ASE$ and $\\triangle CSF$ are also similar $(\\angle SAE = \\angle SCD)$. Therefore, $\\angle DSE = \\angle CSF$.\nBy similarity, we have\n$$\n\\frac{SE}{SF} = \\frac{SA}{SC} = \\frac{AB}{CD} = \\frac{PE}{PF},\n$$\nwhich means that $SP$ is the bisector of angle $S$ in $\\triangle FSE$. This implies that $\\angle ESP = \\angle FSP$ and hence $\\angle ASP = \\angle BSP$, so $SP$ is also the bisector of angle $S$ in $\\triangle ASB$. This means that $P$ is equidistant from the lines $AD$ and $BC$. Thus\n$$\n[APD]/[BPC] = AD/BC,\n$$\nwhich is a constant (we use the notation $[ABC]$ for the area of $\\triangle ABC$).\n\nFor the second case, assume that $AD$ and $BC$ are parallel. Then $ABCD$ is an isosceles trapezoid with $AB = CD$, and we have $BE = DF$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. Then $ME = NF$ and $E$ and $F$ are equidistant from the line $MN$. Thus $P$, the midpoint of $EF$, lies on $MN$. This implies that $P$ is equidistant from $AD$ and $BC$, and hence\n$$\n[APD]/[BPC] = AD/BC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that $1993^{1993} + 1994^{1994} + 1995^{1995} + 1996^{1996}$ is divisible by $10$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $5^{2} = 25$ ends in $5$ again, and $6^{2} = 36$ ends in $6$ again, no matter to what power we raise $5$ or $6$, the resulting numbers will end in $5$ or $6$, respectively. Thus, $1995^{1995}$ ends in $5$ and $1996^{1996}$ ends in $6$.\n\nFurther, $4^{1} = 4$, $4^{2} = 16$, $4^{3} = 64$, $4^{4} = 256$, etc., i.e., an odd power of $4$ ends in $4$, and an even power of $4$ ends in $6$ (why?). Hence, $1994^{1994}$ ends in $6$.\n\nSo, the last digit of $1994^{1994} + 1995^{1995} + 1996^{1996}$ is the same as the last digit of $6 + 5 + 6 = 17$, i.e., it ends in $7$.\n\nIt remains to show that $1993^{1993}$ ends in $3$ (so that $3 + 7 = 10$).\n\nIndeed, $3^{1} = 3$, $3^{2} = 9$, $3^{3} = 27$, $3^{4} = 81$, $3^{5} = 243$, and the pattern will be $3, 9, 7, 1, 3, 9, 7, 1, \\ldots$, etc. So we need to figure out into which of these slots the $1993$rd power of $3$ will fall. Equivalently, we need to find the remainder of $1993$ when divided by $4$ (the length of the pattern above is $4$). Now, $1993 = 4 \\cdot 498 + 1$: $3^{1993}$ will complete $498$ cycles of the pattern, and will end in $3$.\n\nThis shows that $1993^{1993}$ ends in $3$. Therefore, the last digit of the total sum is the last digit of $3 + 6 + 5 + 6 = 20$, $0$, which makes the sum divisible by $10$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A$ and $B$ be two different hospitals that treat exactly the same number of patients during a year. Each patient suffers from one of two diseases, $X$ or $Y$. Hospital $A$ cures a greater percentage of its patients than hospital $B$. Is it possible that hospital $B$ cures both a greater percentage of $X$-patients than $A$, and a greater percentage of $Y$-patients than $A$?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nThis is the well-known Simpson's Paradox: just make $B$ specialize in a riskier disease. For example, let $B$ treat 90 cancer patients and 10 acne patients, with respective cure rates of $50\\%$ and $100\\%$. Let $A$ treat 10 cancer and 90 acne patients, with cure rates of $0\\%$ and $70\\%$, respectively.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBildert works in a cubicle in an office which consists of 27 cubicles arranged in a $3 \\times 3 \\times 3$ cube. Any two cubicles sharing a wall have a connecting door on this wall; for example, the corner cubicles have exactly 3 doors, while the center cubicle has 6 doors: one on each wall, one on the floor, and one on the ceiling. If Bildert starts at the central cubicle, can he visit each of the other 26 cubicles exactly once (i.e. without revisiting any cubicles)?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nSolution I. The answer is no. Denote the central cubicle by $C$, and denote the vertex, edge and face cubicles by $V, E$ and $F$, respectively. The trip must start with $C$ and include every one of the $8 V$'s, $6 F$'s, and $12 E$'s. The sequence must begin with $C F E$. Each cubicle $V$ is adjacent only to $E$ cubicles, and each $F$ cubicle except for the very first one, is adjacent only to $E$ cubicles. This means that for the remaining $13~V$'s and $F$'s that follow the initial $C F E$, at least 12 new $E$'s are needed. This means that we need at least $13 E$'s, impossible.\n\n\nSolution II. Divide the cubicles into two subsets depending on the parity of the sums of coordinates of each cubicle. (Thus, in the above notation, one subset consists of cubicles $C$ and $E$'s, and the other subset consists of $F'$'s and $V'$'s.) Each move alternates between the two subsets. The starting subset has 13 cubicles, while the other subset has 14 cubicles - obviously we cannot keep alternating, because we'll run short of cubicles in the starting subset. Hence, Bildert cannot visit all cubicles without repetitions.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKing Arthur and his 100 knights are having a feast at a round table. Each person has a glass with white or red wine in front of him. Exactly at midnight, every person moves his glass in front his left neighbor if the glass has white wine, or in front of his right neighbor if the glass has red wine. It is known that there is at least one glass with red and one glass with white wine.\n\na. Prove that after midnight there will be at least one person without a glass of wine in front of him.\n\nb. If Lady Guinivera, the Queen, calls the King off the table before midnight, will the conclusion of part (a) still going to be correct?",
    "options": [],
    "answer": "a) There will be at least one person without a glass after the move. b) No; with an even number of people an alternating assignment of colors lets everyone still have a glass.",
    "solution": "Solution:\n\na. Suppose not. Label all seats as $1,2, \\ldots, 101$ by going around the table anticlockwise. If each person still has a glass after midnight, then no one has received glasses from both of his neighbors, i.e. any two people with exactly one person between must have the same color wine. Thus, the following seats must correspond to the same color wine: $1,3,5, \\ldots, 99,101,2,4, \\ldots, 98,100$. But these are all seats - this contradicts the hypothesis that there are glasses with red and glasses with white wine! Therefore, the supposition is wrong, and at least one person will be left without a glass after midnight.\n\nb. The conclusion won't be correct. For example, give white wine to all odd seated people $1,3,5, \\ldots, 99$, and red wine to all even seated people $2,4,6, \\ldots, 100$. After midnight the two groups will switch the colors of their wines.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c, d, e, f$ be positive integers, each at least $2$, whose sum is $S$. Prove that\n$$\na(a-1)+b(b-1)+c(c-1)+d(d-1)+e(e-1)+f(f-1) \\leq (S-10)(S-11)+10.\n$$\nWhen is equality achieved?",
    "options": [],
    "answer": "Equality holds if and only if five of the numbers are 2 and the sixth is S − 10.",
    "solution": "Solution:\nSolution I. Adding $-3S+24$ to both sides makes the inequality equivalent to\n$$\n(a-2)^2 + (b-2)^2 + (c-2)^2 + (d-2)^2 + (e-2)^2 + (f-2)^2 \\leq (S-12)^2.\n$$\nSubstituting $A = a-2$, $B = b-2$, etc., this is the same as\n$$\nA^2 + B^2 + C^2 + D^2 + E^2 + F^2 \\leq (A+B+C+D+E+F)^2.\n$$\nOn the left side we have the sum of squares of nonnegative numbers and on the right side we have the square of the sum. The latter is always larger except when all pairwise products $AB, AC, DF, \\ldots$ are zeros. This happens when all but one of $A, B, C, D, E, F$ are zero, correspondingly, when all but one of $a, b, c, d, e, f$ are $2$.\n\n\nSolution II. The given inequality can be written as\n$$\na^2 + b^2 + c^2 + d^2 + e^2 + f^2 \\leq (S-10)^2 + 20.\n$$\nNow, for any two numbers $x, y \\geq z$, we have\n$$\nx^2 + y^2 \\leq z^2 + (x+y-z)^2.\n$$\nIndeed, this is equivalent to\n$$\nx^2 - z^2 \\leq (x+y-z)^2 - y^2 \\Leftrightarrow (x-z)(x+z) \\leq (x-z)(x+2y-z) \\Leftrightarrow 0 \\leq 2(x-z)(y-z),\n$$\nand the last is true because $x \\geq z$ and $y \\geq z$. Note that (1) replaces the numbers $(x, y)$ by $(z, x+y-z)$ without changing the sum of the two numbers, but increases the sum of their squares. In the original problem, we do this for $a, b \\geq 2$: we replace $(a, b)$ by $(2, a+b-2)$:\n$$\na^2 + b^2 \\leq 2^2 + (a+b-2)^2.\n$$\nWe then do the same for $(a+b-2, c)$: replace them by $(2, a+b+c-4)$, and so on. In the end, we will have replaced five of the original numbers by $2$'s, and the last by $a+b+c+d+e+f-10 = S-10$.\n$$\n\\Rightarrow a^2 + b^2 + c^2 + d^2 + e^2 + f^2 \\leq 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + (S-10)^2 = 20 + (S-10)^2.\n$$\nEquality is achieved if and only if there are equalities each time we apply (1), i.e., five of the given numbers are $2$'s, and the remaining number is therefore $S-10$.\n\n\nSolution III. We first show the following inequality:\n$$\nx(x-1) + y(y-1) \\leq (x+y-2)(x+y-3) + 2, \\quad x, y \\geq 2\n$$\nConsider two complete graphs, with $x$ and $y$ vertices, respectively. (A complete graph has all possible edges drawn.) Thus we have $\\binom{x}{2}$ and $\\binom{y}{2}$ edges in the two graphs. If we glue the graphs together on an edge, we produce a new graph with $x+y-2$ vertices. Count edges: the original configuration had $\\binom{x}{2} + \\binom{y}{2}$ edges, while the new configuration has at most $\\binom{x+y-2}{2}$ edges. Since we lost an edge when we glued the two graphs together, we conclude that\n$$\n\\binom{x}{2} + \\binom{y}{2} \\leq \\binom{x+y-2}{2} + 1\n$$\nThis is equivalent to (2) after multiplying by $2$. Equality is attained if and only if the new graph is also complete, i.e., one of the original graphs must have been just an edge ($x=2$ or $y=2$). From here, apply consecutively (2) to the desired inequality. Again, maximum is attained if and only if five of the given numbers are $2$'s.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn the bases $AB$ and $CD$ of a trapezoid $ABCD$ draw two squares externally to $ABCD$. Let $O$ be the intersection point of the diagonals $AC$ and $BD$, and let $O_{1}$ and $O_{2}$ be the centers of the two squares. Prove that $O_{1}$, $O$ and $O_{2}$ lie on a line (i.e. they are collinear; see Figure.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe idea is to show that $\\triangle OCO_{2} \\sim \\triangle OAO_{1}$. Indeed, first notice that $\\triangle AO_{1}B \\sim \\triangle CO_{2}D$ — both are right isosceles triangles. Therefore, $AO_{1} : CO_{2} = AB : CD$. But $\\triangle AOB \\sim \\triangle COD$ ($AB \\parallel CD \\Rightarrow$ all three angles are the same), so $AB : CD = AO : CO$. This implies $AO_{1} : CO_{2} = AO : CO$. Further, $\\angle O_{1}AO = \\angle O_{1}AB + \\angle BAO = 45^{\\circ} + \\angle DCO = \\angle OCO_{2}$, and we finally conclude that $\\triangle OCO_{2} \\sim \\triangle OAO_{1}$.\n\nHence, $\\angle AOO_{1} = \\angle COO_{2}$. Since $AOC$ is a line, then $O_{2}OO_{1}$ is also a line.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe real positive numbers $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ satisfy the relation $a_{n+1}^{2}=a_{n}+1$ for all $n=1,2, \\ldots$ Prove that at least one of the $a_{i}$'s must be an irrational number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nArgue by contradiction. Suppose that for all $i$, $a_{i}=p_{i} / q_{i}$ for some relatively prime positive integers $p_{i}, q_{i}$. Substitute in the given relation:\n$$\n\\frac{p_{n+1}^{2}}{q_{n+1}^{2}}=\\frac{p_{n}+q_{n}}{q_{n}} \\Rightarrow p_{n+1}^{2} q_{n}=q_{n+1}^{2}\\left(p_{n}+q_{n}\\right)\n$$\nBut $p_{n+1}^{2}$ and $q_{n+1}^{2}$ are relatively prime, and $q_{n}$ and $q_{n}+p_{n}$ are also relatively prime (why?). This means that $p_{n+1}^{2}=p_{n}+q_{n}$ and $q_{n}=q_{n+1}^{2}$. In particular, $q_{n+1}=\\sqrt{q_{n}}=\\sqrt[4]{q_{n-1}}=\\cdots=\\sqrt[2 n]{q_{1}}$. Unless $q_{1}=1$, we will eventually run out of perfect squares in $q_{1}$ and $q_{n}$ will not be an integer, a contradiction. Thus, the only possibility is $q_{1}=1$, so that all $q_{n}=1$, and our numbers $a_{n}$ are after all integers.\n\nBut this is absurd! Indeed, $a_{n+1}=\\sqrt{a_{n}+1}<a_{n}$ (because $a_{n}+1<a_{n}^{2}$ for all integers $a_{n}>1$, and $a_{1}=1$ implies an immediate contradiction for $a_{2}=\\sqrt{2}$ is irrational), so that we obtain an infinite strictly decreasing sequence of positive integers, and there isn't simply such a thing! This shows that our supposition was wrong and at least one of the $a_{n}$'s will be irrational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the $O$-$E$ game, a round starts with player $A$ paying $c$ cents to player $B$. Then $A$ secretly arranges the numbers $1, 3, 5, 7, 9, 11, 13$ in some order as a sequence $a_{1}, a_{2}, \\ldots, a_{7}$, and $B$ secretly arranges $2, 4, 6, 8, 10, 12, 14$ as a sequence $b_{1}, b_{2}, \\ldots, b_{7}$. Finally, the players show their sequences and $B$ pays $A$ one cent for each $i$ in $X=\\{1,2,3,4,5,6,7\\}$ such that $a_{i}<b_{i}$. This finishes the round. What number $c$ would make the game fair? (The game is fair if the total payments by $A$ to $B$ equals the total payments by $B$ to $A$ after all possible distinct rounds are played exactly once.)",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nLet $k$ be in $X$. There are $7!\\,6!\\,(8-k)$ choices of the sequences $a_{1}, a_{2}, \\ldots, a_{7}$ and $b_{1}, b_{2}, \\ldots, b_{7}$ for which $2k-1$ is an $a_{j}$ with $a_{j}<b_{j}$. Indeed, the $a$'s can be any of the $7!$ permutations of the 7 odd integers; then $j$ is the subscript such that $a_{j}=2k-1$, and $b_{j}$ must be one of the $8-k$ numbers in $\\{2k, 2k+2, \\ldots, 14\\}$; the remaining 6 even integers can be arranged in $6!$ ways.\n\nThe total of the payments by $B$ to $A$ for the $(7!)^{2}$ possible rounds is then\n$$\n\\sum_{k=1}^{7} 7!\\,6!\\,(8-k) = 7!\\,6!\\,(7+6+\\cdots+1) = 7!\\,6!\\,28 = (7!)^{2} 4\n$$\n$A$ pays to $B$ a total of $(7!)^{2}c$; so $c=4$ makes the game fair.\nSolution:\n\nFor those who know that the sum of expected values (averages) is the expected value of the sum, we note that in the first spot, $1, 3, 5, 7, 9, 11, 13$ are equally likely, and hence the average payment is the average of the payments for each of these numbers, $7/7, 6/7, 5/7, 4/7, 3/7, 2/7, 1/7$. Hence in the first spot the average payment is\n$$\n\\frac{(7/7 + 6/7 + 5/7 + 4/7 + 3/7 + 2/7 + 1/7)}{7} = 4/7 \\text{ of a cent. }\n$$\nSince there are 7 spots, with an average payment of $4/7$ cent each, the total payment averages 4 cents; so $c=4$ makes the game fair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$\\triangle ABC$ is inscribed in a circle $k$ with center $O$ so that $\\angle ACB = 120^\\circ$.\n\na. If $H$ is the orthocenter of $\\triangle ABC$, prove that $A, B, O, H$ lie on a circle with center the midpoint of the arc $ACB$. (The orthocenter of $\\triangle ABC$ is the intersection point of its three altitudes.)\n\nb. If $G$ is the centroid of $\\triangle ABC$, and $I$ is the incenter of $\\triangle ABH$, prove that the points $O, G, I, H$ lie on a line. (The centroid of $\\triangle ABC$ is the intersection point of its three medians: a median connects a vertex of $\\triangle ABC$ with the midpoint of the opposite side; the incenter of $\\triangle ABC$ is the intersection of its three angle bisectors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $O_1$ be the midpoint of the arc $ACB$ and let $R$ be the radius of $k$.\n\n$\\triangle AOO_1$ and $\\triangle BOO_1$ are equilateral ($\\angle ACB = 120^\\circ \\Rightarrow \\angle AOB = 120^\\circ$). The segments $AB$ and $OO_1$ intersect each other in their midpoint, $D$. If line $AO$ intersects $k$ in point $P$, then $\\angle ABP = \\angle ACP = 90^\\circ$, i.e. $PB \\parallel CH$ and $PC \\parallel BH$, and thus $PBHC$ is a parallelogram. From here, $CH = PB$. Since $OD$ is a midsegment in $\\triangle ABP$, then $PB = 2OD = OO_1$ and $CH = OO_1 = R$. Again, $OCHO_1$ is a parallelogram; moreover, it is a rhombus for $OC = R$. Thus, $O_1H = R$ and the points $A, B, O, H$ lie on a circle $k_1$ with center $O_1$ and radius $R$.\n\nLook at the quadrilateral $CMHN$: it contains two right angles (at $M$ and $N$), and the angle at $C$ is $120^\\circ$, so the angle at $H$ is $60^\\circ$: $\\angle AHB = 60^\\circ$. But $\\angle AOB = 120^\\circ$ (as above), so $AHBO$ do lie on the same circle. Since $A, O, B$ lie on a circle with center $O_1$ (as above), $H$ is forced to lie on the same circle.\n\nb.\n\nIn $k_1$, $\\angle BO_1H$ is central, and $\\angle BAH$ is inscribed, so that $\\angle BO_1H = 2\\angle BAH$ and $\\angle BHO_1 = 90^\\circ - \\angle BAH = \\angle AHC$. In the rhombus $OCHO_1$ the diagonal $OH$ is the bisector of $\\angle CHO_1$. Hence, $\\angle AHO = \\angle AHC + \\angle CHO = \\angle BHO_1 + \\angle OHO_1 = \\angle BHO \\Rightarrow OH$ is the angle bisector also of $\\angle AHB$. This means that point $I$ lies on $OH$.\n\nLet $OH$ intersect the median $CD$ in point $G$. We will show that $G$ is the centroid of $\\triangle ABC$ by showing first that $CG = 2GD$. Indeed, if $E$ and $F$ are the midpoints of $CG$ and $HG$, then $EF$ is a midsegment in $\\triangle CHG$ with $EF = CH/2 = OD$ and $EF \\parallel OD$ (from (a)). Then $\\triangle EFG \\cong \\triangle DOG$, and $EG = GD$, $CG = 2EG = 2GD$. Thus, $G$ is the centroid of $\\triangle ABC$, and the points $I$ and $G$ lie on the line $OH$.\n\nFor those who know about the Euler line: points $O, G, H$ lie on the Euler line of $\\triangle ABC$ (we even know the ratio $OG : GH = 1 : 2$). Thus, it remains to show that $I$ lies on the line $OH$, or equivalently, that $OH$ is the angle bisector of $\\angle AHB$. Recall that $k_1$ was the circle described around $AHBO$ from part (a). Since $OA = OB$, the corresponding arcs $OA$ and $OB$ on $k_1$ are equal, and hence the inscribed angles are equal: $\\angle AHO = \\angle BHO$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $\\triangle ABC$ be inscribed in a circle $k$, and let $M$ be an arbitrary point on $k$ different from $A$, $B$, $C$. Prove that the feet of the three perpendiculars from $M$ to the sides of $\\triangle ABC$ are collinear. (Note: You may have to extend some sides to find these feet: see Figure.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIt will suffice to show that $\\angle AQR = \\angle CQP$ (compare with Problem 3.) Note that both quadrilaterals $MQAR$ and $MQPC$ are cyclic: $\\angle MRA = 90^{\\circ} = \\angle MQA$, and $\\angle MQC = 90^{\\circ} = \\angle MPC$. Using inscribed angles in these quadrilaterals, we obtain that the two angles in question equal correspondingly to:\n$$\n\\angle AQR = \\angle AMR, \\quad \\angle CQP = \\angle CMR.\n$$\nLook carefully at $\\triangle MAR$ and $\\triangle MCP$: both have one right angle, and we hope to show that two more angles are the same. Equivalently, we want to show that these two triangles are similar. Can we do that? Yes. Their third angles are the same: $\\angle MCP = \\angle MAR$ because $\\angle MAR = 180^{\\circ} - \\angle MAB = \\angle MCB$ from the inscribed quadrilateral $MCBA$.\n\nThus, we conclude that $\\triangle MAR \\sim \\triangle MCP$ (two same angles), so their third angles are also the same: $\\angle CMP = \\angle RMA$. Combining this with (1), we finally obtain $\\angle AQR = \\angle CQP$, and therefore $P, Q, R$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $a @ b = \\frac{a^{3} - b^{3}}{a - b}$, for how many real values of $a$ does $a @ 1 = 0$?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nIf $\\frac{a^{3} - 1}{a - 1} = 0$, then $a^{3} - 1 = 0$, or $(a - 1)(a^{2} + a + 1) = 0$.\n\nThus $a = 1$, which is an extraneous solution since that makes the denominator of the original expression $0$, or $a$ is a root of $a^{2} + a + 1$.\n\nBut this quadratic has no real roots, in particular its roots are $\\frac{-1 \\pm \\sqrt{-3}}{2}$.\n\nTherefore there are no such real values of $a$, so the answer is $\\mathbf{0}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPyramid $EARLY$ is placed in $(x, y, z)$ coordinates so that $E = (10, 10, 0)$, $A = (10, -10, 0)$, $R = (-10, -10, 0)$, $L = (-10, 10, 0)$, and $Y = (0, 0, 10)$. Tunnels are drilled through the pyramid in such a way that one can move from $(x, y, z)$ to any of the 9 points $(x, y, z-1)$, $(x \\pm 1, y, z-1)$, $(x, y \\pm 1, z-1)$, $(x \\pm 1, y \\pm 1, z-1)$. Sean starts at $Y$ and moves randomly down to the base of the pyramid, choosing each of the possible paths with probability $\\frac{1}{9}$ each time. What is the probability that he ends up at the point $(8, 9, 0)$?",
    "options": [],
    "answer": "550/9^10",
    "solution": "Solution:\n\nStart by figuring out the probabilities of ending up at each point on the way down the pyramid. Obviously we start at the top vertex with probability $1$, and each point on the next level down with probability $1/9$. Since each probability after $n$ steps will be some integer over $9^{n}$, we will look only at those numerators. The third level down has probabilities as shown below. Think of this as what you would see if you looked at the pyramid from above, and peeled off the top two layers.\n\n12321\n24642\n36963\n24642\n12321\n\nWhat we can observe here is not only the symmetry along vertical, horizontal, and diagonal axes, but also that each number is the product of the numbers at the ends of its row and column (e.g. $6 = 2 \\cdot 3$). This comes from the notion of independence of events, i.e. that if we east and then south, we end up in the same place as if we had moved south and then east. Since we are only looking for the probability of ending up at $(8, 9, 0)$, we need only know that this is true for the top two rows of the square of probabilities, which depend only on the top two rows of the previous layer. This will follow from the calculation of the top row of each square, which we can do via an algorithm similar to Pascal's triangle. In the diagram below, each element is the sum of the 3 above it.\n\n```\n1\n1 1 1\n1 2 3 3 1\n1 3 6 7 6 1\n1 4 10 16 19 16 10 4 1\n```\n\n151530455145301551\nNow observe that the first 3 numbers in row $n$, where the top is row $0$, are $1, n, \\frac{n(n+1)}{2}$. This fact is easily proved by induction on $n$, so the details are left to the reader. Now we can calculate the top two rows of each square via another induction argument, or by independence, to establish that the second row is always $n$ times the first row. Therefore the probability of ending up at the point $(8, 9, 0)$ is $\\frac{550}{9^{10}}$.\nSolution:\n\nAt each move, the $x$ and $y$ coordinates can each increase by $1$, decrease by $1$, or stay the same. The $y$ coordinate must increase $9$ times and stay the same $1$ time, the $x$ coordinate can either increase $8$ times and stay the same $1$ time or decrease $1$ time and increase $9$ times. Now we consider every possible case. First consider the cases where the $x$ coordinate decreases once. If the $x$ coordinate decreases while the $y$ coordinate increases, then we have $8$ moves that are the same and $2$ that are different, which can be done in $\\frac{10!}{8!} = 90$ ways. If the $x$ coordinate decreases while the $y$ coordinate stays the same, then we have $9$ moves that are the same and $1$ other, which can be done in $\\frac{10!}{9!} = 10$ ways. Now consider the cases where the $x$ coordinate stays the same twice. If the $y$ coordinate stays the same while the $x$ coordinate increases, then we have $7$ moves that are the same, $2$ that are the same, and $1$ other, which can be done in $\\frac{10!}{7!2!} = 360$ ways. If the $y$ coordinate stays the same while the $x$ coordinate stays the same, then we have $8$ moves that are the same and $2$ that are different, which can be done in $\\frac{10!}{8!} = 90$ ways. Therefore there are $360 + 90 + 90 + 10 = 550$ paths to $(8, 9, 0)$, out of $9^{10}$ possible paths to the bottom, so the probability of ending up at the point $(8, 9, 0)$ is $\\frac{550}{9^{10}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor what single digit $n$ does 91 divide the 9-digit number $12345 n 789$?",
    "options": [],
    "answer": "7",
    "solution": "Solution 1:\n\n$123450789$ leaves a remainder of $7$ when divided by $91$, and $1000$ leaves a remainder of $90$, or $-1$, so adding $7$ multiples of $1000$ will give us a multiple of $91$.\n\n\nSolution 2:\n\nFor those who don't like long division, there is a quicker way. First notice that $91 = 7 \\cdot 13$, and $7 \\cdot 11 \\cdot 13 = 1001$. Observe that $12345 n 789 = 123 \\cdot 1001000 + 45 n \\cdot 1001 - 123 \\cdot 1001 + 123 - 45 n + 789$. It follows that $91$ will divide $12345 n 789$ iff $91$ divides $123 - 45 n + 789 = 462 - n$. The number $462$ is divisible by $7$ and leaves a remainder of $7$ when divided by $13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlex is stuck on a platform floating over an abyss at $1\\ \\mathrm{ft} / \\mathrm{s}$. An evil physicist has arranged for the platform to fall in (taking Alex with it) after traveling $100\\ \\mathrm{ft}$. One minute after the platform was launched, Edward arrives with a second platform capable of floating all the way across the abyss. He calculates for $5$ seconds, then launches the second platform in such a way as to maximize the time that one end of Alex's platform is between the two ends of the new platform, thus giving Alex as much time as possible to switch. If both platforms are $5\\ \\mathrm{ft}$ long and move with constant velocity once launched, what is the speed of the second platform (in $\\mathrm{ft} / \\mathrm{s}$)?",
    "options": [],
    "answer": "3 ft/s",
    "solution": "Solution:\n\nThe slower the second platform is moving, the longer it will stay next to the first platform. However, it needs to be moving fast enough to reach the first platform before it's too late. Let $v$ be the velocity of the second platform. It starts $65$ feet behind the first platform, so it reaches the back of the first platform at $\\frac{60}{v-1}$ seconds, and passes the front at $\\frac{70}{v-1}$ seconds, so the time to switch is $\\frac{10}{v-1}$. Hence we want $v$ to be as small as possible while still allowing the switch before the first platform falls. Therefore the time to switch will be maximized if the back of the second platform lines up with the front of the first platform at the instant that the first platform has travelled $100\\ \\mathrm{ft}$, which occurs after $100$ seconds. Since the second platform is launched $65$ seconds later and has to travel $105$ feet, its speed is $105 / 35 = \\mathbf{3\\ ft} / \\mathrm{s}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all possible values of $\\frac{d}{a}$ where $a^{2}-6 a d+8 d^{2}=0$, $a \\neq 0$.",
    "options": [],
    "answer": "1/2, 1/4",
    "solution": "Solution:\n\nDividing $a^{2}-6 a d+8 d^{2}=0$ by $a^{2}$, we get $1-6 \\frac{d}{a}+8\\left(\\frac{d}{a}\\right)^{2}=0$.\n\nThe roots of this quadratic are $\\frac{1}{2}, \\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou are trapped in a room with only one exit, a long hallway with a series of doors and land mines. To get out you must open all the doors and disarm all the mines. In the room is a panel with 3 buttons, which conveniently contains an instruction manual. The red button arms a mine, the yellow button disarms two mines and closes a door, and the green button opens two doors. Initially 3 doors are closed and 3 mines are armed. The manual warns that attempting to disarm two mines or open two doors when only one is armed/closed will reset the system to its initial state. What is the minimum number of buttons you must push to get out?",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nClearly we do not want to reset the system at any time. After pressing the red button $r$ times, the yellow button $y$ times, and the green button $g$ times, there will be $3 + r - 2y$ armed mines and $3 + y - 2g$ closed doors, so we want the values of $r$, $y$, and $g$ that make both of these quantities $0$ while minimizing $r + y + g$.\n\nFrom the number of doors we see that $y$ must be odd, from the number of mines we see $y = (3 + r)/2 \\geq 3/2$, so $y \\geq 3$. Then $g = (3 + y)/2 \\geq 3$, and $r = 2y - 3 \\geq 3$, so $r + y + g \\geq 9$.\n\nCall the red, yellow, and green buttons $1$, $2$, and $3$ respectively for notational convenience, then a sequence of buttons that would get you out is $123123123$. Another possibility is $111222333$, and of course there are others. Therefore the answer is $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCarl and Bob can demolish a building in $6$ days, Anne and Bob can do it in $3$, Anne and Carl in $5$. How many days does it take all of them working together if Carl gets injured at the end of the first day and can't come back? Express your answer as a fraction in lowest terms.",
    "options": [],
    "answer": "59/20",
    "solution": "Solution:\n\nLet $a$ be the portion of the work that Anne does in one day, similarly $b$ for Bob and $c$ for Carl. Then what we are given is the system of equations $b + c = 1/6$, $a + b = 1/3$, and $a + c = 1/5$.\n\nThus in the first day they complete $a + b + c = \\frac{1}{2}(1/6 + 1/3 + 1/5) = 7/20$, leaving $13/20$ for Anne and Bob to complete.\n\nThis takes $\\frac{13/20}{1/3} = 39/20$ days, for a total of $\\frac{59}{20}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMatt has somewhere between 1000 and 2000 pieces of paper he's trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries $2,3,4,5,6,7$, and $8$ piles but ends up with one sheet left over each time. How many piles does he need?",
    "options": [],
    "answer": "41",
    "solution": "Solution:\n\nThe number of sheets will leave a remainder of $1$ when divided by the least common multiple of $2,3,4,5,6,7$, and $8$, which is $8 \\cdot 3 \\cdot 5 \\cdot 7 = 840$. Since the number of sheets is between $1000$ and $2000$, the only possibility is $1681$. The number of piles must be a divisor of $1681 = 41^{2}$, hence it must be $41$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $f(x)$ is a monic quartic polynomial such that $f(-1) = -1$, $f(2) = -4$, $f(-3) = -9$, and $f(4) = -16$, find $f(1)$.",
    "options": [],
    "answer": "23",
    "solution": "Solution:\n\nThe given data tells us that the roots of $f(x) + x^{2}$ are $-1$, $2$, $-3$, and $4$. Combining with the fact that $f$ is monic and quartic, we get\n$$\nf(x) + x^{2} = (x + 1)(x - 2)(x + 3)(x - 4).\n$$\nHence\n$$\nf(1) = (1 + 1)(1 - 2)(1 + 3)(1 - 4) - 1 = (2)(-1)(4)(-3) - 1 = 2 \\times (-1) \\times 4 \\times (-3) - 1 = 2 \\times (-1) = -2,\\ -2 \\times 4 = -8,\\ -8 \\times (-3) = 24,\\ 24 - 1 = 23.\n$$\nSo $f(1) = \\boxed{23}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways are there to cover a $3 \\times 8$ rectangle with 12 identical dominoes?",
    "options": [],
    "answer": "153",
    "solution": "Solution:\n\nTrivially there is 1 way to tile a $3 \\times 0$ rectangle, and it is not hard to see there are 3 ways to tile a $3 \\times 2$. Let $T_{n}$ be the number of tilings of a $3 \\times n$ rectangle, where $n$ is even. From the diagram below we see the recursion $T_{n}=3 T_{n-2}+2\\left(T_{n-4}+T_{n-6}+\\ldots+T_{2}+T_{0}\\right)$. Given that, we can just calculate $T_{4}=11, T_{6}=41$, and $T_{8}$ is $\\mathbf{153}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOne of the receipts for a math tournament showed that 72 identical trophies were purchased for $\\$-99.9-$, where the first and last digits were illegible. How much did each trophy cost?",
    "options": [],
    "answer": "$11.11",
    "solution": "Solution:\n\nThe price must be divisible by $8$ and $9$. Thus the last $3$ digits must be divisible by $8$, so the price ends with $992$, and the first digit must be $7$ to make the total divisible by $9$. $\\$799.92 / 72 = \\$11.11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the minimum possible value of the largest of $x y$, $1-x-y+x y$, and $x+y-2 x y$ if $0 \\leq x \\leq y \\leq 1$.",
    "options": [],
    "answer": "4/9",
    "solution": "Solution:\n\nI claim the answer is $\\frac{4}{9}$. Let $s = x + y$, $p = x y$, so $x$ and $y$ are $\\frac{s \\pm \\sqrt{s^{2} - 4p}}{2}$. Since $x$ and $y$ are real, $s^{2} - 4p \\geq 0$.\n\nIf one of the three quantities is less than or equal to $\\frac{1}{9}$, then at least one of the others is at least $\\frac{4}{9}$ by the pigeonhole principle since they add up to $1$.\n\nAssume that $s - 2p < \\frac{4}{9}$, then $s^{2} - 4p < (\\frac{4}{9} + 2p)^{2} - 4p$, and since the left side is non-negative we get\n$$\n0 \\leq p^{2} - \\frac{5}{9} p + \\frac{4}{81} = \\left(p - \\frac{1}{9}\\right)\\left(p - \\frac{4}{9}\\right).\n$$\nThis implies that either $p \\leq \\frac{1}{9}$ or $p \\geq \\frac{4}{9}$, and either way we're done.\n\nThis minimum is achieved if $x$ and $y$ are both $\\frac{1}{3}$, so the answer is $\\frac{4}{9}$, as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nStacy has $d$ dollars. She enters a mall with 10 shops and a lottery stall. First she goes to the lottery and her money is doubled, then she goes into the first shop and spends 1024 dollars. After that she alternates playing the lottery and getting her money doubled (Stacy always wins) then going into a new shop and spending $1024$. When she comes out of the last shop she has no money left. What is the minimum possible value of $d$?",
    "options": [],
    "answer": "1023",
    "solution": "Solution:\n\nWork backwards. Before going into the last shop she had $1024$, before the lottery she had $512$, then $1536$, $768$, $2304$, $1152$, and so on. We can easily prove by induction that if she ran out of money after $n$ shops, $0 \\leq n \\leq 10$, she must have started with $1024 - 2^{10-n}$ dollars. Therefore $d$ is $1023$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn unfair coin has the property that when flipped four times, it has the same probability of turning up 2 heads and 2 tails (in any order) as 3 heads and 1 tail (in any order). What is the probability of getting a head in any one flip?",
    "options": [],
    "answer": "3/5",
    "solution": "Solution:\n\nLet $p$ be the probability of getting a head in one flip. There are $6$ ways to get $2$ heads and $2$ tails, each with probability $p^{2}(1-p)^{2}$, and $4$ ways to get $3$ heads and $1$ tail, each with probability $p^{3}(1-p)$. We are given that\n$$\n6 p^{2}(1-p)^{2} = 4 p^{3}(1-p).\n$$\nClearly $p$ is not $0$ or $1$, so we can divide by $p^{2}(1-p)$ to get\n$$\n6(1-p) = 4p.\n$$\nTherefore $p$ is $\\frac{3}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou are given 16 pieces of paper numbered $16,15, \\ldots, 2,1$ in that order. You want to put them in the order $1,2, \\ldots, 15,16$ switching only two adjacent pieces of paper at a time. What is the minimum number of switches necessary?",
    "options": [],
    "answer": "120",
    "solution": "Solution:\n\nPiece $16$ has to move to the back $15$ times, piece $15$ has to move to the back $14$ times, $\\ldots$. Piece $2$ has to move to the back $1$ time, piece $1$ has to move to the back $0$ times. Since only one piece can move back in each switch, we must have at least $15+14+\\ldots+1=\\mathbf{120}$ switches.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMatt has somewhere between 1000 and 2000 pieces of paper he's trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries $2,3,4,5,6,7$, and $8$ piles but ends up with one sheet left over each time. How many piles does he need?",
    "options": [],
    "answer": "41",
    "solution": "Solution:\n\nThe number of sheets will leave a remainder of $1$ when divided by the least common multiple of $2,3,4,5,6,7$, and $8$, which is $8 \\cdot 3 \\cdot 5 \\cdot 7 = 840$. Since the number of sheets is between $1000$ and $2000$, the only possibility is $1681$. The number of piles must be a divisor of $1681 = 41^{2}$, hence it must be $41$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind an ordered pair $(a, b)$ of real numbers for which $x^{2}+a x+b$ has a non-real root whose cube is 343.",
    "options": [],
    "answer": "(7, 49)",
    "solution": "Solution:\n\nThe cube roots of $343$ are the roots of $x^{3}-343$, which is $(x-7)(x^{2}+7x+49)$. Therefore the ordered pair we want is $(7, 49)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $C$ be a circle with two diameters intersecting at an angle of $30$ degrees. A circle $S$ is tangent to both diameters and to $C$, and has radius $1$. Find the largest possible radius of $C$.",
    "options": [],
    "answer": "1 + sqrt(2) + sqrt(6)",
    "solution": "Solution:\n\nFor $C$ to be as large as possible we want $S$ to be as small as possible. It is not hard to see that this happens in the situation shown below. Then the radius of $C$ is $1 + \\csc 15 = \\mathbf{1} + \\sqrt{\\mathbf{2}} + \\sqrt{\\mathbf{6}}$. The computation of $\\sin 15$ can be done via the half angle formula.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs part of his effort to take over the world, Edward starts producing his own currency. As part of an effort to stop Edward, Alex works in the mint and produces 1 counterfeit coin for every 99 real ones. Alex isn't very good at this, so none of the counterfeit coins are the right weight. Since the mint is not perfect, each coin is weighed before leaving. If the coin is not the right weight, then it is sent to a lab for testing. The scale is accurate $95\\%$ of the time, $5\\%$ of all the coins minted are sent to the lab, and the lab's test is accurate $90\\%$ of the time. If the lab says a coin is counterfeit, what is the probability that it really is?",
    "options": [],
    "answer": "19/28",
    "solution": "Solution:\n\n$5\\%$ of the coins are sent to the lab, and only $95\\%$ of the coins are sent to the lab and counterfeit, so there is a $19\\%$ chance that a coin sent to the lab is counterfeit and an $81\\%$ chance that it is real. The lab could correctly detect a counterfeit coin or falsely accuse a real one of being counterfeit, so the probability that a coin the lab says is counterfeit really is counterfeit is\n$$\n\\frac{19/100 \\cdot 9/10}{19/100 \\cdot 9/10 + 81/100 \\cdot 1/10} = \\frac{19}{28}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all twice differentiable functions $f(x)$ such that $f^{\\prime \\prime}(x)=0$, $f(0)=19$, and $f(1)=99$.",
    "options": [],
    "answer": "f(x) = 80x + 19",
    "solution": "Solution:\nSince $f^{\\prime \\prime}(x)=0$ we must have $f(x)=a x + b$ for some real numbers $a, b$.\n\nThus $f(0) = b = 19$ and $f(1) = a + 19 = 99$, so $a = 80$.\n\nTherefore $f(x) = 80 x + 19$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A_{n}$ be the area outside a regular $n$-gon of side length $1$ but inside its circumscribed circle, let $B_{n}$ be the area inside the $n$-gon but outside its inscribed circle. Find the limit as $n$ tends to infinity of $\\frac{A_{n}}{B_{n}}$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nThe radius of the inscribed circle is $\\frac{1}{2} \\cot \\frac{\\pi}{n}$, the radius of the circumscribed circle is $\\frac{1}{2} \\csc \\frac{\\pi}{n}$, and the area of the $n$-gon is $\\frac{n}{4} \\cot \\frac{\\pi}{n}$. The diagram below should help you verify that these are correct.\n\n![](attached_image_1.png)\n\nThen $A_{n} = \\pi \\left( \\frac{1}{2} \\csc \\frac{\\pi}{n} \\right)^{2} - \\frac{n}{4} \\cot \\frac{\\pi}{n}$, and $B_{n} = \\frac{n}{4} \\cot \\frac{\\pi}{n} - \\pi \\left( \\frac{1}{2} \\cot \\frac{\\pi}{n} \\right)^{2}$, so\n$$\n\\frac{A_{n}}{B_{n}} = \\frac{\\pi \\left( \\csc \\frac{\\pi}{n} \\right)^{2} - n \\cot \\frac{\\pi}{n}}{n \\cot \\frac{\\pi}{n} - \\pi \\left( \\cot \\frac{\\pi}{n} \\right)^{2}}\n$$\nLet $s$ denote $\\sin \\frac{\\pi}{n}$ and $c$ denote $\\cos \\frac{\\pi}{n}$. Multiply numerator and denominator by $s^{2}$ to get\n$$\n\\frac{A_{n}}{B_{n}} \\sim \\frac{\\pi - n c s}{n c s - \\pi c^{2}}\n$$\nNow use Taylor series to replace $s$ by $\\frac{\\pi}{n} - \\frac{\\left( \\frac{\\pi}{n} \\right)^{3}}{6} + \\ldots$ and $c$ by $1 - \\frac{\\left( \\frac{\\pi}{n} \\right)^{2}}{2} + \\ldots$. By l'Hôpital's rule it will suffice to take just enough terms so that the highest power of $n$ in the numerator and denominator is determined, and that turns out to be $n^{-2}$ in each case. In particular, we get the limit\n$$\n\\frac{A_{n}}{B_{n}} = \\frac{\\pi - n \\frac{\\pi}{n} + n \\frac{2}{3} \\left( \\frac{\\pi}{n} \\right)^{3} + \\ldots}{n \\frac{\\pi}{n} - n \\frac{2}{3} \\left( \\frac{\\pi}{n} \\right)^{3} - \\pi + \\pi \\left( \\frac{\\pi}{n} \\right)^{2} + \\ldots} = \\frac{\\frac{2}{3} \\frac{\\pi^{3}}{n^{2}} + \\ldots}{\\frac{1}{3} \\frac{\\pi^{3}}{n^{2}} + \\ldots} \\rightarrow \\mathbf{2}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA rectangle has sides of length $\\sin x$ and $\\cos x$ for some $x$. What is the largest possible area of such a rectangle?",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\nWe wish to maximize $\\sin x \\cdot \\cos x = \\frac{1}{2} \\sin 2x$. But $\\sin 2x \\leq 1$, with equality holding for $x = \\pi/4$, so the maximum is $\\frac{1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind\n$$\n\\int_{-4 \\pi \\sqrt{2}}^{4 \\pi \\sqrt{2}}\\left(\\frac{\\sin x}{1+x^{4}}+1\\right) d x\n$$",
    "options": [],
    "answer": "8*pi*sqrt(2)",
    "solution": "Solution:\nThe function $\\frac{\\sin x}{1+x^{4}}$ is odd, so its integral over this interval is $0$. Thus we get the same answer if we just integrate $d x$, namely, $8 \\pi \\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$f$ is a continuous real-valued function such that $f(x+y)=f(x) f(y)$ for all real $x, y$. If $f(2)=5$, find $f(5)$.",
    "options": [],
    "answer": "25 sqrt(5)",
    "solution": "Solution:\n\nSince $f(n x)=f(x)^{n}$ for all integers $n$, $f(5)=f(1)^{5}$ and $f(2)=f(1)^{2}$, so $f(5)=f(2)^{5 / 2}=25 \\sqrt{5}$.\nSolution:\n\nMore generally, since $f(n x)=f(x)^{n}$ for all integers $n$, $f(1)=c=f(1 / n)^{n}$ for some constant $c$ and all integers $n$. Thus $f(k / n)=f(1 / n)^{k}=f(1)^{k / n}=c^{k / n}$ for all rational numbers $k / n$. By continuity, it follows that $f(x)=c^{x}$ for all real numbers $x$. Since $f(2)=5$, $c=\\sqrt{5}$, so $f(5)=25 \\sqrt{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = x + \\frac{1}{2x + \\frac{1}{2x + \\frac{1}{2x + \\cdots}}}$ for $x > 0$. Find $f(99) f'(99)$.",
    "options": [],
    "answer": "99",
    "solution": "Solution:\nAssume that the continued fraction converges (it does) so that $f(x)$ is well defined. Notice that\n$$\nf(x) - x = \\frac{1}{x + f(x)}\n$$\nso\n$$\nf(x)^2 - x^2 = 1,\n$$\nor\n$$\nf(x) = \\sqrt{1 + x^2}\n$$\n(we need the positive square root since $x > 0$).\n\nThus\n$$\nf'(x) = \\frac{x}{\\sqrt{1 + x^2}},\n$$\nso\n$$\nf(x) f'(x) = x.\n$$\nIn particular,\n$$\nf(99) f'(99) = 99.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate $\\frac{d}{d x}\\left(\\sin x-\\frac{4}{3} \\sin ^{3} x\\right)$ when $x=15$.",
    "options": [],
    "answer": "sqrt(2)/2",
    "solution": "Solution:\n\nOf course this problem can be done by brute force, differentiating and then using the half angle formula to find $\\sin$ and $\\cos$ of $15$, but there is a quicker way. $e^{i x}=\\cos x+i \\sin x$, so $\\sin (3 x)$ is the imaginary part of $(\\cos x+i \\sin x)^{3}$, which is $3 \\cos ^{2} x \\sin x-\\sin ^{3} x=3 \\sin x-4 \\sin ^{3} x$, so the expression we are differentiating is just $\\frac{1}{3} \\sin (3 x)$. Hence the derivative is $\\cos (3 x)$, and $\\cos 45=\\frac{\\sqrt{2}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf a right triangle is drawn in a semicircle of radius $1/2$ with one leg (not the hypotenuse) along the diameter, what is the triangle's maximum possible area?",
    "options": [],
    "answer": "3√3/32",
    "solution": "Solution:\n\nIt is easy to see that we will want one vertex of the triangle to be where the diameter meets the semicircle, so the diameter is divided into segments of length $x$ and $1 - x$, where $x$ is the length of the leg on the diameter. The other leg of the triangle will be the geometric mean of these two numbers, $\\sqrt{x(1 - x)}$. Therefore the area of the triangle is $\\frac{x \\sqrt{x(1 - x)}}{2}$, so it will be maximized when $\\frac{d}{dx}\\left(x^{3} - x^{4}\\right) = 3x^{2} - 4x^{3} = 0$, or $x = 3/4$. Therefore the maximum area is $\\frac{3 \\sqrt{3}}{32}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle is randomly chosen in a circle of radius $1$ in the sense that a point is randomly chosen for its center, then a radius is chosen at random so that the new circle is contained in the original circle. What is the probability that the new circle contains the center of the original circle?",
    "options": [],
    "answer": "1 - ln 2",
    "solution": "Solution:\nIf the center of the new circle is more than $1/2$ away from the center of the original circle then the new circle cannot possibly contain the center of the original one. Let $x$ be the distance between the centers (by symmetry this is all we need to consider), then for $0 \\leq x \\leq 1/2$ the probability of the new circle containing the center of the original one is $1-\\frac{x}{1-x}$. Hence we need to compute\n$$\n\\int_{0}^{1/2}\\left(1-\\frac{x}{1-x}\\right) dx = \\frac{1}{2} - \\int_{0}^{1/2} \\frac{x}{1-x} dx.\n$$\nTo evaluate the integral, we can integrate by parts to get\n$$\n-\\left. x \\ln (1-x) \\right|_{0}^{1/2} - \\int_{0}^{1/2} -\\ln (1-x) dx = -\\frac{1}{2} \\ln \\left(\\frac{1}{2}\\right) - [(1-x) \\ln (1-x) - (1-x)]_{0}^{1/2} = \\ln 2 - \\frac{1}{2}.\n$$\nAlternatively, we can use polynomial division to find that $\\frac{x}{1-x} = -1 + \\frac{1}{1-x}$, so\n$$\n\\int_{0}^{1/2} \\frac{x}{1-x} dx = \\int_{0}^{1/2} \\left(-1 + \\frac{1}{1-x}\\right) dx = \\ln 2 - \\frac{1}{2}.\n$$\nTherefore the probability is $\\frac{1}{2} - (\\ln 2 - \\frac{1}{2}) = 1 - \\ln 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat fraction of the Earth's volume lies above the $45$ degrees north parallel? You may assume the Earth is a perfect sphere. The volume in question is the smaller piece that we would get if the sphere were sliced into two pieces by a plane.",
    "options": [],
    "answer": "(8 - 5√2) / 16",
    "solution": "Solution:\nWithout loss of generality, look at a sphere of radius $1$ centered at the origin. If you like cartesian coordinates, then you can slice the sphere into discs with the same $z$ coordinate, which have radius $\\sqrt{1-z^{2}}$, so the region we are considering has volume\n$$\n\\int_{\\sqrt{2} / 2}^{1} \\pi\\left(1-z^{2}\\right) d z = \\pi\\left(\\frac{2}{3}-\\frac{5 \\sqrt{2}}{12}\\right)\n$$\nand dividing by $4 \\pi / 3$ we get\n$$\n\\frac{8-5 \\sqrt{2}}{16}\n$$\nSolution:\nFor those who prefer spherical coordinates, we can find the volume of the spherical cap plus a cone whose vertex is the center of the sphere. This region is where $r$ ranges from $0$ to $1$, $\\theta$ ranges from $0$ to $2 \\pi$, and $\\phi$ ranges from $0$ to $\\pi / 4$. Remembering we need to subtract off the volume of the cone, which has height $\\frac{1}{\\sqrt{2}}$ and a circular base of radius $\\frac{1}{\\sqrt{2}}$, then divide by $\\frac{4}{3} \\pi$ to get the fraction of the volume of the sphere, we find that we need to evaluate\n$$\n\\frac{\\int_{0}^{1} \\int_{0}^{2 \\pi} \\int_{0}^{\\pi / 4} r^{2} \\sin \\phi \\, d\\phi \\, d\\theta \\, dr - \\frac{1}{3} \\pi\\left(\\frac{1}{\\sqrt{2}}\\right)^{2} \\frac{1}{\\sqrt{2}}}{4 \\pi / 3}\n$$\nThe integral is just\n$$\n2 \\pi \\frac{1}{3}(-\\cos \\frac{\\pi}{4} + \\cos 0) = \\frac{4 \\pi - 2 \\pi \\sqrt{2}}{6}\n$$\nPutting this back in the answer and simplifying yields\n$$\n\\frac{8-5 \\sqrt{2}}{16}\n$$\nSolution:\nCavalieri's Principle states that if two solids have the same cross-sectional areas at every height, then they have the same volume. This principle is very familiar in the plane, where we know that the area of a triangle depends only on the base and height, not the precise position of the apex. To apply it to a sphere, consider a cylinder with radius $1$ and height $1$. Now cut out a cone whose base is the upper circle of the cylinder and whose apex is the center of the lower circle. Then at a height $z$ the area is $\\pi\\left(1^{2}-z^{2}\\right)$, exactly the same as for the upper hemisphere! The portion lying above the $45$ degrees north parallel is that which ranges from height $\\frac{1}{\\sqrt{2}}$ to $1$. The volume of the cylinder in this range is\n$$\n\\pi \\cdot 1^{2}\\left(1-\\frac{1}{\\sqrt{2}}\\right)\n$$\nThe volume of the cone in this range is the volume of the entire cone minus the portion from height $0$ to $\\frac{1}{\\sqrt{2}}$, i.e.,\n$$\n\\frac{1}{3} \\pi\\left(1^{2} \\cdot 1-\\left(\\frac{1}{\\sqrt{2}}\\right)^{2} \\frac{1}{\\sqrt{2}}\\right)\n$$\nTherefore the fraction of the Earth's volume that lies above the $45$ degrees north parallel is\n$$\n\\frac{\\pi\\left(1-\\frac{1}{\\sqrt{2}}\\right)-\\frac{1}{3} \\pi\\left(1-\\frac{1}{2 \\sqrt{2}}\\right)}{4 \\pi / 3} = \\frac{8 - 5 \\sqrt{2}}{16}\n$$\nSolution:\nAnother way to approach this problem is to integrate the function $\\sqrt{1-x^{2}-y^{2}}$ over the region $x^{2}+y^{2} \\leq \\frac{1}{\\sqrt{2}}$, subtract off a cylinder of radius and height $\\frac{1}{\\sqrt{2}}$, then divide by the volume of the sphere. One could also use the nontrivial fact that the surface area of a portion of a sphere of radius $r$ between two parallel planes separated by a distance $z$ is $2 \\pi r^{2} z$, so in particular the surface area of this cap is $2 \\pi\\left(1-\\frac{1}{\\sqrt{2}}\\right)$. Now, the ratio of the surface area of the cap to the surface area of the sphere is the same as the ratio of the volume of the cap plus the cone considered in Solution 2 to the volume of the whole sphere, so this allows us to avoid integration entirely.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA semicircle is inscribed in a semicircle of radius $2$ as shown. Find the radius of the smaller semicircle.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "sqrt(2)",
    "solution": "Solution:\nDraw a line from the center of the smaller semicircle to the center of the larger one, and a line from the center of the larger semicircle to one of the other points of intersection of the two semicircles. We now have a right triangle whose legs are both the radius of the smaller semicircle and whose hypotenuse is $2$, therefore the radius of the smaller semicircle is $\\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a cube with side length $6$, what is the volume of the tetrahedron formed by any vertex and the three vertices connected to that vertex by edges of the cube?",
    "options": [],
    "answer": "36",
    "solution": "Solution:\n\nWe have a tetrahedron whose base is half a face of the cube and whose height is the side length of the cube, so its volume is\n$$\n\\frac{1}{3} \\cdot \\left(\\frac{1}{2} \\cdot 6^{2}\\right) \\cdot 6 = \\mathbf{36}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cross-section of a river is a trapezoid with bases $10$ and $16$ and slanted sides of length $5$. At this section the water is flowing at $\\pi\\ \\mathrm{mph}$. A little ways downstream is a dam where the water flows through $4$ identical circular holes at $16\\ \\mathrm{mph}$. What is the radius of the holes?",
    "options": [],
    "answer": "sqrt(13)/4",
    "solution": "Solution:\nThe volume of water going through any cross-section of the river in an hour (assuming the cross-sections are parallel) is the area times the velocity. The trapezoid has height $4$, hence area $52$, so the volume of water going through at any hour is $52\\pi$. Let $r$ be the radius of the holes, then the total area is $4\\pi r^{2}$, so the volume of water is $64\\pi r^{2}$. Therefore $64\\pi r^{2} = 52\\pi$, so $r = \\frac{\\sqrt{13}}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sphere of radius $1$ is covered in ink and rolling around between concentric spheres of radii $3$ and $5$. If this process traces a region of area $1$ on the larger sphere, what is the area of the region traced on the smaller sphere?",
    "options": [],
    "answer": "9/25",
    "solution": "Solution:\nThe figure drawn on the smaller sphere is just a scaled down version of what was drawn on the larger sphere, so the ratio of the areas is the ratio of the surface area of the spheres. This is the same as the ratio of the squares of the radii, which is $\\frac{9}{25}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA dart is thrown at a square dartboard of side length $2$ so that it hits completely randomly. What is the probability that it hits closer to the center than any corner, but within a distance $1$ of a corner?",
    "options": [],
    "answer": "(π - 2)/4",
    "solution": "Solution:\n\nBy symmetry it will suffice to consider one quarter of the dartboard, which is a square of side length $1$. Therefore the probability is the area of the desired region in this square. The desired region is the part of the circle of radius $1$ centered at a corner that is closer to the opposite corner. The points closer to the opposite corner are those that are on the other side of the diagonal through the other two corners, so the desired region is a quarter of a circle of radius $1$ minus a right triangle with legs of length $1$. Therefore the area (and hence the probability) is $\\frac{\\pi-2}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSquares $ABKL$, $BCMN$, $CAOP$ are drawn externally on the sides of a triangle $ABC$. The line segments $KL$, $MN$, $OP$, when extended, form a triangle $A'B'C'$. Find the area of $A'B'C'$ if $ABC$ is an equilateral triangle of side length $2$.",
    "options": [],
    "answer": "12 + 13*sqrt(3)",
    "solution": "Solution:\n\nTriangle $ABC$ has area $\\sqrt{3}$, and each of the three squares has area $4$. The three remaining regions are congruent, so just consider the one that includes vertex $B$. Triangle $KBN$ has two sides of length $2$ and an angle of $120^{\\circ}$ between them, so bisecting that angle we get two halves of an equilateral triangle of side length $2$, so the area is again $\\sqrt{3}$. The remaining region is an equilateral triangle of side length $2\\sqrt{3}$, so its area is $(2\\sqrt{3})^{2} \\sqrt{3} / 4 = 3\\sqrt{3}$. Therefore the area of $A'B'C'$ is $\\sqrt{3} + 3 \\cdot 4 + 3 \\cdot \\sqrt{3} + 3 \\cdot 3\\sqrt{3} = \\mathbf{12} + \\mathbf{13}\\sqrt{\\mathbf{3}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA regular tetrahedron has two vertices on the body diagonal of a cube with side length $12$. The other two vertices lie on one of the face diagonals not intersecting that body diagonal. Find the side length of the tetrahedron.",
    "options": [],
    "answer": "4√3",
    "solution": "Solution:\nLet $ABCD$ be a tetrahedron of side $s$. We want to find the distance between two of its opposite sides. Let $E$ be the midpoint of $AD$, $F$ the midpoint of $BC$. Then $AE = s/2$, $AF = s\\sqrt{3}/2$, and $\\angle AEF = 90^{\\circ}$. So the distance between the two opposite sides is $EF = \\sqrt{AF^2 - AE^2} = \\sqrt{3s^2/4 - s^2/4} = s/\\sqrt{2}$.\n\nNow we find the distance between a body diagonal and a face diagonal of a cube of side $a$. Let $O$ be the center of the cube and $P$ be the midpoint of the face diagonal. Then the plane containing $P$ and the body diagonal is perpendicular to the face diagonal. So the distance between the body and face diagonals is the distance between $P$ and the body diagonal, which is $\\frac{a}{2} \\sqrt{\\frac{2}{3}}$ (the altitude from $P$ of right triangle $OPQ$, where $Q$ is the appropriate vertex of the cube). So now $\\frac{s}{\\sqrt{2}} = \\frac{a}{2} \\sqrt{\\frac{2}{3}}$, thus $s = a/\\sqrt{3} = 12/\\sqrt{3} = 4\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nStart with an angle of $60^{\\circ}$ and bisect it, then bisect the lower $30^{\\circ}$ angle, then the upper $15^{\\circ}$ angle, and so on, always alternating between the upper and lower of the previous two angles constructed. This process approaches a limiting line that divides the original $60^{\\circ}$ angle into two angles. Find the measure (degrees) of the smaller angle.",
    "options": [],
    "answer": "20°",
    "solution": "Solution:\n\nThe fraction of the original angle is $\\frac{1}{2} - \\frac{1}{4} + \\frac{1}{8} - \\cdots$. This is just a geometric series with first term $\\frac{1}{2}$ and ratio $-\\frac{1}{2}$, so the sum is $\\frac{1}{3}$. Therefore the smaller angle is $20^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$A, B, C, D$, and $E$ are relatively prime integers (i.e., have no single common factor) such that the polynomials $5A x^{4} + 4B x^{3} + 3C x^{2} + 2D x + E$ and $10A x^{3} + 6B x^{2} + 3C x + D$ together have 7 distinct integer roots. What are all possible values of $A$?\n\nYour team has been given a sealed envelope that contains a hint for this problem. If you open the envelope, the value of this problem decreases by 20 points. To get full credit, give the sealed envelope to the judge before presenting your solution.",
    "options": [],
    "answer": "-3, -1, 1, 3",
    "solution": "Solution:\n\nCall the negatives of the roots of the first polynomial $a, b, c, d$, and the negatives of the roots of the second polynomial $e, f, g$ (using the negatives avoids negative signs for the rest of the proof, thus preventing the possibility of dropping a sign). Then\n$$\n5A x^{4} + 4B x^{3} + 3C x^{2} + 2D x + E = 5A(x+a)(x+b)(x+c)(x+d)\n$$\nand\n$$\n10A x^{3} + 6B x^{2} + 3C x + D = 10A(x+e)(x+f)(x+g).\n$$\nThus\n$$\nB = \\frac{5}{4}A(a+b+c+d) = \\frac{5}{3}A(e+f+g),\n$$\n$$\nC = \\frac{5}{3}A(ab+ac+ad+bc+bd+cd) = \\frac{10}{3}A(ef+eg+fg),\n$$\n$$\nD = \\frac{5}{2}A(abc+abd+acd+bcd) = 10A(efg),\n$$\n$$\nE = 5A(abcd).\n$$\nFrom these equations we see that since all the variables are integers, it must be the case that $A \\mid B$, $A \\mid 3C$, $A \\mid D$, and $A \\mid E$, therefore $A, B, C, D$, and $E$ can only be relatively prime if $A$ is $\\pm 1$ or $\\pm 3$.\n\nNow we need numerical examples to show that both of these are possible. Without loss of generality let $g=0$, so $D=0$. Then $e$ and $f$ are\n$$\n\\frac{-3B \\pm \\sqrt{9B^{2} - 30AC}}{10A},\n$$\nso $9B^{2} - 30AC$ must be a perfect square. Let $B=0$ in the hope that solutions will still exist to this simplified problem.\n\nFirst let us try to find an example with $A=-1$, so we need $30C$ to be a perfect square. This first happens for $C=30$, and in that case $e$ and $f$ are $\\pm 3$. We need $a+b+c+d=0$, so let's try to look for $a=-b, c=-d$. This doesn't work for $C=30$ since $3C/5=18$ is not the sum of two distinct squares. For that we will need to try $C=5^{2} \\cdot 30 = 750$, for which we get $e, f = \\pm 15$, $a, b = \\pm 3$, and $c, d = \\pm 21$. Thus for $A=\\pm 1$ we can use $B=D=0$, $C=\\mp 750$, and $E=\\pm 19845$.\n\nSimilarly we can find $A=\\pm 3$, $B=D=0$, $C=\\mp 250$, and $E=\\pm 735$, for which $(a, b, c, d, e, f, g) = (1, -1, 7, -7, 5, -5, 0)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlex, Pei-Hsin, and Edward got together before the contest to send a mailing to all the invited schools. Pei-Hsin usually just stuffs the envelopes, but if Alex leaves the room she has to lick them as well and has a $25\\%$ chance of dying from an allergic reaction before he gets back. Licking the glue makes Edward a bit psychotic, so if Alex leaves the room there is a $20\\%$ chance that Edward will kill Pei-Hsin before she can start licking envelopes. Alex leaves the room and comes back to find Pei-Hsin dead. What is the probability that Edward was responsible?",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nThere are two possibilities: either Edward killed Pei-Hsin or the envelopes did. The envelope could only be responsible if Edward was not, so the chances of that would be $4/5 \\cdot 1/4 = 1/5$. This is the same as the probability that Edward killed her, so the events are equally likely and the answer is $\\mathbf{50\\%}$, or $\\mathbf{1/2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x$, $y$, and $z$ are distinct positive integers such that $x^{2} + y^{2} = z^{3}$, what is the smallest possible value of $x + y + z$.",
    "options": [],
    "answer": "18",
    "solution": "Solution:\nWithout loss of generality let $x > y$. We must have $z^{3}$ expressible as the sum of two squares, and this first happens when $z = 5$. Then $x$ and $y$ can be $10$ and $5$ or $11$ and $2$. If $z > 5$ then $z \\geq 10$ for $z^{3}$ to be a sum of two distinct squares, so $x^{2} > 500$, $x > 22$, so $x + y + z > 32$. Thus the smallest possible value of $x + y + z$ is $11 + 2 + 5 = \\mathbf{18}$.\nSolution:\nIf $z > 5$, then $z \\geq 6$, so $z^{3} \\geq 216$. Now $x^{2} + y^{2} \\geq 216$, so $x \\geq 11$ and $y \\geq 1$, thus $x + y + z \\geq 18$. Since $x = 11$, $y = 1$, $z = 6$ does not work, we must have $x + y + z > 18$, and the solution given is the best possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate $\\sum_{n=0}^{\\infty} \\frac{\\cos n \\theta}{2^{n}}$, where $\\cos \\theta=\\frac{1}{5}$.",
    "options": [],
    "answer": "6/7",
    "solution": "Solution:\n$\\cos n \\theta$ is the real part of $e^{i n \\theta}$, so the sum is the real part of $\\sum_{n=0}^{\\infty} \\frac{e^{i n \\theta}}{2^{n}}$.\n\nThis is a geometric series with initial term $1$ and ratio $\\frac{e^{i \\theta}}{2}$, so its sum is $\\frac{1}{1- e^{i \\theta}/2}$.\n\nWe are given $\\cos \\theta = \\frac{1}{5}$, so $\\sin \\theta = \\pm \\frac{2 \\sqrt{6}}{5}$.\n\nThus the sum is $\\frac{10}{10-1 \\mp 2 i \\sqrt{6}} = \\frac{90 \\pm 20 i \\sqrt{6}}{105}$, and the real part is $\\frac{6}{7}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $r$ be the radius of the inscribed circle of triangle $ABC$. Take a point $D$ on side $BC$, and let $r_{1}$ and $r_{2}$ be the inradii of triangles $ABD$ and $ACD$. Prove that $r$, $r_{1}$, and $r_{2}$ can always be the side lengths of a triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe must show that $r$, $r_{1}$, and $r_{2}$ satisfy the triangle inequality, i.e. that the sum of any two of them exceeds the third. Clearly $r$ is the largest of the three, so we need only verify that $r_{1} + r_{2} > r$.\n\nLet $K$ and $s$ be the area and semiperimeter of triangle $ABC$. Similarly define $K_{1}$, $K_{2}$, $s_{1}$, and $s_{2}$. Observe that $s$ is larger than $s_{1}$ or $s_{2}$ and that $K_{1} + K_{2} = K$. While these facts are almost trivial to verify, they must be stated.\n\nThen $r = K / s$, $r_{1} = K_{1} / s_{1}$, and $r_{2} = K_{2} / s_{2}$, so\n$$\n  r_{1} + r_{2} = \\frac{K_{1}}{s_{1}} + \\frac{K_{2}}{s_{2}} > \\frac{K_{1}}{s} + \\frac{K_{2}}{s} = \\frac{K}{s} = r.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou want to sort the numbers $54321$ using block moves. In other words, you can take any set of numbers that appear consecutively and put them back in at any spot as a block. For example, $653421 \\rightarrow 426531$ is a valid block move for 6 numbers. What is the minimum number of block moves necessary to get $12345$?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nHere is a sequence of 3 moves that works: $54321 \\rightarrow 32541 \\rightarrow 34125 \\rightarrow 12345$.\n\nBut how do we know we can't do it in 2 moves? From any position there are 20 possible permutations via block moves, 16 from moving a block of size 1 and 4 from moving a block of size 2. One could simply write the 20 permutations of $54321$ and the 20 permutations of $12345$ and try to see that they have nothing in common, which would suffice since the inverse of a block move is also a block move. A more clever method is to notice that if we could sort $54321$ in 2 moves then we could sort $4321$ in 2 moves as well by simply deleting the 5 from each step. But $4321$ has only 10 permutations from block moves, namely $3421,3241,3214,4231,4213,2431,4312,1432,4132$, and $2143$. The 10 permutations of $1234$ are $2134,2314,2341,1324,1342,3124,1243,4123,1423$, and $3412$. These two sets of permutations have nothing in common, thus it takes at least 3 moves to sort $4321$, and hence at least 3 moves to sort $54321$.\nSolution:\n\nThere is a more elegant way to show we need at least 3 moves. Given a permutation of $\\{1,2,3,4,5\\}$ (or any ordered set), define a descent to be an adjacent pair of numbers in the permutation such that the left number is greater than the right one. For example, $12345$, $34215$, and $54321$ have $0$, $2$, and $4$ descents, respectively. Any permutation obtained from $12345$ by one block move has (at most) one descent, at the left edge of the moved block. Similarly, any permutation obtained from $54321$ by one block move has (at least) three descents, so that we can't get from $54321$ to $12345$ by two block moves.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate $\\sum_{n=1}^{\\infty} \\frac{n^{5}}{n!}$.",
    "options": [],
    "answer": "52e",
    "solution": "Solution:\nWe start by noticing that $\\sum_{n=1}^{\\infty} \\frac{n}{n!}=\\sum_{n=1}^{\\infty} \\frac{1}{(n-1)!}=\\sum_{n=0}^{\\infty} \\frac{1}{n!}=e$.\n\nNext we see that $\\sum_{n=1}^{\\infty} \\frac{n^{2}}{n!}=\\sum_{n=1}^{\\infty} \\frac{n}{(n-1)!}=\\sum_{n=0}^{\\infty} \\frac{1+n}{n!}=\\sum_{n=0}^{\\infty} \\frac{1}{n!}+\\sum_{n=0}^{\\infty} \\frac{n}{n!}=e+e=2e$.\n\nLet $f(k)=\\sum_{n=1}^{\\infty} \\frac{n^{k}}{n!}$, then $\\sum_{n=1}^{\\infty} \\frac{n^{k}}{n!}=\\sum_{n=1}^{\\infty} \\frac{n^{k-1}}{(n-1)!}=\\sum_{n=0}^{\\infty} \\frac{(1+n)^{k-1}}{n!}$, so by the binomial theorem $f(k)=\\sum_{j=0}^{k-1}\\binom{k-1}{j} \\cdot f(j)$.\n\nArmed with this formula, we can easily compute\n\n$f(3)=f(0)+2 f(1)+f(2)=e+2e+2e=5e$,\n\n$f(4)=1 \\cdot e+3 \\cdot e+3 \\cdot 2e+1 \\cdot 5e=15e$,\n\nand\n\n$f(5)=1 \\cdot e+4 \\cdot e+6 \\cdot 2e+4 \\cdot 5e+1 \\cdot 15e=52e$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the smallest square-free composite number that can divide a number of the form $4242 \\ldots 42 \\pm 1$ ?",
    "options": [],
    "answer": "319",
    "solution": "Solution:\nIt is easy to see that such a number can never be divisible by $2,3,5$, or $7$. They can be divisible by $11$, the smallest example being $4242424241 = 11 \\cdot 547 \\cdot 705073$.\n\nWhat makes this problem hard is finding the next prime that can divide such a number. Let $T_{n} = \\sum_{i=0}^{n} 42 \\cdot 10^{2i}$. Then the numbers $T_{n}$ modulo a prime $p$ will always be periodic, since $T_{n} = 100 T_{n-1} + 42$, so we just need to compute one period and see if it contains $\\pm 1$.\n\nThus we find that modulo $13$ we get $3,4,0,3, \\ldots$, modulo $17$ we get $8,9,7,11,3,2,4,0,8, \\ldots$, modulo $19$ we get $4,5,10,16,8,6,15,3,0,4, \\ldots$, and modulo $23$ we get $19,10,7,6,21,3,20,18,2,12,0,19, \\ldots$, so none of these primes can ever divide $T_{n} \\pm 1$.\n\nBut $424241 = 29 \\cdot 14629$, so $29$ can also divide numbers of this form. Therefore the smallest composite number that can divide $T_{n} \\pm 1$ for some $n$ is $\\mathbf{319}$, and the smallest such $n$ is $83$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou are somewhere on a ladder with 5 rungs. You have a fair coin and an envelope that contains either a double-headed coin or a double-tailed coin, each with probability $1/2$. Every minute you flip a coin. If it lands heads you go up a rung, if it lands tails you go down a rung. If you move up from the top rung you win, if you move down from the bottom rung you lose. You can open the envelope at any time, but if you do then you must immediately flip that coin once, after which you can use it or the fair coin whenever you want. What is the best strategy (i.e. on what rung(s) should you open the envelope)?",
    "options": [],
    "answer": "Open the envelope only on the bottom rung.",
    "solution": "Solution:\nFirst consider the probability of winning if you never open the envelope. Let $q(n)$ be the probability of winning from the $n$\\text{th}$ rung with just the fair coin, then $q(n) = \\frac{q(n-1) + q(n+1)}{2}$, so it is not hard to calculate that $q(n) = n/6$.\n\nIf we open the envelope, then there's a $1/2$ chance that it is heads and we win, and a $1/2$ chance that it is tails and we end up one rung down with just the fair coin (obviously we keep using the double sided coin iff it is double headed).\n\nLet us start by analyzing rung 1. If we don't open the envelope, then we have a $1/2$ chance of losing and a $1/2$ chance of ending up on rung 2 with the envelope. If we do open the envelope, then we have a $1/2$ chance of losing and a $1/2$ chance of winning, which is a better outcome, so we should open the envelope on rung 1.\n\nNext we look at rung 5. If we don't open the envelope, then we have a $1/2$ chance of winning and a $1/2$ chance of moving down to rung 4 with the envelope. If we do open the envelope, then we have a $1/2$ chance of winning and a $1/2$ chance of moving down to rung 4 without the envelope. Let $p(n)$ be the probability of winning from rung $n$ if we are there with the envelope still unopened. Then clearly $p(n) \\geq q(n)$ for all $n$ if we're using optimal strategy, so we should not open the envelope on rung 5.\n\nNext we look at rung 4. If we open the envelope, then our chance of winning is $1/2 + q(3)/2 = 3/4$. If we don't, then our chance of winning is $p(5)/2 + p(3)/2$. We do know that $p(5) = 1/2 + p(4)/2$, but this is not enough to tell us what to do on rung 4.\n\nLooking at rung 3, we can open the envelope for a probability $1/2 + q(2)/2 = 2/3$ of winning, and we can not open the envelope for a probability $p(4)/2 + p(2)/2$ of winning.\n\nOn rung 2, we can open the envelope for a probability $1/2 + q(1)/2 = 7/12$ of winning, and we can not open the envelope for a probability $p(3)/2 + p(1)/2 = p(3)/2 + 1/4$ of winning.\n\nNow we can use all this information together for the complete answer. We know $p(2) \\geq 7/12$, therefore $p(3) \\geq p(4)/2 + 7/24$, and we know $p(4) \\geq p(5)/2 + p(3)/2 \\geq \\frac{1/2 + p(4)/2}{2} + p(3)/2 \\geq \\frac{1/2 + p(4)/2}{2} + \\frac{p(4)/2 + 7/24}{2}$. Isolating $p(4)$ in this inequality, we get $p(4) \\geq 1/2 + 7/24 > 3/4$, therefore we should not open the envelope on rung 4.\n\nNow from $p(4) = p(5)/2 + p(3)/2$ and $p(5) = 1/2 + p(4)/2$ we have $p(4) = \\frac{1/2 + p(4)/2}{2} + p(3)/2$, so $p(3) = 3p(4)/2 - 1/2 \\geq 11/16 > 2/3$, so we should not open the envelope on rung 3.\n\nNow $p(2) \\geq p(3)/2 + 1/4 \\geq 19/32 > 7/12$, so we should not open the envelope on rung 2.\n\nTherefore the best strategy is to open the envelope iff we are on the bottom rung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA combination lock has a 3 number combination, with each number an integer between $0$ and $39$ inclusive. Call the numbers $n_{1}$, $n_{2}$, and $n_{3}$. If you know that $n_{1}$ and $n_{3}$ leave the same remainder when divided by $4$, and $n_{2}$ and $n_{1}+2$ leave the same remainder when divided by $4$, how many possible combinations are there?",
    "options": [],
    "answer": "4000",
    "solution": "Solution:\nThere are $40$ choices for the last number, and for each of these we have $10$ choices for each of the first two numbers, thus giving us a total of $4000$ possible combinations. It is interesting to note that these restrictions are actually true for Master locks.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf 5 points are placed in the plane at lattice points (i.e. points $(x, y)$ where $x$ and $y$ are both integers) such that no three are collinear, then there are 10 triangles whose vertices are among these points. What is the minimum possible number of these triangles that have area greater than $1/2$?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nBy the pigeonhole principle, the 5 points cannot all be distinct modulo 2, so two of them must have a midpoint that is also a lattice point. This midpoint is not one of the 5 since no 3 are collinear. Pick's theorem states that the area of a polygon whose vertices are lattice points is $B/2 + I - 1$ where $B$ is the number of lattice points on the boundary and $I$ is the number in the interior. Thus those two points form the base of 3 triangles whose area will be greater than $1/2$ by Pick's theorem since there are 4 lattice points on the boundary.\n\nNow it also turns out that at least one of the triangles must contain a lattice point, thus giving us a fourth triangle with area greater than $1/2$. This is actually pretty easy to show with the aid of a picture or some visualization. Suppose we have 4 points and we're trying to find a 5th one so that no triangle will contain an interior lattice point. The 4 lattice points must form a quadrilateral of area 1, so in fact it is a parallelogram (think deeply about it). Draw the four sides, extending them throughout the plane. Each vertex is now the tip of an infinite triangular region of the plane, and if the 5th lattice point is chosen in that region then the triangle formed by the 5th point and the two vertices of the parallelogram adjacent to the one we are considering will form a triangle containing the vertex we are considering. But the part of the plane that isn't in one of these 4 regions contains no lattice points or else we could draw a parallelogram congruent to the first one with lattice point vertices and containing that lattice point, but that would violate Pick's theorem since the parallelogram has area 1. Therefore we must have a fourth triangle with area greater than $1/2$ (one must justify that this really is in addition to the 3 triangles we already knew we'd get).\n\nAn example that achieves this minimum is the points $(0,0), (1,0), (1,1), (2,1)$, and $(2,-1)$. Therefore the minimum possible number of these triangles that have area greater than $1/2$ is 4.\n\nA less trivial example that achieves the minimum is $(0,0), (1,1), (2,1), (3,2)$, and $(7,5)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCircles $C_{1}, C_{2}, C_{3}$ have radius $1$ and centers $O, P, Q$ respectively. $C_{1}$ and $C_{2}$ intersect at $A$, $C_{2}$ and $C_{3}$ intersect at $B$, $C_{3}$ and $C_{1}$ intersect at $C$, in such a way that $\\angle A P B = 60^{\\circ}$, $\\angle B Q C = 36^{\\circ}$, and $\\angle C O A = 72^{\\circ}$. Find angle $A B C$ (degrees).",
    "options": [],
    "answer": "90°",
    "solution": "Solution:\n\nUsing a little trig, we have $BC = 2 \\sin 18$, $AC = 2 \\sin 36$, and $AB = 2 \\sin 30$ (see left diagram). Call these $a, b$, and $c$, respectively. By the law of cosines, $b^{2} = a^{2} + c^{2} - 2 a c \\cos ABC$, therefore\n$$\n\\cos ABC = \\frac{\\sin^{2} 18 + \\sin^{2} 30 - \\sin^{2} 36}{2 \\sin 18 \\sin 30}.\n$$\nIn the right diagram below we let $x = 2 \\sin 18$ and see that $x + x^{2} = 1$, hence $\\sin 18 = \\frac{-1 + \\sqrt{5}}{4}$. Using whatever trig identities you prefer you can find that $\\sin^{2} 36 = \\frac{5 - \\sqrt{5}}{4}$, and of course $\\sin 30 = \\frac{1}{2}$. Now simplification yields $\\sin^{2} 18 + \\sin^{2} 30 - \\sin^{2} 36 = 0$, so $\\angle ABC = \\mathbf{90}^{\\circ}$.\n\nNote that this means that if a regular pentagon, hexagon, and decagon are inscribed in a circle, then we can take one side from each and form a right triangle.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA fair coin is flipped every second and the results are recorded with $1$ meaning heads and $0$ meaning tails. What is the probability that the sequence $10101$ occurs before the first occurrence of the sequence $010101$?",
    "options": [],
    "answer": "21/32",
    "solution": "Solution:\nCall it a win if we reach $10101$, a loss if we reach $010101$. Let $x$ be the probability of winning if the first flip is a $1$, let $y$ be the probability of winning if the first flip is a $0$. Then the probability of winning is $(x + y) / 2$ since the first flip is $1$ or $0$, each with probability $1/2$. If we ever get two $1$'s in a row, that is the same as starting with a $1$ as far as the probability is concerned. Similarly, if we get two $0$'s in a row, then we might as well have started with a single $0$.\n\n![](attached_image_1.png)\n\nFrom the tree of all possible sequences, shown above, in which the probability of moving along any particular line is $1/2$, we see that $x = x(1/2 + 1/8) + y(1/4 + 1/16) + 1/16$, and $y = x(1/4 + 1/16) + y(1/2 + 1/8 + 1/32)$. Solving these two equations in two unknowns we get $x = 11/16$ and $y = 5/8$. Therefore the probability that the sequence $10101$ occurs before the first occurrence of the sequence $010101$ is $21/32$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA ladder is leaning against a house with its lower end 15 feet from the house. When the lower end is pulled 9 feet farther from the house, the upper end slides 13 feet down. How long is the ladder (in feet)?",
    "options": [],
    "answer": "25",
    "solution": "Solution:\nOf course the house makes a right angle with the ground, so we can use the Pythagorean theorem. Let $x$ be the length of the ladder and $y$ be the original height at which it touched the house. Then we are given\n$$\nx^{2} = 15^{2} + y^{2} = 24^{2} + (y - 13)^{2}.\n$$\nIsolating $y$ in the second equation we get $y = 20$, thus $x$ is $25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many non-empty subsets of $\\{1,2,3,4,5,6,7,8\\}$ have exactly $k$ elements and do not contain the element $k$ for some $k=1,2, \\ldots, 8$.",
    "options": [],
    "answer": "127",
    "solution": "Solution:\nProbably the easiest way to do this problem is to count how many non-empty subsets of $\\{1,2, \\ldots, n\\}$ have $k$ elements and do contain the element $k$ for some $k$. The element $k$ must have $k-1$ other elements with it to be in a subset of $k$ elements, so there are $\\binom{n-1}{k-1}$ such subsets. Now $\\sum_{k=1}^{n} \\binom{n-1}{k-1} = (1+1)^{n-1} = 2^{n-1}$, so that is how many non-empty sets contain some $k$ and have $k$ elements. The set $\\{1,2, \\ldots, n\\}$ has $2^{n}$ subsets (each element either is or is not in a particular subset), one of which is the empty set, so the number of non-empty subsets of $\\{1,2,3,4,5,6,7,8\\}$ have exactly $k$ elements and do not contain the element $k$ for some $k$ is $2^{n} - 2^{n-1} - 1 = 2^{n-1} - 1$. In the case $n=8$, this yields $\\mathbf{127}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the equation $F O R T Y + T E N + T E N = S I X T Y$, where each of the ten letters represents a distinct digit from $0$ to $9$. Find all possible values of $SIXTY$.",
    "options": [],
    "answer": "31486",
    "solution": "Solution:\n\nSince $Y + N + N$ ends in $Y$, $N$ must be $0$ or $5$. But if $N = 5$ then $T + E + E + 1$ ends in $T$, which is impossible, so $N = 0$ and $E = 5$. Since $F \\neq S$ we must have $O = 9$, $R + T + T + 1 > 10$, and $S = F + 1$. Now $I \\neq 0$, so it must be that $I = 1$ and $R + T + T + 1 > 20$. Thus $R$ and $T$ are $6$ and $7$, $6$ and $8$, or $7$ and $8$ in some order. But $X$ can't be $0$ or $1$ since those are taken, and $X$ cannot be $3$ since $F$ and $S$ have to be consecutive, so it must be that $R + T + T + 1$ is $21$ or $23$. This is satisfied only for $R = 7$, $T = 8$, so $F = 2$, $S = 3$, and $Y = 6$. Thus $S I X T Y = \\mathbf{3 1 4 8 6}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $a$ and $b$ are randomly selected real numbers between $0$ and $1$, find the probability that the nearest integer to $\\frac{a-b}{a+b}$ is odd.",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n\nThe only reasonable way to do this problem is geometrically (yes, you can use integrals to find the areas of the triangles involved, but I don't consider that reasonable).\n\nFirst, let us find the points $(a, b)$ in the plane for which the nearest integer to $\\frac{a-b}{a+b}$ is $0$, i.e. $-\\frac{1}{2} \\leq \\frac{a-b}{a+b} \\leq \\frac{1}{2}$.\n\nTaking the inequalities one at a time, $-\\frac{1}{2} \\leq \\frac{a-b}{a+b}$ implies that $a+b \\geq 2(b-a)$, or $b \\leq 3a$, so these points must lie below the line $y=3x$.\n\nSimilarly, $\\frac{a-b}{a+b} \\leq \\frac{1}{2}$ implies that $(a, b)$ must lie above the line $y=\\frac{1}{3}x$.\n\nNow we can look for the points $(a, b)$ for which the nearest integer to $\\frac{a-b}{a+b}$ is $1$, i.e. $\\frac{1}{2} \\leq \\frac{a-b}{a+b} \\leq \\frac{3}{2}$, and we find that all points in the first quadrant that lie above the line $y=3x$ satisfy this inequality.\n\nSimilarly, the closest integer to $\\frac{a-b}{a+b}$ is $-1$ for all points in the first quadrant below the line $y=\\frac{1}{3}x$.\n\nFor $a$ and $b$ between $0$ and $1$, the locus of points $(a, b)$ for which the nearest integer to $\\frac{a-b}{a+b}$ is odd is two right triangles with legs of length $1$ and $\\frac{1}{3}$, so together they have area $\\frac{1}{3}$.\n\nThe locus of all points $(a, b)$ with $a$ and $b$ between $0$ and $1$ is a square of side length $1$, and thus has area $1$.\n\nTherefore, the probability that the nearest integer to $\\frac{a-b}{a+b}$ is odd is $\\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReduce the number $\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nObserve that $\\left(\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}\\right)^{3} = (2+\\sqrt{5}) - 3\\sqrt[3]{2+\\sqrt{5}} - 3\\sqrt[3]{2-\\sqrt{5}} + (2-\\sqrt{5}) = 4 - 3\\left(\\sqrt[3]{2+\\sqrt{5}} + \\sqrt[3]{2-\\sqrt{5}}\\right)$.\n\nHence $\\sqrt[3]{2+\\sqrt{5}} + \\sqrt[3]{2-\\sqrt{5}}$ is a root of the cubic $x^{3} + 3x - 4 = (x-1)(x^{2} + x + 4)$.\n\nThe roots of $x^{2} + x + 4$ are imaginary, so $\\sqrt[3]{2+\\sqrt{5}} + \\sqrt[3]{2-\\sqrt{5}} = \\mathbf{1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\frac{1}{1-x-x^{2}-x^{3}}=\\sum_{i=0}^{\\infty} a_{n} x^{n}$, for what positive integers $n$ does $a_{n-1}=n^{2}$?",
    "options": [],
    "answer": "1, 9",
    "solution": "Solution:\nMultiplying both sides by $1-x-x^{2}-x^{3}$ the right hand side becomes $a_{0}+(a_{1}-a_{0}) x+(a_{2}-a_{1}-a_{0}) x^{2}+\\ldots$, and setting coefficients of $x^{n}$ equal to each other we find that $a_{0}=1$, $a_{1}=1$, $a_{2}=2$, and $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$ for $n \\geq 3$. Thus the sequence of $a_{n}$'s starts $1,1,2,4,7,13,24,44,81,149,\\ldots$. So we now see that $a_{0}=1^{2}$ and $a_{8}=9^{2}$. What makes it impossible for this to happen again is that the sequence is growing exponentially. It will suffice to show that $a_{n}>1.5^{n}$ for $n>2$, since $n^{2}/(n-1)^{2}<1.5$ for $n \\geq 6$, thus when $a_{n-1}$ exceeds $n^{2}$ at $n=10$ there can be no more solutions to $a_{n-1}=n^{2}$. Observe that $a_{n}>1.5 a_{n-1}$ for $n=3,4,5$. By way of induction, assume it for $n-2$, $n-1$, and $n$, then $a_{n+1}=a_{n}+a_{n-1}+a_{n-2}>1.5^{n}+1.5^{n-1}+1.5^{n-2}=1.5^{n-2}(1+1.5+1.5^{2})>1.5^{n+1}$. Thus, by induction, $a_{n}>1.5^{n}$ for $n>2$, so the only solutions are $\\mathbf{1, 9}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all the roots of $\\left(x^{2}+3 x+2\\right)\\left(x^{2}-7 x+12\\right)\\left(x^{2}-2 x-1\\right)+24=0$.",
    "options": [],
    "answer": "0, 2, 1 - sqrt(6), 1 + sqrt(6), 1 - 2*sqrt(2), 1 + 2*sqrt(2)",
    "solution": "Solution:\nWe re-factor as $(x+1)(x-3)(x+2)(x-4)\\left(x^{2}-2 x-1\\right)+24$, or $\\left(x^{2}-2 x-3\\right)\\left(x^{2}-2 x-8\\right)\\left(x^{2}-2 x-1\\right)+24$, and this becomes $(y-4)(y-9)(y-2)+24$ where $y=(x-1)^{2}$.\n\nNow, $(y-4)(y-9)(y-2)+24=(y-8)(y-6)(y-1)$, so $y$ is $1, 6$, or $8$.\n\nThus the roots of the original polynomial are $0, 2, 1 \\pm \\sqrt{6}, 1 \\pm 2 \\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate $\\sum_{n=2}^{17} \\frac{n^{2}+n+1}{n^{4}+2 n^{3}-n^{2}-2 n}$.",
    "options": [],
    "answer": "592/969",
    "solution": "Solution:\n\nObserve that the denominator $n^{4}+2 n^{3}-n^{2}-2 n = n(n-1)(n+1)(n+2)$. Thus we can rewrite the fraction as\n$$\n\\frac{n^{2}-n+1}{n^{4}+2 n^{3}-n^{2}-2 n} = \\frac{a}{n-1} + \\frac{b}{n} + \\frac{c}{n+1} + \\frac{d}{n+2}\n$$\nfor some real numbers $a$, $b$, $c$, and $d$. This method is called partial fractions. Condensing the right hand side as a fraction over $n^{4}+2 n^{3}-n^{2}-2 n$ we get\n$$\nn^{2}-n+1 = a\\left(n^{3}+3 n^{2}+2 n\\right) + b\\left(n^{3}+2 n^{2}-n-2\\right) + c\\left(n^{3}+n^{2}-2 n\\right) + d\\left(n^{3}-n\\right).\n$$\nComparing coefficients of each power of $n$ we get\n$$\na + b + c + d = 0,\\\\\n3a + 2b + c = 2,\\\\\n2a - b - 2c - d = 2,\\\\\n-2b = 2.\n$$\nThis is a system of 4 equations in 4 variables, and its solution is $a=1/2$, $b=-1/2$, $c=1/2$, and $d=-1/2$.\n\nThus the summation becomes\n$$\n\\frac{1}{2}\\left(1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{5}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\frac{1}{6}+\\cdots+\\frac{1}{16}-\\frac{1}{17}+\\frac{1}{18}-\\frac{1}{19}\\right).\n$$\nNotice that almost everything cancels to leave us with\n$$\n\\frac{1}{2}\\left(1+\\frac{1}{3}-\\frac{1}{17}-\\frac{1}{19}\\right) = \\frac{592}{969}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nI have three opaque jars filled with coins. One contains only nickels; one contains only dimes; one contains some nickels and some dimes. Also, the three jars are labeled \"Nickels,\" \"Dimes,\" and \"Nickels and Dimes,\" but it is known that all of the jars are labeled incorrectly. You are allowed to choose one jar and look at one coin randomly drawn from it. How can you determine which jar is which?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince the jar labeled \"Nickels and Dimes\" is mislabeled, it contains either only nickels or only dimes. This suggests drawing a coin from this jar.\n\nIf the coin drawn is a nickel, then the \"Nickels and Dimes\" jar actually contains only nickels. Then, the \"Dimes\" jar (because it is also mislabeled) must contain nickels and dimes, and the \"Nickels\" jar contains only dimes. If, on the other hand, the coin drawn is a dime, then the \"Nickels and Dimes\" jar contains only dimes; this forces the \"Nickels\" jar to contain coins of both types, and the \"Dimes\" jar then contains nickels. Thus, either way, it is possible to identify all three jars correctly.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2000 distinct positive integers are written down, and it so happens that the product of any 3 different numbers from this list is a square. Prove that each one of them is a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $a$ be any number in the list; we wish to prove that $a$ is a square. Let $b$, $c$, $d$ be any three other elements of the list, all different from each other. Then, the given implies that there are integers $p$, $q$, $r$ with $a b c = p^{2}$, $a b d = q^{2}$, $a c d = r^{2}$. Multiplying these together gives $a^{3} b^{2} c^{2} d^{2} = p^{2} q^{2} r^{2}$; equivalently, \n$$\na = \\frac{p^{2} q^{2} r^{2}}{a^{2} b^{2} c^{2} d^{2}} = \\left(\\frac{p q r}{a b c d}\\right)^{2}.\n$$\nSo $a$ is the square of a rational number, but since it is an integer, we conclude that it is the square of an integer, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo sets of points in the coordinate plane are given: $\\{(-1,1),(-1,2), \\ldots,(-1,2000)\\}$ and $\\{(1,1),(1,2), \\ldots,(1,2000)\\}$. $2000$ line segments are drawn connecting these points so that each point in the first set is connected to exactly one point in the second set, and vice versa. Find, with proof, the sum of the $y$-intercepts of the segments.",
    "options": [],
    "answer": "2001000",
    "solution": "Solution:\n\nNote that, for any real numbers $a$ and $b$, the segment connecting $(-1, a)$ and $(1, b)$ has midpoint $\\left(0, \\frac{a+b}{2}\\right)$, so its $y$-intercept is $\\frac{a+b}{2}$. Now suppose that our given segments connect $(-1,1)$ to $\\left(1, y_{1}\\right)$, $(-1,2)$ to $\\left(1, y_{2}\\right)$, $(-1,3)$ to $\\left(1, y_{3}\\right)$, $\\ldots$, $(-1,2000)$ to $\\left(1, y_{2000}\\right)$. The sum of the $y$-intercepts is then\n\n$\\frac{1+y_{1}}{2}+\\frac{2+y_{2}}{2}+\\frac{3+y_{3}}{2}+\\cdots+\\frac{2000+y_{2000}}{2}=\\frac{1}{2}\\left[(1+2+3+\\cdots+2000)+\\left(y_{1}+y_{2}+y_{3}+\\cdots+y_{2000}\\right)\\right]$.\n\nBut it follows from the given that $y_{1}, y_{2}, \\ldots, y_{2000}$ are just some reordering of $1,2, \\ldots, 2000$, so our sum simplifies to\n$$\n\\frac{1}{2}[(1+2+3+\\cdots+2000)+(1+2+3+\\cdots+2000)]=1+2+3+\\cdots+2000\n$$\nBy the formula for the sum of an arithmetic progression, this equals $2000 \\cdot 2001 / 2 = 2,001,000$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDo there exist positive integers $x, y, z$ such that $28x + 30y + 31z = 365$? What is the largest positive integer $n$ such that the equation $28x + 30y + 31z = n$ cannot be solved in positive integers $x, y, z$?",
    "options": [],
    "answer": "Yes; for example x=1, y=4, z=7. The largest n is 370.",
    "solution": "Solution:\n\nA glance at a calendar confirms that $x = 1$, $y = 4$, $z = 7$ solves the first equation.\n\nFor the second part, we claim the maximum value of $n$ is $370$. To see that this $n$ cannot be obtained, suppose $28x + 30y + 31z = 370$ and consider remainders modulo $28$ to observe that $2y + 3z \\equiv 6 \\pmod{28}$. Also, $30(y + z) < 30y + 31z = 370 - 28x \\leq 342 \\Rightarrow y + z < 12$. Then $2y + 3z < 3(y + z) \\leq 33$ so that the only possible value for $2y + 3z$ is $6$ (from the congruence), but one easily sees that this has no solutions in positive integers.\n\nTo see that any greater value is obtainable, we use induction of step $28$. Observe that\n$$\n\\begin{gathered}\n28(10-t) + 30(2 + t) + 31(1) \\quad (t = 0, 1, \\ldots, 9) \\text{ covers the odd numbers } 371, 373, \\ldots, 389 \\\\\n28(10-t) + 30(1 + t) + 31(2) \\quad (t = 0, 1, \\ldots, 9) \\text{ covers the even numbers } 372, 374, \\ldots, 390 ; \\\\\n28(1) + 30(10-t) + 31(2 + t) \\quad (t = 1, \\ldots, 8) \\text{ covers the numbers } 391, 392, \\ldots, 398\n\\end{gathered}\n$$\nIf $n > 398$, then (either by the induction hypothesis or by one of the above base cases) we can write $n - 28 = 28x + 30y + 31z$ for positive integers $x, y, z$; then $n = 28(x + 1) + 30y + 31z$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven $n+1$ distinct integers from the set $\\{1,2, \\ldots, 2n\\}$ ($n \\geq 1$), prove that some two of them are relatively prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAs is well known, any two consecutive integers are relatively prime. (Proof: let $d$ be the greatest common divisor of $k$ and $k+1$ for some integer $k$; then, since $k+1$ and $k$ are both multiples of $d$, their difference, $1$, is also a multiple of $d$; so $d=1$.)\n\nNow partition the set $\\{1,2, \\ldots, 2n\\}$ into $n$ pairs $\\{1,2\\}, \\{3,4\\}, \\{5,6\\}, \\ldots, \\{2n-1,2n\\}$. Since we have $n+1$ integers but only $n$ pairs, the Pigeonhole Principle tells us that some two integers must belong to the same pair. But that means that they are consecutive, so they are relatively prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEquilateral triangle $A B C$ is inscribed in a circle. Let $D$ be the midpoint of $A B$, and let $E$ be the midpoint of $A C$. The ray $\\overrightarrow{D E}$ meets the circle at $P$. Prove that $D E^{2} = D P \\cdot P E$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf $Q$ is the other intersection of the circle with line $D E$, then one sees by symmetry that line $D E$ is parallel to $B C$ and $Q D = P E$. Then, by power of a point at $D$,\n$$\nP E \\cdot P D = Q D \\cdot P D = B D \\cdot A D = (A B / 2)^{2} = (B C / 2)^{2} = D E^{2}.\n$$\n(The last equality holds since $\\triangle A D E$ is the image of $\\triangle A B C$ under a homothety with ratio $1 / 2$.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that any integer greater than or equal to $7$ can be written as a sum of two relatively prime integers, both greater than $1$. (Two integers are relatively prime if they share no common positive divisor other than $1$. For example, $22$ and $15$ are relatively prime, and thus $37=22+15$ represents the number $37$ in the desired way.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst note that if $d$ divides two integers $a$ and $b$, then $d$ must also divide their difference $a-b$. Therefore, consecutive positive integers are always relatively prime (difference is $1$). Likewise, if $a$ and $b$ are both odd with a difference of $2$ or $4$, then $a$ and $b$ are relatively prime.\n\nNow let $n$ be greater than $7$.\n\n- If $n$ is odd, then $n$ has the form $n=2k+1$, where $k \\geq 3$ is an integer. Thus we write the sum $n=k+(k+1)$.\n\n- If $n$ is even, then $n$ has the form $n=2k$, where $k \\geq 4$ is an integer.\n\n    - If $k$ is even, we write $n=(k-1)+(k+1)$.\n    \n    - If $k$ is odd, we write $n=(k-2)+(k+2)$.\nSolution:\n\n(sketch): Recall that $(x, y)$ denotes the greatest common divisor of the natural numbers $x$ and $y$. To write $n$ as $a+b$ with $(a, b)=1$ and $a, b>1$ is the same as to find $a$ satisfying $(a, n-a)=1$ and $1<a<n-1$. But $(a, n-a)=(a, n)$, so for $n>2$, the number of such $a$ is $\\phi(n)-2$, where $\\phi(n)$ is the Euler $\\phi$-function counting integers $a$ between $1$ and $n-1$ inclusive with $(a, n)=1$. Then the problem reduces to showing that $\\phi(n)>2$ for $n \\geq 7$, and this can be done fairly easily by looking at the well-known formula for $\\phi(n)$:\n\nIf $n=p_{1}^{e_{1}} \\ldots p_{r}^{e_{r}}$ where the $p_{i}$ are distinct primes and $e_{i} \\geq 1$, then\n$$\n\\phi(n)=\\prod_{i=1}^{r} p_{i}^{e_{i}-1}\\left(p_{i}-1\\right)\n$$\nSolution:\n\nWe will prove the stronger statement that any $n \\geq 7$ can be expressed as $p+m$ where $p$ and $m$ are relatively prime integers greater than $1$, and $p$ is prime. If $p$ is any prime less than $n-1$, then $p>1$, $n-p>1$ and $p+(n-p)=n$. Moreover $p$ and $n-p$ will be relatively prime, unless $p$ divides $n-p$, or equivalently $p$ divides $n$. Therefore it suffices to show that there is at least one prime less than $n-1$ that does not divide $n$.\n\nSuppose not. Then $2 \\cdot 3 \\cdot 5 \\cdots \\ell$ divides $n$, where $\\ell$ is the largest prime less than $n-1$. In particular, $2 \\cdot 3 \\cdot 5 \\cdots \\ell \\leq n$. Consider the odd integer $k:=(3 \\cdot 5 \\cdots \\ell)-2$. Since $n \\geq 7$, $k \\geq 3 \\cdot 5-2>1$, and hence $k$ has an odd prime factor $q$. Then $q \\leq k \\leq n/2-2 < n-1$, so $q$ appears in the product $3 \\cdot 5 \\cdots \\ell = k+2$. Now $q$ divides $k+2$ as well as $k$, so $q$ divides $2$, contradicting the fact that $q$ is an odd prime.\n\nHence at least one of $2,3,5, \\ldots, \\ell$ does not divide $n$, and that prime may be used as $p$.\nSolution:\n\nAnother correct solution uses \"Bertrand's postulate\" that if $m>3$, there exists a prime $p$ with $m<p<2m-2$. (This was first proved by Chebyshev: see p. 373 of Hardy and Wright, An Introduction to the Theory of Numbers, 5th edition, Clarendon Press, 1979.) Applying this with $m=\\lceil n/2 \\rceil$, the smallest integer greater than or equal to $n/2$, produces a prime $p$ with $n/2 < p < n-1$. Such $p$ cannot divide $n$, since $p < n < 2p$. Hence as in the previous solution, the partition $n=p+(n-p)$ works.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMartha is a chicken who lives in a coop whose finitely many residents have a definite \"pecking order\": for every pair of distinct chickens, exactly one pecks the other. (However, it is not necessarily true that if $X$ pecks $Y$ and $Y$ pecks $Z$, then $X$ pecks $Z$.) A chicken $X$ is called a \"leader\" of the coop if every other chicken is pecked by $X$ or pecked by a chicken who is pecked by $X$. Prove that if no one in the coop pecks more chickens than Martha, then Martha is a leader.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose that no one pecks more chickens than $M$, yet $M$ is not a leader; we will seek a contradiction. Let $X_{1}, X_{2}, \\ldots, X_{n}$ be the chickens pecked by $M$. Since $M$ is not a leader, there is a chicken $N$ who is neither pecked by $M$ nor by any of the $X_{i}$. But it is given that of any two distinct chickens, one pecks the other; hence, $N$ pecks all of $M, X_{1}, X_{2}, \\ldots, X_{n}$. Thus, we have found at least $n+1$ chickens pecked by $N$. But $M$ only pecks $n$ chickens. Since we assumed nobody pecked more chickens than $M$, we have a contradiction, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree grasshoppers are on a straight line. Every second one grasshopper jumps. It jumps across one (but not across two) of the other grasshoppers. Prove that after 1999 seconds the grasshoppers cannot be in the initial position.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume that the grasshoppers are initially in the order $(A, B, C)$ on the line. Call $(A, B, C)$, $(B, C, A)$, $(C, A, B)$ \"even\" orders and $(A, C, B)$, $(B, A, C)$, $(C, B, A)$ \"odd\" orders. We claim that if the grasshoppers are in an even order, they will be in an odd order the next second, and vice versa. This can be proved by simply checking all the possible situations. For example, if we start from the even order $(A, B, C)$, then either $A$ jumps across $B$, $B$ jumps across $A$, $B$ jumps across $C$, or $C$ jumps across $B$. These yield the respective orders $(B, A, C)$, $(B, A, C)$, $(A, C, B)$, $(A, C, B)$, which are odd. A similar analysis starting from each of the other five positions confirms that we switch parity with every jump.\n\nNow we claim that after an even number of seconds, the grasshoppers are in an even order, and after an odd number of seconds, they are in an odd order. This is a simple induction: after $n=0$ seconds, they are in an order defined to be even. If $n>0$, suppose $n$ is even. Then $n-1$ is odd, so the grasshoppers are in an odd order after $n-1$ seconds (by induction) and will be in an even order after $n$ seconds. Likewise, if $n$ is odd, then the grasshoppers are in an even order after $n-1$ seconds and therefore in an odd order after $n$ seconds. This completes the induction step.\n\nBut this means that after 1999 seconds, the grasshoppers are in an odd order; since they initially were in an even order, they cannot be in the same position.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x_{1} = x_{2} = 1$ and $x_{n+2} = 2 x_{n+1} + 8 x_{n} - 1$ for $n \\geq 1$. Prove that $x_{n}$ is a square for all $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $y_{1} = y_{2} = 1$ and $y_{n+2} = y_{n+1} + 2 y_{n}$ for $n \\geq 1$. Clearly $y_{n}$ is always an integer; we shall show inductively that $x_{n} = y_{n}^{2}$.\n\nFirst we demonstrate that $y_{n+1} - 2 y_{n} = \\pm 1$ for all $n$ by induction. This holds for $n = 1$, and if it holds for some $n$, then we have\n$$\ny_{n+2} - 2 y_{n+1} = (y_{n+1} + 2 y_{n}) - 2 y_{n+1} = - (y_{n+1} - 2 y_{n}) = \\pm 1\n$$\n(of opposite sign from $y_{n+1} - 2 y_{n}$), so it holds for $n+1$.\n\nNow to see that $x_{n} = y_{n}^{2}$, observe that this is true for $n = 1, 2$; if it holds for $n$ and $n+1$, we have\n$$\ny_{n+2}^{2} = (y_{n+1} + 2 y_{n})^{2} = (y_{n+1} + 2 y_{n})^{2} + (y_{n+1} - 2 y_{n})^{2} - 1\n$$\n(since we know the second term equals $1$)\n$$\n= (y_{n+1}^{2} + 4 y_{n} y_{n+1} + 4 y_{n}^{2}) + (y_{n+1}^{2} - 4 y_{n} y_{n+1} + 4 y_{n}^{2}) - 1\n$$\n$$\n= 2 y_{n+1}^{2} + 8 y_{n}^{2} - 1 = 2 x_{n+1} + 8 x_{n} - 1 = x_{n+2}\n$$\ncompleting the induction step and the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn $n \\times n$ matrix of integers is called \"golden\" if, for every row and every column, their union contains all of the numbers $1,2,3, \\ldots, 2 n-1$. Find all golden matrices (of all sizes).",
    "options": [],
    "answer": "Only the 1x1 matrix with the single entry 1.",
    "solution": "Solution:\n\nThe $1 \\times 1$ matrix whose only entry is $1$ is certainly golden; we claim there are no others. To see this, note that each of the numbers $1,2, \\ldots, 2 n-1$ must appear at least $n$ times in the matrix. (Proof: Suppose some number $k$ appears less than $n$ times. Because there are $n$ columns, some column does not contain $k$. Likewise, some row does not contain $k$; then the union of this row and this column fails to contain $k$, contrary to the given). So we have at least $n(2 n-1)$ entries in the matrix. But an $n \\times n$ matrix has $n^{2}$ entries, so $n^{2} \\geq n(2 n-1) \\Rightarrow n \\geq 2 n-1 \\Rightarrow 1 \\geq n$, and only the $1 \\times 1$ matrix can be golden.\nSolution:\n\nAs before, the $1 \\times 1$ matrix consisting only of a $1$ is golden. Now, note that the union of any row and any column in an $n \\times n$ golden matrix contains exactly $2 n-1$ entries, so each of the numbers $1,2, \\ldots, 2 n-1$ must appear exactly once in this union. Therefore, there is no room in the matrix for any numbers other than $1,2, \\ldots, 2 n-1$. Moreover, if any of these entries appeared twice, say once at the intersection of the $i$th row and $j$th column and once at the $i'$th row and $j'$th column, then the union of the $i$th row and $j'$th column would contain this number twice, not leaving enough room for the remaining $2 n-2$ numbers. So each number can appear only once in the matrix, meaning it can have only $2 n-1$ entries. Thus $n^{2}=2 n-1 \\Rightarrow n^{2}-2 n+1=0 \\Rightarrow n=1$, and only the $1 \\times 1$ matrix is golden.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 46527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f$ from the set $\\mathbf{R}$ of real numbers to itself such that $f(x y+1)=x f(y)+2$ for all $x, y \\in \\mathbf{R}$.",
    "options": [],
    "answer": "f(x) = 2x",
    "solution": "Solution:\nThe only such function is $f(x)=2 x$. First, we note that this function really is a solution of the equation, since $2(x y+1)=2 x y+2=x(2 y)+2$ for all $x, y$.\n\nNow let $f$ be any function satisfying the equation. First put $x=y=0$; the equation gives $f(0 \\cdot 0+1)=0 f(0)+2$ or $f(1)=2$. Now suppose $y=1$ (and $x$ may be anything); we have $f(1 x+1)=x f(1)+2$ or $f(x+1)=2 x+2$ for all $x \\in \\mathbf{R}$. Equivalently, $f(x)=f((x-1)+1)=2(x-1)+2=2 x$ for all $x$, so we have verified that this is the only function satisfying the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p(x)$ be a polynomial of degree exactly $3$, with real coefficients. For each real number $a$, let $q_{a}(x)$ be the unique polynomial of degree $2$ or less such that $p(x)-q_{a}(x)$ is divisible by $(x-a)^{3}$. Prove that for $a \\neq b$, the graphs of $q_{a}(x)$ and $q_{b}(x)$ do not intersect. Can you find a generalization to polynomials $p(x)$ of odd degree?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe prove the general case: if $p$ has real coefficients and odd degree $d$, let $q_{a}(x)$ be the polynomial of degree at most $d-1$ such that $p(x)-q_{a}(x)$ is divisible by $(x-a)^{d}$, i.e. $q_{a}(x)$ is the remainder upon dividing $p(x)$ by $(x-a)^{d}$. We will show, for $a \\neq b$, that $q_{a}(x) \\neq q_{b}(x)$ whenever $x$ is a real number.\n\nLet $c$ be the leading coefficient of $p$, and consider any $a$. Since $q_{a}$ has degree less than $d$, $p(x)-q_{a}(x)$ must also have leading coefficient $c$ and therefore be equal to $c(x-a)^{d}$, by consideration of leading coefficients and degrees. Now suppose, for some real $x$, that $q_{a}(x)=q_{b}(x)$. Then $p(x)-q_{a}(x)=p(x)-q_{b}(x)$, i.e. $c(x-a)^{d}=c(x-b)^{d}$, giving $(x-a)^{d}=(x-b)^{d}$ since $c$ is nonzero. But since $d$ is odd, we can take real $d$th roots uniquely, so that $x-a=x-b$ and $a=b$. Thus, the graphs of $q_{a}$ and $q_{b}$ cannot intersect unless $a=b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDo there exist three different prime numbers such that the sum of any two of them is a square?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. We work modulo $4$ (i.e., we look at remainders on division by $4$). Note that any square is $\\equiv 0$ or $1 \\bmod 4$. Indeed, if $n=2k$ is even, $n^{2}=4k^{2} \\equiv 0$, while if $n=2k+1$ is odd, $n^{2}=4k^{2}+4k+1 \\equiv 1$.\n\nNow suppose some three such primes exist. If they are all odd, then each of them is $\\equiv 1$ or $3 \\pmod{4}$; some two must then have the same value modulo $4$ (both $1$ or both $3$), and then their sum is $\\equiv 2 \\pmod{4}$, so it cannot be a square.\n\nThus, the primes are not all odd, so one of them is $2$. Since they are all different, the other two are odd. If these two are both $\\equiv 3$, then again their sum is $\\equiv 2$ and cannot be a square. So one of them is $\\equiv 1$, but then this prime plus $2$ is congruent to $3$, so that sum cannot be a square. So, once again, we can find two of the three primes whose sum is not a square.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle and $D, E, F, G, H, I, J$ points on sides $BC, CA, AB, BC, CA, AB, BC$ respectively, such that $EF$ and $HI$ are parallel to $BC$; $IJ$ and $FG$ are parallel to $CA$; and $DE, GH$ are parallel to $AB$. Prove that $D = J$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst we prove the following lemma: If $PQR$ is a triangle with points $S, T$ on sides $PQ, QR$ such that $ST$ is parallel to $RP$, then (a) $SQ / PQ = TQ / RQ$ and (b) $PS / PQ = RT / RQ$.\n\nProof: First, observe that triangles $PQR, SQT$ are similar (because parallel lines imply equal angles). Equation (a) then follows because the ratios of sides in similar triangles are equal. Manipulating this equation gives\n$$\n\\frac{SQ}{PQ} = \\frac{TQ}{RQ} \\Rightarrow 1 - \\frac{SQ}{PQ} = 1 - \\frac{TQ}{RQ} \\Rightarrow \\frac{PQ - SQ}{PQ} = \\frac{RQ - TQ}{RQ} \\Rightarrow \\frac{PS}{PQ} = \\frac{RT}{RQ}\n$$\nwhich is equation (b).\n\nNow we simply apply this lemma repeatedly. We obtain\n$$\n\\frac{BD}{BC} = \\frac{AE}{AC} \\text{ (by part (b), using line $DE$)} = \\frac{AF}{AB} \\text{ (part (a), line $EF$)} = \\frac{CG}{CB} \\text{ ((b), line $FG$)}\n$$\n$$\n= \\frac{CH}{CA} \\text{ ((a), line $GH$)} = \\frac{BI}{BA} \\text{ ((b), line $HI$)} = \\frac{BJ}{BC} \\text{ ((a), line $IJ$)}.\n$$\nThus $BD / BC = BJ / BC$, or $BD = BJ$. But there is only one point on segment $BC$ at distance $BD$ from $B$; thus we conclude $D = J$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn equilateral triangle in the coordinate plane has vertices $(a, b)$, $(c, d)$, $(e, f)$. Prove that $a, b, c, d, e, f$ cannot all be integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose otherwise. If the triangle has side length $s$, then $s^{2} = (a-c)^{2} + (b-d)^{2}$ is a positive integer, so we may choose the triangle for which $s^{2}$ is as small as possible; this makes $s$ as small as possible. We may further assume that $e = f = 0$, since otherwise this assumption can be made true by translating the triangle by the vector $(-e, -f)$, an operation which does not change the triangle's side length. Then we have\n$$\na^{2} + b^{2} = c^{2} + d^{2} = (a-c)^{2} + (b-d)^{2} = s^{2}.\n$$\nWe consider this chain of equalities modulo $4$, i.e., we look at remainders on dividing by $4$. Note that $n^{2} \\equiv 0$ modulo $4$ if $n$ is even (since $n = 2k \\Rightarrow n^{2} = 4k^{2}$), and $n^{2} \\equiv 1$ when $n$ is odd ($n = 2k + 1 \\Rightarrow n^{2} = 4(k^{2} + k) + 1$). So a sum of two squares is either $0$, $1$, or $2$ modulo $4$, according as the two squares are both even, one even and one odd, or both odd, respectively.\n\nIf $s^{2} \\equiv 0 \\bmod 4$, then the above equation forces $a, b, c, d$ all to be even. Then scaling our triangle by a factor of $1/2$ gives us an equilateral triangle with vertices $(a/2, b/2)$, $(c/2, d/2)$, $(0, 0)$. These coordinates are still integers, and the new triangle is smaller than the previous one, contradicting our assumption of minimality.\n\nIf $s^{2} \\equiv 1 \\bmod 4$, then each of the pairs $\\{a, b\\}$, $\\{c, d\\}$, $\\{a-c, b-d\\}$ contains one even and one odd number. Adding up these six numbers, we get that the even number $2a + 2b$ is a sum of three even and three odd numbers, so it should be odd. So we have a contradiction here as well.\n\nIf $s^{2} \\equiv 2 \\bmod 4$, then $a, b, c, d, a-c, b-d$ are all odd. But if $a$ and $c$ are both odd, then $a-c$ should be even, another contradiction. So in every possible case we have a contradiction; hence, our initial assumption—that all coordinates were integers—must be wrong.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCircles $k_{1}$, $k_{2}$, $k_{3}$ intersect as follows: $k_{1} \\cap k_{2} = \\{A, D\\}$, $k_{1} \\cap k_{3} = \\{B, E\\}$, $k_{2} \\cap k_{3} = \\{C, F\\}$. Also, $A B C D E F$ is a non-self-intersecting hexagon. Prove that\n$$\nA B \\cdot C D \\cdot E F = B C \\cdot D E \\cdot F A.\n$$\n(Hint: first prove that $A D$, $B E$, $C F$ meet at some point $O$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe take the hint. This statement is actually the \"Radical Axis Theorem.\" To prove it, we let $O_{1}$, $O_{2}$, $O_{3}$ be the centers of the circles and $r_{1}$, $r_{2}$, $r_{3}$ their respective radii. We shall consider the set of points $P$ satisfying $P O_{1}^{2} - P O_{2}^{2} = r_{1}^{2} - r_{2}^{2}$. For any $P$, let $Q$ be the foot of the perpendicular from $P$ to $O_{1} O_{2}$, and use directed distances on line $O_{1} O_{2}$ (i.e. distances may be positive or negative according to the order of the points). Then, $P O_{1}^{2} - P O_{2}^{2}$ equals\n$$\n\\left(P Q^{2} + Q O_{1}^{2}\\right) - \\left(P Q^{2} + Q O_{2}^{2}\\right) = Q O_{1}^{2} - Q O_{2}^{2} = \\left(Q O_{1} - Q O_{2}\\right)\\left(Q O_{1} + Q O_{2}\\right) = O_{2} O_{1}\\left(O_{2} O_{1} + 2 Q O_{2}\\right),\n$$\na linear function of $Q$; hence there is exactly one position of $Q$ for which $P O_{1}^{2} - P O_{2}^{2} = r_{1}^{2} - r_{2}^{2}$, and the desired set of $P$ is a line perpendicular to $O_{1} O_{2}$ and passing through this $Q$. But, since $A O_{1} = r_{1}$, $A O_{2} = r_{2}$, we see that $A$ lies on this line; similarly, so does $D$. So the line is line $A D$; i.e. $P O_{1}^{2} - P O_{2}^{2} = r_{1}^{2} - r_{2}^{2}$ iff $P$ is on line $A D$. Likewise, one sees that $P O_{1}^{2} - P O_{3}^{2} = r_{1}^{2} - r_{3}^{2}$ iff $P$ is on line $B E$, and $P O_{2}^{2} - P O_{3}^{2} = r_{2}^{2} - r_{3}^{2}$ iff $P$ is on line $C F$. But then, if we take $P$ to be the intersection of $A D$ and $C F$, we have $P O_{1}^{2} - P O_{3}^{2} = \\left(P O_{1}^{2} - P O_{2}^{2}\\right) + \\left(P O_{2}^{2} - P O_{3}^{2}\\right) = \\left(r_{1}^{2} - r_{2}^{2}\\right) + \\left(r_{2}^{2} - r_{3}^{2}\\right) = r_{1}^{2} - r_{3}^{2}$, so $P$ lies on $B E$. Thus, $A D$, $B E$, $C F$ are concurrent.\n\nNow the quadrilateral $A B D E$ is cyclic with diagonals meeting at $O$, so triangles $A B O$, $E D O$ are similar and $A B / D E = B O / O D$. Similarly, using two other quadrilaterals, we find that $C D / F A = D O / O F$, $E F / B C = F O / O B$. Multiplying together gives\n$$\n\\frac{A B \\cdot C D \\cdot E F}{D E \\cdot F A \\cdot B C} = \\frac{B O}{D O} \\cdot \\frac{D O}{F O} \\cdot \\frac{F O}{B O} = 1,\n$$\nwhich is what we need.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a, b, c, d, e, f$ are real numbers such that $a+b+c+d+e+f=0$ and $a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+f^{3}=0$, prove that $(a+c)(a+d)(a+e)(a+f)=(b+c)(b+d)(b+e)(b+f)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst, let $S$ be the sum of all possible products of three distinct variables from the set $\\{a, b, c, d, e, f\\}$, i.e. $S=abc+abd+abe+\\cdots+def$. Now, we assert that\n$$\n\\begin{aligned}\n& (a+b+c+d+e+f)^3 + 2\\left(a^3+b^3+c^3+d^3+e^3+f^3\\right) \\\\\n& \\quad - 3(a+b+c+d+e+f)\\left(a^2+b^2+c^2+d^2+e^2+f^2\\right) = 6S\n\\end{aligned}\n$$\nTo see this, note that, when the left-hand side is expanded, the terms obtained come in three forms: $a^3$ (i.e. cubes of one variable), $a^2b$ (squares of one variable times another), and $abc$ (products of three different variables). Each term of the form $a^3$ has coefficient $1$ in $(a+\\cdots+f)^3$, $2$ in $2(a^3+\\cdots+f^3)$, and $-3$ in $-3(a+\\cdots+f)(a^2+\\cdots+f^2)$, for a total coefficient of $0$; the terms of the form $a^2b$ have respective coefficients $3, 0, -3$, for a total of $0$; those of the form $abc$ have coefficients $6, 0, 0$, for a total of $6$. Since the right side of (1) consists of the $abc$-type terms with coefficient $6$, the assertion is proven.\n\nHowever, the given tells us that the left-hand side of (1) is zero, so $S=0$. Now, let $p=ab-(cd+ce+cf+de+df+ef)$. We claim that the following holds for all numbers $x$:\n$$\n(x+c)(x+d)(x+e)(x+f) = (x-a)(x-b)(x^2-p) + abp + cdef.\n$$\nWe can verify this by comparing coefficients for corresponding powers of $x$.\n\nThe coefficient of $x^4$ on both sides is $1$; the coefficient of $x^3$ is $c+d+e+f$ on the left and $-(a+b)$ on the right, but the given implies that these are equal. The coefficient of $x^2$ is $cd+ce+cf+de+df+ef$ on the left and $ab-p$ on the right; these are equal by the definition of $p$. The coefficient of $x$ is $cde+cdf+cef+def$ on the left and $ap+bp$ on the right. To see that these are equal, observe that\n$$\n\\begin{gathered}\ncde+cdf+cef+def-(a+b)p = cde+cdf+cef+def-(a+b)ab+(a+b)(cd+ce+cf+de+df+ef) \\\\\n= (S - [abc+abd+abe+abf]) - (a+b)ab = -ab(c+d+e+f+a+b) = 0\n\\end{gathered}\n$$\nwhere we have used the definition of $S$ for the second equality and the fact that $S=0$ for the third. Finally, the constant term equals $cdef$ on both sides. Thus, equation (2) holds. Plugging in $x=a$ and $x=b$, we find\n$$\n\\begin{aligned}\n& (a+c)(a+d)(a+e)(a+f) = (a-a)(a-b)(a^2-p) + abp + cdef = abp + cdef \\\\\n& (b+c)(b+d)(b+e)(b+f) = (b-a)(b-b)(b^2-p) + abp + cdef = abp + cdef\n\\end{aligned}\n$$\nSince the right-hand sides match, we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x_{1}, x_{2}, \\ldots, x_{n}$ be positive numbers, with $n \\geq 2$. Prove that\n$$\n\\left(x_{1}+\\frac{1}{x_{1}}\\right)\\left(x_{2}+\\frac{1}{x_{2}}\\right) \\cdots\\left(x_{n}+\\frac{1}{x_{n}}\\right) \\geq\\left(x_{1}+\\frac{1}{x_{2}}\\right)\\left(x_{2}+\\frac{1}{x_{3}}\\right) \\cdots\\left(x_{n-1}+\\frac{1}{x_{n}}\\right)\\left(x_{n}+\\frac{1}{x_{1}}\\right) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst we will prove a simple lemma involving only two variables: For all positive $a, b$,\n$$\n\\left(a^{2}+1\\right)\\left(b^{2}+1\\right) \\geq (a b+1)^{2}\n$$\nTo see why this is true, multiply out, and after simplifying, we have\n$$\na^{2}+b^{2} \\geq 2 a b\n$$\nThis is equivalent to\n$$\na^{2}-2 a b+b^{2}=(a-b)^{2} \\geq 0\n$$\nwhich of course is true (in fact, for any real numbers $a$ and $b$).\n\nNow we shall attack the problem. Multiplying both sides by $x_{1} x_{2} \\cdots x_{n}$ produces the equivalent inequality\n$$\n\\left(x_{1}^{2}+1\\right)\\left(x_{2}^{2}+1\\right) \\cdots\\left(x_{n}^{2}+1\\right) \\geq\\left(x_{1} x_{2}+1\\right)\\left(x_{2} x_{3}+1\\right) \\cdots\\left(x_{n} x_{1}+1\\right) .\n$$\nApplying the lemma repeatedly yields\n$$\n\\begin{aligned}\n&\\left(x_{1}^{2}+1\\right)\\left(x_{2}^{2}+1\\right) \\geq\\left(x_{1} x_{2}+1\\right)^{2}, \\\\\n&\\left(x_{2}^{2}+1\\right)\\left(x_{3}^{2}+1\\right) \\geq\\left(x_{2} x_{3}+1\\right)^{2}, \\\\\n& \\vdots \\\\\n&\\left(x_{n}^{2}+1\\right)\\left(x_{1}^{2}+1\\right) \\geq\\left(x_{n} x_{1}+1\\right)^{2} .\n\\end{aligned}\n$$\nMultiplying these yields the square of the desired inequality.\nSolution:\nWe shall use induction. Even though the problem begins with $n=2$, we can start by noting that for $n=1$, the statement is merely the trivial\n$$\nx_{1}+1 / x_{1} \\geq x_{1}+1 / x_{1}\n$$\nIn general, suppose without loss of generality that $x_{1}$ is the largest among the given $n$ numbers. The right-hand side products containing $x_{1}$ are: $\\left(x_{1}+1 / x_{2}\\right)\\left(x_{n}+1 / x_{1}\\right)$. We claim that this product will not decrease if we swap the places of $x_{2}$ from the first multiple and $x_{1}$ from the second multiple, i.e.\n$$\n\\left(x_{1}+1 / x_{2}\\right)\\left(x_{n}+1 / x_{1}\\right) \\leq\\left(x_{1}+1 / x_{1}\\right)\\left(x_{n}+1 / x_{2}\\right) .\n$$\nThis inequality is easy to prove: after a little algebra, it becomes\n$$\nx_{1} x_{2}+x_{1} x_{n} \\leq x_{1} x_{1}+x_{2} x_{n}\n$$\nwhich is equivalent to\n$$\n\\left(x_{1}-x_{n}\\right)\\left(x_{1}-x_{2}\\right) \\geq 0\n$$\nand this is true because $x_{1}$ was the largest number among the given $n$ numbers.\n\nNotice that after performing the \"swap,\" we may cancel $\\left(x_{1}+1 / x_{1}\\right)$ from both sides, and what we are left with is the same problem but for the $(n-1)$ numbers $x_{2}, x_{3}, \\ldots x_{n}$. This completes the inductive step.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are 30 members of the math faculty at UC Berkeley. The evil Gastropod places a red or green stamp on each member's forehead, then places them all in separate rooms. Each day, the Gastropod convenes all the faculty in one room and asks any member who has conclusively determined that his/her stamp is red to raise his/her hand. No other communication between them is allowed. As it turns out, everyone has a red stamp (but since each can only see the others' stamps, none of them realize this at first). The Gastropod remarks on the first day: \"At least one of you has a red stamp.\" Given that all Berkeley faculty members are perfect reasoners, and each knows all the others are perfect reasoners, prove that everyone will eventually determine that his/her stamp is red.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe prove the following generalization: if exactly $n$ faculty members were given red stamps and the rest have green stamps (and the procedure is as before), no hands will be raised for the first $n-1$ days, and all $n$ faculty members will raise their hands on the $n$th day. This will be shown by induction.\n\nIf $n=1$, the one faculty member sees on the first day that everyone else has green stamps and realizes that he/she must be red.\n\nIf $n>1$, then each of the red faculty sees $n-1$ red stamps. (The people with green stamps can be disregarded since they cannot offer any useful information: they can never raise their hands, and all the red people already know they are green.) Suppose $A$ has a red stamp. Using the induction hypothesis, $A$ reasons as follows: \"If I am green, then there are $n-1$ red stamps, so nothing will happen for the first $n-2$ days, and all those with red stamps will raise their hands on the $n-1$ th day.\" Also, by symmetry, all the red faculty reason (and act) identically.\n\nConsider the $i$ th day for $i=1,2, \\ldots, n-1$; we claim no hands are raised on any of these days. This holds by strong induction on $i$ : For each such value of $i$, on the $i$ th day, $A$ remembers that no hands were raised on any previous day (by the induction hypothesis), which is consistent (from $A$ 's point of view) with the hypothesis that $A$ is green. So $A$ will not raise his hand on the $i$ th day, because he still believes he could be green; by symmetry, none of the other faculty raise their hands.\n\nOn the $n$th day, however, $A$ notices that no hands were raised on the $n-1$ th day, which contradicts the behavior expected if $A$ were green. So $A$ realizes he is red and raises his hand. Likewise, all the other red faculty raise their hands. This completes the induction step.\n\nIn the given case $n=30$, we see that all the faculty will be silent for 29 days, and they will all realize on the 30th day that their stamps are red.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a cyclic quadrilateral (i.e. inscribed in a circle) such that $DC = AD + BC$. Prove that the intersection point of the angle bisectors of angle $A$ and angle $B$ lies on $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $E$ lie on segment $CD$ so that $CE = BC$. Then, $DE = DC - CE = AD$. So, $\\triangle BCE$ and $\\triangle ADE$ are both isosceles. Let $\\alpha = \\angle DAE = \\angle AED$, and let $\\beta = \\angle CEB = \\angle EBC$. Then $\\angle CDA = \\angle EDA = \\pi - 2\\alpha$, but since the quadrilateral is cyclic, this gives $\\angle ABC = 2\\alpha$. Likewise, $\\angle BCD = \\pi - 2\\beta$ and $\\angle DAB = 2\\beta$.\n\nNow, let the circumcircle of triangle $ABE$ intersect line $CD$ again at $F$. We will assume that $F$ lies on the same side of $E$ as $D$ (the other case is completely analogous). Since quadrilateral $ABEF$ is cyclic, we have $\\angle ABF = \\angle AEF = \\alpha = \\angle ABC / 2$, and $\\angle FAB = \\pi - \\angle BEF = \\angle CEB = \\beta = \\angle DAB / 2$. So $F$ lies on the bisectors of angles $DAB$ and $ABC$, which means it is precisely the intersection point specified in the problem statement. Since $F$ was constructed to lie on $CD$, we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p$ be an odd prime and $a, b$ positive integers satisfying $(p+1)^{a} - p^{b} = 1$. Show that $a = b = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nConsider remainders modulo $p+1$. We know that $1 + (-1)^{b} \\equiv 1 + p^{b} = (p+1)^{a} \\equiv 0$, so (since $p+1 > 2$) $b$ is odd.\n\nNow suppose that $a > 1$. Then\n$$\n0 \\equiv (p+1)^{a-1} = \\frac{p^{b} + 1}{p+1} = p^{b-1} - p^{b-2} + p^{b-3} - \\cdots + 1 \\equiv 1 - (-1) + 1 - \\cdots + 1\n$$\n(there are $b$ terms) $= b$, where the congruences are modulo $p+1$. But then, $b$ is divisible by $p+1$ and so is even, a contradiction.\n\nTherefore $a = 1$, and $p^{b} = (p+1)^{a} - 1 = p \\Rightarrow b = 1$ also.\n\n(The fact that $p$ was prime was not needed.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S=\\{1,2, \\ldots, n\\}$, and let $T$ be the set consisting of all nonempty subsets of $S$. The function $f: T \\rightarrow S$ is \"garish\" if there do not exist sets $A, B \\in T$ such that $A$ is a proper subset of $B$ and $f(A)=f(B)$. Determine, with proof, how many garish functions exist.",
    "options": [],
    "answer": "n!",
    "solution": "Solution:\n\nThere are $n!$ such functions. If $g$ is any bijective map from $S$ to itself (i.e. a permutation of $S$), then the function $f: T \\rightarrow S$ defined by $f(A)=g(|A|)$ (here $|A|$ is the cardinality of set $A$) is garish. To see this, just note that if $A$ is a proper subset of $B$, then $|A|<|B|$, so $g(|A|) \\neq g(|B|)$. There are $n!$ possible choices of the map $g$, and all of them give different maps $f$ (this follows from the fact that every element of $S$ is the cardinality of some set in $T$), so we get $n!$ garish functions this way. We now wish to show that every garish function is of this form.\n\nLet $f$ be garish; our crucial observation is the following: If $a_{1}, a_{2}, \\ldots, a_{n}$ are the elements of $S$ in some order, we shall say that this ordering \"produces\" a sequence of $n$ values $f\\left(\\{a_{1}\\}\\right), f\\left(\\{a_{1}, a_{2}\\}\\right), \\ldots, f\\left(\\{a_{1}, a_{2}, \\ldots, a_{n}\\}\\right)$. These values are all different, since, of any two of these sets, one is properly contained in the other. But there are exactly $n$ possible values for $f$, namely $1,2, \\ldots, n$, so any produced sequence consists of exactly these in some order.\n\nNow choose any integer $k$, $1 \\leq k \\leq n$. We will show that $f$ has the same value on all $k$-element sets. We can assume $k<n$ (otherwise there is only one $k$-element set). First consider any two sets that differ by at most one element; let them be $\\{a_{1}, a_{2}, \\ldots, a_{k}\\}$ and $\\{a_{1}, a_{2}, \\ldots, a_{k-1}, b\\}$. Assume $a_{k} \\neq b$. Let the remaining elements of $S$ (if there are any) be $a_{k+2}, a_{k+3}, \\ldots, a_{n}$ in any order. Consider the sequences produced from the two orderings $a_{1}, a_{2}, \\ldots, a_{k-1}, a_{k}, b, a_{k+2}, \\ldots, a_{n}$ and $a_{1}, a_{2}, \\ldots, a_{k-1}, b, a_{k}, a_{k+2}, \\ldots, a_{n}$. From the above, each produced sequence contains every element of $S$ exactly once. But these two sequences are identical except that the first has $f\\left(\\{a_{1}, \\ldots, a_{k}\\}\\right)$ where the second has $f\\left(\\{a_{1}, \\ldots, a_{k-1}, b\\}\\right)$, so we conclude that these two values of $f$ are identical. Also, if $a_{k}=b$, then the two values are equal (trivially).\n\nThis shows that $f$ has the same value on two $k$-element sets differing by $\\leq 1$ element. Now if $\\{a_{1}, \\ldots, a_{k}\\}$ and $\\{b_{1}, \\ldots, b_{k}\\}$ are any two sets in $T$, we have (applying this repeatedly) that\n$$\nf\\left(\\{a_{1}, a_{2}, \\ldots, a_{k}\\}\\right)=f\\left(\\{b_{1}, a_{2}, \\ldots, a_{k}\\}\\right)=f\\left(\\{b_{1}, b_{2}, a_{3}, \\ldots, a_{k}\\}\\right)=\\cdots=f\\left(\\{b_{1}, b_{2}, \\ldots, b_{k}\\}\\right).\n$$\nSo $f$ is constant over all $k$-element sets. Thus we can define the function $g: S \\rightarrow S$ by letting $g(k)$ be the value of $f$ on any $k$-element set, and $g$ uniquely determines $f$ since $f(A)=g(|A|)$ for all $A \\in T$. Moreover, if $g(k)=g(l)$ for some $k<l$, then letting $A=\\{1,2, \\ldots, k\\}, B=\\{1,2, \\ldots, l\\}$, we have $f(A)=f(B)$, a contradiction since $A$ is a proper subset of $B$. Thus, $g$ is one-one; but since it maps the finite set $S$ to itself, it is actually bijective. Thus all garish functions are indeed in the form claimed above.\nAlternate Solution:\n\nWe observe that $n!$ garish functions can be constructed, precisely as described in the first solution, and we wish to show that all garish functions are of this form. We use induction on $n$. In the base case $n=1$, there is only one possible function, given by $f(\\{1\\})=1$, and it is garish. Now consider any $n$, and suppose the statement is true for $n-1$. If $f$ is a garish function, let $a_{k}=f(\\{1,2, \\ldots, k\\})$ for each $k=1,2, \\ldots, n$. Then all $a_{k}$ are different: if $a_{k}=a_{l}$ for some $k<l$, then $f$ would not be garish, since $\\{1,2, \\ldots, k\\}$ is a proper subset of $\\{1,2, \\ldots, l\\}$. Thus, the elements $a_{1}, \\ldots, a_{n}$ equal $1, \\ldots, n$ in some order. Partition $T$ into three subsets: let $T_{1}$ consist of the nonempty subsets of $\\{1,2, \\ldots, n-1\\}$, let $T_{2}=\\{S\\}$, and let $T_{3}$ be the rest of $T$ (the proper subsets of $S$ containing $n$). We claim that, if $A \\in T_{1}$, then $f(A)=a_{|A|}$. This follows from the induction hypothesis. Formally, we first observe that $f(A) \\neq a_{n}$ for $A \\in T_{1}$ since $A \\subset S$. Then, if we define $h:\\{a_{1}, a_{2}, \\ldots, a_{n-1}\\} \\rightarrow \\{1,2, \\ldots, n-1\\}$ by $h(a_{k})=k$ (and this is well-defined since $a_{1}, \\ldots, a_{n-1}$ are distinct), then the composite $h \\circ f$ is a garish function from $T_{1}$ to $\\{1,2, \\ldots, n-1\\}$, so, by the induction hypothesis, it takes the same value on all $k$-element sets; since $h$ is injective, $f$ takes the same value on all $k$-element sets, as claimed. Also, if $A \\in T_{2}$ (so $A=S$) then $f(A)=a_{n}$, by assumption.\n\nNow consider any set $A \\in T_{3}$; we have $|A|<n$. We claim that $f(A)=a_{|A|}$; this will be shown by strong downward induction on the cardinality $|A|$. If $|A|=n-1$, then $A \\subset S$ and $f(S)=a_{n}$. On the other hand, the set $A-\\{n\\}$ has $n-2$ elements and is contained in $T_{1}$; it certainly has subsets $A_{n-2}, A_{n-3}, \\ldots, A_{1}$ of respective cardinalities $n-2, n-3, \\ldots, 1$, all of which lie in $T_{1}$; thus, $f(A_{i})=a_{i}$ from the previous paragraph. From this, garishness gives $f(A) \\neq a_{n}$ and $f(A) \\neq a_{1}, a_{2}, \\ldots, a_{n-2}$, so $f(A)=a_{n-1}$.\n\nThe induction step is similar: suppose we have proven that, for all $A \\in T_{3}$ with $|A|>m$, $f(A)=a_{|A|}$, and we wish to move to the case $|A|=m$. By successively adding elements to $A$, we can construct sets $A_{m+1}, A_{m+2}, \\ldots, A_{n}=S$ of respective cardinalities $m+1, m+2, \\ldots, n$, which all lie in $T_{3}$, except for $A_{n} \\in T_{2}$. Also, if $m>1$, then $A-\\{n\\}$ has $m-1$ elements and lies in $T_{1}$; repeatedly removing elements, we get sets $A_{m-1}, A_{m-2}, \\ldots, A_{1} \\in T_{1}$ of respective sizes $m-1, m-2, \\ldots, 1$. So we have $f(A_{k})=a_{k}$ for $k>m$ by the induction hypothesis, and $f(A_{k})=a_{k}$ for $k<m$ because these sets lie in $T_{1}$. But each of the $A_{k}$ either contains $A$ or is contained in $A$, so $f(A)$ cannot equal any of these $a_{k}$; hence, it must equal $a_{m}$, completing the induction step.\n\nAt this point, we have shown that $f(A)=a_{|A|}$ for each $A \\in T$, and one sees as in the previous solution that $f(A)=g(|A|)$ for some bijective function $g: S \\rightarrow S$, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the minimum of the function\n$$\nf(x, y)=\\sqrt{(x+1)^{2}+(2 y+1)^{2}}+\\sqrt{(2 x+1)^{2}+(3 y+1)^{2}}+\\sqrt{(3 x-4)^{2}+(5 y-6)^{2}},\n$$\ndefined for all real $x, y>0$.",
    "options": [],
    "answer": "10",
    "solution": "Solution:\nNote that $\\sqrt{(x+1)^{2}+(2 y+1)^{2}}$ is the distance in the coordinate plane from $(0,0)$ to $(x+1,2 y+1)$; $\\sqrt{(2 x+1)^{2}+(3 y+1)^{2}}$ is the distance from $(x+1,2 y+1)$ to $(3 x+2,5 y+2)$; $\\sqrt{(3 x-4)^{2}+(5 y-6)^{2}}$ is the distance from $(3 x+2,5 y+2)$ to $(6,8)$. Thus, $f(x, y)$ is the distance along the polygonal path from $(0,0)$ to $(x+1,2 y+1)$ to $(3 x+2,5 y+2)$ to $(6,8)$, which, by the triangle inequality, cannot be less than the distance from $(0,0)$ to $(6,8)$. Thus $f(x, y) \\geq 10$. One then calculates $f(1 / 4,1 / 3)=10$. So 10 is in fact attainable, and it is our minimum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $f, g$ be two functions from the set $\\mathbf{R}$ of real numbers to itself, such that $f(x)<g(x)$ for all $x \\in \\mathbf{R}$. Prove that there exists an infinite subset $S \\subseteq \\mathbf{R}$ such that $f(x)<g(y)$ for all $x, y \\in S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that, for every $x$, we can choose a rational number $h(x)$ such that $f(x)<h(x)<g(x)$. Proof: since $g(x)-f(x)>0$, we can choose an integer $n$ larger than $1/(g(x)-f(x))$. Then the open interval $(n f(x), n g(x))$ has width $n(g(x)-f(x))>1$, so it contains some integer $m$. So we have $f(x)<m/n<g(x)$, and we can take $h(x)=m/n$.\n\nNow $h$ is a function mapping an uncountably infinite set, $\\mathbf{R}$, to a countably infinite set, $\\mathbf{Q}$ (the set of rational numbers). For any $z \\in \\mathbf{Q}$, consider $h^{-1}(z)$, the set of elements of $\\mathbf{R}$ whose image under $h$ is $z$. Certainly $\\mathbf{R}$ is the union of all such sets, since, for any $x \\in \\mathbf{R}$, $x$ lies in $h^{-1}(h(x))$. If the set $h^{-1}(z)$ is finite (or even countably infinite) for all $z \\in \\mathbf{Q}$, then $\\mathbf{R}$ consists of a union of countably many countable sets, so it is countable. But this is false, so some $h^{-1}(z)$ must be uncountably infinite. Let $S=h^{-1}(z)$ accordingly. Then, for all $x, y \\in S$, we have $f(x)<h(x)=z=h(y)<g(y)$, and $S$ is infinite, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that there exists a set $S$ of $3^{1000}$ points in the plane such that for each point $P$ in $S$, there are at least 2000 points in $S$ whose distance to $P$ is exactly 1 inch.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet us define a $k$-configuration to be a finite set of points on the plane such that for each point $P$ in the set, there are at least $k$ points of the set 1 inch from $P$. Then the problem is asking us to show that there is a 2000-configuration with $3^{1000}$ points.\n\nNotice that an equilateral triangle is a 2-configuration which has 3 points. Now, let $A$ be a $k$-configuration with $N$ points, and let $T$ be an equilateral triangle with unit side length. We shall show that it is possible to \"add\" $A$ and $T$ to create a $(k+2)$-configuration with $3N$ points:\n\nDefine the set $S$ by\n$$\nS = \\{ a + t \\mid a \\in A,\\ t \\in T \\}\n$$\nwhere we treat the points as vectors. In other words, $S$ consists of the vector sums of every point in $A$ with every point in $T$. Since $A$ has $N$ points, and $T$ has 3 points, the set $S$ will have $3N$ points as long as all of these sums are distinct. For the time being, let us assume that the sums are all distinct.\n\nNow we will show that $S$ is a $(k+2)$-configuration. Consider any point $a + t$ in $S$, where $a \\in A$ and $t \\in T$. Since $A$ is a $k$-configuration, there are $k$ points $a_1, \\ldots, a_k \\in A$ that are 1 inch away from $a$. Likewise, there are two points $t_1, t_2$ in $T$ which are each 1 inch away from $t$. It is easy to check that the $k+2$ points\n$$\nt + a_1, \\ldots, t + a_k ;\\ a + t_1, a + t_2\n$$\nare each 1 inch away from $a + t$. Thus $S$ is a $(k+2)$-configuration.\n\nBut how do we ensure that all $3N$ sums are distinct? The sums fail to be distinct only if there are pairs $a, a' \\in A$ and $t, t' \\in T$ with $a + t = a' + t'$, which in turn is true if and only if $a - a' = t' - t$. To ensure that this does not happen, it suffices to rotate one of the two sets (say, $T$) so that the slopes of all of the lines connecting all pairs of points in $T$ do not equal any of the slopes in $A$ (easy to do since there are finitely many points).\n\nFor example, in the following diagram, we attempt to \"add\" two equilateral triangles (the second triangle is outlined), but because of equal slopes, the sum contains only 6 points.\n![](attached_image_1.png)\n\nOn the other hand, if we rotate the second triangle (in this case, by 30 degrees), the resulting sum contains 9 points (and you should check that this new set is indeed a 4-configuration).\n![](attached_image_2.png)\n\nClearly, we can continue this summation process, adding additional copies of equilateral triangles (making sure to rotate so that no slopes are equal). For each triangle that we add, the new set will have three times as many points. Thus if we add 1000 triangles, we will get a set with $3^{1000}$ points which is a $2 + 2 + \\cdots + 2 = 2000$-configuration.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the UC Berkeley bureaucratic hierarchy, certain administrators report to certain other administrators. It so happens that if $A$ reports to $B$ and $B$ reports to $C$, then $C$ reports to $A$. Also, administrators do not report to themselves. Prove that all the administrators can be divided into three disjoint groups $X, Y, Z$ so that, if $A$ reports to $B$, then either $A$ is in $X$ and $B$ is in $Y$, or $A$ is in $Y$ and $B$ is in $Z$, or $A$ is in $Z$ and $B$ is in $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nForm a graph whose vertices represent the administrators, with an edge between $A$ and $B$ if one of $A, B$ reports to the other. We will temporarily assume that the graph is connected. Consider any walk on this graph (i.e. a finite sequence of vertices, any two consecutive members of which are connected by an edge). Whenever this walk traverses an edge from $A$ to $B$, assign this edge a weight of 1 if $A$ reports to $B$ and -1 if $B$ reports to $A$; then let the \"value\" of the walk be the sum of the weights of its edges. This definition is unambiguous, i.e. we cannot have both $A$ reporting to $B$ and $B$ reporting to $A$, since then we would also have to have $A$ reporting to $A$, which is prohibited.\n\nWe claim that if a walk is a cycle (it starts and ends at the same vertex), its value is divisible by 3. If not, choose a counterexample that uses as few edges as possible. Let its value be $v$. We consider the possible weights of successive edges in our cycle. First, suppose there are two consecutive 1's, representing some movements from $A$ to $B$ and from $B$ to $C$. So $A$ reports to $B$, and $B$ reports to $C$; hence, from the given, $C$ reports to $A$, and we can replace these two edges $A B C$ by one edge $A C$, whose weight will be -1. So the new cycle we have obtained has one edge less than the old one, and its value is $v-3$, which is not divisible by 3 (since $v$ wasn't). This contradicts the minimality of the original cycle, so this situation is impossible. Similarly, if there were two consecutive -1's, corresponding to edges $C B A$, we could replace them with an edge $C A$ of weight 1, and we would have a new cycle, shorter than the previous one, of value $v+3$, which again is not divisible by 3. So this, too, is impossible.\n\nWe conclude that the edge weights of our cycle must alternate between 1 and -1. The initial and final edges must also have opposite weights, or else we can replace them with one edge to obtain a shorter counterexample, precisely as in the previous paragraph. Actually, we avoid this problem if the initial and final edges are the same, so that our minimal cycle consists of only one edge, but then this edge runs from $A$ to $A$, which is illegal. So, the only remaining possibility is that our cycle weights are of the form $1,-1,1,-1, \\ldots,-1$, or $-1,1,-1,1, \\ldots, 1$, but either way, the sum is 0, which is in fact a multiple of 3. So we have ruled out all the cases and found that a counterexample to our claim is impossible. Thus, the claim holds.\n\nNow choose any vertex $P$. For $Q$ any other vertex, we can define the value of $Q$ to be the element of the set $\\{0,1,2\\}$ which is congruent mod 3 to the value of a walk from $P$ to $Q$. (Such a walk exists, since the graph is connected). This is well-defined as long as we know that the value of $Q$ is independent of the walk chosen. But if there exist two walks from $P$ to $Q$ with respective values $v, w$, then traversing the first walk from $P$ to $Q$ and traversing the second walk backwards from $Q$ to $P$ gives a cycle with value $v-w$. The above claim then shows that $v-w$ is divisible by 3, so both walks do give the same value, as required.\n\nNow just let $X$ consist of all vertices of value 0 (including $P$), $Y$ all vertices of value 1, and $Z$ those of value 2. If $A B$ is an edge (with $A$ reporting to $B$), then it is clear that a walk from $P$ to $A$, followed by the edge $A B$, gives a walk from $P$ to $B$. So, the value of $B$ equals the value of $A$ plus $1 (\\bmod 3)$. Consequently: if $A \\in X$ then $B \\in Y$; if $A \\in Y$ then $B \\in Z$; if $A \\in Z$ then $B \\in X$.\n\nThus, the problem is solved in the connected case. But if our administrator graph was disconnected, let the components be $C_{i}$ for some set of indices $i$. For each $C_{i}$, partition it into sets $X_{i}, Y_{i}, Z_{i}$ by the connected case; then, let $X=\\cup X_{i}, Y=\\cup Y_{i}, Z=\\cup Z_{i}$. Since any two administrators connected by an edge must lie in the same $C_{i}$, it easily follows that these sets $X, Y, Z$ do the trick.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice plays the following game of solitaire on a $20 \\times 20$ chessboard. She begins by placing 100 pennies, 100 nickels, 100 dimes, and 100 quarters on the board so that each of the 400 squares contains exactly one coin. She then chooses 59 of these coins and removes them from the board. After that, she removes coins, one at a time, subject to the following rules:\n- A penny may be removed only if there are four squares of the board adjacent to its square (up, down, left, and right) that are vacant (do not contain coins). Squares \"off the board\" do not count towards this four: for example, a non-corner square bordering the edge of the board has three adjacent squares, so a penny in such a square cannot be removed under this rule, even if all three adjacent squares are vacant.\n- A nickel may be removed only if there are at least three vacant squares adjacent to its square. (And again, \"off the board\" squares do not count.)\n- A dime may be removed only if there are at least two vacant squares adjacent to its square (\"off the board\" squares do not count).\n- A quarter may be removed only if there is at least one vacant square adjacent to its square (\"off the board\" squares do not count).\nAlice wins if she eventually succeeds in removing all the coins. Prove that it is impossible for her to win.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume that the squares have unit length. Consider, at any time during the game, the perimeter of the region of empty squares (this region may or may not be connected). For example, suppose that at some time 64 squares are empty. If the empty squares are packed together forming an $8 \\times 8$ square, the perimeter will equal $4 \\cdot 8=32$. On the other hand, if none of the 64 empty squares are adjacent to one another, the perimeter will equal $4 \\cdot 64=256$ (each square has perimeter 4).\nAt the start of the game, there are $k=59$ empty squares. At this point the perimeter is at most $4k$. Now consider what happens as different coins are removed.\n\nIn order for a penny to be removed, it has to be surrounded on all 4 sides by empty squares. The following diagram illustrates the situation. The shaded squares are empty, the squares containing \"$X$\" are occupied by arbitrary coins, and \"$P$\" of course denotes a penny. The thick lines indicate the perimeter.\n\n| $X$ | $X$ | $X$ | $X$ | $X$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $X$ |  |  | $X$ | $X$ |\n| $X$ |  | $P$ |  | $X$ |\n| $X$ | $X$ |  | $X$ | $X$ |\n| $X$ | $X$ |  | $X$ | $X$ |\n\n$\\longrightarrow$\n\n| $X$ | $X$ | $X$ | $X$ | $X$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $X$ |  |  | $X$ | $X$ |\n| $X$ |  |  |  | $X$ |\n| $X$ | $X$ |  | $X$ | $X$ |\n| $X$ | $X$ |  | $X$ | $X$ |\n\nBefore the removal, the perimeter included the boundary of the square containing the penny. After the removal, this boundary is gone. Thus, whenever a penny is removed, the perimeter decreases by 4.\n\n| $X$ | $X$ | $X$ | $X$ | $X$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $X$ |  |  | $X$ | $X$ |\n| $X$ |  | $N$ |  |  |\n| $X$ |  | $X$ | $X$ | $X$ |\n| $X$ |  | $X$ | $X$ | $X$ |\n\n$\\longrightarrow$\n\n| $X$ | $X$ | $X$ | $X$ | $X$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $X$ |  |  | $X$ | $X$ |\n| $X$ |  |  |  |  |\n| $X$ |  | $X$ | $X$ | $X$ |\n| $X$ |  | $X$ | $X$ | $X$ |\n\nIn this case, the perimeter decreases by 2. Thus, no matter what, after a nickel is removed, the perimeter decreases by at least 2.\nBy similar reasoning (draw diagrams!) we conclude that if a dime is removed, the perimeter does not increase (it also could decrease by 2 or 4), and when a quarter is removed, the perimeter may increase, but by at most 2 (it also could not change or decrease by 2 or 4).\n\nNow we will show that it is impossible for the board to evolve so that no coins are left. Note that this would cause the perimeter to equal $4 \\cdot 20=80$. Suppose that at the start, $p$ pennies, $n$ nickels, $d$ dimes, and $q$ quarters were removed (so $p+n+d+q=k$). Let $t=10$. If the board evolved so that all the remaining coins were removed, then $t^{2}-p$ pennies would be removed, decreasing the perimeter by exactly $4\\left(t^{2}-p\\right)$. Likewise, $t^{2}-n$ nickels would be removed, but this will decrease the perimeter by at least $2\\left(t^{2}-n\\right)$. The $t^{2}-d$ dimes that would be removed decrease the perimeter by at least zero. And finally, $t^{2}-q$ quarters are removed and these could make the perimeter increase by at most $2\\left(t^{2}-q\\right)$. Recall that the starting value for the perimeter is at most $4k$. Thus, if all coins were removed, the final value of the perimeter would be at most\n$$\n4k-4\\left(t^{2}-p\\right)-2\\left(t^{2}-n\\right)+2\\left(t^{2}-q\\right)=4k-4t^{2}+4p+2n-2q.\n$$\nBut $4p+2n-2q$ is certainly less than or equal to $4(p+n+d+q)=4k$, so the final value of the perimeter is at most\n$$\n4k-4t^{2}+4k=8k-4t^{2}=8 \\cdot 59-4 \\cdot 10^{2}=72,\n$$\nContradicting the fact that the perimeter must equal 80.\n\n\nSolution 2:\n\nDefine $P$ to be perimeter of empty region (as defined in the previous solution) and let $q, d, n$ be respectively the number of quarters, dimes, and nickels on the board at a given time. Then the quantity\n$$\nP-4q-2d+2n\n$$\nis a monovariant: it is non-increasing (no matter how the coins are legally removed). The initial value of this quantity is at most 72, yet the value of the empty board is 80. Hence the board will never be empty.\n\n\nSolution 3:\n\nFor simplicity, give each coin the value 1, 2, 3, 4 (pennies, nickels, dimes, quarters) according to the minimum number of adjacent squares that are required for their removal. If $x$ is a configuration of coins let $f(x)$ denote the sum of the coins plus the number of pairs of adjacent empty squares on the board. If $x'$ is a configuration obtained from $x$ by (legally) removing some coin, then $f(x') \\geq f(x)$ by the rules (e.g., removing a dime decreases the total coin value by 3 but increases the number of orthogonally adjacent pairs by at least 3). In other words, $f(x)$ is a monovariant. A simple count shows that $f$ (empty) $=760$ and\n$$\nf(\\text{initial})=764=100 \\cdot 1+100 \\cdot 2+100 \\cdot 3+41 \\cdot 4\n$$\nsince the smallest initial $f()$ is obtained by removing 59 pennies (and not creating any adjacent squares). Since we can't move upwards from 764 and end up at 760 the desired sequences of moves doesn't exist.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46544,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, \\ldots, a_{2000}$ be real numbers in the interval $[0,1]$. Find the maximum possible value of\n$$\n\\sum_{1 \\leq i<j \\leq 2000}(j-i)\\left|a_{j}-a_{i}\\right|\n$$",
    "options": [],
    "answer": "1000000000",
    "solution": "Solution:\nThe answer is $1,000,000,000$. First, note that the desired sum $S$ is convex as a function of each $a_{i}$. (Indeed, $\\left|a_{i}-a\\right|$ is convex for any real number $a$, and a sum of convex functions is itself convex.) Consequently, it attains a maximum at some point where each variable lies at an endpoint of its interval, i.e. $S$ has a maximum with each $a_{i}$ equal to either $0$ or $1$. So we may restrict our attention to these cases.\n\nThen $\\left|a_{j}-a_{i}\\right|=0$ if $a_{i}=a_{j}$ and $1$ otherwise, so $S$ is really the sum of all values of $j-i$ where $a_{i} \\neq a_{j}$. Equivalently, it equals $\\sum_{1 \\leq i<j \\leq 2000}(j-i)$ minus the sum of $j-i$ over pairs for which $a_{i}=a_{j}$. So our goal is equivalent to minimizing the latter sum. Fix a positive integer $k$, and suppose there are $k$ values of $i$ with $a_{i}=0$ (and $2000-k$ values with $a_{i}=1$). If we consider the sum of $j-i$ over all pairs $i<j$ with $a_{i}=a_{j}=0$, this sum is clearly minimized when all the $i$'s with $a_{i}=0$ are consecutive; likewise the sum over all pairs with $a_{i}=a_{j}=1$ is minimized when all the $1$'s are consecutive. So we may restrict our attention to the case where $a_{1}=a_{2}=\\cdots=a_{k}=0$ and $a_{k+1}=\\cdots=a_{2000}=1$. (The case where the block of $1$'s precedes the block of $0$'s is symmetrically equivalent.)\n\nIn this case, the sum $S$ comes out to be\n$$\n\\begin{aligned}\n& \\sum_{\\substack{1 \\leq i<j \\leq 2000 \\\\\\ a_{i} \\neq a_{j}}}(j-i)=\\sum_{i=1}^{k} \\sum_{j=k+1}^{2000}(j-i)=\\sum_{j=k+1}^{2000}\\left(\\sum_{i=1}^{k} j\\right)-\\sum_{i=1}^{k}\\left(\\sum_{j=k+1}^{2000} i\\right)=\\sum_{j=k+1}^{2000} k j-\\sum_{i=1}^{k}(2000-k) i \\\\\n= & \\frac{(2000-k)(2000+[k+1])}{2}-(2000-k) \\frac{k(k+1)}{2}=1000 k(2000-k)=1000\\left(1000^{2}-[k-1000]^{2}\\right)\n\\end{aligned}\n$$\nwhich is at most $1000^{3}$, with equality holding iff $k=1000$. So $1000^{3}$ is our desired maximum, attained when $a_{i}=0$ for $i \\leq 1000$ and $a_{i}=1$ for $i>1000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEquilateral triangles $A B C_{1}$, $B C A_{1}$, $C A B_{1}$ are constructed outwards on the sides of a triangle $A B C$. Prove:\n\na. The centers of these three triangles form another equilateral triangle.\n\nb. Segments $A A_{1}$, $B B_{1}$, $C C_{1}$ are concurrent.\n\nc. Segments $A A_{1}$, $B B_{1}$, $C C_{1}$ have equal lengths.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution: (All angles will be directed modulo $\\pi$ for simplicity.)\n\na. Let the respective centers of triangles $A B C_{1}$, $B C A_{1}$, $C A B_{1}$ be denoted $O_{C}$, $O_{A}$, $O_{B}$. A rotation about $A$ by an angle of $\\pi / 6$, together with a scaling (about $A$) by a factor of $\\sqrt{3} / 3$, takes $B$ into $O_{C}$. Let $P$ be the image of $C$ under this same transformation, so that triangles $A B C$, $A O_{C} P$ are similar. Also, since $A O_{B} = A C \\sqrt{3} / 3 = A P$ and $\\angle O_{B} A P = \\angle O_{B} A C + \\angle C A P = \\pi / 6 + \\pi / 6 = \\pi / 3$, triangle $O_{B} A P$ is equilateral and $O_{B} P = O_{B} A = O_{B} C$. Also, $P O_{C} = C B \\sqrt{3} / 3$ (by similarity) $= C O_{A}$, and, moreover, $\\angle O_{C} P O_{B} = \\angle O_{C} P A + \\pi / 3 = \\angle B C A + \\pi / 3 = \\pi / 6 + \\angle B C A + \\pi / 6 = \\angle O_{A} C O_{B}$. These facts tell us that triangles $O_{C} P O_{B}$, $O_{A} C O_{B}$ are congruent, so $O_{C} O_{B} = O_{A} O_{B}$. The proof that $O_{B} O_{A} = O_{C} O_{A}$ is completely analogous.\n\nb. Performing a rotation by angle $\\pi / 3$ about $C$ turns $A_{1}$ into $B$ and $A$ into $B_{1}$. Hence, it takes segment $A_{1} A$ into $B B_{1}$, and these lines must then have an angle $\\pi / 3$ between them; i.e. if $Q$ is their intersection, then $\\angle A Q B_{1} = \\pi / 3 = \\angle A C B_{1}$, so that $Q$ lies on the circumcircle of triangle $C A B_{1}$. Likewise, $\\angle A_{1} P B = \\angle A_{1} C B$ and $Q$ is on the circumcircle of triangle $B C A_{1}$. These imply $\\angle A Q C = \\angle A B_{1} C = -\\pi / 3$ (remembering that angles are directed) and $\\angle C Q B = \\angle C A_{1} B = -\\pi / 3$, so that $\\angle B Q A = -\\pi / 3 = \\angle B C_{1} A$ and $Q$ lies on the circumcircle of triangle $A B C_{1}$. At this point we have shown that the intersection $Q$ of $A A_{1}$, $B B_{1}$ is also the second intersection of the circumcircles of $\\triangle C A B_{1}$, $\\triangle B C A_{1}$, and that it is also the second intersection of the circumcircles of $\\triangle B C A_{1}$, $\\triangle A B C_{1}$. But similar reasoning would show that the intersection of $A A_{1}$, $C C_{1}$ is the second intersection of the circumcircles of $\\triangle B C A_{1}$, $\\triangle A B C_{1}$, so this intersection must be $Q$. Hence $A A_{1}$, $B B_{1}$, $C C_{1}$ are concurrent at $Q$.\n\nc. As shown in the solution to the previous part, a rotation takes segment $A A_{1}$ to segment $B B_{1}$, so these segments have equal lengths. It is analogously seen that $B B_{1} = C C_{1}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA permutation of the numbers $1,2, \\ldots, n$ is called \"bad\" if it contains a subsequence of 10 numbers in decreasing order, and \"good\" otherwise. For example, for $n=15$,\n$$\n15,13,1,12,7,11,9,8,10,6,5,4,3,2,14\n$$\nis a bad permutation, because it contains the subsequence\n$$\n15,13,12,11,10,6,5,4,3,2\n$$\nProve that, for each $n$, the number of good permutations is at most $81^{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider any permutation of $1,2, \\ldots, n$. Let the \"height\" of a number $i$ in the permutation be the length of the longest decreasing subsequence ending in $i$. Then, the permutation is bad if and only if some number has height at least 10. Also note that all numbers with the same height must be arranged in increasing order. To see that this is so, suppose that two numbers $i, j$ both have height $h$, where $i<j$, but $j$ precedes $i$ in the permutation. Then there exists a decreasing subsequence of the permutation having length $h$, ending in $j$. By attaching $i$ to the end of this, we get a decreasing sequence of length $h+1$ ending in $i$; this contradicts the assumption that $i$ had height $h$.\n\nNow we know that, in any good permutation, every number has a height in the range $1,2, \\ldots, 9$. We also know that, for each height $h$, the numbers with height $h$ occur in the permutation in increasing order. This means that, if we know which of the numbers $1,2, \\ldots, n$ have height $h$ and which positions in the permutation are occupied by numbers of height $h$ for each $h=1,2, \\ldots, 9$, we can uniquely reconstruct the permutation from this information. Since every possible good permutation can be represented in this form, the number of good permutations is at most equal to the number of possible assignments of heights $1,2, \\ldots, 9$ to the $n$ numbers and the $n$ positions in the permutation. There are $9^{n}$ ways of assigning heights from this range to $1,2, \\ldots, n$, and another $9^{n}$ ways of assigning heights to the $n$ positions; altogether this makes for at most $\\left(9^{n}\\right)^{2}=81^{n}$ possible good permutations.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $O$ be the center of a circle $k$. Points $A, B, C, D, E, F$ on $k$ are chosen such that the triangles $OAB$, $OCD$, $OEF$ are equilateral. Let $L, M, N$ be the midpoints of $BC$, $DE$, $FA$ respectively. Prove that triangle $LMN$ also is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe simplest way to do this is using complex numbers. Let $\\omega = (-1 + i \\sqrt{3}) / 2$, a cube root of unity, and we have $1 + \\omega = -\\omega^{2}$. Then a triangle $XYZ$ is equilateral (with vertices labeled in counterclockwise order) iff segment $ZX$ is the rotation image of $YZ$ through angle $2\\pi/3$; representing the points by their values in the complex plane, this says $X - Z = \\omega(Z - Y)$ or, equivalently, $X + \\omega Y + \\omega^{2} Z = 0$. In particular, when $Y = 0$, we get $X = -\\omega^{2} Z$.\n\nNow, position the given points so that $O$ is at the origin; then the given yields $B = -\\omega^{2} A$, $D = -\\omega^{2} C$, $F = -\\omega^{2} E$. Also, clearly $K = (B + C)/2$, $L = (D + E)/2$, $M = (F + A)/2$. Thus\n$$\nK + \\omega L + \\omega^{2} M = \\left(\\omega^{2} A + B + C + \\omega D + \\omega E + \\omega^{2} F\\right)/2 = \\left(\\omega^{2} A - \\omega^{2} A + C - C + \\omega E - \\omega E\\right)/2 = 0\n$$\ngiving what we need.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a quadrilateral, and let $O$ be the intersection of $AC$ and $BD$. Quadrilateral $A'B'C'D'$ is obtained by rotating $ABCD$ about $O$ by some angle. Let $A_1, B_1, C_1, D_1$ be the intersection points of the lines $A'B'$ and $AB$, $B'C'$ and $BC$, $C'D'$ and $CD$, $D'A'$ and $DA$, respectively. Prove that quadrilateral $A_1B_1C_1D_1$ is cyclic if and only if $AC$ is perpendicular to $BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst we prove a lemma: Suppose some line $ST$ is rotated by angle $\\theta$ about $O$ to obtain line $S'T'$. Let $ST$ and $S'T'$ intersect at $S_1$, and let $S_2$ be the foot of the perpendicular from $O$ to $ST$. Then a rotation about $O$ of angle $\\theta/2$, together with a homothety about $O$ of ratio $\\sec(\\theta/2)$, sends $S_2$ to $S_1$.\n\nProof: If $P$ is the image of $S_2$ under this transformation, then $\\triangle PS_2O$ is a right triangle (right angle at $S_2$), so $P$ lies on $ST$. Let $Q$ be the reflection of $P$ across line $S_2O$; then $Q$ also lies on $ST$, $OQ = OP$, and $\\angle QOP = \\theta$. Thus, a rotation about $O$ through angle $\\theta$ sends $Q$ to $P$. But, since $Q \\in ST$, we have $P \\in S'T'$, so that $P = S_1$ as desired.\n\nNow, in the given situation, let $A_2, B_2, C_2, D_2$ denote the feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ respectively; we have shown that a rotation and homothety about $O$ sends $A_2, B_2, C_2, D_2$ to $A_1, B_1, C_1, D_1$. Thus, quadrilaterals $A_1B_1C_1D_1$ and $A_2B_2C_2D_2$ are similar, so it suffices to show that the latter is cyclic iff $AC$ is perpendicular to $BD$.\n\nFor this, note that $\\angle OA_2B = \\pi/2 = \\angle OB_2B$, so quadrilateral $OA_2BB_2$ is cyclic; similarly, quadrilaterals $OB_2CC_2$, $OC_2DD_2$, $OD_2AA_2$ are cyclic. From these cyclicities we get\n$$\n\\begin{gathered}\n\\angle A_2B_2O = \\angle A_2BO ; \\quad \\angle OB_2C_2 = \\angle OCC_2 ; \\quad \\angle C_2D_2O = \\angle C_2DO ; \\quad \\angle OD_2A_2 = \\angle OAA_2 \\\\\n\\Rightarrow \\angle A_2B_2C_2 + \\angle C_2D_2A_2 = \\angle A_2BO + \\angle OCC_2 + \\angle C_2DO + \\angle OAA_2 = (\\angle ABO + \\angle OAB) + (\\angle OCD + \\angle CDO) \\\\\n= (\\pi - \\angle BOA) + (\\pi - \\angle DOC) = 2\\pi - 2\\angle DOC.\n\\end{gathered}\n$$\nFrom this, we see that $\\angle A_2B_2C_2 + \\angle C_2D_2A_2 = \\pi$ iff $\\angle DOC = \\pi/2$; i.e. quadrilateral $A_2B_2C_2D_2$ is cyclic iff $AC$ and $BD$ are perpendicular.\nAlternate Solution:\n\nObserve that $\\angle OAA_1 = \\angle OAB = \\angle OA'B' = \\angle OA'A_1$, so that quadrilateral $OAA'A_1$ is cyclic. Likewise, $\\angle OAD_1 = \\angle OAD = \\angle OA'D' = \\angle OA'D_1$, so quadrilateral $OA'AD_1$ is cyclic. Thus, $A_1, D_1$ both lie on the circumcircle of $\\triangle OA'A$, so the pentagon $OD_1AA'A_1$ is cyclic. Similar reasoning gives that pentagons $OA_1BB'B_1$, $OB_1CC'C_1$, and $OC_1DD'D_1$ are cyclic.\n\nBut now we simply compute:\n$$\n\\begin{gathered}\n\\angle A_1B_1C_1 + \\angle C_1D_1A_1 = \\angle A_1B_1O + \\angle OB_1C_1 + \\angle C_1D_1O + \\angle OD_1A_1 \\\\\n= \\angle A_1BO + \\angle OCC_1 + \\angle C_1DO + \\angle OAA_1 \\quad \\text{(because of the previously cited cyclicities)} \\\\\n= \\angle ABO + \\angle OCD + \\angle CDO + \\angle OAB = (\\pi - \\angle BOA) + (\\pi - \\angle DOC) = 2\\pi - 2\\angle DOC,\n\\end{gathered}\n$$\nand now the result follows exactly as in the previous solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_n\\}_{n \\ge 0}$ be a sequence of real numbers such that $a_{n+1} \\ge a_n^2 + \\frac{1}{5}$ for all $n \\ge 0$. Prove that $\\sqrt{a_{n+5}} \\ge a_{n-5}^2$ for all $n \\ge 5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to prove that $a_{n+5} \\ge a_n^2$ for $n \\ge 0$, for then we would have $\\sqrt{a_{n+5}} \\ge a_n$ and $a_n \\ge a_{n-5}^2$ for all $n \\ge 5$, which implies the desired result. Adding the inequalities\n$$\n\\begin{align*}\na_{n+5} &\\ge a_{n+4}^2 + \\frac{1}{5}, \\\\\na_{n+4} &\\ge a_{n+3}^2 + \\frac{1}{5}, \\\\\na_{n+3} &\\ge a_{n+2}^2 + \\frac{1}{5}, \\\\\na_{n+2} &\\ge a_{n+1}^2 + \\frac{1}{5}, \\\\\na_{n+1} &\\ge a_n^2 + \\frac{1}{5}\n\\end{align*}\n$$\nyields\n$$\n\\begin{align*}\na_{n+5} &\\ge \\sum_{k=1}^{4} (a_{n+k}^2 - a_{n+k}) + 1 + a_n^2 \\\\\n&= \\sum_{k=1}^{4} \\left( a_{n+k}^2 - a_{n+k} + \\frac{1}{4} \\right) + a_n^2 \\\\\n&= \\sum_{k=1}^{4} \\left( a_{n+k} - \\frac{1}{2} \\right)^2 + a_n^2 \\ge a_n^2,\n\\end{align*}\n$$\nas desired.\n(by Alison Miller) For any $k$, we have the following inequality:\n$$\na_{k+1} + \\frac{1}{20} \\ge a_k^2 + \\frac{1}{5} + \\frac{1}{20} = a_k^2 + \\frac{1}{4} \\ge a_k,\n$$\nbecause $(a_k - \\frac{1}{2})^2 \\ge 0$. Summing up this inequalities over $k = n + 1, \\dots, n + 4$, we have\n$$\na_{n+5} + \\frac{1}{5} \\ge a_{n+1} \\ge a_n^2 + \\frac{1}{5},\n$$\nand similarly $a_n \\ge a_{n-5}^2$, so\n$$\na_{n+5} \\ge a_n^2 \\ge a_{n-5}^4,\n$$\nimplying the desired result.\nDefine the function $f(x) = x^2 + \\frac{1}{5}$, which is increasing for $x \\ge 0$. Also define the function $g(x) = f(x) - x = (x - r_1)(x - r_2)$, where\n$$\nr_1 = \\frac{5 - \\sqrt{5}}{10} \\quad \\text{and} \\quad r_2 = \\frac{5 + \\sqrt{5}}{10}.\n$$\nObserve that $0 < r_1 < r_2 < 1$. So, for nonnegative $x$, $g(x)$ is zero for $x = r_1, r_2$, negative for $x \\in (r_1, r_2)$, and positive otherwise. Therefore, for nonnegative $x$, $f(x) = x$ for $x = r_1, r_2$, $f(x) < x$ for $x \\in (r_1, r_2)$, and $f(x) > x$ otherwise.\nFix $n \\ge 5$, and write $a = |a_{n-5}|$. It suffices to prove that $a_{n+5} \\ge a_{n-5}^4$. We consider the following cases.\n(i) $a \\le r_1$. For $x \\in [0, r_1]$, observe that $x \\le f(x) \\le f(r_1) = r_1$. An easy proof by induction on $k = 0, 1, \\dots, 9$ then shows that $f^{(k)}(x) \\le f^{(k+1)}(x) \\le r_1$. Therefore, $a \\le f(a) \\le f^{(2)}(a) \\le \\dots \\le f^{(10)}(a) = a_{n+5}$. Because $a < 1$, we have $a_{n+5} \\ge a > a^4 = a_{n-5}^4$, as desired.\n(ii) $a \\in (r_2, 1]$. For $x > r_2$, observe that $f(x) > x$ and hence $f(x) > x > r_2$. We can repeat this argument to show that $f^{(k+1)}(x) > f^{(k)}(x) > r_2$ for $k = 0, 1, \\dots, 9$. Therefore, $a_{n+5} = f^{(10)}(a) > f^{(9)}(a) > \\dots > a \\ge a^4 = a_{n-5}^4$, as desired.\n(iii) $a > 1$. Notice that for all $k$, $a_{k+1} > a_k^2$. Therefore, $a_{n+5} > a_{n-5}^{10} = a^6 \\cdot a_{n-5}^4 > a_{n-5}^4$, as desired.\n(iv) $a \\in (r_1, r_2)$. For $x \\in (r_1, r_2)$, observe that $r_1 = f(r_1) < f(x) < f(r_2) = r_2$. Therefore, $a_{n+5} = f^{(10)}(a) > r_1$. Hence, if we can verify that $r_1 > r_2^4$, then we would have\n$$\na_{n+5} > r_1 > r_2^4 > a^4 = a_{n-5}^4,\n$$\nas desired. Note that\n$$\nr_2^4 = \\frac{\\sum_{k=0}^{4} \\binom{4}{k} (\\sqrt{5})^k 5^{4-k}}{10000},\n$$\nwhich expands to\n$$\n\\frac{5^4 + 4 \\cdot 5^3 \\cdot \\sqrt{5} + 6 \\cdot 5^2 \\cdot (\\sqrt{5})^2 + 4 \\cdot 5 \\cdot (\\sqrt{5})^3 + (\\sqrt{5})^4}{10000}.\n$$\nHence $r_2^4 = (7 + 3\\sqrt{5})/50$. Then\n$$\n\\begin{aligned}\n320 < 324 &\\implies 8\\sqrt{5} < 18 \\\\\n&\\implies 7 + 3\\sqrt{5} < 5(5 - \\sqrt{5}) \\\\\n&\\implies \\frac{7 + 3\\sqrt{5}}{50} < \\frac{5 - \\sqrt{5}}{10} \\\\\n&\\implies r_2^4 < r_1,\n\\end{aligned}\n$$\nas needed. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with $O$ as its circumcenter. Let $P$ on line $BC$ be the foot of the altitude from $A$. Assume that $\\angle BCA \\ge \\angle ABC + 30^\\circ$. Prove that $\\angle CAB + \\angle COP < 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** (By Gabriel Carroll and Tiankai Liu) Because $\\angle AOC = 2\\beta$, we have $\\angle CAO = 90^\\circ - \\beta$. Note that in right triangle $APC$, $\\angle CAP = 90^\\circ - \\gamma$, so\n$$\n\\begin{aligned}\n\\angle PAO &= \\angle CAO - \\angle CAP \\\\\n&= \\gamma - \\beta.\n\\end{aligned}\n$$\nBy (1), $\\sin \\angle PAO \\ge \\frac{1}{2}$.\n![](attached_image_1.png)\nLet $N$ be the foot of perpendicular from $O$ to segment $AP$. Then $MPNO$ is a rectangle, so\n$$\n\\begin{aligned}\n\\frac{MP}{OA} &= \\frac{ON}{OA} \\\\\n&= \\sin \\angle PAO \\\\\n&\\ge \\frac{1}{2},\n\\end{aligned}\n$$\nor $2MP \\ge OA = OC$. In right triangles $OCM$ and $OPM$, $OC > CM$ and $OP > MP$. Therefore,\n$$\n\\begin{aligned}\nPC - MP &= MC - 2MP \\\\\n&\\le MC - OC \\\\\n&< MC - MC \\\\\n&= 0.\n\\end{aligned}\n$$\nWe obtain $OP > MP > PC$, as desired.\n\n\n**Second Solution.**\n![](attached_image_2.png)\nLet $K$ and $Q$ be the reflections of $A$ and $P$, respectively, across line $OM$. Then $KQPA$ is a rectangle and $QP = KA$. Note that\n$$\n\\begin{aligned}\n\\angle KOA &= \\angle BOA - \\angle BOK = \\angle BOA - \\angle COA \\\\\n&= 2\\gamma - 2\\beta \\ge 60^\\circ.\n\\end{aligned}\n$$\nBecause $K$ and $A$ are reflections of each other across line $OM$, we have $OK = OA$ and thus $K$ lies on $\\omega$. Hence, $QP = KA = 2R \\sin(\\frac{1}{2}\\angle KOA) \\ge R$. By the **Triangle Inequality**, we have\n$$\n\\begin{aligned}\nOP + R &= Q + OC > QC \\\\\n&= QP + PC \\ge R + PC,\n\\end{aligned}\n$$\nwhich implies (*).\n\n\n**Third Solution.** By (1) and the **Product-to-Sum formulas**,\n$$\n\\begin{aligned}\n4 \\sin \\beta \\cos \\gamma &= 2[\\sin(\\gamma + \\beta) - \\sin(\\gamma - \\beta)] \\\\\n&= 2 \\sin \\alpha - 2 \\sin(\\gamma - \\beta) < 2 - 1 = 1.\n\\end{aligned}\n$$\nApplying the **Extended Law of Sines** gives\n$$\nCP \\cdot CB = (AC \\cos \\gamma)(CB) = (2R \\sin \\beta \\cos \\gamma)(2R \\sin \\alpha) < R^2. \\quad (2)\n$$\nWe present two proofs of (*) based on (2).\n---\n* We choose a point $J$ on ray $CB$ so that $CJ \\cdot CP = R^2$. It follows from this equation and from (2) that $CJ > CB$, so that $\\angle OBC > \\angle OJC$. Since $OC/CJ = PC/CO$ and $\\angle JCO = \\angle PCO$, triangles $JCO$ and $OCP$ are similar and $\\angle OJC = \\angle COP$. It follows that\n$$\n\\begin{aligned}\n\\angle COP &= \\angle OJC < \\angle OBC \\\\\n&= 90^\\circ - \\angle BAC,\n\\end{aligned}\n$$\nor $\\angle CAB + \\angle COP < 90^\\circ$.\n* The power of $P$ with respect to circle $\\omega$ is $BP \\cdot PC = R^2 - OP^2$. By (2), we obtain\n$$\n\\begin{aligned}\nOP^2 &= R^2 - BP \\cdot PC \\\\\n&> PC \\cdot CB - BP \\cdot PC \\\\\n&= PC^2.\n\\end{aligned}\n$$\nThus, $OP > PC$, as desired.\n\n\n**Fourth Solution.** By the Extended Law of Sines, $AB = 2R \\sin \\gamma$ and $AC = 2R \\sin \\beta$. Therefore, by (1) and the **Addition and Subtraction formulas**, we obtain\n$$\n\\begin{aligned}\nBP - PC &= AB \\cos \\beta - AC \\cos \\gamma \\\\\n&= 2R(\\sin \\gamma \\cos \\beta - \\sin \\beta \\cos \\gamma) \\\\\n&= 2R \\sin(\\gamma - \\beta) \\geq R.\n\\end{aligned}\n$$\nBy the Triangle inequality,\n$$\nR + OP = BO + OP > BP \\geq R + PC,\n$$\nwhich implies that $OP > CP$, as desired.\n\n\n**Fifth Solution.** (By Zhiqiang Zhang, China) By the Extended Law of Sines and the Product-to-Sum formulas, we have\n$$\n\\begin{aligned} CP &= AC \\cos \\gamma = 2R \\sin \\beta \\cos \\gamma \\\\\n&= R(\\sin(\\beta + \\gamma) - \\sin(\\gamma - \\beta)) \\\\\n&< R(1 - \\sin 30^\\circ) = \\frac{OC}{2}, \\end{aligned}\n$$\nor $2CP < OC$. On the other hand, applying the **Law of Cosines** in triangle $COP$ yields\n$$\nCP^2 + OC^2 - 2(CP)(OC) \\cos \\angle OCB = OP^2.\n$$\nNow (*) follows from the inequality\n$$\nOP^2 > CP^2 + OC^2 - OC \\cdot OC \\cdot 1 = CP^2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEach vertex of a regular 17-gon is colored red, blue, or green in such a way that no two adjacent vertices have the same color. Call a triangle \"multicolored\" if its vertices are colored red, blue, and green, in some order. Prove that the 17-gon can be cut along nonintersecting diagonals to form at least two multicolored triangles.\n\n(A diagonal of a polygon is a line segment connecting two nonadjacent vertices. Diagonals are called nonintersecting if each pair of them either intersect in a vertex or do not intersect at all.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenote the colors by $1, 2, 3$. Notice that all three colors must be used. This is true because if only two colors were used, the vertex coloring would be of the type $121212 \\ldots$ which is impossible, since $17$ is odd (this would make two adjacent vertices the same color). Hence there are three consecutive vertices colored (without loss of generality) $1, 2, 3$, respectively. (For otherwise, if $3$ consecutive vertices had coloration of the form $a b a$, the next three vertices [overlapping two vertices] would have to be colored $b a b$, etc., forcing the pattern $a b a b a b \\ldots$ which is not possible with an odd number of vertices.)\n\nThus we have $4$ cases depending on the colors of the vertices adjacent to this segment: $21231$, $21232$, $31231$, $31232$. It is easy to see that the desired construction can be done in each case; the figure below illustrates this.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many digits are in the base two representation of $10!$ (factorial)?",
    "options": [],
    "answer": "21",
    "solution": "Solution:\n\nWe write $10! = 2^{8} \\cdot 3^{4} \\cdot 5^{2} \\cdot 7$. The number of digits (base 2) of $10!$ is equal to $\\left[ \\log_{2} 10! \\right] = 8 + \\log_{2} \\left( 3^{4} \\cdot 5^{2} \\cdot 7 \\right)$. Since $2^{13} < 3^{4} \\cdot 5^{2} \\cdot 7 < 2^{14}$, the number of digits is $8 + 13 = 21$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the last digit of $17^{103} + 5$?",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nLet $\\langle N \\rangle$ be the last digit of $N$. $\\left\\langle 17^{2} \\right\\rangle = 9$, $\\left\\langle 17^{3} \\right\\rangle = 3$, $\\left\\langle 17^{4} \\right\\rangle = 1$, and $\\left\\langle 17^{5} \\right\\rangle = 7$. Since this pattern keeps on repeating itself, $\\left\\langle 17^{4N} \\right\\rangle = 1$ for any integer $N$. Thus $\\left\\langle 17^{2 \\cdot 25} \\right\\rangle = \\left\\langle 17^{100} \\right\\rangle = 1$, so $\\left\\langle 17^{103} \\right\\rangle = 3$, and $\\left\\langle 17^{103} + 5 \\right\\rangle = \\langle 3 + 5 \\rangle = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow that there exist infinitely many natural numbers $n$ with the following property: the sum of all the positive divisors of $n$, excluding $n$ itself, equals $n+12$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $p$ be any prime number greater than $3$; we show that $n = 6p$ has the desired property. The positive divisors of $6p = 2 \\cdot 3 \\cdot p$ are $1, 2, 3, p, 2 \\cdot 3, 2 \\cdot p, 3 \\cdot p$, and $2 \\cdot 3 \\cdot p$. The sum of all the factors other than $6p$ is equal to $6p + 12$, as needed.\n\nIt is well known that there are infinitely many prime numbers. Since each value of $p$ gives a different value for $n = 6p$, we obtain infinitely many values for $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle of radius $3$ crosses the center of a square of side length $2$. Find the difference between the areas of the nonoverlapping portions of the figures.",
    "options": [],
    "answer": "9π - 4",
    "solution": "Solution:\nCall the area of the square $s$, the area of the circle $c$, and the area of the overlapping portion $x$. The area of the circle not overlapped by the square is $c - x$ and the area of the square not overlapped by the circle is $s - x$, so the difference between these two is $(c - x) - (s - x) = c - s = 9\\pi - 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind $x-y$, given that $x^{4} = y^{4} + 24$, $x^{2} + y^{2} = 6$, and $x + y = 3$.",
    "options": [],
    "answer": "4/3",
    "solution": "Solution:\n\n$\\frac{24}{6 \\cdot 3} = \\frac{x^{4} - y^{4}}{(x^{2} + y^{2})(x + y)} = \\frac{(x^{2} + y^{2})(x + y)(x - y)}{(x^{2} + y^{2})(x + y)} = x - y = \\frac{4}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46557,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence of ants walk from $(0,0)$ to $(1,0)$ in the plane. The $n$th ant walks along $n$ semicircles of radius $\\frac{1}{n}$ with diameters lying along the line from $(0,0)$ to $(1,0)$. Let $L_{n}$ be the length of the path walked by the $n$th ant. Compute $\\lim_{n \\rightarrow \\infty} L_{n}$.",
    "options": [],
    "answer": "π",
    "solution": "Solution:\nA semicircle of radius $\\frac{1}{n}$ has length $\\frac{1}{2} \\pi \\left(\\frac{2}{n}\\right) = \\frac{\\pi}{n}$, so $n$ such semicircles have total length $\\pi$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlex picks his favorite point $(x, y)$ in the first quadrant on the unit circle $x^{2}+y^{2}=1$, such that a ray from the origin through $(x, y)$ is $\\theta$ radians counterclockwise from the positive $x$-axis. He then computes $\\cos^{-1}\\left(\\frac{4x+3y}{5}\\right)$ and is surprised to get $\\theta$. What is $\\tan(\\theta)$?",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n\n$x = \\cos(\\theta),\\ y = \\sin(\\theta)$. By the trig identity you never thought you'd need, $\\frac{4x+3y}{5} = \\cos(\\theta-\\phi)$, where $\\phi$ has sine $3/5$ and cosine $4/5$. Now $\\theta-\\phi=\\theta$ is impossible, since $\\phi \\neq 0$, so we must have $\\theta-\\phi=-\\theta$, hence $\\theta=\\phi/2$. Now use the trusty half-angle identities to get $\\tan(\\theta)=\\frac{1}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$A$ is the center of a semicircle, with radius $A D$ lying on the base. $B$ lies on the base between $A$ and $D$, and $E$ is on the circular portion of the semicircle such that $E B A$ is a right angle. Extend $E A$ through $A$ to $C$, and put $F$ on line $C D$ such that $E B F$ is a line. Now $E A = 1$, $A C = \\sqrt{2}$, $B F = \\frac{2 - \\sqrt{2}}{4}$, $C F = \\frac{2 \\sqrt{5} + \\sqrt{10}}{4}$, and $D F = \\frac{2 \\sqrt{5} - \\sqrt{10}}{4}$. Find $D E$.",
    "options": [],
    "answer": "sqrt(2 - sqrt(2))",
    "solution": "Solution:\n\nLet $\\theta = \\angle A E D$ and $x = D E$. By the law of cosines on triangle $A D E$, we have\n$$\n1 = 1 + x^{2} - 2x \\cos \\theta \\implies 2x \\cos \\theta = x^{2}.\n$$\nThen by the law of cosines on triangle $C D E$ (note that $C D = \\sqrt{5}$), we have\n$$\n5 = (1 + \\sqrt{2})^{2} + x^{2} - 2(1 + \\sqrt{2}) x \\cos \\theta = (1 + \\sqrt{2})^{2} + x^{2} - (1 + \\sqrt{2}) x^{2}.\n$$\nSolving the quadratic equation gives $x = \\sqrt{2 - \\sqrt{2}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the real solutions of $(2x+1)(3x+1)(5x+1)(30x+1) = 10$.",
    "options": [],
    "answer": "x = (-4 + sqrt(31)) / 15, x = (-4 - sqrt(31)) / 15",
    "solution": "Solution:\n$(2x+1)(3x+1)(5x+1)(30x+1) = [(2x+1)(30x+1)][(3x+1)(5x+1)] = (60x^2 + 32x + 1)(15x^2 + 8x + 1) = (4y+1)(y+1) = 10$, where $y = 15x^2 + 8x$.\n\nThe quadratic equation in $y$ yields $y = 1$ and $y = -\\frac{9}{4}$.\n\nFor $y = 1$, we have $15x^2 + 8x - 1 = 0$, so $x = \\frac{-4 \\pm \\sqrt{31}}{15}$.\n\nFor $y = -\\frac{9}{4}$, we have $15x^2 + 8x + \\frac{9}{4} = 0$, which yields only complex solutions for $x$.\n\nThus the real solutions are $\\frac{-4 \\pm \\sqrt{31}}{15}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate the definite integral\n$$\\int_{-1}^{+1} \\frac{2 u^{332}+u^{998}+4 u^{1664} \\sin u^{691}}{1+u^{666}} \\, \\mathrm{d}u.$$",
    "options": [],
    "answer": "2/333 * (1 + π/4)",
    "solution": "Solution:\n\nThe term $\\frac{4 u^{1664} \\sin u^{691}}{1+u^{666}}$ is odd in $u$, so its integral is $0$.\n\nNow make the substitution $u = v^{1/333} \\Rightarrow \\mathrm{d}u = \\frac{1}{333} v^{-332/333} \\mathrm{d}v$ to find that\n$$\n\\int_{-1}^{+1} \\frac{2 u^{332} + u^{998}}{1 + u^{666}} \\, \\mathrm{d}u = \\frac{1}{333} \\int_{-1}^{+1} \\frac{2 + v^{2}}{1 + v^{2}} \\, \\mathrm{d}v = \\frac{1}{333} \\int_{-1}^{+1} \\left(1 + \\frac{1}{1 + v^{2}}\\right) \\mathrm{d}v.\n$$\n\nThis equals\n$$\n\\frac{2}{333} \\int_{0}^{1} \\left(1 + \\frac{1}{1 + v^{2}}\\right) \\mathrm{d}v = \\frac{2}{333} \\left(1 + \\int_{0}^{1} \\frac{1}{1 + v^{2}} \\, \\mathrm{d}v\\right) = \\frac{2}{333} \\left(1 + \\tan^{-1} 1\\right) = \\frac{2}{333} \\left(1 + \\frac{\\pi}{4}\\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine a monic irreducible polynomial with integral coefficients to be a polynomial with leading coefficient 1 that cannot be factored, and the prime factorization of a polynomial with leading coefficient 1 as the factorization into monic irreducible polynomials. How many not necessarily distinct monic irreducible polynomials are there in the prime factorization of $\\left(x^{8}+x^{4}+1\\right)\\left(x^{8}+x+1\\right)$ (for instance, $(x+1)^{2}$ has two prime factors)?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\n$x^{8}+x^{4}+1=\\left(x^{8}+2 x^{4}+1\\right)-x^{4}=\\left(x^{4}+1\\right)^{2}-\\left(x^{2}\\right)^{2}=\\left(x^{4}-x^{2}+1\\right)\\left(x^{4}+x^{2}+1\\right)=\\left(x^{4}-x^{2}+1\\right)\\left(x^{2}+x+1\\right)\\left(x^{2}-x+1\\right)$, and $x^{8}+x+1=\\left(x^{2}+x+1\\right)\\left(x^{6}-x^{5}+x^{3}-x^{2}+1\\right)$. If an integer polynomial $f(x)=a_{n} x^{n}+\\cdots+a_{0}(\\bmod p)$, where $p$ does not divide $a_{n}$, has no zeros, then $f$ has no rational roots. Taking $p=2$, we find $x^{6}-x^{5}+x^{3}-x^{2}+1$ is irreducible. The prime factorization of our polynomial is thus $\\left(x^{4}-x^{2}+1\\right)\\left(x^{2}-x+1\\right)\\left(x^{2}+x+1\\right)^{2}\\left(x^{6}-x^{5}+x^{3}-x^{2}+1\\right)$, so the answer is 5 .",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the 6-digit number beginning and ending in the digit 2 that is the product of three consecutive even integers.",
    "options": [],
    "answer": "287232",
    "solution": "Solution:\n\nBecause the last digit of the product is $2$, none of the three consecutive even integers end in $0$. Thus they must end in $2, 4, 6$ or $4, 6, 8$, so they must end in $4, 6, 8$ since $2 \\cdot 4 \\cdot 6$ does not end in $2$. Call the middle integer $n$. Then the product is $(n-2) n (n+2) = n^{3} - 4n$, so $n > \\sqrt[3]{200000} = \\sqrt[3]{200 \\cdot 10^{3}} \\approx 60$, but clearly $n < \\sqrt[3]{300000} = \\sqrt[3]{300 \\cdot 10^{3}} < 70$. Thus $n = 66$, and the product is $66^{3} - 4 \\cdot 66 = 287232$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAll subscripts in this problem are to be considered modulo $6$, that means for example that $\\omega_{7}$ is the same as $\\omega_{1}$. Let $\\omega_{1}, \\ldots, \\omega_{6}$ be circles of radius $r$, whose centers lie on a regular hexagon of side length $1$. Let $P_{i}$ be the intersection of $\\omega_{i}$ and $\\omega_{i+1}$ that lies further from the center of the hexagon, for $i=1, \\ldots, 6$. Let $Q_{i}$, $i=1 \\ldots 6$, lie on $\\omega_{i}$ such that $Q_{i}, P_{i}, Q_{i+1}$ are colinear. Find the number of possible values of $r$.",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\nConsider two consecutive circles $\\omega_{i}$ and $\\omega_{i+1}$. Let $Q_{i}, Q_{i}'$ be two points on $\\omega_{i}$ and $Q_{i+1}, Q_{i+1}'$ on $\\omega_{i+1}$ such that $Q_{i}, P_{i}$ and $Q_{i+1}$ are colinear and also $Q_{i}', P_{i}$ and $Q_{i+1}'$. Then $Q_{i} Q_{i}' = 2 \\angle Q_{i} P_{i} Q_{i}' = 2 \\angle Q_{i+1} P_{i} Q_{i+1}' = \\angle Q_{i+1} Q_{i+1}'$. Refer to the center of $\\omega_{i}$ as $O_{i}$. The previous result shows that the lines $O_{i} Q_{i}$ and $O_{i+1} Q_{i+1}$ meet at the same angle as the lines $O_{i} Q_{i}'$ and $O_{i+1} Q_{i+1}'$, call this angle $\\psi_{i}$. $\\psi_{i}$ is a function solely of the circles $\\omega_{i}$ and $\\omega_{i+1}$ and the distance between them (we have just showed that any two points $Q_{i}$ and $Q_{i}'$ on $\\omega_{i}$ give the same value of $\\psi_{i}$, so $\\psi_{i}$ can't depend on this.) Now, the geometry of $\\omega_{i}$ and $\\omega_{i+1}$ is the same for every $i$, so $\\psi_{i}$ is simply a constant $\\psi$ which depends only on $r$. We know $6 \\psi = 0 \\bmod 2\\pi$ because $Q_{7} = Q_{1}$.\n\nWe now compute $\\psi$. It suffices to do the computation for some specific choice of $Q_{i}$. Take $Q_{i}$ to be the intersection of $O_{i} O_{i+1}$ and $\\omega_{i}$ which is further from $O_{i+1}$. We are to compute the angle between $O_{i} Q_{i}$ and $O_{i+1} Q_{i+1}$ which is the same as $\\angle O_{i} O_{i+1} Q_{i+1}$. Note the triangle $\\triangle O_{i} P_{i} O_{i+1}$ is isosceles, call the base angle $\\xi$. We have\n\n$\\angle O_{i} O_{i+1} Q_{i+1} = \\angle O_{i} O_{i+1} P_{i} + \\angle P_{i} O_{i+1} Q_{i+1} = \\xi + \\left(\\pi - 2 \\angle O_{i+1} P_{i} Q_{i+1}\\right) = \\xi + \\left(\\pi - 2\\left(\\pi - \\angle Q_{i} O_{i+1} P_{i} - \\angle P_{i} Q_{i} O_{i+1}\\right)\\right) = \\xi - \\pi + 2\\left(\\xi + (1/2) \\angle P_{i} O_{i} O_{i+1}\\right) = \\xi - \\pi + 2(\\xi + (1/2) \\xi) = 4\\xi - \\pi$.\n\nSo we get $6(4\\xi - \\pi) = 0 \\bmod 2\\pi$. Noting that $\\xi$ must be acute, $\\xi = \\pi/12, \\pi/6, \\pi/4, \\pi/3$ or $5\\pi/12$. $r$ is uniquely determined as $(1/2) \\sec \\xi$ so there are $5$ possible values of $r$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 46565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle and let $\\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ = D_2P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** The key observation is the following Lemma.\n**Lemma.** *Segment* $D_1Q$ *is a diameter of circle* $\\omega$.\n*Proof.* Let $I$ be the center of circle $\\omega$, that is, $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\\omega$ again at $Q'$, and extend segment $AQ'$ through $Q'$ to intersect segment $BC$ at $D'$. We show that $D_2 = D'$, which in turn implies that $Q = Q'$, that is, $D_1Q$ is a diameter of $\\omega$.\n![](attached_image_1.png)\nLet $\\ell$ be the line tangent to circle $\\omega$ at $Q'$, and let $\\ell$ intersect segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\\omega$ is an **excircle** of triangle $AB_1C_1$. Let $\\mathbf{H}_1$ denote the **dilation** with center $A$ and ratio $AD'/AQ'$. Since $\\ell \\perp D_1Q'$ and $BC \\perp D_1Q$, $\\ell \\parallel BC$. Hence, $AB/AB_1 = AC/AC_1 = AD'/AQ'$. Thus, $\\mathbf{H}_1(Q') = D'$, $\\mathbf{H}_1(B_1) = B$, and $\\mathbf{H}_1(C_1) = C$. It also follows that the excircle $\\Omega$ of triangle $ABC$ opposite vertex $A$ is tangent to side $BC$ at $D'$.\n\nIt is well known that\n$$\nCD_1 = \\frac{1}{2}(BC + CA - AB). \\qquad (1)\n$$\nWe compute $BD'$. Let $X$ and $Y$ denote the points of tangency of circle $\\Omega$ with rays $AB$ and $AC$, respectively. Then by equal tangents, $AX = AY$, $BD' = BX$, and $D'C = YC$. Hence,\n$$\n\\begin{align*}\nAX &= AY = \\frac{1}{2}(AX + AY) \\\\\n&= \\frac{1}{2}(AB + BX + YC + CA) \\\\\n&= \\frac{1}{2}(AB + BC + CA).\n\\end{align*}\n$$\nIt follows that\n$$\nBD' = BX = AX - AB = \\frac{1}{2}(BC + CA - AB). \\qquad (2)\n$$\nCombining (1) and (2) yields $BD' = CD_1$. Thus,\n$$\nBD_2 = BD_1 - D_2D_1 = D_2C - D_2D_1 = D_1C = BD',\n$$\nthat is, $D' = D_2$, as desired. ■\n\nNow we prove our main result. Let $M_1$ and $M_2$ be the midpoints of segments $BC$ and $CA$, respectively. Then $M_1$ is also the midpoint of segment $D_1D_2$, from which it follows that $IM_1$ is a midline of triangle $D_1QD_2$. Hence,\n$$\nQD_2 = 2IM_1 \\qquad (3)\n$$\nand $AD_2 \\parallel M_1I$. Similarly, we can prove that $BE_2 \\parallel M_2I$.\n![](attached_image_2.png)\nLet $G$ be the centroid of triangle $ABC$. Thus, segments $AM_1$ and $BM_2$ intersect at $G$. Define transformation $\\mathbf{H}_2$ as the dilation with center $G$ and ratio $-1/2$. Then $\\mathbf{H}_2(A) = M_1$ and $\\mathbf{H}_2(B) = M_2$. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since $AD_2 \\parallel M_1I$ and $BE_2 \\parallel M_2I$, $\\mathbf{H}$ maps lines $AD_2$ and $BE_2$ to lines $M_1I$ and $M_2I$, respectively. It also follows that $\\mathbf{H}_2(P) = I$ and that\n$$\n\\frac{IM_1}{AP} = \\frac{GM_1}{AG} = \\frac{1}{2}\n$$\nor\n$$\nAP = 2IM_1. \\qquad (4)\n$$\nCombining (3) and (4) yields\n$$\nAQ = AP - QP = 2IM_1 - QP = QD_2 - QP = PD_2,\n$$\nas desired.\n\n\n**Second Solution.** From the Lemma, we have\n$$\n\\frac{AQ}{AD_2} = \\frac{r}{r_a},\n$$\nwhere $r$ and $r_a$ are the radii of circles $\\omega$ and $\\Omega$, respectively. Note that\n$$\nr(AB + BC + CA) = 2[ABC]\n$$\nand that\n$$\n\\begin{aligned} r_a(AB + AC - BC) &= 2[I_aAB] + 2[I_aAC] - 2[I_aBC] \\\\\n&= 2[I_aBAC] - 2[I_aBC] = 2[ABC], \\end{aligned}\n$$\nwhere $I_a$ is the center of $\\Omega$ and $[\\mathcal{R}]$ is the area of region $\\mathcal{R}$. Thus,\n$$\n\\frac{AQ}{AD_2} = \\frac{AB + AC - BC}{AB + BC + CA}. \\qquad (5)\n$$\n![](attached_image_3.png)\nApplying **Menelaus's Theorem** to triangle $AD_2C$ and line $BE_2$ gives\n$$\n\\frac{AP}{PD_2} \\cdot \\frac{D_2B}{BC} \\cdot \\frac{CE_2}{E_2A} = 1,\n$$\nor\n$$\n\\frac{AP}{PD_2} = \\frac{BC \\cdot E_2A}{D_2B \\cdot CE_2} = \\frac{BC \\cdot CE_1}{CD_1 \\cdot AE_1} \\\\\n= \\frac{BC}{AE_1} = \\frac{2BC}{AB + AC - BC}.\n$$\nHence,\n$$\n\\frac{AD_2}{PD_2} = 1 + \\frac{AP}{PD_2} = \\frac{AB + AC + BC}{AB + AC - BC},\n$$\nor\n$$\n\\frac{PD_2}{AD_2} = \\frac{AB + AC - BC}{AB + AC + BC}. \\qquad (6)\n$$\nThe desired result now follows from (5) and (6).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Express\n$$\n\\sum_{k=0}^{n} (-1)^k (n-k)!(n+k)!\n$$\nin closed form.",
    "options": [],
    "answer": "\\sum_{k=0}^{n} (-1)^k (n-k)!(n+k)! = \\frac{(n!)^2}{2} + \\frac{(-1)^n (2n+1)!}{2(n+1)}",
    "solution": "**First Solution.** (By Tiankai Liu) Let\n$$\nf(k) = (n+1-k)!(n+k)!\n$$\nfor integers $0 \\le k \\le n + 1$. Note that\n$$\n\\begin{aligned}\nf(k) + f(k + 1) &= (n+1-k)!(n+k)! + (n-k)!(n+k+1)! \\\\\n&= (n + 1 - k + n + k + 1)(n-k)!(n+k)! \\\\\n&= 2(n + 1)(n-k)!(n+k)!.\n\\end{aligned}\n$$\nTherefore,\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n} (-1)^k (n-k)!(n+k)! &= \\frac{1}{2(n+1)} \\sum_{k=0}^{n} (-1)^k [f(k) + f(k+1)] \\\\\n&= \\frac{f(0) + (-1)^n f(n+1)}{2(n+1)} \\\\\n&= \\frac{(n+1)!n! + (-1)^n 0!(2n+1)!}{2(n+1)} \\\\\n&= \\frac{(n!)^2}{2} + \\frac{(-1)^n (2n+1)!}{2(n+1)}.\n\\end{aligned}\n$$\n\n\n**Second Solution.** (by Gabriel Carroll) Let\n$$\nf(x) = \\sum_{k=0}^{\\infty} k!x^k \\quad \\text{and} \\quad g(x) = f(x)f(-x).\n$$\nNote that\n$$\n\\begin{align*}\nf'(x) &= \\sum_{k=0}^{\\infty} (k+1)!(k+1)x^k \\\\\n&= \\sum_{k=0}^{\\infty} (k+2)!x^k - \\sum_{k=0}^{\\infty} (k+1)!x^k \\\\\n&= \\frac{f(x) - x - 1}{x^2} - \\frac{f(x) - 1}{x} \\\\\n&= \\frac{f(x)(1-x) - 1}{x^2}\n\\end{align*}\n$$\nand\n$$\n\\begin{align*}\ng'(x) &= f(-x)f'(x) - f(x)f'(-x) \\\\\n&= \\frac{f(-x)f(x)(1-x) - f(-x)}{x^2} - \\frac{f(x)f(-x)(1+x) - f(x)}{x^2} \\\\\n&= \\frac{f(x) - f(-x)}{x^2} - \\frac{2g(x)}{x}.\n\\end{align*}\n$$\nDenote by $[x^k]h(x)$ the coefficient of $x^k$ in a power series $h(x)$. We have\n$$\n\\begin{align*}\n[x^{2n}]g(x) &= \\frac{1}{2n}[x^{2n-1}]g'(x) \\\\\n&= \\frac{1}{2n}[x^{2n-1}]\\left(\\frac{f(x)-f(-x)}{x^2} - \\frac{2g(x)}{x}\\right) \\\\\n&= \\frac{[x^{2n+1}](f(x)-f(-x))}{2n} - \\frac{[x^{2n}]g(x)}{n}.\n\\end{align*}\n$$\nIt follows that\n$$\n[x^{2n}]g(x) \\left(1 + \\frac{1}{n}\\right) = \\frac{2(2n+1)!}{2n},\n$$\nor\n$$\n[x^{2n}]g(x) = \\frac{(2n+1)!}{n+1}. \\qquad (1)\n$$\n---\nOn the other hand,\n$$\n\\begin{aligned}\ng(x) &= f(x)f(-x) = \\sum_{k=0}^{\\infty} k!x^k \\sum_{k=0}^{\\infty} (-1)^k k!x^k \\\\\n&= \\sum_{n=0}^{\\infty} \\left( \\sum_{k=0}^{n} (-1)^{n-k} k!(n-k)! \\right) x^n.\n\\end{aligned}\n$$\nIt follows that\n$$\n\\begin{aligned}\n[x^{2n}]g(x) &= \\sum_{k=0}^{2n} (-1)^{2n-k} k!(2n-k)! = \\sum_{k=0}^{2n} (-1)^k k!(2n-k)! \\\\\n&= 2 \\sum_{k=0}^{n} (-1)^k k!(2n-k)! - (-1)^n n!^2 \\\\\n&= 2 \\sum_{m=0}^{n} (-1)^{n-m} (n-m)!(n+m)! - (-1)^n n!^2.\n\\end{aligned}\n$$\nThe last step can be seen easily by taking $m = n - k$. Thus,\n$$\n[x^{2n}]g(x) = 2 \\sum_{k=0}^{n} (-1)^{n-k} (n-k)!(n+k)! - (-1)^n n!^2,\n$$\nor\n$$\n(-1)^n [x^{2n}] g(x) + n!^2 = 2 \\sum_{k=0}^{n} (-1)^{-k} (n-k)!(n+k)! \\quad (2)\n$$\nCombining (1) and (2) yields\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n} (-1)^k (n-k)!(n+k)! &= \\frac{(-1)^n [x^{2n}] g(x) + (n!)^2}{2} \\\\\n&= \\frac{(-1)^n (2n+1)!}{2(n+1)} + \\frac{(n!)^2}{2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall three sides of an opaque cube adjacent if someone can see them all at once. Draw a plane through the centers of each triple of adjacent sides of a cube with edge length $1$. Find the volume of the closed figure bounded by the resulting planes.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nThe volume of the figure is half the volume of the cube (which can be seen by cutting the cube into $8$ equal cubes and realizing that the planes cut each of these cubes in half), namely $\\frac{1}{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind $x+y$, given that $x^{2}-y^{2}=10$ and $x-y=2$.",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n$x^{2}-y^{2}=(x-y)(x+y)=2(x+y)=10$, so $x+y=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $J H I Z$ be a rectangle, and let $A$ and $C$ be points on sides $Z I$ and $Z J$, respectively. The perpendicular from $A$ to $C H$ intersects line $H I$ in $X$, and the perpendicular from $C$ to $A H$ intersects line $H J$ in $Y$. Prove that $X, Y$ and $Z$ are collinear (lie on the same line).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nObserve that $\\angle X A I = \\angle X H C = \\angle H C J$. Hence $\\triangle X A I \\sim \\triangle H C J$, and thus $X I / H J = A I / C J$. Likewise, $Y J / H I = C J / A I$. Putting these together yields $X I / H J = H I / Y J$, and hence\n$$\n\\frac{X I}{Z I} = \\frac{Z J}{Y J} \\Rightarrow \\triangle X Z I \\sim \\triangle Z Y J\n$$\nSince $\\angle J Z I = 90^{\\circ}$, this immediately implies $\\angle Y Z X = 180^{\\circ}$, and $X, Y, Z$ are collinear.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46570,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall three sides of an opaque cube adjacent if someone can see them all at once. Draw a plane through the centers of each triple of adjacent sides of a cube with edge length $1$. Find the volume of the closed figure bounded by the resulting planes.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nThe volume of the figure is half the volume of the cube (which can be seen by cutting the cube into $8$ equal cubes and realizing that the planes cut each of these cubes in half), namely $\\frac{1}{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind $\\log_{n}\\left(\\frac{1}{2}\\right) \\log_{n-1}\\left(\\frac{1}{3}\\right) \\cdots \\log_{2}\\left(\\frac{1}{n}\\right)$ in terms of $n$.",
    "options": [],
    "answer": "(-1)^{n-1}",
    "solution": "Solution:\n\nUsing $\\log \\frac{1}{x} = -\\log x$ and $\\log_{b} a = \\frac{\\log a}{\\log b}$, we get that the product equals\n$$\n\\frac{(-\\log 2)(-\\log 3) \\cdots (-\\log n)}{\\log n \\cdots \\log 3 \\log 2} = (-1)^{n-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe polynomial $3 x^{5}-250 x^{3}+735 x$ is interesting because it has the maximum possible number of relative extrema and points of inflection at integer lattice points for a quintic polynomial. What is the sum of the $x$-coordinates of these points?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nThe first derivative is $15 x^{4}-750 x^{2}+735$, whose roots (which give the relative extrema) sum to $750 / 15 = 50$. The second derivative is $60 x^{3}-1500 x$, whose roots (which give the points of inflection) sum to $1500 / 60 = 25$, for a grand total of $75$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe UC Berkeley math department is about to move into a new, one-story building consisting of a $2001 \\times 2001$ square grid of rooms. They would like to install doors between adjacent rooms so that each room has exactly two doors. Prove that this cannot be done.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose that it can be done. Color the building in checkerboard fashion, and suppose that we obtain $a$ white rooms and $b$ black rooms. Each door connects a white room with a black room. So, if we consider, for each white room, the number of doors adjoining it, we will count each door exactly once. Since every room is to have 2 doors, the total number of doors will be $2a$ by this count. But similarly, if we consider, for each black room, the number of adjoining doors, each door will be counted once, so that the total number of doors is also equal to $2b$. So, $2a = 2b$, or $a = b$. It follows that the total number of rooms is $a + b = 2a$, an even number. But we also know the number of rooms is $2001^2$, an odd number - contradiction. Hence, the desired condition cannot be met.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n5 married couples gather at a party. As they come in and greet each other, various people exchange handshakes - but, of course, people never shake hands with themselves or with their own respective spouses. At the end of the party, one woman goes around asking people how many hands they shook, and she gets nine different answers. How many hands did she herself shake?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nSuppose that there were $n$ couples, and the woman asked all $2n-1$ other attendees how many hands they shook and received $2n-1$ different answers. We will show that she herself shook $n-1$ hands; hence, in our particular case, the answer is $4$.\n\nWe work by induction. When $n=1$, there is one couple, and no handshakes can occur, proving the base case. Now suppose the result holds for $n$ couples; we will prove it is valid for $n+1$ couples. With $n+1$ couples present, the woman receives $2n+1$ different answers to her question. But no person $P$ can shake more than $2n$ hands (for $2n+2$ people, minus $P$ and $P$'s spouse); hence, these $2n+1$ numbers must be exactly $0,1,2, \\ldots, 2n$ in some order. In particular, one of these people, $A$, shook everyone else's hand except $A$'s own spouse (that accounts for the \"$2n$\" answer), and another, $B$, shook no hands (the \"$0$\" answer). Because $B$ did not shake $A$'s hand, $A$ and $B$ must be married to each other. The remaining $2n$ people include the woman who asked the question, together with those who answered $1,2, \\ldots, 2n-1$ to her question. Now pretend that $A$ and $B$ had not attended the party, so we are left with $n$ couples. Each of these people shook hands with $A$ and not with $B$; therefore, when $A$ and $B$ are removed, their handshake counts become $0,1,2, \\ldots, 2n-2$. Hence, by the induction hypothesis, the questioner shook $n-1$ hands. But now, if we put $A$ and $B$ back in, we note that the woman shook $A$'s hand as well (and not $B$'s). So, altogether, she shook $n$ hands. This completes the induction step, and now the proof is done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind $(x+1)\\left(x^{2}+1\\right)\\left(x^{4}+1\\right)\\left(x^{8}+1\\right) \\cdots$, where $|x|<1$.",
    "options": [],
    "answer": "1/(1 - x)",
    "solution": "Solution:\n\nLet $S = (x+1)\\left(x^{2}+1\\right)\\left(x^{4}+1\\right)\\left(x^{8}+1\\right) \\cdots = 1 + x + x^{2} + x^{3} + \\cdots$.\n\nSince $x S = x + x^{2} + x^{3} + x^{4} + \\cdots$, we have $(1-x) S = 1$, so $S = \\frac{1}{1-x}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn a certain unidirectional highway, trucks move steadily at $60$ miles per hour spaced $1/4$ of a mile apart. Cars move steadily at $75$ miles per hour spaced $3$ seconds apart. A lone sports car weaving through traffic at a steady forward speed passes two cars between each truck it passes. How quickly is it moving in miles per hour?",
    "options": [],
    "answer": "90",
    "solution": "Solution:\n\nThe cars are $1/8$ of a mile apart. Consider the reference frame in which the trucks move at $0$ velocity (and the cars move at $15$). Call the speed of the sports car in this reference frame $v$. The amount of time for the sports car to move from one truck to the next is $\\frac{1/4 \\text{ miles}}{v}$, and the amount of time for two regular cars to pass the truck is $\\frac{1/8 \\text{ miles}}{15\\ \\mathrm{mph}}$. Equating these, we get $v=30$, and $v+60=90\\ \\mathrm{mph}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that\n\na. each contestant solved at most six problems, and\n\nb. for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy.\n\nProve that there is a problem that was solved by at least three girls and at least three boys.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** (By Zhiqiang Zhang, China) We proceed indirectly. Assume that no problem is both girl-easy and boy-easy.\n\n**Lemma 1.** Let $A = \\{p_1, p_2, \\dots, p_k\\}$ denote the set of all girl-hard problems. Let $B = \\{p_{k+1}, p_{k+2}, \\dots, p_{k+m}\\}$ denote the set of all boy-hard problems that are not in $A$. Then $k \\ge 11$, $m \\ge 11$, and consequently,\n$$\nk + m \\geq 22. \\tag{1}\n$$\n*Proof.* By our assumption, each problem is either girl-hard, boy-hard, or both. Thus, if $P$ is the set of all the problems, then $A \\cup B = P$. A boy contestant $b$ can solve at most 6 problems in set $A$, so there are at most $2 \\cdot 6 = 12$ girls who each solved a problem in $A$ that $b$ solved. By condition (b), there are at least $21 - 12$ girls who each solved a problem in $B$ that $b$ solved. It follows that each boy must solve some problems in set $B$. But each problem in set $B$ can be solved by at most 2 boys, implying that $21 \\le 2m$ and hence $m \\ge 11$. In exactly the same way, we can prove that $k \\ge 11$. ■\n\nFor $1 \\le i \\le k+m$, let $x_i$ be the number of contestants who solved problem $p_i$. In the light of condition (a), we have\n$$\nx_1 + x_2 + \\dots + x_{k+m} \\le 6 \\times 42. \\tag{2}\n$$\nFor each $p_i \\in P = A \\cup B$, let $q_i$ be the number of girl and boy pairs $(g, b)$ such that girl $g$ and boy $b$ both solved $p_i$. By condition (b), $\\sum_{i=1}^{k+m} q_i \\ge 21^2$. For each $i = 1, 2, \\dots, k+m$, note that $x_i \\ge 2$ because each problem was solved by at least one girl and one boy. Therefore,\n$$\nq_i \\le \\max\\{(x_i - 1) \\times 1, (x_i - 2) \\times 2\\} \\le 2x_i - 3.\n$$\nHence, by (2) we obtain\n$$\n\\begin{align*} \n21^2 &\\le \\sum_{i=1}^{k+m} q_i \\le (2x_1 - 3) + (2x_2 - 3) + \\dots + (2x_{k+m} - 3) \\\\ \n&= 2(x_1 + x_2 + \\dots + x_{k+m}) - 3(k+m) \\\\ \n&\\le 2 \\times 6 \\times 42 - 3(k+m), \n\\end{align*}\n$$\nor\n$k+m \\le 21,$\ncontradicting (1).\n\nThus, our assumption was wrong and there is a problem solved by at least three girls and three boys.\n\n\n**Second Solution.** (By Gabriel Carroll and Liang Xiao, China) Let $p$ be the number of problems. For the sake of contradiction, we assume that no problem is both girl-easy and boy-easy. As in the first solution, we prove that this assumption implies both that $p \\ge 22$ and that $p \\le 21$, which is impossible.\n\n**Lemma 2.** *There are at least 11 girl-easy problems and 11 boy-easy problems.*\n\n*Proof.* Suppose some girl solves no boy-easy problems. Then every problem she solves was solved by at most two boys. Since she solves at most six problems, at most 12 boys solve a problem that she solves, contradicting condition (b). Thus, every girl solved a boy-easy problem. Since each boy-easy problem was solved by at most two girls, there must be at least $\\lceil \\frac{21}{2} \\rceil = 11$ boy-easy problems. Likewise, there are at least 11 girl-easy problems. ■\n\nBy our assumption, there is no problem that is both boy-easy and girl easy. Thus, there are at least 22 problems, that is,\n$$\np \\ge 22. \\tag{3}\n$$\nAssume that the $i$th problem was solved by $g_i$ girls and $b_i$ boys. For each $i$, either $g_i \\in \\{1,2\\}$, or $g_i > 2$ and $b_i \\le 2$. Consider the quantity\n$$\nQ = (g_i - 2)(b_i - 2).\n$$\nIf $g_i = 2$, then $Q = 0$; if $g_i = 1$, then $Q = 2 - b_i \\le 1$ since $b_i \\ge 1$ by assumption; finally, if $g_i > 2$ and $b_i \\le 2$, then $Q \\le 0$. In any case, we have $Q = (g_i - 2)(b_i - 2) \\le 1$, that is,\n$$\ng_i b_i \\le 2g_i + 2b_i - 3. \\quad (4)\n$$\nNow $g_i b_i$ counts the number of pairs $(g, b)$ in which girl $g$ and boy $b$ both solved problem $i$. It is given that every possible pair $(g, b)$ is counted for some $i$, so\n$$\n\\sum_{i=1}^{p} g_i b_i \\ge 21^2. \\quad (5)\n$$\nOn the other hand, everyone solved at most six problems, so $\\sum_{i=1}^{p} g_i \\le 21 \\cdot 6$, and likewise $\\sum_{i=1}^{p} b_i \\le 21 \\cdot 6$. Thus, by (4) and (5), we obtain\n$$\n\\begin{aligned} 21^2 &\\le \\sum_{i=1}^{p} g_i b_i \\le \\sum_{i=1}^{p} (2g_i + 2b_i - 3) \\\\ &= 2 \\left( \\sum_{i=1}^{p} g_i + \\sum_{i=1}^{p} b_i \\right) - 3p \\\\ &\\le 2(21 \\cdot 6 + 21 \\cdot 6) - 3p = 21 \\cdot 24 - 3p. \\end{aligned}\n$$\nIt follows that $3p \\le 21 \\cdot 24 - 21^2 = 21 \\cdot 3$, or\n$$\np \\le 21, \\quad (6)\n$$\ncontradicting (3).\n\nThus, our initial assumption was wrong, and there is a problem that was solved by three boys and three girls.\n\n\n**Third Solution.** (Based on work by Ian Le) We provide an alternative proof of (6).\n\nFor each $i = 1, 2, \\dots, p$, let $\\epsilon_i = \\min\\{g_i, b_i\\}$, and let $M_i = \\max\\{g_i, b_i\\}$. Note that $\\epsilon_i$ is either 1 or 2. Assume without loss of generality that $\\epsilon_1 = \\epsilon_2 = \\dots = \\epsilon_r = 2$ and that $\\epsilon_{r+1} = \\epsilon_{r+2} = \\dots = \\epsilon_p = 1$. Then, by (5),\n$$\n\\begin{aligned} 441 &\\le \\sum_{i=1}^{p} g_i b_i = \\sum_{i=1}^{p} \\epsilon_i M_i = 2 \\sum_{i=1}^{r} M_i + \\sum_{i=r+1}^{p} M_i \\\\ &= 2 \\sum_{i=1}^{p} M_i - \\sum_{i=r+1}^{p} M_i. \\end{aligned}\n$$\nNote that $M_i \\ge 1$. It follows that\n$$\n441 \\le 2 \\sum_{i=1}^{p} M_i - p + r. \\qquad (7)\n$$\nOn the other hand, since each girl or boy solved at most 6 problems, we must have $\\sum_{i=1}^{p}(g_i + b_i) \\le 2 \\cdot 6 \\cdot 21 = 252$, or\n$$\n252 \\ge \\sum_{i=1}^{p} (g_i + b_i) = \\sum_{i=1}^{p} (\\epsilon_i + M_i) = p + r + \\sum_{i=1}^{p} M_i.\n$$\nMultiplying by 2 on both sides of this inequality and rearranging gives\n$$\n2 \\sum_{i=1}^{p} M_i - p + r \\le 2 \\cdot 252 - 3p - r = 504 - 3p - r. \\quad (8)\n$$\nCombining (7) and (8), we obtain $441 \\le 504 - 3p - r$, or\n$$\np \\le (504 - 441)/3 = 21,\n$$\nas desired.\n\n\n**Fourth Solution.** We introduce the following symbols: $G$ is the set of girls at the competition, $B$ is the set of boys, $P$ is the set of problems, $P(g)$ is the set of problems solved by $g \\in G$, and $P(b)$ is the set of problems solved by $b \\in B$. Finally, $G(p)$ is the set of girls that solve $p \\in P$ and $B(p)$ is the set of boys that solve $p$. In terms of this notation, we have that for all $g \\in G$ and $b \\in B$,\n(a) $|P(g)| \\le 6$, $|P(b)| \\le 6$, and (b) $P(g) \\cap P(b) \\ne \\emptyset$\n\nHence, a problem is boy-easy (resp., boy-hard) if and only if $|B(p)| \\ge 3$ ($|B(p)| \\le 2$); a problem is girl-easy (resp., girl-hard) if and only if $|G(p)| \\ge 3$ (resp., $|G(p)| \\le 2$). We wish to prove that some $p \\in P$ satisfies $|G(p)| \\ge 3$ and $|B(p)| \\ge 3$. For sake of contradiction, assume instead that every problem is either boy-hard or girl-hard or both, that is, for each $p \\in P$, either $|G(p)| \\le 2$ or $|B(p)| \\le 2$ or both. We shall reach a contradiction by counting, in two different ways, all ordered triples $(p, g, b)$ such that $p \\in P(g) \\cap P(b)$. With\n$$\nT = \\{ (p, g, b) \\mid p \\in P(g) \\cap P(b) \\},\n$$\ncondition (b) yields\n$$\n|T| = \\sum_{g \\in G} \\sum_{b \\in B} |P(g) \\cap P(b)| \\ge |G| \\cdot |B| = 21^2. \\qquad (9)\n$$\nWe continue by noting that\n$$\n\\sum_{p \\in P} |G(p)| = \\sum_{g \\in G} |P(g)| \\leq 6|G| \\quad \\text{and} \\quad \\sum_{p \\in P} |B(p)| \\leq 6|B|. \\quad (10)\n$$\nThe equality in (10) is obtained by counting the total number of problems, including multiplicity, solved by all the girls. (That is, if a problem solved by *n* girls, then it is counted *n* times.)\n\nLet $P_{ge}$ and $P_{gh}$ denote the sets of all girl-easy and girl-hard problems, respectively. By our assumption, if $p \\in P_{ge}$, then it must be boy-hard, i.e. $|B(p)| \\leq 2$. Hence,\n$$\n\\begin{aligned} |T| &= \\sum_{p \\in P} |G(p)| \\cdot |B(p)| \\\\ &= \\sum_{p \\in P_{ge}} |G(p)| \\cdot |B(p)| + \\sum_{p \\in P_{gh}} |G(p)| \\cdot |B(p)|, \\end{aligned}\n$$\nor\n$$\n|T| \\leq 2 \\sum_{p \\in P_{ge}} |G(p)| + 2 \\sum_{p \\in P_{gh}} |B(p)|. \\quad (11)\n$$\n**Lemma 3.** We have\n$$\n\\sum_{p \\in P_{gh}} |G(p)| \\geq |G|; \\quad \\sum_{p \\in P_{ge}} |G(p)| \\leq 5|G|.\n$$\nand\n$$\n\\sum_{p \\in P_{ge}} |B(p)| \\geq |B|; \\quad \\sum_{p \\in P_{gh}} |B(p)| \\leq 5|B|.\n$$\n*Proof.* Let $g \\in G$ be arbitrary. By the Pigeonhole Principle, conditions (a) and (b) imply that $g$ solves some problem $p$ that is solved by at least \\lceil 21/6\\rceil = 4 boys. By assumption, $p$ is boy-easy, implying that $p$ is girl-hard. Thus, every girl solves at least one problem in $P_{gh}$. Hence,\n$$\n\\sum_{p \\in P_{gh}} |G(p)| \\geq |G|. \\quad (12)\n$$\nIn view of (10) and (12) we have\n$$\n\\sum_{p \\in P_{ge}} |G(p)| = \\sum_{p \\in P} |G(p)| - \\sum_{p \\in P_{gh}} |G(p)| \\leq 5|G|.\n$$\nAlso, each boy solves a problem that is solved by at least four girls, so each boy solves a girl-easy problem. Thus,\n$$\n\\sum_{p \\in P_{ge}} |B(p)| \\geq |B|.\n$$\nIt follows from this inequality and (10) that\n$$\n\\sum_{p \\in P_{gh}} |B(p)| \\leq 5|B|\n$$\nas well.\n\nUsing Lemma 3 and equality (11), we find\n$$\nT \\leq 10|G| + 10|B| = 20 \\times 21.\n$$\nThis contradicts (9), so the proof is complete.\n\n\n**Fifth Solution.** Keep the notation of the fifth solution, and again assume that no problem is both girl-easy and boy-easy. For each $p \\in P$, color $p$ red if it is boy-easy, color $p$ blue if it is girl-easy, and color $p$ either color if it is both boy- and girl-hard.\n\nConsider a chessboard with 21 rows, each representing one of the girls, and 21 columns, each representing one of the boys. For each $g \\in G$ and $b \\in B$, by condition (b) there exists a problem $p$ in $P(g) \\cap P(b)$; and assign $p$'s color to the square corresponding to $(g, b)$. By the **Pigeonhole Principle**, one of the two colors is assigned to at least \\lceil 441/2\\rceil = 221 squares, and thus some row has at least \\lceil 221/21\\rceil = 11 blue squares or some column has at least 11 red squares.\n\nFirst suppose that there is a row, corresponding to $g \\in G$, which has at least 11 blue squares. Then for each of the 11 squares, the blue problem that was chosen in assigning the color was solved by at most 2 boys. Thus, we account for at least \\lceil 11/2\\rceil = 6 distinct problems solved by $g$. In view of condition (a), $g$ solves only these problems. But then at most 12 boys solve a problem also solved by $g$, in violation of condition (b). A similar proof shows that it is impossible for any column to contain 11 red squares.\n\nHence, our original assumption was wrong and there are some $p \\in P$ satisfying $|G(p)| \\geq 3$ and $|B(p)| \\geq 3$.\n\n\n**Sixth Solution.** (By Reid Barton) Assign each problem a unique letter, and also number the boys 1, 2, ..., 21 and number the girls 1, 2, ..., 21. Construct a 21 × 21 matrix of letters as follows: in the $i$th row and $j$th column, write the letter of any problem that both the $i$th girl and the $j$th boy solved — at least one such problem exists by condition (b). If we consider the $i$th row, each letter in that row corresponds to a problem that the $i$th girl solved. Since each girl solved at most six problems, each row contains at most 6 distinct letters. Similarly, each column contains at most 6 distinct letters.\n\n**Lemma 4.** In each row (resp., column), consider the letters which appear at least three times. At least 11 squares in the row (resp., column) contain one of these letters.\n\n*Proof.* There are at most 6 different letters, and they cannot all appear at most twice, since there are $21 > 12$ letters total. So at most 5 different letters appear at most twice, giving a total of at most 10 squares containing letters appearing at most twice. Then at least 11 other squares each contain a letter that appears at least three times. ■\n\nIn the matrix, color all the squares which contain letters appearing at least three times in the same row (resp., column) in red (resp., blue). By Lemma 4, each row contains at least 11 red squares, so the total number of red squares is at least $21 \\times 11$. Similarly, each column contains at least 11 blue squares, so the total number of blue squares is at least $21 \\times 11$. Since there are only $21 \\times 21 < 21 \\times 11 + 21 \\times 11$ total squares, some square is colored both red and blue. Because the letter in this square appears in three different columns and three different rows, at least three boys and three girls solved the corresponding problem. Thus, we find the problem satisfying the desired property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$, and $c$ be nonnegative real numbers such that\n$$\na^2 + b^2 + c^2 + abc = 4.\n$$\nProve that\n$$\n0 \\le ab + bc + ca - abc \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** (By Richard Stong) From the condition, at least one of $a, b$, and $c$ does not exceed $1$, say $a \\le 1$. Then\n$$\nab + bc + ca - abc = a(b + c) + bc(1 - a) \\ge 0.\n$$\nTo obtain equality, we have $a(b + c) = bc(1 - a) = 0$. If $a = 1$, then $b + c = 0$ or $b = c = 0$, which contradicts the given condition $a^2 + b^2 + c^2 + abc = 4$. Hence $1 - a \\ne 0$ and only one of $b$ and $c$ is $0$. Without loss of generality, say $b = 0$. Therefore $b + c > 0$ and $a = 0$. Plugging $a = b = 0$ back into the given condition gives $c = 2$. By permutation, the lower bound holds if and only if $(a, b, c)$ is one of the triples $(2, 0, 0)$, $(0, 2, 0)$, and $(0, 0, 2)$.\n\nNow we prove the upper bound. Let us note that some two of the three numbers $a, b$, and $c$ are both greater than or equal to $1$ or less than or equal to $1$. Without loss of generality, we assume that the numbers with this property are $b$ and $c$. Then we have\n$$\n(1 - b)(1 - c) \\ge 0. \\qquad (1)\n$$\nThe given equality $a^2 + b^2 + c^2 + abc = 4$ and the inequality $b^2 + c^2 \\ge 2bc$ imply\n$$\na^2 + 2bc + abc \\le 4, \\quad \\text{or} \\quad bc(2 + a) \\le 4 - a^2.\n$$\nDividing both sides of the last inequality by $2 + a$ yields\n$$\nbc \\le 2 - a. \\qquad (2)\n$$\nCombining (1) and (2) gives\n$$\n\\begin{aligned}\nab + bc + ac - abc &\\le ab + 2 - a + ac(1 - b) \\\\\n&= 2 - a(1 + bc - b - c) \\\\\n&= 2 - a(1 - b)(1 - c) \\le 2,\n\\end{aligned}\n$$\nas desired.\n\nThe last equality holds if and only if $b = c$ and $a(1 - b)(1 - c) = 0$. Hence, equality for the upper bound holds if and only if $(a, b, c)$ is one of the triples $(1, 1, 1)$, $(0, \\sqrt{2}, \\sqrt{2})$, $(\\sqrt{2}, 0, \\sqrt{2})$, and $(\\sqrt{2}, \\sqrt{2}, 0)$.\n\n\n**Second Solution.** (by Oaz Nir) We prove only the upper bound here. Either two of $a, b, c$ are less than or equal to $1$, or two are greater than or equal to $1$. Assume $b$ and $c$ have this property. Then\n$$\nb + c - bc = 1 - (1 - b)(1 - c) \\le 1. \\qquad (3)\n$$\nViewing the given equality as a quadratic equation in $a$ and solving for $a$ yields\n$$\na = \\frac{-bc \\pm \\sqrt{(b^2 - 4)(c^2 - 4)}}{2}.\n$$\nNote that\n$$(b^2 - 4)(c^2 - 4) = b^2c^2 - 4(b^2 + c^2) + 16$$\n$$\\leq b^2c^2 - 8bc + 16 = (4 - bc)^2.$$\nFor the given equality to hold, we must have $b, c \\leq 2$ so that $4 - bc \\geq 0$. Hence,\n$$\na \\leq \\frac{-bc + |4 - bc|}{2} = \\frac{-bc + 4 - bc}{2} = 2 - bc,\n$$\nor\n$$\n2 - bc \\geq a. \\qquad (4)\n$$\nCombining (3) and (4) gives\n$$\n2 - bc \\geq a(b + c - bc) = ab + ac - abc,\n$$\nor\n$$\nab + ac + bc - abc \\leq 2,\n$$\nas desired.\n\n\n**Third Solution.** We prove only the upper bound here. Define functions $f, g$ as\n$$f(x, y, z) = x^2 + y^2 + z^2 + xyz = (x + y)^2 + z^2 - (2 - z)xy,$$\n$$g(x, y, z) = xy + yz + zx - xyz = z(x + y) + (1 - z)xy$$\nfor all nonnegative numbers $x, y, z$. Observe that if $z \\leq 1$, then both $f$ and $g$ are unbounded, increasing functions of $x$ and $y$.\n\nAssume that $f(a, b, c) = 4$ and, without loss of generality, that $a \\geq b \\geq c \\geq 0$. Then $c \\leq 1$.\nLet $a' = (a+b)/2$. Because $a+b = a'+a'$ and $ab \\leq (\\frac{a-b}{2})^2 + ab = a'^2$, we have\n$$\nf(a', a', c) \\leq f(a, b, c) = 4 \\quad \\text{and} \\quad g(a', a', c) \\geq g(a, b, c).\n$$\nNow increase $a'$ to $e \\geq 0$ such that $f(e, e, c) = 4$. Note that $g(e, e, c) \\geq g(a', a', c)$. It suffices to prove that $g(e, e, c) \\leq 2$.\n\nSince $f(e, e, c) = 2e^2 + c^2 + e^2c = 4$, $e^2 = (4 - c^2)/(2 + c) = 2 - c$.\nWe obtain that\n$$\n\\begin{aligned}\ng(e, e, c) - 2 &= 2ec + (1-c)e^2 - 2 \\\\\n&\\le 2e(2-e^2) + (e^2-1)e^2 - 2 \\\\\n&= e^4 - 2e^3 - e^2 + 4e - 2 \\\\\n&= (e-1)^2(e^2-1) \\le 0,\n\\end{aligned}\n$$\nsince $1 \\le e \\le \\sqrt{2}$.\n\n\n**Fourth Solution.** We prove only the upper bound here. It is clear that $a, b, c \\le 2$. If $a = 0$, then the given equality reduces to $b^2 + c^2 = 4$. Then $ab + bc + ca - abc = bc \\le \\frac{(b-c)^2}{2} + bc = \\frac{b^2+c^2}{2} = 2$. Similarly, the inequality is true if $b$ or $c$ equals $0$.\n\nSuppose that $a, b, c > 0$. Solving for $a$ yields\n$$\n\\begin{aligned}\na &= \\frac{-bc + \\sqrt{b^2c^2 - 4(b^2 + c^2 - 4)}}{2} \\\\\n&= \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.\n\\end{aligned}\n$$\nWe make the trigonometric substitution $b = 2 \\sin u$ and $c = 2 \\sin v$, where $0^\\circ < u, v < 90^\\circ$. Then\n$$\na = 2(-\\sin u \\sin v + \\cos u \\cos v) = 2 \\cos (u + v).\n$$\nSet $u = B/2$, $v = C/2$, and $A = 180^\\circ - B - C$. Then $a = 2 \\cos(u+v) = 2 \\sin(A/2)$, $b = 2 \\sin(B/2)$, and $c = 2 \\sin(C/2)$, where $A, B, C$ are the angles of a triangle. We have\n$$\n\\begin{aligned}\nab &= 4 \\sin \\frac{A}{2} \\sin \\frac{B}{2} = 2 \\sqrt{\\sin A \\tan \\frac{A}{2} \\sin B \\tan \\frac{B}{2}} \\\\\n&= 2 \\sqrt{\\sin A \\tan \\frac{B}{2} \\sin B \\tan \\frac{A}{2}}\n\\end{aligned}\n$$\nBy the **AM-GM Inequality**, this is at most\n$$\n\\begin{aligned}\n\\sin A \\tan \\frac{B}{2} + \\sin B \\tan \\frac{A}{2} \\\\\n&= \\sin A \\cot \\frac{A+C}{2} + \\sin B \\cot \\frac{B+C}{2}.\n\\end{aligned}\n$$\nLikewise,\n$$\nbc \\le \\sin B \\cot \\frac{B+A}{2} + \\sin C \\cot \\frac{C+A}{2},\n$$\n$$\nca \\le \\sin C \\cot \\frac{C+B}{2} + \\sin A \\cot \\frac{A+B}{2}.\n$$\nTherefore, applying the **Sum-to-product**, **Product-to-sum**, and **Double-angle formulas**, we have\n$$\n\\begin{align*}\nab + bc + ca &\\le (\\sin A + \\sin B) \\cot \\frac{A+B}{2} + (\\sin B + \\sin C) \\cot \\frac{B+C}{2} \\\\\n&\\quad + (\\sin C + \\sin A) \\cot \\frac{C+A}{2} \\\\\n&= 2 \\cos \\frac{A-B}{2} \\cos \\frac{A+B}{2} + 2 \\cos \\frac{B-C}{2} \\cos \\frac{B+C}{2} \\\\\n&\\quad + 2 \\cos \\frac{C-A}{2} \\cos \\frac{C+A}{2} \\\\\n&= 2(\\cos A + \\cos B + \\cos C) \\\\\n&= 6 - 4 \\left( \\sin^2 \\frac{A}{2} + \\sin^2 \\frac{B}{2} + \\sin^2 \\frac{C}{2} \\right) \\\\\n&= 6 - (a^2 + b^2 + c^2).\n\\end{align*}\n$$\nUsing the given equality, this last quantity equals $2 + abc$. It follows that\n$$\nab + bc + ca \\le 2 + abc,\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46579,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a set $S$, let $|S|$ denote the number of elements in $S$. Let $A$ be a set of positive integers with $|A| = 2001$. Prove that there exists a set $B$ such that\n(i) $B \\subseteq A$;\n(ii) $|B| \\geq 668$;\n(iii) for any $u, v \\in B$ (not necessarily distinct), $u + v \\notin B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** (By Reid Barton) For a positive integer $n$, let $Z_n$ denote the set of residues modulo $n$. Let $\\phi(n)$ be the **Euler function** which is defined to be the number of integers between $1$ and $n$ relatively prime to $n$. Call a set $A$ of residues modulo $3^n$ **sum-free** if for any $a, b \\in A$, $a+b$ is not (congruent to) an element of $A$.\n\nLemma. For any $n \\ge 1$, there exist $3^n - 1$ sum-free sets of $3^{n-1}$ residues modulo $3^n$ such that every nonzero residue modulo $3^n$ appears in exactly $3^{n-1}$ of the subsets.\n\n*Proof.* We construct the desired subsets inductively. For $n = 1$ we take the sets $\\{1\\}$ and $\\{2\\}$.\n\nLet $n \\ge 1$, and suppose that the statement holds for $n$, that is, we have $3^n - 1$ sum-free subsets $A_1, A_2, \\dots, A_{3^{n-1}}$ of $Z_{3^n}$ such that every nonzero element of $Z_{3^n}$ belongs to exactly $3^{n-1}$ of the $A_i$. Construct sets $B_1, B_2, \\dots, B_{3^{n-1}}$ by\n$$\nB_i = \\{ x \\in Z_{3^{n+1}} \\mid x \\equiv m' \\pmod{3^n}, m' \\in A_i \\}.\n$$\nThen each $B_i$ contains $3|A_i| = 3^n$ residues, and the $B_i$ are sum-free, because if $a, b \\in B_i$, $(a+b) \\pmod{3^n}$ is not in $A_i$ so $a+b$ is not in $B_i$. Moreover, each element $x$ of $Z_{3^{n+1}}$ which is not $0$ modulo $3^n$ (i.e., all elements except $0, 3^n, 2 \\cdot 3^n$) is in exactly $3^{n-1}$ of the $B_i$, namely those corresponding to the $A_i$ containing $x \\pmod{3^n}$.\n\nNow define sets\n$$\nC = \\{3^n, 3^n + 1, \\dots, 2 \\cdot 3^n - 1\\}\n$$\nand\n$$\nU = \\{ x \\in Z_{3^{n+1}} \\mid \\gcd(x, 3) = 1 \\}.\n$$\nThen $C \\subset Z_{3^{n+1}}$, $|C| = 3^n$. Note that $C$ is sum-free, because if $a, b \\in C$ with $3^n \\le a, b < 2 \\cdot 3^n$ then $2 \\cdot 3^n \\le a+b < 4 \\cdot 3^n$ so $a+b$ is not congruent modulo $3^{n+1}$ to an element of $C$. For each $y \\in U$, let $C_y = yC = \\{yx \\mid x \\in C\\}$. Then $C_y$ is also sum-free for every $y \\in U$, because if we had $ya$ and $yb$ in $C_y$ with $ya + yb \\in C_y$, then $a$ and $b$ would be elements of $C$ summing to an element of $C$. Also, every $C_y$ contains $|C| = 3^n$ residues because multiplication by $y$ is invertible. Since $|U| = \\phi(3^{n+1}) = 2 \\times 3^n$, there are $2 \\cdot 3^n$ of sets $C_y$.\n\nConsider the sets\n$B_1, B_2, \\dots, B_{3^{n-1}}, C_1, C_2, C_4, C_5, C_7, \\dots, C_{3^{n+1}-2}, C_{3^{n+1}-1}$.\nThere are $3^n - 1 + 2 \\cdot 3^n = 3^{n+1} - 1$ of these sets, so it suffices to check that every nonzero residue modulo $3^{n+1}$ appears in exactly $3^n$ of them.\n\nLet $m$ be a nonzero residue modulo $3^{n+1}$, and write $m = 3^k s$, $0 \\le k \\le n$, $\\gcd(s, 3) = 1$. We consider two cases.\n\n(i) $k < n$. Then $m$ is a nonzero residue modulo $3^n$, so $m$ is contained in exactly $3^{n-1}$ of the sets $B_i$, namely those which correspond to $A_i$ with $m' \\in A_i$ (where $m \\equiv m' \\pmod{3^n}$). The number of sets $C_i$ containing $m$ is the number of solutions to $y^{-1}m \\in C$ with $y \\in U$, or the number of $z \\in U$ such that $zm \\in C$. Since $m \\neq 0$, as $z$ ranges over $Z_{3^{n+1}}$, one third of the residues $zm$ are in $C$; likewise, since $3m \\neq 0$, one third of the residues $3zm$ are in $C$. Since $U = Z_{3^{n+1}} \\setminus 3Z_{3^{n+1}}$, $zm \\in C$ for one third of the values $z \\in U$. So $m$ is in one third of the sets $C_y$, giving $\\frac{1}{3}|U| = 2 \\cdot 3^{n-1}$ more sets containing $m$. The total number of sets containing $m$ is then $3^{n-1} + 2 \\cdot 3^{n-1} = 3^n$.\n\n(ii) $k=n$. Then $m = 3^n$ or $2 \\cdot 3^n$ ($s=1$ or $2$ respectively). Then $m \\bmod 3^n = 0$, so $m$ does not appear in any of the sets $A_i$. However, $m$ appears in every set $C_y$ with $y \\equiv s \\pmod 3$, so $m$ appears in $3^n$ of the $C_y$. Thus the total number of sets containing $m$ is again $3^n$.\n\nThus the sets $\\{B_i\\}$, $\\{C_y\\}$ have the desired properties and the Lemma holds by induction.\n$\\blacksquare$\n\nNow let $3^n$ be a power of $3$ larger than the sum of any two elements of $A$. By the Lemma, there exist $3^n - 1$ sets $S_1, \\dots, S_{3^{n-1}}$ of $3^{n-1}$ residues modulo $3^n$ such that every nonzero residue modulo $3^n$ appears in exactly $3^{n-1}$ of the $S_i$. Let $n_i$ be the number of elements of $A$ contained in $S_i$. Since every element of $A$ appears $3^{n-1}$ times,\n$$\n\\sum_{i=1}^{3^n-1} n_i = 3^{n-1} |A|\n$$\nso some $n_i$ is at least\n$$\n\\frac{3^{n-1}|A|}{3^n-1} > \\frac{1}{3}|A| = \\frac{2001}{3} = 667.\n$$\nLet $B$ be the set of elements of $A$ contained in $S_i$. Then $|B| \\ge 668$, and if $u, v \\in B$, then $u + v \\notin B$, because $S_i$ is sum-free. Thus the set $B$ has the desired properties.\n**Second Solution.** Let the elements of $A$ be $a_1, \\dots, a_{2001}$. Let $p$ be a prime number such that $p \\equiv 2 \\pmod 3$ and $p$ is larger than all the $a_i$. Such a prime $p$ exists by **Dirichlet's Theorem**, although the result can also be easily proven directly. There is at least one prime congruent to $2$ modulo $3$ (namely, $2$). Suppose there were only finitely many primes congruent to $2$ modulo $3$, and let their product be $P$. Then $3P-1$, which is larger than $P$ and congruent to $2$ modulo $3$, must have another prime divisor congruent to $2$ modulo $3$, contradiction. Thus, the original assumption was wrong, and there are infinitely many odd primes that are congruent to $2$ modulo $3$. Specifically, one such prime is larger than all $a_i$.\n\nAll elements of $S$ are distinct and nonzero modulo $p$. Call a number $n$ **mediocre** if the least positive residue of $n$ modulo $p$ lies in $[(p+1)/3, (2p-1)/3]$. For any $1 \\le i \\le 2001$, there are exactly $(p+1)/3$ integer values of $k \\in [1, p-1]$ such that $ka_i$ is mediocre. Thus, there are\n$$\n\\frac{2001(p+1)}{3} = 667(p+1)\n$$\npairs of $(k, i)$ such that $ka_i$ is mediocre. By the **Pigeonhole Principle**, there exists some $k$ for which the set\n$$\nB = \\{a_i \\mid ka_i \\text{ is mediocre}\\}\n$$\nhas at least $668$ elements.\n\nWe now claim that this $B$ satisfies the desired properties. It suffices to show that $k$ times the sum of any two elements of $B$ is not mediocre and hence cannot equal $k$ times any element of $B$. To that end, note that $k$ times the sum of any two elements of $B$ cannot be mediocre because it is congruent modulo $p$ to some number in $[2(p+1)/3, 2(2p-1)/3]$ or, equivalently, to some number in $[0, (p-2)/3] \\cup [(2p+2)/3, p-1]$, which is a set containing no mediocre numbers. Thus, the set $B$ satisfies the desired properties.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are some red and blue marbles in a box. We are told that there are twelve more red marbles than blue marbles, and we experimentally determine that when we pick a marble randomly we get a blue marble one quarter of the time. How many marbles are there in the box?",
    "options": [],
    "answer": "24",
    "solution": "Solution:\n\nCall the number of blue marbles $x$, so the number of red marbles is $x + 12$ and the total number of marbles is $2x + 12$. The probability of picking a blue marble is $\\frac{x}{2x + 12} = \\frac{1}{4}$. Solving, $x = 6$, so $2x + 12 = 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46581,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many times does $24$ divide into $100!$ (factorial)?",
    "options": [],
    "answer": "32",
    "solution": "Solution:\n\nWe first determine the number of times $2$ and $3$ divide into $100! = 1 \\cdot 2 \\cdot 3 \\cdots 100$. Let $\\langle N \\rangle_{n}$ be the number of times $n$ divides into $N$ (i.e., we want to find $\\langle 100! \\rangle_{24}$).\n\nSince $2$ only divides into even integers, $\\langle 100! \\rangle_{2} = \\langle 2 \\cdot 4 \\cdot 6 \\cdots 100 \\rangle$. Factoring out $2$ once from each of these multiples, we get that $\\langle 100! \\rangle_{2} = \\left\\langle 2^{50} \\cdot 1 \\cdot 2 \\cdot 3 \\cdots 50 \\right\\rangle_{2}$. Repeating this process, we find that $\\langle 100! \\rangle_{2} = \\left\\langle 2^{50+25+12+6+3+1} \\cdot 1 \\right\\rangle_{2} = 97$.\n\nSimilarly, $\\langle 100! \\rangle_{3} = \\left\\langle 3^{33+11+3+1} \\right\\rangle_{3} = 48$.\n\nNow $24 = 2^{3} \\cdot 3$, so for each factor of $24$ in $100!$ there needs to be three multiples of $2$ and one multiple of $3$ in $100!$. Thus $\\langle 100! \\rangle_{24} = \\min\\left( \\left[ \\langle 100! \\rangle_{2} / 3 \\right], \\langle 100! \\rangle_{3} \\right ) = 32$, where $[N]$ is the greatest integer less than or equal to $N$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA balloon that blows up in the shape of a perfect cube is being blown up at a rate such that at time $t$ fortnights, it has surface area $6 t$ square furlongs. At how many cubic furlongs per fortnight is the air being pumped in when the surface area is $144$ square furlongs?",
    "options": [],
    "answer": "3 sqrt 6",
    "solution": "Solution:\n\nThe surface area at time $t$ is $6 t$, so the volume is $t^{3/2}$. Hence the air is being pumped in at a rate of $\\frac{3}{2} \\sqrt{t}$. When the surface area is $144$, $t=24$, so the answer is $3 \\sqrt{6}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(n)$ be a function satisfying the following three conditions for all positive integers $n$:\n(a) $f(n)$ is a positive integer,\n(b) $f(n+1) > f(n)$,\n(c) $f(f(n)) = 3n$.\nFind $f(2001)$.",
    "options": [],
    "answer": "3816",
    "solution": "Solution:\nWe will show that $f(2001)$ must equal $3816$. We start by proving a lemma which gives us some of the values of $f(n)$.\n\nLemma: For $n = 0, 1, 2, \\ldots$,\n(a) $f\\left(3^{n}\\right) = 2 \\cdot 3^{n}$; and\n(b) $f\\left(2 \\cdot 3^{n}\\right) = 3^{n+1}$.\n\nProof: We use induction. For $n = 0$, note that $f(1) \\neq 1$, otherwise $3 = f(f(1)) = f(1) = 1$, which is impossible. Since $f(k)$ is a positive integer for all positive integers $k$, we conclude that $f(1) > 1$. Since $f(n+1) > f(n)$, $f$ is increasing. Thus $1 < f(1) < f(f(1)) = 3$ or $f(1) = 2$. Hence $f(2) = f(f(1)) = 3$.\n\nSuppose that for some positive integer $n \\geq 1$,\n$$\nf\\left(3^{n}\\right) = 2 \\cdot 3^{n} \\text{ and } f\\left(2 \\cdot 3^{n}\\right) = 3^{n+1}\n$$\nThen,\n$$\nf\\left(3^{n+1}\\right) = f\\left(f\\left(2 \\cdot 3^{n}\\right)\\right) = 2 \\cdot 3^{n+1}\n$$\nand\n$$\nf\\left(2 \\cdot 3^{n+1}\\right) = f\\left(f\\left(3^{n+1}\\right)\\right) = 3^{n+2}\n$$\nas desired. This completes the induction, and establishes the lemma.\n\nContinuing with our solution, there are $3^{n} - 1$ integers $m$ such that $3^{n} < m < 2 \\cdot 3^{n}$ and there are $3^{n} - 1$ integers $m'$ such that\n$$\nf\\left(3^{n}\\right) = 2 \\cdot 3^{n} < m' < 3^{n+1} = f\\left(2 \\cdot 3^{n}\\right).\n$$\nSince $f$ is an increasing function,\n$$\nf\\left(3^{n} + m\\right) = 2 \\cdot 3^{n} + m\n$$\nfor $0 \\leq m \\leq 3^{n}$. Therefore\n$$\nf\\left(2 \\cdot 3^{n} + m\\right) = f\\left(f\\left(3^{n} + m\\right)\\right) = 3\\left(3^{n} + m\\right)\n$$\nfor $0 \\leq m \\leq 3^{n}$. Hence\n$$\nf(2001) = f\\left(2 \\cdot 3^{6} + 543\\right) = 3\\left(3^{6} + 543\\right) = 3816.\n$$\nSolution:\n(Sketch) Andrew Dudzik's insight was to recognize that $f(n)$ deals in a very simple way, with the base-3 representation of $n$. Let $n = a_{1} a_{2} \\cdots a_{t}$ be the base-3 digits of $n$. For example, if $n = 50$ in base 10, we would write $n = 1212$ in base 3 and hence $a_{1} = 1, a_{2} = 2, a_{3} = 1, a_{4} = 2$. Dudzik proved the following:\n\n1. If $a_{1} = 1$, then $f(n) = 2 a_{2} a_{3} \\cdots a_{t}$.\n2. If $a_{1} = 2$, then $f(n) = 1 a_{2} a_{3} \\cdots a_{t} 0$\n\nThese two statements can be proven easily with induction; we leave this as an exercise for the reader. Note that Dudzik's formulas allow us to immediately compute $f(2001)$. In base-3 notation, $2001$ is equal to $2202010$. Thus, by formula #2,\n$$\nf(2202010) = 12020100 = 1 \\cdot 3^{2} + 2 \\cdot 3^{4} + 2 \\cdot 3^{6} + 3^{7},\n$$\nand this equals $3816$ in base-10 notation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a square and $E$ a point on side $CD$. The circle inscribed in triangle $ADE$ touches $DE$ at $F$, and the circle inside quadrilateral $ABCE$, tangent to sides $AB$, $BC$, $EA$, touches $AB$ at $G$. Prove that lines $AE$, $BD$, and $FG$ meet in a point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nExtend lines $BC$ and $AE$ to intersect at $H$. Then the circle inside quadrilateral $ABCE$, tangent to $AB$, $BC$, and $EA$, is really the inscribed circle of $\\triangle HBA$. (Actually, this is only true if the circle lies inside $\\triangle HBA$ rather than outside it. However, the fact that $CE$ is parallel to $AB$ with $CE < CD = AB$ readily implies that $E$ lies between $H$ and $A$, and $C$ lies between $H$ and $B$, so that the whole quadrilateral $ABCE$ lies within $\\triangle HBA$, so the circle drawn inside it does too.)\n\nNow let $P$ be the intersection point of lines $BD$ and $AE$. Consider the homothety (scaling) about $P$ that sends point $D$ to point $B$. Since homotheties preserve directions of lines, this map takes line $AD$ to the line through $B$ and parallel to $AD$, namely line $HB$. Similarly, it takes line $DE$ to line $BA$. And line $EA$ passes through $P$, the center of the homothety, so it goes to itself.\n\nThus, our homothety takes lines $AD$, $DE$, $EA$ to lines $HB$, $BA$, $AH(=AE)$, respectively, so it takes $\\triangle ADE$ to $\\triangle HBA$. Consequently, the incircle of $\\triangle ADE$ is mapped to the incircle of $\\triangle HBA$, and the map also matches corresponding tangency points: $F$ goes to $G$. But if a homothety about $P$ takes $F$ to $G$, then $P$, $F$, $G$ must be collinear. We now know that $P$ lies on lines $AE$, $BD$, and $FG$, which is what we need.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the 18th digit after the decimal point of $\\frac{10000}{9899}$?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n$\\frac{10000}{9899}$ satisfies $100(x-1) = 1.01 x$, so each pair of adjacent digits is generated by adding the previous two pairs of digits. So the decimal is $1.01020305081321345590\\ldots$, and the 18th digit is $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCalculate the sum of the coefficients of $P(x)$ if $(20 x^{27} + 2 x^{2} + 1) P(x) = 2001 x^{2001}$.",
    "options": [],
    "answer": "87",
    "solution": "Solution:\nThe sum of coefficients of $f(x)$ is the value of $f(1)$ for any polynomial $f$. Plugging in $1$ to the above equation, $P(1) = \\frac{2001}{23} = 87$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe manager of Chez Gastropod wants to write a menu consisting of 15 dishes. A \"meal\" is defined to be a subset of this menu (possibly empty), but some meals are legal and others are not. The manager may choose which meals are legal, but there is a requirement that the intersection of any two legal meals should still be legal. He wants there to be exactly 2001 legal meals. Can he do it?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nThe answer is yes. Take any arbitrary 15-element set (menu), and call a collection of subsets (meals) \"valid\" if the intersection of any two sets in the collection is again in the collection. Thus, the objective is to show that there exists a valid collection containing exactly 2001 sets. We will show, by downward induction, that there exists a valid collection with exactly $n$ sets for each $n$, $0 \\leq n \\leq 2^{15} = 32768$.\n\nThe base case $n = 2^{15}$ is clear: the collection of all subsets of the menu is certainly valid. For the induction step, suppose that there is a valid collection $C$ of $n$ subsets ($1 \\leq n \\leq 32768$); we will show that there is a valid collection of $n-1$ subsets. Choose a subset in $C$ of maximum possible size, and remove it; let $C'$ denote the remaining collection, so that it consists of $n-1$ subsets. We claim $C'$ is still valid. Indeed, if $S, T \\in C'$, then the removed subset cannot be contained within either $S$ or $T$ (because of maximality), hence it certainly cannot equal their intersection. But $S \\cap T$ was in $C$; hence, it is also in $C'$, as needed. Thus, the claim holds. Now, let $n = 2001$, and the problem is solved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSquare $ABCD$ is drawn. Isosceles triangle $CDE$ is drawn with $E$ a right angle. Square $DEFG$ is drawn. Isosceles triangle $FGH$ is drawn with $H$ a right angle. This process is repeated infinitely so that no two figures overlap each other. If square $ABCD$ has area $1$, compute the area of the entire figure.",
    "options": [],
    "answer": "5/2",
    "solution": "Solution:\n\nLet the area of the $n$th square drawn be $S_{n}$ and the area of the $n$th triangle be $T_{n}$. Since the hypotenuse of the $n$th triangle is of length $\\sqrt{S_{n}}$, its legs are of length $l=\\sqrt{\\frac{S_{n}}{2}}$, so $S_{n+1}=l^{2}=\\frac{S_{n}}{2}$ and $T_{n}=\\frac{l^{2}}{2}=\\frac{S_{n}}{4}$. Using the recursion relations, $S_{n}=\\frac{1}{2^{n-1}}$ and $T_{n}=\\frac{1}{2^{n+1}}$, so $S_{n}+T_{n}=\\frac{1}{2^{n-1}}+\\frac{1}{2^{n+1}}=\\left(\\frac{1}{2}+2\\right) \\frac{1}{2^{n}}=\\frac{5}{2} \\frac{1}{2^{n}}$. Thus the total area of the figure is $\\sum_{n=1}^{\\infty} S_{n}+T_{n}=\\frac{5}{2} \\sum_{n=1}^{\\infty} \\frac{1}{2^{n}}=\\frac{5}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point in the plane of triangle $ABC$ such that the segments $PA$, $PB$, and $PC$ are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to $PA$. Prove that $\\angle BAC$ is acute.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the **Cauchy-Schwarz Inequality**,\n$$\n\\sqrt{PB^2 + PC^2} \\sqrt{AC^2 + AB^2} \\ge PB \\cdot AC + PC \\cdot AB.\n$$\nApplying the **(Generalized) Ptolemy's Inequality** to quadrilateral $ABPC$ yields\n$$\nPB \\cdot AC + PC \\cdot AB \\geq PA \\cdot BC.\n$$\n![](attached_image_1.png)\nBecause $PA$ is the longest side of an obtuse triangle with side lengths $PA$, $PB$, $PC$, we have $PA > \\sqrt{PB^2 + PC^2}$ and hence\n$$\nPA \\cdot BC \\geq \\sqrt{PB^2 + PC^2} \\cdot BC.\n$$\nCombining these three inequalities yields $\\sqrt{AB^2 + AC^2} > BC$, implying that angle $BAC$ is acute.\nLet $D$ and $Q$ be the feet of the perpendiculars from $B$ and $P$ to line $AC$, respectively. Then $DQ \\leq BP$. Furthermore, the given conditions imply that $AP^2 > BP^2 + PC^2$, which can be written as $AP^2 - PC^2 > BP^2$. Hence,\n$$\n\\begin{align*}\nAQ^2 &\\geq AQ^2 - QC^2 \\\\\n&= (AP^2 - PQ^2) - (CP^2 - PQ^2) \\\\\n&= AP^2 - PC^2 \\\\\n&> BP^2 \\\\\n&\\geq DQ^2.\n\\end{align*}\n$$\nLet $\\ell$ be the ray $AC$ minus the point $A$. Note that, since $PA > PC$, $Q$ lies on ray $\\ell$. If $D$ did not lie on $\\ell$, then $AQ$ would be less than or equal to $DQ$, a contradiction. Thus, $D$ lies on $\\ell$, and angle $BAC$ is acute.\n![](attached_image_2.png)\nSet up a coordinate system on the plane with $A = (0,0)$, $B = (a,0)$, $C = (b,c)$, and $P = (x,y)$. Without loss of generality, we may assume that $a > 0$ and that $c > 0$. Proving that angle $BAC$ is acute is equivalent to proving that $b > 0$. Since $PA^2 > PB^2 + PC^2$,\n$$\nx^2 + y^2 > (x-a)^2 + y^2 + (x-b)^2 + (y-c)^2.\n$$\nHence,\n$$\n0 > (x-a)^2 - 2bx + b^2 + (y-c)^2 \\geq -2bx.\n$$\nSince $PA > PB$, we have $x > \\frac{a}{2} > 0$. It follows that $b > 0$, as desired.\nWe first prove the following Lemma.\n\n**Lemma.** For any four points $W$, $X$, $Y$, and $Z$ in the plane,\n$$\nWY^2 + XZ^2 \\leq WX^2 + XY^2 + YZ^2 + ZW^2.\n$$\n*Proof.* Pick an arbitrary origin $O$ and let $w$, $x$, $y$, $z$ denote the vectors from $O$ to $W$, $X$, $Y$, $Z$, respectively. Then\n$$\n\\begin{align*}\nWX^2 + XY^2 + YZ^2 + ZW^2 - WY^2 - XZ^2 \\\\\n&= |w-x|^2 + |x-y|^2 + |y-z|^2 + |z-w|^2 - |w-y|^2 - |x-z|^2 \\\\\n&= w \\cdot w + x \\cdot x + y \\cdot y + z \\cdot z \\\\\n&\\quad -2(w \\cdot x + x \\cdot y + y \\cdot z + z \\cdot w - w \\cdot y - x \\cdot z) \\\\\n&= |w+y-x-z|^2,\n\\end{align*}\n$$\nwhich is always nonnegative. Equality holds if and only if $w + y = x + z$, which is true if and only if $WXYZ$ is a (possibly degenerate) parallelogram.\n\nApplying the Lemma to points $A$, $B$, $C$, and $P$ gives\n$$\n\\begin{align*}\n0 \\le AB^2 + BP^2 + PC^2 + CA^2 - AP^2 - BC^2 \\\\\n&= (PB^2 + PC^2 - PA^2) + (AB^2 + AC^2 - BC^2) \\\\\n&< 0 + (AB^2 + AC^2 - BC^2) \\\\\n&= AB^2 + AC^2 - BC^2.\n\\end{align*}\n$$\nTherefore, angle $BAC$ is acute.\nIn this solution, $\\sin^{-1}$ takes on values between $0^\\circ$ and $90^\\circ$. Note that $\\angle PAB < 90^\\circ$, since $PB < PA$. Applying the **Law of Sines** to triangle $PAB$ yields\n$$\n\\sin \\angle PAB = \\frac{PB}{PA} \\sin \\angle ABP \\le \\frac{PB}{PA}.\n$$\nIt follows that\n$$\n\\angle PAB \\le \\sin^{-1} \\frac{PB}{PA}.\n$$\nSince $PA^2 > PB^2 + PC^2$, we have similarly\n$$\n\\angle PAC \\le \\sin^{-1} \\frac{PC}{PA} < \\sin^{-1} \\frac{\\sqrt{PA^2 - PB^2}}{PA}.\n$$\nThus,\n$$\n\\begin{align*}\n\\angle BAC &\\le \\angle BAP + \\angle PAC \\\\\n&< \\sin^{-1} \\frac{PB}{PA} + \\sin^{-1} \\frac{\\sqrt{PA^2 - PB^2}}{PA}.\n\\end{align*}\n$$\nIf $\\theta = \\sin^{-1} \\frac{PB}{PA}$, then $\\sin(90^\\circ - \\theta) = \\cos \\theta = \\sqrt{1 - \\sin^2 \\theta} = \\frac{\\sqrt{PA^2 - PB^2}}{PA}$.\nHence,\n$$\n\\angle BAC < \\sin^{-1} \\frac{PB}{PA} + \\sin^{-1} \\frac{\\sqrt{PA^2 - PB^2}}{PA} = 90^\\circ,\n$$\nand angle $BAC$ is acute.\n![](attached_image_3.png)\nNote that $PA^2 > PB^2 + PC^2$. Regard $P$ as fixed and $A, B$, and $C$ as free to rotate on circles of radii $PA, PB$, and $PC$ about $P$, respectively. As $A, B, C$ vary, $\\angle BAC$ will be maximized when $B$ and $C$ are on opposite sides of line $PA$ and $\\angle ABP$ and $\\angle ACP$ are right angles, i.e., lines $AB$ and $AC$ are tangent to the circles passing through $B$ and $C$.\nWithout loss of generality, we assume that $PA > PB \\ge PC$. In this case, $ABPC$ is cyclic and $AB^2 = PA^2 - PB^2 > PC^2$, and similarly $AC^2 > PB^2$. Hence, on the circumcircle of $ABPC$, arcs $AB$ and $AC$ are bigger than arcs $PC$ and $PB$, respectively. Thus, $\\angle BPC > \\angle BAC$. Since these two angles are supplementary, angle $BAC$ is acute.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46590,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 51 senators in a senate. The senate needs to be divided into $n$ committees so that each senator is on one committee. Each senator hates exactly three other senators. (If senator $A$ hates senator $B$, then senator $B$ does *not* necessarily hate senator $A$.) Find the smallest $n$ such that it is always possible to arrange the committees so that no senator hates another senator on his or her committee.",
    "options": [],
    "answer": "7",
    "solution": "The smallest such number is $7$. Assume that there are $7$ senators $A_1, \\dots, A_7$ such that each $A_i$ hates $A_{i+1}$, $A_{i+2}$, and $A_{i+3}$ (where indices are taken modulo $7$). In this situation, for any different $A_i, A_j$, either $A_i$ hates $A_j$ or vice versa. The senators $A_1, \\dots, A_7$ must be placed on seven different committees. Thus, $n \\ge 7$.\n\nIn order to show that $n \\le 7$, we will prove the following stronger statement by induction on $k \\ge 1$: for $k$ senators, each of whom hates *at most* $3$ others, it is possible to arrange the senators into $7$ committees so that no senator hates another senator on his or her committee.\n\nFor the base case $k = 1$, note that we can have $7$ committees, $6$ of which are empty and $1$ of which contains the sole senator.\n\nNow assume the claim is true for all $k \\le m-1$. Suppose we are given $m$ senators, each of whom hates at most $3$ others. If each of those $m$ senators is hated by more than $3$ others, then the total number of acts of hating must be greater than $3m$, but this is not possible since each senator hates at most $3$ others. Therefore, there must be at least one senator $A$ who is hated by at most $3$ others. By the induction hypothesis, we can split the $m-1$ other senators into $7$ committees satisfying the property that no senator hates another senator on the same committee. By the **Pigeonhole Principle**, one of those committees contains neither a person whom $A$ hates nor a person who hates $A$. We can therefore place $A$ in that committee. The induction is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an odd integer greater than $1$ and let $c_1, c_2, \\dots, c_n$ be integers. For each permutation $a = (a_1, a_2, \\dots, a_n)$ of $\\{1, 2, \\dots, n\\}$, define $S(a) = \\sum_{i=1}^n c_i a_i$. Prove that there exist permutations $b$ and $c$, $b \\neq c$, such that $n!$ divides $S(b) - S(c)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** Let $\\sum_a$ denote the sum over all $n!$ permutations $a = (a_1, a_2, \\dots, a_n)$. We compute $\\sum_a S(a) \\mod n!$ in two ways, one of which assuming that the desired conclusion is false, and reach a contradiction.\n\nSuppose, for the sake of contradiction, that the claim is false. Then each $S(a)$ must have a different remainder $\\mod n!$. Since there are exactly $n!$ such permutations $a$, there exists exactly one permutation $a$ such that $S(a) \\equiv s \\pmod{n!}$ for each $s = 1, 2, \\dots, n!$. Since $n > 1$, $n!$ is even and $n! + 1$ is odd. Hence,\n$$\n\\sum_a S(a) \\equiv \\sum_{s=1}^{n!} s \\equiv \\frac{n!}{2} \\cdot (n! + 1) \\pmod{n!},\n$$\nor\n$$\n\\sum_a S(a) \\equiv \\frac{n!}{2} \\pmod{n!}. \\qquad (1)\n$$\nOn the other hand, for $i, k \\in \\{1, \\dots, n\\}$, we have $a_i = k$ in exactly $(n-1)!$ permutations $a$. Thus, for $1 \\le i \\le n$,\n$$\n\\sum_a a_i = (n-1)!(1+2+\\dots+n) = n! \\cdot \\frac{n+1}{2}.\n$$\nHence,\n$$\n\\sum_a S(a) = \\sum_a \\sum_{i=1}^n c_i a_i = \\sum_a \\sum_{i=1}^n c_i a_i = \\sum_{i=1}^n \\left( c_i \\sum_a a_i \\right).\n$$\nBecause $n+1$ is even, $n!$ divides $\\sum_a a_i = n! \\cdot \\frac{n+1}{2}$ for each $i$. It follows that $n!$ divides $\\sum_a S(a)$, contradicting (1). Therefore, the initial assumption was false, and there do exist distinct permutations $b$ and $c$ such that $n!$ is a divisor of $S(b) - S(c)$.\n\n\n**Second Solution.** (by Liang Xiao, China) Let $n = 2m + 1$, and let $r$ be the number of indices $i$ such that $c_i$ is odd. For a permutation $a = (a_1, a_2, \\dots, a_n)$ of $\\{1, 2, \\dots, n\\}$, let\n$$\nT(a) = \\sum_{i=1}^{n} i c_{a_i} = c_{a_1} + 2c_{a_2} + \\dots + nc_{a_n},\n$$\nand let $\\lambda_a$ be the number of indices $i$ such that both $i$ and $c_{a_i}$ are odd. Note that $T(a)$ is equal $S(a)$ in a different order.\nFor any given positive integer $\\lambda$, $0 \\le \\lambda \\le r$, we count the number of permutations $a$ of $\\{1, 2, \\dots, n\\}$ with $\\lambda_a = \\lambda$. There are $\\binom{m+1}{\\lambda}$ ways to choose $\\lambda$ numbers $i_1, i_2, \\dots, i_\\lambda$ from the set $\\{1, 3, \\dots, 2m+1\\}$ and $\\binom{m}{r-\\lambda}$ ways to choose $r-\\lambda$ numbers $j_1, j_2, \\dots, j_{r-\\lambda}$ from the set $\\{2, 4, \\dots, 2m\\}$. Once the numbers $i_1, \\dots, i_\\lambda, j_1, \\dots, j_{r-\\lambda}$ have been chosen, there are $r!(n-r)!$ permutations $a = (a_1, a_2, \\dots, a_n)$ of $\\{1, 2, \\dots, n\\}$ such that $c_{a_{i_1}}, \\dots, c_{a_{i_\\lambda}}$ and $c_{a_{j_1}}, \\dots, c_{a_{j_{r-\\lambda}}}$ are all odd. Thus, there are\n$$\nr!(n-r)! \\binom{m+1}{\\lambda} \\binom{m}{r-\\lambda}\n$$\npermutations $a$ of $\\{1, 2, \\dots, n\\}$ such that $\\lambda_a = \\lambda$.\n\nLet $E$ and $O$ be the number of permutations $a$ of $\\{1, 2, \\dots, n\\}$ such that $S(a)$ is even and odd, respectively. Then\n$$E = r!(n-r)! \\sum_{2|\\lambda} \\binom{m+1}{\\lambda} \\binom{m}{r-\\lambda},$$\n$$O = r!(n-r)! \\sum_{2|\\lambda} \\binom{m+1}{\\lambda} \\binom{m}{r-\\lambda},$$\nand\n$$\nE - O = r!(n-r)! \\sum_{\\lambda=0}^{r} (-1)^{\\lambda} \\binom{m+1}{\\lambda} \\binom{m}{r-\\lambda}.\n$$\nDenote by $[x^k]h(x)$ the coefficient of $x^k$ in a power series $h(x)$. Let $g_1(x) = (1-x)^{m+1}$, $g_2(x) = (1+x)^m$, and $f(x) = g_1(x)g_2(x)$. Then\n$$[x^r]f(x) = \\sum_{\\lambda=0}^{r} ([x^\\lambda]g_1(x)) ([x^{r-\\lambda}]g_2(x))$$\n$$= \\sum_{\\lambda=0}^{r} (-1)^\\lambda \\binom{m+1}{\\lambda} \\binom{m}{r-\\lambda}.$$\nOn the other hand,\n$$f(x) = (1 - x^2)^m (1 - x) = \\sum_{i=0}^{m} \\binom{m}{i} (-1)^i x^{2i} (1 - x)$$\n$$= \\sum_{i=0}^{m} \\binom{m}{i} ((-1)^i x^{2i} + (-1)^{i+1} x^{2i+1}).$$\nTherefore,\n$$\n[x^r]f(x) = \\left\\lfloor \\frac{m}{\\frac{r+1}{2}} \\right\\rfloor \\cdot (-1)^{\\left\\lfloor \\frac{r+1}{2} \\right\\rfloor} \\neq 0,\n$$\nimplying that $E - O \\neq 0$.\nIn order for the desired result to be false, for each number $i$ from $1$ to $n!$, there must be exactly one permutation $a$ of $1, 2, \\dots, n$ such that $S(a) \\equiv i \\pmod{n!}$. As $n!$ is even, this implies that $E = O$, contradicting that fact $E - O \\neq 0$. Thus, we have proven the desired conclusion.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSome students are taking a math contest, in which each student takes one of four tests. One third of the students take one test, one fourth take another test, one fifth take the next test, and $26$ students take the last test. How many students are taking the contest in total?",
    "options": [],
    "answer": "120",
    "solution": "Solution:\n\nCall the total number of students $n$. We know\n$$\n n = \\frac{n}{3} + \\frac{n}{4} + \\frac{n}{5} + 26\n$$\nso $n = 120$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$P$ is a polynomial. When $P$ is divided by $x-1$, the remainder is $-4$. When $P$ is divided by $x-2$, the remainder is $-1$. When $P$ is divided by $x-3$, the remainder is $4$. Determine the remainder when $P$ is divided by $x^{3}-6x^{2}+11x-6$.",
    "options": [],
    "answer": "x^2 - 5",
    "solution": "Solution:\n\nThe remainder polynomial is simply the order two polynomial that goes through the points $(1, -4)$, $(2, -1)$, and $(3, 4)$: $x^{2} - 5$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven that $7,999,999,999$ has at most two prime factors, find its largest prime factor.",
    "options": [],
    "answer": "4002001",
    "solution": "Solution:\n$7,999,999,999 = 8 \\cdot 10^{9} - 1 = 2000^{3} - 1 = (2000 - 1)\\left(2000^{2} + 2000 + 1\\right)$, so $\\left(2000^{2} + 2000 + 1\\right) = 4,002,001$ is its largest prime factor.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the size of the largest rectangle that can be drawn inside of a $3$-$4$-$5$ right triangle with one of the rectangle's sides along one of the legs of the triangle?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nClearly one vertex of the rectangle will be at the right angle. Position the triangle with the leg of length $4$ along the $x$-axis and the leg of length $3$ along the $y$-axis. Then the hypotenuse is along the line $y = 3 - (3/4)x$.\n\nSuppose the rectangle has a side of length $y$ along the leg of length $3$. Then the area is $y \\left(\\frac{4}{3}\\right)(3-y) = 4y - \\frac{4}{3}y^{2}$. The derivative of this is $0$ when $4 - \\frac{8}{3}y = 0$, or $y = \\frac{3}{2}$, giving an area of $3$.\n\nOr, if you prefer, suppose the rectangle has a side of length $x$ along the leg of length $4$. Then the area is $x(3 - (3/4)x) = 3x - \\frac{3}{4}x^{2}$. The derivative of this is $0$ when $3 - \\frac{3}{2}x = 0$, or $x = 2$, again giving an area of $3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a line segment $A B$, construct a segment half as long as $A B$ using only a compass. Construct a segment one-third as long as $A B$ using only a compass.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst, we provide (part of) an algorithm for circular inversion. Suppose we are given point $O$ and a circle centered at $O$ of some radius $r$. If $P$ is a point outside the circle, we wish to construct point $Q$ on ray $O P$, satisfying $O P \\cdot O Q = r^{2}$. Let the circle centered at $P$, with radius $O P$, intersect the given circle centered at $O$ at points $X$ and $Y$. Then let the circle with center $X$, radius $r$ (which is constructible since $r = O X$), and the circle with center $Y$, radius $r$, intersect at $O$ and $Q$. We claim this point $Q$ is what we want. Indeed, it is clear from symmetry that $O P$ is the perpendicular bisector of $X Y$. Since $X Q = r = Y Q$, $Q$ lies on this bisector as well—that is, on line $O P$. Moreover, let $H$ denote the intersection of $X Y$ with $O P$; then $P H < P X = P O \\Rightarrow H$ lies on ray $O P$, and, since $Q$ is the reflection of $O$ across $X Y$, $Q$ will lie on the opposite side of $H$ from $O$. This shows that $Q$ lies on ray $O P$, not just on line $O P$. Finally, observe that $P X = P O$ and $X O = X Q \\Rightarrow \\angle O X P = \\angle P O X = \\angle Q O X = \\angle X Q O$, so, by equal angles, $\\triangle P O X \\sim \\triangle X Q O$. Thus, $O P / O X = Q X / Q O \\Rightarrow O P \\cdot O Q = O X \\cdot Q X = r^{2}$, as needed.\n\nNow that this is done, we return to the original problem. By scaling, assume $A B = 1$. By drawing circles centered at $A$ and $B$ of radius $1$, and letting $C$ be one of their intersection points, we obtain an equilateral $\\triangle A B C$. Similarly, we successively construct equilateral triangles $B C D$, $B D E$ (with $D \\neq A$, $E \\neq C$). We have $\\angle A B E = \\angle A B C + \\angle C B D + \\angle D B E = 3(\\pi / 3) = \\pi$, so $A, B, E$ are collinear, and $A E = A B + B E = 2$. Then, since we have drawn the circle with center $A$ and radius $1$, we can invert $E$ across it according to the paragraph above, obtaining $F$ such that $A F = 1 / 2$. In fact, $F$ lies on the given segment $A B$, so the segment $A F$ is fully drawn.\n\nSimilarly, to construct a segment of length $1 / 3$, let $A B$ be given; extend $A B$ to $E$ as above so that $B E = 1$, and then repeat the process, extending $B E$ to $G$ so that $E G = 1$. Then $A G = 3$, and inverting $G$ across our circle (center $A$, radius $1$) will give what we need.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA kingdom consists of 12 cities located on a one-way circular road. A magician comes on the 13th of every month to cast spells. He starts at the city which was the 5th down the road from the one that he started at during the last month (for example, if the cities are numbered 1-12 clockwise, and the direction of travel is clockwise, and he started at city #9 last month, he will start at city #2 this month). At each city that he visits, the magician casts a spell if the city is not already under the spell, and then moves on to the next city. If he arrives at a city which is already under the spell, then he removes the spell from this city, and leaves the kingdom until the next month. Last Thanksgiving the capital city was free of the spell. Prove that it will be free of the spell this Thanksgiving as well.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNumber the cities from $0$ to $11$. Encode the current state of affairs by starting at city $11$, writing \"1\" if the city is not under a spell, and a \"0\" if it is under a spell. Do the same thing for each city, from city $10$ down to city $0$. Interpret the resulting list as the binary expansion of an integer $X$ between $0$ and $m=2^{12}-1$.\nSome thought (using modular arithmetic modulo $m$, and the way that addition works in base 2) shows that the rules can be interpreted as saying that if the magician starts at city $k$ then $2^{k}$ is added to $Y$, where $Y$ encodes the state as above. After 12 months the magician starts once at all cities, and this has the net effect of adding\n$$\n1+2+2^{2}+2^{3}+\\ldots+2^{11}=2^{12}-1=m\n$$\nto the original $X$ which obviously leaves the state unchanged.\n\n\nSolution:\nWe shall denote the state of the kingdom by a 12-tuple of $0$'s and $1$'s, with $0$ and $1$ meaning \"free of spell\" and \"under spell,\" respectively. For example, suppose that the initial state is $(0,1,0,0,0,0,1,0,0,0,0,0)$. This means that city #1 is free of the spell, city #2 is under the spell, city #3 is free, etc. Next, define $S_{j}$, for $j=1,2, \\ldots, 12$ to be the transformation of the kingdom after the magician makes a visit that starts at city $j$. For example,\n$$\nS_{1}(0,1,0,0,0,0,1,0,0,0,0,0)=(1,0,0,0,0,0,1,0,0,0,0,0) \\text{,}\n$$\nsince she puts a spell on city #1, then comes to city #2 which is already under the spell, frees it, and stops. Likewise,\n$$\nS_{8}(0,1,0,0,0,0,1,0,0,0,0,0)=(1,0,0,0,0,0,1,1,1,1,1,1) \\text{,}\n$$\nbecause she puts cities #8-12 under spell, then continues around the circle to #1, puts it under spell, then finally stops after freeing city #2. One more example:\n$$\nS_{3}(0,0,0,0,0,0,0,0,0,0,0,0)=(1,1,0,1,1,1,1,1,1,1,1,1) \\text{,}\n$$\nbecause she puts every city under the spell, starting with #3, until she comes to a city that is already under the spell, and this happens when she comes to city #3 for the second time.\nThe key to this problem is the remarkable fact that the $S_{j}$ transformations commute; i.e.\n$$\nS_{i} S_{j}=S_{j} S_{i} \\quad \\text{ for all } i, j \\in\\{1,2,3, \\ldots, 12\\}\n$$\nLet us assume this fact, to see how it quickly solves the problem. Without loss of generality, suppose the magician starts a visit at city #1. Then during the course of the year, her subsequent visits will start at the following cities, in this order:\n$$\n6,11,4,9,2,7,12,5,10,3,8\n$$\nIn other words, during the year she will make 12 visits, and each visit will start at a different city. In other words, the kingdom will be transformed, in order, by $S_{1}, S_{6}, S_{11}, \\ldots$ But since these transformations are commutative, we can rearrange the order that we perform the transformations. So, during the course of the year, the kingdom will be transformed by $S_{1}, S_{2}, S_{3}, \\ldots, S_{12}$, where we may choose the order in any way that we like.\nConsider the state of the kingdom last Thanksgiving. Suppose cities $a, b, c, \\ldots$ are spell-free (where city $a$ is the capital) and cities $A, B, \\ldots$ are under the spell. Now we shall perform the transformations, in this order:\n- $S_{A}, S_{B}, \\ldots$ : Each of these are \"single-city\" transformations that merely free the starting city and then stop. In other words, after these transformations have been performed, all the cities which were under the spell are now free, and hence at this point, all cities in the kingdom are free. (It may be that no cities were under the spell in the first place, so this step may not take place.)\n- $S_{a}$ : Starting with $a$, all cities get put under the spell (since they were previously all free), until the magician returns to $a$, frees it, and stops. Now all cities are under the spell, except for the capital city $a$.\n- The remaining transformations $S_{b}, S_{c} \\ldots$ Since all cities except $a$ are under the spell, each of these are again \"single-city\" transformations which end up freeing cities $b, c, \\ldots$.\nThus, after the 12 visits, city $a$ is still free. In fact, we have shown more, namely that the state of each of the 12 cities of the kingdom will repeat every 12 months. For example, if city #5 was free last Thanksgiving, it will be free this Thanksgiving; if city #7 was under the spell last Thanksgiving, it will also be under the spell this Thanksgiving, etc.\n\nIt remains to prove the key fact that the $S_{j}$ transformations are commutative. Specifically, we will show that\n$$\nS_{j}\\left(S_{i}\\left(x_{1}, x_{2}, \\ldots, x_{12}\\right)\\right)=S_{i}\\left(S_{j}\\left(x_{1}, x_{2}, \\ldots, x_{12}\\right)\\right)\n$$\nfor any distinct $i, j$ and for any 12-tuple $(x_{1}, x_{2}, \\ldots, x_{12})$. We can do this by examining several cases which depend on the 12-tuple.\n1) All the coordinates are zero. In other words, all towns are currently free of the spell. It is easy to check that performing $S_{i}$ on this will make all $x_{k}=1$ except $x_{i}=0$. Then performing $S_{j}$ will change $x_{j}$ to $0$. The net result is that all values are $1$ except for $x_{i}=x_{j}=0$. Clearly, we will have the same result if we perform $S_{j}$ first, and then $S_{i}$.\n2) Exactly one of the coordinates is $1$.\n(a) Suppose $x_{i}=1$ and all others are zero. Performing $S_{i}$ makes $x_{i}=0$ and hence now all values are zero. Then $S_{j}$ results in all values equalling $1$ except $x_{j}=0$. Conversely, performing $S_{j}$ first will put all cities from $j$ to $i-1$ under the spell (travelling clockwise) until the magician reaches city $i$ which she then frees of the spell and stops. So now there is an \"arc\" of $1$'s from $x_{j}$ to $x_{i-1}$ and all other values are $0$. Performing $S_{i}$ on this turns the values of the arc from $i$ to $j-1$ into $1$'s, until she reaches city $j$, and then she frees this city and stops. The net result again is that all values are $1$ except $x_{j}=0$. The figure below illustrates this. We assume clockwise travel and denote free/spell by white/black, respectively.\n![](attached_image_1.png)\n\n(b) Suppose $x_{j}=1$ and all others are zero. This case is exactly the same as 2(a) above; just interchange $i$ and $j$.\n(c) Suppose that $x_{k}=1$ and all others are zero, with $k \\neq i$ and $k \\neq j$. The figure below illustrates the case where $k$ lies on the clockwise arc between $i$ and $j$. The case where $k$ lies between $j$ and $i$ has a similar picture.\n![](attached_image_2.png)\n\n3) At least two coordinates equal $1$.\n(a) Suppose $x_{k}=x_{\\ell}=1$, where $k$ lies on the clockwise arc from $i$ to $j$ and $\\ell$ lies on the arc from $j$ to $i$ (it is possible that $k$ or $\\ell$ may equal one or both of $i$ and $j$.) Then the order of $S_{i}$ and $S_{j}$ will not matter-each transformation alters non-overlapping arcs: $S_{i}$ only affects the arc from $i$ to $k$, while $S_{j}$ affects the arc from $j$ to $\\ell$.\n(b) One of the arcs described in 3(a) contains only zeros. Without loss of generality, suppose that the arc from $i$ to $j$ is all zeros (including $i, j$ ) and let $k$ be the first coordinate clockwise from $j$ such that $x_{k}=1$. The picture below handles this case. The cities between $k$ and $i$ are not shaded, but they could be black or white; their state is not relevant, since they will be unaffected by $S_{i}$ or $S_{j}$.\n![](attached_image_3.png)\n\nThese cases encompass all possibilities, so we conclude that $S_{i}$ and $S_{j}$ commute for all 12-tuples and the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBoris was given a Connect Four game set for his birthday, but his color-blindness makes it hard to play the game. Still, he enjoys the shapes he can make by dropping checkers into the set. If the number of shapes possible modulo (horizontal) flips about the vertical axis of symmetry is expressed as $9(1+2+\\cdots+n)$, find $n$.\n\n(Note: the board is a vertical grid with seven columns and eight rows. A checker is placed into the grid by dropping it from the top of a column, and it falls until it hits either the bottom of the grid or another checker already in that column. Also, $9(1+2+\\cdots+n)$ is the number of shapes possible, with two shapes that are horizontal flips of each other counted as one. In other words, the shape that consists solely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers in the leftmost row are to be considered the same shape.)",
    "options": [],
    "answer": "729",
    "solution": "Solution:\n\nThere are $9^{7}$ total shapes possible, since each of the 7 columns can contain anywhere from 0 to 8 checkers. The number of shapes symmetric with respect to a horizontal flip is the number of shapes of the leftmost four columns, since the configuration of these four columns uniquely determines the configuration of the remaining columns if it is known the shape is symmetric: $9^{4}$. Now we know there are $9^{7}-9^{4}$ non-symmetric shapes, so there are $\\frac{9^{7}-9^{4}}{2}$ non-symmetric shapes modulo flips. Thus the total number of shapes modulo flips is $9^{4}+\\frac{9^{7}-9^{4}}{2}=9^{4}\\left(1+\\frac{9^{3}-1}{2}\\right)=9^{4}\\left(\\frac{9^{3}+1}{2}\\right)=\\frac{3^{8}\\left(3^{6}-1\\right)}{2}=9\\left(1+2+\\cdots+3^{6}\\right)$, so $n=3^{6}=729$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 46599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA circle has two parallel chords of length $x$ that are $x$ units apart. If the part of the circle included between the chords has area $2 + \\pi$, find $x$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nLet $C$ be the area of the circle, $S$ be the area of the square two of whose edges are the chords, and $A$ be the area of the part of the circle included between the chords. The radius of the circle is $\\frac{\\sqrt{2}}{2} x$, so $C = \\frac{\\pi}{2} x^{2}$, and $S = x^{2}$. Then the area $A$ is the area of the square plus one half of the difference between the areas of the circle and square:\n\n$$\nA = \\frac{C - S}{2} + S = \\frac{C + S}{2} = \\frac{1 + \\frac{\\pi}{2}}{2} x^{2}\n$$\n\nso\n\n$$\nx = \\sqrt{\\frac{2A}{1 + \\frac{\\pi}{2}}} = 2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are $3,999,999$ cities in Antarctica, and some pairs of them are connected by roads. It is known that, given any two cities, there is a sequence of roads leading from one to the other. Prove that the cities can be divided into $1999$ groups (of $2001$ cities each) such that, given any two cities in the same group, it is possible to get from one to the other using at most $4000$ roads.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst, we provide some relevant graph-theoretic background. Any finite, connected graph can be turned into a tree (a connected graph without cycles) by removing some edges. Proof: If our graph has a cycle, any edge of that cycle can be removed without disconnecting the graph. So remove this edge, leaving a new graph. If it has a cycle, we can again remove an edge; continuing in this manner, we must eventually stop, since there are only finitely many edges to remove. We then have a graph with no cycles; since no edge removal ever disconnected the graph, it must still be connected.\n\nAlso, given a tree, we can choose a root vertex $r$. Then, for any vertex $v$, there is a unique path from $v$ to $r$, never repeating a vertex (uniqueness follows from the absence of cycles). We call $v$ a descendant of $w$ if this path goes through $w$. Every vertex is considered to be a descendant of itself and of $r$. Suppose $v$ is a descendant of $w$; then the path from $v$ to $r$ consists of the path from $v$ to $w$ followed by the path from $w$ to $r$. It follows that descent is transitive: if $w$ in turn is a descendant of $u$, then $v$ is a descendant of $u$. It also follows that $d(v, r)=d(v, w)+d(w, r)$, where $d(x, y)$ denotes the distance (i.e. number of edges in the path) from $x$ to $y$. Finally, a vertex $v$ with no descendants can be removed and the graph will remain connected. Proof: every other vertex is connected to $r$ by a path that does not pass through $v$, so these vertices will remain connected to $r$ - and hence to each other - when $v$ is removed.\n\nNow we can solve our original problem. We state the graph-theoretic translation: given a connected graph $G$ on $k n$ vertices $(k \\geq 0, n \\geq 1)$, these vertices can be partitioned into $k$ sets of size $n$ such that $d(v, w) \\leq 2 n-2$ whenever $v, w$ are in the same set. (In our case, $k=1999, n=2001$.) We prove this by induction on $k$. If $k=0$, we form no vertex sets, and the statement is vacuously true. Now suppose the statement holds for $k-1$, where $k \\geq 1$, and we have a graph $G$ on $k n$ vertices. It suffices to prove the result when $G$ is a tree, since otherwise we can make it a tree by removing some edges and partition the vertices of this tree appropriately. The same partition will then work for the original graph $G$, since the distance between two vertices cannot increase when we put the deleted edges back.\n\nSo suppose $G$ is a tree, and arbitrarily choose a root $r$. Now let $a$ be a vertex for which $d(a, r)$ is maximal. Let $a=v_{1}, v_{2}, \\ldots, v_{q}=r$ be the path from $a$ to $r$, and choose the smallest $i$ such that $v_{i}$ has at least $n$ descendants. (Some such $i$ surely exists, since $r$ has $k n$ descendants.) Let $S$ be the set of descendants of $v_{i}$; note that if $v \\in S$, then every descendant of $v$ is in $S$, by transitivity. Notice that $v_{1}, v_{2}, \\ldots, v_{i-1}$ are all descendants of $v_{i-1}$, so the minimality of $i$ implies $i-1<n$. Thus, $d\\left(a, v_{i}\\right)=i-1 \\leq n-1$. Now we claim the distance between any two elements of $S$ is at most $2 n-2$. Indeed, suppose $b, c \\in S$. We have $d\\left(b, v_{i}\\right)=d(b, r)-d\\left(v_{i}, r\\right) \\leq d(a, r)-d\\left(v_{i}, r\\right)$ (by choice of $a$ ) $=d\\left(a, v_{i}\\right) \\leq n-1$. Similarly, $d\\left(c, v_{i}\\right) \\leq n-1$, and so $d(b, c) \\leq d\\left(b, v_{i}\\right)+d\\left(v_{i}, c\\right) \\leq 2(n-1)$, as claimed.\n\nNow let $a_{1}$ be an element of $S$ at maximal distance from $r$. (For example, take $a_{1}=a$.) Then $a_{1}$ can have no descendants (except itself) in $G$, since if $b$ were a descendant of $a_{1}$, we would have $d(b, r)=d\\left(a_{1}, b\\right)+d\\left(a_{1}, r\\right)$, contradicting maximality. Thus, we can remove $a_{1}$ from $G$ to leave a graph $G_{1}$, which is still connected - in fact, still a tree with root $r$. Now a similar argument shows that, if $a_{2} \\in S \\cap G_{1}$ is chosen to have maximal distance from $r$, then $a_{2}$ can have no descendants in $G_{1}$ : any descendant would lie in $S$, by transitivity, and it would also be farther from $r$ than $a_{2}$, violating maximality. So, deleting $a_{2}$ from $G_{1}$ gives another rooted tree, $G_{2}$. Then, we can choose $a_{3} \\in S \\cap G_{2}$ to be maximally distant from $r$, and so forth. We continue removing vertices in this manner; since $S$ has at least $n$ elements, we can remove $n$ vertices. Thus, we choose distinct vertices $a_{1}, a_{2}, \\ldots, a_{n}$, all of which lie in $S$; this means that any two of these vertices are at distance $\\leq 2 n-2$ from each other, and the remaining graph, $G-\\left\\{a_{1}, \\ldots, a_{n}\\right\\}$, is still a tree. Now using the induction hypothesis, this remaining graph can be partitioned to form the remaining $k-1$ sets of vertices, and the desired partition of $G$ is accomplished.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind $a+b+c+d+e$ if\n$$\n\\begin{gathered}\n3 a+2 b+4 d=10, \\\\\n6 a+5 b+4 c+3 d+2 e=8, \\\\\na+b+2 c+5 e=3, \\\\\n2 c+3 d+3 e=4, \\text{ and } \\\\\na+2 b+3 c+d=7 .\n\\end{gathered}\n$$",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nAdding the first, third, and fifth equations, we get $5 a+5 b+5 c+5 d+5 e=10+3+7 \\Rightarrow a+b+c+d+e=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a set of integers (not necessarily positive) such that\n\na. there exist $a, b \\in S$ with $\\gcd(a, b) = \\gcd(a - 2, b - 2) = 1$;\n\nb. if $x$ and $y$ are elements of $S$ (possibly equal), then $x^2 - y$ also belongs to $S$.\n\nProve that $S$ is the set of all integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In the solution below we use the expression $S$ is stable under $x \\mapsto f(x)$ to mean that if $t$ belongs to $S$, then $f(t)$ also belongs to $S$. If $c, d \\in S$, then by condition (b), $S$ is stable under $x \\mapsto c^2 - x$ and $x \\mapsto d^2 - x$. Hence, it is stable under $x \\mapsto c^2 - (d^2 - x) = x + (c^2 - d^2)$. Similarly, $S$ is stable under $x \\mapsto x + (d^2 - c^2)$. Hence, $S$ is stable under $x \\mapsto x + n$ and $x \\mapsto x - n$, whenever $n$ is an integer linear combination of finitely many numbers in $T = \\{c^2 - d^2 \\mid c, d \\in S\\}$.\n\nBy condition (a), $S \\neq \\emptyset$ and hence $T \\neq \\emptyset$ as well. For the sake of contradiction, assume that some $p$ divides every element in $T$. Then $c^2 - d^2 \\equiv 0 \\pmod{p}$ for all $c, d \\in S$. In other words, for each $c, d \\in S$, either $d \\equiv c \\pmod{p}$ or $d \\equiv -c \\pmod{p}$. Given $c \\in S$, $c^2 - c \\in S$, by condition (b), so $c^2 - c \\equiv c \\pmod{p}$ or $c^2 - c \\equiv -c \\pmod{p}$. Hence,\n$$\nc \\equiv 0 \\pmod{p} \\text{ or } c \\equiv 2 \\pmod{p} \\quad (*)\n$$\nfor each $c \\in S$. By condition (a), there exist some $a$ and $b$ in $S$ such that $\\gcd(a, b) = 1$, that is, at least one of $a$ or $b$ cannot be divisible by $p$. Denote such an element of $S$ by $\\alpha$; thus, $\\alpha \\not\\equiv 0 \\pmod{p}$. Similarly, by condition (a), $\\gcd(a - 2, b - 2) = 1$, so $p$ cannot divide both $a - 2$ and $b - 2$. Thus, there is an element of $S$, call it $\\beta$, such that $\\beta \\not\\equiv 2 \\pmod{p}$. By (*), $\\alpha \\equiv 2 \\pmod{p}$ and $\\beta \\equiv 0 \\pmod{p}$. By condition (b), $\\beta^2 - \\alpha \\in S$. Taking $c = \\beta^2 - \\alpha$ in (*) yields either $-2 \\equiv 0 \\pmod{p}$ or $-2 \\equiv 2 \\pmod{p}$, so $p = 2$. Now (*) says that all elements of $S$ are even, contradicting condition (a). Hence, our assumption is false and no prime divides every element in $T$.\n\nIt follows that $T \\neq \\{0\\}$. Let $x$ be an arbitrary nonzero element of $T$. For each prime divisor of $x$, there exists an element in $T$ which is not divisible by that prime. The set $A$ consisting of $x$ and each of these elements is finite. By construction, $\\text{gcd}\\{y \\mid y \\in A\\} = 1$, and $1$ can be written as an integer linear combination of finitely many elements in $A$ and hence in $T$. Therefore, $S$ is stable under $x \\mapsto x+1$ and $x \\mapsto x-1$. Because $S$ is nonempty, it follows that $S$ is the set of all integers.\nDefine $T$, $a$, and $b$ as in the first solution. We present another proof that no prime divides every element in $T$. Suppose, for sake of contradiction, that such a prime $p$ does exist. By condition (b), $a^2 - a, b^2 - b \\in S$. Therefore, $p$ divides $a^2 - b^2$, $x_1 = (a^2 - a)^2 - a^2$, and $x_2 = (b^2 - b)^2 - b^2$. Because $\\text{gcd}(a, b) = 1$, both $\\text{gcd}(a^2 - b^2, a^3)$ and $\\text{gcd}(a^2 - b^2, b^3)$ equal 1, so $p$ does not divide $a^3$ or $b^3$. But $p$ does divide $x_1 = a^3(a - 2)$ and $x_2 = b^3(b - 2)$, so it must divide $a - 2$ and $b - 2$. Because $\\text{gcd}(a - 2, b - 2) = 1$ by condition (a), this implies $p \\mid 1$, a contradiction. Therefore our original assumption was false, and no such $p$ exists.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46603,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, $\\angle B = 2\\angle C$. Let $P$ and $Q$ be points on the perpendicular bisector of segment $BC$ such that rays $AP$ and $AQ$ trisect $\\angle A$. Prove that $PQ < AB$ if and only if $\\angle B$ is obtuse.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $ABQ_1P$ be a rhombus and let segments $AQ_1$ and $BP$ intersect at $Q_2$. Let $X$ be a point on ray $Q_2Q_1$. Because lines $PQ_2$ and $Q_1Q_2$ are perpendicular, $PX$ increases as $X$ moves away from $Q_2$. Note that $AQ_1$ bisects angle $BAP$, that is, $Q$ is on the ray $Q_2Q_1$. Because $\\angle ABC > \\angle BCA$, $A$ and $B$ are on the same side of line $PQ$. Because $PQ_1 \\parallel AB$ and $PQ \\perp BC$, $\\angle CBA$ is obtuse if and only if the segment $PQ_1$ and $B$ are on different sides of line $PQ$. It follows that $\\angle CBA$ is obtuse if and only if $Q_1$ is farther from $Q_2$ on ray $Q_2Q_1$ than $Q$, that is, $PQ < PQ_1 = AB$.\n\n\nSecond Solution. (by Reid Barton) We may assume without loss of generality that $P$ is closer to $A$ than is $Q$. Let $X$ be the intersection of segment $AC$ and ray $QP$, and let $A_1$ be the reflection of $A$ across the line $PQ$. By symmetry, $A_1$ lies on the ray $BX$ and $AA_1 \\parallel BC$.\n\nLet $\\angle BCA = x$. Then $\\angle A_1AC = x = \\angle XBC$. Since $\\angle ABC = 2x$, $\\angle ABA_1 = x$. It follows that\n$$\n\\begin{aligned}\n\\angle PXA &= \\angle PXB + \\angle BXA \\\\\n&= 90^\\circ - \\angle XBC + \\angle BXA \\\\\n&= 90^\\circ + \\angle BCX = 90^\\circ + x.\n\\end{aligned}\n$$\nSince $\\angle CAB = 180^\\circ - 3x$, it follows that $\\angle XAP = \\angle CAP = 60^\\circ - x$. Hence,\n$$\n\\angle APX = 180^\\circ - \\angle PXA - \\angle XAP = 30^\\circ.\n$$\nHence, $\\angle APA_1 = 2\\angle APX = 60^\\circ$. Because $A_1P = AP$, this implies that triangle $PAA_1$ is equilateral and that $AA_1 = AP$. Note also that $\\angle BA_1A = \\angle A_1BC = \\angle ABA_1$. Hence, $AB = AA_1 = AP$. Therefore, $PQ < AB$ if and only if $PQ < AP$. In triangle $APQ$, $PQ < AP$ if and only if $\\angle PAQ < \\angle AQP$, that is,\n$$\n\\angle PAQ < \\angle APX - \\angle PAQ = 30^\\circ - \\angle PAQ,\n$$\nor $2\\angle PAQ < 30^\\circ$. Because $\\angle PAQ = 60^\\circ - x$, we conclude that $PQ < AB$ if and only if\n$$\n90^\\circ < 2x = \\angle ABC,\n$$\nas desired.\n\n\nThird Solution.\nAs in the first two solutions, let $\\angle BCA = x$ and assume that $\\angle CAB = 3\\angle CAP$. Also, let $\\angle PBC = y$. Then $\\angle CAB = 180^\\circ - 3x$, $\\angle PAB = 120^\\circ - 2x$, $\\angle ABP = 2x - y$, and $\\angle BPA = 60^\\circ + y$. From the **Law of Sines** in triangles $ABC$, $ABP$, and $BCP$, we obtain\n$$\n\\begin{aligned}\n\\frac{BC}{\\sin(180^\\circ - 3x)} &= \\frac{AB}{\\sin x}, \\\\\n\\frac{AB}{\\sin(60^\\circ + y)} &= \\frac{BP}{\\sin(120^\\circ - 2x)}, \\\\\n\\frac{PC}{\\sin y} &= \\frac{BC}{\\sin(180^\\circ - 2y)}.\n\\end{aligned}\n$$\nMultiplying the above relations and taking into account that $BP = PC$ yields\n$$\n\\sin 3x \\sin (60^\\circ + y) \\sin y = \\sin x \\sin (60^\\circ + 2x) \\sin 2y,\n$$\nor\n$$\n\\frac{\\sin 3x}{\\sin x} \\sin (60^\\circ + y) = \\frac{\\sin 2y}{\\sin y} \\sin (60^\\circ + 2x).\n$$\nBy the **Triple-angle formulas**, $\\sin 3x = \\sin x(3 - 4\\sin^2 x)$. By the **Double-angle formulas**, $\\sin 2y = \\sin y \\cdot 2\\cos y$. It follows that\n$$\n(3 - 4 \\sin^2 x) \\sin (60^\\circ + y) = 2 \\cos y \\sin (60^\\circ + 2x),\n$$\nor, by the **Double-angle formulas**,\n$$\n(1 + 2 \\cos 2x) \\sin (60^\\circ + y) = 2 \\cos y \\sin (60^\\circ + 2x).\n$$\nApplying the **Product-to-Sum formulas** gives\n$$\n\\begin{aligned}\n& \\sin (60^\\circ + y) + \\sin (60^\\circ + y + 2x) + \\sin (60^\\circ + y - 2x) \\\\\n&= \\sin (60^\\circ + 2x + y) + \\sin (60^\\circ + 2x - y).\n\\end{aligned}\n$$\nWe obtain\n$$\n\\sin (60^\\circ + y) = \\sin [60^\\circ + (2x - y)] - \\sin [60^\\circ - (2x - y)],\n$$\nor, using the **Difference-to-Product formulas**,\n$$\n\\sin (60^\\circ + y) = 2 \\cos 60^\\circ \\sin(2x - y) = \\sin (2x - y).\n$$\nNote that $60^\\circ + y > 0$, $2x - y > 0$, and that $(60^\\circ + y) + (2x - y) = 60^\\circ + 2x < 180^\\circ$ (as $3x = 180^\\circ - \\angle BAC < 180^\\circ$). It follows that $60^\\circ + y = 2x - y$, i.e., $\\angle BPA = \\angle ABP$.\nHence, triangle $ABP$ is isosceles and $AP = AB$. It follows that $PQ < AB$ is equivalent to $PQ < AP$. In triangle $APQ$, this is equivalent to\n$$\n60^\\circ - x = \\angle PAQ < \\angle AQP = x - 30^\\circ,\n$$\nthat is, $\\angle B = 2x > 90^\\circ$. It follows that $PQ > AB$ if and only if $\\angle B$ is obtuse.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46604,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$, let segment $AP$ bisect $\\angle BAC$, with $P$ on side $BC$, and let segment $BQ$ bisect $\\angle ABC$, with $Q$ on side $CA$. It is known that $\\angle BAC = 60^\\circ$ and that $AB + BP = AQ + QB$. What are the possible angles of triangle $ABC$?",
    "options": [],
    "answer": "∠ABC = 80°, ∠BCA = 40°, ∠CAB = 60°",
    "solution": "**First Solution.** (by Reid Barton and Gabriel Carroll) Extend segment $AB$ through $B$ to $R$ so that $BR = BP$, and construct $S$ on ray $AQ$ so that $AS = AR$.\n\n![](attached_image_1.png)\n\n**Lemma 1.** Points $B$, $P$, $S$ are collinear, and consequently, $S$ coincides with $C$.\n\n*Proof.* Because $BR = BP$, triangle $BPR$ is isosceles with base angles\n$$\n\\angle BRP = \\angle RPB = (180^\\circ - \\angle PBR)/2 = x = \\angle QBP. \\quad (2)\n$$\nNote that $AS = AR$ and $\\angle RAS = \\angle BAC = 60^\\circ$, implying that triangle $ARS$ is equilateral. Since line $AP$ bisects $\\angle RAS$, $R$ and $S$ are symmetric with respect to line $AP$. Thus,\n$$\nPR = PS \\qquad (3)\n$$\nand $\\angle ARP = \\angle PSA$, or $\\angle BRP = \\angle PSQ$. By (2), we have\n$$\n\\angle QBP = \\angle BRP = \\angle PSQ. \\qquad (4)\n$$\n\nBecause $AQ + QS = AB + BR = AB + BP = AQ + QB$, $QS = QB$. Hence, triangle $BQS$ is isosceles with\n$$\n\\angle BSQ = \\angle QBS. \\qquad (5)\n$$\n\nNow, assume to the contrary that triangle $BPS$ is nondegenerate. Then either $AC < AS$, as in the first diagram below, or $AC > AS$, as in the second diagram.\n\n![](attached_image_2.png)\n![](attached_image_3.png)\n\nIn either diagram, combining (4) and (5) gives\n$$\n\\angle PBS = |\\angle QBP - \\angle QBS| = |\\angle PSQ - \\angle BSQ| = \\angle PSB,\n$$\nthat is, triangle $PBS$ is isosceles with $PB = PS$. By (3), it follows that $PB = PS = PR$. Hence, triangle $BPR$ is equilateral. But then $\\angle ABC = 180^\\circ - \\angle CBR = 120^\\circ$, and by (1), $y = 0^\\circ$, which is absurd. Therefore, our assumption was wrong and $B, P, S$ are collinear. Consequently, $S = C$. ■\n\n---\n\n**Second Solution.** (By Zhiqiang Zhang, China) Construct $R$ on ray $AB$ beyond $B$ such that $BR = BP$, and construct $S$ on ray $AQ$ beyond $Q$ such that $QS = QB$. Let lines $BS$ and $AP$ intersect at $D$. Note that $AR = AB + BP = AQ + QB = AS$. Then both triangles $BPR$ and $BQS$ are isosceles with\n$$\n\\angle BRP = \\angle RPB = \\frac{\\angle ABC}{2} = x = \\angle QBP \\qquad (6)\n$$\nand\n$$\n\\angle QBS = \\angle BSQ = \\frac{\\angle BQA}{2} = \\frac{180^\\circ - 60^\\circ - x}{2} = 60^\\circ - \\frac{x}{2}. \\quad (7)\n$$\n\nBecause $y > 0^\\circ$, equation (1) implies $x < 60^\\circ$. Because $AR = AS$ and $\\angle RAS = 60^\\circ$, we also have that $60^\\circ = \\angle ARS = \\angle BRS$. Using these relations as well as (6), we find that $\\angle BRP = x < 60^\\circ = \\angle BRS$, implying that $P$ is inside triangle $ARS$. We now consider two cases.\n\ni. $C = S$. Since $BQ = QC$, $y = \\angle BCQ = \\angle QBC = x$, so by (1), $\\angle ABC = 80^\\circ$ and $\\angle BCA = 40^\\circ$. We need to check whether such a triangle has the desired properties. Setting $x = y$, in isosceles triangle $BPR$ we have $\\angle BRP = \\angle BPR = x$. Hence, $\\angle ARP = \\angle BRP = x = y = \\angle PCA$. This equality, along with the equations $\\angle PAR = \\angle CAP$ and $AP = AP$, implies that triangles $APD$ and $APC$ are congruent. Thus, $AR = AC$, or $AB + BP = AQ + BQ$. Therefore, $\\angle ABC = 80^\\circ$, $\\angle BCA = 40^\\circ$, and $\\angle CAB = 60^\\circ$ are possible angles of triangle $ABC$.\n\nii. $C \\neq S$.\n\n![](attached_image_4.png)\n\nSince $AR = AS$ and $AP$ is the angle bisector of angles $RAS$ and $BAC$, $R$ and $S$ are symmetric with respect to line $AP$, and $\\angle BRD = \\angle DSQ$. By (7), we have\n$$\n\\angle BRD = \\angle DSQ = \\angle BSQ = \\angle QBS = \\angle QBD. \\quad (8)\n$$\nBy subtraction, from (6) and (8) we obtain\n$$\n\\begin{aligned}\n\\angle PRD &= \\angle BRD - \\angle BRP \\\\\n&= \\angle QBD - \\angle QBP = \\angle PBD.\n\\end{aligned}\n$$\nBecause $P, D$ are on the same side of line $BR$ (specifically, between rays $AR$ and $CS$), quadrilateral $BPDR$ must be cyclic. Hence, $\\angle BRD = \\angle BPA$. Because angle $BPA$ is an exterior angle of triangle $APC$, by using (1), we have $\\angle BPA = 30^\\circ + y = 150^\\circ - 2x$. On the other hand, by (8) and (7), $\\angle BRD = \\angle DSQ = \\angle BSQ =$\n$$\n60^\\circ - \\frac{x}{2}.\n$$\nNow the relation $\\angle BRD = \\angle BPA$ reads\n$$\n150^\\circ - 2x = 60^\\circ - \\frac{x}{2},\n$$\nor $x = 60^\\circ$. But then $y = 0^\\circ$, which is impossible. Thus, there is no solution under the assumption $C \\neq S$.\n\n---\n\n**Third Solution.** (By Liang Xiao, China) Construct $R$ on ray $AB$ beyond $B$ such that $BR = BP$ and $S$ on ray $AQ$ beyond $Q$ such that $QS = QB$. Note that $AR = AB + BP = AQ + QB = AS$ and hence, because $\\angle SAR = 60^\\circ$, triangle $ARS$ is equilateral. Now we calculate a few angles:\n$$\n\\begin{aligned}\n\\angle BPA &= 150^\\circ - 2x, \\\\\n\\angle BQA &= 120^\\circ - x, \\\\\n\\angle BSQ &= \\angle QBS = \\frac{\\angle BQA}{2} = 60^\\circ - \\frac{x}{2}, \\\\\n\\angle ABS &= \\angle ABQ + \\angle QBS = 60^\\circ + \\frac{x}{2}, \\\\\n\\angle RSB &= 60^\\circ - \\angle BSQ = \\frac{x}{2}, \\\\\n\\angle SBR &= 180^\\circ - \\angle ABS = 120^\\circ - \\frac{x}{2}.\n\\end{aligned}\n$$\nNoting that $BR = BP$ and $AS = RS$, we apply the **Law of Sines** to triangles $BRS$, $ABP$, and $ABS$ to obtain\n$$\n\\begin{aligned}\n\\frac{\\sin \\angle RSB}{\\sin \\angle SBR} &= \\frac{BR}{RS} = \\frac{BP}{AS} = \\frac{BP}{AB} \\cdot \\frac{AB}{AS} \\\\\n&= \\frac{\\sin \\angle PAB}{\\sin \\angle BPA} \\cdot \\frac{\\sin \\angle BSA}{\\sin \\angle ABS}.\n\\end{aligned}\n$$\nWriting these angles in terms of $x$, we have\n$$\n\\frac{\\sin\\left(\\frac{x}{2}\\right)}{\\sin\\left(120^\\circ - \\frac{x}{2}\\right)} = \\frac{\\sin 30^\\circ}{\\sin(150^\\circ - 2x)} \\cdot \\frac{\\sin\\left(60^\\circ - \\frac{x}{2}\\right)}{\\sin\\left(60^\\circ + \\frac{x}{2}\\right)}.\n$$\nNow, $x = (120^\\circ - y)/2 < 60^\\circ$ and hence $\\sin(120^\\circ - \\frac{x}{2}) = \\sin(60^\\circ + \\frac{x}{2}) \\neq 0$. Cancelling the common terms and clearing the denominators of the above equation, we obtain\n$$\n2 \\sin \\left( \\frac{x}{2} \\right) \\sin(30^\\circ + 2x) = \\cos \\left( 30^\\circ + \\frac{x}{2} \\right).\n$$\nUsing the **Product-to-sum formulas** gives\n$$\n\\cos \\left( 30^\\circ + \\frac{3x}{2} \\right) - \\cos \\left( 30^\\circ + \\frac{5x}{2} \\right) = \\cos \\left( 30^\\circ + \\frac{x}{2} \\right),\n$$\nor\n$$\n\\cos \\left( 30^\\circ + \\frac{3x}{2} \\right) = \\cos \\left( 30^\\circ + \\frac{5x}{2} \\right) + \\cos \\left( 30^\\circ + \\frac{x}{2} \\right).\n$$\nApplying the **Sum-to-product formulas** yields\n$$\n\\cos \\left( 30^\\circ + \\frac{3x}{2} \\right) = 2 \\cos \\left( 30^\\circ + \\frac{3x}{2} \\right) \\cos x.\n$$\nHence, either $\\cos x = \\frac{1}{2}$ or $\\cos(30^\\circ + \\frac{3x}{2}) = 0$. The first case is impossible because $0^\\circ < x < 60^\\circ$. In the second, $0^\\circ < 30^\\circ + \\frac{3x}{2} < 120^\\circ$, implying that $30^\\circ + \\frac{3x}{2} = 90^\\circ$ and $x = 40^\\circ$. Consequently, it is easy to check that $\\angle ABS = \\angle ABC = 80^\\circ$ and $S = C$. Therefore, $\\angle ABC = 80^\\circ$, $\\angle BCA = 40^\\circ$, and $\\angle CAB = 60^\\circ$.\n\n---\n\n**Fourth Solution.** (By Hui Zheng, China) Applying the Law of Sines to triangles $ABP$ and $ABQ$ yields\n$$\n\\frac{AB + BP}{AB} = 1 + \\frac{\\sin 30^\\circ}{\\sin(2x + 30^\\circ)}\n$$\nand\n$$\n\\frac{AQ + QB}{AB} = \\frac{\\sin x + \\sin 60^\\circ}{\\sin(x + 60^\\circ)}.\n$$\nThus, the condition $AB + BP = AQ + BQ$ implies that\n$$\n1 + \\frac{\\sin 30^\\circ}{\\sin(2x + 30^\\circ)} = \\frac{\\sin x + \\sin 60^\\circ}{\\sin(x + 60^\\circ)}, \\quad (12)\n$$\nor\n$$\n\\frac{\\sin(x + 60^\\circ)}{2 \\sin(2x + 30^\\circ)} = \\sin x + \\sin 60^\\circ - \\sin(x + 60^\\circ).\n$$\nBy (1), $y > 0^\\circ$ and $0^\\circ < x < 60^\\circ$. Applying the Difference-to-product formulas twice yields\n$$\n\\begin{aligned}\n\\frac{\\sin(x + 60^\\circ)}{2 \\sin(2x + 30^\\circ)} &= \\sin 60^\\circ - [\\sin(x + 60^\\circ) - \\sin x] \\\\\n&= \\sin 60^\\circ - 2 \\sin 30^\\circ \\cos(x + 30^\\circ) \\\\\n&= \\sin 60^\\circ - \\sin(60^\\circ - x) \\\\\n&= 2 \\sin \\frac{x}{2} \\cos \\left(60 - \\frac{x}{2}\\right).\n\\end{aligned}\n$$\nBy the **Double-angle formulas**, the numerator on the left-hand side is\n$$\n\\sin(x + 60^\\circ) = \\sin(120^\\circ - x) = 2 \\sin\\left(60^\\circ - \\frac{x}{2}\\right) \\cos\\left(60^\\circ - \\frac{x}{2}\\right).\n$$\nPlugging in this expression and then dividing by $\\cos(60^\\circ - \\frac{x}{2})$ — which is nonzero because $0^\\circ < x < 60^\\circ$ — we find that\n$$\n\\frac{\\sin(60^\\circ - \\frac{x}{2})}{\\sin(2x + 30^\\circ)} = 2 \\sin(\\frac{x}{2}).\n$$\nClearing the denominator and applying the Product-to-sum formulas, the last equation becomes\n$$\n\\begin{aligned}\n\\sin\\left(60^\\circ - \\frac{x}{2}\\right) &= 2 \\sin\\left(\\frac{x}{2}\\right) \\sin(2x + 30^\\circ) \\\\\n&= 2 \\sin\\left(\\frac{x}{2}\\right) \\cos(2x - 60^\\circ) \\\\\n&= \\sin\\left(\\frac{5x}{2} - 60^\\circ\\right) + \\sin\\left(60^\\circ - \\frac{3x}{2}\\right),\n\\end{aligned}\n$$\nor\n$$\n\\begin{aligned}\n\\sin\\left(60 - \\frac{3x}{2}\\right) &= \\sin\\left(60^\\circ - \\frac{x}{2}\\right) - \\sin\\left(\\frac{5x}{2} - 60^\\circ\\right) \\\\\n&= 2 \\sin\\left(60^\\circ - \\frac{3x}{2}\\right) \\cos x\n\\end{aligned}\n$$\nby applying the Difference-to-product formulas. Hence,\n$$\n\\sin(60^\\circ - \\frac{3x}{2}) = 0 \\quad \\text{or} \\quad \\cos x = \\frac{1}{2}.\n$$\nThe only solution to either equation in the range $0^\\circ < x < 60^\\circ$ is $x = 40^\\circ$. Thus, $\\angle ABC = 80^\\circ$, $\\angle BCA = 40^\\circ$, and $\\angle CAB = 60^\\circ$.\n\n---\n\n**Fifth Solution.** (By Kiran Kedlaya) We use complex numbers in this solution. Rewrite (12) as\n$$\n1 + \\frac{1}{2\\cos(2x - 60^\\circ)} - \\frac{\\sin x + \\sin 60^\\circ}{\\sin(x + 60^\\circ)} = 0. \\quad (13)\n$$\nSet $z = \\cos x + i \\sin x$ and $\\omega = e^{2\\pi i/6} = \\cos 60^\\circ + i \\sin 60^\\circ$. Because $0^\\circ < x < 60^\\circ$, we know that $-1, 1 \\neq z\\omega$. Now, $\\bar{z} = \\frac{1}{z}$ and $\\bar{\\omega} = \\frac{1}{\\omega}$. Also,\n$$\n\\omega^3 = -1 \\quad \\text{and} \\quad \\omega - \\omega^2 = 1. \\quad (14)\n$$\nIt follows that $2i \\sin x = z - \\bar{z} = z - \\frac{1}{z}$, $2i \\sin 60^\\circ = \\omega - \\frac{1}{\\omega}$, $2i \\sin(x + 60^\\circ) = z\\omega - \\frac{1}{z\\omega}$, and $2 \\cos(2x - 60^\\circ) = \\frac{z^2}{\\omega} + \\frac{\\omega}{z^2}$. We rewrite (13) in terms of $z$:\n$$\n1 + \\frac{1}{\\frac{z^2}{\\omega} + \\frac{\\omega}{z^2}} - \\frac{z - \\frac{1}{z} + \\omega - \\frac{1}{\\omega}}{z\\omega - \\frac{1}{z\\omega}} = 0. \\quad (15)\n$$\nThe second term on the left-hand side of (15) is $\\frac{z^2\\omega}{z^4+\\omega^2}$ and the third term simplifies to\n$$\n\\frac{(z + \\omega) - \\frac{z+\\omega}{z\\omega}}{\\frac{(zw)^2-1}{z\\omega}} = \\frac{(z + \\omega)(z\\omega - 1)}{(z\\omega + 1)(z\\omega - 1)} = \\frac{z + \\omega}{z\\omega + 1}.\n$$\nThus, (15) simplifies to\n$$\n1 + \\frac{z^2\\omega}{z^4 + \\omega^2} - \\frac{z + \\omega}{z\\omega + 1} = 0.\n$$\nClearing the denominators in this equations yields\n$$\n(z^4 + \\omega^2)(z\\omega + 1) + z^2\\omega(z\\omega + 1) - (z + \\omega)(z^4 + \\omega^2) = 0.\n$$\nIf $x \\approx 60^\\circ$, then triangles $ABQ$ and $PBQ$ are approximately equilateral and $AB + BP \\approx AQ + QB$. Thus, we suspect that $z = \\omega$ is a solution\n\n---\n\nof the above equation. With this in mind, we factor (14) as follows:\n$$\n\\begin{align*}\n0 &= (z^4 + \\omega^2)(z\\omega + 1) + z^2\\omega(z\\omega + 1) - (z + \\omega)(z^4 + \\omega^2) \\\\\n&= z^5\\omega + z\\omega^3 + z^4 + \\omega^2 + z^3\\omega^2 + z^2\\omega - z^5 - z\\omega^2 - z^4\\omega - \\omega^3 \\\\\n&= (\\omega - 1)z^5 + (1 - \\omega)z^4 + \\omega^2 z^3 + \\omega z^2 + (\\omega^3 - \\omega^2)z + (\\omega^2 - \\omega^3) \\\\\n&= \\omega^2 z^5 - \\omega^2 z^4 + \\omega^2 z^3 + \\omega z^2 - \\omega z + \\omega \\\\\n&= \\omega^2 z^3 (z^2 - z + 1) + \\omega (z^2 - z + 1) \\\\\n&= \\omega^2 (z^3 - \\omega^2) (z^2 - z + 1).\n\\end{align*}\n$$\nHence, $z^3 = \\omega^2 = \\cos 120^\\circ + i \\sin 120^\\circ$, so $z = \\cos 40^\\circ + i \\sin 40^\\circ$, $\\cos 160^\\circ + i \\sin 160^\\circ$, or $\\cos 280^\\circ + i \\sin 280^\\circ$. yielding $x = 40^\\circ, 160^\\circ$, or $280^\\circ$. The only value in the range $(0^\\circ, 60^\\circ)$ is $x = 40^\\circ$, so $\\angle ABC = 80^\\circ$, $\\angle BCA = 40^\\circ$, and $\\angle CAB = 60^\\circ$.\n\n---\n\n**Sixth Solution.** (By Belur Jana Venkatachala, India) Let $a = BC$, $b = CA$, $c = AB$, and $2s = a + b + c$. By the **Angle Bisector Theorem**, we obtain\n$$\nBP = \\frac{ca}{b+c} \\quad \\text{and} \\quad AQ = \\frac{cb}{c+a}. \\qquad (16)\n$$\n**Lemma 2.** We have\n$$\nBQ = \\frac{2ca}{c+a} \\sqrt{\\frac{s(s-b)}{ca}}.\n$$\n*Proof.* Let $[R]$ denote the area of region $R$. Then,\n$$\n\\begin{align*}\n2ac \\sin x \\cos x &= ac \\sin 2x = 2[ABC] = 2[ABQ] + 2[QBC] \\\\\n&= c \\cdot BQ \\sin x + a \\cdot BQ \\sin x \\\\\n&= BQ \\cdot (a+c) \\sin x,\n\\end{align*}\n$$\nor\n$$\nBQ = \\frac{2ac}{a+c} \\cos x. \\qquad (17)\n$$\nBy the **Law of Cosines**, we obtain\n$$\n\\cos 2x = \\cos \\angle ABC = \\frac{a^2 + c^2 - b^2}{2ac}.\n$$\nApplying the Double-angle formulas yields\n$$\n\\begin{aligned}\n2 \\cos^2 x &= \\cos 2x + 1 = \\frac{a^2 + c^2 - b^2}{2ac} + 1 \\\\\n&= \\frac{(a^2 + 2ac + c^2) - b^2}{2ac} = \\frac{(a+c)^2 - b^2}{2ac} \\\\\n&= \\frac{(a+c+b)(a+c-b)}{2ac} = \\frac{2s(s-b)}{ac},\n\\end{aligned}\n$$\nor $\\cos x = \\sqrt{\\frac{s(s-b)}{ac}}$. Substituting this into (17) yields the desired result. (The result also follows directly from the Angle Bisector Theorem and the **Stewart's Theorem**.) ■\n\nBy lemma 2, the given condition reads\n$$\nc + \\frac{ca}{b+c} = \\frac{cb}{c+a} + \\frac{2ca}{c+a} \\sqrt{\\frac{s(s-b)}{ca}},\n$$\nor\n$$\n\\frac{a+b+c}{b+c} = \\frac{b}{a+c} + \\frac{2a}{c+a} \\sqrt{\\frac{s(s-b)}{ca}}.\n$$\nClearing the denominators yields\n$$\n(a+b+c)(c+a)\\sqrt{c} = b(b+c)\\sqrt{c} + 2\\sqrt{a}(b+c)\\sqrt{s(s-b)}. \\quad (18)\n$$\nSince\n$$\n\\begin{aligned}\n(a+b+c)(c+a) - b(b+c) &= (a+b+c)(c+a) - b(a+b+c) + ab \\\\\n&= (a+b+c)(a+c-b) + ab \\\\\n&= 4s(s-b) + ab,\n\\end{aligned}\n$$\n(18) can be rewritten as\n$$\n4\\sqrt{cs(s-b)} + ab\\sqrt{c} = 2\\sqrt{a}(b+c)\\sqrt{s(s-b)}.\n$$\nTaking $t = \\sqrt{s(s-b)}$, we obtain a quadratic in $t$:\n$$\n4\\sqrt{ct^2 - 2\\sqrt{a}(b+c)t + ab\\sqrt{c}} = 0,\n$$\nor $(2t - \\sqrt{ac})(2\\sqrt{ct} - \\sqrt{ab}) = 0$. Thus,\n$$\nt = \\frac{\\sqrt{ac}}{2} \\quad \\text{or} \\quad t = \\frac{\\sqrt{ab}}{2\\sqrt{c}}.\n$$\nIf $t = \\frac{\\sqrt{ac}}{2}$, then $s(s-b) = \\frac{ac}{4}$, leading to\n$$\n\\frac{(a+c+b)(a+c-b)}{4} = \\frac{(a+c)^2 - b^2}{4} = \\frac{ac}{4},\n$$\nor $a^2 + ac + c^2 = b^2$. This equation and the Law of Cosines imply that $\\angle ABC = 120^\\circ$, which is absurd.\n\nIf $t = \\frac{\\sqrt{ab}}{2\\sqrt{c}}$, then $s(s-b) = \\frac{ab^2}{4c}$, leading to\n$$\n(a+c)^2 - b^2 = \\frac{ab^2}{c}.\n$$\nHence,\n$$\n(a+c)^2 = \\frac{(a+c)b^2}{c},\n$$\nor\n$$\n\\frac{c}{b} = \\frac{cb}{a+c} \\cdot \\frac{1}{c}.\n$$\nBy the second part of (16), the last equation is\n$$\n\\frac{AB}{AC} = \\frac{AQ}{AB},\n$$\nwhich, along with $\\angle BAC = \\angle QAB$, implies that triangles $ABQ$ and $ACB$ are similar. Hence, $\\angle ACB = \\angle ABQ = \\angle BQC$, that is, $\\angle ABC = 80^\\circ$, $\\angle BCA = 40^\\circ$, and $\\angle CAB = 60^\\circ$.\n\n---\n\nTherefore, $\\angle ABC = 80^\\circ$, $\\angle BCA = 40^\\circ$, and $\\angle CAB = 60^\\circ$ is the only solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46605,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all $x$ between $-\\frac{\\pi}{2}$ and $\\frac{\\pi}{2}$ such that $1-\\sin^{4} x-\\cos^{2} x=\\frac{1}{16}$.",
    "options": [],
    "answer": "x = ±π/12, ±5π/12",
    "solution": "Solution:\n$1-\\sin^{4} x-\\cos^{2} x = \\frac{1}{16} \\Rightarrow (16-16 \\cos^{2} x) - \\sin^{4} x - 1 = 0 \\Rightarrow 16 \\sin^{4} x - 16 \\sin^{2} x + 1 = 0$.\n\nUse the quadratic formula in $\\sin x$ to obtain $\\sin^{2} x = \\frac{1}{2} \\pm \\frac{\\sqrt{3}}{4}$.\n\nSince $\\cos 2x = 1 - 2 \\sin^{2} x = \\pm \\frac{\\sqrt{3}}{2}$, we get $x = \\pm \\frac{\\pi}{12}, \\pm \\frac{5\\pi}{12}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a_{1}=3$ and define $a_{n+1} = \\left(3 a_{n}^{2} + 1\\right) / 2 - a_{n}$ for $n \\geq 1$. If $n$ is a power of $3$, prove that $a_{n}$ is divisible by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe main trick is finding a closed-form expression for $a_{n}$, which requires some experimentation. We will show that $a_{n} = \\left(2^{2^{n}+1} + 1\\right) / 3$ for all $n$ by induction. It is easy to check that the formula holds for $n=1$. And if it holds for some $n$, then\n\n$$\n\\begin{aligned}\na_{n+1} & = \\frac{3 a_{n}^{2} + 1}{2} - a_{n} \\\\\n& = \\frac{1}{2}\\left(3\\left[\\frac{2^{2^{n}+1} + 1}{3}\\right]^{2} + 1\\right) - \\frac{2^{2^{n}+1} + 1}{3} \\\\\n& = \\frac{1}{2}\\left(\\frac{\\left(2^{2^{n}+1}\\right)^{2} + 2\\left(2^{2^{n}+1}\\right) + 1}{3} + 1\\right) - \\frac{2^{2^{n}+1} + 1}{3} \\\\\n& = \\frac{1}{2}\\left(\\frac{2^{2^{n+1}+2} + 2\\left(2^{2^{n}+1}\\right) + 4}{3}\\right) - \\frac{2^{2^{n}+1} + 1}{3} \\\\\n& = \\frac{2^{2^{n+1}+1} + 2^{2^{n}+1} + 2}{3} - \\frac{2^{2^{n}+1} + 1}{3} \\\\\n& = \\frac{2^{2^{n+1}+1} + 1}{3},\n\\end{aligned}\n$$\n\ngiving the induction step.\n\nNow, by Euler's theorem, $3^{k}$ divides $2^{2 \\cdot 3^{k-1}} - 1$, for any nonnegative integer $k$, since $\\phi\\left(3^{k}\\right)$ (i.e. the number of integers in $\\{1,2, \\ldots, 3^{k}\\}$ relatively prime to $3^{k}$) equals $2 \\cdot 3^{k-1}$. But notice that $2^{2 \\cdot 3^{k-1}} - 1 = \\left(2^{3^{k-1}} - 1\\right)\\left(2^{3^{k-1}} + 1\\right)$, and $3^{k-1}$ is odd $\\Rightarrow 2^{3^{k-1}} \\equiv 2 \\pmod{3} \\Rightarrow 2^{3^{k-1}} - 1$ is relatively prime to $3^{k}$, so, in fact, $3^{k}$ divides $2^{3^{k-1}} + 1$. Also, for any integers $c, d$ with $d$ odd, $2^{c} + 1 \\mid 2^{c d} + 1$. We conclude that $3^{k} \\mid 2^{a} + 1$ whenever $3^{k-1} \\mid a$ and $a$ is odd.\n\nApplying this result twice in succession, we find that $3^{k} \\mid 2^{3^{k}} + 1$ and then that $3^{k+1} \\mid 2^{2^{3^{k}} + 1} + 1$, so that $3^{k} \\mid \\left(2^{2^{3^{k}} + 1} + 1\\right) / 3 = a_{3^{k}}$ for any integer $k \\geq 0$, and this is what we wanted to prove.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46607,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor each positive integer $n$, let $a_{n}$ be the number of permutations $\\tau$ of $\\{1,2, \\ldots, n\\}$ such that $\\tau(\\tau(\\tau(x)))=x$ for $x=1,2, \\ldots, n$. The first few values are\n$$\na_{1}=1,\\ a_{2}=1,\\ a_{3}=3,\\ a_{4}=9.\n$$\nProve that $3^{334}$ divides $a_{2001}$.\n\n(A permutation of $\\{1,2, \\ldots, n\\}$ is a rearrangement of the numbers $\\{1,2, \\ldots, n\\}$, or equivalently, a one-to-one and onto function from $\\{1,2, \\ldots, n\\}$ to $\\{1,2, \\ldots, n\\}$. For example, one permutation of $\\{1,2,3\\}$ is the rearrangement $\\{2,1,3\\}$, which is equivalent to the function $\\sigma:\\{1,2,3\\} \\rightarrow\\{1,2,3\\}$ defined by $\\sigma(1)=2,\\ \\sigma(2)=1,\\ \\sigma(3)=3$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider the permutations $\\tau$ of $\\{1,2, \\ldots, n\\}$ such that $\\tau(\\tau(\\tau(x)))=x$ for $x=1,2, \\ldots, n$. Then for each $x \\in\\{1,2, \\ldots, n\\}$, there are only two possibilities. Either $x$ is a fixed point; i.e., $\\tau(x)=x$, or else $x$ is a member of a 3-cycle $(xyz)$; i.e. $\\tau(x)=y,\\ \\tau(y)=z,\\ \\tau(z)=x$, where $x, y, z$ are 3 distinct numbers.\n\nWe can thus partition these permutations into two cases: Either 1 is a fixed point, or it is not. In the first case, the remaining elements $\\{2, \\ldots, n\\}$ can be permuted in $a_{n-1}$ ways. In the second case, 1 is part of a 3-cycle $(1\\ i\\ j)$, where $i \\neq j$ and $i, j \\in\\{2, \\ldots, n\\}$. There are $(n-1)(n-2)$ such 3-cycles (order counts). Then the remaining $n-3$ elements can be permuted in $a_{n-3}$ ways. Hence we have (for $n>3$)\n$$\na_{n}=a_{n-1}+(n-1)(n-2) a_{n-3}.\n$$\nDefine $b_{n}$ to be the highest power of 3 which divides $a_{n}$; i.e.\n$$\n3^{b_{n}} \\mid\\mid a_{n}\n$$\nIf $n \\not \\equiv 0 \\pmod{3}$, then $(n-1)(n-2)$ will be a multiple of 3, so equation (1) yields $a_{n}=a_{n-1}+3k a_{n-3}$, where $k$ is a positive integer. Hence\n$$\nb_{n} \\geq \\min \\left(b_{n-1},\\ b_{n-3}+1\\right),\\quad n \\not \\equiv 0 \\pmod{3}\n$$\nIf $n \\equiv 0 \\pmod{3}$, then $(n-1)(n-2) \\equiv -1 \\pmod{3}$, but both $(n-2)(n-3)$ and $(n-3)(n-4)$ are multiples of 3. Plugging into (1), we have\n$$\n\\begin{aligned}\na_{n}-a_{n-1} &= (n-1)(n-2) a_{n-3} = (3u-1) a_{n-3}, \\\\\na_{n-1}-a_{n-2} &= (n-2)(n-3) a_{n-4} = 3v a_{n-4}, \\\\\na_{n-2}-a_{n-3} &= (n-3)(n-4) a_{n-5} = 3w a_{n-5},\n\\end{aligned}\n$$\nfor positive integers $u, v, w$. Adding these, we get\n$$\na_{n}-a_{n-3} = (3u-1) a_{n-3} + 3v a_{n-4} + 3w a_{n-5},\n$$\nso\n$$\na_{n} = 3\\left(u a_{n-3} + v a_{n-4} + w a_{n-5}\\right).\n$$\nThis implies that\n$$\nb_{n} \\geq 1 + \\min\\left(b_{n-3},\\ b_{n-4},\\ b_{n-5}\\right),\\quad n \\equiv 0 \\pmod{3},\\ n>5\n$$\nWe know that $b_{1}=0,\\ b_{2}=0,\\ b_{3}=1$. Employing (2), we get $b_{4} \\geq 1,\\ b_{5} \\geq 1$. Then (3) yields $b_{6} \\geq 1$. Applying (2) again yields $b_{7} \\geq 1,\\ b_{8} \\geq 1$. But now when (3) is applied, we have $b_{9} \\geq 2$. It is evident that continuing this process, (2) and (3) will increment $b_{n}$ by at least one as $n$ increases by 6. In other words,\n$$\nb_{6k+3} \\geq k+1\n$$\nSince $2001=6 \\cdot 333+3$, we are done.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the area of a square inscribed in a semicircle of radius $1$, with one of its sides flush with the diameter of the semicircle?",
    "options": [],
    "answer": "4/5",
    "solution": "Solution:\n\nCall the center of the semicircle $O$, a point of contact of the square and the circular part of the semicircle $A$, the closer vertex of the square on the diameter $B$, and the side length of the square $x$. We know $OA = 1$, $AB = x$, $OB = \\frac{x}{2}$, and $\\angle ABO$ is right. By the Pythagorean theorem, $x^{2} = \\frac{4}{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the volume of the tetrahedron with vertices $(5,8,10)$, $(10,10,17)$, $(4,45,46)$, $(2,5,4)$.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nEach vertex $(x, y, z)$ obeys $x + y = z + 3$, so all the vertices are coplanar and the volume of the tetrahedron is $0$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the sum of the coefficients of the expansion $(x+2y-1)^6$?",
    "options": [],
    "answer": "64",
    "solution": "Solution:\nThe sum of the coefficients of a polynomial is that polynomial evaluated at $1$, which for the question at hand is $(1+2 \\cdot 1-1)^6 = 2^6 = 64$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, a_{3}, \\ldots$ be a sequence of positive integers with the following property: if $S$ is any nonempty set of positive integers, there exists $s \\in S$ with $a_{s} \\leq \\operatorname{gcd}(S)$. Prove that $n!$ is divisible by $a_{1} a_{2} \\cdots a_{n}$ for every positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFix $n$. Arrange the integers $a_{1}, \\ldots, a_{n}$ in nonincreasing order, $a_{i_{1}} \\geq a_{i_{2}} \\geq \\cdots \\geq a_{i_{n}}$. We claim there exists a bijective function $f:\\{1,2, \\ldots, n\\} \\rightarrow \\{1,2, \\ldots, n\\}$ such that $a_{i_{k}} \\mid f(k)$ for each $k=1,2, \\ldots, n$. To demonstrate this, we construct $f$ inductively. Suppose that $f(k)$ has been defined for all values of $k$ less than some $j$, and we wish to define $f(j)$. Let $d$ be the greatest common divisor of $i_{1}, i_{2}, \\ldots, i_{j}$. By the hypothesis, there exists some $i \\in \\{i_{1}, i_{2}, \\ldots, i_{j}\\}$ such that $a_{i} \\leq d$; since $a_{i_{j}} = \\min \\{a_{i_{1}}, a_{i_{2}}, \\ldots, a_{i_{j}}\\}$, we have $a_{i_{j}} \\leq a_{i} \\leq d$. Now, we know of $j$ distinct multiples of $d$ lying in the set $\\{1,2, \\ldots, n\\}$ (namely, $i_{1}, i_{2}, \\ldots, i_{j}$); this many multiples can only exist if $j d \\leq n$. Then, the numbers $a_{i_{j}}, 2 a_{i_{j}}, \\ldots, j a_{i_{j}}$ are also all in $\\{1,2, \\ldots, n\\}$, since $j a_{i_{j}} \\leq j d \\leq n$. At most $j-1$ of these can have been used up by the previously defined values $f(1), f(2), \\ldots, f(j-1)$, so some value is left over; we define $f(j)$ to be such a value. Then $a_{i_{j}} \\mid f(j)$, as required.\n\nSo we can recursively define $f(1), f(2), \\ldots, f(n)$ by the above method, and our construction ensures that $f$ is injective. Since it maps the finite set $\\{1,2, \\ldots, n\\}$ to itself, it must actually be bijective. Since $a_{i_{k}} \\mid f(k)$ for each $k$, we have\n$$\na_{1} a_{2} \\cdots a_{n} = a_{i_{1}} a_{i_{2}} \\cdots a_{i_{n}} \\mid f(1) f(2) \\cdots f(n) = 1 \\cdot 2 \\cdots n = n!$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the 6-digit number beginning and ending in the digit 2 that is the product of three consecutive even integers.",
    "options": [],
    "answer": "287232",
    "solution": "Solution:\nBecause the last digit of the product is $2$, none of the three consecutive even integers end in $0$. Thus they must end in $2, 4, 6$ or $4, 6, 8$, so they must end in $4, 6, 8$ since $2 \\cdot 4 \\cdot 6$ does not end in $2$. Call the middle integer $n$. Then the product is $(n-2) n (n+2) = n^{3} - 4n$, so $n > \\sqrt[3]{200000} = \\sqrt[3]{200 \\cdot 10^{3}} \\approx 60$, but clearly $n < \\sqrt[3]{300000} = \\sqrt[3]{300 \\cdot 10^{3}} < 70$. Thus $n = 66$, and the product is $66^{3} - 4 \\cdot 66 = 287232$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a > b > c > d$ be positive integers and suppose\n$$\nac + bd = (b + d + a - c)(b + d - a + c).\n$$\nProve that $ab + cd$ is not prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** For the sake of contradiction, assume that $ab + cd$ is prime. Note that\n$$\nab + cd = (a + d)c + (b - c)a = m \\cdot \\gcd(a + d, b - c)\n$$\nfor some positive integer $m$. Writing $g = \\gcd(a+d, b-c)$, we have\n$$\nm = \\frac{a+d}{g} \\cdot c + \\frac{b-c}{g} \\cdot a \\geq c+a > 1.\n$$\nTherefore, because $ab+cd$ is prime, $g=1$.\nSubstituting $ac+bd = (a+d)b - (b-c)a$ for the left-hand side of the given condition, we obtain\n$$\n(a+d)b - (b-c)a = (a+d)(b+d-a+c) + (b-c)(b+d-a+c),\n$$\nor\n$$\n(a+d)(a-c-d) = (b-c)(b+c+d).\n$$\nHence, there exists a positive integer $k$ such that\n$$\na-c-d = k(b-c),\n$$\n$$\nb+c+d = k(a+d).\n$$\nAdding these equations, we obtain $a+b = k(a+b-c+d)$ and thus $k(c-d) = (k-1)(a+b)$. Recall that $a > b > c > d > 0$. If $k=1$, then $c=d$, a contradiction. If $k \\ge 2$, then\n$$\n2 \\ge \\frac{k}{k-1} = \\frac{a+b}{c-d} > \\frac{2b}{c} > 2,\n$$\na contradiction.\nTherefore, our original assumption was wrong, and $ab+cd$ is not prime.\n\n\n**Second Solution.** (By Yonggao Chen, China) We give a proof by contradiction. Assume that $p = ab+cd$ is prime. Then $ab \\equiv -cd \\pmod p$. By (1),\n$$\nb^2(b^2 + bd + d^2) = b^2(a^2 - ac + c^2) = (ab)^2 - ab(bc) + b^2c^2.\n$$\nIt follows that\n$$\n\\begin{aligned} b^2(b^2 + bd + d^2) &\\equiv (ab)^2 - ab(bc) + b^2c^2 \\\\ &\\equiv (cd)^2 + cd(bc) + b^2c^2 \\\\ &\\equiv c^2(b^2 + bd + d^2) \\pmod p, \\end{aligned}\n$$\nimplying that $p \\mid (b^2 - c^2)(b^2 + bd + d^2)$. Observe that $0 < b^2 - c^2 < b^2 < ab < p$. Thus, $p$ and $b^2 - c^2$ must be relatively prime, so\n$$\np \\mid (b^2 + bd + d^2). \\qquad (2)\n$$\nBecause\n$$\n0 < b^2 + bd + d^2 < ab + ab + cd = 2ab + cd < 2p,\n$$\nwe must have $b^2 + bd + d^2 = p = ab + cd$ or, equivalently,\n$$\nb(b+d-a) = d(c-d).\n$$\nBecause $ab + cd$ is prime, $b$ must be relatively prime to $d$, so $b \\mid c - d$. This is impossible, because $0 < c - d < b$.\n\n\n**Third Solution.** (By Zhiqiang Zhang, China) Let $x = a - c$, $y = a + c$, $u = b - d$, and $v = b + d$. By the given condition, we have\n$$\n\\begin{aligned} y^2 - x^2 + v^2 - u^2 &= 4(ac + bd) \\\\ &= 4[(b+d) + (a-c)][(b+d) - (a-c)] \\\\ &= 4(v+x)(v-x) = 4(v^2 - x^2), \\end{aligned}\n$$\nor\n$$\ny^2 - u^2 = 3(v^2 - x^2). \\qquad (3)\n$$\nLet $s = a + b + c + d$, $x_1 = s - 2d$, $x_2 = s - 2c$, $x_3 = s - 2b$, and $x_4 = s - 2a$. Then $x_1 = y + u$, $x_2 = v + x$, $x_3 = y - u$, $x_4 = v - x$. Because $a > b > c > d$, $x_1 > x_2 > x_3 > x_4$. Now (3) reads\n$$\nx_1x_3 = 3x_2x_4. \\qquad (4)\n$$\nBecause\n$$\nxu + vy = (a - c)(b - d) + (a + c)(b + d) = 2(ab + cd)\n$$\nand\n$$\nx_1x_2 + x_3x_4 = (y + u)(v + x) + (y - u)(v - x) = 2(xu + yv),\n$$\nwe have\n$$\nab + cd = \\frac{1}{4}(x_1x_2 + x_3x_4). \\qquad (5)\n$$\nLet $g = \\gcd(x_1, x_4)$. It is clear that $s \\equiv x_i \\pmod 2$ for $i = 1, 2, 3, 4$. We consider the following cases.\n(i) $s \\equiv 1 \\pmod 2$. First suppose that $g = 1$. Then by (4), there exists some positive integer $k$ such that $x_3 = kx_4$ and, consequently, $kx_1 = 3x_2$. Because $x_3 > x_4$, $k > 1$; because $x_1 > x_2$, $k < 3$. Therefore, $k = 2$. But then $x_3$ is even, contradicting the assumption that each $x_i$ is odd.\nIt follows that $g > 1$, and that $g$ divides $x_1x_2+x_3x_4 = 4(ab+cd)$.\nBecause $x_1$ and $x_4$ are odd, $g$ is odd as well, implying that $g \\mid (ab+cd)$. Also observe that $x_1x_2+x_3x_4 \\ge 3x_1+2x_4 \\ge 3g+2g = 5g$, so that $g < ab+cd$. Therefore, $ab+cd$ is divisible by a number strictly between 1 and $ab+cd$, implying that it is composite.\n(ii) $s \\equiv x_i \\equiv 0 \\pmod{2}$. Let $x_i' = x_i/2$ for $i = 1, 2, 3, 4$, and let $g' = \\gcd(x_1', x_2')$. Then $g = 2g' \\ge 2$ and $x_1' > x_2' > x_3' > x_4'$. Note also that (4) and (5) become\n$$\nx_1'x_3' = 3x_2'x_4' \\quad \\text{and} \\quad ab + cd = x_1'x_2' + x_3x_4', \\quad (4')\n$$\nrespectively.\nIf $g' > 1$, then $g' \\mid (ab + cd)$. Since $ab + cd > x_2' \\ge g'$, $ab + cd$ must be composite.\nIf $g' = 1$, then by (4'), $x_3' = kx_4'$ and $kx_1' = 3x_2'$ for some positive integer $k$. Then again by $x_3' > x_4'$ and $x_1' > x_2'$, $k = 2$. Hence 2 divides both $x_2'$ and $x_3'$ implying that $ab+cd$ is even (by (4')). Since $ab+cd > a > 2$, $ab+cd$ is composite.\nFrom the above arguments, we conclude that $ab + cd$ is not prime.\n\n\n**Fourth Solution.** (By Andrei Vorobiev, Russia) Let $x = b+d+a-c$. It is clear that $x > 1$. We have $c \\equiv a+b+d \\pmod x$ and $d \\equiv c-a-b \\pmod x$. These congruences, combined with the given condition, yield\n$$\n\\begin{aligned}\n0 &\\equiv ac + bd \\equiv a(a + b + d) + bd \\\\\n&\\equiv (a + b)(a + d) \\pmod x\n\\end{aligned}\n$$\nand\n$$\n\\begin{aligned}\n0 &\\equiv ac + bd \\equiv ac + b(c - a - b) \\\\\n&\\equiv (a + b)(c - b) \\pmod x.\n\\end{aligned}\n$$\nHence, $x \\mid (a+b)(a+d)$ and $x \\mid (a+b)(c-b)$.\nBecause $a+b > (a+b) - (c-d) = x$ and $2x = 2[a + (b-c) + d] > 2a > a+b$, $a+b$ is not divisible by $x$. Thus, there is a prime $p$ that divides each of $x$, $(a+d)$, and $(c-b)$. To finish, we only need to prove that $p$ is a proper divisor of $ab+cd$. In fact, $ab+cd > a+d \\ge p$ and\n$$\np \\mid (a+d)b + (c-b)d = ab + cd,\n$$\nas desired.\n\n\n**Fifth Solution.** Let $ABCD$ be the quadrilateral with $AB = a, BC = d, CD = b, AD = c, \\angle BAD = 60^\\circ$, and $\\angle BCD = 120^\\circ$. Such a quadrilateral exists in view of (1) and the **Law of Cosines**; the common value in (1) is $BD^2$. Let $\\angle ABC = \\alpha$, so that $\\angle CDA = 180^\\circ - \\alpha$. Applying the Law of Cosines to triangles $ABC$ and $ACD$ gives\n$$\na^2 + d^2 - 2ad \\cos \\alpha = AC^2 = b^2 + c^2 + 2bc \\cos \\alpha.\n$$\nHence, $2 \\cos \\alpha = (a^2 + d^2 - b^2 - c^2)/(ad + bc)$, and\n$$\nAC^2 = a^2 + d^2 - ad \\frac{a^2 + d^2 - b^2 - c^2}{ad + bc} = \\frac{(ab + cd)(ac + bd)}{ad + bc}.\n$$\nBecause $ABCD$ is cyclic, the **Ptolemy's Theorem** yields\n$$\n(AC \\cdot BD)^2 = (AB \\cdot CD + AD \\cdot BD)^2 = (ab + cd)^2\n$$\nIt follows that\n$$\n(ac + bd)(a^2 - ac + c^2) = (ab + cd)(ad + bc). \\quad (6)\n$$\n(Note that straightforward algebra can also be used to obtain (6) from (1).) Observe that\n$$\nab + cd > ac + bd > ad + bc. \\quad (7)\n$$\nThe first inequality follows from $(a-d)(b-c) > 0$, and the second from $(a-b)(c-d) > 0$.\nNow assume that $ab+cd$ is prime. It then follows from (7) that $ab+cd$ and $ac+bd$ are relatively prime. Hence, from (6), it must be true that $ac+bd$ divides $ad+bc$. However, this is impossible by (7). Thus, $ab+cd$ must not be prime.\n\n\n**Sixth Solution.** (By Reid Barton and Gabriel Carroll) Let $\\omega = e^{\\frac{2\\pi i}{3}}$. Then\n$$\n\\omega^3 = 1 \\quad \\text{and} \\quad 1 + \\omega + \\omega^2 = 0. \\qquad (8)\n$$\nWe are going to use two fundamental facts about the ring $\\mathbb{Z}[\\omega]$:\n* Fact 1. $\\mathbb{Z}[\\omega]$ is a unique factorization domain (UFD);\n* Fact 2. the units in $\\mathbb{Z}[\\omega]$ are $\\pm 1, \\pm\\omega, \\pm\\omega^2$.\nFactoring (1) in $\\mathbb{Z}[\\omega]$ gives\n$$\n(c + \\omega a)(c + \\omega^2 a) = (b - \\omega d)(b - \\omega^2 d). \\qquad (9)\n$$\n**Lemma 1.** If $a > b > c > d$ are positive integers satisfying (9), and $ab + cd$ is prime, then $c + \\omega a$ and $b - \\omega d$ are not relatively prime.\n*Proof.* Assume for the sake of contradiction that $c + \\omega a$ and $b - \\omega d$ are relatively prime. Since complex conjugation is an automorphism of $\\mathbb{Z}[\\omega]$ sending $\\omega$ to $\\omega^2$, $c + \\omega^2 a$ and $b - \\omega^2 d$ must also be relatively prime. From the two facts, we conclude that $c + \\omega a = u(b - \\omega^2 d)$ for some unit $u \\in \\{\\pm 1, \\pm \\omega, \\pm \\omega^2\\}$.\nIf $u = \\pm 1$, then $c + \\omega a = \\pm (b - \\omega^2 d) = \\pm (b + d) \\pm \\omega d$ (by the second part of (8)), contradicting $a \\neq \\pm d$.\nIf $u = \\pm \\omega$, then $c + \\omega a = \\pm \\omega(b - \\omega^2 d) = \\mp d \\pm \\omega b$ (by the first part of (8), contradicting both $a \\neq \\pm b$ and $c \\neq \\mp d$).\nIf $u = \\pm \\omega^2$, then $c + \\omega a = \\pm \\omega^2(b - \\omega^2 d) = \\pm(\\omega^2 b - \\omega d) = \\mp b \\mp (b+d)\\omega$ (by (8)), contradicting $c \\neq \\mp b$.\nIn all cases, we reach a contradiction, so $c + \\omega a$ and $b - \\omega d$ are not relatively prime. ■\n**Lemma 2.** If $a > b > c > d$ are positive integers satisfying (1) and $ab + cd$ is prime, then $ad = cb + cd$.\n*Proof.* Since $a, b, c, d$ satisfy (1), they also satisfy (9), so by Lemma 1, there exists some prime $p = q + r\\omega \\in \\mathbb{Z}[\\omega]$ such that $p \\mid c + \\omega a$ and $p \\mid b - \\omega d$. Then $\\overline{p} = q + r\\omega^2 \\mid b - \\omega^2 d$, so $p\\overline{p} \\mid (c + \\omega a)(b - \\omega^2 d)$. Note that\n$$\nN(p) = p\\overline{p} = q^2 - qr + r^2\n$$\nand that\n$$\n(c + \\omega a)(b - \\omega^2 d) = bc + \\omega ab - \\omega^2 dc - \\omega^3 ad \\\\ = (-ad + bc + dc) + (ab + cd)\\omega.\n$$\nTherefore, $N(p) \\mid [(-ad + bc + dc) + (ab + cd)\\omega]$. Since $N(p) \\in \\mathbb{Z}$, we must have $N(p) \\mid (-ad + bc + dc)$ and $N(p) \\mid ab + cd$. Since $ab + cd$ is prime, $N(p) = ab + cd$, and so $ab + cd \\mid (-ad + bc + cd)$. But\n$$\nab + cd - (-ad + bc + dc) = (ab - bc) + ad > 0\n$$\nand\n$$\nab + cd + (-ad + bc + dc) > ab - ad > 0,\n$$\nso $|-ad+bc+cd| < ab+cd$. Hence, we must have $-ad+bc+cd = 0$, that is, $ad = cb + cd$.\nNow suppose that $a > b > c > d > 0$ are integers satisfying (1). Then\n$$\n\\begin{aligned}\n(a-c)^2 + (a-c)c + c^2 &= a^2 - 2ac + c^2 + ac - c^2 + c^2 \\\\\n&= a^2 - ac + c^2 \\\\\n&= b^2 + bd + d^2.\n\\end{aligned}\n$$\nSince $c > d > 0$, we must have $a - c < b$ implying $(a - c)d < bd < bc$, or $ad < cb + cd$. By Lemma 2, $ab + cd$ cannot be prime, and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be positive real numbers such that\n$$\na + b + c \\ge abc.\n$$\nProve that at least two of the inequalities\n$$\n\\frac{2}{a} + \\frac{3}{b} + \\frac{6}{c} \\ge 6, \\quad \\frac{2}{c} + \\frac{3}{a} + \\frac{6}{b} \\ge 6,\n$$\nare true.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** Assume, for the sake of contradiction, that at least two of the numbers\n$$\n\\frac{2}{a} + \\frac{3}{b} + \\frac{6}{c}, \\quad \\frac{2}{b} + \\frac{3}{c} + \\frac{6}{a}, \\quad \\frac{2}{c} + \\frac{3}{a} + \\frac{6}{b}\n$$\nare less than $6$. Without loss of generality, we may further assume that the first and the last are less than $6$. Then\n$$\n\\frac{5}{a} + \\frac{9}{b} + \\frac{8}{c} < 12.\n$$\nAlso, because $b + c \\ge a(bc - 1)$, we have\n$$\n\\frac{1}{a} \\ge \\frac{bc - 1}{b + c}.\n$$\nIt follows that\n$$\n\\frac{5(bc - 1)}{b + c} + \\frac{9}{b} + \\frac{8}{c} < 12,\n$$\nor\n$$\n5b^{2}c^{2} + 12bc - 12b^{2}c - 12c^{2}b + 9c^{2} + 8b^{2} < 0. \\quad (1)\n$$\nCompleting squares yields\n$$\n(2bc - 2b - 3c)^2 + b^2(c - 2)^2 < 0,\n$$\na contradiction. This proves the conclusion. To obtain equalities, that is two of the members\n$$\n\\frac{2}{a} + \\frac{3}{b} + \\frac{6}{c}, \\quad \\frac{2}{b} + \\frac{3}{c} + \\frac{6}{a}, \\quad \\frac{2}{c} + \\frac{3}{a} + \\frac{6}{b}\n$$\nare equal to $6$, we must have $c - 2 = 0$ and $2bc - 2b - 3c = 0$, that is $c = 2$, $b = 3$, $a = 1$. Therefore the equalities hold if and only if $(a, b, c)$ is one of the triples $(1, 3, 2)$, $(3, 2, 1)$, $(2, 1, 3)$.\n\n**Or:** Rewriting (1) as a quadratic form in $b$ yields\n$$\n(5c^2 - 12c + 8)b^2 - 12c(c - 1)b + 9c^2 < 0,\n$$\nwhich is impossible, because the leading coefficient\n$$\n5c^2 - 12c + 8 = 5\\left(c - \\frac{6}{5}\\right)^2 + \\frac{4}{5}\n$$\nis always positive and the discriminant\n$$\n\\begin{aligned} \\Delta &= [12c(c-1)]^2 - 36c^2(5c^2 - 12c + 8) \\\\ &= 36c^2[4(c-1)^2 - (5c^2 - 12c + 8)] = -36c^2(c-2)^2 \\end{aligned}\n$$\nis always nonpositive.\n\n\n**Second Solution.** (by David Shin) Perform the substitutions $x = 1/a$, $y = 1/b$, and $z = 1/c$. It suffices to prove that at least two of the inequalities\n$$\n2x + 3y + 6z \\le 6, \\quad 2y + 3z + 6x \\le 6, \\quad 2z + 3x + 6y \\le 6\n$$\nare true, where\n$$\nx, y, z > 0 \\quad \\text{and} \\quad xy + yz + zx \\ge 1.\n$$\nAssume, for the sake of contradiction, that at least two of the given inequalities are false. Without loss of generality, we may assume that $2x + 3y + 6z < 6$ and $2y + 3z + 6x < 6$. Then\n$$\n\\begin{aligned} & 144 > [(2x + 3y + 6z) + (2y + 3z + 6x)]^2 \\\\ &= (8x + 5y + 9z)^2 \\\\ &= 64x^2 + 80xy + 25y^2 + 81z^2 + 90yz + 144zx \\\\ &= 64x^2 - 64xy + 16y^2 + 9y^2 - 54yz + 81z^2 + 144(xy + yz + zx) \\\\ &= (8x - 4y)^2 + (3y - 9z)^2 + 144 \\ge 144, \\end{aligned}\n$$\na contradiction. Thus, our assumptions is false and at least two of the desired inequalities must be true. To obtain equalities, that is two of the numbers\n$$\n\\frac{2}{a} + \\frac{3}{b} + \\frac{6}{c}, \\quad \\frac{2}{b} + \\frac{3}{c} + \\frac{6}{a}, \\quad \\frac{2}{c} + \\frac{3}{a} + \\frac{6}{b}\n$$\nare equal to $6$, we must have $8x - 4y = 3y - 9z = 0$, or $a : b : c = (1/x) : (1/y) : (1/z) = 2 : 1 : 3$. Therefore the equalities hold if and only if $(a, b, c) = (1, 3, 2)$, $(3, 2, 1)$, $(2, 1, 3)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the last digit of $1^{1} + 2^{2} + 3^{3} + \\cdots + 100^{100}$?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nLet $L(d, n)$ be the last digit of a number ending in $d$ to the $n$th power. For $n \\geq 1$, we know that $L(0, n) = 0$, $L(1, n) = 1$, $L(5, n) = 5$, $L(6, n) = 6$. All numbers ending in odd digits in this series are raised to odd powers; for odd $n$, $L(3, n) = 3$ or $7$, $L(7, n) = 3$ or $7$, $L(9, n) = 9$. All numbers ending in even digits are raised to even powers; for even $n$, $L(2, n) = 4$ or $6$, $L(4, n) = L(6, n) = 6$, $L(8, n) = 6$ or $4$. Further, for each last digit that has two possible values, the possible values will be present equally as often. Now define $S(d)$ such that $S(0) = 0$ and for $1 \\leq d \\leq 9$, $S(d) = L(d, d) + L(d, d+10) + L(d, d+20) + L(d, d+30) + \\cdots + L(d, d+90)$, so that the sum we want to calculate becomes $S(0) + S(1) + S(2) + \\cdots + S(9)$. But by the above calculations all $S(d)$ are divisible by $10$, so their sum is divisible by $10$, which means its last digit is $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA right triangle has a hypotenuse of length $2$, and one of its legs has length $1$. The altitude to its hypotenuse is drawn. What is the area of the rectangle whose diagonal is this altitude?",
    "options": [],
    "answer": "3√3/16",
    "solution": "Solution:\n\nCall the triangle $ABC$, with $AC = 2$ and $BC = 1$. By the Pythagorean theorem, $AB = \\sqrt{3}$. Call the point at which the altitude intersects the hypotenuse $D$. Let $E \\neq B$ be the vertex of the rectangle on $AB$ and $F \\neq B$ be the vertex of the rectangle on $BC$. Triangle $BDC$ is similar to triangle $ABC$, so $BD = \\frac{\\sqrt{3}}{2}$. Triangle $DBF$ is similar to triangle $ABC$, so $DF = \\frac{\\sqrt{3}}{4}$ and $BF = \\frac{3}{4}$. The area of the rectangle is thus $\\frac{\\sqrt{3}}{4} \\cdot \\frac{3}{4} = \\frac{3\\sqrt{3}}{16}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are two red, two black, two white, and a positive but unknown number of blue socks in a drawer. It is empirically determined that if two socks are taken from the drawer without replacement, the probability they are of the same color is $\\frac{1}{5}$. How many blue socks are there in the drawer?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nLet the number of blue socks be $x > 0$. Then the probability of drawing a red sock from the drawer is $\\frac{2}{6 + x}$ and the probability of drawing a second red sock from the drawer is $\\frac{1}{6 + x - 1} = \\frac{1}{5 + x}$, so the probability of drawing two red socks from the drawer without replacement is $\\frac{2}{(6 + x)(5 + x)}$. This is the same as the probability of drawing two black socks from the drawer and the same as the probability of drawing two white socks from the drawer. The probability of drawing two blue socks from the drawer, similarly, is $\\frac{x(x - 1)}{(6 + x)(5 + x)}$.\n\nThus the probability of drawing two socks of the same color is the sum of the probabilities of drawing two red, two black, two white, and two blue socks from the drawer:\n\n$$\n3 \\cdot \\frac{2}{(6 + x)(5 + x)} + \\frac{x(x - 1)}{(6 + x)(5 + x)} = \\frac{x^2 - x + 6}{(6 + x)(5 + x)} = \\frac{1}{5}.\n$$\n\nCross-multiplying and distributing gives $5x^2 - 5x + 30 = x^2 + 11x + 30$, so $4x^2 - 16x = 0$, and $x = 0$ or $4$. But since $x > 0$, there are $4$ blue socks.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe graph of $x^{2}-(y-1)^{2}=1$ has one tangent line with positive slope that passes through $(x, y)=(0,0)$. If the point of tangency is $(a, b)$, find $\\sin ^{-1}\\left(\\frac{a}{b}\\right)$ in radians.",
    "options": [],
    "answer": "pi/4",
    "solution": "Solution:\nDifferentiating both sides of the equation, we find that $2x - 2(y-1) \\frac{\\mathrm{d}y}{\\mathrm{d}x} = 0$, and so $\\frac{\\mathrm{d}y}{\\mathrm{d}x} = \\frac{x}{y-1} = \\frac{a}{b-1}$. The line passing through $(0,0)$ and $(a, b)$ has slope $\\frac{b}{a}$, so $\\frac{b}{a} = \\frac{a}{b-1}$. Solving simultaneously with $a^{2} - (b-1)^{2} = 1$, we get $b^{2} - b - \\left[(b-1)^{2} + 1\\right] = 0$, and so $b = 2,\\ a = \\sqrt{2}$. Finally, $\\sin^{-1}\\left(\\frac{a}{b}\\right) = \\frac{\\pi}{4}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 46619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the radius of the smallest sphere in which 4 spheres of radius $1$ will fit?",
    "options": [],
    "answer": "(2 + sqrt(6)) / 2",
    "solution": "Solution:\n\nThe centers of the smaller spheres lie on a tetrahedron. Let the points of the tetrahedron be $(1,1,1)$, $(-1,-1,1)$, $(-1,1,-1)$, and $(1,-1,-1)$. These points have distance $\\sqrt{3}$ from the center, and $\\sqrt{2}$ from each other, so the radius of the smallest sphere in which $4$ spheres of radius $\\sqrt{2}$ will fit is $\\sqrt{2} + \\sqrt{3}$. Scale this to the correct answer by dividing by $\\sqrt{2}$: $\\frac{2 + \\sqrt{6}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou take a wrong turn on the way to MIT and end up in Transylvania, where $99\\%$ of the inhabitants are vampires and the rest are regular humans. For obvious reasons, you want to be able to figure out who's who. On average, nine-tenths of the vampires are correctly identified as vampires and nine-tenths of humans are correctly identified as humans. What is the probability that someone identified as a human is actually a human?",
    "options": [],
    "answer": "1/12",
    "solution": "Solution:\nConsider a sample of $1000$ inhabitants. On average, $990$ are vampires and $10$ are people. $99$ vampires are identified as human and $9$ humans are identified as human. So out of the $108$ who pass, only $\\frac{1}{12}$ are human.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $\\angle ABC = \\angle ADC = 135^\\circ$ and\n$$\nAC^2 \\cdot BD^2 = 2AB \\cdot BC \\cdot CD \\cdot DA.\n$$\nProve that the diagonals of quadrilateral $ABCD$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.**\n![](attached_image_1.png)\nLet $\\alpha$ be the angle formed by diagonals $AC$ and $BD$, where $0^\\circ < \\alpha \\le 90^\\circ$. We calculate the area of quadrilateral $ABCD$ in two ways. We obtain\n$$\n[ABCD] = \\frac{1}{2}(AC \\cdot BD \\sin \\alpha)\n$$\nand\n$$\n\\begin{align*} \n[ABCD] &= [ABC] + [CDA] \\\\ \n&= \\frac{1}{2}(AB \\cdot BC \\sin 135^\\circ) + \\frac{1}{2}(CD \\cdot DA \\sin 135^\\circ) \\\\ \n&= \\frac{1}{2\\sqrt{2}}(AB \\cdot BC + CD \\cdot DA). \n\\end{align*}\n$$\nCombining the above two equations yields\n$$\nAC^2 \\cdot BD^2 \\sin^2 \\alpha = \\frac{1}{2}(AB \\cdot BC + CD \\cdot DA)^2.\n$$\nFor real numbers $a$ and $b$, $(a + b)^2 \\ge 4ab$, hence\n$$\nAC^2 \\cdot BD^2 \\sin^2 \\alpha \\ge 2AB \\cdot BC \\cdot CD \\cdot DA.\n$$\nIt follows that $\\sin^2 \\alpha \\ge 1$. Therefore, $\\sin \\alpha = 1$ and $\\alpha = 90^\\circ$, as desired.\n\n\n**Second Solution.**\n![](attached_image_2.png)\nConsider an **inversion** of center $B$ and radius $1$. Let $A', C', D'$ denote the images of $A, C, D$ under the inversion, respectively. Then $\\angle A'D'B = \\angle BAD$ and $\\angle BD'C' = \\angle DCB$. It follows that\n$$\n\\angle A'D'C' = \\angle BAD + \\angle BCD = 360^\\{\\circ\\} - 2 \\cdot 135^\\{\\circ\\} = 90^\\{\\circ\\}.\n$$\nApplying the **Inversive Distance Formula** we obtain\n$$\nA'C' = \\frac{AC}{AB \\cdot BC}, \\quad C'D' = \\frac{CD}{BD \\cdot BC}, \\quad D'A' = \\frac{DA}{AB \\cdot BD}.\n$$\nApplying the **Pythagorean Theorem** to right triangle $A'C'D'$ yields\n$$\n\\left( \\frac{AC}{AB \\cdot BC} \\right)^2 = \\left( \\frac{CD}{BD \\cdot BC} \\right)^2 + \\left( \\frac{DA}{AB \\cdot BD} \\right)^2,\n$$\nor\n$$\n(AC \\cdot BD)^2 = (AB \\cdot CD)^2 + (BC \\cdot DA)^2.\n$$\nRewriting this as\n$$\n(AB \\cdot CD - BC \\cdot DA)^2 = 0,\n$$\nwe have $AB \\cdot CD = BC \\cdot DA$; that is, $AB/BC = AD/CD$. Because $\\angle ABC = \\angle CDA$, it follows that triangles $ABC$ and $ADC$ are similar. These two triangle share the corresponding side $AC$, so they are congruent. This implies that $ABCD$ is a **kite** with segment $AC$ as its symmetry axis. Therefore, $AC \\perp BD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEquilateral triangle $A B C$ with side length $1$ is drawn. A square is drawn such that its vertex at $A$ is opposite to its vertex at the midpoint of $B C$. Find the area enclosed within the intersection of the insides of the triangle and square. Hint: $\\sin 75 = \\frac{\\sqrt{2}(\\sqrt{3}+1)}{4}$.",
    "options": [],
    "answer": "3/(4(√3+1))",
    "solution": "Solution:\n\nLet $D$ be the midpoint of $B C$, $F \\neq A$ be the point of intersection of the square and triangle lying on $A C$, $b$ be the length of $F C$, $x$ be the side length of the triangle, and $y$ be the length of $A D$.\n\nBy the law of sines on triangle $C D F$, we have\n$$\n\\frac{2 \\sin 75}{x} = \\frac{\\sin 45}{b},\n$$\nso\n$$\nb = \\frac{x \\sin 45}{2 \\sin 75} = \\frac{\\sqrt{2} x}{4 \\sin 75}.\n$$\n\nThe area of the desired figure can easily be seen to be $\\frac{1}{2}(x-b) y$ since it can be seen as two triangles of width $y$ and height $\\frac{x-b}{2}$.\n\nThis reduces to\n$$\n\\frac{1}{2}\\left(1-\\frac{\\sqrt{2}}{4 \\sin 75}\\right) x y.\n$$\n\nThen by the Pythagorean theorem on triangle $A B D$, $x^{2} = \\left(\\frac{x}{2}\\right)^{2} + y^{2}$, so $y = \\frac{\\sqrt{3}}{2} x$, and the area becomes\n$$\n\\frac{\\sqrt{3}}{4}\\left(1-\\frac{\\sqrt{2}}{4 \\sin 75}\\right) x^{2} = \\frac{\\sqrt{3}}{4}\\left(1-\\frac{\\sqrt{2}}{4 \\sin 75}\\right) = \\frac{3}{4(\\sqrt{3}+1)}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the coefficient of $x^{12}$ in the Maclaurin series (i.e. Taylor series around $x=0$) for $\\frac{1}{1-3x+2x^{2}}$.",
    "options": [],
    "answer": "2047",
    "solution": "Solution:\nIf you know formal power series, then this is not such a hard question, but since this is a calculus test... Use partial fractions to get\n$$\n\\frac{1}{1-3x+2x^{2}} = \\frac{1/2}{1-2x} - \\frac{1}{1-x}.\n$$\nNow each of these can be expanded as a geometric series (or take derivatives and get the same result) to get\n$$\n\\frac{1}{2}\\left(1 + 2x + 4x^{2} + 8x^{3} + \\cdots\\right) - \\left(1 + x + x^{2} + x^{3} + \\cdots\\right),\n$$\nso the coefficient of $x^{n}$ is $2^{n-1} - 1$. When $n=12$, that's $2047$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOrder these four numbers from least to greatest: $5^{56}, 10^{51}, 17^{35}, 31^{28}$.",
    "options": [],
    "answer": "5^{56}, 31^{28}, 17^{35}, 10^{51}",
    "solution": "Solution:\n$10^{51} > 9^{51} = 3^{102} = 27^{34} > 17^{35} > 16^{35} = 32^{28} > 31^{28} > 25^{28} = 5^{56}$, so the ordering is $5^{56}, 31^{28}, 17^{35}, 10^{51}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind $(x+1)\\left(x^{2}+1\\right)\\left(x^{4}+1\\right)\\left(x^{8}+1\\right) \\cdots$, where $|x|<1$.",
    "options": [],
    "answer": "1/(1 - x)",
    "solution": "Solution:\nLet $S = (x+1)\\left(x^{2}+1\\right)\\left(x^{4}+1\\right)\\left(x^{8}+1\\right) \\cdots = 1 + x + x^{2} + x^{3} + \\cdots$.\n\nSince $xS = x + x^{2} + x^{3} + x^{4} + \\cdots$, we have $(1-x)S = 1$, so $S = \\frac{1}{1-x}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA real number $x$ is randomly chosen in the interval $\\left[-15 \\frac{1}{2}, 15 \\frac{1}{2}\\right]$. Find the probability that the closest integer to $x$ is odd.",
    "options": [],
    "answer": "16/31",
    "solution": "Solution:\nBy using a graphical method, we can see that, for real $x$ on $\\left[-n-\\frac{1}{2}, n+\\frac{1}{2}\\right]$, $n$ an even integer, the probability that the closest integer to $x$ is odd is $\\frac{n}{2 n+1}$. The desired probability is $\\left(\\frac{15}{31}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe Fibonacci numbers are defined by $F_{1} = F_{2} = 1$ and $F_{n+2} = F_{n+1} + F_{n}$ for $n \\geq 1$. The Lucas numbers are defined by $L_{1} = 1$, $L_{2} = 2$, and $L_{n+2} = L_{n+1} + L_{n}$ for $n \\geq 1$. Calculate\n$$\n\\frac{\\prod_{n=1}^{15} \\frac{F_{2n}}{F_{n}}}{\\prod_{n=1}^{13} L_{n}}\n$$",
    "options": [],
    "answer": "1149852",
    "solution": "Solution:\n\nIt is easy to show that $L_{n} = \\frac{F_{2n}}{F_{n}}$, so the product above is $L_{14} L_{15} = 843 \\cdot 1364 = 1149852$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA polynomial $P$ has four roots, $\\frac{1}{4}, \\frac{1}{2}, 2, 4$. The product of the roots is $1$, and $P(1) = 1$. Find $P(0)$.",
    "options": [],
    "answer": "8/9",
    "solution": "Solution:\nA polynomial $Q$ with $n$ roots, $x_{1}, \\ldots, x_{n}$, and $Q\\left(x_{0}\\right) = 1$ is given by\n$$\nQ(x) = \\frac{(x - x_{1})(x - x_{2}) \\cdots (x - x_{n})}{(x_{0} - x_{1})(x_{0} - x_{2}) \\cdots (x_{0} - x_{n})}\n$$\nso\n$$\nP(0) = \\frac{1}{\\frac{3}{4} \\cdot \\frac{1}{2} \\cdot (-1) \\cdot (-3)} = \\frac{8}{9}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of nonnegative integers $(m, n)$ such that\n$$\n(m + n - 5)^2 = 9mn.\n$$",
    "options": [],
    "answer": "All solutions are given (up to order) by {5 F_{2k}^2, 5 F_{2k+2}^2} and {L_{2k-1}^2, L_{2k+1}^2} for nonnegative integers k, where F denotes the Fibonacci numbers and L denotes the Lucas numbers (extended with L_{-1} = −1 and L_0 = 2).",
    "solution": "**First Solution.** The equation is symmetric in $m$ and $n$. The solutions are the unordered pairs\n$$\n\\{5F_{2k}^2, 5F_{2k+2}^2\\}, \\quad \\{L_{2k-1}^2, L_{2k+1}^2\\},\n$$\nwhere $k$ is a nonnegative integer and $\\{F_j\\}$, $\\{L_j\\}$ are the **Fibonacci** and **Lucas sequences**, respectively — that is, the sequences defined by $F_1 = F_2 = 1$, $L_1 = 1$, $L_2 = 3$, and the recursive relations $F_{j+2} = F_{j+1} + F_j$ and $L_{j+2} = L_{j+1} + L_j$ for $j \\ge 1$. Note that we amended the Lucas sequence by considering $L_1 = -1$ and $L_0 = 2$.\nLet $g = \\gcd(m, n)$ and write $m = gm_1$ and $n = gn_1$. Because $9mn$ is a perfect square, $m_1$ and $n_1$ are perfect squares. Let $m_1 = x^2$ and $n_1 = y^2$. The given condition becomes\n$$\n(gx^2 + gy^2 - 5)^2 = 9g^2x^2y^2.\n$$\nTaking the square root on both sides yields\n$$\ng(x^2 + y^2) - 5 = \\pm 3gxy,\n$$\nor\n$$\ng(x^2 + y^2 \\pm 3xy) = 5.\n$$\nIf $g(x^2 + y^2 + 3xy) = 5$, then $x^2 + y^2 + 3xy \\le 5$, implying that $x = y = g = 1$ and $(m, n) = (1, 1)$.\nOtherwise, $g(x^2 + y^2 - 3xy) = 5$ and $g = 1$ or $5$. Fix $g$ equal to one of these values, so that\n$$\nx^2 - 3xy + y^2 = \\frac{5}{g}. \\qquad (1)\n$$\nLet $\\{a, b\\}$ denote an unordered pair of numbers. We call $\\{a, b\\}$ a *g-pair* if $\\{x, y\\} = \\{a, b\\}$ satisfies (1) and *a* and *b* are positive integers. Also, we call $\\{p, q\\}$ *smaller* (respectively, *larger*) than $\\{r, s\\}$ if $p + q$ is smaller (respectively, larger) than $r + s$.\nSuppose that $\\{a, b\\}$ is a *g-pair*. View (1) as a monic quadratic in $x$ with $y = b$ constant. The coefficient of $x$ in a monic quadratic function $(x - r_1)(x - r_2)$ equals $-(r_1 + r_2)$, implying that $\\{3b - a, b\\}$ should also satisfy (1). Indeed,\n$$\nb^2 - 3b(3b - a) + (3b - a)^2 = a^2 - 3ab + b^2 = \\frac{5}{g}.\n$$\nAlso, if $b > 2$, note that\n$$\na^2 - 3ab + b^2 = \\frac{5}{g} < b^2.\n$$\nIt follows that $a^2 - 3ab < 0$ and so $3b - a > 0$. Thus, if $\\{a, b\\}$ is a $g$-pair with $b > 2$, then $\\{b, 3b - a\\}$ is a $g$-pair as well. Also note that for $a' = b$ and $b' = 3b - a$, $\\{a', 3b' - a'\\} = \\{a, b\\}$.\nFurthermore, if $a \\ge b$, note that $a \\ne b$ because otherwise $-a^2 = 5/g > 0$, which is impossible. Thus, $a > b$ and\n$$\na^2 - 3ab + b^2 = \\frac{5}{g} > b^2 - a^2,\n$$\nwhich implies that $a(2a - 3b) > 0$ and hence $a + b > b + (3b - a)$ and also $3b - a > b$. Thus, $\\{a', b'\\}$ is a smaller $g$-pair than $(a, b)$ with $a' \\ge b'$.\nGiven any $g$-pair $\\{a, b\\}$ with $b \\le a$, if $b \\le 2$ then $a$ must equal $r(g)$, where $r(5) = 3$ if $r(1) = 4$. Otherwise, according to the above observation we can repeatedly reduce it to a smaller $g$-pair until $\\min(a, b) \\le 2$ — that is, to the $g$-pair $(r(g), 1)$.\nBeginning with $\\{r(g), 1\\}$, we reverse the reducing process so that $\\{x, y\\}$ is replaced by the larger $g$-pair $\\{3x - y, x\\}$. Moreover, this must generate all $g$-pairs since all $g$-pairs can be reduced to $\\{r(g), 1\\}$. We may express these possible pairs in terms of the Fibonacci and Lucas numbers; for $g = 1$, observe that $L_2 = 1$, $L_3 = 4 = r(1)$, and that\n$$\n\\begin{aligned}\nL_{2k+3} &= L_{2k+2} + L_{2k+1} = (L_{2k+1} + L_{2k}) + L_{2k+1} \\\\\n&= (L_{2k+1} + (L_{2k+1} - L_{2k-1})) + L_{2k+1} \\\\\n&= 3L_{2k+1} - L_{2k-1}\n\\end{aligned}\n$$\nfor $k \\ge 0$. For $g = 5$, the Fibonacci numbers satisfy an analogous recursive relation, and $F_2 = 1$, $F_4 = 3 = r(5)$. Therefore, $\\{m, n\\} = \\{L_{2k-1}^2, L_{2k+1}^2\\}$ and $\\{m, n\\} = \\{5F_{2k}^2, 5F_{2k+2}^2\\}$ for $k \\ge 0$.\n**Second Solution.** As we have seen in the first solutions, it suffices to solve the equations\n$$\nx^2 + y^2 - 3xy = 1 \\qquad (2)\n$$\nand\n$$\nx^2 + y^2 - 3xy = 5. \\qquad (3)\n$$\nThe first equation is equivalent to $(2x - 3y)^2 - 5y^2 = 4$, which is a *Pell's* equation of the form\n$$\nu^2 - dv^2 = 4. \\qquad (4)\n$$\nAll the solutions $(u, v) = (u_k, v_k)$, $k \\ge 1$, of equation (4) are given by\n$$\n\\frac{1}{2} (u_k + v_k \\sqrt{d}) = \\left( \\frac{u_1 + v_1 \\sqrt{d}}{2} \\right)^k, \\quad k \\ge 1, \\qquad (5)\n$$\nwhere $(u, v) = (u_1, v_1)$ is the solution to (4) with least positive $u$ value.\nFrom (5) it is not difficult to see that\n$$\nu_k = \\left( \\frac{u_1 + v_1 \\sqrt{d}}{2} \\right)^k + \\left( \\frac{u_1 - v_1 \\sqrt{d}}{2} \\right)^k\n$$\nand\n$$\nv_k = \\frac{1}{\\sqrt{d}} \\left[ \\left( \\frac{u_1 + v_1 \\sqrt{d}}{2} \\right)^k - \\left( \\frac{u_1 - v_1 \\sqrt{d}}{2} \\right)^k \\right]\n$$\nfor all $k \\ge 1$. In our case, $(u_1, v_1) = (3, 1)$. Hence\n$$\n\\begin{aligned}\ny_k &= v_k = \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{3+\\sqrt{5}}{2} \\right)^k - \\left( \\frac{3-\\sqrt{5}}{2} \\right)^k \\right] \\\\\n&= \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{1+\\sqrt{5}}{2} \\right)^{2k} - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^{2k} \\right] = F_{2k},\n\\end{aligned}\n$$\nwhere $\\{F_n\\}_n^\\infty$ is the Fibonacci sequence, and\n$$\n2x_k - 3y_k = u_k = \\left( \\frac{3+\\sqrt{5}}{2} \\right)^k - \\left( \\frac{3-\\sqrt{5}}{2} \\right)^k,\n$$\nimplying\n$$\n\\begin{aligned}\nx_k &= \\frac{1}{2} (u_k + 3v_k) \\\\\n&= \\frac{1}{2} \\left[ \\left( \\frac{3}{\\sqrt{5}} + 1 \\right) \\left( \\frac{3+\\sqrt{5}}{2} \\right)^k - \\left( \\frac{3}{\\sqrt{5}} - 1 \\right) \\left( \\frac{3-\\sqrt{5}}{2} \\right)^k \\right] \\\\\n&= \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{3+\\sqrt{5}}{2} \\right)^{k+1} - \\left( \\frac{3-\\sqrt{5}}{2} \\right)^{k+1} \\right] \\\\\n&= \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{1+\\sqrt{5}}{2} \\right)^{2k+2} - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^{2k+2} \\right] = F_{2k+2}.\n\\end{aligned}\n$$\nIt follows that $m_k = 5x_k^2 = 5F_{2k+2}^2$ and $n_k = 5y_k^2 = 5F_{2k}^2$, for $k \\ge 1$.\n\nEquation (3) is equivalent to $(2x-3y)^2 - 5y^2 = 20$. Setting $2x-3y = 5v$ and $y = u$, the equation reduces to $u^2 - 5v^2 = -4$. This is a Pell's equation of the form\n$$\nu^2 - dv^2 = -4, \\qquad (6)\n$$\nwhose solutions $(u, v) = (u_k, v_k)$, $k = 1, 3, 5, \\dots$, are given by\n$$\n\\frac{1}{2} (u_k + v_k \\sqrt{d}) = \\left( \\frac{u_1 + v_1 \\sqrt{d}}{2} \\right)^k, \\quad k = 1, 3, 5, \\dots,\n$$\nwhere $(u, v) = (u_1, v_1)$ is the solution to (6) with least positive $u$ value. It follows that\n$$\nu_k = \\left( \\frac{u_1 + v_1 \\sqrt{d}}{2} \\right)^k + \\left( \\frac{u_1 - v_1 \\sqrt{d}}{2} \\right)^k\n$$\nand\n$$\nv_k = \\frac{1}{\\sqrt{d}} \\left[ \\left( \\frac{u_1 + v_1 \\sqrt{d}}{2} \\right)^k - \\left( \\frac{u_1 - v_1 \\sqrt{d}}{2} \\right)^k \\right]\n$$\nfor $k = 1, 3, 5, \\dots$. In our case, $(u_1, v_1) = (1, 1)$. Hence\n$$\nu_k = \\left( \\frac{1 + \\sqrt{5}}{2} \\right)^k + \\left( \\frac{1 - \\sqrt{5}}{2} \\right)^k = L_k,\n$$\nwhere $\\{L_n\\}^\\infty_{n=1}$ is the Lucas sequence, and\n$$\nv_k = \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{1 + \\sqrt{5}}{2} \\right)^k - \\left( \\frac{1 - \\sqrt{5}}{2} \\right)^k \\right] = F_k,\n$$\nfor $k = 1, 3, 5, \\dots$. Hence $y_k = L_k$, $k = 1, 3, 5, \\dots$, and\n$$\n\\begin{align*}\nx_k &= \\frac{1}{2}(3L_k + 5F_k) \\\\\n&= \\frac{1}{2}\\left[ (3 + \\sqrt{5})\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^k + (3 - \\sqrt{5})\\left(\\frac{1 - \\sqrt{5}}{2}\\right)^k \\right] \\\\\n&= \\left(\\frac{1 + \\sqrt{5}}{2}\\right)^{k+2} + \\left(\\frac{1 - \\sqrt{5}}{2}\\right)^{k+2} \\\\\n&= L_{k+2}.\n\\end{align*}\n$$\nTherefore $m_k = 5L_{k+2}^2$, $n_k = 5L_k^2$, $k = -1, 1, 3, 5, \\dots$. Here we extended the definition of Lucas number to $L_0 = 2$ and $L_{-1} = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46630,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExpress $\\frac{\\sin 10+\\sin 20+\\sin 30+\\sin 40+\\sin 50+\\sin 60+\\sin 70+\\sin 80}{\\cos 5 \\cos 10 \\cos 20}$ without using trigonometric functions.",
    "options": [],
    "answer": "4√2",
    "solution": "Solution:\n\nWe will use the identities $\\cos a + \\cos b = 2 \\cos \\frac{a+b}{2} \\cos \\frac{a-b}{2}$ and $\\sin a + \\sin b = 2 \\sin \\frac{a+b}{2} \\cos \\frac{a-b}{2}$.\n\nThe numerator is\n$$(\\sin 10 + \\sin 80) + (\\sin 20 + \\sin 70) + (\\sin 30 + \\sin 60) + (\\sin 40 + \\sin 50)$$\n\n$= 2 \\sin 45 (\\cos 35 + \\cos 25 + \\cos 15 + \\cos 5)$\n\n$= 2 \\sin 45 ((\\cos 35 + \\cos 5) + (\\cos 25 + \\cos 15))$\n\n$= 4 \\sin 45 \\cos 20 (\\cos 15 + \\cos 5)$\n\n$= 8 \\sin 45 \\cos 20 \\cos 10 \\cos 5$\n\nSo the fraction equals $8 \\sin 45 = 4 \\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate $\\sum_{n=0}^{\\infty} \\cot^{-1}\\left(n^{2}+n+1\\right)$.",
    "options": [],
    "answer": "pi/2",
    "solution": "Solution:\n\n$\\sum_{n=0}^{\\infty} \\cot^{-1} \\left(n^{2}+n+1\\right) = \\sum_{n=0}^{\\infty} \\arctan \\left(\\frac{1}{n^{2}+n+1}\\right) = \\sum_{n=0}^{\\infty} \\arctan (n+1) - \\arctan (n)$ by the sum/difference formula for tangent. This sum, taken out to $n=N$, telescopes to $-\\arctan (0) + \\arctan (N+1)$. So as $N$ goes to infinity, the sum goes to $\\pi / 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many integers between $1$ and $2000$ inclusive share no common factors with $2001$?",
    "options": [],
    "answer": "1232",
    "solution": "Solution:\nTwo integers are said to be relatively prime if they share no common factors, that is if there is no integer greater than $1$ that divides evenly into both of them. Note that $1$ is relatively prime to all integers. Let $\\varphi(n)$ be the number of integers less than $n$ that are relatively prime to $n$. Since $\\varphi(m n)=\\varphi(m) \\varphi(n)$ for $m$ and $n$ relatively prime, we have $\\varphi(2001)=\\varphi(3 \\cdot 23 \\cdot 29)=(3-1)(23-1)(29-1)=1232$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of positive integer solutions to $n^{x} + n^{y} = n^{z}$ with $n^{z} < 2001$.",
    "options": [],
    "answer": "10",
    "solution": "Solution:\nIf $n = 1$, the relation can not hold, so assume otherwise. If $x > y$, the left hand side factors as $n^{y}(n^{x-y} + 1)$ so $n^{x-y} + 1$ is a power of $n$. But it leaves a remainder of 1 when divided by $n$ and is greater than 1, a contradiction. We reach a similar contradiction if $y > x$. So $y = x$ and $2 n^{x} = n^{z}$, so 2 is a power of $n$ and $n = 2$. So all solutions are of the form $2^{x} + 2^{x} = 2^{x+1}$, which holds for all $x$. $2^{x+1} < 2001$ implies $x < 11$, so there are 10 solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA point on a circle inscribed in a square is $1$ and $2$ units from the two closest sides of the square. Find the area of the square.",
    "options": [],
    "answer": "100",
    "solution": "Solution:\n\nCall the point in question $A$, the center of the circle $O$, and its radius $r$. Consider a right triangle $BOA$ with hypotenuse $OA$: $OA$ has length $r$, and $BO$ and $BA$ have lengths $r-1$ and $r-2$. By the Pythagorean theorem,\n$$(r-1)^2 + (r-2)^2 = r^2$$\nwhich gives\n$$r^2 - 6r + 5 = 0$$\nso $r = 5$ since $r > 4$. The area of the square is $(2r)^2 = 100$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoint $D$ is drawn on side $B C$ of equilateral triangle $A B C$, and $A D$ is extended past $D$ to $E$ such that angles $E A C$ and $E B C$ are equal. If $B E = 5$ and $C E = 12$, determine the length of $A E$.",
    "options": [],
    "answer": "17",
    "solution": "Solution:\n\nBy construction, $A B E C$ is a cyclic quadrilateral. Ptolemy's theorem says that for cyclic quadrilaterals, the sum of the products of the lengths of the opposite sides equals the product of the lengths of the diagonals. This yields $$(B C)(A E) = (B A)(C E) + (B E)(A C).$$ Since $A B C$ is equilateral, $B C = A C = A B$, so dividing out by this common value we get $$A E = C E + B E = 17.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a finite set of positive integers. Prove that there exists a finite set $B$ of positive integers such that $A \\subseteq B$ and\n$$\n\\prod_{x \\in B} x = \\sum_{x \\in B} x^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any finite set $S$ of positive integers, let\n$$\nD(S) = \\prod_{x \\in S} x - \\sum_{x \\in S} x^2.\n$$\nIf $D(A) = 0$, then we take $B = A$.\n\nIf $D(A) < 0$, then let $m = \\max A$. Write $A'_k = A \\cup \\{m+1, m+2, \\dots, m+k\\}$. Then there is a positive integer $k$ such that\n$$\n\\begin{aligned}\n-D(A) &< (m+1)^k - (k^3 + 2mk^2 + m^2k) \\\\\n&= (m+1)^k - k(m+k)^2 \\le D(A'_k) - D(A)\n\\end{aligned}\n$$\nand hence $D(A'_k) > 0$. Thus, it suffices to find a finite set $B$ containing $A'_k$ such that $D(B) = 0$, because then $B$ contains $A$ as well. This reduces the problem to the next and final case.\n\nAssume that $D(A) > 0$, and write $A_0 = A$. Define $A_{k+1} = A_k \\cup \\{\\prod_{x \\in A_k} x - 1\\}$ recursively for $k = 0, 1, \\dots, D(A) - 1$. If $D(A_k) > 0$, we have $A_k \\ne \\{1\\}$ and hence\n$$\n\\max A_k < \\sum_{x \\in A_k} x^2 = \\prod_{x \\in A_k} x - D(A_k) < \\prod_{x \\in A_k} x.\n$$\nTherefore, $\\prod_{x \\in A_k} x - 1 > \\max A_k$ and $A_{k+1}$ has one more element than $A_k$. It follows that\n$$\n\\begin{align*}\nD(A_{k+1}) &= \\prod_{x \\in A_{k+1}} x - \\sum_{x \\in A_{k+1}} x^2 \\\\\n&= \\prod_{x \\in A_k} x (\\prod_{x \\in A_k} x - 1) - \\sum_{x \\in A_k} x^2 - (\\prod_{x \\in A_k} x - 1)^2 \\\\\n&= \\prod_{x \\in A_k} x - \\sum_{x \\in A_k} x^2 - 1 \\\\\n&= D(A_k) - 1.\n\\end{align*}\n$$\nBecause $D(A_0) > 0$, it follows that $D(A_k) = D(A) - k > 0$ for $k < D(A)$ and that $D(A_{D(A)}) = 0$. Taking $B = A_{D(A)}$ completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nParallelogram $A E C F$ is inscribed in square $A B C D$. It is reflected across diagonal $A C$ to form another parallelogram $A E^{\\prime} C F^{\\prime}$. The region common to both parallelograms has area $m$ and perimeter $n$. Compute the value of $\\frac{m}{n^{2}}$ if $A F: A D=1: 4$.",
    "options": [],
    "answer": "7/400",
    "solution": "Solution:\n\nBy symmetry, the region is a rhombus, $AXCY$, centered at the center of the square, $O$. Consider isoceles right triangle $ACD$. By the technique of mass points, we find that $DO: YO=7: 1$. Therefore, the rhombus is composed of four triangles, whose sides are in the ratio $1: 7: 5 \\sqrt{2}$. The perimeter of the rhombus is $20 \\sqrt{2} N$, and the area is $14 N^{2}$. The required ratio is thus $\\frac{7}{400}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo circles are concentric. The area of the ring between them is $A$. In terms of $A$, find the length of the longest chord contained entirely within the ring.",
    "options": [],
    "answer": "2 * sqrt(A / pi)",
    "solution": "Solution:\n\nLet the radii of the circles be $r$ and $R > r$, so $A = \\pi (R^2 - r^2)$. By the Pythagorean theorem, the length of the chord is $2 \\sqrt{R^2 - r^2} = 2 \\sqrt{\\frac{A}{\\pi}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the real solutions of $(2x+1)(3x+1)(5x+1)(30x+1) = 10$.",
    "options": [],
    "answer": "(-4 + sqrt(31))/15 and (-4 - sqrt(31))/15",
    "solution": "Solution:\n$(2x+1)(3x+1)(5x+1)(30x+1) = [(2x+1)(30x+1)][(3x+1)(5x+1)] = (60x^2 + 32x + 1)(15x^2 + 8x + 1) = (4y+1)(y+1) = 10$, where $y = 15x^2 + 8x$.\n\nThe quadratic equation in $y$ yields $y = 1$ and $y = -\\frac{9}{4}$.\n\nFor $y = 1$, we have $15x^2 + 8x - 1 = 0$, so $x = \\frac{-4 \\pm \\sqrt{31}}{15}$.\n\nFor $y = -\\frac{9}{4}$, we have $15x^2 + 8x + \\frac{9}{4} = 0$, which yields only complex solutions for $x$.\n\nThus the real solutions are $\\frac{-4 \\pm \\sqrt{31}}{15}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute $\\sum_{i=1}^{\\infty} \\frac{a i}{a^{i}}$ for $a>1$.",
    "options": [],
    "answer": "a^2/(a-1)^2",
    "solution": "Solution:\n\nThe sum $S = a + a x + a x^{2} + a x^{3} + \\cdots$ for $x < 1$ can be determined by realizing that $x S = a x + a x^{2} + a x^{3} + \\cdots$ and $(1-x) S = a$, so $S = \\frac{a}{1-x}$.\n\nUsing this, we have\n$$\n\\sum_{i=1}^{\\infty} \\frac{a i}{a^{i}} = a \\sum_{i=1}^{\\infty} \\frac{i}{a^{i}} = a\\left[\\frac{1}{a} + \\frac{2}{a^{2}} + \\frac{3}{a^{3}} + \\cdots\\right]\n$$\n$$\na\\left[\\left(\\frac{1}{a} + \\frac{1}{a^{2}} + \\frac{1}{a^{3}} + \\cdots\\right) + \\left(\\frac{1}{a^{2}} + \\frac{1}{a^{3}} + \\frac{1}{a^{4}} + \\cdots\\right) + \\cdots\\right]\n$$\n$$\na\\left[\\frac{1}{1-a} + \\frac{1}{a} \\frac{1}{1-a} + \\frac{1}{a^{2}} \\frac{1}{1-a} + \\cdots\\right] = \\frac{a}{1-a}\\left[1 + \\frac{1}{a} + \\frac{1}{a^{2}} + \\cdots\\right] = \\left(\\frac{a}{1-a}\\right)^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBoris was given a Connect Four game set for his birthday, but his color-blindness makes it hard to play the game. Still, he enjoys the shapes he can make by dropping checkers into the set. If the number of shapes possible modulo (horizontal) flips about the vertical axis of symmetry is expressed as $9(1+2+\\cdots+n)$, find $n$.\n\n(Note: the board is a vertical grid with seven columns and eight rows. A checker is placed into the grid by dropping it from the top of a column, and it falls until it hits either the bottom of the grid or another checker already in that column. Also, $9(1+2+\\cdots+n)$ is the number of shapes possible, with two shapes that are horizontal flips of each other counted as one. In other words, the shape that consists solely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers in the leftmost row are to be considered the same shape.)",
    "options": [],
    "answer": "729",
    "solution": "Solution:\n\nThere are $9^{7}$ total shapes possible, since each of the 7 columns can contain anywhere from 0 to 8 checkers. The number of shapes symmetric with respect to a horizontal flip is the number of shapes of the leftmost four columns, since the configuration of these four columns uniquely determines the configuration of the remaining columns if it is known the shape is symmetric: $9^{4}$. Now we know there are $9^{7} - 9^{4}$ non-symmetric shapes, so there are $\\frac{9^{7} - 9^{4}}{2}$ non-symmetric shapes modulo flips. Thus the total number of shapes modulo flips is $9^{4} + \\frac{9^{7} - 9^{4}}{2} = 9^{4}\\left(1 + \\frac{9^{3} - 1}{2}\\right) = 9^{4}\\left(\\frac{9^{3} + 1}{2}\\right) = \\frac{3^{8}(3^{6} - 1)}{2} = 9 \\cdot \\frac{3^{6}(3^{6} + 1)}{2} = 9(1 + 2 + \\cdots + 3^{6})$, so $n = 3^{6} = 729$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 46642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn the planet Lemniscate, the people use the elliptic table of elements, a far more advanced version of our periodic table. They're not very good at calculus, though, so they've asked for your help. They know that Kr is somewhat radioactive and deteriorates into Pl, a very unstable element that deteriorates to form the stable element As. They started with a block of Kr of size 10 and nothing else. (Their units don't translate into English, sorry.) At time $t$, they let $x(t)$ be the amount of Kr, $y(t)$ the amount of Pl, and $z(t)$ the amount of As. They know that $x'(t) = -x$, and that, in the absence of Kr, $y'(t) = -2y$. Your job is to find at what time $t$ the quantity of Pl will be largest. You should assume that the entire amount of Kr that deteriorates has turned into Pl.",
    "options": [],
    "answer": "ln 2",
    "solution": "Solution:\n\nThis problem is long-winded since it's giving an autonomous linear system of differential equations without using any such language (and it includes a number of subtle references). The system we have is $x' = -x$, $y' = x - 2y$. It's not hard to see that $x = 10 e^{-t}$ satisfies the first equation and the initial condition. Plugging this into the second equation and using the integrating factor $e^{2t}$ (or using eigenvalues and eigenvectors to solve the system directly, though I don't want to begin to explain what that means) lets us solve for $y$. More precisely, we want to solve $y' + 2y = 10 e^{-t}$. Multiply by $e^{2t}$ and simplify the left hand side to get $(y e^{2t})' = 10 e^{t}$. Integrating both sides with respect to $t$ then yields $y e^{2t} = 10 e^{t} + C$, or $y = 10 e^{-t} + C e^{-2t}$. Since $y(0) = 0$, we find $C = -10$. Now to maximize $y$, we solve $y'(t) = 0$, or $-10 e^{-t} + 20 e^{-2t} = 0$, or $t = \\ln 2$.",
    "topic": "Geometry",
    "subtopic": "Non-Euclidean Geometry"
  },
  {
    "id": 46643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of positive integer solutions to $n^{x} + n^{y} = n^{z}$ with $n^{z} < 2001$.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\nIf $n = 1$, the relation cannot hold, so assume otherwise. If $x > y$, the left hand side factors as $n^{y} (n^{x-y} + 1)$ so $n^{x-y} + 1$ is a power of $n$. But it leaves a remainder of $1$ when divided by $n$ and is greater than $1$, a contradiction. We reach a similar contradiction if $y > x$. So $y = x$ and $2 n^{x} = n^{z}$, so $2$ is a power of $n$ and $n = 2$. So all solutions are of the form $2^{x} + 2^{x} = 2^{x+1}$, which holds for all $x$. $2^{x+1} < 2001$ implies $x < 11$, so there are $10$ solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46644,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo circles have centers that are $d$ units apart, and each has diameter $\\sqrt{d}$. For any $d$, let $A(d)$ be the area of the smallest circle that contains both of these circles. Find $\\lim_{d \\rightarrow \\infty} \\frac{A(d)}{d^{2}}$.",
    "options": [],
    "answer": "π/4",
    "solution": "Solution:\n\nThis equals $\\lim_{d \\rightarrow \\infty} \\frac{\\pi\\left(\\frac{d+\\sqrt{d}}{2}\\right)^{2}}{d^{2}} = \\frac{\\pi}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn $(l, a)$-design of a set is a collection of subsets of that set such that each subset contains exactly $l$ elements and that no two of the subsets share more than $a$ elements. How many $(2,1)$-designs are there of a set containing 8 elements?",
    "options": [],
    "answer": "2^28",
    "solution": "Solution:\n\nThere are $\\binom{8}{2} = 28$ 2-element subsets. Any two distinct such subsets have at most 1 common element; hence, for each subset, we can decide independently whether or not it belongs to the design, and we thus obtain $2^{28}$ designs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A B C$ be a right triangle with right angle at $B$. Let $A C D E$ be a square drawn exterior to triangle $A B C$. If $M$ is the center of this square, find the measure of $\\angle M B C$.",
    "options": [],
    "answer": "45°",
    "solution": "Solution:\n\n![](attached_image_1.png)\nNote that triangle $M C A$ is a right isosceles triangle with $\\angle A C M=90^{\\circ}$ and $\\angle M A C=45^{\\circ}$. Since $\\angle A B C=90^{\\circ}$, there is a circle $k$ with diameter $A C$ which also passes through points $B$ and $C$. As inscribed angles, $\\angle M A C=\\angle A B C$, thus the measure of $\\angle M B C=45^{\\circ}$.\n\n\nSolution 2: Place 3 copies of triangle $A B C$ on the square as shown below.\n![](attached_image_2.png)\nClearly the new diagram is a large square (which can be proven easily by looking at the angles of the copies of the triangle). The diagonals of this large square meet at $M$. By symmetry, $\\angle M B C=45^{\\circ}$.\n\n\nSolution 3: Place triangle $A B C$ on coordinate axes so that $A=(0,2 c), B=(0,0), C=(2 a, 0)$. Let $N$ be the midpoint of hypotenuse $A C$ and draw line $\\ell$ through $N$ parallel to $B C$ which meets $A B$ at $F$. Drop a perpendicular from $M$ to $B C$ which meets $\\ell$ at $G$. It is evident that triangles $A F N$ and $N G M$ are congruent.\nThus $N G=A F=c$ and $M G=F N=a$. Consequently, the coordinates of $M$ are $(a, c)+(c, a)=(a+c, a+c)$, so the line from $M$ to the origin (at $B$) has a slope of 1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA man, standing on a lawn, is wearing a circular sombrero of radius $3$ feet. Unfortunately, the hat blocks the sunlight so effectively that the grass directly under it dies instantly. If the man walks in a circle of radius $5$ feet, what area of dead grass will result?",
    "options": [],
    "answer": "60π ft^2",
    "solution": "Solution:\n$60\\pi\\ \\mathrm{ft}^2$\n\nLet $O$ be the center of the man's circular trajectory. The sombrero kills all the grass that is within $3$ feet of any point that is $5$ feet away from $O$—i.e., all the grass at points $P$ with $2 \\leq OP \\leq 8$. The area of this annulus is then $\\pi\\left(8^2 - 2^2\\right) = 60\\pi$ square feet.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the number of palindromes that are less than $1000$.",
    "options": [],
    "answer": "108",
    "solution": "Solution:\n\nEvery one-digit number (there are nine) is a palindrome.\n\nThe two-digit palindromes have the form $\\underline{a}\\,\\underline{a}$ for a nonzero digit $a$, so there are nine of them.\n\nA three-digit palindrome is $\\underline{a}\\,\\underline{b}\\,\\underline{a}$ with $a$ a nonzero digit and $b$ any digit, so there are $9 \\times 10 = 90$ of these.\n\nThus the number of palindromes less than $1000$ is $9 + 9 + 90 = \\mathbf{108}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNine nonnegative numbers have average $10$. What is the greatest possible value for their median?",
    "options": [],
    "answer": "18",
    "solution": "Solution:\n\nIf the median is $m$, then the five highest numbers are all at least $m$, so the sum of all the numbers is at least $5m$. Thus $90 \\geq 5m \\Rightarrow m \\leq 18$. Conversely, we can achieve $m=18$ by taking four $0$'s and five $18$'s.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the maximum number of lattice points (i.e. points with integer coordinates) in the plane that can be contained strictly inside a circle of radius $1$?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\n$4$. The circle centered at $(1/2, 1/2)$ shows that $4$ is achievable. On the other hand, no two points within the circle can be at a mutual distance of $2$ or greater. If there are more than four lattice points, classify all such points by the parity of their coordinates: (even, even), (even, odd), (odd, even), or (odd, odd). Then some two points lie in the same class. Since they are distinct, this means either their first or second coordinates must differ by at least $2$, so their distance is at least $2$, a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe squares of a chessboard are numbered from left to right and top to bottom (so that the first row reads $1, 2, \\ldots, 8$, the second reads $9, 10, \\ldots, 16$, and so forth). The number $1$ is on a black square. How many black squares contain odd numbers?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\n$16$. The black squares in the $n$th row contain odd numbers when $n$ is odd and even numbers when $n$ is even; thus there are four rows where the black squares contain odd numbers, and each such row contributes four black squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEight knights are randomly placed on a chessboard (not necessarily on distinct squares). A knight on a given square attacks all the squares that can be reached by moving either (1) two squares up or down followed by one square left or right, or (2) two squares left or right followed by one square up or down. Find the probability that every square, occupied or not, is attacked by some knight.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\n$0$. Since every knight attacks at most eight squares, the event can only occur if every knight attacks exactly eight squares. However, each corner square must be attacked, and some experimentation readily finds that it is impossible to place a knight so as to attack a corner and seven other squares as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches, and tomato and cheese sandwiches. It is common for one meal to include multiple types of sandwiches. On a certain day, it was found that 80 customers had meals which contained both ham and cheese; 90 had meals containing both ham and tomatoes; 100 had meals containing both tomatoes and cheese. 20 customers' meals included all three ingredients. How many customers were there?",
    "options": [],
    "answer": "230",
    "solution": "Solution: 230. Everyone who ate just one sandwich is included in exactly one of the first three counts, while everyone who ate more than one sandwich is included in all four counts. Thus, to count each customer exactly once, we must add the first three figures and subtract the fourth twice: $80+90+100-2 \\cdot 20=230$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $\\triangle ABC$ be equilateral, and let $D, E, F$ be points on sides $BC, CA, AB$ respectively, with $FA=9$, $AE=EC=6$, $CD=4$. Determine the measure (in degrees) of $\\angle DEF$.",
    "options": [],
    "answer": "60",
    "solution": "Solution:\n\n$H, I$ be the respective midpoints of sides $BC, AB$, and also extend $CB$ and $EF$ to intersect at $J$. By equal angles, $\\triangle EIF \\sim \\triangle JBF$. However, $BF=12-9=3=9-6=IF$, so in fact $\\triangle EIF \\cong \\triangle JBF$, and then $JB=6$. Now let $HI$ intersect $EF$ at $K$, and notice that $\\triangle EIK \\sim \\triangle JHK \\Rightarrow IK/HK=EI/JH=6/12=1/2 \\Rightarrow HK=4$, since $IK+HK=HI=6$. Now consider the $60^{\\circ}$ rotation about $E$ carrying triangle $CHE$ to triangle $HIE$; we see that it also takes $D$ to $K$, and thus $\\angle DEF=\\angle DEK=60^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOne fair die is rolled; let $a$ denote the number that comes up. We then roll $a$ dice; let the sum of the resulting $a$ numbers be $b$. Finally, we roll $b$ dice, and let $c$ be the sum of the resulting $b$ numbers. Find the expected (average) value of $c$.",
    "options": [],
    "answer": "343/8",
    "solution": "Solution:\n\n$343 / 8$\n\nThe expected result of an individual die roll is $(1+2+3+4+5+6)/6 = 7/2$. For any particular value of $b$, if $b$ dice are rolled independently, then the expected sum is $(7/2) b$. Likewise, when we roll $a$ dice, the expected value of their sum $b$ is $(7/2) a$, so the expected value of $c$ is $(7/2)^2 a$. Similar reasoning again shows us that the expected value of $a$ is $7/2$ and so the expected value of $c$ overall is $(7/2)^3 = 343/8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA continuous real function $f$ satisfies the identity $f(2x) = 3f(x)$ for all $x$. If $\\int_{0}^{1} f(x) dx = 1$, what is $\\int_{1}^{2} f(x) dx$?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nLet $S = \\int_{1}^{2} f(x) dx$. By setting $u = 2x$, we see that\n$$\n\\int_{1/2}^{1} f(x) dx = \\int_{1/2}^{1} \\frac{f(2x)}{3} dx = \\int_{1}^{2} \\frac{f(u)}{6} du = S/6.\n$$\nSimilarly, $\\int_{1/4}^{1/2} f(x) dx = S/36$, and in general\n$$\n\\int_{1/2^{n}}^{1/2^{n-1}} f(x) dx = S/6^{n}.\n$$\nAdding finitely many of these, we have\n$$\n\\int_{1/2^{n}}^{1} f(x) dx = S/6 + S/36 + \\cdots + S/6^{n} = S \\cdot \\frac{1 - 1/6^{n}}{5}.\n$$\nTaking the limit as $n \\to \\infty$, we have\n$$\n\\int_{0}^{1} f(x) dx = S/5.\n$$\nThus $S = 5$, the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe mathematician John is having trouble remembering his girlfriend Alicia's 7-digit phone number. He remembers that the first four digits consist of one $1$, one $2$, and two $3$'s. He also remembers that the fifth digit is either a $4$ or $5$. While he has no memory of the sixth digit, he remembers that the seventh digit is $9$ minus the sixth digit. If this is all the information he has, how many phone numbers does he have to try if he is to make sure he dials the correct number?",
    "options": [],
    "answer": "240",
    "solution": "Solution:\n\nThere are $\\frac{4!}{2!} = 12$ possibilities for the first four digits. There are two possibilities for the fifth digit. There are $10$ possibilities for the sixth digit, and this uniquely determines the seventh digit. So he has to dial $12 \\cdot 2 \\cdot 10 = 240$ numbers.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the value of\n\n$$\n2002+\\frac{1}{2}\\left(2001+\\frac{1}{2}\\left(2000+\\cdots+\\frac{1}{2}\\left(3+\\frac{1}{2} \\cdot 2\\right)\\right) \\cdots\\right) .\n$$",
    "options": [],
    "answer": "4002",
    "solution": "Solution:\n4002. We can show by induction that\n$$\nn+\\frac{1}{2}\\left([n-1]+\\frac{1}{2}\\left(\\cdots+\\frac{1}{2} \\cdot 2\\right) \\cdots\\right)=2(n-1).\n$$\nFor $n=3$ we have $3+\\frac{1}{2} \\cdot 2=4$, giving the base case, and if the result holds for $n$, then\n$$(n+1)+\\frac{1}{2} 2(n-1)=2 n=2(n+1)-2.$$\nThus the claim holds, and now plug in $n=2002$.\n\nAlternate Solution:\nExpand the given expression as\n$$\n2002+\\frac{2001}{2}+\\frac{2000}{2^{2}}+\\cdots+\\frac{2}{2^{2000}}.\n$$\nLetting $S$ denote this sum, we have\n$$\n\\frac{S}{2}=\\frac{2002}{2}+\\frac{2001}{2^{2}}+\\cdots+\\frac{2}{2^{2001}},\n$$\nso\n$$\nS-\\frac{S}{2}=2002-\\left(\\frac{1}{2}+\\frac{1}{4}+\\cdots+\\frac{1}{2^{2000}}\\right)-\\frac{2}{2^{2001}}=2002-\\left(1-\\frac{1}{2^{2000}}\\right)-\\frac{1}{2^{2000}}=2001,\n$$\nso $S=4002$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many real solutions are there to the equation\n$$\n||||x|-2|-2|-2|=||||x|-3|-3|-3| ?\n$$",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n6. The graphs of the two sides of the equation can be graphed on the same plot to reveal six intersection points.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThis question forms a three question multiple choice test. After each question, there are 4 choices, each preceded by a letter. Please write down your answer as the ordered triple (letter of the answer of Question $\\# 1$, letter of the answer of Question $\\# 2$, letter of the answer of Question \\#3). If you find that all such ordered triples are logically impossible, then write \"no answer\" as your answer. If you find more than one possible set of answers, then provide all ordered triples as your answer.\nWhen we refer to \"the correct answer to Question $X$ \" it is the actual answer, not the letter, to which we refer. When we refer to \"the letter of the correct answer to question $X$ \" it is the letter contained in parentheses that precedes the answer to which we refer.\nYou are given the following condition: No two correct answers to questions on the test may have the same letter.\n\nQuestion 1. If a fourth question were added to this test, and if the letter of its correct answer were (C), then:\n(A) This test would have no logically possible set of answers.\n(B) This test would have one logically possible set of answers.\n(C) This test would have more than one logically possible set of answers.\n(D) This test would have more than one logically possible set of answers.\n\nQuestion 2. If the answer to Question 2 were \"Letter (D)\" and if Question 1 were not on this multiple-choice test (still keeping Questions 2 and 3 on the test), then the letter of the answer to Question 3 would be:\n(A) Letter (B)\n(B) Letter (C)\n(C) Letter (D)\n(D) Letter (A)\n\nQuestion 3. Let $P_{1}=1$. Let $P_{2}=3$. For all $i>2$, define $P_{i}=P_{i-1} P_{i-2}-P_{i-2}$. Which is a factor of $P_{2002}$ ?\n(A) 3\n(B) 4\n(C) 7\n(D) 9",
    "options": [],
    "answer": "(A, C, D)",
    "solution": "Solution:\n\n(A, C, D).\n\nQuestion 2: In order for the answer to be consistent with the condition, \"If the answer to Question 2 were Letter (D),\" the answer to this question actually must be \"Letter (D).\" The letter of this answer is (C).\n\nQuestion 1: If a fourth question had an answer with letter (C), then at least two answers would have letter (C) (the answers to Questions 2 and 4). This is impossible. So, (A) must be the letter of the answer to Question 1.\n\nQuestion 3: If we inspect the sequence $P_{i}$ modulo $3,4,7$, and $9$ (the sequences quickly become periodic), we find that $3,7$, and $9$ are each factors of $P_{2002}$. We know that letters (A) and (C) cannot be repeated, so the letter of this answer must be (D).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA domino is a 1-by-2 or 2-by-1 rectangle. A domino tiling of a region of the plane is a way of covering it (and only it) completely by nonoverlapping dominoes. For instance, there is one domino tiling of a 2-by-1 rectangle and there are 2 tilings of a 2-by-2 rectangle (one consisting of two horizontal dominoes and one consisting of two vertical dominoes). How many domino tilings are there of a 2-by-10 rectangle?",
    "options": [],
    "answer": "89",
    "solution": "Solution:\nThe number of tilings of a 2-by-$n$ rectangle is the $n$th Fibonacci number $F_{n}$, where $F_{0} = F_{1} = 1$ and $F_{n} = F_{n-1} + F_{n-2}$ for $n \\geq 2$. (This is not hard to show by induction.) The answer is $89$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn omino is a 1-by-1 square or a 1-by-2 horizontal rectangle. An omino tiling of a region of the plane is a way of covering it (and only it) by ominoes. How many omino tilings are there of a 2-by-10 horizontal rectangle?",
    "options": [],
    "answer": "7921",
    "solution": "Solution:\n\nThere are exactly as many omino tilings of a 1-by-$n$ rectangle as there are domino tilings of a 2-by-$n$ rectangle. Since the rows don't interact at all, the number of omino tilings of an $m$-by-$n$ rectangle is the number of omino tilings of a 1-by-$n$ rectangle raised to the $m$th power, $F_{n}^{m}$. The answer is $89^{2} = 7921$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many sequences of $0$s and $1$s are there of length $10$ such that there are no three $0$s or $1$s consecutively anywhere in the sequence?",
    "options": [],
    "answer": "178",
    "solution": "Solution:\nWe can have blocks of either $1$ or $2$ $0$s and $1$s, and these blocks must be alternating between $0$s and $1$s. The number of ways of arranging blocks to form a sequence of length $n$ is the same as the number of omino tilings of a $1$-by-$n$ rectangle, and we may start each sequence with a $0$ or a $1$, making $2 F_{n}$ or, in this case, $178$ sequences.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDivide an $m$-by-$n$ rectangle into $mn$ nonoverlapping $1$-by-$1$ squares. A polyomino of this rectangle is a subset of these unit squares such that for any two unit squares $S, T$ in the polyomino, either\n\n(1) $S$ and $T$ share an edge or\n\n(2) there exists a positive integer $n$ such that the polyomino contains unit squares $S_{1}, S_{2}, S_{3}, \\ldots, S_{n}$ such that $S$ and $S_{1}$ share an edge, $S_{n}$ and $T$ share an edge, and for all positive integers $k < n$, $S_{k}$ and $S_{k+1}$ share an edge.\n\nWe say a polyomino of a given rectangle spans the rectangle if for each of the four edges of the rectangle the polyomino contains a square whose edge lies on it.\n\nWhat is the minimum number of unit squares a polyomino can have if it spans a $128$-by-$343$ rectangle?",
    "options": [],
    "answer": "470",
    "solution": "Solution:\n\nTo span an $a \\times b$ rectangle, we need at least $a + b - 1$ squares. Indeed, consider a square of the polyomino bordering the left edge of the rectangle and one bordering the right edge. There exists a path connecting these squares; suppose it runs through $c$ different rows. Then the path requires at least $b-1$ horizontal and $c-1$ vertical steps, so it uses at least $b + c - 1$ different squares. However, since the polyomino also hits the top and bottom edges of the rectangle, it must run into the remaining $a - c$ rows as well, so altogether we need at least $a + b - 1$ squares. On the other hand, this many squares suffice - just consider all the squares bordering the lower or right edges of the rectangle. So, in our case, the answer is $128 + 343 - 1 = 470$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of pentominoes (5-square polyominoes) that span a $3$-by-$3$ rectangle, where polyominoes that are flips or rotations of each other are considered the same polyomino.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nBy enumeration, the answer is $6$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor how many integers $a$ ($1 \\leq a \\leq 200$) is the number $a^{a}$ a square?",
    "options": [],
    "answer": "107",
    "solution": "Solution:\n\nIf $a$ is even, we have $a^{a} = \\left(a^{a / 2}\\right)^{2}$, which is always a square.\n\nIf $a$ is odd, $a^{a} = \\left(a^{(a-1) / 2}\\right)^{2} \\cdot a$, which is a square precisely when $a$ is a square.\n\nThus we have 100 even values of $a$ and 7 odd square values ($1^{2}, 3^{2}, \\ldots, 13^{2}$) for a total of 107.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$p$ and $q$ are primes such that the numbers $p+q$ and $p+7q$ are both squares. Find the value of $p$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nWriting $x^2 = p + q$, $y^2 = p + 7q$, we have $6q = y^2 - x^2 = (y - x)(y + x)$. Since $6q$ is even, one of the factors $y - x$, $y + x$ is even, and then the other is as well; thus $6q$ is divisible by $4 \\Rightarrow q$ is even $\\Rightarrow q = 2$ and $6q = 12$. We may assume $x, y$ are both taken to be positive; then we must have $y - x = 2$, $y + x = 6 \\Rightarrow x = 2$, so $p + 2 = 2^2 = 4 \\Rightarrow p = 2$ also.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEight knights are randomly placed on a chessboard (not necessarily on distinct squares). A knight on a given square attacks all the squares that can be reached by moving either (1) two squares up or down followed by one square left or right, or (2) two squares left or right followed by one square up or down. Find the probability that every square, occupied or not, is attacked by some knight.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\n$0$. Since every knight attacks at most eight squares, the event can only occur if every knight attacks exactly eight squares. However, each corner square must be attacked, and some experimentation readily finds that it is impossible to place a knight so as to attack a corner and seven other squares as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA lattice point in the plane is a point of the form $(n, m)$, where $n$ and $m$ are integers. Consider a set $S$ of lattice points. We construct the transform of $S$, denoted by $S'$, by the following rule: the pair $(n, m)$ is in $S'$ if and only if any of $(n, m-1)$, $(n, m+1)$, $(n-1, m)$, $(n+1, m)$, and $(n, m)$ is in $S$. How many elements are in the set obtained by successively transforming $\\{(0,0)\\}$ 14 times?",
    "options": [],
    "answer": "421",
    "solution": "Solution:\n\nTransforming it $k \\geq 1$ times yields the \"diamond\" of points $(n, m)$ such that $|n| + |m| \\leq k$. The diamond contains $(k+1)^2 + k^2$ lattice points (this can be seen by rotating the plane 45 degrees and noticing the lattice points in the transforms form two squares, one of which is contained in the other), so the answer is $421$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind $\\lim_{h \\rightarrow 0} \\frac{x^{2}-(x+h)^{2}}{h}$.",
    "options": [],
    "answer": "-2x",
    "solution": "Solution:\nThis equals $\\lim_{h \\rightarrow 0} \\frac{x^{2}-x^{2}-h^{2}-2 h x}{h} = \\lim_{h \\rightarrow 0} -h - 2x = -2x$.\n\nAlternate Solution: This is the definition of the derivative of $-x^{2}$ with respect to $x$, which is $-2x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou are in a completely dark room with a drawer containing 10 red, 20 blue, 30 green, and 40 khaki socks. What is the smallest number of socks you must randomly pull out in order to be sure of having at least one of each color?",
    "options": [],
    "answer": "91",
    "solution": "Solution:\n\n91. The maximum number of socks that can be pulled out without representing every color is $20$ blue $+30$ green $+40$ khaki $=90$, so $91$ is the minimum needed to ensure that this doesn't happen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the number of four-digit integers $n$ such that $n$ and $2 n$ are both palindromes.",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nLet $n = \\underline{a} \\underline{b} \\underline{b} \\underline{a}$. If $a, b \\leq 4$ then there are no carries in the multiplication $n \\times 2$, and $2 n = (2 a)(2 b)(2 b)(2 a)$ is a palindrome. We shall show conversely that if $n$ and $2 n$ are palindromes, then necessarily $a, b \\leq 4$. Hence the answer to the problem is $4 \\times 5 = \\mathbf{20}$ (because $a$ cannot be zero).\n\nIf $a \\geq 5$ then $2 n$ is a five-digit number whose most significant digit is 1, but because $2 n$ is even, its least significant digit is even, contradicting the assumption that $2 n$ is a palindrome. Therefore $a \\leq 4$. Consequently $2 n$ is a four-digit number, and its tens and hundreds digits must be equal. Because $a \\leq 4$, there is no carry out of the ones place in the multiplication $n \\times 2$, and therefore the tens digit of $2 n$ is the ones digit of $2 b$. In particular, the tens digit of $2 n$ is even. But if $b \\geq 5$, the carry out of the tens place makes the hundreds digit of $2 n$ odd, which is impossible. Hence $b \\leq 4$ as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches, and tomato and cheese sandwiches. It is common for one meal to include multiple types of sandwiches. On a certain day, it was found that 80 customers had meals which contained both ham and cheese; 90 had meals containing both ham and tomatoes; 100 had meals containing both tomatoes and cheese. 20 customers' meals included all three ingredients. How many customers were there?",
    "options": [],
    "answer": "230",
    "solution": "Solution:\n230. Everyone who ate just one sandwich is included in exactly one of the first three counts, while everyone who ate more than one sandwich is included in all four counts. Thus, to count each customer exactly once, we must add the first three figures and subtract the fourth twice: $80+90+100-2 \\cdot 20=230$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe Antarctican language has an alphabet of just 16 letters. Interestingly, every word in the language has exactly 3 letters, and it is known that no word's first letter equals any word's last letter (for instance, if the alphabet were $\\{a, b\\}$ then $a a b$ and $a a a$ could not both be words in the language because $a$ is the first letter of a word and the last letter of a word; in fact, just $a a a$ alone couldn't be in the language). Given this, determine the maximum possible number of words in the language.",
    "options": [],
    "answer": "1024",
    "solution": "Solution:\n\n1024\n\nEvery letter can be the first letter of a word, or the last letter of a word, or possibly neither, but not both. If there are $a$ different first letters and $b$ different last letters, then we can form $a \\cdot 16 \\cdot b$ different words (and the desired conditions will be met). Given the constraints $0 \\leq a, b ; a+b \\leq 16$, this product is maximized when $a=b=8$, giving the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe Dyslexian alphabet consists of consonants and vowels. It so happens that a finite sequence of letters is a word in Dyslexian precisely if it alternates between consonants and vowels (it may begin with either). There are 4800 five-letter words in Dyslexian. How many letters are in the alphabet?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nSuppose there are $c$ consonants, $v$ vowels. Then there are $c \\cdot v \\cdot c \\cdot v \\cdot c + v \\cdot c \\cdot v \\cdot c \\cdot v = (c v)^2 (c+v)$ five-letter words. Thus, $c+v = 4800 / (c v)^2 = 3 \\cdot (40 / c v)^2$, so $c v$ is a divisor of $40$. If $c v \\leq 10$, we have $c+v \\geq 48$, impossible for $c, v$ integers; if $c v = 40$, then $c+v = 3$ which is again impossible. So $c v = 20$, giving $c+v = 12$, the answer. As a check, this does have integer solutions: $(c, v) = (2, 10)$ or $(10, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA path of length $n$ is a sequence of points $(x_{1}, y_{1}),(x_{2}, y_{2}), \\ldots,(x_{n}, y_{n})$ with integer coordinates such that for all $i$ between $1$ and $n-1$ inclusive, either\n(1) $x_{i+1}=x_{i}+1$ and $y_{i+1}=y_{i}$ (in which case we say the $i$th step is rightward) or\n(2) $x_{i+1}=x_{i}$ and $y_{i+1}=y_{i}+1$ (in which case we say that the $i$th step is upward).\nThis path is said to start at $(x_{1}, y_{1})$ and end at $(x_{n}, y_{n})$. Let $P(a, b)$, for $a$ and $b$ nonnegative integers, be the number of paths that start at $(0,0)$ and end at $(a, b)$.\nFind $\\sum_{i=0}^{10} P(i, 10-i)$.",
    "options": [],
    "answer": "1024",
    "solution": "Solution:\nThis is just the number of paths of length $10$. The $i$th step can be either upward or rightward, so there are $2^{10}=1024$ such paths.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA restricted path of length $n$ is a path of length $n$ such that for all $i$ between $1$ and $n-2$ inclusive, if the $i$th step is upward, the $i+1$st step must be rightward.\nFind the number of restricted paths that start at $(0,0)$ and end at $(7,3)$.",
    "options": [],
    "answer": "56",
    "solution": "Solution:\n\nThis is equal to the number of lattice paths from $(0,0)$ to $(7,3)$ that use only rightward and diagonal (upward+rightward) steps plus the number of lattice paths from $(0,0)$ to $(7,2)$ that use only rightward and diagonal steps, which is equal to the number of paths (as defined above) from $(0,0)$ to $(4,3)$ plus the number of paths from $(0,0)$ to $(5,2)$, or $\\binom{4+3}{3} + \\binom{5+2}{2} = 56$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA math professor stands up in front of a room containing 100 very smart math students and says, \"Each of you has to write down an integer between 0 and 100, inclusive, to guess 'two-thirds of the average of all the responses.' Each student who guesses the highest integer that is not higher than two-thirds of the average of all responses will receive a prize.\" If among all the students it is common knowledge that everyone will write down the best response, and there is no communication between students, what single integer should each of the 100 students write down?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nSince the average cannot be greater than $100$, no student will write down a number greater than $\\frac{2}{3} \\cdot 100$. But then the average cannot be greater than $\\frac{2}{3} \\cdot 100$, and, realizing this, each student will write down a number no greater than $\\left(\\frac{2}{3}\\right)^{2} \\cdot 100$. Continuing in this manner, we eventually see that no student will write down an integer greater than $0$, so this is the answer.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAnother professor enters the same room and says, \"Each of you has to write down an integer between 0 and 200. I will then compute $X$, the number that is 3 greater than half the average of all the numbers that you will have written down. Each student who writes down the number closest to $X$ (either above or below $X$ ) will receive a prize.\" One student, who misunderstood the question, announces to the class that he will write the number 107. If among the other 99 students it is common knowledge that all 99 of them will write down the best response, and there is no further communication between students, what single integer should each of the 99 students write down?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nUse the same logic to get 7. Note 6 and 8 do not work.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the two hands of an analog clock, each of which moves with constant angular velocity. Certain positions of these hands are possible (e.g. the hour hand halfway between the 5 and 6 and the minute hand exactly at the 6), while others are impossible (e.g. the hour hand exactly at the 5 and the minute hand exactly at the 6). How many different positions are there that would remain possible if the hour and minute hands were switched?",
    "options": [],
    "answer": "143",
    "solution": "Solution:\n\n143 We can look at the twelve-hour cycle beginning at midnight and ending just before noon, since during this time, the clock goes through each possible position exactly once. The minute hand has twelve times the angular velocity of the hour hand, so if the hour hand has made $t$ revolutions from its initial position $(0 \\leq t < 1)$, the minute hand has made $12 t$ revolutions. If the hour hand were to have made $12 t$ revolutions, the minute hand would have made $144 t$. So we get a valid configuration by reversing the hands precisely when $144 t$ revolutions land the hour hand in the same place as $t$ revolutions - i.e. when $143 t = 144 t - t$ is an integer, which clearly occurs for exactly 143 values of $t$ corresponding to distinct positions on the clock ($144 - 1 = 143$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCount how many 8-digit numbers there are that contain exactly four nines as digits.",
    "options": [],
    "answer": "433755",
    "solution": "Solution:\n\nThere are $\\binom{8}{4} \\cdot 9^{4}$ sequences of 8 numbers with exactly four nines. A sequence of digits of length 8 is not an 8-digit number, however, if and only if the first digit is zero. There are $\\binom{7}{4} 9^{3}$-digit sequences that are not 8-digit numbers. The answer is thus $\\binom{8}{4} \\cdot 9^{4} - \\binom{7}{4} 9^{3} = 433755$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence $s_{0}, s_{1}, s_{2}, s_{3}, \\ldots$ is defined by $s_{0}=s_{1}=1$ and, for every positive integer $n$, $s_{2 n}=s_{n}$, $s_{4 n+1}=s_{2 n+1}$, $s_{4 n-1}=s_{2 n-1}+s_{2 n-1}^{2} / s_{n-1}$. What is the value of $s_{1000}$?",
    "options": [],
    "answer": "720",
    "solution": "Solution:\nSome experimentation with small values may suggest that $s_{n}=k!$, where $k$ is the number of ones in the binary representation of $n$, and this formula is in fact provable by a straightforward induction. Since $1000_{10}=1111101000_{2}$, with six ones, $s_{1000}=6!=720$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA game is played with two players and an initial stack of $n$ pennies ($n \\geq 3$). The players take turns choosing one of the stacks of pennies on the table and splitting it into two stacks. The winner is the player who makes a move that causes all stacks to be of height 1 or 2. For which starting values of $n$ does the player who goes first win, assuming best play by both players?",
    "options": [],
    "answer": "The first player wins exactly when the starting size is even or equals three; for all other odd starting sizes greater than three, the second player wins.",
    "solution": "Solution:\n\nPlayer 1 wins if and only if $n=3$ or $n$ is even; player 2 wins for all odd $n>3$. We can easily check this for the first few cases, say up to $n=6$, and then we can proceed by induction.\n\n- If $n>6$ is even, player 1 creates a pile of size 1 and a pile of size $n-1$. Since $n-1$ is odd, player 1 will win by the inductive hypothesis (since player 1 is now going second and the \"starting position\" has an odd number of pennies).\n\n- If $n \\geq 7$ is odd, player 1 will create one odd and one even pile. Player 2 can then divide the even pile into two odd piles. Continuing in this way, player 2 can always answer player 1's move and present player 1 with only odd piles. When the piles reach size 1, they are irrelevant. The critical value is size 3: the only way that player 2 can lose is if player 2 presents player 1 with a single 3-pile (and many 1-piles). But for this to happen, player 1 would have to have produced either a single 2-pile and one 3-pile, or a single 4-pile. In either case, player 2 wins on the next move by reducing, in the first case, the 3-pile to a 1- and 2-pile, and in the second case, breaking the 4-pile into two 2-piles. In sum, player 2's winning strategy is to always produce only odd piles unless this will produce all 1's and a single 3; in which case the \"terminal\" strategy above is employed.\nSolution:\n\nAt each stage of the game, let $S$ be the sum of one less than each pile height. For example, if the piles are $8,6,2,2,1$, then $S=7+5+1+1+0$. Observe that $S$ is also equal to the total number of pennies minus the number of piles, and thus, $S$ always decreases by 1 each turn. We shall analyze the parity of $S$.\n\nThe only way a player will win is if on their turn, one pile has 3 or 4 pennies, with the other piles (if there are any) of height 2 or 1. We call such a position \"penultimate.\" For example, 4, 2, 2, 2, 1 is penultimate (and the winning move will split the 4 into two piles).\n\nIf a position is not penultimate, but can be turned into a penultimate position in one move, we call it \"antepenultimate.\" For example, 4, 3, 2, 2, 2, 1 is antepenultimate (if the 4 is split). Notice that an antepenultimate position doesn't always have to become a penultimate position in one move. For example, with the position above, instead of splitting the 4, we could split one of the 2-piles.\n\nThe antepenultimate positions fall into 2 cases:\n- A 5 or 6, and the rest (if any) are $2$'s and $1$'s.\n- Two piles that are 3 or 4, and the rest (if any) are $2$'s and $1$'s.\n\nWe make two observations:\n- The number of 1's are irrelevant (since they cannot be changed, and also they do not alter the value of $S$).\n- Without loss of generality, the number of $2$'s in a pile (if there are any) is either 1 or 2. If a position had more than two 2's, they can be removed and it would not change the parity of $S$, nor would it affect who wins the game (since the only thing that can be done with a 2-pile is either to leave it alone or split it into two 1-piles).\n\nThus there are only a few cases to check, and in all of them, we discover that:\nIf a position is antepenultimate and $S$ is odd, then the player who has this position will win. If a position is antepenultimate and $S$ is even, then the player who has this position will lose.\n\nFor example, 5, 2, 1 is antepenultimate and $S=5$. The player splits the 2, and then her opponent has no choice but to split the 5, handing the penultimate position to the first player, who wins.\n\nNow we can use this analysis to look at what happens from the very start of the game.\n- If $n=3$ or $n=4$, the first player already has a penultimate position and wins.\n- If $n=5$, then the starting position is antepenultimate and $S$ is even. Clearly the second player wins.\n- If $n=6$, then the starting position is antepenultimate and $S$ is odd. Clearly the first player wins.\n- If $n>7$, then the starting position is not antepenultimate. If $S$ is odd to start, it will always be odd when it is player 1's turn (since $S$ drops by 1 each time). So at some point, player 1 will either get the penultimate position and will win, or will get the antepenultimate position with odd $S$, and will win. By similar reasoning, if $n$ is odd, $S$ starts out even, and player 1 will lose.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many four-digit numbers are there in which at least one digit occurs more than once?",
    "options": [],
    "answer": "4464",
    "solution": "Solution:\n\nThere are $9000$ four-digit numbers altogether. If we consider how many four-digit numbers have all their digits distinct, there are $9$ choices for the first digit (since we exclude leading zeroes), and then $9$ remaining choices for the second digit, then $8$ for the third, and $7$ for the fourth, for a total of $9 \\cdot 9 \\cdot 8 \\cdot 7 = 4536$. Thus the remaining $9000 - 4536 = 4464$ numbers have a repeated digit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve for $x$ in $3 = \\sqrt{x + \\sqrt{x + \\sqrt{x + \\cdots}}}$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nLet $y = \\sqrt{x + \\sqrt{x + \\sqrt{x + \\cdots}}}$. Then $y = \\sqrt{x + y}$.\n\nGiven $y = 3$, so $3 = \\sqrt{x + 3}$.\n\nSquaring both sides:\n$$\n9 = x + 3\n$$\nSubtracting $x$ from both sides:\n$$\n9 - x = 3\n$$\nSo $x = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA square and a regular hexagon are drawn with the same side length. If the area of the square is $\\sqrt{3}$, what is the area of the hexagon?",
    "options": [],
    "answer": "9/2",
    "solution": "Solution:\nThe hexagon is composed of six equilateral triangles each of side length $\\sqrt[4]{3}$ (with base $b = \\sqrt[4]{3}$ and height $\\frac{\\sqrt{3}}{2} b$), so the total area is $\\frac{9}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46687,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nReal numbers $a, b, c$ satisfy the equations $a+b+c=26,\\ 1/a+1/b+1/c=28$. Find the value of\n\n$$\n\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+\\frac{a}{c}+\\frac{c}{b}+\\frac{b}{a} .\n$$",
    "options": [],
    "answer": "725",
    "solution": "Solution:\nMultiplying the two given equations gives\n\n$$\n\\frac{a}{a}+\\frac{a}{b}+\\frac{a}{c}+\\frac{b}{a}+\\frac{b}{b}+\\frac{b}{c}+\\frac{c}{a}+\\frac{c}{b}+\\frac{c}{c}=26 \\cdot 28=728,\n$$\n\nand subtracting $3$ from both sides gives the answer, $725$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that a positive integer $n$ has the property that $n, 2 n, 3 n, \\ldots, 9 n$ are all palindromes. Prove that the decimal digits of $n$ are all zeros or ones.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst consider the ones digit $a$ of $n$; we claim that $a=1$. Certainly $a$ cannot be even, for then $5 n$ would be divisible by $10$. If $a$ is $5$, $7$, or $9$, then $2 n$ has an even ones digit, while its most significant digit is $1$. If $a$ is $3$, then $4 n$ has an even ones digit but most significant digit $1$. Thus $a=1$ is the only possibility. Moreover $9 n$ has the same number of digits as $n$, for otherwise $9 n$ would have most significant digit $1$ but least significant digit $9$, which is forbidden.\n\nNow suppose $n$ has at least one digit that is neither a zero nor a one. Let $b$ be the leftmost (i.e., most significant) such digit, so that the left end of the decimal representation of $n$ looks like\n$$\n\\underline{a_{1}} \\cdots \\underline{a_{r}} \\underline{b} \\cdots\n$$\nfor some $r \\geq 1$ and digits $a_{i} \\in \\{0,1\\}$. When $n$ is multiplied by $9$, there will be a carry out of the column containing $b$. In particular, the $r^{\\text{th}}$ digit from the left in $9 n$ will not be $9 a_{r}$. But the right end of the decimal representation of $n$ is\n$$\n\\cdots \\underline{a_{r}} \\cdots \\underline{a_{1}} ;\n$$\nbecause each $a_{i}$ is $0$ or $1$, there are no carries out of the first $r-1$ columns, so the $r^{\\text{th}}$ digit from the right in $9 n$ will be $9 a_{r}$. Thus $9 n$ is not a palindrome, a contradiction. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWe are given the values of the differentiable real functions $f$, $g$, $h$, as well as the derivatives of their pairwise products, at $x=0$:\n$$\nf(0)=1 ; \\quad g(0)=2 ; \\quad h(0)=3 ; \\quad (g h)'(0)=4 ; \\quad (h f)'(0)=5 ; \\quad (f g)'(0)=6 .\n$$\nFind the value of $(f g h)'(0)$.",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n16\n\nBy the product rule,\n$$(f g h)' = f' g h + f g' h + f g h' = \\left( (f g)' h + (g h)' f + (h f)' g \\right) / 2.$$\nEvaluated at $0$, this gives $16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46690,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many triples $(A, B, C)$ of positive integers (positive integers are the numbers $1,2,3,4, \\ldots$) are there such that $A+B+C=10$, where order does not matter (for instance the triples $(2,3,5)$ and $(3,2,5)$ are considered to be the same triple) and where two of the integers in a triple could be the same (for instance $(3,3,4)$ is a valid triple).",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nThe triples can merely be enumerated: $(1,1,8)$, $(1,2,7)$, $(1,3,6)$, $(1,4,5)$, $(2,2,6)$, $(2,3,5)$, $(2,4,4)$, and $(3,3,4)$. There are 8 elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA conical flask contains some water. When the flask is oriented so that its base is horizontal and lies at the bottom (so that the vertex is at the top), the water is 1 inch deep. When the flask is turned upside-down, so that the vertex is at the bottom, the water is 2 inches deep. What is the height of the cone?",
    "options": [],
    "answer": "1/2 + sqrt(93)/6",
    "solution": "Solution:  \nLet $h$ be the height, and let $V$ be such that $V h^{3}$ equals the volume of the flask. When the base is at the bottom, the portion of the flask not occupied by water forms a cone similar to the entire flask, with a height of $h-1$; thus its volume is $V(h-1)^{3}$. When the base is at the top, the water occupies a cone with a height of $2$, so its volume is $V \\cdot 2^{3}$. Since the water's volume does not change,\n$$\n\\begin{gathered}\nV h^{3}-V(h-1)^{3}=8 V \\\\\n\\Rightarrow h^{3}-(h-1)^{3}=8 \\\\\n\\Rightarrow 3 h^{2}-3 h+1=8 \\\\\n\\Rightarrow 3 h^{2}-3 h-7=0 .\n\\end{gathered}\n$$\nSolving via the quadratic formula and taking the positive root gives $h=\\frac{1}{2}+\\frac{\\sqrt{93}}{6}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nExpress, as concisely as possible, the value of the product\n$$\n\\left(0^{3}-350\\right)\\left(1^{3}-349\\right)\\left(2^{3}-348\\right)\\left(3^{3}-347\\right) \\cdots\\left(349^{3}-1\\right)\\left(350^{3}-0\\right)\n$$",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n0 . One of the factors is $7^{3}-343=0$, so the whole product is zero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo circles have radii $13$ and $30$, and their centers are $41$ units apart. The line through the centers of the two circles intersects the smaller circle at two points; let $A$ be the one outside the larger circle. Suppose $B$ is a point on the smaller circle and $C$ a point on the larger circle such that $B$ is the midpoint of $AC$. Compute the distance $AC$.",
    "options": [],
    "answer": "12 \\sqrt{13}",
    "solution": "Solution:\n\n$12 \\sqrt{13}$\n\nCall the large circle's center $O_1$. Scale the small circle by a factor of $2$ about $A$; we obtain a new circle whose center $O_2$ is at a distance of $41 - 13 = 28$ from $O_1$, and whose radius is $26$. Also, the dilation sends $B$ to $C$, which thus lies on circles $O_1$ and $O_2$. So points $O_1$, $O_2$, $C$ form a $26$-$28$-$30$ triangle. Let $H$ be the foot of the altitude from $C$ to $O_1 O_2$; we have $CH = 24$ and $HO_2 = 10$. Thus, $HA = 36$, and $AC = \\sqrt{24^2 + 36^2} = 12 \\sqrt{13}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe expression $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$. Find the value of\n$$\n\\left\\lfloor\\frac{2002!}{2001!+2000!+1999!+\\cdots+1!}\\right\\rfloor .\n$$",
    "options": [],
    "answer": "2000",
    "solution": "Solution:\n\n2000\n\nWe break up $2002! = 2002 \\cdot 2001!$ as\n$$\n\\begin{gathered}\n2000(2001!)+2 \\cdot 2001(2000!)=2000(2001!)+2000(2000!)+2002 \\cdot 2000(1999!) \\\\\n>2000(2001!+2000!+1999!+\\cdots+1!)\n\\end{gathered}\n$$\nOn the other hand,\n$$\n2001(2001!+2000!+\\cdots+1!)>2001(2001!+2000!)=2001(2001!)+2001!=2002!\n$$\nThus we have $2000<\\dfrac{2002!}{2001!+\\cdots+1!}<2001$, so the answer is $2000$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoints $P$ and $Q$ are $3$ units apart. A circle centered at $P$ with a radius of $\\sqrt{3}$ units intersects a circle centered at $Q$ with a radius of $3$ units at points $A$ and $B$. Find the area of quadrilateral $APBQ$.",
    "options": [],
    "answer": "3√11/2",
    "solution": "Solution:\n\nThe area is twice the area of triangle $APQ$, which is isosceles with side lengths $3, 3, \\sqrt{3}$. By Pythagoras, the altitude to the base has length $\\sqrt{3^{2} - (\\sqrt{3} / 2)^{2}} = \\sqrt{33} / 2$, so the triangle has area $\\frac{\\sqrt{99}}{4}$. Double this to get $\\frac{3 \\sqrt{11}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $a, b, c, d$ are real numbers such that\n$$\n|a-b|+|c-d|=99 ; \\quad|a-c|+|b-d|=1\n$$\nDetermine all possible values of $|a-d|+|b-c|$.",
    "options": [],
    "answer": "99",
    "solution": "Solution:\n99 If $w \\geq x \\geq y \\geq z$ are four arbitrary real numbers, then $|w-z|+|x-y|=|w-y|+|x-z|=w+x-y-z \\geq w-x+y-z=|w-x|+|y-z|$. Thus, in our case, two of the three numbers $|a-b|+|c-d|,|a-c|+|b-d|,|a-d|+|b-c|$ are equal, and the third one is less than or equal to these two. Since we have a 99 and a 1, the third number must be 99.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the set consisting of all real values of $x$ such that the three numbers $2^{x}, 2^{x^{2}}, 2^{x^{3}}$ form a non-constant arithmetic progression (in that order).",
    "options": [],
    "answer": "∅",
    "solution": "Solution:\nThe empty set, $\\varnothing$. Trivially, $x=0,1$ yield constant arithmetic progressions; we show that there are no other possibilities. If these numbers do form a progression, then, by the AM-GM (arithmetic mean-geometric mean) inequality,\n$$\n\\begin{aligned}\n& 2 \\cdot 2^{x^{2}}=2^{x}+2^{x^{3}} \\geq 2 \\sqrt{2^{x} \\cdot 2^{x^{3}}} \\\\\n\\Rightarrow & 2^{x^{2}} \\geq 2^{\\left(x+x^{3}\\right) / 2} \\Rightarrow x^{2} \\geq\\left(x+x^{3}\\right) / 2 \\\\\n\\Rightarrow & x(x-1)^{2}=x^{3}-2 x^{2}+x \\leq 0 .\n\\end{aligned}\n$$\nAssuming $x \\neq 0,1$, we can divide by $(x-1)^{2}>0$ and obtain $x<0$. However, then $2^{x}, 2^{x^{3}}$ are less than 1, while $2^{x^{2}}$ is more than 1, so the given sequence cannot possibly be an arithmetic progression.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a positive integer \"mild\" if its base-3 representation never contains the digit $2$. How many values of $n$ ($1 \\leq n \\leq 1000$) have the property that $n$ and $n^{2}$ are both mild?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nSuch a number, which must consist entirely of $0$'s and $1$'s in base $3$, can never have more than one $1$. Indeed, if $n = 3^{a} + 3^{b} +$ higher powers where $b > a$, then $n^{2} = 3^{2a} + 2 \\cdot 3^{a+b} +$ higher powers which will not be mild. On the other hand, if $n$ does just have one $1$ in base $3$, then clearly $n$ and $n^{2}$ are mild. So the values of $n \\leq 1000$ that work are $3^{0}, 3^{1}, \\ldots, 3^{6}$; there are $7$ of them.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMassachusetts Avenue is ten blocks long. One boy and one girl live on each block. They want to form friendships such that each boy is friends with exactly one girl and vice versa. Nobody wants a friend living more than one block away (but they may be on the same block). How many pairings are possible?",
    "options": [],
    "answer": "89",
    "solution": "Solution:\n\n89\n\nLet $a_n$ be the number of pairings if there are $n$ blocks; we have $a_1 = 1$, $a_2 = 2$, and we claim the Fibonacci recurrence is satisfied. Indeed, if there are $n$ blocks, either the boy on block 1 is friends with the girl on block 1, leaving $a_{n-1}$ possible pairings for the people on the remaining $n-1$ blocks, or he is friends with the girl on block 2, in which case the girl on block 1 must be friends with the boy on block 2, and then there are $a_{n-2}$ possibilities for the friendships among the remaining $n-2$ blocks. So $a_n = a_{n-1} + a_{n-2}$, and we compute: $a_3 = 3$, $a_4 = 5$, $\\ldots$, $a_{10} = 89$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the $x$-$y$ plane, draw a circle of radius $2$ centered at $(0,0)$. Color the circle red above the line $y=1$, color the circle blue below the line $y=-1$, and color the rest of the circle white. Now consider an arbitrary straight line at distance $1$ from the circle. We color each point $P$ of the line with the color of the closest point to $P$ on the circle. If we pick such an arbitrary line, randomly oriented, what is the probability that it contains red, white, and blue points?",
    "options": [],
    "answer": "2/3",
    "solution": "Solution:\n\nLet $O = (0,0)$, $P = (1,0)$, and $H$ the foot of the perpendicular from $O$ to the line. If $\\angle P O H$ (as measured counterclockwise) lies between $\\pi / 3$ and $2\\pi / 3$, the line will fail to contain blue points; if it lies between $4\\pi / 3$ and $5\\pi / 3$, the line will fail to contain red points. Otherwise, it has points of every color. Thus, the answer is $1 - \\frac{2\\pi}{3} / 2\\pi = \\frac{2}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways are there of using diagonals to divide a regular $6$-sided polygon into triangles such that at least one side of each triangle is a side of the original polygon and that each vertex of each triangle is a vertex of the original polygon?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nThe number of ways of triangulating a convex $(n+2)$-sided polygon is $\\binom{2 n}{n} \\frac{1}{n+1}$, which is $14$ in this case. However, there are two triangulations of a hexagon which produce one triangle sharing no sides with the original polygon, so the answer is $14-2=12$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the area of the region in the first quadrant $x>0, y>0$ bounded above the graph of $y=\\arcsin (x)$ and below the graph of $y=\\arccos (x)$.",
    "options": [],
    "answer": "2 - sqrt(2)",
    "solution": "Solution:\nWe can integrate over $y$ rather than $x$. In particular, the solution is\n$$\n\\int_{0}^{\\pi / 4} \\sin y \\, dy + \\int_{\\pi / 4}^{\\pi / 2} \\cos y \\, dy = \\left(1 - \\frac{1}{\\sqrt{2}}\\right) 2 = 2 - \\sqrt{2}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 46703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe call a set of professors and committees on which they serve a university if\n(1) given two distinct professors there is one and only one committee on which they both serve,\n(2) given any committee, $C$, and any professor, $P$, not on that committee, there is exactly one committee on which $P$ serves and no professors on committee $C$ serve, and\n(3) there are at least two professors on each committee; there are at least two committees. What is the smallest number of committees a university can have?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nLet $C$ be any committee. Then there exists a professor $P$ not on $C$ (or else there would be no other committees). By axiom 2, $P$ serves on a committee $D$ having no common members with archetype $C$. Each of these committees has at least two members, and for each $Q \\in C, R \\in D$, there exists (by axiom 1) a committee containing $Q$ and $R$, which (again by axiom 1) has no other common members with $C$ or $D$. Thus we have at least\n$$2 + 2 \\cdot 2 = 6$$\ncommittees. This minimum is attainable - just take four professors and let any two professors form a committee.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan is holding one end of a 26 inch long piece of light string that has a heavy bead on it with each hand (so that the string lies along two straight lines). If he starts with his hands together at the start and leaves his hands at the same height, how far does he need to pull his hands apart so that the bead moves upward by 8 inches?",
    "options": [],
    "answer": "24 inches",
    "solution": "Solution:\n\nAfter he pulls the bead is 5 inches below his hands, and it is 13 inches from each hand. Using the Pythagorean theorem, his hands must be $2 \\cdot 12 = 24$ inches apart.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor $n \\geq 1$, let $a_{n}$ be the largest odd divisor of $n$, and let $b_{n}=a_{1}+a_{2}+\\cdots+a_{n}$. Prove that $b_{n} \\geq \\frac{n^{2}+2}{3}$, and determine for which $n$ equality holds. For example,\n$$\na_{1}=1,\\ a_{2}=1,\\ a_{3}=3,\\ a_{4}=1,\\ a_{5}=5,\\ a_{6}=3\n$$\nthus\n$$\nb_{6}=1+1+3+1+5+3=14 \\geq \\frac{6^{2}+2}{3}=12 \\frac{2}{3} .\n$$",
    "options": [],
    "answer": "Equality holds if and only if n is a power of two.",
    "solution": "Solution:\n\nInspecting a few values leads one to the correct guess that equality holds if and only if $n$ is a power of $2$. The idea is to use induction by \"doubling\" to prove the result. When $n=1$, we have $1=1$ as desired. Now suppose $n \\geq 2$. If $n=2k$ with $k \\geq 1$, then\n$$\n\\begin{aligned}\nb_{n} & =\\left(a_{1}+a_{3}+\\cdots+a_{2k-1}\\right)+\\left(a_{2}+a_{4}+\\cdots+a_{2k}\\right) \\\\\n& =1+3+\\cdots+(2k-1)+\\left(a_{1}+a_{2}+\\cdots+a_{k}\\right) \\\\\n& =k^{2}+b_{k} \\\\\n& \\geq k^{2}+\\left(k^{2}+2\\right) / 3 \\quad \\text{ (by the inductive hypothesis) } \\\\\n& =\\left(n^{2}+2\\right) / 3,\n\\end{aligned}\n$$\nwith equality if and only if $k$ is a power of $2$, i.e., if and only if $n$ is a power of $2$. If instead $n=2k+1$ with $k \\geq 1$, then\n$$\n\\begin{aligned}\nb_{n} & =\\left(a_{1}+a_{3}+\\cdots+a_{2k+1}\\right)+\\left(a_{2}+a_{4}+\\cdots+a_{2k}\\right) \\\\\n& =1+3+\\cdots+(2k+1)+\\left(a_{1}+a_{2}+\\cdots+a_{k}\\right) \\\\\n& =(k+1)^{2}+b_{k} \\\\\n& \\geq(k+1)^{2}+\\left(k^{2}+2\\right) / 3 \\quad \\text{ (by the inductive hypothesis) } \\\\\n& =\\frac{(2k+1)^{2}+2}{3}+\\frac{2k+2}{3} \\\\\n& >\\left(n^{2}+2\\right) / 3\n\\end{aligned}\n$$\nso strict inequality holds, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf a positive integer multiple of $864$ is picked randomly, with each multiple having the same probability of being picked, what is the probability that it is divisible by $1944$?",
    "options": [],
    "answer": "1/9",
    "solution": "Solution:\nThe probability that a multiple of $864 = 2^{5} 3^{3}$ is divisible by $1944 = 2^{3} 3^{5}$ is the same as the probability that a multiple of $2^{2}$ is divisible by $3^{2}$, which since $4$ and $9$ are relatively prime is $\\frac{1}{9}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo fair coins are simultaneously flipped. This is done repeatedly until at least one of the coins comes up heads, at which point the process stops. What is the probability that the other coin also came up heads on this last flip?",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n\n$\\frac{1}{3}$. Let the desired probability be $p$. There is a $\\frac{1}{4}$ chance that both coins will come up heads on the first toss. Otherwise, both can come up heads simultaneously only if both are tails on the first toss, and then the process restarts as if from the beginning; thus this situation occurs with probability $\\frac{p}{4}$. Thus $p = \\frac{1}{4} + \\frac{p}{4}$; solving, $p = \\frac{1}{3}$.\n\nAlternate Solution:\n\nThe desired event is equivalent to both coins coming up tails for $n$ successive turns (for some $n \\geq 0$), then both coins coming up heads. For any fixed value of $n$, the probability of this occurring is $\\frac{1}{4^{n+1}}$. Since all these events are disjoint, the total probability is $\\frac{1}{4} + \\frac{1}{4^{2}} + \\frac{1}{4^{3}} + \\cdots = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be an integer. Prove that\n$$\n\\left\\lfloor\\frac{n}{2}\\right\\rfloor+\\left\\lfloor\\frac{n+1}{2}\\right\\rfloor=n .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose $n=2m$ is even. Then $\\lfloor n/2\\rfloor=\\lfloor m\\rfloor=m$ and $\\lfloor(n+1)/2\\rfloor=\\lfloor m+1/2\\rfloor=m$, whose sum is $m+m=2m=n$.\n\nOtherwise $n=2m+1$ is odd. In this case $\\lfloor n/2\\rfloor=\\lfloor m+1/2\\rfloor=m$ and $\\lfloor(n+1)/2\\rfloor=\\lfloor m+1\\rfloor=m+1$, whose sum is $m+(m+1)=2m+1=n$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the volume of the three-dimensional solid given by the inequality $\\sqrt{x^{2}+y^{2}} + |z| \\leq 1$.",
    "options": [],
    "answer": "2π/3",
    "solution": "Solution:\n\n$2 \\pi / 3$. The solid consists of two cones, one whose base is the circle $x^{2}+y^{2}=1$ in the $xy$-plane and whose vertex is $(0,0,1)$, and the other with the same base but vertex $(0,0,-1)$. Each cone has a base area of $\\pi$ and a height of $1$, for a volume of $\\pi / 3$, so the answer is $2 \\pi / 3$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor any integer $n$, define $\\lfloor n\\rfloor$ as the greatest integer less than or equal to $n$. For any positive integer $n$, let\n$$\nf(n) = \\lfloor n \\rfloor + \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{3} \\right\\rfloor + \\cdots + \\left\\lfloor \\frac{n}{n} \\right\\rfloor.\n$$\nFor how many values of $n$, $1 \\leq n \\leq 100$, is $f(n)$ odd?",
    "options": [],
    "answer": "55",
    "solution": "Solution:\n\nNotice that, for fixed $a$, $\\lfloor n / a \\rfloor$ counts the number of integers $b \\in \\{1, 2, \\ldots, n\\}$ which are divisible by $a$; hence, $f(n)$ counts the number of pairs $(a, b)$, $a, b \\in \\{1, 2, \\ldots, n\\}$ with $b$ divisible by $a$. For any fixed $b$, the number of such pairs is $d(b)$ (the number of divisors of $b$), so the total number of pairs $f(n)$ equals $d(1) + d(2) + \\cdots + d(n)$. But $d(b)$ is odd precisely when $b$ is a square, so $f(n)$ is odd precisely when there are an odd number of squares in $\\{1, 2, \\ldots, n\\}$. This happens for $1 \\leq n < 4; 9 \\leq n < 16; \\ldots; 81 \\leq n < 100$. Adding these up gives 55 values of $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all the integers $n>1$ with the following property: the numbers $1,2, \\ldots, n$ can be arranged in a line so that, of any two adjacent numbers, one is divisible by the other.",
    "options": [],
    "answer": "2, 3, 4, 6",
    "solution": "Solution:\n$2,3,4,6$\n\nThe values $n=2,3,4,6$ work, as shown by respective examples $1,2$; $2,1,3$; $2,4,1,3$; $3,6,2,4,1,5$. We shall show that there are no other possibilities.\n\nIf $n=2k+1$ is odd, then none of the numbers $k+1, k+2, \\ldots, 2k+1$ can divide any other, so no two of these numbers are adjacent. This is only possible if they occupy the 1st, 3rd, $\\ldots,(2k+1)$th positions in the line, which means every number $\\leq k$ is adjacent to two of these and hence divides two of them. But $k$ only divides one of these numbers when $k \\geq 2$. Thus no odd $n \\geq 5$ works.\n\nIf $n=2k$ is even, the numbers $k+1, k+2, \\ldots, 2k$ again must be mutually nonadjacent, but now this means we can have up to two numbers $\\leq k$ each of which is adjacent to only one number $>k$, and if there are two such numbers, they must be adjacent. If $k \\geq 4$, then each of $k-1, k$ divides only one of the numbers $k+1, \\ldots, 2k$, so $k-1, k$ must be adjacent, but this is impossible. Thus no even $k \\geq 8$ works, and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\ngiven that $a, b, c$ are positive integers satisfying\n$$\na+b+c=\\operatorname{gcd}(a, b)+\\operatorname{gcd}(b, c)+\\operatorname{gcd}(c, a)+120\n$$\ndetermine the maximum possible value of $a$.",
    "options": [],
    "answer": "240",
    "solution": "Solution:\n\n240. Notice that $(a, b, c) = (240, 120, 120)$ achieves a value of 240. To see that this is maximal, first suppose that $a > b$. Notice that\n$$\na + b + c = \\operatorname{gcd}(a, b) + \\operatorname{gcd}(b, c) + \\operatorname{gcd}(c, a) + 120 \\leq \\operatorname{gcd}(a, b) + b + c + 120,\n$$\nor $a \\leq \\operatorname{gcd}(a, b) + 120$. However, $\\operatorname{gcd}(a, b)$ is a proper divisor of $a$, so $a \\geq 2 \\cdot \\operatorname{gcd}(a, b)$. Thus, $a - 120 \\leq \\operatorname{gcd}(a, b) \\leq a / 2$, yielding $a \\leq 240$.\n\nNow, if instead $a \\leq b$, then either $b > c$ and the same logic shows that $b \\leq 240 \\Rightarrow a \\leq 240$, or $b \\leq c, c > a$ (since $a, b, c$ cannot all be equal) and then $c \\leq 240 \\Rightarrow a \\leq b \\leq c \\leq 240$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe unknown real numbers $x$, $y$, $z$ satisfy the equations\n$$\n\\frac{x+y}{1+z}=\\frac{1-z+z^{2}}{x^{2}-x y+y^{2}} ; \\quad \\frac{x-y}{3-z}=\\frac{9+3 z+z^{2}}{x^{2}+x y+y^{2}}\n$$\nFind $x$.",
    "options": [],
    "answer": "14^(1/3)",
    "solution": "Solution:\n\n$\\sqrt[3]{14}$\n\nCross-multiplying in both equations, we get, respectively,\n$$\nx^{3}+y^{3}=1+z^{3}, \\quad x^{3}-y^{3}=27-z^{3}.\n$$\nNow adding gives $2 x^{3}=28$, or $x=\\sqrt[3]{14}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of sequences $a_{1}, a_{2}, \\ldots, a_{10}$ of positive integers with the property that $a_{n+2}=a_{n+1}+a_{n}$ for $n=1,2, \\ldots, 8$, and $a_{10}=2002$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nLet $a_{1}=a$, $a_{2}=b$; we successively compute $a_{3}=a+b$, $a_{4}=a+2b$, $\\ldots$, $a_{10}=21a+34b$. The equation $2002=21a+34b$ has three positive integer solutions, namely $(84,7)$, $(50,28)$, $(16,49)$, and each of these gives a unique sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoints $A$, $B$, $C$ in the plane satisfy $\\overline{A B} = 2002$, $\\overline{A C} = 9999$. The circles with diameters $A B$ and $A C$ intersect at $A$ and $D$. If $\\overline{A D} = 37$, what is the shortest distance from point $A$ to line $B C$?",
    "options": [],
    "answer": "37",
    "solution": "Solution:\n\n$\\angle A D B = \\angle A D C = \\pi / 2$ since $D$ lies on the circles with $A B$ and $A C$ as diameters, so $D$ is the foot of the perpendicular from $A$ to line $B C$, and the answer is the given $37$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe real function $f$ has the property that, whenever $a$, $b$, $n$ are positive integers such that $a + b = 2^{n}$, the equation $f(a) + f(b) = n^{2}$ holds. What is $f(2002)$?",
    "options": [],
    "answer": "96",
    "solution": "Solution:\n\nWe know $f(a) = n^{2} - f\\left(2^{n} - a\\right)$ for any $a$, $n$ with $2^{n} > a$; repeated application gives\n$$\n\\begin{gathered}\nf(2002) = 11^{2} - f(46) = 11^{2} - \\left(6^{2} - f(18)\\right) = 11^{2} - \\left(6^{2} - \\left(5^{2} - f(14)\\right)\\right) \\\\\n= 11^{2} - \\left(6^{2} - \\left(5^{2} - \\left(4^{2} - f(2)\\right)\\right)\\right) .\n\\end{gathered}\n$$\nBut $f(2) = 2^{2} - f(2)$, giving $f(2) = 2$, so the above simplifies to $11^{2} - \\left(6^{2} - \\left(5^{2} - \\left(4^{2} - 2\\right)\\right)\\right) = 96$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA permutation of a finite set is a one-to-one function from the set to itself; for instance, one permutation of $\\{1,2,3,4\\}$ is the function $\\pi$ defined such that $\\pi(1)=1, \\pi(2)=3$, $\\pi(3)=4$, and $\\pi(4)=2$. How many permutations $\\pi$ of the set $\\{1,2, \\ldots, 10\\}$ have the property that $\\pi(i) \\neq i$ for each $i=1,2, \\ldots, 10$, but $\\pi(\\pi(i))=i$ for each $i$ ?",
    "options": [],
    "answer": "945",
    "solution": "Solution:\nFor each such $\\pi$, the elements of $\\{1,2, \\ldots, 10\\}$ can be arranged into pairs $\\{i, j\\}$ such that $\\pi(i)=j ; \\pi(j)=i$. Choosing a permutation $\\pi$ is thus tantamount to choosing a partition of $\\{1,2, \\ldots, 10\\}$ into five disjoint pairs. There are 9 ways to pair off the number 1, then 7 ways to pair off the smallest number not yet paired, and so forth, so we have $9 \\cdot 7 \\cdot 5 \\cdot 3 \\cdot 1=945$ partitions into pairs.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo integers are relatively prime if they don't share any common factors, i.e. if their greatest common divisor is $1$. Define $\\varphi(n)$ as the number of positive integers that are less than $n$ and relatively prime to $n$. Define $\\varphi_{d}(n)$ as the number of positive integers that are less than $d n$ and relatively prime to $n$.\nWhat is the least $n$ such that $\\varphi_{x}(n)=64000$, where $x=\\varphi_{y}(n)$, where $y=\\varphi(n)$ ?",
    "options": [],
    "answer": "41",
    "solution": "Solution:\n\nFor fixed $n$, the pattern of integers relatively prime to $n$ repeats every $n$ integers, so $\\varphi_{d}(n)=d \\varphi(n)$. Therefore the expression in the problem equals $\\varphi(n)^{3}$. The cube root of $64000$ is $40$. $\\varphi(p)=p-1$ for any prime $p$. Since $40$ is one less than a prime, the least $n$ such that $\\varphi(n)=40$ is $41$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo $4 \\times 4$ squares are randomly placed on an $8 \\times 8$ chessboard so that their sides lie along the grid lines of the board. What is the probability that the two squares overlap?",
    "options": [],
    "answer": "529/625",
    "solution": "Solution:\n\n$529/625$. Each square has $5$ horizontal $\\cdot$ $5$ vertical $=25$ possible positions, so there are $625$ possible placements of the squares. If they do not overlap, then either one square lies in the top four rows and the other square lies in the bottom four rows, or one square lies in the left four columns and the other lies in the right four columns. The first possibility can happen in $2 \\cdot 5 \\cdot 5=50$ ways (two choices of which square goes on top, and five horizontal positions for each square); likewise, so can the second. However, this double-counts the $4$ cases in which the two squares are in opposite corners, so we have $50+50-4=96$ possible non-overlapping arrangements $\\Rightarrow 25^{2}-96=529$ overlapping arrangements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA square and a regular hexagon are drawn with the same side length. If the area of the square is $\\sqrt{3}$, what is the area of the hexagon?",
    "options": [],
    "answer": "9/2",
    "solution": "Solution:\n\nThe hexagon is composed of six equilateral triangles each of side length $\\sqrt[4]{3}$ (with base $b = \\sqrt[4]{3}$ and height $\\frac{\\sqrt{3}}{2} b$), so the total area is $\\frac{9}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProfessor Moriarty has designed a \"prime-testing trail.” The trail has 2002 stations, labeled $1, \\ldots, 2002$. Each station is colored either red or green, and contains a table which indicates, for each of the digits $0, \\ldots, 9$, another station number. A student is given a positive integer $n$, and then walks along the trail, starting at station $1$. The student reads the first (leftmost) digit of $n$, and looks this digit up in station $1$'s table to get a new station location. The student then walks to this new station, reads the second digit of $n$ and looks it up in this station's table to get yet another station location, and so on, until the last (rightmost) digit of $n$ has been read and looked up, sending the student to his or her final station. Here is an example that shows possible values for some of the tables. Suppose that $n=19$ :\n\n| Station | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 (red) | 15 | 29 | 314 | 16 | 2002 | 97 | 428 | 1613 | 91 | 24 |\n| $\\vdots$ |  |  |  |  |  |  |  |  |  |  |\n| $\\vdots$ |  |  |  |  |  |  |  |  |  |  |\n| 29 (red) | 98 | 331 | 1918 | 173 | 15 | 41 | 17 | 631 | 1211 | 1429 |\n| $\\vdots$ |  |  |  |  |  |  |  |  |  |  |\n| 1429 (green) | 7 | 18 | 31 | 43 | 216 | 1709 | 421 | 53 | 224 | 1100 |\n\nUsing these tables, station $1$, digit $1$ leads to station $29$; station $29$, digit $9$ leads to station $1429$; and station $1429$ is green.\n\nProfessor Moriarty claims that for any positive integer $n$, the final station (in the example, $1429$) will be green if and only if $n$ is prime. Is this possible?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNo, this is impossible. Suppose on the contrary that such a trail has been designed. Since there are infinitely many primes, we can choose a prime $p$ with more than $2002$ decimal digits. When we test $p$ on the trail, we will visit some station more than once. Write $p$ as $10^{N} A + 10^{m} B + C$ so that the station after the digits of $A$ is the same as after the digits of $B$. Then we may reach the same station by running through the digits of $A$, then running through the digits of $B$ any number of times, then running through the digits of $C$.\n\nThat is, for all $k \\geq 0$, the numbers\n$$\n\\begin{gathered}\nC + 10^{m} B \\left(1 + 10^{N-m} + 10^{2(N-m)} + \\cdots + 10^{k(N-m)}\\right) + 10^{(k+1)(N-m)+m} A \\\\\n= C - \\frac{10^{m} B}{10^{N-m}-1} + 10^{(k+1)(N-m)} \\left(10^{m} A + \\frac{10^{m} B}{10^{N-m}-1}\\right)\n\\end{gathered}\n$$\nall lead to the same terminal station, and so must all be prime. For simplicity, write this expression as $p_{k} = r + t^{k} s$, and choose $k$ large enough so that $p_{k}$ does not divide the denominators of $r, s, t$. Then $t^{k} \\equiv t^{k+p_{k}-1} \\pmod{p_{k}}$ by Fermat's little theorem, so $p_{k+p_{k}-1}$ is divisible by $p_{k}$ and hence is not prime, contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the minimum vertical distance between the graphs of $2+\\sin (x)$ and $\\cos (x)$?",
    "options": [],
    "answer": "2 - sqrt(2)",
    "solution": "Solution:\n\nThe derivative of $2+\\sin (x)-\\cos (x)$ is $\\cos x+\\sin x$, which in the interval $0 \\leq x<2 \\pi$ is zero at $x=\\frac{3 \\pi}{4}, \\frac{7 \\pi}{4}$. At $\\frac{7 \\pi}{4}$, when $\\sin (x)$ is negative and $\\cos (x)$ is positive, the distance reaches its minimal value of $2-\\sqrt{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 46723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider a square of side length $1$. Draw four lines that each connect a midpoint of a side with a corner not on that side, such that each midpoint and each corner is touched by only one line. Find the area of the region completely bounded by these lines.",
    "options": [],
    "answer": "1/5",
    "solution": "Solution:\n\nIn unit square $ABCD$, denote by $E, F, G, H$ the respective midpoints of sides $AB, BC, CD, DA$. Let $I$ be the intersection of $AF$ and $DE$, let $J$ be the intersection of $BG$ and $AF$, let $K$ be the intersection of $CH$ and $BG$, and let $L$ be the intersection of $DE$ and $CH$. We want to find the area of square $IJKL$.\n\nThe area of $ABF$ is $\\frac{1}{4}$, which is equal to $\\frac{1}{2} AF \\cdot BJ = \\frac{\\sqrt{5}}{4} BJ$, so $BJ = \\frac{1}{\\sqrt{5}}$.\n\nUsing similar triangles, $GK = JF = \\frac{1}{2} BJ$.\n\nThus the length of a side of $IJKL$ is $JK = \\frac{\\sqrt{5}}{2} - \\frac{1}{\\sqrt{5}} - \\frac{1}{2} \\frac{1}{\\sqrt{5}} = \\frac{1}{\\sqrt{5}}$, and the area of $IJKL$ is $\\frac{1}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46724,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the greatest common divisor of the numbers $2002+2, 2002^{2}+2, 2002^{3}+2, \\ldots$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nNotice that $2002+2$ divides $2002^{2}-2^{2}$, so any common divisor of $2002+2$ and $2002^{2}+2$ must divide $(2002^{2}+2)-(2002^{2}-2^{2})=6$. On the other hand, every number in the sequence is even, and the $n$th number is always congruent to $1^{n}+2 \\equiv 0$ modulo $3$. Thus, $6$ divides every number in the sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve for integers $n$ that\n$$\n\\left\\lfloor\\frac{n}{2}\\right\\rfloor\\left\\lfloor\\frac{n+1}{2}\\right\\rfloor=\\left\\lfloor\\frac{n^{2}}{4}\\right\\rfloor .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose $n=2m$ is even; then $\\lfloor n / 2\\rfloor = \\lfloor m\\rfloor = m$ and $\\lfloor(n+1) / 2\\rfloor = \\lfloor m+1 / 2\\rfloor = m$, whose product is $m^{2} = \\left\\lfloor m^{2}\\right\\rfloor = \\left\\lfloor (2m)^{2} / 4 \\right\\rfloor$.\n\nOtherwise $n=2m+1$ is odd, so that $\\lfloor n / 2\\rfloor = \\lfloor m+1 / 2\\rfloor = m$ and $\\lfloor(n+1) / 2\\rfloor = \\lfloor m+1\\rfloor = m+1$, whose product is $m^{2} + m$.\n\nOn the other side, we find that\n$$\n\\left\\lfloor\\frac{n^{2}}{4}\\right\\rfloor = \\left\\lfloor\\frac{4m^{2} + 4m + 1}{4}\\right\\rfloor = \\left\\lfloor m^{2} + m + \\frac{1}{4} \\right\\rfloor = m^{2} + m,\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the number of subsets $S$ of $\\{1,2,3, \\ldots, 10\\}$ with the following property: there exist integers $a < b < c$ with $a \\in S$, $b \\notin S$, $c \\in S$.",
    "options": [],
    "answer": "968",
    "solution": "Solution:\n\nThere are $2^{10} = 1024$ subsets of $\\{1,2, \\ldots, 10\\}$ altogether. Any subset without the specified property must be either the empty set or a block of consecutive integers. To specify a block of consecutive integers, we either have just one element (10 choices) or a pair of distinct endpoints $\\left(\\binom{10}{2} = 45\\right.$ choices$)$. So the number of sets with our property is $1024 - (1 + 10 + 45) = 968$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGive the set of all positive integers $n$ such that $\\varphi(n) = 2002^{2} - 1$.",
    "options": [],
    "answer": "empty set",
    "solution": "Solution:\nThe empty set, $\\varnothing$. If $m$ is relatively prime to $n$ and $m < n$, then $n - m$ must likewise be relatively prime to $n$, and these are distinct for $n > 2$ since $n / 2, n$ are not relatively prime. Therefore, for all $n > 2$, $\\varphi(n)$ must be even. $2002^{2} - 1$ is odd, and $\\varphi(2) = 1 \\neq 2002^{2} - 1$, so no numbers $n$ fulfill the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine $\\varphi^{k}(n)$ as the number of positive integers that are less than or equal to $n / k$ and relatively prime to $n$. Find $\\varphi^{2001}\\left(2002^{2}-1\\right)$. (Hint: $\\varphi(2003)=2002$.)",
    "options": [],
    "answer": "1233",
    "solution": "Solution:\n\n$\\varphi^{2001}\\left(2002^{2}-1\\right)=\\varphi^{2001}(2001 \\cdot 2003)=$ the number of $m$ that are relatively prime to both $2001$ and $2003$, where $m \\leq 2003$. Since $\\varphi(n)=n-1$ implies that $n$ is prime, we must only check for those $m$ relatively prime to $2001$, except for $2002$, which is relatively prime to $2002^{2}-1$. So $\\varphi^{2001}\\left(2002^{2}-1\\right)=\\varphi(2001)+1=\\varphi(3 \\cdot 23 \\cdot 29)+1=(3-1)(23-1)(29-1)+1=1233$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a quadrilateral, and let $E, F, G, H$ be the respective midpoints of $AB, BC, CD, DA$. If $EG = 12$ and $FH = 15$, what is the maximum possible area of $ABCD$?",
    "options": [],
    "answer": "180",
    "solution": "Solution:\nThe area of $EFGH$ is $EG \\cdot FH \\sin \\theta / 2$, where $\\theta$ is the angle between $EG$ and $FH$. This is at most $90$. However, we claim the area of $ABCD$ is twice that of $EFGH$. To see this, notice that $EF = AC / 2 = GH$, $FG = BD / 2 = HE$, so $EFGH$ is a parallelogram. The half of this parallelogram lying inside triangle $DAB$ has area $(BD / 2)(h / 2)$, where $h$ is the height from $A$ to $BD$, and triangle $DAB$ itself has area $BD \\cdot h / 2 = 2 \\cdot (BD / 2)(h / 2)$. A similar computation holds in triangle $BCD$, proving the claim. Thus, the area of $ABCD$ is at most $180$. And this maximum is attainable—just take a rectangle with $AB = CD = 15$, $BC = DA = 12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABC$ is a triangle with points $E, F$ on sides $AC, AB$, respectively. Suppose that $BE, CF$ intersect at $X$. It is given that $AF / FB = (AE / EC)^2$ and that $X$ is the midpoint of $BE$. Find the ratio $CX / XF$.",
    "options": [],
    "answer": "sqrt(5)",
    "solution": "Solution:\n\nLet $x = AE / EC$. By Menelaus's theorem applied to triangle $ABE$ and line $CXF$,\n$$\n1 = \\frac{AF}{FB} \\cdot \\frac{BX}{XE} \\cdot \\frac{EC}{CA} = \\frac{x^2}{x+1}.\n$$\nThus, $x^2 = x + 1$, and $x$ must be positive, so $x = (1 + \\sqrt{5}) / 2$.\n\nNow apply Menelaus to triangle $ACF$ and line $BXE$, obtaining\n$$\n1 = \\frac{AE}{EC} \\cdot \\frac{CX}{XF} \\cdot \\frac{FB}{BA} = \\frac{CX}{XF} \\cdot \\frac{x}{x^2 + 1},\n$$\nso $CX / XF = (x^2 + 1) / x = (2x^2 - x) / x = 2x - 1 = \\sqrt{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many pairs of integers $(a, b)$, with $1 \\leq a \\leq b \\leq 60$, have the property that $b$ is divisible by $a$ and $b+1$ is divisible by $a+1$?",
    "options": [],
    "answer": "106",
    "solution": "Solution:\n\nThe divisibility condition is equivalent to $b-a$ being divisible by both $a$ and $a+1$, or, equivalently (since these are relatively prime), by $a(a+1)$. Any $b$ satisfying the condition is automatically $\\geq a$, so it suffices to count the number of values $b-a \\in \\{1-a, 2-a, \\ldots, 60-a\\}$ that are divisible by $a(a+1)$ and sum over all $a$. The number of such values will be precisely $60 / [a(a+1)]$ whenever this quantity is an integer, which fortunately happens for every $a \\leq 5$; we count:\n\n$a=1$ gives 30 values of $b$;\n$a=2$ gives 10 values of $b$;\n$a=3$ gives 5 values of $b$;\n$a=4$ gives 3 values of $b$;\n$a=5$ gives 2 values of $b$;\n$a=6$ gives 2 values ($b=6$ or $48$);\nany $a \\geq 7$ gives only one value, namely $b=a$, since $b>a$ implies $b \\geq a+a(a+1)>60$.\n\nAdding these up, we get a total of 106 pairs.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$x$, $y$ are positive real numbers such that $x + y^{2} = x y$. What is the smallest possible value of $x$?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nNotice that $x = \\dfrac{y^{2}}{y-1} = 2 + (y-1) + \\dfrac{1}{y-1} \\geq 2 + 2 = 4$. Conversely, $x = 4$ is achievable, by taking $y = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways, without taking order into consideration, can $2002$ be expressed as the sum of $3$ positive integers (for instance, $1000+1000+2$ and $1000+2+1000$ are considered to be the same way)?",
    "options": [],
    "answer": "334000",
    "solution": "Solution:\n\nCall the three numbers that sum to $2002$ $A$, $B$, and $C$. In order to prevent redundancy, we will consider only cases where $A \\leq B \\leq C$. Then $A$ can range from $1$ to $667$, inclusive. For odd $A$, there are $1000-\\frac{3(A-1)}{2}$ possible values for $B$. For each choice of $A$ and $B$, there can only be one possible $C$, since the three numbers must add up to a fixed value. We can add up this arithmetic progression to find that there are $167167$ possible combinations of $A, B, C$, for odd $A$. For each even $A$, there are $1002-\\frac{3A}{2}$ possible values for $B$. Therefore, there are $166833$ possible combinations for even $A$. In total, this makes $334000$ possibilities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA sequence is defined by $a_{0}=1$ and $a_{n}=2^{a_{n-1}}$ for $n \\geq 1$. What is the last digit (in base 10) of $a_{15}$?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nCertainly $a_{13} \\geq 2$, so $a_{14}$ is divisible by $2^{2}=4$. Writing $a_{14}=4k$, we have $a_{15}=2^{4k}=16^{k}$. But every power of $16$ ends in $6$, so this is the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the value of\n$$\n1 \\cdot 2 - 2 \\cdot 3 + 3 \\cdot 4 - 4 \\cdot 5 + \\cdots + 2001 \\cdot 2002.\n$$",
    "options": [],
    "answer": "2004002",
    "solution": "Solution:\n$2004002$. Rewrite the expression as\n$$\n2 + 3 \\cdot (4 - 2) + 5 \\cdot (6 - 4) + \\cdots + 2001 \\cdot (2002 - 2000)\n$$\n$$\n= 2 + 6 + 10 + \\cdots + 4002.\n$$\nThis is an arithmetic progression with $(4002 - 2)/4 + 1 = 1001$ terms and average $2002$, so its sum is $1001 \\cdot 2002 = 2004002$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor positive integers $L$, let $S_{L} = \\sum_{n=1}^{L} \\lfloor n / 2 \\rfloor$. Determine all $L$ for which $S_{L}$ is a square number.",
    "options": [],
    "answer": "L is even or L = 1",
    "solution": "Solution:\n\nWe distinguish two cases depending on the parity of $L$.\n\nSuppose first that $L = 2k - 1$ is odd, where $k \\geq 1$. Then\n$$\nS_{L} = \\sum_{1 \\leq n \\leq 2k-1} \\left\\lfloor \\frac{n}{2} \\right\\rfloor = 2 \\sum_{0 \\leq m < k} m = 2 \\cdot \\frac{k(k-1)}{2} = k(k-1).\n$$\nIf $k = 1$, this is the square number $0$. If $k > 1$ then $(k-1)^2 < k(k-1) < k^2$, so $k(k-1)$ is not square.\n\nNow suppose $L = 2k$ is even, where $k \\geq 1$. Then $S_{L} = S_{L-1} + k = k^2$ is always square.\n\nHence $S_{L}$ is square exactly when $L = 1$ or $L$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of the even positive divisors of $1000$.",
    "options": [],
    "answer": "2184",
    "solution": "Solution:\n$2184$. Notice that $2k$ is a divisor of $1000$ if and only if $k$ is a divisor of $500$, so we need only find the sum of the divisors of $500$ and multiply by $2$. This can be done by enumerating the divisors individually, or simply by using the formula: \n$$\n\\sigma\\left(2^{2} \\cdot 5^{3}\\right) = (1+2+2^{2})(1+5+5^{2}+5^{3}) = 1092,\n$$\nand then doubling gives $2184$.\n\nAlternate Solution: The sum of all the divisors of $1000$ is \n$$\n(1+2+2^{2}+2^{3})(1+5+5^{2}+5^{3}) = 2340.\n$$\nThe odd divisors of $1000$ are simply the divisors of $125$, whose sum is $1+5+5^{2}+5^{3}=156$; subtracting this from $2340$, we are left with the sum of the even divisors of $1000$, which is $2184$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn how many ways can the numbers $1,2, \\ldots, 2002$ be placed at the vertices of a regular $2002$-gon so that no two adjacent numbers differ by more than $2$? (Rotations and reflections are considered distinct.)",
    "options": [],
    "answer": "4004",
    "solution": "Solution:\n\n$4004$. There are $2002$ possible positions for the $1$. The two numbers adjacent to the $1$ must be $2$ and $3$; there are two possible ways of placing these. The positions of these numbers uniquely determine the rest: for example, if $3$ lies clockwise from $1$, then the number lying counterclockwise from $2$ must be $4$; the number lying clockwise from $3$ must be $5$; the number lying counterclockwise from $4$ must now be $6$; and so forth. Eventually, $2002$ is placed adjacent to $2000$ and $2001$, so we do get a valid configuration. Thus there are $2002 \\cdot 2$ possible arrangements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf we pick (uniformly) a random square of area $1$ with sides parallel to the $x$- and $y$-axes that lies entirely within the $5$-by-$5$ square bounded by the lines $x=0$, $x=5$, $y=0$, $y=5$ (the corners of the square need not have integer coordinates), what is the probability that the point $(x, y) = (4.5, 0.5)$ lies within the square of area $1$?",
    "options": [],
    "answer": "1/64",
    "solution": "Solution:\n\nThe upper-left corner of the unit square is picked uniformly from the square $0 \\leq x \\leq 4$; $1 \\leq y \\leq 5$, and for it to contain the desired point it must lie in the square $3.5 \\leq x \\leq 4$; $1 \\leq y \\leq 1.5$. The answer is the ratio of the squares' areas, $\\frac{1}{4} / 16 = \\frac{1}{64}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $x$ that satisfy $x = 1 - x + x^{2} - x^{3} + x^{4} - x^{5} + \\cdots$ (be careful; this is tricky).",
    "options": [],
    "answer": "(-1 + sqrt(5))/2",
    "solution": "Solution:\nMultiplying both sides by $1 + x$ gives $(1 + x)x = 1$, or $x = \\frac{-1 \\pm \\sqrt{5}}{2}$. However, the series only converges for $|x| < 1$, so only the answer $x = \\frac{-1 + \\sqrt{5}}{2}$ makes sense.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the positive value of $a$ such that the parabola $y = x^{2} + 1$ bisects the area of the rectangle with vertices $(0,0)$, $(a, 0)$, $(0, a^{2} + 1)$, and $(a, a^{2} + 1)$.",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "Solution:\n$\\sqrt{3}$\n\nThe area of the rectangle is $a^{3} + a$. The portion under the parabola has area $\\int_{0}^{a} x^{2} + 1\\, dx = a^{3} / 3 + a$. Thus we wish to solve the equation $a^{3} + a = 2\\left(a^{3} / 3 + a\\right)$; dividing by $a$ and rearranging gives $a^{2} / 3 = 1$, so $a = \\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA man, standing on a lawn, is wearing a circular sombrero of radius $3$ feet. Unfortunately, the hat blocks the sunlight so effectively that the grass directly under it dies instantly. If the man walks in a circle of radius $5$ feet, what area of dead grass will result?",
    "options": [],
    "answer": "60π",
    "solution": "Solution:\n$60\\pi\\ \\mathrm{ft}^2$\n\nLet $O$ be the center of the man's circular trajectory. The sombrero kills all the grass that is within $3$ feet of any point that is $5$ feet away from $O$—i.e., all the grass at points $P$ with $2 \\leq OP \\leq 8$. The area of this annulus is then\n$$\n\\pi\\left(8^2 - 2^2\\right) = 60\\pi\n$$\nsquare feet.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $5 \\times 5$ square grid has the number $-3$ written in the upper-left square and the number $3$ written in the lower-right square. In how many ways can the remaining squares be filled in with integers so that any two adjacent numbers differ by $1$, where two squares are adjacent if they share a common edge (but not if they share only a corner)?",
    "options": [],
    "answer": "250",
    "solution": "Solution:\n\nIf the square in row $i$, column $j$ contains the number $k$, let its \"index\" be $i + j - k$. The constraint on adjacent squares now says that if a square has index $r$, the squares to its right and below it each have index $r$ or $r + 2$. The upper-left square has index $5$, and the lower-right square has index $7$, so every square must have index $5$ or $7$.\n\nThe boundary separating the two types of squares is a path consisting of upward and rightward steps; it can be extended along the grid's border so as to obtain a path between the lower-left and upper-right corners. Conversely, any such path uniquely determines each square's index and hence the entire array of numbers—except that the two paths lying entirely along the border of the grid fail to separate the upper-left from the lower-right square and thus do not create valid arrays (since these two squares should have different indices).\n\nEach path consists of $5$ upward and $5$ rightward steps, so there are $\\binom{10}{5} = 252$ paths, but two are impossible, so the answer is $250$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBob Barker went back to school for a PhD in math, and decided to raise the intellectual level of The Price is Right by having contestants guess how many objects exist of a certain type, without going over. The number of points you will get is the percentage of the correct answer, divided by 10, with no points for going over (i.e. a maximum of 10 points).\n\nLet's see the first object for our contestants...a table of shape $(5, 4, 3, 2, 1)$ is an arrangement of the integers $1$ through $15$ with five numbers in the top row, four in the next, three in the next, two in the next, and one in the last, such that each row and each column is increasing (from left to right, and top to bottom, respectively). For instance:\n\n```\n1\n6\n10 11 12\n13 14\n15\n```\n\nis one table. How many tables are there?",
    "options": [],
    "answer": "292864",
    "solution": "Solution:\n\n$15! / \\left(3^{4} \\cdot 5^{3} \\cdot 7^{2} \\cdot 9\\right) = 292864$. These are Standard Young Tableaux.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOur next object up for bid is an arithmetic progression of primes. For example, the primes $3$, $5$, and $7$ form an arithmetic progression of length $3$. What is the largest possible length of an arithmetic progression formed of positive primes less than $1,000,000$? Be prepared to justify your answer.",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\n$12$. We can get $12$ with $110437$ and difference $13860$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOur third and final item comes to us from Germany, I mean Geometry. It is known that a regular $n$-gon can be constructed with straightedge and compass if $n$ is a prime that is $1$ plus a power of $2$. It is also possible to construct a $2n$-gon whenever an $n$-gon is constructible, or a $p_{1} p_{2} \\cdots p_{m}$-gon where the $p_{i}$'s are distinct primes of the above form. What is really interesting is that these conditions, together with the fact that we can construct a square, is that they give us all constructible regular $n$-gons. What is the largest $n$ less than $4,300,000,000$ such that a regular $n$-gon is constructible?",
    "options": [],
    "answer": "4294967296",
    "solution": "Solution:\nThe known primes of this form (Fermat primes) are $3$, $5$, $17$, $257$, and $65537$, and the result is due to Gauss (German). If there are other such primes (unknown), then they are much bigger than $10^{10}$. So for each product of these primes, we can divide $4.3 \\cdot 10^{9}$ by that number and take $\\log_{2}$ to find the largest power of $2$ to multiply by, then compare the resulting numbers. There are $32$ cases to check, or just observe that $2^{32} = 4,294,967,296$ is so close that there's likely a shortcut. Note that $2^{32} + 1$ is divisible by $641$, and hence not prime. $3 \\cdot 5 \\cdot 17 \\cdot 257 \\cdot 65537 = 2^{32} - 1$ is smaller; replacing any of the factors by the closest power of $2$ only decreases the product, and there's not enough room to squeeze in an extra factor of $2$ without replacing all of them, and that gives us $2^{32}$, so indeed that it is the answer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $T_{L} = \\sum_{n=1}^{L} \\left\\lfloor n^{3} / 9 \\right\\rfloor$ for positive integers $L$. Determine all $L$ for which $T_{L}$ is a square number.",
    "options": [],
    "answer": "L in {1, 2}",
    "solution": "Solution:\nSince $T_{L}$ is square if and only if $9 T_{L}$ is square, we may consider $9 T_{L}$ instead of $T_{L}$.\nIt is well known that $n^{3}$ is congruent to $0, 1$, or $8$ modulo $9$ according as $n$ is congruent to $0, 1$, or $2$ modulo $3$. (Proof: $(3m + k)^{3} = 27m^{3} + 3(9m^{2})k + 3(3m)k^{2} + k^{3} \\equiv k^{3} \\pmod{9}$.) Therefore\n$n^{3} - 9\\left\\lfloor n^{3} / 9 \\right\\rfloor$ is $0, 1$, or $8$ according as $n$ is congruent to $0, 1$, or $2$ modulo $3$. We find therefore that\n$$\n\\begin{aligned}\n9 T_{L} &= \\sum_{1 \\leq n \\leq L} 9\\left\\lfloor \\frac{n^{3}}{9} \\right\\rfloor \\\\\n&= \\sum_{1 \\leq n \\leq L} n^{3} - \\#\\{1 \\leq n \\leq L : n \\equiv 1 \\pmod{3}\\} - 8\\#\\{1 \\leq n \\leq L : n \\equiv 2 \\pmod{3}\\} \\\\\n&= \\left( \\frac{1}{2} L(L+1) \\right)^{2} - \\left\\lfloor \\frac{L+2}{3} \\right\\rfloor - 8\\left\\lfloor \\frac{L+1}{3} \\right\\rfloor .\n\\end{aligned}\n$$\nClearly $9 T_{L} < (L(L+1)/2)^{2}$ for $L \\geq 1$. We shall prove that $9 T_{L} > (L(L+1)/2 - 1)^{2}$ for $L \\geq 4$, whence $9 T_{L}$ is not square for $L \\geq 4$. Because\n$$\n(L(L+1)/2 - 1)^{2} = (L(L+1)/2)^{2} - L(L+1) + 1,\n$$\nwe need only show that\n$$\n\\left\\lfloor \\frac{L+2}{3} \\right\\rfloor + 8\\left\\lfloor \\frac{L+1}{3} \\right\\rfloor \\leq L^{2} + L - 2 .\n$$\nBut the left-hand side of this is bounded above by $3L + 10/3$, and the inequality $3L + 10/3 \\leq L^{2} + L - 2$ means exactly $L^{2} - 2L - 16/3 \\geq 0$ or $(L-1)^{2} \\geq 19/3$, which is true for $L \\geq 4$, as desired.\nHence $T_{L}$ is not square for $L \\geq 4$. By direct computation we find $T_{1} = T_{2} = 0$ and $T_{3} = 3$, so $T_{L}$ is square only for $L \\in \\{\\mathbf{1}, \\mathbf{2}\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDenote by $\\langle x\\rangle$ the fractional part of the real number $x$ (for instance, $\\langle 3.2\\rangle=0.2$ ). A positive integer $N$ is selected randomly from the set $\\{1,2,3, \\ldots, M\\}$, with each integer having the same probability of being picked, and $\\left\\langle\\frac{87}{303} N\\right\\rangle$ is calculated. This procedure is repeated $M$ times and the average value $A(M)$ is obtained. What is $\\lim _{M \\rightarrow \\infty} A(M)$ ?",
    "options": [],
    "answer": "50/101",
    "solution": "Solution:\nThis method of picking $N$ is equivalent to uniformly randomly selecting a positive integer. Call this the average value of $\\left\\langle\\frac{87}{303} N\\right\\rangle$ for $N$ a positive integer. In lowest terms, $\\frac{87}{303}=\\frac{29}{101}$, so the answer is the same as the average value of $\\frac{0}{101}, \\frac{1}{101}, \\ldots, \\frac{100}{101}$, which is $\\frac{1+2+\\cdots+100}{101 \\cdot 101}=\\frac{100 \\cdot 101 / 2}{101 \\cdot 101}=\\frac{50}{101}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA rubber band is 4 inches long. An ant begins at the left end. Every minute, the ant walks one inch rightwards along the rubber band, but then the band is stretched (uniformly) by one inch. For what value of $n$ will the ant reach the right end during the $n$th minute?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\nThe ant traverses $1/4$ of the band's length in the first minute, $1/5$ of the length in the second minute (the stretching does not affect its position as a fraction of the band's length), $1/6$ of the length in the third minute, and so on. Since\n$$\n1/4 + 1/5 + \\cdots + 1/9 < 0.25 + 0.20 + 0.167 + 0.143 + 0.125 + 0.112 = 0.997 < 1\n$$\nthe ant does not cover the entire band in six minutes. However,\n$$\n1/4 + \\cdots + 1/10 > 0.25 + 0.20 + 0.16 + 0.14 + 0.12 + 0.11 + 0.10 = 1.08 > 1\n$$\nso seven minutes suffice.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe real numbers $x, y, z, w$ satisfy\n\n\\begin{aligned}\n2x + y + z + w &= 1 \\\\\nx + 3y + z + w &= 2 \\\\\nx + y + 4z + w &= 3 \\\\\nx + y + z + 5w &= 25.\n\\end{aligned}\n\nFind the value of $w$.",
    "options": [],
    "answer": "11/2",
    "solution": "Solution:\n\n$11/2$. Multiplying the four equations by $12, 6, 4, 3$ respectively, we get\n\n\\begin{aligned}\n24x + 12y + 12z + 12w &= 12 \\\\\n6x + 18y + 6z + 6w &= 12 \\\\\n4x + 4y + 16z + 4w &= 12 \\\\\n3x + 3y + 3z + 15w &= 75\n\\end{aligned}\n\nAdding these yields $37x + 37y + 37z + 37w = 111$, or $x + y + z + w = 3$. Subtract this from the fourth given equation to obtain $4w = 22$, or $w = 11/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA manufacturer of airplane parts makes a certain engine that has a probability $p$ of failing on any given flight. There are two planes that can be made with this sort of engine, one that has 3 engines and one that has 5. A plane crashes if more than half its engines fail. For what values of $p$ do the two plane models have the same probability of crashing?",
    "options": [],
    "answer": "0, 1/2, 1",
    "solution": "Solution:\nThey have the same probability of failing if $\\binom{5}{2} p^{3}(1-p)^{2}+\\binom{5}{1} p^{4}(1-p)+p^{5}=\\binom{3}{1} p^{2}(1-p)+p^{3}$, which is true iff $p^{2}\\left(6 p^{3}-15 p^{2}+12 p-3\\right)=0$. This is clearly true for $p=0$. We know it is true for $p=1$, since both probabilities would be 1 in this case, so we know $p-1$ is a factor of $6 p^{3}-15 p^{2}+12 p-3$. Thus, factoring gives that the engines have the same probability of failing if $p^{2}(p-1)\\left(6 p^{2}-9 p+3\\right)=0$. By the quadratic formula (or by factoring), the quadratic has roots $p=\\frac{1}{2}, 1$, so the answer is $0, \\frac{1}{2}, 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEquilateral triangle $A B C$ of side length $2$ is drawn. Three squares external to the triangle, $A B D E$, $B C F G$, and $C A H I$, are drawn. What is the area of the smallest triangle that contains these squares?",
    "options": [],
    "answer": "12 + 7√3",
    "solution": "Solution:\n\nThe equilateral triangle with sides lying on lines $D G$, $E H$, and $F I$ has minimal area. (The only other reasonable candidate is the triangle with sides along $D E$, $F G$, $H I$, but a quick sketch shows that it is larger.) Let $J$, $K$, and $L$ be the vertices of this triangle closest to $D$, $H$, and $F$, respectively. Clearly, $K I = F L = 2$. Triangle $F C I$ is a $30^{\\circ}$-$30^{\\circ}$-$120^{\\circ}$ triangle, so we can calculate the length of $F I$ as $2 \\sqrt{3}$, making the side length of $\\triangle J K L$ $4 + 2 \\sqrt{3}$, and its area $12 + 7 \\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA circle is inscribed in a square dartboard. If a dart is thrown at the dartboard and hits the dartboard in a random location, with all locations having the same probability of being hit, what is the probability that it lands within the circle?",
    "options": [],
    "answer": "π/4",
    "solution": "Solution:\n\nThe answer is the area of the circle over the area of the square, which is $\\frac{\\pi}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTwo fair coins are simultaneously flipped. This is done repeatedly until at least one of the coins comes up heads, at which point the process stops. What is the probability that the other coin also came up heads on this last flip?",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n$\\frac{1}{3}$. Let the desired probability be $p$. There is a $\\frac{1}{4}$ chance that both coins will come up heads on the first toss. Otherwise, both can come up heads simultaneously only if both are tails on the first toss, and then the process restarts as if from the beginning; thus this situation occurs with probability $\\frac{p}{4}$. Thus $p = \\frac{1}{4} + \\frac{p}{4}$; solving, $p = \\frac{1}{3}$.\n\nAlternate Solution:\nThe desired event is equivalent to both coins coming up tails for $n$ successive turns (for some $n \\geq 0$), then both coins coming up heads. For any fixed value of $n$, the probability of this occurring is $\\frac{1}{4^{n+1}}$. Since all these events are disjoint, the total probability is $\\frac{1}{4} + \\frac{1}{4^2} + \\frac{1}{4^3} + \\cdots = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$$\n\\frac{3}{1^{2} \\cdot 2^{2}}+\\frac{5}{2^{2} \\cdot 3^{2}}+\\frac{7}{3^{2} \\cdot 4^{2}}+\\cdots+\\frac{29}{14^{2} \\cdot 15^{2}}.\n$$",
    "options": [],
    "answer": "224/225",
    "solution": "Solution:\n\nThe sum telescopes as\n\n$$\n\\left(\\frac{1}{1^{2}}-\\frac{1}{2^{2}}\\right)+\\left(\\frac{1}{2^{2}}-\\frac{1}{3^{2}}\\right)+\\cdots+\\left(\\frac{1}{14^{2}}-\\frac{1}{15^{2}}\\right)=\\frac{1}{1^{2}}-\\frac{1}{15^{2}}=\\frac{224}{225}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDraw a square of side length $1$. Connect its sides' midpoints to form a second square. Connect the midpoints of the sides of the second square to form a third square. Connect the midpoints of the sides of the third square to form a fourth square. And so forth. What is the sum of the areas of all the squares in this infinite series?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nThe area of the first square is $1$, the area of the second is $\\frac{1}{2}$, the area of the third is $\\frac{1}{4}$, etc., so the answer is $1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\cdots = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a $9 \\times 9$ chess board, we consider all the rectangles whose edges lie along grid lines (the board consists of 81 unit squares, and the grid lines lie on the borders of the unit squares). For each such rectangle, we put a mark in every one of the unit squares inside it. When this process is completed, how many unit squares will contain an even number of marks?",
    "options": [],
    "answer": "56",
    "solution": "Solution:\n\n56. Consider the rectangles which contain the square in the $i$\\text{th} row and $j$\\text{th} column. There are $i$ possible positions for the upper edge of such a rectangle, $10-i$ for the lower edge, $j$ for the left edge, and $10-j$ for the right edge; thus we have $i(10-i) j(10-j)$ rectangles altogether, which is odd iff $i, j$ are both odd, i.e. iff $i, j \\in \\{1,3,5,7,9\\}$. There are thus 25 unit squares which lie in an odd number of rectangles, so the answer is $81-25=56$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCount the number of triangles with positive area whose vertices are points whose $(x, y)$-coordinates lie in the set $\\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\\}$.",
    "options": [],
    "answer": "76",
    "solution": "Solution:\n\nThere are $\\binom{9}{3} = 84$ triples of points. 8 of them form degenerate triangles (the ones that lie on a line), so there are $84 - 8 = 76$ nondegenerate triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate $\\int_{0}^{(\\sqrt{2}-1) / 2} \\frac{\\mathrm{~d} x}{(2 x+1) \\sqrt{x^{2}+x}}$.",
    "options": [],
    "answer": "π/4",
    "solution": "Solution:\n\nLet $u=\\sqrt{x^{2}+x}$. Then $d u=\\frac{2 x+1}{2 \\sqrt{x^{2}+x}} d x$. So the integral becomes $2 \\int \\frac{d u}{4 x^{2}+4 x+1}$, or $2 \\int \\frac{d u}{4 u^{2}+1}$. This is $\\tan^{-1}(2 u)$, yielding a final answer of $\\tan^{-1}\\left(2 \\sqrt{x^{2}+x}\\right)+C$ for the indefinite integral. The definite integral becomes $\\tan^{-1}(1)-\\tan^{-1}(0)=\\frac{\\pi}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $x$ with $0 \\leq x < 2\\pi$ that satisfy $\\sin x + \\cos x = \\sqrt{2}$.",
    "options": [],
    "answer": "π/4",
    "solution": "Solution:\nSquaring both sides gives $\\sin^2 x + \\cos^2 x + 2 \\sin x \\cos x = 1 + \\sin 2x = 2$, so $x = \\frac{\\pi}{4}, \\frac{5\\pi}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor any positive integer $n$, let $f(n)$ denote the number of 1's in the base-2 representation of $n$. For how many values of $n$ with $1 \\leq n \\leq 2002$ do we have $f(n) = f(n+1)$?",
    "options": [],
    "answer": "501",
    "solution": "Solution:\n\n501. If $n$ is even, then $n+1$ is obtained from $n$ in binary by changing the final 0 to a 1; thus $f(n+1) = f(n) + 1$. If $n$ is odd, then $n+1$ is obtained by changing the last 0 to a 1, the ensuing string of 1's to 0's, and then changing the next rightmost 0 to a 1. This produces no net change in the number of 1's iff $n$ ends in 01 in base 2. Thus,\n$f(n+1) = f(n)$ if and only if $n$ is congruent to $1 \\bmod 4$, and there are 501 such numbers in the specified range.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $f$ is a differentiable real function such that $f(x) + f'(x) \\leq 1$ for all $x$, and $f(0) = 0$. What is the largest possible value of $f(1)$? (Hint: consider the function $e^{x} f(x)$.)",
    "options": [],
    "answer": "1 - 1/e",
    "solution": "Solution:\n\n$1 - 1/e$\n\nLet $g(x) = e^{x} f(x)$; then $g'(x) = e^{x} (f(x) + f'(x)) \\leq e^{x}$. Integrating from $0$ to $1$, we have\n$$\ng(1) - g(0) = \\int_{0}^{1} g'(x) \\, dx \\leq \\int_{0}^{1} e^{x} \\, dx = e - 1.\n$$\nBut $g(1) - g(0) = e \\cdot f(1)$, so we get $f(1) \\leq (e - 1)/e$.\n\nThis maximum is attained if we actually have $g'(x) = e^{x}$ everywhere; this entails the requirement $f(x) + f'(x) = 1$, which is met by $f(x) = 1 - e^{-x}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$A$ and $B$ are two points on a circle with center $O$, and $C$ lies outside the circle, on ray $A B$. Given that $A B = 24$, $B C = 28$, $O A = 15$, find $O C$.",
    "options": [],
    "answer": "41",
    "solution": "Solution:\n\nLet $M$ be the midpoint of $A B$; then $\\triangle O M B$ is a right triangle with $O B = 15$, $M B = 12$, so $O M = 9$.\n\nNow $\\triangle O M C$ is a right triangle with $O M = 9$, $M C = 40$, so $O C = 41$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven that $a, b, c$ are positive real numbers and $\\log_{a} b + \\log_{b} c + \\log_{c} a = 0$, find the value of $\\left(\\log_{a} b\\right)^{3} + \\left(\\log_{b} c\\right)^{3} + \\left(\\log_{c} a\\right)^{3}$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nLet $x = \\log_{a} b$ and $y = \\log_{b} c$; then $\\log_{c} a = -(x + y)$. Thus we want to compute the value of $x^{3} + y^{3} + (-(x + y))^{3} = x^{3} + y^{3} - (x + y)^{3} = -3x^{2}y - 3x y^{2} = -3 x y (x + y)$. On the other hand, $-x y (x + y) = \\left(\\log_{a} b\\right) \\left(\\log_{b} c\\right) \\left(\\log_{c} a\\right) = 1$, so the answer is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46765,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a pair of integers $a$ and $b$, with $0 < a < b < 1000$, the set $S \\subseteq \\{1, 2, \\dots, 2003\\}$ is called a *skipping set* for $(a, b)$ if for any pair of elements $s_1, s_2 \\in S$, $|s_1 - s_2| \\notin \\{a, b\\}$. Let $f(a, b)$ be the maximum size of a skipping set for $(a, b)$. Determine the maximum and minimum values of $f$.",
    "options": [],
    "answer": "Maximum f is 1334; minimum f is 668.",
    "solution": "The maximum and minimum values of $f$ are $1334$ and $668$, respectively.\n\na. First, we will show that the maximum value of $f$ is $1334$. The set $S = \\{1, 2, \\dots, 667\\} \\cup \\{1336, 1337, \\dots, 2002\\}$ is a skipping set for $(a, b) = (667, 668)$, so $f(667, 668) \\ge 1334$.\n\nNow we prove that for any $0 < a < b < 1000$, $f(a, b) \\le 1334$. Because $a \\ne b$, we can choose $d \\in \\{a, b\\}$ such that $d \\ne 668$. We assume first that $d \\ge 669$. Then consider the $2003 - d \\le 1334$ sets $\\{1, d+1\\}, \\{2, d+2\\}, \\dots, \\{2003-d, 2003\\}$. Each can contain at most one element of $S$, so $|S| \\le 1334$.\n\nWe assume second that $d \\le 667$ and that $\\lceil \\frac{2003}{a} \\rceil$ is even, that is, $\\lceil \\frac{2003}{a} \\rceil = 2k$ for some positive integer $k$. Then each of the congruence classes of $1, 2, \\dots, 2003$ modulo $a$ contains at most $2k$ elements. Therefore at most $k$ members of each of these congruence classes can belong to $S$. Consequently,\n$$\n\\begin{aligned}\n|S| &\\le ka < \\frac{1}{2} \\left( \\frac{2003}{a} + 1 \\right) a = \\frac{2003 + a}{2} \\\\\n&\\le 1335,\n\\end{aligned}\n$$\nimplying that $|S| \\le 1334$.\n\nFinally, we assume that $d \\le 667$ and that $\\lceil \\frac{2003}{a} \\rceil$ is odd, that is, $\\lceil \\frac{2003}{a} \\rceil = 2k + 1$ for some positive integer $k$. Then, as before, $S$ can contain at most $k$ elements from each congruence class of $\\{1, 2, \\dots, 2ka\\}$ modulo $a$. Then\n$$\n\\begin{align*}\n|S| &\\le ka + (2003 - 2ka) = 2003 - ka \\\\\n&= 2003 - \\left( \\frac{\\left\\lfloor \\frac{2003}{a} \\right\\rfloor}{2} - 1 \\right) a \\\\\n&\\le 2003 - \\left( \\frac{\\frac{2003}{a} - 1}{2} \\right) a \\\\\n&= \\frac{2003 + a}{2} \\le 1335.\n\\end{align*}\n$$\nThe last inequality holds if and only if $a = 667$. But if $a = 667$, then $\\frac{2003}{a}$ is not an integer, and so the second inequality is strict. Thus, $|S| \\le 1334$. Therefore the maximum value of $f$ is $1334$.\n\nb. We will now show that the minimum value of $f$ is $668$. First, we will show that $f(a, b) \\ge 668$ by constructing a skipping set $S$ for any $(a, b)$ with $|S| \\ge 668$. Note that if we add $x$ to $S$, then we are not allowed to add $x$, $x+a$, or $x+b$ to $S$ at any later time. Then at each step, let us add to $S$ the smallest element of $\\{1, 2, \\dots, 2003\\}$ that is not already in $S$ and that has not already been disallowed from being in $S$. Then since adding this element prevents at most three elements from being added at any future time, we can always perform this step $\\lceil \\frac{2003}{3} \\rceil = 668$ times. Thus, $|S| \\ge 668$, so $f(a, b) \\ge 668$.\n\nNow notice that if we let $a = 1, b = 2$, then at most one element from each of the $668$ sets $\\{1, 2, 3\\}, \\{4, 5, 6\\}, \\dots, \\{1999, 2000, 2001\\}, \\{2002, 2003\\}$ can belong to $S$. This implies that $f(1, 2) = 668$, so indeed the minimum value of $f$ is $668$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We proceed by induction. The property is clearly true for $n=1$. Assume that $N = a_1a_2\\dots a_n$ is divisible by $5^n$ and has only odd digits. Consider the numbers\n$$\nN_1 = 1a_1a_2\\dots a_n = 1 \\cdot 10^n + 5^nM = 5^n(1 \\cdot 2^n + M),\n$$\n$$\nN_2 = 3a_1a_2\\dots a_n = 3 \\cdot 10^n + 5^nM = 5^n(3 \\cdot 2^n + M),\n$$\n$$\nN_3 = 5a_1a_2\\dots a_n = 5 \\cdot 10^n + 5^nM = 5^n(5 \\cdot 2^n + M),\n$$\n$$\nN_4 = 7a_1a_2\\dots a_n = 7 \\cdot 10^n + 5^nM = 5^n(7 \\cdot 2^n + M),\n$$\n$$\nN_5 = 9a_1a_2\\dots a_n = 9 \\cdot 10^n + 5^nM = 5^n(9 \\cdot 2^n + M).\n$$\nThe numbers $1 \\cdot 2^n + M, 3 \\cdot 2^n + M, 5 \\cdot 2^n + M, 7 \\cdot 2^n + M, 9 \\cdot 2^n + M$ give distinct remainders when divided by $5$. Otherwise the difference of some two of them would be a multiple of $5$, which is impossible, because neither $2^n$ is a multiple of $5$, nor is the difference of any two of the numbers $1, 3, 5, 7, 9$. It follows that one of the numbers $N_1, N_2, N_3, N_4, N_5$ is divisible by $5^n \\cdot 5$, and the induction is complete.\nFor an $m$ digit number $a$, where $m \\ge n$, let $\\ell(a)$ denote the $m-n$ leftmost digits of $a$. (That is, we consider $\\ell(a)$ as an $(m-n)$-digit number.) It is clear that we can choose a large odd number $k$ such that $a_0 = 5^n \\cdot k$ has at least $n$ digits. Assume that $a_0$ has $m_0$ digits, where $m_0 \\ge n$. Note that the $a_0$ is an odd multiple of $5$. Hence the unit digit of $a_0$ is $5$.\n\nIf the $n$ rightmost digits of $a_0$ are all odd, then the number $b_0 = a_0 - \\ell(a_0) \\cdot 10^n$ satisfies the conditions of the problem, because $b_0$ has only odd digits (the same as those $n$ leftmost digits of $a_0$) and that $b_0$ is the difference of two multiples of $5^n$.\n\nIf there is an even digit among the $n$ rightmost digits of $a_0$, assume that $i_1$ is the smallest positive integer such that the $i_1$th rightmost digit of $a_0$ is even. Then number $a_1 = a_0 + 5^n \\cdot 10^{i_1-1}$ is a multiple of $5^n$ with at least $n$ digits. The $(i-1)$th rightmost digit is the same as that of $a_0$ and the $i_1$th rightmost digit of $a_1$ is odd. If the $n$ rightmost digits of $a_1$ are all odd, then $b_1 = a_1 - \\ell(a_1) \\cdot 10^n$ satisfies the conditions of the problem. If there is an even digit among the $n$ rightmost digits of $a_1$, assume that $i_2$ is the smallest positive integer such that the $i_2$th rightmost digit of $a_1$ is even. Then $i_2 > i_1$. Set $a_2 = a_1 + 5^n \\cdot 10^{i_2-1}$. We can repeat the above process of checking the rightmost digits of $a_2$ and eliminate the rightmost even digits of $a_2$, if there is such a digit among the $n$ rightmost digits of $a_2$. This process can be repeated for at most $n-1$ times because the unit digit of $a_0$ is $5$. Thus, we can obtain a number $a_k$, for some nonnegative integer $k$, such that $a_k$ is a multiple of $5^n$ with its $n$ rightmost digits all odd. Then $b_k = a_k - \\ell(a_k) \\cdot 10^n$ is a number that satisfies the conditions of the problem.\nConsider all the nonnegative multiples of $5^n$ that have no more than $n$ digits. There are $2^n$ such multiples, namely, $m_0 = 0, m_1 = 5^n, m_2 = 2 \\cdot 5^n, \\dots, m_{2^{n-1}} = (2^n - 1)5^n$. For each $m_i$, we define an $n$-digit binary string $s(m_i)$. If $m_i$ is a $k_i$-digit number, the leftmost $n-k_i$ digits of $s(m_i)$ are all $0$'s, and the $j$th digit, $1 \\le j \\le k_i$, of $s(m_i)$ is $1$ (or $0$) if the $j$th rightmost digit of $m_i$ is odd (or even). (For example, for $n=4$, $m_0 = 0, m_1 = 625, m_2 = 1250$, and $s(m_0) = s(m_1) = s(m_2) = 1010$.) There are $2^n$ $n$-digit binary strings. It suffices to show that $s$ is one-to-one, that is, $s(m_i) \\ne s(m_j)$ for $i \\ne j$. Because then there must be a $m_i$ with $s(m_i)$ being a string of $n$ $1$'s, that is, $m_i$ has $n$ digits and all of them are odd.\n\nWe write $m_i$ and $m_j$ in binary system. Then there is a smallest positive integer $k$ such that the $k$th rightmost digit in the binary representations of $m_i$ and $m_j$ are different. Without loss of generality, we assume that those $k$th digits for $m_i$ and $m_j$ are $1$ and $0$, respectively. Then $m_i = s_i + 2^k + t$ and $m_j = s_j + t$, where $s_i, s_j, t$ are positive integers such that $2^{k+1}$ divides both $s_i$ and $s_j$ and that $0 \\le t \\le 2^k - 1$. Note that adding $2^k \\cdot 5^n$ to $t \\cdot 5^n$ will change the parity of the $(k+1)$th rightmost digit of $t \\cdot 5^n$ while not affect the $k$ rightmost digits of $t \\cdot 5^n$. Note also that adding $s_i \\cdot 5^n$ (or $s_j \\cdot 5^n$) to $(2^k + t) \\cdot 5^n$ (or $t \\cdot 5^n$) will not affect the last $k+1$ digits of $(2^k + t) \\cdot 5^n$ (or $t \\cdot 5^n$). Hence the $(k+1)$th rightmost digits in the decimal representations of $m_i \\cdot 5^n$ and $m_j \\cdot 5^n$ have different parities. Thus $s(m_i) \\neq s(m_j)$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn integer is a perfect number if and only if it is equal to the sum of all of its divisors except itself. For example, $28$ is a perfect number since $28 = 1 + 2 + 4 + 7 + 14$.\n\nLet $n!$ denote the product $1 \\cdot 2 \\cdot 3 \\cdots n$, where $n$ is a positive integer. An integer is a factorial if and only if it is equal to $n!$ for some positive integer $n$. For example, $24$ is a factorial number since $24 = 4! = 1 \\cdot 2 \\cdot 3 \\cdot 4$.\n\nFind all perfect numbers greater than $1$ that are also factorials.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nThe only perfect factorial is $6 = 3!$. Certainly, $2! = 2$ is not perfect. For $n > 3$, note that $n! = 6k$, where $k > 1$, and thus the factors of $n!$ will include $1, k, 2k, 3k$. This sums to $6k + 1$, showing that $n!$ is not perfect.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) One Sunday, Zvezda wrote 14 numbers in a circle, so that each number is equal to the sum of its two neighbors. Prove that the sum of all 14 numbers is 0.\n\nb) On the next Sunday, Zvezda wrote 21 numbers in a circle, and this time each number was equal to half the sum of its two neighbors. What is the sum of all 21 numbers, if one of the numbers is 3?",
    "options": [],
    "answer": "a) 0; b) 63",
    "solution": "Solution:\na)\nDenoting the numbers $a_{1}, a_{2}, \\ldots, a_{14}$, and their sum as $S$ we have $a_{i} = a_{i-1} + a_{i+1}$ for $i = 1, \\ldots, 14$ (we take $a_{15} = a_{1}, a_{0} = a_{14}$). Summing all these equalities we get $S = 2S$ (since each $a_{i}$ appears exactly once on the left and exactly twice on the right). Therefore $S = 0$.\n\nb)\nAgain, let the numbers be $a_{1}, \\ldots, a_{21}$. Let $a_{i}$ be maximal among the numbers. Then we have $a_{i} \\geq a_{i-1}, a_{i} \\geq a_{i+1}$ and so $a_{i} \\geq \\frac{a_{i-1} + a_{i+1}}{2}$. But $a_{i} = \\frac{a_{i-1} + a_{i+1}}{2}$, so both inequalities are actually equalities. Next, considering now $a_{i-1}$ we conclude $a_{i+1} = a_{i+2}$. Proceeding in this way, we conclude that all the numbers are equal. As one of them is $3$, the sum of all numbers is $3 \\times 21 = 63$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest value of $x$ such that $a \\geq 14 \\sqrt{a} - x$ for all nonnegative $a$.",
    "options": [],
    "answer": "49",
    "solution": "Solution:\nWe want to find the smallest value of $x$ such that $x \\geq 14 \\sqrt{a} - a$ for all $a$. This is just the maximum possible value of $14 \\sqrt{a} - a = 49 - (\\sqrt{a} - 7)^2$, which is clearly $49$, achieved when $a = 49$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n10 people are playing musical chairs with $n$ chairs in a circle. They can be seated in $7!$ ways (assuming only one person fits on each chair, of course), where different arrangements of the same people on chairs, even rotations, are considered different. Find $n$.",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nThe number of ways 10 people can be seated on $n$ chairs is $n!$ multiplied by the number of ways one can choose $n$ people out of 10. Hence we must solve $7! = n! \\cdot \\dfrac{10!}{n! \\cdot (10-n)!}$. This is equivalent to $(10-n)! = \\dfrac{10!}{7!} = 8 \\cdot 9 \\cdot 10 = 720 = 6!$. We therefore have $n = 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$AD$ and $BC$ are both perpendicular to $AB$, and $CD$ is perpendicular to $AC$. If $AB = 4$ and $BC = 3$, find $CD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "20/3",
    "solution": "Solution:\n\nBy Pythagoras in $\\triangle ABC$, $AC = 5$. But $\\angle CAD = 90^{\\circ} - \\angle BAC = \\angle ACB$, so right triangles $CAD$ and $BCA$ are similar, and $\\frac{CD}{AC} = \\frac{BA}{CB} = \\frac{4}{3} \\Rightarrow CD = \\frac{20}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou have $2003$ switches, numbered from $1$ to $2003$, arranged in a circle. Initially, each switch is either ON or OFF, and all configurations of switches are equally likely. You perform the following operation: for each switch $S$, if the two switches next to $S$ were initially in the same position, then you set $S$ to ON; otherwise, you set $S$ to OFF. What is the probability that all switches will now be ON?",
    "options": [],
    "answer": "1/2^{2002}",
    "solution": "Solution:\n$1 / 2^{2002}$\n\nThere are $2^{2003}$ equally likely starting configurations. All switches end up ON if and only if switches $1,3,5,7, \\ldots, 2003,2,4, \\ldots, 2002$—i.e., all $2003$ of them—were initially in the same position. This initial position can be ON or OFF, so this situation occurs with probability $2 / 2^{2003} = 1 / 2^{2002}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSimplify $\\sqrt[2003]{2 \\sqrt{11}-3 \\sqrt{5}} \\cdot \\sqrt[4006]{89+12 \\sqrt{55}}$.",
    "options": [],
    "answer": "-1",
    "solution": "Solution:\nNote that $(2 \\sqrt{11}+3 \\sqrt{5})^{2}=89+12 \\sqrt{55}$. So, we have\n$$\n\\begin{aligned}\n\\sqrt[2003]{2 \\sqrt{11}-3 \\sqrt{5}} \\cdot \\sqrt[4006]{89+12 \\sqrt{55}} & =\\sqrt[2003]{2 \\sqrt{11}-3 \\sqrt{5}} \\cdot \\sqrt[2003]{2 \\sqrt{11}+3 \\sqrt{5}} \\\\\n& =\\sqrt[2003]{(2 \\sqrt{11})^{2}-(3 \\sqrt{5})^{2}}=\\sqrt[2003]{-1}=-1\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA point is chosen randomly with uniform distribution in the interior of a circle of radius $1$. What is its expected distance from the center of the circle?",
    "options": [],
    "answer": "2/3",
    "solution": "Solution:\nThe probability of the point falling between a distance $r$ and $r + d r$ from the center is the ratio of the area of the corresponding annulus to the area of the whole circle: $$\\frac{\\pi\\left[(r + d r)^2 - r^2\\right]}{\\pi} \\rightarrow \\frac{2 \\pi r d r}{\\pi} = 2 r d r$$ for small values of $d r$.\n\nThen the expected distance is\n$$\n\\int_{0}^{1} r \\cdot 2 r d r = 2 \\int_{0}^{1} r^2 d r = 2 \\left[ \\frac{r^3}{3} \\right]_0^1 = 2 \\cdot \\frac{1}{3} = \\frac{2}{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA compact disc has the shape of a circle of diameter 5 inches with a 1-inch-diameter circular hole in the center. Assuming the capacity of the CD is proportional to its area, how many inches would need to be added to the outer diameter to double the capacity?",
    "options": [],
    "answer": "2 inches",
    "solution": "Solution:\nDoubling the capacity is equivalent to doubling the area, which is initially $\\pi\\left[(5 / 2)^2 - (1 / 2)^2\\right] = 6\\pi$. Thus we want to achieve an area of $12\\pi$, so if the new diameter is $d$, we want $\\pi\\left[(d / 2)^2 - (1 / 2)^2\\right] = 12\\pi \\Rightarrow d = 7$. Thus we need to add 2 inches to the diameter.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeveral positive integers are given, not necessarily all different. Their sum is $2003$. Suppose that $n_{1}$ of the given numbers are equal to $1$, $n_{2}$ of them are equal to $2$, $\\ldots$, $n_{2003}$ of them are equal to $2003$. Find the largest possible value of\n$$\nn_{2}+2 n_{3}+3 n_{4}+\\cdots+2002 n_{2003}\n$$",
    "options": [],
    "answer": "2002",
    "solution": "Solution:\n\nThe sum of all the numbers is $n_{1}+2 n_{2}+\\cdots+2003 n_{2003}$, while the number of numbers is $n_{1}+n_{2}+\\cdots+n_{2003}$. Hence, the desired quantity equals\n$$\n\\begin{gathered}\n\\left(n_{1}+2 n_{2}+\\cdots+2003 n_{2003}\\right)-\\left(n_{1}+n_{2}+\\cdots+n_{2003}\\right) \\\\\n=(\\text{sum of the numbers})-(\\text{number of numbers})\n\\end{gathered}\n$$\n$$\n=2003-\\text{(number of numbers)},\n$$\nwhich is maximized when the number of numbers is minimized. Hence, we should have just one number, equal to $2003$, and then the specified sum is $2003-1=2002$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $P(x)$ is a polynomial such that $P(1) = 1$ and\n\n$$\n\\frac{P(2x)}{P(x+1)} = 8 - \\frac{56}{x+7}\n$$\n\nfor all real $x$ for which both sides are defined. Find $P(-1)$.",
    "options": [],
    "answer": "-5/21",
    "solution": "Solution:\nCross-multiplying gives $(x+7) P(2x) = 8(x+7) P(x+1) - 56 P(x+1) = 8(x+7) P(x+1) - 56 P(x+1)$. But more simply, rearrange:\n\n$\\frac{P(2x)}{P(x+1)} = 8 - \\frac{56}{x+7}$\n\n$\\Rightarrow \\frac{P(2x)}{P(x+1)} = \\frac{8(x+7) - 56}{x+7} = \\frac{8x + 56 - 56}{x+7} = \\frac{8x}{x+7}$\n\nSo:\n\n$\\frac{P(2x)}{P(x+1)} = \\frac{8x}{x+7}$\n\nCross-multiplied:\n\n$(x+7) P(2x) = 8x P(x+1)$\n\nSuppose $P(x)$ has degree $n$ and leading coefficient $c$. Then $P(2x)$ has leading term $c (2x)^n = c 2^n x^n$, and $P(x+1)$ has leading term $c x^n$. So the left side's leading term is $x \\cdot c 2^n x^n = c 2^n x^{n+1}$, and the right side's leading term is $8x \\cdot c x^n = 8c x^{n+1}$. Equate:\n\n$c 2^n = 8c \\implies 2^n = 8 \\implies n = 3$.\n\nSo $P(x)$ is cubic.\n\nNow, if $x=0$, the right side is $0$, so the left side is $7 P(0) = 0 \\implies P(0) = 0$.\n\nSo $P(x) = x Q(x)$ for some quadratic $Q(x)$.\n\nPlug into the original equation:\n\n$P(2x) = 2x Q(2x)$, $P(x+1) = (x+1) Q(x+1)$\n\nSo:\n\n$\\frac{2x Q(2x)}{(x+1) Q(x+1)} = \\frac{8x}{x+7}$\n\nCross-multiplied:\n\n$(x+7) 2x Q(2x) = 8x (x+1) Q(x+1)$\n\nIf $x=0$, both sides are $0$.\n\nDivide both sides by $x$ (for $x \\neq 0$):\n\n$2(x+7) Q(2x) = 8(x+1) Q(x+1)$\n\nOr:\n\n$(x+7) Q(2x) = 4(x+1) Q(x+1)$\n\nNow, try $x = -1$:\n\n$(-1+7) Q(-2) = 4(0) Q(0) \\implies 6 Q(-2) = 0 \\implies Q(-2) = 0$\n\nSo $Q(x)$ has root at $x = -2$, so $Q(x) = (x+2) R(x)$ for some linear $R(x)$.\n\nNow, $Q(2x) = (2x+2) R(2x)$, $Q(x+1) = (x+3) R(x+1)$\n\nPlug into the previous equation:\n\n$(x+7) (2x+2) R(2x) = 4(x+1)(x+3) R(x+1)$\n\nDivide both sides by $2$:\n\n$(x+7)(x+1) R(2x) = 2(x+1)(x+3) R(x+1)$\n\nIf $x = -3$:\n\n$(-3+7)(-3+1) R(-6) = 2(-3+1)(-3+3) R(-2)$\n\n$(4)(-2) R(-6) = 2(-2)(0) R(-2) = 0$\n\nSo $R(-6) = 0$\n\nSo $R(x)$ has root at $x = -6$, so $R(x) = (x+6) S(x)$, $S(x)$ constant.\n\nTherefore,\n\n$P(x) = x(x+2)(x+6) S$\n\nSince $P(x)$ is cubic, $S$ is a constant.\n\nGiven $P(1) = 1$:\n\n$P(1) = 1 \\cdot 3 \\cdot 7 \\cdot S = 21 S = 1 \\implies S = \\frac{1}{21}$\n\nTherefore,\n\n$P(x) = \\frac{1}{21} x(x+2)(x+6)$\n\nSo,\n\n$P(-1) = \\frac{1}{21} \\cdot (-1) \\cdot 1 \\cdot 5 = \\frac{-5}{21}$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that every consistent 2-configuration of order 4 on a finite set $A$ has a subset that is a consistent 2-configuration of order 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst, assume the 2-configuration has just one cell. We claim there exists a sequence $a_{0}, a_{1}, \\ldots, a_{n}$ of elements of $A$ (not necessarily all distinct) such that the list\n$$\n\\{a_{0}, a_{1}\\}, \\{a_{1}, a_{2}\\}, \\ldots, \\{a_{n-1}, a_{n}\\}, \\{a_{n}, a_{0}\\}\n$$\ncontains each element of the 2-configuration exactly once. To see this, consider the longest sequence such that $\\{a_{0}, a_{1}\\}, \\ldots, \\{a_{n-1}, a_{n}\\}, \\{a_{n}, a_{0}\\}$ are all distinct elements of the 2-configuration. (We may take $n=0$ if necessary. Note that the finiteness condition ensures such a maximal sequence exists.) Each element of $A$ occurs an even number of times among these pairs (since each occurrence in the sequence contributes to two pairs). If every element occurs 4 times or 0 times, then the elements occurring in the sequence form a cell, since they cannot occur in any other pairs in the 2-configuration. Hence, they are all of $A$, and our sequence uses all the pairs in the 2-configuration, so the claim follows. Otherwise, there is some element $a_{i}$ occurring exactly twice. Choose $b_{1}$ so that $\\{a_{i}, b_{1}\\}$ is one of the two pairs in the 2-configuration not used by our sequence. Then choose $b_{2}$ so that $\\{b_{1}, b_{2}\\}$ is another pair not used thus far. Continue in this manner, choosing new elements $b_{k}$ with $\\{b_{k}, b_{k+1}\\}$ a pair not already used, until we reach a point where finding another unused pair is impossible. Now, our pairs so far are\n$$\n\\begin{gathered}\n\\{a_{0}, a_{1}\\}, \\ldots, \\{a_{n-1}, a_{n}\\}, \\{a_{n}, a_{0}\\}, \\\\\n\\{a_{i}, b_{1}\\}, \\{b_{1}, b_{2}\\}, \\ldots, \\{b_{k-1}, b_{k}\\} .\n\\end{gathered}\n$$\nEvery element is used in an even number of these pairs, except possibly $a_{i}$, which is used in three pairs, and $b_{k}$, which is used in an odd number of pairs (so one or three) unless $a_{i}=b_{k}$, in which case this element occurs four times. But since it is impossible to continue the sequence, $b_{k}$ must indeed have been used four times, so $b_{k}=a_{i}$.\nBut now we can construct the following sequence of distinct elements of the 2-configuration:\n$$\n\\begin{gathered}\n\\{a_{0}, a_{1}\\}, \\ldots, \\{a_{i-1}, a_{i}\\}, \\{a_{i}, b_{1}\\}, \\{b_{1}, b_{2}\\}, \\ldots, \\{b_{k-1}, a_{i}\\}, \\\\\n\\{a_{i}, a_{i+1}\\}, \\ldots, \\{a_{n-1}, a_{n}\\}, \\{a_{n}, a_{0}\\} .\n\\end{gathered}\n$$\nThis contradicts the maximality of our original sequence. This contradiction means that our original sequence must have used all the pairs in the 2-configuration, after all.\nSo we can express the 2-configuration via such a sequence of pairs, where each pair's second element equals the first element of the next pair. If $A$ has $n$ elements, then (since each element appears in four pairs) we have $2n$ pairs. So we can choose the 1st, 3rd, 5th, $\\ldots, (2n-1)$th pairs, and then each element of $A$ belongs to just two of these pairs, because each occurrence of the element as an $a_{i}$ contributes to two consecutive pairs from our original sequence (or the first and last such pairs). Thus, we have our consistent 2-configuration of order 2, as desired.\nFinally, if $A$ consists of more than one cell, then the pairs within any given cell form a consistent 2-configuration of order 4 on that cell. So we simply apply the above procedure to obtain a consistent 2-configuration of order 2 on each cell, and then combining these gives a consistent 2-configuration of order 2 on $A$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate\n\n$$\n\\int_{-\\infty}^{\\infty} \\frac{1-x^{2}}{1+x^{4}} d x\n$$",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\n$0$\n\nLet $S=\\int_{0}^{\\infty} \\frac{1}{x^{4}+1} d x$; note that the integral converges absolutely. Substituting $x=1/u$, so that $d x=-1/u^{2} d u$, we have\n\n$$\n\\begin{gathered}\nS=\\int_{0}^{\\infty} \\frac{1}{1+x^{4}} d x=\\int_{\\infty}^{0} \\frac{1}{1+u^{-4}} \\frac{d u}{-u^{2}}=\\int_{\\infty}^{0} \\frac{-u^{2}}{u^{4}+1} d u \\\\\n=\\int_{0}^{\\infty} \\frac{u^{2}}{1+u^{4}} d u=\\int_{0}^{\\infty} \\frac{x^{2}}{1+x^{4}} d x\n\\end{gathered}\n$$\n\n(the manipulations are justified by absolute convergence), from which we see that\n$$\n\\int_{0}^{\\infty}\\frac{1-x^{2}}{1+x^{4}} d x=0.\n$$\nSince the integrand is an even function, it follows that the integral from $-\\infty$ to $\\infty$ is zero as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that $A, B, C, D$ are four points in the plane, and let $Q, R, S, T, U, V$ be the respective midpoints of $A B, A C, A D, B C, B D, C D$. If $Q R = 2001$, $S U = 2002$, $T V = 2003$, find the distance between the midpoints of $Q U$ and $R V$.",
    "options": [],
    "answer": "2001",
    "solution": "Solution:\nThis problem has far more information than necessary: $Q R$ and $U V$ are both parallel to $B C$, and $Q U$ and $R V$ are both parallel to $A D$. Hence, $Q U V R$ is a parallelogram, and the desired distance is simply the same as the side length $Q R$, namely $2001$. (See figure, next page.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA calculator has a display, which shows a nonnegative integer $N$, and a button, which replaces $N$ by a random integer chosen uniformly from the set $\\{0,1, \\ldots, N-1\\}$, provided that $N>0$. Initially, the display holds the number $N=2003$. If the button is pressed repeatedly until $N=0$, what is the probability that the numbers $1,10,100$, and $1000$ will each show up on the display at some point?",
    "options": [],
    "answer": "1/2224222",
    "solution": "Solution:\nFirst, we claim that if the display starts at some $N$, the probability that any given number $M<N$ will appear at some point is $1/(M+1)$. We can show this by induction on $N$.\n\nIf $N = M+1$ (the base case), $M$ can only be reached if it appears after the first step, and this occurs with probability $1/N = 1/(M+1)$.\n\nIf $N > M+1$ and the claim holds for $N-1$, then there are two possibilities starting from $N$. If the first step leads to $N-1$ (this occurs with probability $1/N$), the probability of seeing $M$ subsequently is $1/(M+1)$ by the induction hypothesis. If the first step leads to something less than $N-1$ (probability $(N-1)/N$), then it leads to any of the integers $\\{0,1, \\ldots, N-2\\}$ with equal probability. But this is exactly what the first step would have been if we had started from $N-1$; hence, the probability of seeing $M$ is again $1/(M+1)$ by induction. Thus, the overall probability of seeing $M$ is\n$$\n\\frac{1}{N} \\cdot \\frac{1}{M+1} + \\frac{N-1}{N} \\cdot \\frac{1}{M+1} = \\frac{1}{M+1},\n$$\nproving the induction step and the claim.\n\nNow let $P(N, M)$ ($M<N$) be the probability of eventually seeing the number $M$ if we start at $N$; note that this is the same as the conditional probability of seeing $M$ given that we see $N$. Hence, the desired probability is\n$$\nP(2003,1000) \\cdot P(1000,100) \\cdot P(100,10) \\cdot P(10,1) = \\frac{1}{1001} \\cdot \\frac{1}{101} \\cdot \\frac{1}{11} \\cdot \\frac{1}{2} = \\frac{1}{2224222}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConvex quadrilateral $M A T H$ is given with $H M / M T = 3 / 4$, and $\\angle A T M = \\angle M A T = \\angle A H M = 60^{\\circ}$. $N$ is the midpoint of $M A$, and $O$ is a point on $T H$ such that lines $M T$, $A H$, $N O$ are concurrent. Find the ratio $H O / O T$.",
    "options": [],
    "answer": "9/16",
    "solution": "Solution:\n\n$\\triangle M A T$ is equilateral, so $H M / A T = H M / M T = 3 / 4$. Also, $\\angle A H M = \\angle A T M$, so the quadrilateral is cyclic. Now, let $P$ be the intersection of $M T$, $A H$, $N O$. Extend $M H$ and $N O$ to intersect at point $Q$. Then by Menelaus's theorem, applied to triangle $A H M$ and line $Q N P$, we have\n$$\n\\frac{H Q}{Q M} \\cdot \\frac{M N}{N A} \\cdot \\frac{A P}{P H} = 1\n$$\nwhile applying the same theorem to triangle $T H M$ and line $Q P O$ gives\n$$\n\\frac{H Q}{Q M} \\cdot \\frac{M P}{P T} \\cdot \\frac{T O}{O H} = 1\n$$\nCombining gives $H O / O T = (M P / P T) \\cdot (A N / N M) \\cdot (H P / P A) = (M P / P A) \\cdot (H P / P T)$ (because $A N / N M = 1$). But since $M A T H$ is cyclic, $\\triangle A P T \\sim \\triangle M P H$, so $M P / P A = H P / P T = H M / A T = 3 / 4$, and the answer is $(3 / 4)^2 = 9 / 16$. (See figure.)\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBessie the cow is trying to navigate her way through a field. She can travel only from lattice point to adjacent lattice point, can turn only at lattice points, and can travel only to the east or north. (A lattice point is a point whose coordinates are both integers.) $(0,0)$ is the southwest corner of the field. $(5,5)$ is the northeast corner of the field. Due to large rocks, Bessie is unable to walk on the points $(1,1)$, $(2,3)$, or $(3,2)$. How many ways are there for Bessie to travel from $(0,0)$ to $(5,5)$ under these constraints?",
    "options": [],
    "answer": "32",
    "solution": "Solution: 32\nIn the figure, each point is labeled with the number of ways to reach that point. The numbers are successively computed as follows: The point $(0,0)$ can trivially be reached in 1 way. When Bessie reaches any subsequent point $(x, y)$ (other than a rock), she can arrive either via a northward or an eastward step, so the number of ways she can reach that point equals the number of ways of reaching $(x-1, y)$ plus the number of ways of reaching $(x, y-1)$. By iterating this calculation, we eventually find that $(5,5)$ can be reached in 32 ways.\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x y = 5$ and $x^{2} + y^{2} = 21$, compute $x^{4} + y^{4}$.",
    "options": [],
    "answer": "391",
    "solution": "Solution:\nWe have $441 = (x^{2} + y^{2})^{2} = x^{4} + y^{4} + 2(x y)^{2} = x^{4} + y^{4} + 50$, yielding $x^{4} + y^{4} = 391$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA positive integer will be called \"sparkly\" if its smallest (positive) divisor, other than $1$, equals the total number of divisors (including $1$). How many of the numbers $2, 3, \\ldots, 2003$ are sparkly?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nSuppose $n$ is sparkly; then its smallest divisor other than $1$ is some prime $p$. Hence, $n$ has $p$ divisors. However, if the full prime factorization of $n$ is $p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{r}^{e_{r}}$, the number of divisors is $\\left(e_{1}+1\\right)\\left(e_{2}+1\\right) \\cdots\\left(e_{r}+1\\right)$. For this to equal $p$, only one factor can be greater than $1$, so $n$ has only one prime divisor - namely $p$ - and we get $e_{1}=p-1 \\Rightarrow n=p^{p-1}$. Conversely, any number of the form $p^{p-1}$ is sparkly. There are just three such numbers in the desired range $\\left(2^{1}, 3^{2}, 5^{4}\\right)$, so the answer is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe product of the digits of a 5-digit number is $180$. How many such numbers exist?",
    "options": [],
    "answer": "360",
    "solution": "Solution:\nLet the digits be $a, b, c, d, e$. Then $a b c d e = 180 = 2^{2} \\cdot 3^{2} \\cdot 5$. We observe that there are 6 ways to factor $180$ into digits $a, b, c, d, e$ (ignoring differences in ordering):\n$$\n180 = 1 \\cdot 1 \\cdot 4 \\cdot 5 \\cdot 9 = 1 \\cdot 1 \\cdot 5 \\cdot 6 \\cdot 6 = 1 \\cdot 2 \\cdot 2 \\cdot 5 \\cdot 9 = 1 \\cdot 2 \\cdot 3 \\cdot 5 \\cdot 6 = 1 \\cdot 3 \\cdot 3 \\cdot 4 \\cdot 5 = 2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 5.\n$$\nThere are (respectively) $60, 30, 60, 120, 60$, and $30$ permutations of these breakdowns, for a total of $360$ numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat fraction of the area of a regular hexagon of side length $1$ is within distance $\\frac{1}{2}$ of at least one of the vertices?",
    "options": [],
    "answer": "π√3/9",
    "solution": "Solution:\n$\\text{The hexagon has area } 6\\left(\\frac{\\sqrt{3}}{4}\\right)(1)^2 = \\frac{3\\sqrt{3}}{2}$. The region we want consists of six $120^\\circ$ arcs of circles of radius $\\frac{1}{2}$, which can be reassembled into two circles of radius $\\frac{1}{2}$. So its area is $\\frac{\\pi}{2}$, and the ratio of areas is $\\frac{\\pi \\sqrt{3}}{9}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are 10 cities in a state, and some pairs of cities are connected by roads. There are 40 roads altogether. A city is called a \"hub\" if it is directly connected to every other city. What is the largest possible number of hubs?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nIf there are $h$ hubs, then $\\binom{h}{2}$ roads connect the hubs to each other, and each hub is connected to the other $10-h$ cities; we thus get $\\binom{h}{2} + h(10-h)$ distinct roads. So, $40 \\geq \\binom{h}{2} + h(10-h) = -h^{2}/2 + 19h/2$, or $80 \\geq h(19-h)$. The largest $h \\leq 10$ satisfying this condition is $h=6$, and conversely, if we connect each of 6 cities to every other city and place the remaining $40 - [\\binom{6}{2} + 6(10-6)] = 1$ road wherever we wish, we can achieve 6 hubs. So 6 is the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of the reciprocals of all the (positive) divisors of $144$.",
    "options": [],
    "answer": "403/144",
    "solution": "Solution:\nAs $d$ ranges over the divisors of $144$, so does $144 / d$, so the sum of $1 / d$ is $1 / 144$ times the sum of the divisors of $144$. Using the formula for the sum of the divisors of a number (or just counting them out by hand), we get that this sum is $403$, so the answer is $403/144$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $r, s, t$ be the solutions to the equation $x^{3}+a x^{2}+b x+c=0$. What is the value of $(r s)^{2}+(s t)^{2}+(r t)^{2}$ in terms of $a, b$, and $c$?",
    "options": [],
    "answer": "b^2 - 2 a c",
    "solution": "Solution:\n$b^{2}-2 a c$\n\nWe have $(x-r)(x-s)(x-t)=x^{3}+a x^{2}+b x+c$, so\n$$\na=-(r+s+t), \\quad b=r s+s t+r t, \\quad c=-r s t .\n$$\nSo we have\n$$\n(r s)^{2}+(s t)^{2}+(r t)^{2}=(r s+s t+r t)^{2}-2 r s t(r+s+t)=b^{2}-2 a c .\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46791,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex polygon $P$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $P$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\mathcal{P} = A_1A_2\\dots A_n$, where $n$ is an integer with $n \\ge 3$. The problem is trivial for $n=3$ because there are no diagonals and thus no dissections. We assume that $n \\ge 4$. Our proof is based on the following Lemma.\n\n**Lemma** Let $ABCD$ be a convex quadrilateral such that all its sides and diagonals have rational lengths. If segments $AC$ and $BD$ meet at $P$, then segments $AP$, $BP$, $CP$, $DP$ all have rational lengths.\n\n![](attached_image_1.png)\n\nIt is clear by the Lemma that the desired result holds when $\\mathcal{P}$ is a convex quadrilateral. Let $A_iA_j$ ($1 \\le i < j \\le n$) be a diagonal of $\\mathcal{P}$. Assume that $C_1, C_2, \\dots, C_m$ are the consecutive division points on diagonal $A_iA_j$ (where point $C_1$ is the closest to vertex $A_i$ and $C_m$ is the closest to $A_j$). Then the segments $C_\\ell C_{\\ell+1}$, $1 \\le \\ell \\le m-1$, are the sides of all polygons in the dissection. Let $C_\\ell$ be the point where diagonal $A_iA_j$ meets diagonal $A_sA_t$. Then quadrilateral $A_iA_sA_jA_t$ satisfies the conditions of the Lemma. Consequently, segments $A_iC_\\ell$ and $C_\\ell A_j$ have rational lengths. Therefore, segments $A_iC_1, A_iC_2, \\dots, A_jC_m$ all have rational lengths. Thus, $C_\\ell C_{\\ell+1} = AC_{\\ell+1} - AC_\\ell$ is rational. Because $i, j, \\ell$ are arbitrarily chosen, we proved that all sides of all polygons in the dissection are also rational numbers.\n\nNow we present two proofs of the Lemma to finish our proof.\n\n*   **First approach** We show only that segment $AP$ is rational, the proof for the others being similar. Introduce Cartesian coordinates with $A = (0, 0)$ and $C = (c, 0)$. Put $B = (a, b)$ and $D = (d, e)$. Then by hypothesis, the numbers\n$$\n\\begin{aligned}\nAB &= \\sqrt{a^2 + b^2}, & AC &= c, & AD &= \\sqrt{d^2 + e^2}, \\\\\nBC &= \\sqrt{(a-c)^2 + b^2}, & BD &= \\sqrt{(a-d)^2 + (b-e)^2}, \\\\\nCD &= \\sqrt{(d-c)^2 + e^2},\n\\end{aligned}\n$$\nare rational. In particular,\n$$\nBC^2 - AB^2 - AC^2 = (a-c)^2 + b^2 - (a^2 + b^2) - c^2 = -2ac\n$$\nis rational. Because $c \\neq 0$, $a$ is rational. Likewise, $d$ is rational.\n\nNow we have that $b^2 = AB^2 - a^2$, $e^2 = AD^2 - d^2$, and $(b-e)^2 = BD^2 - (a-d)^2$ are rational, and so that $2be = b^2 + e^2 - (b-e)^2$ is rational. Because quadrilateral $ABCD$ is convex, $b$ and $e$ are nonzero and have opposite sign. Hence $b/e = 2be/2b^2$ is rational.\n\nWe now calculate\n$$\nP = \\left( \\frac{bd - ae}{b - e}, 0 \\right),\n$$\nso\n$$\nAP = \\frac{\\frac{b}{e} \\cdot d - a}{\\frac{b}{e} - 1}\n$$\nis rational.\n\n* **Second approach** To prove the Lemma, we set $\\angle DAP = A_1$ and $\\angle BAP = A_2$. Applying the **Law of Cosines** to triangles $ADC$, $ABC$, $ABD$ shows that angles $A_1, A_2, A_1+A_2$ all have rational cosine values. By the Addition formula, we have\n$$\n\\sin A_1 \\sin A_2 = \\cos A_1 \\cos A_2 - \\cos(A_1 + A_2),\n$$\nimplying that $\\sin A_1 \\sin A_2$ is rational.\nThus\n$$\n\\frac{\\sin A_2}{\\sin A_1} = \\frac{\\sin A_2 \\sin A_1}{\\sin^2 A_1} = \\frac{\\sin A_2 \\sin A_1}{1 - \\cos^2 A_1}\n$$\nis rational.\n\nNote that the ratio between the areas of triangles *ADP* and *ABP* is equal to $\\frac{PD}{BP}$. Therefore\n$$\n\\frac{BP}{PD} = \\frac{[ABP]}{[ADP]} = \\frac{\\frac{1}{2}AB \\cdot AP \\cdot \\sin A_2}{\\frac{1}{2}AD \\cdot AP \\cdot \\sin A_1} = \\frac{AB}{AD} \\cdot \\frac{\\sin A_2}{\\sin A_1},\n$$\nimplying that $\\frac{PD}{BP}$ is rational. Because $BP + PD = BD$ is rational, both $BP$ and $PD$ are rational. Similarly, $AP$ and $PC$ are rational, proving the Lemma.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46792,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of positive integers $(a, b)$ such that\n$$\n\\frac{a^2}{2ab^2 - b^3 + 1}\n$$\nis a positive integer.",
    "options": [],
    "answer": "(a, b) = (2t, 1), (t, 2t), (8t^4 - t, 2t) for all positive integers t",
    "solution": "**First Solution.** (Based on work by Anders Kaseorg) Rewrite equation $(*)$ as $a^2 - 2ab^2k = -b^3k + k$. Adding $b^4k^2$ to both sides completes the square on the left-hand side and gives\n$$\n(kb^2 - a)^2 = b^4k^2 - b^3k + k,\n$$\nor\n$$\n(2kb^2 - 2a)^2 = (2b^2k)^2 - 2b(2b^2k) + 4k.\n$$\nCompleting the square on the right-hand side gives\n$$\n(2kb^2 - 2a)^2 = (2b^2k - b)^2 + 4k - b^2.\n$$\nor,\n$$\ny^2 - x^2 = 4k - b^2,\n$$\nwhere $x = 2kb^2 - b$ and $y = 2kb^2 - 2a$.\nIf $4k = b^2$, then either $x = y$ or $x = -y$. In the former case, $b = 2a$; in the latter case, $4kb^2 - b = 2a$, that is, $b^4 - b = 2a$. Because $k = b^2/4$ is an integer if and only if $b$ is even, we get the solutions\n$$\n\\left(\\frac{b}{2}, b\\right) \\quad \\text{and} \\quad \\left(\\frac{b^4 - b}{2}, b\\right)\n$$\nfor any even $b$; that is, $(a, b) = (t, 2t)$ and $(a, b) = (8t^4 - t, 2t)$ for all positive integers $t$.\nIf $4k < b^2$, then $y^2 < x^2$, so $y^2 \\le (x-1)^2$ (since $x$ is clearly positive). Thus,\n$$\n4k - b^2 \\le (x - 1)^2 - x^2 = -2x + 1 = -4kb^2 + 2b + 1,\n$$\nor, $4k(b^2 + 1) \\le b^2 + 2b + 1 < 3b^2 + 1$. Because $4k > 3$, this is a contradiction.\nSimilarly, if $4k > b^2$, then $y^2 > x^2$, so $y^2 \\ge (x+1)^2$. Thus,\n$$\n4k - b^2 \\ge (x + 1)^2 - x^2 = 2x + 1 = 4kb^2 - 2b + 1,\n$$\nor $4k(b^2 - 1) + (b - 1)^2 \\le 0$. We must have $b = 1$ and\n$$\nk = \\frac{a^2}{2a - 1 + 1} = \\frac{a}{2}.\n$$\nThis is an integer whenever $a$ is even, so we get the solutions $(a, b) = (2t, 1)$ for all positive integers $t$.\n\n\n**Second Solution.** (Based on work by Po-Ru Loh) Assume that $b = 1$. Then\n$$\n\\frac{a^2}{2ab^2 - b^3 + 1} = \\frac{a^2}{2a} = \\frac{a}{2}\n$$\nis a positive integer if and only if $a$ is even. Thus, $(a, b) = (2t, 1)$ are solutions of the problem for all positive integers $t$.\nNow we assume that $b > 1$. Viewing equation $(*)$ as a quadratic in $a$, replace $a$ by $x$ to consider the equation\n$$\nx^2 - 2b^2kx + (b^3 - 1)k = 0 \\quad (*)'\n$$\nfor fixed positive integers $b$ and $k$. Its roots are\n$$\nx = \\frac{2b^2k \\pm \\sqrt{4b^4k^2 - 4b^3k + 4k}}{2} = b^2k \\pm \\sqrt{b^4k^2 - b^3k + k}.\n$$\nAssume that $x_1 = a$ is an integer root of equation $(*')$. Then $b^4k^2 - b^3k + k$ must be a perfect square. We claim that\n$$\n\\left(b^2k - \\frac{b}{2} - \\frac{1}{2}\\right)^2 < b^4k^2 - b^3k + k < \\left(b^2k - \\frac{b}{2} + \\frac{1}{2}\\right)^2.\n$$\nNote first that\n$$\n\\begin{aligned}\n\\left(b^2 k - \\frac{b}{2} - \\frac{1}{2}\\right)^2 &= b^4 k^2 - 2b^2 k \\left(\\frac{b}{2} + \\frac{1}{2}\\right)^2 + \\frac{1}{4}(b+1)^2 \\\\\n&= b^4 k^2 - b^3 k - b^2 k + \\frac{1}{4}(b+1)^2.\n\\end{aligned}\n$$\nTo establish the first inequality in our claim, it suffices to show that\n$$\n-b^2 k + \\frac{1}{4}(b+1)^2 < k,\n$$\nor, $(b+1)^2 < 4(b^2+1)k$, which is evident as $(b+1)^2 < 2(b^2+1)$ and $k \\ge 1$.\nNote also that\n$$\n\\begin{aligned}\n\\left(b^2 k - \\frac{b}{2} + 1\\right)^2 &= b^4 k^2 + 2b^2 k \\left(\\frac{1}{2} - \\frac{b}{2}\\right) + \\frac{1}{4}(1-b)^2 \\\\\n&> b^4 k^2 - b^3 k + b^2 k \\\\\n&> b^4 k^2 - b^3 k + k, \\quad \\text{as } b > 1,\n\\end{aligned}\n$$\nwhich establishes the second inequality in our claim.\nBecause all of $b, k$, and $\\sqrt{b^4 k^2 - b^3 k + k}$ are positive integers, we conclude from our claim that $b^2 k - \\frac{b}{2}$ is an integer and that\n$$\nb^4 k^2 - b^3 k + k = \\left(b^2 k - \\frac{b}{2}\\right)^2 = b^4 k - b^3 k + \\frac{b^2}{4},\n$$\nand so $k = b^2/4$. Thus, $b = 2t$ for some positive integer $t$. The two solutions of the equation $(*)$ becomes\n$$\nx = b^2 k \\pm \\left(b^2 k - \\frac{b}{2}\\right),\n$$\nthat is. $x = t$ or $x = 8t^4 - t$. Hence, $(a, b) = (t, 2t)$ and $(a, b) = (8t^4 - t, 2t)$ are the possible solutions of the problem, in addition to the solutions $(2t, 1)$.\n\n\n**Third Solution.** Because both $k$ and $a^2$ are positive, $2ab^2 - b^3 + 1 > 0$, or,\n$$\n2a > b - \\frac{1}{b^2}.\n$$\nBecause $a$ and $b$ are positive integers, we have $2a \\ge b$. Because $k$ is a positive integer, $a^2 \\ge 2ab^2 - b^3 + 1$, or, $a^2 \\ge b^2(2a - b) + 1$. Because\n$$\na^2 > b^2(2a - b) \\geq 0,\n$$\nwe have\n$$\na > b \\quad \\text{or} \\quad 2a = b. \\qquad (\\dagger)\n$$\nWe consider again the quadratic equation $(*)$ for fixed positive integers $b$ and $k$, and assume that $x_1 = a$ is an integer root of equation $(*)$. Then the other root $x_2$ is also an integer because $x_1 + x_2 = 2b^2k$. Without loss of generality, we assume that $x_1 \\geq x_2$. Then $x_1 \\geq b^2k > 0$. Furthermore, because $x_1x_2 = (b^3 - 1)k$, we obtain\n$$\n0 \\leq x_2 = \\frac{(b^3 - 1)k}{x_1} \\leq \\frac{(b^3 - 1)k}{b^2k} < b.\n$$\nIf $x_2 = 0$, then $b^3 - 1 = 0$, and so $x_1 = 2k$ and $(a, b)$ can be written in the form of $(2t, 1)$ for some integers $t$.\nIf $x_2 > 0$, then $(a, b) = (x_2, b)$ is a pair of positive integers satisfying the equations $(\\dagger)$ and $(*)$. We conclude that $2x_2 = b$, and so\n$$\nk = \\frac{x_2^2}{2x_2b^2 - b^3 + 1} = x_2^2 = \\frac{b^2}{4},\n$$\nand $x_1 = b^4/2 - b/2$. Thus, $(a, b)$ can be written in the form of either $(t, 2t)$ or $(8t^3 - t, 2t)$ for some positive integers $t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFive mathematicians find a bag of 100 gold coins in a room. They agree to split up the coins according to the following plan:\n- The oldest person in the room proposes a division of the coins among those present. (No coin may be split.) Then all present, including the proposer, vote on the proposal.\n- If at least $50\\%$ of those present vote in favor of the proposal, the coins are distributed accordingly and everyone goes home. (In particular, a proposal wins on a tie vote.)\n- If fewer than $50\\%$ of those present vote in favor of the proposal, the proposer must leave the room, receiving no coins. Then the process is repeated: the oldest person remaining proposes a division, and so on.\n- There is no communication or discussion of any kind allowed, other than what is needed for the proposer to state his or her proposal, and the voters to cast their vote.\nAssume that each person is equally intelligent and each behaves optimally to maximize his or her share. How much will each person get?\n\nProblem:\n\nFive mathematicians find a bag of 100 gold coins in a room. They agree to split up the coins according to the following plan:\n- The oldest person in the room proposes a division of the coins among those present. (No coin may be split.) Then all present, including the proposer, vote on the proposal.\n- If at least $50\\%$ of those present vote in favor of the proposal, the coins are distributed accordingly and everyone goes home. (In particular, a proposal wins on a tie vote.)\n- If fewer than $50\\%$ of those present vote in favor of the proposal, the proposer must leave the room, receiving no coins. Then the process is repeated: the oldest person remaining proposes a division, and so on.\n- There is no communication or discussion of any kind allowed other than what is needed for the proposer to state his or her proposal, and the voters to cast their vote.\nAssume that each person wishes to maximize his or her share of the coins and behaves optimally. How much will each person get?",
    "options": [],
    "answer": "(98, 0, 1, 0, 1) or (98, 0, 0, 1, 1) or (98, 0, 1, 1, 0)",
    "solution": "Solution:\n\nThe first (oldest) person will get 98 coins, the second will get no coins, and one coin apiece will go to two of the last three people. If we write the distribution as an ordered 5-tuple, with leftmost being oldest, etc., then the three possible solutions are $(98, 0, 1, 0, 1)$, $(90, 0, 0, 1, 1)$, and $(98, 0, 1, 1, 0)$.\n\nTo see why, we first mention a few principles. Since the mathematicians are all intelligent, and equally intelligent, but only out for themselves and unable to cooperate or negotiate, each person must assume the worst. No one will approve a plan that gives them zero coins. But furthermore, no one will vote against a plan that gives them a positive number of coins, if it may be possible that they could end up with fewer coins. This is not a game of chance; a person will settle, for example, for just one coin, if by rejecting this offer he or she may end up with no coins.\n\nThis explains why the common wrong answer \"everyone votes against the oldest in order to decrease the number of players\" will not happen with rational players. If the number of players dropped to 4 people, then the first person would have the advantage, and need only propose something that one other person would agree to. For example, the proposal may be $(50,0,0,50)$. This would give players #2, #3 (originally #3, #4) no coins at all, but only 2 votes are needed for the proposal to pass. On the other hand, the proposal could be $(50,0,50,0)$. In other words, when there are 5 people present, it is not in the best interest of #3, #4, or #5 to reject the proposal, if they are offered non-zero amounts of coins, since they may end up with zero later, and have absolutely no control over what happens!\n\nIndeed, by analyzing smaller cases, it becomes clear how much advantage the first person has, and how the second person is in the worst situation. With just two people, $(100,0)$ wins all for the first person. With three people, $(99,0,1)$ will be accepted, since #3 dare not reject it, for then #3 would be the loser in the 2-person situation. Consequently, with 4 people, the winning proposals will be $(99,0,1,0)$ or $(99,0,0,1)$. In both cases, the people with zero will vote against the proposal, and the one person who is offered a single coin will accept, since in no circumstances will they do better than this. (Conceivably, the youngest person can vote against $(99,0,0,1)$, reasoning that the next proposal will be $(99,0,1)$ which also offers him or her one coin; knowing this, the most likely 4-person proposal will be $(99,0,1,0)$ which will definitely not be rejected by person #3.)\n\nFinally, looking at 5 people, it is now clear that the first person need only \"pay off\" two others, with the least amount. Person #2 cannot be paid off, since he or she will always be better off being the proposer. Knowing this, person #1 always offers nothing to person #2. Persons #3, #4, #5 all realize that they may end up with no coins at all, by the analysis above. Consequently, if any two of them are offered just a single coin, they will accept, realizing that failing to accept will result in just 4 players, with absolutely no guarantee of any coins at all. In particular, player #3 (in the 5-person situation) realizes that with 4 players, he will surely be offered (and will receive) no coins, so she will gladly accept one coin in the 5-person situation. Persons #4 and #5 (in the 5-person case) are in the situation of not knowing for sure whether they will be offered 1 or 0 coins in the 4-person case. Consequently, if either of them is offered just one coin, the offer is accepted, since a certainty of getting one coin is superior to a possibility of getting no coins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $A$ is a set with $n$ elements, and $k$ is a divisor of $n$. Find the number of consistent $k$-configurations of $A$ of order 1.",
    "options": [],
    "answer": "n! / ((n/k)! (k!)^{n/k})",
    "solution": "Solution:\n\nGiven such a $k$-configuration, we can write out all the elements of one of the $k$-element subsets, then all the elements of another subset, and so forth, eventually obtaining an ordering of all $n$ elements of $A$. Conversely, given any ordering of the elements of $A$, we can construct a consistent $k$-configuration of order 1 from it by grouping together the first $k$ elements, then the next $k$ elements, and so forth. In fact, each consistent $k$-configuration of order 1 corresponds to $(n / k)! (k!)^{n / k}$ different such orderings, since the elements of $A$ within each of the $n / k$ $k$-element subsets can be ordered in $k!$ ways, and the various subsets can also be ordered with respect to each other in $(n / k)!$ different ways. Thus, since there are $n!$ orderings of the elements of $A$, we get\n$$\n\\frac{n!}{(n / k)! (k!)^{n / k}}\n$$\ndifferent consistent $k$-configurations of order 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$OPEN$ is a square, and $T$ is a point on side $NO$, such that triangle $TOP$ has area $62$ and triangle $TEN$ has area $10$. What is the length of a side of the square?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\n$62 = PO \\cdot OT / 2$ and $10 = EN \\cdot TN / 2 = PO \\cdot TN / 2$, so adding gives $72 = PO \\cdot (OT + TN) / 2 = PO \\cdot ON / 2 = PO^2 / 2 \\Rightarrow PO = 12$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute $\\frac{\\tan^{2}\\left(20^{\\circ}\\right)-\\sin^{2}\\left(20^{\\circ}\\right)}{\\tan^{2}\\left(20^{\\circ}\\right) \\sin^{2}\\left(20^{\\circ}\\right)}$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nIf we multiply top and bottom by $\\cos^{2}\\left(20^{\\circ}\\right)$, the numerator becomes $\\sin^{2}\\left(20^{\\circ}\\right) \\cdot (1-\\cos^{2} 20^{\\circ}) = \\sin^{4}\\left(20^{\\circ}\\right)$, while the denominator becomes $\\sin^{4}\\left(20^{\\circ}\\right)$ also. So they are equal, and the ratio is $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46797,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAs shown, $U$ and $C$ are points on the sides of triangle $M N H$ such that $M U = s$, $U N = 6$, $N C = 20$, $C H = s$, $H M = 25$. If triangle $U N C$ and quadrilateral $M U C H$ have equal areas, what is $s$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nUsing brackets to denote areas, we have $[M C H] = [U N C] + [M U C H] = 2[U N C]$. On the other hand, triangles with equal altitudes have their areas in the same ratio as their bases, so\n$$\n2 = \\frac{[M N H]}{[U N C]} = \\frac{[M N H]}{[M N C]} \\cdot \\frac{[M N C]}{[U N C]} = \\frac{N H}{N C} \\cdot \\frac{M N}{U N} = \\frac{s+20}{20} \\cdot \\frac{s+6}{6}.\n$$\nClearing the denominator gives $(s+20)(s+6) = 240$, and solving the quadratic gives $s = 4$ or $-30$. Since $s > 0$, we must have $s = 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA particle moves along the $x$-axis in such a way that its velocity at position $x$ is given by the formula $v(x) = 2 + \\sin x$. What is its acceleration at $x = \\frac{\\pi}{6}$?",
    "options": [],
    "answer": "5√3/4",
    "solution": "Solution:\nAcceleration is given by\n$$\na = \\frac{dv}{dt} = \\frac{dv}{dx} \\cdot \\frac{dx}{dt} = \\frac{dv}{dx} \\cdot v = \\cos x \\cdot (2 + \\sin x)\n$$\nAt $x = \\frac{\\pi}{6}$:\n$$\na = \\cos\\left(\\frac{\\pi}{6}\\right) \\cdot \\left(2 + \\sin\\left(\\frac{\\pi}{6}\\right)\\right) = \\frac{\\sqrt{3}}{2} \\cdot \\left(2 + \\frac{1}{2}\\right) = \\frac{\\sqrt{3}}{2} \\cdot \\frac{5}{2} = \\frac{5\\sqrt{3}}{4}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou have a list of real numbers, whose sum is $40$. If you replace every number $x$ on the list by $1-x$, the sum of the new numbers will be $20$. If instead you had replaced every number $x$ by $1+x$, what would the sum then be?",
    "options": [],
    "answer": "100",
    "solution": "Solution:\n\nLet $n$ be the number of numbers on the list. If each initial number is replaced by its negative, the sum will then be $-40$, and adding $1$ to every number on this list increases the sum by $n$, so $n - 40 = 20 \\Rightarrow n = 60$. Then, if we had simply added $1$ to each of the initial numbers (without negating first), the sum would increase to $40 + n = 40 + 60 = 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe graph of $x^{4}=x^{2} y^{2}$ is a union of $n$ different lines. What is the value of $n$?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nThe equation $x^{4}-x^{2} y^{2}=0$ factors as $x^{2}(x+y)(x-y)=0$, so its graph is the union of the three lines $x=0$, $x+y=0$, and $x-y=0$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 46801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA grasshopper lives on a coordinate line. It starts off at $1$. It can jump either $1$ unit or $5$ units either to the right or to the left. However, the coordinate line has holes at all points with coordinates divisible by $4$ (e.g. there are holes at $-4, 0, 4, 8$ etc.), so the grasshopper can not jump to any of those points. Can it reach point $3$ after $2003$ jumps?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\nEach jump changes the parity of grasshopper's coordinate. After $2003$ jumps the grasshopper will be at an even point on the coordinate line, and therefore can not be at $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the smallest number of regular hexagons of side length $1$ needed to completely cover a disc of radius $1$?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nFirst, we show that two hexagons do not suffice. Specifically, we claim that a hexagon covers less than half of the disc's boundary. First, a hexagon of side length $1$ may be inscribed in a circle, and this covers just $6$ points. Translating the hexagon vertically upward (regardless of its orientation) will cause it to no longer touch any point on the lower half of the circle, so that it now covers less than half of the boundary. By rotational symmetry, the same argument applies to translation in any other direction, proving the claim. Then, two hexagons cannot possibly cover the disc.\n\nThe disc can be covered by three hexagons as follows. Let $P$ be the center of the circle. Put three non-overlapping hexagons together at point $P$. This will cover the circle, since each hexagon will cover a $120^{\\circ}$ sector of the circle.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$r$ and $s$ are integers such that\n$$\n3 r \\geq 2 s - 3 \\text{ and } 4 s \\geq r + 12.\n$$\nWhat is the smallest possible value of $r / s$?",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nWe simply plot the two inequalities in the $s r$-plane and find the lattice point satisfying both inequalities such that the slope from it to the origin is as low as possible. We find that this point is $(2,4)$ (or $(3,6)$), as circled in the figure, so the answer is $2 / 4 = 1 / 2$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThere are 100 houses in a row on a street. A painter comes and paints every house red. Then, another painter comes and paints every third house (starting with house number 3) blue. Another painter comes and paints every fifth house red (even if it is already red), then another painter paints every seventh house blue, and so forth, alternating between red and blue, until 50 painters have been by. After this is finished, how many houses will be red?",
    "options": [],
    "answer": "52",
    "solution": "Solution:\nHouse $n$ ends up red if and only if the largest odd divisor of $n$ is of the form $4k+1$. We have 25 values of $n=4k+1$; 13 values of $n=2(4k+1)$ (given by $k=0,1,2,\\ldots,12$); 7 values of $n=4(4k+1)$ ($k=0,1,\\ldots,6$); 3 values of $n=8(4k+1)$ ($k=0,1,2$); 2 of the form $n=16(4k+1)$ (for $k=0,1$); 1 of the form $n=32(4k+1)$; and 1 of the form $n=64(4k+1)$. Thus we have a total of $25+13+7+3+2+1+1=52$ red houses.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many lattice points are enclosed by the triangle with vertices $(0,99)$, $(5,100)$, and $(2003, 500)$? Don't count boundary points.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nUsing the determinant formula, we get that the area of the triangle is\n$$\n\\left|\\begin{array}{cc}\n5 & 1 \\\\\n2003 & 401\n\\end{array}\\right| / 2 = 1\n$$\nThere are 4 lattice points on the boundary of the triangle (the three vertices and $(1004, 300)$), so it follows from Pick's Theorem that there are 0 in the interior.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute the radius of the inscribed circle of a triangle with sides $15$, $16$, and $17$.",
    "options": [],
    "answer": "sqrt(21)",
    "solution": "Solution:\n\nHeron's formula gives that the area is $\\sqrt{24 \\cdot 9 \\cdot 8 \\cdot 7} = 24 \\sqrt{21}$. Then, using the result that the area of a triangle equals the inradius times half the perimeter, we see that the radius is $\\sqrt{21}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an isosceles triangle with apex $A$. Let $I$ be the incenter. If $AI = 3$ and the distance from $I$ to $BC$ is $2$, then what is the length of $BC$?",
    "options": [],
    "answer": "4√5",
    "solution": "Solution:\n\nLet $X$ and $Y$ be the points where the incircle touches $AB$ and $BC$, respectively. Then $AXI$ and $AYB$ are similar right triangles. Since $I$ is the incenter, we have $IX = IY = 2$. Using the Pythagorean theorem on triangle $AXI$, we find $AX = \\sqrt{5}$. By similarity, $AY / AX = BY / IX$. Plugging in the numbers given, $5 / \\sqrt{5} = BY / 2$, so $BY = 2\\sqrt{5}$. $Y$ is the midpoint of $BC$, so $BC = 4\\sqrt{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all integers $m$ such that $m^{2}+6m+28$ is a perfect square.",
    "options": [],
    "answer": "6 and -12",
    "solution": "Solution:\nWe must have $m^{2}+6m+28 = n^{2}$, where $n$ is an integer. Rewrite this as $(m+3)^{2} + 19 = n^{2} \\Rightarrow n^{2} - (m+3)^{2} = 19 \\Rightarrow (n - m - 3)(n + m + 3) = 19$.\n\nLet $a = n - m - 3$ and $b = n + m + 3$, so we want $ab = 19$. This leaves only 4 cases:\n\n- $a = 1$, $b = 19$. Solve the system $n - m - 3 = 1$ and $n + m + 3 = 19$ to get $n = 10$ and $m = 6$, giving one possible solution.\n- $a = 19$, $b = 1$. Solve the system, as above, to get $n = 10$ and $m = -12$.\n- $a = -1$, $b = -19$. We get $n = -10$ and $m = -12$.\n- $a = -19$, $b = -1$. We get $n = -10$ and $m = 6$.\n\nThus the only $m$ are $6$ and $-12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe rational numbers $x$ and $y$, when written in lowest terms, have denominators $60$ and $70$, respectively. What is the smallest possible denominator of $x+y$?",
    "options": [],
    "answer": "84",
    "solution": "Solution:\n\nWrite $x + y = \\dfrac{a}{60} + \\dfrac{b}{70} = \\dfrac{7a + 6b}{420}$. Since $a$ is relatively prime to $60$ and $b$ is relatively prime to $70$, it follows that none of the primes $2, 3, 7$ can divide $7a + 6b$, so we won't be able to cancel any of these factors in the denominator. Thus, after reducing to lowest terms, the denominator will still be at least $2^{2} \\cdot 3 \\cdot 7 = 84$ (the product of the powers of $2, 3$, and $7$ dividing $420$). On the other hand, $84$ is achievable, by taking (e.g.) $\\dfrac{1}{60} + \\dfrac{3}{70} = \\dfrac{25}{420} = \\dfrac{5}{84}$. So $84$ is the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA point in three-space has distances $2, 6, 7, 8, 9$ from five of the vertices of a regular octahedron. What is its distance from the sixth vertex?",
    "options": [],
    "answer": "sqrt(21)",
    "solution": "Solution:\nBy a simple variant of the British Flag Theorem, if $ABCD$ is a square and $P$ any point in space, $AP^{2} + CP^{2} = BP^{2} + DP^{2}$. Four of the five given vertices must form a square $ABCD$, and by experimentation we find their distances to the given point $P$ must be $AP = 2$, $BP = 6$, $CP = 9$, $DP = 7$.\n\nThen $A$, $C$, and the other two vertices $E$, $F$ also form a square $AECF$, so\n$$\n85 = AP^{2} + CP^{2} = EP^{2} + FP^{2} = 8^{2} + FP^{2} \\Rightarrow FP = \\sqrt{21}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA palindrome is a positive integer that reads the same backwards as forwards, such as $82328$. What is the smallest $5$-digit palindrome that is a multiple of $99$?",
    "options": [],
    "answer": "54945",
    "solution": "Solution:\nWrite the number as $X Y Z Y X$. This is the same as $10000 X + 1000 Y + 100 Z + 10 Y + X = 99(101 X + 10 Y + Z) + 20 Y + 2 X + Z$. We thus want $20 Y + 2 X + Z$ to be a multiple of $99$, with $X$ as small as possible. This expression cannot be larger than $20 \\cdot 9 + 2 \\cdot 9 + 9 = 207$, and it is greater than $0$ (since $X \\neq 0$), so for this to be a multiple of $99$, it must equal $99$ or $198$. Consider these two cases.\n\nTo get $198$, we must have $Y = 9$, which then leaves $2 X + Z = 18$. The smallest possible $X$ is $5$, and then $Z$ becomes $8$ and we have the number $59895$.\n\nTo get $99$, we must have $Y = 4$. Then, $2 X + Z = 19$, and, as above, we find the minimal $X$ is $5$ and then $Z = 9$. This gives us the number $54945$. This is smaller than the other number, so it is the smallest number satisfying the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA lattice point is a point $(x, y)$ with both $x$ and $y$ integers. Find, with proof, the smallest $n$ such that every set of $n$ lattice points contains three points that are the vertices of a triangle with integer area. (The triangle may be degenerate, in other words, the three points may lie on a straight line and hence form a triangle with area zero.)",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nClearly, $n=4$ is too small, for the 4 points could be vertices of a unit square, and all the possible triangles will have areas of $1/2$. We will show that $n=5$ works. Every lattice point falls into one of four parity classes: (even, even), (even, odd), (odd, even), and (odd, odd). For example the point $(9,8)$ is in the (odd, even) class. By the pigeonhole principle, among the 5 lattice points there will be at least two points, $A$ and $B$, in the same parity class. We claim that if $C$ is any other lattice point, the triangle $A B C$ will have integral area.\n\nWe will use the following simple lemma:\nIf the vertices of a triangle are all lattice points, and one side is horizontal or vertical with even length, then the triangle must have integer area.\nTo see why the lemma is true, let $P Q R$ be the triangle with $P Q$ even, and, say, horizontal. Let $h$ be the length of the altitude from $R$ to $P Q$. Clearly, $h$ is an integer, since it is equal to the absolute value of the difference in $y$-coordinates of $R$ and $P$. Then $[P Q R] = P Q \\cdot h / 2$, and the lemma follows. (We are using the notation $[P Q R]$ to denote the area of triangle $P Q R$.)\n\nNow we will show that $[A B C]$ must be an integer. If $A B$ is horizontal or vertical, then the length of $A B$ is even (since the endpoints are in the same parity class), so the lemma immediately implies that $[A B C]$ is an integer.\nIf $A B$ is neither horizontal nor vertical, choose the lattice point $D$ such that $A D B$ is a right triangle with right angle at $D$ (there may be more than one choice).\n\n![](attached_image_1.png)\n\nThen\n$$\n[A B C] = [A C B D] - [A D B].\n$$\n\nSince $A D$ and $D B$ both have even length, $[A B C]$ is an integer. Furthermore,\n$$\n[A C B D] = [C A D] + [C D B],\n$$\nand by the same reasoning $[C A D]$ and $[C D B]$ are both integers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all ordered triples of primes $(p, q, r)$ such that\n$$\np \\mid q^r + 1, \\quad q \\mid r^p + 1, \\quad r \\mid p^q + 1.\n$$",
    "options": [],
    "answer": "[(2, 5, 3), (5, 3, 2), (3, 2, 5)]",
    "solution": "We check that this is a solution:\n$$\n2 \\mid 126 = 5^3 + 1, \\quad 5 \\mid 10 = 3^2 + 1, \\quad 3 \\mid 33 = 2^5 + 1.\n$$\nNow let $p, q, r$ be three primes satisfying the given divisibility relations. Since $q$ does not divide $q^r + 1$, $p \\neq q$, and similarly $q \\neq r, r \\neq p$, so $p, q$ and $r$ are all distinct. We now prove a lemma.\n**Lemma.** Let $p, q, r$ be distinct primes with $p \\mid q^r + 1$, and $p > 2$. Then either $2r \\mid p-1$ or $p \\mid q^2 - 1$.\n*Proof.* Since $p \\mid q^r + 1$, we have\n$$\nq^r \\equiv -1 \\not\\equiv 1 \\pmod{p}, \\quad \\text{because } p > 2,\n$$\nbut\n$$\nq^{2r} \\equiv (-1)^2 \\equiv 1 \\pmod{p}.\n$$\nLet $d$ be the order of $q \\pmod{p}$; then from the above congruences, $d$ divides $2r$ but not $r$. Since $r$ is prime, the only possibilities are $d = 2$ or $d = 2r$. If $d = 2r$, then $2r \\mid p-1$ because $d \\mid p-1$. If $d = 2$, then $q^2 \\equiv 1 \\pmod{p}$ so $p \\mid q^2 - 1$. This proves the lemma. ■\nNow let's first consider the case where $p, q$ and $r$ are all odd. Since $p \\mid q^r + 1$, by the lemma either $2r \\mid p-1$ or $p \\mid q^2 - 1$. But $2r \\mid p-1$ is impossible because\n$$\n2r \\mid p-1 \\implies p \\equiv 1 \\pmod{r} \\implies 0 \\equiv p^q + 1 \\equiv 2 \\pmod{r}\n$$\nand $r > 2$. So we must have $p \\mid q^2 - 1 = (q - 1)(q + 1)$. Since $p$ is an odd prime and $q - 1, q + 1$ are both even, we must have\n$$\np \\mid \\frac{q-1}{2} \\quad \\text{or} \\quad p \\mid \\frac{q+1}{2};\n$$\neither way,\n$$\np \\le \\frac{q+1}{2} < q.\n$$\nBut then by a similar argument we may conclude $q < r, r < p$, a contradiction.\nThus, at least one of $p, q, r$ must equal 2. By a cyclic permutation we may assume that $p = 2$. Now $r \\mid 2^q + 1$, so by the lemma, either $2q \\mid r - 1$ or $r \\mid 2^2 - 1$. But $2q \\mid r - 1$ is impossible as before, because $q$ divides $r^2 + 1 = (r^2 - 1) + 2$ and $q > 2$. Hence, we must have $r \\mid 2^2 - 1$. We conclude that $r = 3$, and $q \\mid r^2 + 1 = 10$. Because $q \\ne p$, we must have $q = 5$. Hence (2, 5, 3) and its cyclic permutations are the only solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\neq 0$. For every sequence of integers\n$$\na = a_0, a_1, a_2, \\dots, a_n\n$$\nsatisfying $0 \\le a_i \\le i$, for $i = 0, \\dots, n$, define another sequence\n$$\nt(a) = t(a)_0, t(a)_1, t(a)_2, \\dots, t(a)_n\n$$\nby setting $t(a)_i$ to be the number of terms in the sequence $a$ that precede the term $a_i$ and are different from $a_i$. Show that, starting from any sequence $a$ as above, fewer than $n$ applications of the transformation $t$ lead to a sequence $b$ such that $t(b) = b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** Note first that the transformed sequence $t(a)$ also satisfies the inequalities $0 \\le t(a)_i \\le i$, for $i = 0, \\dots, n$. Call any integer sequence that satisfies these inequalities an *index bounded sequence*.\nWe prove now that $a_i \\le t(a)_i$, for $i = 0, \\dots, n$. Indeed, this is clear if $a_i = 0$. Otherwise, let $x = a_i > 0$ and $y = t(a)_i$. None of the first $x$ consecutive terms $a_0, a_1, \\dots, a_{x-1}$ is greater than $x-1$, so they are all different from $x$ and precede $x$ (see the diagram below). Thus $y \\ge x$, that is, $a_i \\le t(a)_i$, for $i = 0, \\dots, n$.\n\n$$\n\\begin{array}{c|cccccc}\n & 0 & 1 & \\dots & x-1 & \\dots & i \\\\\\hline\n a & a_0 & a_1 & \\dots & a_{x-1} & \\dots & x \\\\\nt(a) & t(a)_0 & t(a)_1 & \\dots & t(a)_{x-1} & \\dots & y\n\\end{array}\n$$\n\nThis already shows that the sequences stabilize after finitely many applications of the transformation $t$, because the value of the index $i$ term in index bounded sequences cannot exceed $i$. Next we prove that if $a_i = t(a)_i$, for some $i = 0, \\dots, n$, then no further applications of $t$ will ever change the index $i$ term. We consider two cases.\n\n* In this case, we assume that $a_i = t(a)_i = 0$. This means that no term on the left of $a_i$ is different from 0, that is, they are all 0. Therefore the first $i$ terms in $t(a)$ will also be 0 and this repeats (see the diagram below).\n\n$$\n\\begin{array}{c|cccc}\n & 0 & 1 & \\dots & i \\\\\\hline\n a & 0 & 0 & \\dots & 0 \\\\\nt(a) & 0 & 0 & \\dots & 0\n\\end{array}\n$$\n\n* In this case, we assume that $a_i = t(a)_i = x > 0$. The first $x$ terms are all different from $x$. Because $t(a)_i = x$, the terms $a_x, a_{x+1}, \\dots, a_{i-1}$ must then all be equal to $x$. Consequently, $t(a)_j = x$ for $j = x, \\dots, i-1$ and further applications of $t$ cannot change the index $i$ term (see the diagram below).\n\n|   | 0 | 1 | ... | $x - 1$ | $x$ | $x + 1$ | ... | $i$ |\n|---|---|---|-----|---------|-----|---------|-----|-----|\n| $a$ | $a_0$ | $a_1$ | ... | $a_{x-1}$ | $x$ | $x$ | ... | $x$ |\n| $t(a)$ | $t(a)_0$ | $t(a)_1$ | ... | $t(a)_{x-1}$ | $x$ | $x$ | ... | $x$ |\n\nFor $0 \\le i \\le n$, the index $i$ entry satisfies the following properties: (i) it takes integer values; (ii) it is bounded above by $i$; (iii) its value does not decrease under transformation $t$; and (iv) once it stabilizes under transformation $t$, it never changes again. This shows that no more than $n$ applications of $t$ lead to a sequence that is stable under the transformation $t$.\n\nFinally, we need to show that no more than $n-1$ applications of $t$ is needed to obtain a fixed sequence from an initial $n+1$-term index bounded sequence $a = (a_0, a_1, \\dots, a_n)$. We induct on $n$.\n\nFor $n=1$, the two possible index bounded sequences $(a_0, a_1) = (0, 0)$ and $(a_0, a_1) = (0, 1)$ are already fixed by $t$ so we need zero applications of $t$.\n\nAssume that any index bounded sequence $(a_0, a_1, \\dots, a_n)$ reach a fixed sequence after no more than $n-1$ applications of $t$. Consider an index bounded sequence $a = (a_0, a_1, \\dots, a_{n+1})$. It suffices to show that $a$ will be stabilized in no more than $n$ applications of $t$. We approach indirectly by assuming on the contrary that $n+1$ applications of transformations are needed. This can happen only if $a_{n+1} = 0$ and each application of $t$ increased the index $n+1$ term by exactly 1. Under transformation $t$, the resulting value of index $i$ term will not be effected by index $j$ term for $i < j$. Hence by the induction hypothesis, the subsequence $a' = (a_0, a_1, \\dots, a_n)$ will be stabilized in no more than $n-1$ applications of $t$. Because index $n$ term is stabilized at value $x \\le n$ after no more than $\\min\\{x, n-1\\}$ applications of $t$ and index $n+1$ term obtains value $x$ after exactly $x$ applications of $t$ under our current assumptions. We conclude that the index $n+1$ term would become equal to the index $n$ term after no more than $n-1$ applications of $t$. However, once two consecutive terms in a sequence are equal they stay equal and stabilize together. Because the index $n$ term needs no more than $n-1$ transformations to be stabilized, $a$ can be stabilized in no more than $n-1$ applications of $t$, which contradicts our assumption of $n+1$ applications needed. Thus our assumption was wrong and we need at most $n$ applications of transformation $t$ to stabilize an $(n + 1)$-term index bounded sequence. This completes our inductive proof.\n**Second Solution.** We prove that for $n \\ge 2$, the claim holds without the initial condition $0 \\le a_i \\le i$. (Of course this does not prove anything stronger, but it's convenient.) We do this by induction on $n$, the case $n=2$ being easy to check by hand as in the first solution.\n\nNote that if $c = (c_0, \\dots, c_n)$ is a sequence in the image of $t$, and $d$ is the sequence $(c_1, \\dots, c_n)$, then the following two statements are true:\n\n(a) If $e$ is the sequence obtained from $d$ by subtracting 1 from each nonzero term, then $t(d) = t(e)$. (If there are no zero terms in $d$, then subtracting 1 clearly has no effect. If there is a zero term in $d$, it must occur at the beginning, and then every nonzero term is at least 2.)\n\n(b) One can compute $t(c)$ by applying $t$ to the sequence $c_1, \\dots, c_n$, adding 1 to each nonzero term, and putting a zero in front.\n\nThe recipe of (b) works for computing $t^i(c)$ for any $i$, by (a) and induction on $i$.\n\nWe now apply the induction hypothesis to $t(a)_1, \\dots, t(a)_n$ to see that it stabilizes after $n-2$ more applications of $t$; by the recipe above, that means $a$ stabilizes after $n-1$ applications of $t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sets $A$ and $B$ form a partition of positive integers if $A \\cap B = \\emptyset$ and $A \\cup B = \\mathbb{N}$. The set $S$ is called prohibited for the partition, if $k + l \\neq s$ for any $k, l \\in A, s \\in S$ and any $k, l \\in B, s \\in S$.\n\na) Define Fibonacci numbers $f_{i}$ by letting $f_{1} = 1$, $f_{2} = 2$ and $f_{i+1} = f_{i} + f_{i-1}$, so that $f_{3} = 3$, $f_{4} = 5$ etc. How many partitions for which the set $F$ of all Fibonacci numbers is prohibited are there? (We count $A, B$ and $B, A$ as the same partition.)\n\nb) How many partitions for which the set $P$ of all powers of $2$ is prohibited are there? What if we require in addition that $P \\subseteq A$ ?",
    "options": [],
    "answer": "a) Exactly one partition. b) Infinitely many partitions; if all powers of two must lie in A, exactly one partition.",
    "solution": "Solution:\n\nb) We prove the following: Given a partition of the set of all powers of $2$ (i.e. two sets $Q$ and $R$ such that each $2^{k}$ is in exactly one of $Q, R$) there exists a unique partition $A, B$ of positive integers with all powers of $2$ prohibited and with $Q \\subseteq A, R \\subseteq B$. Note that this implies that there are infinitely many partitions of integers with powers of $2$ prohibited, and exactly one such partition with $P \\subseteq A$.\n\nFirst we show that the required partition is unique if it exists. Without loss of generality $1$ is in $A$. Suppose now that we have been able to place all integers up to some $k$ unambiguously, i.e. for any $l < k$ we know whether $l$ is in $A$ or in $B$. If $k$ is a power of $2$ we know where to put it. Otherwise let $2^{i}$ be the smallest power of $2$ strictly greater than $k$. Then $d = 2^{i} - k$ is positive and less than $k$. Therefore we know to which of two sets ($A$ or $B$) the number $d$ belongs. Then we have no choice but to place $k$ in the other set (otherwise $k$ and $d$ would be in the same set, and since $k + d = 2^{i}$ this would contradict $A, B$ being acceptable). So there exists no more than one acceptable partition.\n\nWe now show that the desired partition exists. Since we have already defined above a recursive construction producing $A, B$, we just need to check that the resulting partition is in fact acceptable. Suppose not. Then there exist $n < m$ such that $n + m = 2^{i}$ for some $i$ and $m, n$ are in the same set of the partition. Then $m$ is not a power of $2$ (if it were, $n$ would be greater than $m$). Note that $n < m$ implies that $2^{i}$ is the smallest power of $2$ strictly greater than $m$. Then by construction $m$ is in the set different from $n$. Contradiction. So there is no such pair $m, n$. The partition constructed above works. This completes the proof.\n\na) The proof is similar to that of part b. We will call a partition acceptable if $F$ is prohibited for it. First, we show that there exists no more than one acceptable partition. Since we do not distinguish between $A, B$ we may assume without loss of generality that $1$ is in $A$. Suppose now that we have been able to place all integers up to some $k$ unambiguously, i.e. for any $l < k$ we know whether $l$ is in $A$ or in $B$. Let $f_{i}$ be the smallest Fibonacci number strictly greater than $k$. Then $d = f_{i} - k$ is positive and less than $k$ (to see that, note that if $f_{i} - k > k$ then $2k < f_{i}$, $k < f_{i}/2 \\leq f_{i-1}$, contradicting the choice of $f_{i}$).\n\nTherefore we know to which of two sets ($A$ or $B$) the number $d$ belongs. Then we have no choice but to place $k$ in the other set (otherwise $k$ and $d$ would be in the same set, and since $k + d = f_{i}$ this would contradict $A, B$ being acceptable). So there exists no more than one acceptable partition.\n\nWe now show that there actually exists an acceptable partition. Since we have already defined above a recursive construction producing $A, B$, we just need to check that the resulting partition is in fact acceptable. Suppose not. Then there exist a smallest $m$ such that $n + m = f_{i}$ for some $n < m$, and $m, n$ are in the same set (by renaming $A$ and $B$ if necessary we may assume $m, n$ are in $A$), i.e. the first $m$ for which there is a problem with the above recursive construction. Then if $f_{j}$ is as before the smallest Fibonacci number bigger than $m$ we have $f_{j-1} \\leq m < f_{j}$ which together with $n < m$ gives $f_{j-1} < m + n < 2f_{j} < f_{j+2}$. But $m + n = f_{j}$ is excluded by construction (recall that $m$ is assigned to the set other than that of $f_{j} - m$). So it must be that $m + n = f_{j+1}$. On the other hand for $\\hat{m} = f_{j} - m$, $\\hat{n} = f_{j} - n$ we have:\n\n1. $\\hat{n} > \\hat{m} > 0$ and $\\hat{n} = f_{j} - n < m$, so the bigger of $\\hat{n}, \\hat{m}$ is less than the bigger of $m, n$.\n2. $\\hat{m} + \\hat{n} = f_{j} + f_{j} - (m + n) = f_{j} + f_{j} - f_{j+1} = f_{j} - f_{j-1} = f_{j-2}$.\n3. $\\hat{m}$ is in $B$ by construction.\n4. $\\hat{n}$ is in $B$, because otherwise $\\hat{n}, n$ will be a pair of elements of $A$ adding up to a Fibonacci number with the maximal element in the pair less than $m$, contradicting our choice of $m$.\n\nThese observations together mean that $\\hat{n}, \\hat{m}$ is a pair of elements of $B$ adding up to a Fibonacci number with the maximal element in the pair less than $m$, contradicting our choice of $m$. This contradiction shows that $A, B$ constructed above is indeed an acceptable partition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many positive rational numbers less than $\\pi$ have denominator at most $7$ when written in lowest terms? (Integers have denominator $1$.)",
    "options": [],
    "answer": "54",
    "solution": "Solution:\n\nWe can simply list them. The table shows that there are $3+3+6+6+12+6+18=54$.\n\n| Denominator | Values |\n| ---: | :--- |\n| 1 | $\\frac{1}{1}, \\frac{2}{1}, \\frac{3}{1}$ |\n| 2 | $\\frac{1}{2}, \\frac{3}{2}, \\frac{5}{2}$ |\n| 3 | $\\frac{1}{3}, \\frac{2}{3}, \\frac{4}{3}, \\frac{5}{3}, \\frac{7}{3}, \\frac{8}{3}$ |\n| 4 | $\\frac{1}{4}, \\frac{3}{4}, \\frac{5}{4}, \\frac{7}{4}, \\frac{9}{4}, \\frac{11}{4}$ |\n| 5 | $\\frac{1}{5}, \\frac{2}{5}, \\frac{3}{5}, \\frac{4}{5}, \\frac{6}{5}, \\frac{7}{5}, \\frac{8}{5}, \\frac{9}{5}, \\frac{11}{5}, \\frac{12}{5}, \\frac{13}{5}, \\frac{14}{5}$ |\n| 6 | $\\frac{1}{6}, \\frac{5}{6}, \\frac{7}{6}, \\frac{11}{6}, \\frac{13}{6}, \\frac{17}{6}$ |\n| 7 | $\\frac{1}{7}, \\frac{2}{7}, \\ldots, \\frac{6}{7}, \\frac{8}{7}, \\frac{9}{7}, \\ldots, \\frac{13}{7}, \\frac{15}{7}, \\frac{16}{7}, \\ldots, \\frac{20}{7}$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA room is built in the shape of the region between two semicircles with the same center and parallel diameters. The farthest distance between two points with a clear line of sight is $12$ m. What is the area (in $\\mathrm{m}^{2}$) of the room?",
    "options": [],
    "answer": "18π",
    "solution": "Solution:\n$18 \\pi$\nThe maximal distance is as shown in the figure. Call the radii $R$ and $r$, $R > r$. Then $R^{2} - r^{2} = 6^{2}$ by the Pythagorean theorem, so the area is $(\\pi / 2) \\cdot (R^{2} - r^{2}) = 18 \\pi$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the area of the region bounded by the curves $y = x^{2003}$ and $y = x^{1 / 2003}$ and lying above the $x$-axis?",
    "options": [],
    "answer": "1001/1002",
    "solution": "Solution:\nThe two curves intersect at $(0, 0)$ and $(1, 1)$, so the desired area is\n\n$$\n\\int_{0}^{1} \\left( x^{1 / 2003} - x^{2003} \\right) dx = \\left[ \\frac{x^{2004 / 2003}}{2004 / 2003} - \\frac{x^{2004}}{2004} \\right]_{0}^{1} = \\frac{1001}{1002}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest $n$ such that $n!$ ends in 290 zeroes.",
    "options": [],
    "answer": "1170",
    "solution": "Solution:\nEach 0 represents a factor of $10 = 2 \\cdot 5$. Thus, we wish to find the smallest factorial that contains at least 290 2's and 290 5's in its prime factorization. Let this number be $n!$, so the factorization of $n!$ contains 2 to the power $p$ and 5 to the power $q$, where\n$$\np = \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{2^2} \\right\\rfloor + \\left\\lfloor \\frac{n}{2^3} \\right\\rfloor + \\cdots \\text{ and } q = \\left\\lfloor \\frac{n}{5} \\right\\rfloor + \\left\\lfloor \\frac{n}{5^2} \\right\\rfloor + \\left\\lfloor \\frac{n}{5^3} \\right\\rfloor + \\cdots\n$$\n(this takes into account one factor for each single multiple of 2 or 5 that is $\\leq n$, an additional factor for each multiple of $2^2$ or $5^2$, and so on). Naturally, $p \\geq q$ because 2 is smaller than 5. Thus, we want to bring $q$ as low to 290 as possible. If $q = \\left\\lfloor \\frac{n}{5} \\right\\rfloor + \\left\\lfloor \\frac{n}{5^2} \\right\\rfloor + \\left\\lfloor \\frac{n}{5^3} \\right\\rfloor + \\cdots$, we form a rough geometric sequence (by taking away the floor function) whose sum is represented by $290 \\approx \\frac{n / 5}{1 - 1 / 5}$. Hence we estimate $n = 1160$, and this gives us $q = 288$. Adding 10 to the value of $n$ gives the necessary two additional factors of 5, and so the answer is 1170.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDaniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a $60\\%$ chance of winning each point, what is the probability that he will win the game?",
    "options": [],
    "answer": "9/13",
    "solution": "Solution:\n\nConsider the situation after two points. Daniel has a $9/25$ chance of winning, Scott, $4/25$, and there is a $12/25$ chance that the players will be tied. In the latter case, we revert to the original situation. In particular, after every two points, either the game returns to the original situation, or one player wins. If it is given that the game lasts $2k$ rounds, then the players must be at par after $2(k-1)$ rounds, and then Daniel wins with probability $(9/25)/(9/25+4/25) = 9/13$. Since this holds for any $k$, we conclude that Daniel wins the game with probability $9/13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na.\nLet $A_{n} = \\{a_{1}, a_{2}, a_{3}, \\ldots, a_{n}, b\\}$, for $n \\geq 3$, and let $C_{n}$ be the 2-configuration consisting of $\\{a_{i}, a_{i+1}\\}$ for all $1 \\leq i \\leq n-1$, $\\{a_{1}, a_{n}\\}$, and $\\{a_{i}, b\\}$ for $1 \\leq i \\leq n$. Let $S_{e}(n)$ be the number of subsets of $C_{n}$ that are consistent of order $e$. Find $S_{e}(101)$ for $e=1,2$, and $3$.\n\nb.\nLet $A = \\{V, W, X, Y, Z, v, w, x, y, z\\}$. Find the number of subsets of the 2-configuration\n$$\n\\begin{gathered}\n\\{\\{V, W\\},\\{W, X\\},\\{X, Y\\},\\{Y, Z\\},\\{Z, V\\},\\{v, x\\},\\{v, y\\},\\{w, y\\},\\{w, z\\},\\{x, z\\}, \\\\\n\\{V, v\\},\\{W, w\\},\\{X, x\\},\\{Y, y\\},\\{Z, z\\}\\}\n\\end{gathered}\n$$\nthat are consistent of order 1.\n\nc.\nLet $A = \\{a_{1}, b_{1}, a_{2}, b_{2}, \\ldots, a_{10}, b_{10}\\}$, and consider the 2-configuration $C$ consisting of $\\{a_{i}, b_{i}\\}$ for all $1 \\leq i \\leq 10$, $\\{a_{i}, a_{i+1}\\}$ for all $1 \\leq i \\leq 9$, and $\\{b_{i}, b_{i+1}\\}$ for all $1 \\leq i \\leq 9$. Find the number of subsets of $C$ that are consistent of order 1.",
    "options": [],
    "answer": "a) S1(101) = 101, S2(101) = 101, S3(101) = 0; b) 6; c) 89",
    "solution": "Solution:\n\na.\nFor convenience, we assume the $a_{i}$ are indexed modulo $101$, so that $a_{i+1} = a_{1}$ when $a_{i} = a_{101}$.\n\nIn any consistent subset of $C_{101}$ of order $1$, $b$ must be paired with exactly one $a_{i}$, say $a_{1}$. Then, $a_{2}$ cannot be paired with $a_{1}$, so it must be paired with $a_{3}$, and likewise we find we use the pairs $\\{a_{4}, a_{5}\\}, \\{a_{6}, a_{7}\\}, \\ldots, \\{a_{100}, a_{101}\\}$—and this does give us a consistent subset of order 1. Similarly, pairing $b$ with any other $a_{i}$ would give us a unique extension to a consistent configuration of order 1. Thus, we have one such 2-configuration for each $i$, giving $S_{1}(101) = 101$ altogether.\n\nIn a consistent subset of order $2$, $b$ must be paired with two other elements. Suppose one of them is $a_{i}$. Then $a_{i}$ is also paired with either $a_{i-1}$ or $a_{i+1}$, say $a_{i+1}$. But then $a_{i-1}$ needs to be paired up with two other elements, and $a_{i}$ is not available, so it must be paired with $a_{i-2}$ and $b$. Now $b$ has its two pairs determined, so nothing else can be paired with $b$. Thus, for $j \\neq i-1, i$, we have that $a_{j}$ must be paired with $a_{j-1}$ and $a_{j+1}$. So our subset must be of the form\n$$\n\\{\\{b, a_{i}\\}, \\{a_{i}, a_{i+1}\\}, \\{a_{i+1}, a_{i+2}\\}, \\ldots, \\{a_{101}, a_{1}\\}, \\ldots, \\{a_{i-2}, a_{i-1}\\}, \\{a_{i-1}, b\\}\\}\n$$\nfor some $i$. On the other hand, for any $i = 1, \\ldots, 101$, this gives a subset meeting our requirements. So, we have $101$ possibilities, and $S_{2}(101) = 101$.\n\nFinally, in a consistent subset of order $3$, each $a_{i}$ must be paired with $a_{i-1}, a_{i+1}$, and $b$. But then $b$ occurs in $101$ pairs, not just $3$, so we have a contradiction. Thus, no such subset exists, so $S_{3}(101) = 0$.\n\n\nb.\nNo more than two of the pairs $\\{v, x\\}, \\{v, y\\}, \\{w, y\\}, \\{w, z\\}, \\{x, z\\}$ may be included in a 2-configuration of order 1, since otherwise at least one of $v, w, x, y, z$ would occur more than once. If exactly one is included, say $\\{v, x\\}$, then $w, y, z$ must be paired with $W, Y, Z$, respectively, and then $V$ and $X$ cannot be paired. So either none or exactly two of the five pairs above must be used. If none, then $v, w, x, y, z$ must be paired with $V, W, X, Y, Z$, respectively, and we have $1$ 2-configuration arising in this manner. If exactly two are used, we can check that there are $5$ ways to do this without duplicating an element:\n$$\n\\{v, x\\},\\{w, y\\} \\quad \\{v, x\\},\\{w, z\\} \\quad \\{v, y\\},\\{w, z\\} \\quad \\{v, y\\},\\{x, z\\} \\quad \\{w, y\\},\\{x, z\\}\n$$\nIn each case, it is straightforward to check that there is a unique way of pairing up the remaining elements of $A$. So we get $5$ 2-configurations in this way, and the total is $6$.\n\n\nc.\nLet $A_{n} = \\{a_{1}, b_{1}, a_{2}, b_{2}, \\ldots, a_{n}, b_{n}\\}$ for $n \\geq 1$, and consider the 2-configuration $C_{n}$ consisting of $\\{a_{i}, b_{i}\\}$ for all $1 \\leq i \\leq n$, $\\{a_{i}, a_{i+1}\\}$ for all $1 \\leq i \\leq n-1$, and $\\{b_{i}, b_{i+1}\\}$ for all $1 \\leq i \\leq n-1$. Let $N_{n}$ be the number of subsets of $C_{n}$ that are consistent of order 1 (call these \"matchings\" of $C_{n}$). Consider any matching of $C_{n+2}$. Either $a_{n+2}$ is paired with $b_{n+2}$, in which case the remaining elements of our matching form a matching of $C_{n+1}$; or $a_{n+2}$ is paired with $a_{n+1}$, in which case $b_{n+2}$ must be paired with $b_{n+1}$, and the remaining elements form a matching of $C_{n}$. It follows that $N_{n+2} = N_{n+1} + N_{n}$. By direct calculation, $N_{1} = 1$ and $N_{2} = 2$, and now computing successive values of $N_{n}$ using the recurrence yields $N_{10} = 89$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are 16 members on the Height-Measurement Matching Team. Each member was asked, \"How many other people on the team - not counting yourself - are exactly the same height as you?\" The answers included six 1's, six 2's, and three 3's. What was the sixteenth answer? (Assume that everyone answered truthfully.)",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nFor anyone to have answered $3$, there must have been exactly $4$ people with the same height, and then each of them would have given the answer $3$. Thus, we need at least four $3$'s, so $3$ is the remaining answer. (More generally, a similar argument shows that the number of members answering $n$ must be divisible by $n+1$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe sequence $a_{1}, a_{2}, a_{3}, \\ldots$ of real numbers satisfies the recurrence\n$$\na_{n+1}=\\frac{a_{n}^{2}-a_{n-1}+2 a_{n}}{a_{n-1}+1} .\n$$\nGiven that $a_{1}=1$ and $a_{9}=7$, find $a_{5}$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nLet $b_{n}=a_{n}+1$. Then the recurrence becomes $b_{n+1}-1=\\left(b_{n}^{2}-b_{n-1}\\right) / b_{n-1}=b_{n}^{2} / b_{n-1}-1$, so $b_{n+1}=b_{n}^{2} / b_{n-1}$. It follows that the sequence $\\left(b_{n}\\right)$ is a geometric progression, from which $b_{5}^{2}=b_{1} b_{9}=2 \\cdot 8=16 \\Rightarrow b_{5}= \\pm 4$. However, since all $b_{n}$ are real, they either alternate in sign or all have the same sign (depending on the sign of the progression's common ratio); either way, $b_{5}$ has the same sign as $b_{1}$, so $b_{5}=4 \\Rightarrow a_{5}=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cylinder of base radius $1$ is cut into two equal parts along a plane passing through the center of the cylinder and tangent to the two base circles. Suppose that each piece's surface area is $m$ times its volume. Find the greatest lower bound for all possible values of $m$ as the height of the cylinder varies.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nLet $h$ be the height of the cylinder. Then the volume of each piece is half the volume of the cylinder, so it is $\\frac{1}{2} \\pi h$. The base of the piece has area $\\pi$, and the ellipse formed by the cut has area $\\pi \\cdot 1 \\cdot \\sqrt{1+\\frac{h^{2}}{4}}$ because its area is the product of the semiaxes times $\\pi$. The rest of the area of the piece is half the lateral area of the cylinder, so it is $\\pi h$. Thus, the value of $m$ is\n$$\n\\begin{aligned}\n\\frac{\\pi+\\pi \\sqrt{1+h^{2} / 4}+\\pi h}{\\pi h / 2} & =\\frac{2+2 h+\\sqrt{4+h^{2}}}{h} \\\\\n& =\\frac{2}{h}+2+\\sqrt{\\frac{4}{h^{2}}+1}\n\\end{aligned}\n$$\na decreasing function of $h$ whose limit as $h \\rightarrow \\infty$ is $3$. Therefore the greatest lower bound of $m$ is $3$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x$, $y$, and $z$ are real numbers such that $2x^{2} + y^{2} + z^{2} = 2x - 4y + 2xz - 5$, find the maximum possible value of $x - y + z$.",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nThe equation rearranges as $(x-1)^{2} + (y+2)^{2} + (x-z)^{2} = 0$, so we must have $x = 1$, $y = -2$, $z = 1$, giving us $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe are given triangle $ABC$, with $AB = 9$, $AC = 10$, and $BC = 12$, and a point $D$ on $BC$. $B$ and $C$ are reflected in $AD$ to $B'$ and $C'$, respectively. Suppose that lines $BC'$ and $B'C$ never meet (i.e., are parallel and distinct). Find $BD$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nThe lengths of $AB$ and $AC$ are irrelevant. Because the figure is symmetric about $AD$, lines $BC'$ and $B'C$ meet if and only if they meet at a point on line $AD$. So, if they never meet, they must be parallel to $AD$. Because $AD$ and $BC'$ are parallel, triangles $ABD$ and $ADC'$ have the same area. Then $ABD$ and $ADC$ also have the same area. Hence, $BD$ and $CD$ must have the same length, so $BD = \\frac{1}{2} BC = 6$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$O K R A$ is a trapezoid with $O K$ parallel to $R A$. If $O K=12$ and $R A$ is a positive integer, how many integer values can be taken on by the length of the segment in the trapezoid, parallel to $O K$, through the intersection of the diagonals?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nLet $R A = x$. If the diagonals intersect at $X$, and the segment is $P Q$ with $P$ on $K R$, then $\\triangle P K X \\sim \\triangle R K A$ and $\\triangle O K X \\sim \\triangle R A X$ (by equal angles), giving $R A / P X = A K / X K = 1 + A X / X K = 1 + A R / O K = (x+12)/12$, so $P X = 12x/(12+x)$. Similarly $X Q = 12x/(12+x)$ also, so $P Q = 24x/(12+x) = 24 - \\frac{288}{12+x}$. This has to be an integer. $288 = 2^{5} 3^{2}$, so it has $(5+1)(3+1) = 18$ divisors. $12 + x$ must be one of these. We also exclude the 8 divisors that don't exceed 12, so our final answer is 10.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA certain lottery has tickets labeled with the numbers $1,2,3, \\ldots, 1000$. The lottery is run as follows: First, a ticket is drawn at random. If the number on the ticket is odd, the drawing ends; if it is even, another ticket is randomly drawn (without replacement). If this new ticket has an odd number, the drawing ends; if it is even, another ticket is randomly drawn (again without replacement), and so forth, until an odd number is drawn. Then, every person whose ticket number was drawn (at any point in the process) wins a prize.\nYou have ticket number $1000$. What is the probability that you get a prize?",
    "options": [],
    "answer": "1/501",
    "solution": "Solution:\n$1 / 501$\nNotice that the outcome is the same as if the lottery instead draws all the tickets, in random order, and awards a prize to the holder of the odd ticket drawn earliest and each even ticket drawn before it. Thus, the probability of your winning is the probability that, in a random ordering of the tickets, your ticket precedes all the odd tickets. This, in turn, equals the probability that your ticket is the first in the set consisting of your own and all odd-numbered tickets, irrespective of the other even-numbered tickets. Since all $501!$ orderings of these tickets are equally likely, the desired probability is $1 / 501$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA teacher must divide 221 apples evenly among 403 students. What is the minimal number of pieces into which she must cut the apples? (A whole uncut apple counts as one piece.)",
    "options": [],
    "answer": "611",
    "solution": "Solution:\nConsider a bipartite graph, with 221 vertices representing the apples and 403 vertices representing the students; each student is connected to each apple that she gets a piece of. The number of pieces then equals the number of edges in the graph. Each student gets a total of $221 / 403 = 17 / 31$ apple, but each component of the graph represents a complete distribution of an integer number of apples to an integer number of students and therefore uses at least 17 apple vertices and 31 student vertices. Then we have at most $221 / 17 = 403 / 31 = 13$ components in the graph, so there are at least $221 + 403 - 13 = 611$ edges. On the other hand, if we simply distribute in the straightforward manner - proceeding through the students, cutting up a new apple whenever necessary but never returning to a previous apple or student - we can create a graph without cycles, and each component does involve 17 apples and 31 students. Thus, we get 13 trees, and 611 edges is attainable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA quagga is an extinct chess piece whose move is like a knight's, but much longer: it can move 6 squares in any direction (up, down, left, or right) and then 5 squares in a perpendicular direction. Find the number of ways to place 51 quaggas on an $8 \\times 8$ chessboard in such a way that no quagga attacks another. (Since quaggas are naturally belligerent creatures, a quagga is considered to attack quaggas on any squares it can move to, as well as any other quaggas on the same square.)",
    "options": [],
    "answer": "68",
    "solution": "Solution:\nRepresent the 64 squares of the board as vertices of a graph, and connect two vertices by an edge if a quagga can move from one to the other. The resulting graph consists of 4 paths of length 5 and 4 paths of length 3 (given by the four rotations of the two paths shown, next page), and 32 isolated vertices. Each path of length 5 can accommodate at most 3 nonattacking quaggas in a unique way (the first, middle, and last vertices), and each path of length 3 can accommodate at most 2 nonattacking quaggas in a unique way; thus, the maximum total number of nonattacking quaggas we can have is $4 \\cdot 3+4 \\cdot 2+32=52$. For 51 quaggas to fit, then, just one component of the graph must contain one less quagga than its maximum.\n\nIf this component is a path of length 5, there are $\\binom{5}{2}-4=6$ ways to place the two quaggas on nonadjacent vertices, and then all the other locations are forced; the 4 such paths then give us $4 \\cdot 6=24$ possibilities this way. If it is a path of length 3, there are 3 ways to place one quagga, and the rest of the board is forced, so we have $4 \\cdot 3=12$ possibilities here. Finally, if it is one of the 32 isolated vertices, we simply leave this square empty, and the rest of the board is forced, so we have 32 possibilities here. So the total is $24+12+32=68$ different arrangements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Rays $BA$ and $ED$ intersect at $F$ while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB \\cdot MD = MC^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Extend segment $DM$ through $M$ to $G$ such that $FG \\parallel CD$.\n![](attached_image_1.png)\nThen $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD \\parallel CG$. Hence $MC = MF$ if and only if $\\angle GCD = \\angle FDA$, that is, $\\angle FDA + \\angle CGF = 180^\\circ$.\n\nBecause quadrilateral $ABED$ is cyclic, $\\angle FDA = \\angle ABE$. It follows that $MC = MF$ if and only if\n$$\n180^\\circ = \\angle FDA + \\angle CGF = \\angle ABE + \\angle CGF,\n$$\nthat is, quadrilateral $CBFG$ is cyclic, which is equivalent to\n$$\n\\angle CBM = \\angle CBG = \\angle CFG = \\angle DCF = \\angle DCM.\n$$\nBecause $\\angle DMC = \\angle CMB$, $\\angle CBM = \\angle DCM$ if and only if triangles $BCM$ and $CDM$ are similar, that is\n$$\n\\frac{CM}{BM} = \\frac{DM}{CM},\n$$\nor $MB \\cdot MD = MC^2$.\n![](attached_image_2.png)\nWe first assume that $MB \\cdot MD = MC^2$. Because $\\frac{MC}{MD} = \\frac{MB}{MC}$ and $\\angle CMD = \\angle BMC$, triangles $CMD$ and $BMC$ are similar. Consequently, $\\angle MCD = \\angle MBC$. Because quadrilateral $ABED$ is cyclic, $\\angle DAE = \\angle DBE$. Hence\n$$\n\\angle FCA = \\angle MCD = \\angle MBC = \\angle DBE = \\angle DAE = \\angle CAE,\n$$\nimplying that $AE \\parallel CF$, and so $\\angle AEF = \\angle CFE$. Because quadrilateral $ABED$ is cyclic, $\\angle ABD = \\angle AED$. Hence\n$$\n\\angle FBM = \\angle ABD = \\angle AED = \\angle AEF = \\angle CFE = \\angle MFD.\n$$\nBecause $\\angle FBM = \\angle DFM$ and $\\angle FMB = \\angle DMF$, triangles $BFM$ and $FDM$ are similar. Consequently, $\\frac{FM}{DM} = \\frac{BM}{FM}$, or $FM^2 = BM \\cdot DM = CM^2$. Therefore $MC^2 = MB \\cdot MD$ implies $MC = MF$.\n\nNow we assume that $MC = MF$. Applying **Ceva's Theorem** to triangle $BCF$ and **cevians** $BM, CA, FE$ gives\n$$\n\\frac{BA}{AF} \\cdot \\frac{FM}{MC} \\cdot \\frac{CE}{EB} = 1,\n$$\nimplying that $\\frac{BA}{AF} = \\frac{BE}{EC}$, so $AE \\parallel CF$. Thus, $\\angle DCM = \\angle DAE$. Because quadrilateral $ABED$ is cyclic, $\\angle DAE = \\angle DBE$. Hence\n$$\n\\angle DCM = \\angle DAE = \\angle DBE = \\angle CBM.\n$$\nBecause $\\angle CBM = \\angle DCM$ and $\\angle CMB = \\angle DMC$, triangles $BCM$ and $CDM$ are similar. Consequently, $\\frac{CM}{DM} = \\frac{BM}{CM}$, or $CM^2 = BM \\cdot DM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N}$ denote the set of positive integers. Find all functions $f : \\mathbb{N} \\to \\mathbb{N}$ such that\n$$\nf(m+n)f(m-n) = f(m^2)\n$$\nfor all $m, n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "f(n) = 1 for all positive integers n",
    "solution": "Function $f(n) = 1$, for all $n \\in \\mathbb{N}$, is the only function satisfying the conditions of the problem.\n\nNote that\n$$\nf(1)f(2n-1) = f(n^2) \\quad \\text{and} \\quad f(3)f(2n-1) = f((n+1)^2)\n$$\nfor $n \\ge 3$. Thus\n$$\n\\frac{f(3)}{f(1)} = \\frac{f((n+1)^2)}{f(n^2)}.\n$$\nSetting $\\frac{f(3)}{f(1)} = k$ yields $f(n^2) = k^{n-3}f(9)$ for $n \\ge 3$. Similarly, for all $h \\ge 1$,\n$$\n\\frac{f(h+2)}{f(h)} = \\frac{f((m+1)^2)}{f(m^2)}\n$$\nfor sufficiently large $m$ and is thus also $k$. Hence $f(2h) = k^{h-1}f(2)$ and $f(2h+1) = k^h f(1)$.\n\nBut\n$$\n\\frac{f(25)}{f(9)} = \\frac{f(25)}{f(23)} \\cdots \\frac{f(11)}{f(9)} = k^8\n$$\nand\n$$\n\\frac{f(25)}{f(9)} = \\frac{f(25)}{f(16)} \\cdot \\frac{f(16)}{f(9)} = k^2,\n$$\nso $k = 1$ and $f(16) = f(9)$. This implies that $f(2h+1) = f(1) = f(2) = f(2j)$ for all $j, h$, so $f$ is constant. From the original functional equation it is then clear that $f(n) = 1$ for all $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn a certain country, there are 100 senators, each of whom has 4 aides. These senators and aides serve on various committees. A committee may consist either of 5 senators, of 4 senators and 4 aides, or of 2 senators and 12 aides. Every senator serves on 5 committees, and every aide serves on 3 committees. How many committees are there altogether?",
    "options": [],
    "answer": "160",
    "solution": "Solution:\nIf each senator gets a point for every committee on which she serves, and every aide gets $1/4$ point for every committee on which he serves, then the 100 senators get 500 points altogether, and the 400 aides get 300 points altogether, for a total of 800 points. On the other hand, each committee contributes 5 points, so there must be $800 / 5 = 160$ committees.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $A B C$ with area $51$, points $D$ and $E$ trisect $A B$ and points $F$ and $G$ trisect $B C$. Find the largest possible area of quadrilateral $D E F G$.",
    "options": [],
    "answer": "17",
    "solution": "Solution:\n\nAssume $E$ is between $D$ and $B$, and $F$ is between $G$ and $B$ (the alternative is to switch two points, say $D$ and $E$, which clearly gives a non-convex quadrilateral with smaller area).\n\nIf two triangles have their bases on the same line and the same opposite vertex, then it follows from the $\\frac{1}{2} b h$ formula that their areas are in the same ratio as their bases. In particular (brackets denote areas),\n$$\n\\frac{[D B G]}{[A B C]} = \\frac{[D B G]}{[A B G]} \\cdot \\frac{[A B G]}{[A B C]} = \\frac{D B}{A B} \\cdot \\frac{B G}{B C} = \\frac{2}{3} \\cdot \\frac{2}{3} = \\frac{4}{9},\n$$\nand similarly $[E B F] / [A B C] = 1 / 9$. Subtracting gives $[D E F G] / [A B C] = 1 / 3$, so the answer is $[A B C] / 3 = 17$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many 2-digit positive integers have an even number of positive divisors?",
    "options": [],
    "answer": "84",
    "solution": "Solution:\n\nAn integer has an odd number of divisors precisely if it is a square. So we take the 90 2-digit numbers $10, 11, \\ldots, 99$ and remove the 6 squares $4^{2}, 5^{2}, \\ldots, 9^{2}$, for a total of 84.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $z = 1 - 2i$. Find $\\frac{1}{z} + \\frac{2}{z^{2}} + \\frac{3}{z^{3}} + \\cdots$.",
    "options": [],
    "answer": "(2i - 1)/4",
    "solution": "Solution:\nLet $x = \\frac{1}{z} + \\frac{2}{z^{2}} + \\frac{3}{z^{3}} + \\cdots$, so $z \\cdot x = \\left(1 + \\frac{2}{z} + \\frac{3}{z^{2}} + \\frac{4}{z^{3}} + \\cdots\\right)$.\n\nThen $z \\cdot x - x = 1 + \\frac{1}{z} + \\frac{1}{z^{2}} + \\frac{1}{z^{3}} + \\cdots = \\frac{1}{1 - 1/z} = \\frac{z}{z - 1}$.\n\nSolving for $x$ in terms of $z$, we obtain $x = \\frac{z}{(z - 1)^{2}}$.\n\nPlugging in the original value of $z$ produces $x = \\frac{2i - 1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe circle $\\omega$ is drawn through the vertices $A$ and $B$ of the triangle $ABC$. If $\\omega$ intersects $AC$ at point $M$ and $BC$ at point $P$, and the segment $MP$ contains the center of the circle inscribed in $ABC$. Given that $AB = c$, $BC = a$ and $CA = b$, find $MP$.",
    "options": [],
    "answer": "c(a + b)/(a + b + c)",
    "solution": "Solution:\nSince $AMPB$ is cyclic, $\\angle CMP = \\angle CBA$ and $\\angle CPM = \\angle CAB$. Therefore the triangle $CMP$ is similar to the triangle $CBA$.\n\nLet $CM = x \\cdot CB = x a$, where $x$ is the coefficient of similarity. Then $CP = x \\cdot CA = x b$, $MP = x \\cdot AB = x c$.\n\nIf $I$ is the center of the circle inscribed in $ABC$ and $r$ its radius, we have the following equality for the areas:\n$$S_{CMP} = S_{CMI} + S_{CPI}.$$\nBut $S_{CMP} = \\frac{1}{2} x a x b \\sin C$, and $S_{CMI} = \\frac{1}{2} x a r$, $S_{CPI} = \\frac{1}{2} x b r$.\n\nPlugging these in and solving for $x$ we get:\n$$x = \\frac{a + b}{a + b + c}.$$\nTherefore,\n$$MP = x c = \\frac{c(a + b)}{a + b + c}.$$\nSolution:\nKeeping the notation from the first solution, we have\n\n$$MP = MI + IP = r\\left(\\frac{1}{\\sin B} + \\frac{1}{\\sin A}\\right) = r\\left(\\frac{a c}{2 S_{ABC}} + \\frac{b c}{2 S_{ABC}}\\right) = \\frac{c(a + b)}{a + b + c}.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFarmer John is inside of an ellipse with reflective sides, given by the equation $x^{2} / a^{2} + y^{2} / b^{2} = 1$, with $a > b > 0$. He is standing at the point $(3, 0)$, and he shines a laser pointer in the $y$-direction. The light reflects off the ellipse and proceeds directly toward Farmer Brown, traveling a distance of $10$ before reaching him. Farmer John then spins around in a circle; wherever he points the laser, the light reflects off the wall and hits Farmer Brown. What is the ordered pair $(a, b)$?",
    "options": [],
    "answer": "(5, 4)",
    "solution": "Solution:\nThe points where the farmers are standing must be the foci of the ellipse, so they are $(3, 0)$ and $(-3, 0)$. If the total distance traveled is $10$, then $a$ must be half of that, or $5$, since the distance traveled by a ray reflecting off the wall from when it leaves one focus to when it reaches the other focus is $2a$, the length of the major axis. If a focus is at $x = 3$, then we have $3 = \\sqrt{a^{2} - b^{2}} = \\sqrt{25 - b^{2}}$, yielding $b = 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSimplify: $2 \\sqrt{1.5+\\sqrt{2}}-(1.5+\\sqrt{2})$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\nThe given expression equals $\\sqrt{6+4 \\sqrt{2}}-(1.5+\\sqrt{2}) = \\sqrt{6+2 \\sqrt{8}}-(1.5+\\sqrt{2})$.\n\nBut on inspection, we see that $(\\sqrt{2}+\\sqrt{4})^{2} = 6+2 \\sqrt{8}$, so the answer is $(\\sqrt{2}+\\sqrt{4})-(1.5+\\sqrt{2}) = 2-3/2 = 1/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequence of real numbers $x_{1}, x_{2}, x_{3}, \\ldots$ satisfies $\\lim_{n \\rightarrow \\infty}\\left(x_{2 n}+x_{2 n+1}\\right)=315$ and $\\lim_{n \\rightarrow \\infty}\\left(x_{2 n}+x_{2 n-1}\\right)=2003$. Evaluate $\\lim_{n \\rightarrow \\infty}\\left(x_{2 n} / x_{2 n+1}\\right)$.",
    "options": [],
    "answer": "-1",
    "solution": "Solution:\n\nWe have $\\lim_{n \\rightarrow \\infty}\\left(x_{2 n+1}-x_{2 n-1}\\right)=\\lim_{n \\rightarrow \\infty}\\left[\\left(x_{2 n}+x_{2 n+1}\\right)-\\left(x_{2 n}+x_{2 n-1}\\right)\\right]=315-2003=-1688$; it follows that $x_{2 n+1} \\rightarrow -\\infty$ as $n \\rightarrow \\infty$. Then\n\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{x_{2 n}}{x_{2 n+1}}=\\lim_{n \\rightarrow \\infty} \\frac{x_{2 n}+x_{2 n+1}}{x_{2 n+1}}-1=-1\n$$\n\nsince $x_{2 n}+x_{2 n+1} \\rightarrow 315$ while $x_{2 n+1} \\rightarrow -\\infty$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAll the sequences consisting of five letters from the set $\\{T, U, R, N, I, P\\}$ (with repetitions allowed) are arranged in alphabetical order in a dictionary. Two sequences are called \"anagrams\" of each other if one can be obtained by rearranging the letters of the other. How many pairs of anagrams are there that have exactly 100 other sequences between them in the dictionary?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nConvert each letter to a digit in base $6$: $I \\mapsto 0$, $N \\mapsto 1$, $P \\mapsto 2$, $R \\mapsto 3$, $T \\mapsto 4$, $U \\mapsto 5$. Then the dictionary simply consists of all base-$6$ integers from $00000_{6}$ to $55555_{6}$ in numerical order. If one number can be obtained from another by a rearrangement of digits, then the numbers are congruent modulo $5$ (this holds because a number $a^{a b c d e_{6}} = 6^{4} \\cdot a + 6^{3} \\cdot b + 6^{2} \\cdot c + 6 \\cdot d + e$ is congruent modulo $5$ to $a + b + c + d + e$), but if there are $100$ other numbers between them, then their difference is $101$, which is not divisible by $5$. So there are no such pairs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA hotel consists of a $2 \\times 8$ square grid of rooms, each occupied by one guest. All the guests are uncomfortable, so each guest would like to move to one of the adjoining rooms (horizontally or vertically). Of course, they should do this simultaneously, in such a way that each room will again have one guest. In how many different ways can they collectively move?",
    "options": [],
    "answer": "1156",
    "solution": "Solution:\n\nImagine that the rooms are colored black and white, checkerboard-style. Each guest in a black room moves to an adjacent white room (and vice versa). If, for each such guest, we place a domino over the original room and the new room, we obtain a covering of the $2 \\times n$ grid by $n$ dominoes, since each black square is used once and each white square is used once. Applying a similar procedure to each guest who begins in a white room and moves to a black room, we obtain a second domino tiling. Conversely, it is readily verified that any pair of such tilings uniquely determines a movement pattern. Also, it is easy to prove by induction that the number of domino tilings of a $2 \\times n$ grid is the $(n+1)$th Fibonacci number (this holds for the base cases $n=1,2$, and for a $2 \\times n$ rectangle, the two rightmost squares either belong to one vertical domino, leaving a $2 \\times (n-1)$ rectangle to be covered arbitrarily, or to two horizontal dominoes which also occupy the adjoining squares, leaving a $2 \\times (n-2)$ rectangle to be covered freely; hence, the numbers of tilings satisfy the Fibonacci recurrence). So the number of domino tilings of a $2 \\times 8$ grid is $34$, and the number of pairs of such tilings is $34^2 = 1156$, the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA tightrope walker stands in the center of a rope of length $32$ meters. Every minute she walks forward one meter with probability $3/4$ and backward one meter with probability $1/4$. What is the probability that she reaches the end in front of her before the end behind her?",
    "options": [],
    "answer": "3^16 / (3^16 + 1)",
    "solution": "Solution:\nAfter one minute, she is three times as likely to be one meter forward as one meter back. After two minutes she is either in the same place, two meters forward, or two meters back. The chance of being two meters forward is clearly $(3/4)^2$ and thus $3^2 = 9$ times greater than the chance of being two back $((1/4)^2)$. We can group the minutes into two-minute periods and ignore the periods of no net movement, so we can consider her to be moving $2$ meters forward or backward each period, where forward movement is $3^2$ times as likely as backward movement. Repeating the argument inductively, we eventually find that she is $3^{16}$ times more likely to move $16$ meters forward than $16$ meters backward, and thus the probability is $$\\frac{3^{16}}{3^{16} + 1}$$ that she will meet the front end of the rope first.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWrite down an integer $N$ between $0$ and $10$, inclusive. You will receive $N$ points unless some other team writes down the same $N$, in which case you receive nothing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA partition of a number $n$ is a sequence of positive integers, arranged in descending order, whose sum is $n$. For example, $n=4$ has 5 partitions: $1+1+1+1=2+1+1=2+2=3+1=4$. Given two different partitions of the same number, $n=a_{1}+a_{2}+\\cdots+a_{k}=b_{1}+b_{2}+\\cdots+b_{l}$, where $k \\leq l$, the first partition is said to dominate the second if all of the following inequalities hold:\n$$\n\\begin{aligned}\na_{1} & \\geq b_{1} ; \\\\\na_{1}+a_{2} & \\geq b_{1}+b_{2} ; \\\\\na_{1}+a_{2}+a_{3} & \\geq b_{1}+b_{2}+b_{3} ; \\\\\n& \\vdots \\\\\na_{1}+a_{2}+\\cdots+a_{k} & \\geq b_{1}+b_{2}+\\cdots+b_{k} .\n\\end{aligned}\n$$\nFind as many partitions of the number $n=20$ as possible such that none of the partitions dominates any other. Your score will be the number of partitions you find. If you make a mistake and one of your partitions does dominate another, your score is the largest $m$ such that the first $m$ partitions you list constitute a valid answer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind a set $S$ of positive integers such that no two distinct subsets of $S$ have the same sum. Your score will be $\\left\\lfloor 20\\left(2^{n} / r-2\\right)\\right\\rfloor$, where $n$ is the number of elements in the set $S$, and $r$ is the largest element of $S$ (assuming, of course, that this number is nonnegative).",
    "options": [],
    "answer": "An optimal choice is S = {1, 2, 4, ..., 2^{n-1}}, which yields 2^n / r = 2 and thus the maximal score of 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a square, and let $E$ be an internal point on side $AD$. Let $F$ be the foot of the perpendicular from $B$ to $CE$. Suppose $G$ is a point such that $BG = FG$, and the line through $G$ parallel to $BC$ passes through the midpoint of $EF$. Prove that $AC < 2 \\cdot FG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst note that, for given $E, F$, there is only one point $G$ with the required properties: since $BG = FG$, $G$ must lie on the perpendicular bisector of $BF$, and by the second condition, $G$ lies on the line through the midpoint of $EF$ parallel to $BC$; $G$ must thus be the unique intersection of these two lines. We now give an alternative construction for $G$ from which the result will follow.\n\nSince $BA, CD$ are parallel and equal, we may translate $\\triangle CDE$ to give a triangle $\\triangle BAE'$. Then $E', A, E$ are collinear, with $E'E = E'A + AE = ED + AE = AD = BC$, and $E'E, BC$ are parallel, so we may also translate $\\triangle BCF$ to give $\\triangle E'EF'$. Then $F, E, F'$ are collinear. Now $E'F' = BF$, and $\\angle E'F'E = \\angle BFC = 90^\\circ = \\angle BFE$; hence, $BE'F'F$ is a rectangle. Let $G'$ be its center. Certainly $BG' = FG'$. Let the line through $G'$ parallel to $BC$ hit $BE', FF'$ at $M, N$ respectively; then symmetry gives $E'M = FN$. However, by translation, $EN = E'M$, so $EN = FN$, and the parallel to $BC$ through $G'$ bisects $EF$. Thus, by the uniqueness of $G$ already proven, $G = G'$, the center of $BE'F'F$.\n\n![](attached_image_1.png)\n\nNow,\n$$\n2 \\cdot FG = FE' = \\sqrt{BF^2 + BE'^2} > \\sqrt{2 \\cdot BF \\cdot BE'}\n$$\n(by AM-GM, strict because $BF < BC = CD < CE = BE'$)\n\n$$\n= \\sqrt{2[ BFF'E' ]} = \\sqrt{2( [BAE'] + [E'EF'] + [ABFE] )}\n$$\n(here brackets denote areas)\n$$\n\\begin{gathered}\n= \\sqrt{2([CDE] + [BCF] + [ABFE])} = \\sqrt{2[ABCD]} \\\\\n= \\sqrt{2} \\cdot AB = AC.\n\\end{gathered}\nSolution:\n\nThe following solution, due to Philip Sung of Saratoga High School, not only establishes the inequality, but computes the difference.\n\n![](attached_image_2.png)\n\nLet $I$ bisect $\\overline{BF}$ and $J$ bisect $\\overline{EF}$. Extend $AD$; and draw a parallel to $\\overline{FE}$ through $B$; let this meet $AD$ at $H$. Extend $\\overrightarrow{BF}$ so it meets $DC$ at $K$.\n\nWe will show that $FBHE$ is the image of $FIGJ$ under a scaling with ratio $2$ and center $F$. Obviously, $F$ is taken to $F$, $I$ is taken to $B$, and $J$ to $E$. $G$ is the intersection of $IG$ and $JG$, so it is mapped to the intersection of their images, $\\overline{BH}$ is the image of $\\overline{IG}$ because it is parallel to $\\overline{IG}$ and passes through the image of $I$ (i.e., $B$). Similarly, $\\overline{EH}$ is the image of $\\overline{JG}$. $H$ lies on both $\\overline{EH}$ and $\\overline{BH}$, so it is the image of $G$.\n\n$\\angle HBF$ is right, as is $\\angle ABC$. So they cut off equal angles: $\\angle HCA \\cong \\angle KBC$. In addition, because $\\overline{AB} \\cong \\overline{CB}$ and $\\angle HAB \\cong \\angle KCB$ (they are both right), $\\triangle HAB \\cong \\triangle KCB$ and $\\overline{AH} \\cong \\overline{CK}$.\n\n$FH = 2FG$ because $H$ is the image of $G$ under aforementioned dilation. The problem statement then reduces to:\n\n$$\nAC < FH.\n$$\n\nWe will show, equivalently, that $AC^2 < FH^2$ ($AC$ and $FH$ are both positive.)\n\n$$\n\\begin{array}{rlrl}\nFH^2 & = FB^2 + BH^2 & & (\\triangle FBH \\text{ is right}) \\\\\n& = FB^2 + BA^2 + AH^2 & & (\\triangle BAH \\text{ is right}) \\\\\n& = FB^2 + BA^2 + CK^2 & & (AH = CK) \\\\\n& = FB^2 + BA^2 + CF^2 + FK^2 & (\\triangle CK \\text{ is right}) \\\\\n& = (FB^2 + CF^2) + BA^2 + FK^2 & \\\\\n& = BC^2 + BA^2 + FK^2 & & (\\triangle BFC \\text{ is right}) \\\\\n& = AC^2 + FK^2 & & (\\triangle ABC \\text{ is right}) \\\\\n& > AC^2 & & (FK \\text{ is nonzero}) \\\\\nFH^2 & > AC^2 & &\n\\end{array}\n$$\n\nand we're done!\n\n$$\n\\therefore\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n$$\n\\frac{(2a + b + c)^2}{2a^2 + (b+c)^2} + \\frac{(2b + c + a)^2}{2b^2 + (c+a)^2} + \\frac{(2c + a + b)^2}{2c^2 + (a+b)^2} \\le 8.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** (Based on work by Matthew Tang and Anders Kaseorg)\nBy multiplying $a$, $b$, and $c$ by a suitable factor, we reduce the problem to the case when $a + b + c = 3$. The desired inequality reads\n$$\n\\frac{(a+3)^2}{2a^2+(3-a)^2} + \\frac{(b+3)^2}{2b^2+(3-b)^2} + \\frac{(c+3)^2}{2c^2+(3-c)^2} \\le 8.\n$$\nSet\n$$\nf(x) = \\frac{(x+3)^2}{2x^2 + (3-x)^2}\n$$\nIt suffices to prove that $f(a) + f(b) + f(c) \\le 8$. Note that\n$$\n\\begin{aligned}\nf(x) &= \\frac{x^2 + 6x + 9}{3(x^2 - 2x + 3)} = \\frac{1}{3} \\cdot \\frac{x^2 + 6x + 9}{x^2 - 2x + 3} \\\\\n&= \\frac{1}{3} \\left( 1 + \\frac{8x + 6}{x^2 - 2x + 3} \\right) \\\\\n&= \\frac{1}{3} \\left( 1 + \\frac{8x + 6}{(x-1)^2 + 2} \\right) \\le \\frac{1}{3} \\left( 1 + \\frac{8x + 6}{2} \\right) \\\\\n&= \\frac{1}{3}(4x + 4).\n\\end{aligned}\n$$\nHence,\n$$\nf(a) + f(b) + f(c) \\le \\frac{1}{3}(4a + 4 + 4b + 4 + 4c + 4) = 8,\n$$\nas desired, with equality if and only if $a = b = c$.\n\n\n**Second Solution.** (By Liang Qin) Setting $x = a+b$, $y = b+c$, $z = c+a$ gives $2a+b+c = x+z$, hence $2a = x+z-y$ and their analogous forms. The desired inequality becomes\n$$\n\\frac{2(x+z)^2}{(x+z-y)^2+2y^2} + \\frac{2(z+y)^2}{(z+y-x)^2+2x^2} + \\frac{2(y+x)^2}{(y+x-z)^2+2z^2} \\le 8.\n$$\nBecause $2(s^2 + t^2) \\ge (s + t)^2$ for all real numbers $s$ and $t$, we have $2(x + z - y)^2 + 2y^2 \\ge (x + z - y + y)^2 = (x + z)^2$. Hence\n$$\n\\begin{aligned}\n\\frac{2(x+z)^2}{(x+z-y)^2+2y^2} &= \\frac{4(x+z)^2}{2(x+z-y)^2+4y^2} \\le \\frac{4(x+z)^2}{(x+z)^2+2y^2} \\\\\n&= \\frac{4}{1+2 \\cdot \\frac{y^2}{(x+z)^2}} \\le \\frac{4}{1+2 \\cdot \\frac{y^2}{2(x^2+z^2)}} \\\\\n&= \\frac{4(x^2+z^2)}{x^2+y^2+z^2}.\n\\end{aligned}\n$$\nIt is not difficult to see that the desired result follows from summing up the above inequality and its analogous forms.\n\n\n**Third Solution.** (By Richard Stong) Note that\n$$\n\\begin{aligned}\n(2x + y)^2 + 2(x - y)^2 &= 4x^2 + 4xy + y^2 + 2x^2 - 4xy + 2y^2 \\\\\n&= 3(2x^2 + y^2).\n\\end{aligned}\n$$\nSetting $x = a$ and $y = b + c$ yields\n$$\n(2a + b + c)^2 + 2(a - b - c)^2 = 3(2a^2 + (b+c)^2).\n$$\nThus, we have\n$$\n\\begin{aligned} \\frac{(2a + b + c)^2}{2a^2 + (b + c)^2} &= \\frac{3(2a^2 + (b+c)^2) - 2(a-b-c)^2}{2a^2 + (b+c)^2} \\\\\n&= 3 - \\frac{2(a-b-c)^2}{2a^2 + (b+c)^2}. \\end{aligned}\n$$\nand its analogous forms. Thus, the desired inequality is equivalent to\n$$\n\\frac{(a - b - c)^2}{2a^2 + (b + c)^2} + \\frac{(b - a - c)^2}{2b^2 + (c + a)^2} + \\frac{(c - a - b)^2}{2c^2 + (a + b)^2} \\geq \\frac{1}{2}.\n$$\nBecause $(b+c)^2 \\le 2(b^2+c^2)$, we have $2a^2+(b+c)^2 \\le 2(a^2+b^2+c^2)$ and its analogous forms. It suffices to show that\n$$\n\\frac{(a - b - c)^2}{2(a^2 + b^2 + c^2)} + \\frac{(b - a - c)^2}{2(a^2 + b^2 + c^2)} + \\frac{(c - a - b)^2}{2(a^2 + b^2 + c^2)} \\ge \\frac{1}{2},\n$$\nor,\n$$\n(a - b - c)^2 + (b - a - c)^2 + (c - a - b)^2 \\ge a^2 + b^2 + c^2.\n$$\nMultiplying this out, the left-hand side of the last inequality becomes $3(a^2+b^2+c^2)-2(ab+bc+ca)$. Therefore the last inequality is equivalent to $2[a^2 + b^2 + c^2 - (ab + bc + ca)] \\ge 0$, which is evident because\n$$\n2[a^2 + b^2 + c^2 - (ab + bc + ca)] = (a - b)^2 + (b - c)^2 + (c - a)^2.\n$$\nEqualities hold if and only if $(b+c)^2 = 2(b^2+c^2)$ and $(c+a)^2 = 2(c^2+a^2)$, that is, $a = b = c$.\n\n\n**Fourth Solution.** We first convert the inequality into\n$$\n\\frac{2a(a + 2b + 2c)}{2a^2 + (b + c)^2} + \\frac{2b(b + 2c + 2a)}{2b^2 + (c + a)^2} + \\frac{2c(c + 2a + 2b)}{2c^2 + (a + b)^2} \\le 5.\n$$\nSplitting the 5 among the three terms yields the equivalent form\n$$\n\\sum_{\\text{cyc}} \\frac{4a^2 - 12a(b + c) + 5(b + c)^2}{3[2a^2 + (b + c)^2]} \\ge 0, \\quad (1)\n$$\nwhere $\\sum_{\\text{cyc}}$ is the **cyclic sum** of variables $(a, b, c)$. The numerator of the term shown factors as $(2a-x)(2a-5x)$, where $x = b+c$. We will show\n$$\n\\frac{(2a-x)(2a-5x)}{3(2a^2+x^2)} \\geq -\\frac{4(2a-x)}{3(a+x)}. \\qquad (2)\n$$\nIndeed, (2) is equivalent to\n$$\n(2a-x)[(2a-5x)(a+x) + 4(2a^2+x^2)] \\geq 0,\n$$\nwhich reduces to\n$$\n(2a-x)(10a^2 - 3ax - x^2) = (2a-x)^2(5a+x) \\geq 0,\n$$\nwhich is evident. We proved that\n$$\n\\frac{4a^2 - 12a(b+c) + 5(b+c)^2}{3[2a^2 + (b+c)^2]} \\geq -\\frac{4(2a-b-c)}{3(a+b+c)},\n$$\nhence (1) follows. Equality holds if and only if $2a = b + c$, $2b = c + a$, $2c = a + b$, i.e., when $a = b = c$.\n\n\n**Fifth Solution.** Given a function $f$ of $n$ variables, we define the symmetric sum\n$$\n\\sum_{\\text{sym}} f(x_1, \\dots, x_n) = \\sum_{\\sigma} f(x_{\\sigma(1)}, \\dots, x_{\\sigma(n)})\n$$\nwhere $\\sigma$ runs over all permutations of $1, \\dots, n$ (for a total of $n!$ terms). For example, if $n = 3$, and we write $x, y, z$ for $x_1, x_2, x_3$,\n$$\n\\begin{aligned} \\sum_{\\text{sym}} x^3 &= 2x^3 + 2y^3 + 2z^3 \\\\\n\\sum_{\\text{sym}} x^2y &= x^2y + y^2z + z^2x + x^2z + y^2x + z^2y \\\\\n\\sum_{\\text{sym}} xyz &= 6xyz. \\end{aligned}\n$$\nWe combine the terms in the desired inequality over a common denominator and use symmetric sum notation to simplify the algebra. The numerator of the difference between the two sides is\n$$\n2 \\sum_{\\text{sym}} (4a^6 + 4a^5b + a^4b^2 + 5a^4bc + 5a^3b^3 - 26a^3b^2c + 7a^2b^2c^2), \\quad (3)\n$$\nand it suffices to show the the expression in (3) is always greater or equal to 0. By the **Weighted AM-GM Inequality**, we have $4a^6 + b^6 + c^6 \\geq 6a^4bc$, $3a^5b + 3a^5c + b^5a + c^5a \\geq 8a^4bc$, and their analogous forms. Adding those inequalities yields\n$$\n\\sum_{\\text{sym}} 6a^6 \\geq \\sum_{\\text{sym}} 6a^4bc \\quad \\text{and} \\quad \\sum_{\\text{sym}} 8a^5b \\geq \\sum_{\\text{sym}} 8a^4bc.\n$$\nConsequently, we obtain\n$$\n\\sum_{\\text{sym}} 4a^6 + 4a^5b + 5a^4bc \\geq \\sum_{\\text{sym}} 13a^4bc. \\quad (4)\n$$\nAgain by the AM-GM Inequality, we have $a^4b^2 + b^4c^2 + c^4a^2 \\geq 4a^2b^2c^2$, $a^3b^3 + b^3c^3 + c^3a^3 \\geq 3a^2b^2c^2$, and their analogous forms. Thus,\n$$\n\\sum_{\\text{sym}} a^4b^2 + 5a^3b^3 \\geq \\sum_{\\text{sym}} 6a^2b^2c^2,\n$$\nor\n$$\n\\sum_{\\text{sym}} a^4b^2 + 5a^3b^3 + 7a^2b^2c^2 \\geq \\sum_{\\text{sym}} 13a^2b^2c^2. \\quad (5)\n$$\nRecalling **Schur's Inequality**, we have\n$$\n\\begin{aligned} & a^3 + b^3 + c^3 + 3abc - (a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2) \\\\\n& = a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \\geq 0, \\end{aligned}\n$$\nor\n$$\n\\sum_{\\text{sym}} a^3 - 2a^2b + abc \\geq 0.\n$$\nThus\n$$\n\\sum_{\\text{sym}} 13a^4bc - 26a^3b^2c + 13a^2b^2c^2 \\geq 13abc \\sum_{\\text{sym}} a^3 - 2a^2b + abc \\geq 0. \\quad (6)\n$$\nAdding (4), (5), and (6) yields (3).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be real numbers in the interval $(0, \\frac{\\pi}{2})$. Prove that\n$$\n\\frac{\\sin a \\sin(a-b) \\sin(a-c)}{\\sin(b+c)} + \\frac{\\sin b \\sin(b-c) \\sin(b-a)}{\\sin(c+a)} + \\frac{\\sin c \\sin(c-a) \\sin(c-b)}{\\sin(a+b)} \\ge 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the **Product-to-sum formulas** and the **Double-angle formulas**, we have\n$$\n\\begin{aligned}\n\\sin(\\alpha - \\beta) \\sin(\\alpha + \\beta) &= \\frac{1}{2}[\\cos 2\\beta - \\cos 2\\alpha] \\\\\n&= \\sin^2 \\alpha - \\sin^2 \\beta.\n\\end{aligned}\n$$\nHence, we obtain\n$$\n\\begin{aligned}\n& \\sin a \\sin(a-b) \\sin(a-c) \\sin(a+b) \\sin(a+c) \\\\\n&= \\sin c(\\sin^2 a - \\sin^2 b)(\\sin^2 a - \\sin^2 c)\n\\end{aligned}\n$$\nand its analogous forms. Therefore, it suffices to prove that\n$$\nx(x^2 - y^2)(x^2 - z^2) + y(y^2 - z^2)(y^2 - x^2) + z(z^2 - x^2)(z^2 - y^2) \\ge 0,\n$$\nwhere $x = \\sin a$, $y = \\sin b$, and $z = \\sin c$ (hence $x, y, z > 0$). Since the last inequality is symmetric with respect to $x, y, z$, we may assume that $x \\ge y \\ge z > 0$. It suffices to prove that\n$$\nx(y^2 - x^2)(z^2 - x^2) + z(z^2 - x^2)(z^2 - y^2) \\ge y(z^2 - y^2)(y^2 - x^2),\n$$\nwhich is evident as\n$$\nx(y^2 - x^2)(z^2 - x^2) \\ge 0\n$$\nand\n$$\nz(z^2 - x^2)(z^2 - y^2) \\geq z(y^2 - x^2)(z^2 - y^2) \\geq y(z^2 - y^2)(y^2 - x^2).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute the surface area of a cube inscribed in a sphere of surface area $\\pi$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nThe sphere's radius $r$ satisfies $4 \\pi r^{2} = \\pi \\Rightarrow r = 1/2$, so the cube has body diagonal $1$, hence side length $1/\\sqrt{3}$. So, its surface area is $6(1/\\sqrt{3})^{2} = 2$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $B_{k}(n)$ be the largest possible number of elements in a 2-separable $k$-configuration of a set with $2n$ elements ($2 \\leq k \\leq n$). Find a closed-form expression (i.e. an expression not involving any sums or products with a variable number of terms) for $B_{k}(n)$.",
    "options": [],
    "answer": "B_k(n) = \\binom{2n}{k} - 2\\binom{n}{k}",
    "solution": "Solution:\nFirst, a lemma: For any $a$ with $0 \\leq a \\leq 2n$, $\\binom{a}{k} + \\binom{2n-a}{k} \\geq 2\\binom{n}{k}$. (By convention, we set $\\binom{a}{k} = 0$ when $a < k$.)\n\nProof: We may assume $a \\geq n$, since otherwise we can replace $a$ with $2n-a$. Now we prove the result by induction on $a$. For the base case, if $a = n$, then the lemma states that $2\\binom{n}{k} \\geq 2\\binom{n}{k}$, which is trivial. If the lemma holds for some $a > 0$, then by the familiar identity,\n$$\n\\begin{aligned}\n& \\left[\\binom{a+1}{k} + \\binom{2n-a-1}{k}\\right] - \\left[\\binom{a}{k} + \\binom{2n-a}{k}\\right] \\\\\n=\\ & \\left[\\binom{a+1}{k} - \\binom{a}{k}\\right] - \\left[\\binom{2n-a}{k} - \\binom{2n-a-1}{k}\\right] \\\\\n=\\ & \\binom{a}{k-1} - \\binom{2n-a-1}{k-1} > 0\n\\end{aligned}\n$$\n(since $a > 2n-a-1$), so $\\binom{a+1}{k} + \\binom{2n-a-1}{k} > \\binom{a}{k} + \\binom{2n-a}{k} \\geq 2\\binom{n}{k}$, giving the induction step. The lemma follows.\n\nNow suppose that the elements of $A$ are labeled such that $a$ elements of the set $A$ receive the number 1 and $2n-a$ elements receive the number 2. Then the $k$-configuration can include all $k$-element subsets of $A$ except those contained among the $a$ elements numbered 1 or the $2n-a$ elements numbered 2. Thus, we have at most $\\binom{2n}{k} - \\binom{a}{k} - \\binom{2n-a}{k}$ elements in the $k$-configuration, and by the lemma, this is at most\n$$\n\\binom{2n}{k} - 2\\binom{n}{k}\n$$\nOn the other hand, we can achieve $\\binom{2n}{k} - 2\\binom{n}{k}$ via the recipe above—take all the $k$-element subsets of $A$, except those contained entirely within the first $n$ elements or entirely within the last $n$ elements. Then, labeling the first $n$ elements with the number 1 and the last $n$ elements with the number 2 shows that the configuration is 2-separable. So,\n$$\nB_{k}(n) = \\binom{2n}{k} - 2\\binom{n}{k}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe wish to color the integers $1,2,3, \\ldots, 10$ in red, green, and blue, so that no two numbers $a$ and $b$, with $a-b$ odd, have the same color. (We do not require that all three colors be used.) In how many ways can this be done?",
    "options": [],
    "answer": "186",
    "solution": "Solution:\n\nThe condition is equivalent to never having an odd number and an even number in the same color. We can choose one of the three colors for the odd numbers and distribute the other two colors freely among the 5 even numbers; this can be done in $3 \\cdot 2^{5}=96$ ways. We can also choose one color for the even numbers and distribute the other two colors among the 5 odd numbers, again in 96 ways. This gives a total of 192 possibilities. However, we have double-counted the $3 \\cdot 2=6$ cases where all odd numbers are the same color and all even numbers are the same color, so there are actually $192-6=186$ possible colorings.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor which $n$ is it possible to fill the $n$ by $n$ table with $0$'s, $1$'s and $2$'s so that the sums of numbers in rows and columns take all different values from $1$ to $2n$?",
    "options": [],
    "answer": "n is even",
    "solution": "Solution:\nFor odd $n$ it is impossible to create a table like that. Indeed, in such a table the sum of all column sums and row sums would be $1+2+\\ldots+2n = n(2n+1)$, and so would be odd. But it would also be twice the sum of all the numbers in the table (each number counted twice - once in the column sum, once in the row sum), and so would have to be even. This contradiction proves that no such table exists.\n\nFor any even $n$ such a table exists. If one takes a $2k$ by $2k$ table and fills in all the elements above the main diagonal (running from upper left to lower right corner of the table) with $2$'s, all the elements below the main diagonal with $0$'s, first $k$ elements on the diagonal with $1$'s and the last $k$ elements on the main diagonal with $2$'s, one gets a table that satisfies the conditions. Indeed, the sums of elements in the first $k$ rows (top to bottom) are $4k-1, 4k-3, \\ldots, 2k+1$, the next $k$ rows $2k, 2k-2, \\ldots, 2$. The sums of elements in the first $k$ columns (left to right) are $1, 3, \\ldots, 2k-1$, the next $k$ columns $2k+2, \\ldots, 4k$. We see that all the numbers from $1$ to $4k$ appear exactly once, as wanted.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA room is built in the shape of the region between two semicircles with the same center and parallel diameters. The farthest distance between two points with a clear line of sight is $12$ m. What is the area (in $\\mathrm{m}^{2}$) of the room?",
    "options": [],
    "answer": "18π",
    "solution": "Solution:\n$18 \\pi$\nThe maximal distance is as shown in the figure (next page). Call the radii $R$ and $r$, $R > r$. Then $R^{2} - r^{2} = 6^{2}$ by the Pythagorean theorem, so the area is $(\\pi / 2) \\cdot (R^{2} - r^{2}) = 18 \\pi$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou are given a $10 \\times 2$ grid of unit squares. Two different squares are adjacent if they share a side. How many ways can one mark exactly nine of the squares so that no two marked squares are adjacent?",
    "options": [],
    "answer": "36",
    "solution": "Solution:\nSince each row has only two squares, it is impossible for two marked squares to be in the same row. Therefore, exactly nine of the ten rows contain marked squares.\n\nConsider two cases:\n\nCase 1: The first or last row is empty. These two cases are symmetrical, so assume without loss of generality that the first row is empty. There are two possibilities for the second row: either the first square is marked, or the second square is marked. Since the third row must contain a marked square, and it cannot be in the same column as the marked square in the second row, the third row is determined by the second. Similarly, all the remaining rows are determined. This leaves two possibilities if the first row is empty. Thus, there are four possibilities if the first or last row is empty.\n\nCase 2: The empty row is not the first or last. Then, there are two blocks of (one or more) consecutive rows of marked squares. As above, the configuration of the rows in each of the two blocks is determined by the position of the marked square in the first of its rows. That makes $2 \\times 2 = 4$ possible configurations. There are eight possibilities for the empty row, making a total of 32 possibilities in this case.\n\nTogether, there are 36 possible configurations of marked squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the minimum distance from the point $(0, 5/2)$ to the graph of $y = x^{4}/8$.",
    "options": [],
    "answer": "sqrt(17)/2",
    "solution": "Solution:\nWe want to minimize $x^{2} + \\left(\\frac{x^{4}}{8} - \\frac{5}{2}\\right)^{2} = \\frac{x^{8}}{64} - \\frac{5 x^{4}}{8} + x^{2} + \\frac{25}{4}$, which is equivalent to minimizing $\\frac{z^{4}}{4} - 10 z^{2} + 16 z$, where we have set $z = x^{2}$.\n\nThe derivative of this expression is $z^{3} - 20 z + 16$, which is seen on inspection to have $4$ as a root, leading to the factorization $(z - 4)(z + 2 - 2 \\sqrt{2})(z + 2 + 2 \\sqrt{2})$.\n\nSince $z = x^{2}$ ranges over $[0, \\infty)$, the possible minima are at $z = 0$, $z = -2 + 2 \\sqrt{2}$, and $z = 4$. However, the derivative is positive on $(0, -2 + 2 \\sqrt{2})$, so this leaves only $0$ and $4$ to be tried.\n\nWe find that the minimum is in fact achieved at $z = 4$, so the closest point on the graph is given by $x = \\pm 2$, with distance $\\sqrt{2^{2} + \\left(\\frac{2^{4}}{8} - \\frac{5}{2}\\right)^{2}} = \\frac{\\sqrt{17}}{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 46857,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the vertices of a regular hexagon are written six nonnegative integers whose sum is $2003$. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number $0$ appears at all six vertices.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define the *sum* and *maximum* of a position to be the sum and maximum of the six numbers at the vertices. We will show that from any position in which the sum is odd, it is possible to reach the all-zero position.\nOur strategy alternates between two steps:\n\na) from a position with odd sum, move to a position with exactly one odd number;\nb) from a position with exactly one odd number, move to a position with odd sum and strictly smaller maximum, or to the all-zero position.\n\nNote that no move will ever increase the maximum, so this strategy is guaranteed to terminate, because each step of type (b) decreases the maximum by at least one, and it can only terminate at the all-zero position. It suffices to show how each step can be carried out.\n\nFirst, consider a position\n$$\nA \\begin{smallmatrix} B & C \\\\ F & E \\end{smallmatrix} D\n$$\nwith odd sum. Then either $A + C + E$ or $B + D + F$ is odd; assume without loss of generality that $A + C + E$ is odd. If exactly one of $A, C$ and $E$ is odd, say $A$ is odd, we can make the sequence of moves\n$$\n1 \\begin{smallmatrix} B & 0 \\\\ F & 0 \\end{smallmatrix} D \\rightarrow 1 \\begin{smallmatrix} 1 & 0 \\\\ 1 & 0 \\end{smallmatrix} \\mathbf{0} \\rightarrow \\mathbf{0} \\begin{smallmatrix} 1 & 0 \\\\ 1 & 0 \\end{smallmatrix} 0 \\rightarrow 0 \\begin{smallmatrix} 1 & 0 \\\\ \\mathbf{0} & 0 \\end{smallmatrix} 0 \\pmod{2},\n$$\nwhere a letter or number in boldface represents a move at that vertex, and moves that do not affect each other have been written as a single move for brevity. Hence we can reach a position with exactly one odd number. Similarly, if $A, C, E$ are all odd, then the sequence of moves\n$$\n1 \\begin{smallmatrix} B & 1 \\\\ F & 1 \\end{smallmatrix} D \\rightarrow 1 \\begin{smallmatrix} \\mathbf{0} & 1 \\\\ \\mathbf{0} & 1 \\end{smallmatrix} \\mathbf{0} \\rightarrow 1 \\begin{smallmatrix} 0 & \\mathbf{0} \\\\ 0 & \\mathbf{0} \\end{smallmatrix} 0 \\pmod{2},\n$$\nbrings us to a position with exactly one odd number. Thus we have shown how to carry out step (a).\n\nNow assume that we have a position\n$$\nA \\begin{smallmatrix} B & C \\\\ F & E \\end{smallmatrix} D\n$$\nwith $A$ odd and all other numbers even. We want to reach a position with smaller maximum. Let $M$ be the maximum. There are two cases, depending on the parity of $M$.\n\n* In this case, $M$ is even, so one of $B, C, D, E, F$ is the maximum. In particular, $A < M$.\nWe claim after making moves at $B, C, D, E$, and $F$ in that order, the sum is odd and the maximum is less than $M$. Indeed, the following\n$$\n\\begin{array}{r@{\\;}c@{\\;}l@{\\quad}l@{\\quad}l}\n\\begin{array}{cccc}\n& & \\mathbf{1} & \\\\\n& \\mathbf{0} & \\mathbf{0} & \\\\\n\\mathbf{1} & \\mathbf{0} & \\mathbf{0} & \\\\\n\\mathbf{0} & \\mathbf{0} & \\mathbf{0} &\n\\end{array}\n& \\rightarrow &\n\\begin{array}{c}\n\\mathbf{1} \\\\\n\\mathbf{0} \\\\\n\\mathbf{1} \\\\\n\\mathbf{0} \\\\\n\\mathbf{0}\n\\end{array}\n& \\rightarrow &\n\\begin{array}{cccc}\n\\mathbf{1} & \\mathbf{1} & \\mathbf{0} & \\\\\n\\mathbf{0} & \\mathbf{0} & \\mathbf{0} & \\\\\n\\mathbf{1} & \\mathbf{0} & \\mathbf{1} & \\\\\n\\mathbf{0} & \\mathbf{1} & \\mathbf{1} & \\\\\n\\mathbf{0} & \\mathbf{0} & \\mathbf{1} &\n\\end{array}\n& \\rightarrow \\\\\n& &\n\\begin{array}{cccc}\n\\mathbf{1} & \\mathbf{1} & \\mathbf{1} & \\\\\n\\mathbf{0} & \\mathbf{1} & \\mathbf{1} & \\\\\n\\mathbf{1} & \\mathbf{1} & \\mathbf{1} & \\\\\n\\mathbf{0} & \\mathbf{1} & \\mathbf{1} & \\\\\n\\mathbf{1} & \\mathbf{1} & \\mathbf{1} &\n\\end{array}\n& \\rightarrow &\n\\begin{array}{c}\n\\mathbf{1} \\\\\n\\mathbf{0} \\\\\n\\mathbf{1} \\\\\n\\mathbf{0} \\\\\n\\mathbf{1}\n\\end{array}\n& \\pmod{2}.\n\\end{array}\n$$\nshows how the numbers change in parity with each move. Call this new position\n$$\nA' \\begin{matrix} B' \\\\ F' \\end{matrix} C' \\begin{matrix} C' \\\\ E' \\end{matrix} D'.\n$$\nThe sum is odd, since there are five odd numbers. The numbers $A'$, $B'$, $C'$, $D'$, $E'$ are all less than $M$, since they are odd and $M$ is even, and the maximum can never increase. Also, $F' = |A' - E'| \\le \\max\\{A', E'\\} < M$. So the maximum has been decreased.\n\n* In this case, $M$ is odd, so $M = A$ and the other numbers are all less than $M$.\nIf $C > 0$, then we make moves at $B$, $F$, $A$, and $F$, in that order. The sequence of positions is\n$$\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\begin{array}{cccc}\n& & \\mathbf{1} & \\\\\n& \\mathbf{0} & \\mathbf{0} & \\\\\n\\mathbf{1} & \\mathbf{0} & \\mathbf{0} & \\\\\n\\mathbf{0} & \\mathbf{0} & \\mathbf{0} &\n\\end{array}\n& \\rightarrow &\n\\begin{array}{c}\n\\mathbf{1} \\\\\n\\mathbf{0} \\\\\n\\mathbf{1} \\\\\n\\mathbf{0}\n\\end{array}\n& \\rightarrow \\\\\n& &\n\\begin{array}{cccc}\n\\mathbf{1} & \\mathbf{0} & \\mathbf{0} & \\\\\n\\mathbf{0} & \\mathbf{0} & \\mathbf{1} & \\\\\n\\mathbf{0} & \\mathbf{0} & \\mathbf{0} &\n\\end{array}\n& \\rightarrow \\\\\n& &\n\\begin{array}{cccc}\n\\mathbf{0} & \\mathbf{1} & \\mathbf{0} & \\\\\n\\mathbf{1} & \\mathbf{0} & \\mathbf{0} & \\\\\n\\mathbf{0} & \\mathbf{0} & \\mathbf{0} &\n\\end{array}\n& \\pmod{2}.\n\\end{array}\n$$\nCall this new position\n$$\nA' \\begin{matrix} B' \\\\ F' \\end{matrix} C' \\begin{matrix} C' \\\\ E' \\end{matrix} D'.\n$$\nThe sum is odd, since there is exactly one odd number. As before, the only way the maximum could not decrease is if $B' = A$; but this is impossible, since $B' = |A - C| < A$ because $0 < C < M = A$. Hence we have reached a position with odd sum and lower maximum.\nIf $E > 0$, then we apply a similar argument, interchanging $B$ with $F$ and $C$ with $E$.\nIf $C = E = 0$, then we can reach the all-zero position by the following sequence of moves:\n$$\nA \\begin{matrix} B \\\\ F \\end{matrix} \\mathbf{0} \\mathbf{D} \\rightarrow A \\begin{matrix} A \\\\ A \\end{matrix} \\mathbf{0} \\mathbf{0} \\mathbf{0} \\rightarrow \\mathbf{0} \\begin{matrix} A \\\\ A \\end{matrix} \\mathbf{0} \\mathbf{0} \\mathbf{0} \\rightarrow \\mathbf{0} \\begin{matrix} \\mathbf{0} \\\\ \\mathbf{0} \\end{matrix} \\mathbf{0} \\mathbf{0} \\mathbf{0} \\mathbf{0} .\n$$\n(Here $0$ represents zero, not any even number.)\nHence we have shown how to carry out a step of type (b), proving the desired result. The problem statement follows since $2003$ is odd.\n(By Richard Stong) We will show that if there is a pair of opposite vertices with odd sum (which of course is true if the sum of all the vertices is odd), then we can reduce to a position of all zeros.\nFocus on such a pair $\\{a, d\\}$ with smallest possible $\\max\\{a, d\\}$. We will show we can always reduce this smallest maximum of a pair of opposite vertices with odd sum or reduce to the all-zero position. Because the smallest maximum takes nonnegative integer values, we must be able to achieve the all-zero position.\nTo see this assume without loss of generality that $a \\ge d$ and consider an arc $(a, x, y, d)$ of the position\n$$\na^{x \\ y}^{* *}d\n$$\nConsider updating $x$ and $y$ alternately, starting with $x$. If $\\max\\{x, y\\} > a$, then in at most two updates we reduce $\\max\\{x, y\\}$. Thus, we can repeat this *alternate updating* process and we must eventually reach a point when $\\max\\{x, y\\} \\le a$, and hence this will be true from then on.\nUnder this alternate updating process, the arc of the hexagon will eventually enter a unique cycle of length four modulo $2$ in at most one update. Indeed, we have\n$$\n\\begin{pmatrix} 0 & 0 \\\\ * * & 0 \\end{pmatrix} \\to \\begin{pmatrix} 1 & 0 \\\\ * * & 0 \\end{pmatrix} \\to \\begin{pmatrix} 1 & 1 \\\\ * * & 0 \\end{pmatrix} \\to \\begin{pmatrix} 0 & 1 \\\\ * * & 0 \\end{pmatrix} \\to \\begin{pmatrix} 0 & 0 \\\\ * * & 0 \\end{pmatrix} \\pmod{2}\n$$\nand\n$$\n1^{\\begin{smallmatrix} 0 & 0 \\\\ * & * \\end{smallmatrix}} 0 \\to 1^{\\begin{smallmatrix} 0 & 0 \\\\ * & * \\end{smallmatrix}} 0 \\pmod{2}; \\quad 1^{\\begin{smallmatrix} 1 & 0 \\\\ * & * \\end{smallmatrix}} 0 \\to 1^{\\begin{smallmatrix} 1 & 0 \\\\ * & * \\end{smallmatrix}} 0 \\pmod{2}\n$$\n$$\n1^{\\begin{smallmatrix} 1 & 1 \\\\ * & * \\end{smallmatrix}} 0 \\to 1^{\\begin{smallmatrix} 1 & 1 \\\\ * & * \\end{smallmatrix}} 0 \\pmod{2}; \\quad 1^{\\begin{smallmatrix} 0 & 1 \\\\ * & * \\end{smallmatrix}} 0 \\to 1^{\\begin{smallmatrix} 0 & 1 \\\\ * & * \\end{smallmatrix}} 0 \\pmod{2},\n$$\nor\n$$\n0^{\\begin{smallmatrix} 0 & 1 \\\\ * & * \\end{smallmatrix}} 1 \\to 0^{\\begin{smallmatrix} 1 & 1 \\\\ * & * \\end{smallmatrix}} 1 \\to 0^{\\begin{smallmatrix} 1 & 0 \\\\ * & * \\end{smallmatrix}} 1 \\to 0^{\\begin{smallmatrix} 0 & 0 \\\\ * & * \\end{smallmatrix}} 1 \\to 0^{\\begin{smallmatrix} 0 & 1 \\\\ * & * \\end{smallmatrix}} 1 \\pmod{2}\n$$\nand\n$$\n0^{\\begin{smallmatrix} 0 & 0 \\\\ * & * \\end{smallmatrix}} 1 \\to 0^{\\begin{smallmatrix} 0 & 0 \\\\ * & * \\end{smallmatrix}} 1 \\pmod{2}; \\quad 0^{\\begin{smallmatrix} 0 & 1 \\\\ * & * \\end{smallmatrix}} 1 \\to 0^{\\begin{smallmatrix} 0 & 1 \\\\ * & * \\end{smallmatrix}} 1 \\pmod{2}\n$$\n$$\n0^{\\begin{smallmatrix} 1 & 1 \\\\ * & * \\end{smallmatrix}} 1 \\to 0^{\\begin{smallmatrix} 1 & 0 \\\\ * & * \\end{smallmatrix}} 1 \\pmod{2}; \\quad 0^{\\begin{smallmatrix} 1 & 0 \\\\ * & * \\end{smallmatrix}} 1 \\to 0^{\\begin{smallmatrix} 1 & 0 \\\\ * & * \\end{smallmatrix}} 1 \\pmod{2}.\n$$\nFurther note that each possible parity for $x$ and $y$ will occur equally often.\nApplying this alternate updating process to both arcs $(a, b, c, d)$ and $(a, e, f, d)$ of\n$$\na^b c^c d,\n$$\nwe can make the other four entries be at most $a$ and control their parity. Thus we can create a position\n$$\na^{x_1} x_2 d^{x_5} x_4\n$$\nwith $x_i + x_{i+3}$ $(i = 1, 2)$ odd and $M_i = \\max\\{x_i, x_{i+3}\\} \\le a$. In fact, we can have $m = \\min\\{M_1, M_2\\} < a$, as claimed, unless both arcs enter a cycle modulo $2$ where the values congruent to $a$ modulo $2$ are always exactly $a$. More precisely, because the sum of $x_i$ and $x_{i+3}$ is odd, one of them is not congruent to $a$ and so has its value strictly less than $a$. Thus both arcs must pass through the state $(a, a, a, d)$ (modulo $2$, this is either $(0, 0, 0, 1)$ or $(1, 1, 1, 0)$) in a cycle of length four. It is easy to check that for this to happen, $d = 0$. Therefore, we can achieve the position\n$$\na^a a^a 0.\n$$\nFrom this position, the sequence of moves\n$$\na^a a^a 0 \\to a^{\\begin{smallmatrix} 0 & a \\\\ 0 & a \\end{smallmatrix}} 0 \\to \\begin{smallmatrix} 0 & 0 \\\\ 0 & 0 \\end{smallmatrix} 0\n$$\ncompletes the task.\n(By Tiankai Liu) In the beginning, because $A+B+C+D+E+F$ is odd, either $A+C+E$ or $B+D+F$ is odd; assume without loss of generality it is the former. Perform the following steps repeatedly.\n\na. In this case we assume that $A, C, E$ are all nonzero. Suppose without loss of generality that $A \\ge C \\ge E$. Perform the sequence of moves\n$$\nA \\overset{B}{\\underset{F}{C}} D \\to A \\overset{(\\mathbf{A}-\\mathbf{C})}{\\underset{(\\mathbf{A}-\\mathbf{E})}{C}} E \\overset{(\\mathbf{C}-\\mathbf{E})}{\\to} \\mathbf{C} \\overset{(A-C)}{(A-E)} \\overset{C}{\\underset{(\\mathbf{A}-\\mathbf{C})}{\\to}} (C-E),\n$$\nwhich decreases the sum of the numbers in positions $A, C, E$ while keeping that sum odd.\n\nb. In this case we assume that exactly one among $A, C, E$ is zero. Assume without loss of generality that $A \\ge C > E = 0$. Then, because $A+C+E$ is odd, $A$ must be strictly greater than $C$. Therefore, $-A < A - 2C < A$, and the sequence of moves\n$$\nA \\overset{B}{\\underset{F}{C}} 0 D \\to A \\overset{(\\mathbf{A}-\\mathbf{C})}{\\underset{\\mathbf{A}}{C}} 0 \\overset{\\mathbf{C}}{\\to} \\mathbf{C} \\overset{(A-C)}{\\underset{A}{\\to}} (A-C) |\\mathbf{A}-2\\mathbf{C}| \\mathbf{C},\n$$\ndecreases the sum of the numbers in positions $A, C, E$ while keeping that sum odd.\n\nc. In this case we assume that exactly two among $A, C, E$ are zero. Assume without loss of generality that $A > C = E = 0$. Then perform the sequence of moves\n$$\nA \\overset{B}{\\underset{F}{C}} 0 D \\to A \\overset{\\mathbf{A}}{\\underset{\\mathbf{A}}{0}} 0 \\overset{\\mathbf{0}}{\\to} \\mathbf{0} \\overset{A 0}{\\underset{A 0}{0}} \\overset{\\mathbf{0}}{\\to} \\mathbf{0} \\overset{\\mathbf{0}}{\\underset{\\mathbf{0}}{0}} 0.\n$$\nBy repeatedly applying step (a) as long as it applies, then doing the same for step (b) if necessary, and finally applying step (c) if necessary,\n$$\n\\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}\n$$\ncan eventually be achieved.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\overline{AH_1}$, $\\overline{BH_2}$, and $\\overline{CH_3}$ be the altitudes of an acute scalene triangle $ABC$. The incircle of triangle $ABC$ is tangent to $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$ at $T_1, T_2$, and $T_3$, respectively. For $k = 1, 2, 3$, let $P_i$ be the point on line $H_iH_{i+1}$ (where $H_4 = H_1$) such that $H_iT_iP_i$ is an acute isosceles triangle with $H_iT_i = H_iP_i$. Prove that the circumcircles of triangles $T_1P_1T_2$, $T_2P_2T_3$, $T_3P_3T_1$ pass through a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution.** (By Po-Ru Loh) We begin by showing that points $O_3, H_2, T_2$, and $O_3$ lie on a cyclic. We will prove this by establishing $\\angle O_3O_1H_2 = \\angle O_3T_2C = \\angle O_3T_2H_2$. To find $\\angle O_3O_1H_2$, observe that triangles $H_2AH_3$ and $H_2H_1C$ are similar. Indeed, quadrilateral $BH_3H_2C$\n\n![](attached_image_1.png)\n\nis cyclic so $\\angle H_2H_3A = \\angle C$, and likewise $\\angle CH_1H_2 = \\angle A$. Now, $O_1$ and $O_3$ are corresponding incenters of similar triangles, so it follows that triangles $H_2AO_1$ and $H_2H_1O_3$ are also similar, and hence are related by a **spiral similarity** about $H_2$. Thus,\n$$\n\\frac{AH_2}{H_1H_2} = \\frac{O_1H_2}{O_3H_2}\n$$\nand\n$$\n\\begin{aligned} \\angle AH_2H_1 &= \\angle AH_2O_1 + \\angle O_1H_2H_1 \\\\ &= \\angle O_1H_2H_1 + \\angle H_1H_2O_3 = \\angle O_1H_2O_3. \\end{aligned}\n$$\nIt follows that another spiral similarity about $H_2$ takes triangle $H_2AH_1$ to triangle $H_2O_1O_3$. Hence $\\angle O_3O_1H_2 = \\angle H_1AH_2 = 90^\\circ - \\angle C$.\nWe wish to show that $\\angle O_3T_2C = 90^\\circ - \\angle C$ as well, or in other words, $T_2O_3 \\perp BC$. To do this, drop the altitude from $O_3$ to $BC$ and let it intersect $BC$ at $D$. Triangles $ABC$ and $H_1H_2C$ are similar as before, with corresponding incenters $I$ and $O_3$. Furthermore, $IT_2$ and $O_3D$ also correspond. Hence, $CT_2/T_2A = CD/DH_1$, and so $T_2D \\parallel AH_1$. Thus, $T_2D \\perp BC$, and it follows that $T_2O_3 \\perp BC$.\n\n![](attached_image_2.png)\n\nHaving shown that $O_1H_2T_2O_3$ is cyclic, we may now write $\\angle O_1T_2O_3 = \\angle O_1H_2O_3$. Since triangles $H_2AO_1$ and $H_2H_1O_3$ are related by a spiral similarity about $H_2$, we have\n$$\n\\angle O_1H_2O_3 = \\angle AH_2H_1 = 180^\\circ - \\angle B,\n$$\nby noting that $ABH_2H_1$ is cyclic. Likewise,\n$$\n\\angle O_2T_3O_1 = 180^\\circ - \\angle C \\quad \\text{and} \\quad \\angle O_3T_1O_2 = 180^\\circ - \\angle A,\n$$\nand so $\\angle O_1T_2O_3 + \\angle O_2T_3O_1 + \\angle O_3T_1O_2 = 360^\\circ$. Therefore, $\\angle T_3O_1T_2$, $\\angle T_1O_2T_3$, and $\\angle T_2O_3T_1$ of hexagon $O_1T_2O_3T_1O_2T_3$ also sum to $360^\\circ$. Now let $H$ be the intersection of circles $\\omega_1$ and $\\omega_2$. Then $\\angle T_2HT_3 = 180^\\circ - \\frac{1}{2}\\angle T_3O_1T_2$ and $\\angle T_3HT_1 = 180^\\circ - \\frac{1}{2}\\angle T_1O_2T_3$. Therefore,\n$$\n\\begin{aligned} \\angle T_1HT_2 &= 360^\\circ - \\angle T_2HT_3 - \\angle T_3HT_1 \\\\ &= \\frac{1}{2}\\angle T_3O_1T_2 + \\frac{1}{2}\\angle T_1O_2T_3 = 180^\\circ - \\frac{1}{2}\\angle T_1O_3T_2, \\end{aligned}\n$$\nand so $H$ lies on the circle $\\omega_3$ as well. Hence, circles $\\omega_1$, $\\omega_2$, and $\\omega_3$ share a common point, as wanted.\n\n\n**Second Solution.** (By Anders Kaseorg) Note that $AH_2 = AB \\cos \\angle A$ and $AH_3 = AC \\cos \\angle A$, so triangles $AH_2H_3$ and $ABC$ are similar with ratio $\\cos \\angle A$. Thus, since $O_1$ is the incenter of triangle $AH_2H_3$, $AO_1 = AI \\cos \\angle A$. If $X_1$ is the intersection of segments $AI$ and $T_2T_3$,\n\n![](attached_image_3.png)\n\nwe have $\\angle IX_1T_2 = \\angle AT_2I = 90^\\circ$, and so\n$$\n\\begin{aligned}\nX_1I = T_2I \\cos \\angle T_2IA &= AI \\cos^2 \\angle T_2IA = AI \\sin^2 \\frac{\\angle A}{2} \\\\\n&= AI \\cdot \\frac{1 - \\cos \\angle A}{2} = \\frac{AI - AO_1}{2} = \\frac{O_1I}{2}.\n\\end{aligned}\n$$\nHence $O_1X_1 = X_1I$, so $O_1$ is the reflection of $I$ across line $T_2T_3$, and $O_1T_2 = IT_2 = IT_3 = O_1T_3$. Therefore, $O_1T_2IT_3$, and similarly $O_2T_3IT_1$ and $O_3T_1IT_2$, are rhombi with the same side length $r$, implying that circles $\\omega_1, \\omega_2, \\omega$ have the same radius $r$. We also conclude that $O_1T_2 = T_3I = O_2T_1$ and $O_1T_2 \\parallel T_3I \\parallel O_2T_1$, and so $O_1O_2T_1T_2$ is a parallelogram. Hence the midpoints of $O_1T_1$ and $O_2T_2$ (similarly $O_3T_3$) are the same point $P$, and $O_1O_2O_3$ is the reflection of $T_1T_2T_3$ across $P$. If $H$ is the reflection of $I$ across $P$, we have $O_1H = O_2H = O_3H = r$, that is, $H$ is a common point of the three circumcircles.\n\n\n**Third Solution.** We use directed lengths (along line $BC$, with $C$ to $B$ as the positive direction) and directed angles modulo $180^\\circ$ in this proof. (For segments not lying on line $BC$, we assume its direction as the direction of its projection on line $BC$.) We claim that $\\omega_i$, $i = 1, 2, 3$, all pass through $H$, the orthocenter of triangle $T_1T_2T_3$. Without loss of generality, it suffices to prove that $T_1P_1T_2H$ is cyclic. If $AB = AC$, then $T_1 = H_1 = P_1$ and the case is trivial. Let $AB = c$, $BC = a$, $CA = b$, $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$, and $\\angle ACB = \\gamma$.\n\n![](attached_image_4.png)\n\nLet $Q$ be the intersection of lines $HP_1$ and $BC$. Note that\n$$\n\\begin{align*} \n\\angle HT_2T_1 &= 90^\\circ - \\angle T_2T_1T_3 \\\\ \n&= 90^\\circ - [180^\\circ - \\angle T_3T_1B - \\angle CT_1T_2] \\\\ \n&= 90^\\circ - \\left[180^\\circ - \\left(90^\\circ - \\frac{\\beta}{2}\\right) - \\left(90^\\circ - \\frac{C}{2}\\right)\\right] \\\\ \n&= \\frac{\\alpha}{2}. \n\\end{align*}\n$$\n(Likewise, $\\angle T_2T_1H = \\beta/2$.) Thus to prove that $T_1P_1T_2H$ is cyclic is equivalent to prove that $\\angle QP_1T_1 = \\alpha/2$.\nLet $Q_H$ and $Q_P$ be the respective feet of perpendiculars from $H$ and $P_1$ to line $BC$. Because $\\angle AH_1B = \\angle AH_2B = 90^\\circ$, $ABH_1H_2$ is cyclic, and so $\\angle T_1H_1P_1 = \\angle BH_1P_1 = \\alpha$. Thus triangles $AT_3T_2$ and $H_1T_1P_1$\n\nare similar, implying that\n$$\n\\angle Q_P P_1 T_1 = 90^\\circ - \\angle P_1 T_1 H_1 = 90^\\circ - \\left( 90^\\circ - \\frac{\\angle T_1 H_1 P_1}{2} \\right) = \\frac{\\alpha}{2}.\n$$\nTherefore, to prove that $\\angle Q_P P_1 T_1 = \\alpha/2$, we have now reduced to proving that $Q_P = Q_H$, or\n$$\n\\frac{T_1 Q_P}{T_1 H_1} = \\frac{T_1 Q_H}{T_1 H_1}. \\qquad (1)\n$$\nNote that\n$$\nT_1 H_1 = P_1 H_1 \\quad \\text{and} \\quad \\frac{T_1 Q_P}{T_1 H_1} = 1 - \\frac{Q_P H_1}{T_1 H_1},\n$$\nthat is,\n$$\n\\frac{T_1 Q_P}{T_1 H_1} = 1 - \\frac{Q_P H_1}{P_1 H_1} = 1 - \\cos \\angle T_1 H_1 P_1 = 1 - \\cos \\alpha. \\quad (2)\n$$\nOn the other hand, applying the **Law of Cosines** to triangle $ABC$ gives\n$$\n\\begin{aligned} T_1 H_1 &= T_1 C - H_1 C = \\frac{a+b-c}{2} - b \\cos \\gamma \\\\ &= \\frac{a+b-c}{2} - \\frac{a^2+b^2-c^2}{2a} = \\frac{a(b-c) - (b^2-c^2)}{2a}, \\end{aligned}\n$$\nor\n$$\nT_1 H_1 = \\frac{(b-c)(a-b-c)}{2a} = \\frac{(c-b)(b+c-a)}{2a}. \\quad (3)\n$$\nNow we calculate $T_1 Q_H$. Because $H$ is the orthocenter of triangle $T_1 T_2 T_3$,\n$$\n\\begin{aligned} \\angle T_1 H T_2 &= 180^\\circ - \\angle H T_2 T_1 - \\angle T_2 T_1 H \\\\ &= (90^\\circ - \\angle H T_2 T_1) + (90^\\circ - \\angle T_2 T_1 H) \\\\ &= \\angle T_2 T_1 T_3 + \\angle T_3 T_2 T_1 = 180^\\circ - \\angle T_1 T_3 T_2. \\end{aligned}\n$$\nApplying the **Law of Sines** to triangle $T_1 T_2 H$ and applying the **Extended Law of Sines** to triangle $T_1 T_2 T_3$ gives\n$$\n\\frac{T_1 H}{\\sin \\angle H T_2 T_1} = \\frac{T_1 T_2}{\\sin \\angle T_1 H T_2} = \\frac{T_1 T_2}{\\sin \\angle T_1 T_3 T_2} = 2r,\n$$\nand consequently,\n$$\nT_1 H = 2r \\sin \\frac{\\alpha}{2}.\n$$\nBecause\n$$\n\\begin{aligned}\n\\angle Q_H T_1 H &= \\angle CT_1 T_2 + \\angle T_2 T_1 H = (90^\\circ - \\frac{\\gamma}{2}) + \\frac{\\beta}{2} \\\\\n&= 90^\\circ + \\frac{\\beta - \\gamma}{2},\n\\end{aligned}\n$$\nwe obtain\n$$\nT_1 Q_H = T_1 H \\cos \\angle H T_1 Q_H = 2r \\sin \\frac{\\alpha}{2} \\sin \\frac{\\gamma - \\beta}{2}. \\quad (4)\n$$\nCombining equations (1), (2), (3), and (4), we conclude that it suffices to prove that\n$$\n1 - \\cos \\alpha = \\frac{4ar \\sin \\frac{\\alpha}{2} \\sin \\frac{\\gamma-\\beta}{2}}{(c-b)(b+c-a)}. \\quad (5)\n$$\nApplying the fact\n$$\n\\frac{\\sin \\frac{\\alpha}{2}}{\\cos \\frac{\\alpha}{2}} = \\tan \\frac{\\alpha}{2} = \\frac{r}{AT_2} = \\frac{2r}{b+c-a},\n$$\nand applying the Law of Sines to triangle $ABC$, (5) becomes\n$$\n1 - \\cos \\alpha = \\frac{2 \\sin \\alpha \\sin^2 \\frac{\\alpha}{2} \\sin \\frac{\\gamma-\\beta}{2}}{\\cos \\frac{\\alpha}{2} (\\sin \\gamma - \\sin \\beta)}. \\quad (6)\n$$\nBy the **Double-angle formulas**, $1 - \\cos \\alpha = 2 \\sin^2 \\frac{\\alpha}{2}$ and $\\sin \\alpha = 2 \\sin \\frac{\\alpha}{2} \\cos \\frac{\\alpha}{2}$ and so (6) reads\n$$\n\\sin \\gamma - \\sin \\beta = 2 \\sin \\frac{\\alpha}{2} \\sin \\frac{\\gamma - \\beta}{2}.\n$$\nBy the **Difference-to-product formulas**, the last equation reduces to\n$$\n2 \\cos \\frac{\\beta + \\gamma}{2} \\sin \\frac{\\gamma - \\beta}{2} = 2 \\sin \\frac{\\alpha}{2} \\sin \\frac{\\gamma - \\beta}{2},\n$$\nwhich is evident.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn how many ways can 3 bottles of ketchup and 7 bottles of mustard be arranged in a row so that no bottle of ketchup is immediately between two bottles of mustard? (The bottles of ketchup are mutually indistinguishable, as are the bottles of mustard.)",
    "options": [],
    "answer": "22",
    "solution": "Solution:\n22\nConsider the blocks of consecutive bottles of ketchup in such an arrangement. A block of just one bottle must occur at the beginning or the end of the row, or else it would be between two bottles of mustard. However, a block of two or three bottles can occur anywhere. We cannot have three blocks of one bottle each, since there are only two possible locations for such blocks. Thus, we either have a block of one bottle and a block of two, or one block of all three bottles. In the first case, if the single bottle occurs at the beginning of the row, then anywhere from 1 to 7 bottles of mustard may intervene before the block of 2 ketchup bottles, giving 7 possible arrangements. We likewise have 7 arrangements if the single bottle occurs at the end of the row. Finally, if there is just one block of three bottles, anywhere from 0 to 7 mustard bottles may precede it, giving 8 possible arrangements. So, altogether, we have $7+7+8=22$ configurations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor $n$ an integer, evaluate\n$$\n\\lim_{n \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{n^{2}-0^{2}}}+\\frac{1}{\\sqrt{n^{2}-1^{2}}}+\\cdots+\\frac{1}{\\sqrt{n^{2}-(n-1)^{2}}}\\right)\n$$",
    "options": [],
    "answer": "pi/2",
    "solution": "Solution:\n\nNote that $\\frac{1}{\\sqrt{n^{2}-i^{2}}}=\\frac{1}{n} \\cdot \\frac{1}{\\sqrt{1-\\left(\\frac{i}{n}\\right)^{2}}}$, so that the sum we wish to evaluate is just a Riemann sum. Then,\n\n$$\n\\lim_{n \\rightarrow \\infty}\\left(\\frac{1}{n} \\sum_{i=0}^{n-1} \\frac{1}{\\sqrt{1-\\left(\\frac{i}{n}\\right)^{2}}}\\right)=\\int_{0}^{1} \\frac{1}{\\sqrt{1-x^{2}}} \\, dx=\\left[\\sin^{-1} x\\right]_{0}^{1}=\\frac{\\pi}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center of the room is unoccupied. A teacher decides to reassign the seats such that each student will occupy a chair adjacent to his/her present one (i.e. move one desk forward, back, left or right). In how many ways can this reassignment be made?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nColor the chairs red and black in checkerboard fashion, with the center chair black. Then all 18 red chairs are initially occupied. Also notice that adjacent chairs have different colors. It follows that we need 18 black chairs to accommodate the reassignment, but there are only 17 of them. Thus, the answer is 0.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}=1$, and let $a_{n}=\\left\\lfloor n^{3} / a_{n-1}\\right\\rfloor$ for $n>1$. Determine the value of $a_{999}$.",
    "options": [],
    "answer": "999",
    "solution": "Solution:\nWe claim that for any odd $n$, $a_{n}=n$. The proof is by induction. To get the base cases $n=1$, $3$, we compute $a_{1}=1$, $a_{2}=\\left\\lfloor 2^{3} / 1\\right\\rfloor=8$, $a_{3}=\\left\\lfloor 3^{3} / 8\\right\\rfloor=3$.\n\nAnd if the claim holds for odd $n \\geq 3$, then $a_{n+1}=\\left\\lfloor(n+1)^{3} / n\\right\\rfloor=n^{2}+3 n+3$, so $a_{n+2}=\\left\\lfloor(n+2)^{3} /(n^{2}+3 n+3)\\right\\rfloor=\\left\\lfloor(n^{3}+6 n^{2}+12 n+8) /(n^{2}+3 n+2)\\right\\rfloor=\\left\\lfloor n+2+\\frac{n^{2}+3 n+2}{n^{2}+3 n+3}\\right\\rfloor=n+2$.\n\nSo the claim holds, and in particular, $a_{999}=999$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine the Fibonacci numbers by $F_{0}=0$, $F_{1}=1$, $F_{n}=F_{n-1}+F_{n-2}$ for $n \\geq 2$. For how many $n$, $0 \\leq n \\leq 100$, is $F_{n}$ a multiple of $13$?",
    "options": [],
    "answer": "15",
    "solution": "Solution:\n\nThe sequence of remainders modulo $13$ begins $0, 1, 1, 2, 3, 5, 8, 0$, and then we have $F_{n+7} \\equiv 8 F_{n}$ modulo $13$ by a straightforward induction. In particular, $F_{n}$ is a multiple of $13$ if and only if $7 \\mid n$, so there are $15$ such $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTake a clay sphere of radius $13$, and drill a circular hole of radius $5$ through its center. Take the remaining \"bead\" and mold it into a new sphere. What is this sphere's radius?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\nLet $r$ be the radius of the sphere. We take cross sections of the bead perpendicular to the line of the drill and compare them to cross sections of the sphere at the same distance from its center. At a height $h$, the cross section of the sphere is a circle with radius $\\sqrt{r^{2}-h^{2}}$ and thus area $\\pi\\left(r^{2}-h^{2}\\right)$. At the same height, the cross section of the bead is an annulus with outer radius $\\sqrt{13^{2}-h^{2}}$ and inner radius $5$, for an area of $\\pi\\left(13^{2}-h^{2}\\right)-\\pi\\left(5^{2}\\right)=\\pi\\left(12^{2}-h^{2}\\right)$ (since $13^{2}-5^{2}=12^{2}$). Thus, if $r=12$, the sphere and the bead will have the same cross-sectional area $\\pi\\left(12^{2}-h^{2}\\right)$ for $|h| \\leq 12$ and $0$ for $|h|>12$. Since all the cross sections have the same area, the two clay figures then have the same volume. And certainly there is only one value of $r$ for which the two volumes are equal, so $r=12$ is the answer.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers $112$, $121$, $123$, $153$, $243$, $313$, and $322$ are among the rows, columns, and diagonals of a $3 \\times 3$ square grid of digits (rows and diagonals read left-to-right, and columns read top-to-bottom). What 3-digit number completes the list?",
    "options": [],
    "answer": "524",
    "solution": "Solution:\n\n| 1 | 1 | 2 |\n| :--- | :--- | :--- |\n| 5 | 2 | 4 |\n| 3 | 1 | 3 |\n\nThe center digit is the middle digit of 4 numbers (hence at least 3 members of the above list), so it must be $2$. The top-left digit begins at least 2 members of the above list, so it must be $1$ or $3$. If it is $3$, then after placing $313$ we see that we need three more numbers starting with $3$, impossible; hence, it is $1$. So $243$ and $313$ must (in some order) be the last row and the last column, and now it is easy to complete the grid as shown; the answer is $524$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that any 2-configuration containing $e$ elements is $m$-separable for some $m \\leq \\frac{1}{2} + \\sqrt{2e + \\frac{1}{4}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose $m$ is the minimum integer for which the given configuration $C$ on set $A$ is $m$-separable, and fix a corresponding labeling of the elements of $A$. Let $A_{i}$ be the set of all elements with the label $i$. Then, for any $i, j$ with $1 \\leq i < j \\leq m$, there must exist $a_{i} \\in A_{i}, a_{j} \\in A_{j}$ with $\\{a_{i}, a_{j}\\} \\in C$, since otherwise the elements of $A_{j}$ could have been reassigned the label $i$, decreasing the number of distinct labels necessary and thus contradicting the minimality of $m$.\n\nWe thus get at least $\\binom{m}{2}$ different elements of $C$. Therefore, $e \\geq \\binom{m}{2} = \\frac{m(m-1)}{2} = \\frac{\\left(m-\\frac{1}{2}\\right)^2 - \\frac{1}{4}}{2}$, and solving for $m$ gives the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou have infinitely many boxes, and you randomly put 3 balls into them. The boxes are labeled $1,2, \\ldots$. Each ball has probability $1 / 2^{n}$ of being put into box $n$. The balls are placed independently of each other. What is the probability that some box will contain at least 2 balls?",
    "options": [],
    "answer": "5/7",
    "solution": "Solution:\n$$\n\\frac{5}{7}\n$$\nNotice that the answer is the sum of the probabilities that boxes $1,2, \\ldots$, respectively, contain at least 2 balls, since those events are mutually exclusive. For box $n$, the probability of having at least 2 balls is\n$$\n3\\left[\\left(1 / 2^{n}\\right)^{2}\\left(1-1 / 2^{n}\\right)\\right]+\\left(1 / 2^{n}\\right)^{3}=\\frac{3}{2^{2 n}}-\\frac{2}{2^{3 n}}=\\frac{3}{4^{n}}-\\frac{2}{8^{n}}\n$$\nSumming to infinity using the geometric series formula, we get the answer $\\frac{3/4}{1-1/4}-\\frac{2/8}{1-1/8}$, which is equal to $\\frac{5}{7}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor what value of $a > 1$ is\n\n$$\n\\int_{a}^{a^{2}} \\frac{1}{x} \\log \\frac{x-1}{32} \\, dx\n$$\nminimum?",
    "options": [],
    "answer": "a = 3",
    "solution": "Solution:\nLet $f(a) = \\int_{a}^{a^{2}} \\frac{1}{x} \\log \\frac{x-1}{32} \\, dx$. Then we want $\\frac{d f}{d a} = 0$; by the Fundamental Theorem of Calculus and the chain rule, this implies that\n\n$$\n2a \\left( \\frac{1}{a^{2}} \\log \\frac{a^{2}-1}{32} \\right) - \\frac{1}{a} \\log \\frac{a-1}{32} = \\frac{d}{d a} \\left( \\int_{c}^{a^{2}} \\frac{1}{x} \\log \\frac{x-1}{32} \\, dx - \\int_{c}^{a} \\frac{1}{x} \\log \\frac{x-1}{32} \\, dx \\right) = 0,\n$$\nwhere $c$ is any constant with $1 < c < a$. Then $2 \\log \\frac{a^{2}-1}{32} = \\log \\frac{a-1}{32}$, so that $\\left( \\frac{a^{2}-1}{32} \\right)^{2} = \\frac{a-1}{32}$. After canceling factors of $(a-1)/32$ (since $a > 1$), this simplifies to $\\left(a^{2}-1\\right)(a+1) = 32 \\Rightarrow a^{3} + a^{2} - a - 33 = 0$, which in turn factors as $(a-3)(a^{2} + 4a + 11) = 0$. The quadratic factor has no real solutions, so this leaves only $a = 3$. However, we have that $a > 1$, and we can check that $f(1) = 0$, $\\lim_{a \\rightarrow \\infty} f(a) > 0$, and $f(3) < 0$, so the global minimum does occur at $a = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$, $b$, $c$ be the three roots of $p(x) = x^{3} + x^{2} - 333 x - 1001$. Find $a^{3} + b^{3} + c^{3}$.",
    "options": [],
    "answer": "2003",
    "solution": "Solution:\n\nWe know that $x^{3} + x^{2} - 333 x - 1001 = (x - a)(x - b)(x - c) = x^{3} - (a + b + c) x^{2} + (ab + bc + ca) x - abc$.\n\nAlso, $(a + b + c)^{3} - 3(a + b + c)(ab + bc + ca) + 3abc = a^{3} + b^{3} + c^{3}$.\n\nThus,\n$$\na^{3} + b^{3} + c^{3} = (-1)^{3} - 3(-1)(-333) + 3 \\cdot 1001 = 2003.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the real value of $x$ such that $x^{3}+3 x^{2}+3 x+7=0$.",
    "options": [],
    "answer": "-1 - \\sqrt[3]{6}",
    "solution": "Solution:\nRewrite the equation as $(x+1)^{3}+6=0$ to get $(x+1)^{3}=-6 \\Rightarrow x+1=\\sqrt[3]{-6} \\Rightarrow x=-1-\\sqrt[3]{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $R S T U V$ be a regular pentagon. Construct an equilateral triangle $P R S$ with point $P$ inside the pentagon. Find the measure (in degrees) of angle $P T V$.",
    "options": [],
    "answer": "6°",
    "solution": "Solution:\n\nWe have $\\angle P R V=\\angle S R V-\\angle S R P=108^{\\circ}-60^{\\circ}=48^{\\circ}$. Since $P R=R S=R V$, triangle $P R V$ is isosceles, so that $\\angle V P R=\\angle R V P=\\left(180^{\\circ}-\\angle P R V\\right) / 2=66^{\\circ}$. Likewise, we have $\\angle T P S=66^{\\circ}$, so that\n$$\n\\angle T P V=360^{\\circ}-(\\angle V P R+\\angle R P S+\\angle S P T)=360^{\\circ}-\\left(66^{\\circ}+60^{\\circ}+66^{\\circ}\\right)=168^{\\circ}\n$$\nFinally, by symmetry, triangle $P T V$ is isosceles $(P T=T V)$, so $\\angle P T V=\\angle T V P=\\left(180^{\\circ}-\\angle T P V\\right) / 2=6^{\\circ}$. (See the figure.)\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na$ and $b$ are integers such that $a + \\sqrt{b} = \\sqrt{15 + \\sqrt{216}}$. Compute $a / b$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\n$\\frac{1}{2}$\n\nSquaring both sides gives $a^{2} + b + 2a \\sqrt{b} = 15 + \\sqrt{216}$; separating rational from irrational parts, we get $a^{2} + b = 15$, $4a^{2}b = 216$, so $a^{2}$ and $b$ equal $9$ and $6$. $a$ is an integer, so $a^{2} = 9$, $b = 6 \\Rightarrow a / b = 3 / 6 = 1 / 2$. (We cannot have $a = -3$, since $a + \\sqrt{b}$ is positive.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cell of a 2-configuration of a set $A$ is a nonempty subset $C$ of $A$ such that\ni. for any two distinct elements $a, b$ of $C$, there exists a sequence $c_{0}, c_{1}, \\ldots, c_{n}$ of elements of $A$ with $c_{0}=a, c_{n}=b$, and such that $\\left\\{c_{0}, c_{1}\\right\\},\\left\\{c_{1}, c_{2}\\right\\}, \\ldots,\\left\\{c_{n-1}, c_{n}\\right\\}$ are all elements of the 2-configuration, and\nii. if $a$ is an element of $C$ and $b$ is an element of $A$ but not of $C$, there does NOT exist a sequence $c_{0}, c_{1}, \\ldots, c_{n}$ of elements of $A$ with $c_{0}=a, c_{n}=b$, and such that $\\left\\{c_{0}, c_{1}\\right\\},\\left\\{c_{1}, c_{2}\\right\\}, \\ldots,\\left\\{c_{n-1}, c_{n}\\right\\}$ are all elements of the 2-configuration.\nAlso, we define a 2-configuration of $A$ to be barren if there is no subset $\\left\\{a_{0}, a_{1}, \\ldots, a_{n}\\right\\}$ of $A$, with $n \\geq 2$, such that $\\left\\{a_{0}, a_{1}\\right\\},\\left\\{a_{1}, a_{2}\\right\\}, \\ldots,\\left\\{a_{n-1}, a_{n}\\right\\}$ and $\\left\\{a_{n}, a_{0}\\right\\}$ are all elements of the 2-configuration.\nShow that, given any 2-configuration of a set $A$, every element of $A$ belongs to exactly one cell.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst, given $a$, let $C_{a}$ be the set of all $b \\in A$ for which there exists a sequence $a=c_{0}, c_{1}, \\ldots, c_{n}=b$ as in the definition of a cell. Certainly $a \\in C_{a}$ (take $n=0$); we claim that $C_{a}$ is a cell. If $b, b^{\\prime} \\in C_{a}$, then there exist sequences $a=c_{0}, c_{1}, \\ldots, c_{n}=b$ and $a=c_{0}^{\\prime}, c_{1}^{\\prime}, \\ldots, c_{m}^{\\prime}=b^{\\prime}$, so the sequence $b=c_{n}, c_{n-1}, \\ldots, c_{1}, c_{0}, c_{1}^{\\prime}, \\ldots, c_{m}^{\\prime}=b^{\\prime}$ shows that the first condition is met.\n\nFor the second, suppose that there does exist a sequence $b=c_{0}, c_{1}, \\ldots, c_{n}=b^{\\prime}$ with $b \\in C_{a}, b^{\\prime} \\notin C_{a}$. Then, concatenating with our sequence from $a$ to $b$, we get a sequence from $a$ to $b^{\\prime}$, contradicting the assumption $b^{\\prime} \\notin C_{a}$. Thus, the second condition holds, and $C_{a}$ is a cell. So $a$ lies in at least one cell.\n\nBut now, note that if $C$ is a cell containing $a$, then all $b$ for which such a sequence from $a$ to $b$ exists must lie in $C$ (or the second condition is violated), and if no such sequence exists, then $b$ cannot lie in $C$ (or the first condition is violated). Thus, the elements of $C$ are uniquely determined, so there is exactly one cell containing $a$, and the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDaniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a $60\\%$ chance of winning each point, what is the probability that he will win the game?",
    "options": [],
    "answer": "9/13",
    "solution": "Solution:\n\nConsider the situation after two points. Daniel has a $9/25$ chance of winning, Scott, $4/25$, and there is a $12/25$ chance that the players will be tied. In the latter case, we revert to the original situation. In particular, after every two points, either the game returns to the original situation, or one player wins. If it is given that the game lasts $2k$ rounds, then the players must be at par after $2(k-1)$ rounds, and then Daniel wins with probability $(9/25)/(9/25+4/25) = 9/13$. Since this holds for any $k$, we conclude that Daniel wins the game with probability $9/13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor any subset $S \\subseteq \\{1,2, \\ldots, 15\\}$, a number $n$ is called an \"anchor\" for $S$ if $n$ and $n+|S|$ are both members of $S$, where $|S|$ denotes the number of members of $S$. Find the average number of anchors over all possible subsets $S \\subseteq \\{1,2, \\ldots, 15\\}$.",
    "options": [],
    "answer": "13/8",
    "solution": "Solution:\nWe first find the sum of the numbers of anchors of all subsets $S$; this is equivalent to finding, for each $n$, the number of sets for which $n$ is an anchor, and then summing over all $n$. Suppose that $n$ is an anchor for $S$, and $S$ has $k$ elements. Then $n, n+k \\in S \\Rightarrow k \\geq 2$, and also $n+k \\leq 15$, or $k \\leq 15-n$. The remaining $k-2$ elements of $S$ (other than $n$ and $n+k$) may be freely chosen from the remaining 13 members of $\\{1,2, \\ldots, 15\\}$, so we get $\\binom{13}{k-2}$ possible sets $S$. Summing over all allowed values of $k$, we then have $\\binom{13}{0}+\\binom{13}{1}+\\binom{13}{2}+\\cdots+\\binom{13}{13-n}$ sets with $n$ as an anchor. If we sum over all $n=1,2, \\ldots, 13$ (since there are no possible values of $k$ when $n>13$), we get a total of\n$$\n13\\binom{13}{0}+12\\binom{13}{1}+11\\binom{13}{2}+\\cdots+\\binom{13}{12}\n$$\nIf we call this quantity $A$, then, by symmetry, $2A$ equals\n$$\n\\begin{aligned}\n& 13\\binom{13}{0}+12\\binom{13}{1}+11\\binom{13}{2}+\\cdots+\\binom{13}{12} \\\\\n& + \\left(\\binom{13}{1}+2\\binom{13}{2}+\\cdots+12\\binom{13}{12}+13\\binom{13}{13}\\right) = 13 \\cdot 2^{13}.\n\\end{aligned}\n$$\nSo $A=13 \\cdot 2^{12}$ is the total number of anchors over all possible sets $S$. Finally, to find the average number of anchors, we divide by the number of sets, which is $2^{15}$; thus, the answer is $13 \\cdot 2^{12} / 2^{15} = 13 / 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many solutions in nonnegative integers $(a, b, c)$ are there to the equation\n$$\n2^{a} + 2^{b} = c! ?\n$$",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nWe can check that $2^{a} + 2^{b}$ is never divisible by $7$, so we must have $c < 7$. The binary representation of $2^{a} + 2^{b}$ has at most two $1$'s. Writing $0!$, $1!$, $2!$, $\\ldots$, $6!$ in binary, we can check that the only possibilities are $c = 2, 3, 4$, giving solutions $(0, 0, 2)$, $(1, 2, 3)$, $(2, 1, 3)$, $(3, 4, 4)$, $(4, 3, 4)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA right circular cone with a height of 12 inches and a base radius of 3 inches is filled with water and held with its vertex pointing downward. Water flows out through a hole at the vertex at a rate in cubic inches per second numerically equal to the height of the water in the cone. (For example, when the height of the water in the cone is 4 inches, water flows out at a rate of 4 cubic inches per second.) Determine how many seconds it will take for all of the water to flow out of the cone.",
    "options": [],
    "answer": "9π/2 seconds",
    "solution": "Solution:\nWhen the water in the cone is $h$ inches high, it forms a cone similar to the original, so that its base has radius $h / 4$ and its volume is hence $\\pi h^{3} / 48$. The given condition then states that\n$$\n\\frac{d}{d t}\\left(\\frac{\\pi h^{3}}{48}\\right) = -h \\Rightarrow \\frac{\\pi h^{2}}{16} \\cdot \\frac{d h}{d t} = -h \\Rightarrow 2 h \\cdot \\frac{d h}{d t} = -\\frac{32}{\\pi} .\n$$\nIntegrating with respect to $t$, we get that $h^{2} = -\\frac{32 t}{\\pi} + C$; setting $t = 0$, $h = 12$, we get $C = 144$. The cone empties when $h = 0$, so $0 = -\\frac{32 t}{\\pi} + 144 \\Rightarrow t = \\frac{9 \\pi}{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $x \\geq 0$, $y \\geq 0$ are integers, randomly chosen with the constraint $x+y \\leq 10$, what is the probability that $x+y$ is even?",
    "options": [],
    "answer": "6/11",
    "solution": "Solution:\n\nFor each $p \\leq 10$, if $x+y=p$, $x$ can range from $0$ to $p$, yielding $p+1$ ordered pairs $(x, y)$. Thus there are a total of $1+2+3+\\cdots+11$ allowable ordered pairs $(x, y)$, but $1+3+5+\\cdots+11$ of these pairs have an even sum. So the desired probability is\n$$\n\\frac{1+3+5+\\cdots+11}{1+2+3+\\cdots+11}=\\frac{6^{2}}{11 \\cdot 6}=\\frac{6}{11}.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA broken calculator has the $+$ and $\\times$ keys switched. For how many ordered pairs $(a, b)$ of integers will it correctly calculate $a+b$ using the labelled $+$ key?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nWe need $a + b = a b$, or $a = \\frac{b}{b-1} = 1 - \\frac{1}{b-1}$, so $1/(b-1)$ is an integer. Thus $b$ must be $0$ or $2$, and $a$ is $0$ or $2$, respectively. So there are $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be an equilateral triangle of side length $2$. Let $\\omega$ be its circumcircle, and let $\\omega_{A}$, $\\omega_{B}$, $\\omega_{C}$ be circles congruent to $\\omega$ centered at each of its vertices. Let $R$ be the set of all points in the plane contained in exactly two of these four circles. What is the area of $R$?",
    "options": [],
    "answer": "2 sqrt(3)",
    "solution": "Solution:\n$\\omega_{A}$, $\\omega_{B}$, $\\omega_{C}$ intersect at the circumcenter; thus, every point within the circumcircle, and no point outside of it, is in two or more circles. The area inside exactly two circles is shaded in the figure. The two intersection points of $\\omega_{A}$ and $\\omega_{B}$, together with $A$, form the vertices of an equilateral triangle. As shown, this equilateral triangle cuts off a \"lip\" of $\\omega$ (bounded by a $60^{\\circ}$ arc of $\\omega$ and the corresponding chord) and another, congruent lip of $\\omega_{B}$ that is not part of the region of interest. By rotating the first lip to the position of the second, we can reassemble the equilateral triangle. Doing this for each of the 6 such triangles, we see that the desired area equals the area of a regular hexagon inscribed in $\\omega$. The side length of this hexagon is $(2 / 3) \\cdot (\\sqrt{3} / 2) \\cdot 2 = 2 \\sqrt{3} / 3$, so its area is $6 \\cdot (\\sqrt{3} / 4) \\cdot (2 \\sqrt{3} / 3)^{2} = 2 \\sqrt{3}$, and this is the answer.\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $ABC$, $\\angle ABC = 50^{\\circ}$ and $\\angle ACB = 70^{\\circ}$. Let $D$ be the midpoint of side $BC$. A circle is tangent to $BC$ at $B$ and is also tangent to segment $AD$; this circle intersects $AB$ again at $P$. Another circle is tangent to $BC$ at $C$ and is also tangent to segment $AD$; this circle intersects $AC$ again at $Q$. Find $\\angle APQ$ (in degrees).",
    "options": [],
    "answer": "70",
    "solution": "Solution:\n\nSuppose the circles are tangent to $AD$ at $E, F$, respectively; then, by equal tangents, $DE = DB = DC = DF \\Rightarrow E = F$ (as shown). So, by the Power of a Point Theorem, $AP \\cdot AB = AE^{2} = AF^{2} = AQ \\cdot AC \\Rightarrow AP / AQ = AC / AB \\Rightarrow \\triangle APQ \\sim \\triangle ACB$, giving $\\angle APQ = \\angle ACB = 70^{\\circ}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor $x$ a real number, let $f(x)=0$ if $x<1$ and $f(x)=2x-2$ if $x \\geq 1$. How many solutions are there to the equation\n$$\nf(f(f(f(x))))=x ?\n$$",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nCertainly $0, 2$ are fixed points of $f$ and therefore solutions. On the other hand, there can be no solutions for $x<0$, since $f$ is nonnegative-valued; for $0 < x < 2$, we have $0 \\leq f(x) < x < 2$ (and $f(0) = 0$), so iteration only produces values below $x$, and for $x > 2$, $f(x) > x$, and iteration produces higher values. So there are no other solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAt a certain college, there are 10 clubs and some number of students. For any two different students, there is some club such that exactly one of the two belongs to that club. For any three different students, there is some club such that either exactly one or all three belong to that club. What is the largest possible number of students?",
    "options": [],
    "answer": "513",
    "solution": "Solution:\n\nLet $C$ be the set of clubs; each student then corresponds to a subset of $C$ (the clubs to which that student belongs). The two-student condition implies that these subsets must be all distinct. Now (assuming there is more than one student) some student belongs to a nonempty set $S$ of clubs. For every subset $T \\subseteq C$, let $f(T)$ be the subset of $C$ consisting of those clubs that are in exactly one of $S$ and $T$ (so that $f(T) = (S \\cup T) - (S \\cap T)$). It is straightforward to check that $f(f(T)) = T$ and $f(T) \\neq T$, so that the collection of all $2^{10}$ subsets of $C$ is partitioned into pairs $\\{T, f(T)\\}$. Moreover,\nas long as $S$ is distinct from $T$ and $f(T)$, every club is in either none or exactly two of the sets $S, T$, and $f(T)$, so we cannot have a student corresponding to $T$ and another corresponding to $f(T)$. This puts an upper bound of 513 possible students (one for $S$, one for $\\emptyset = f(S)$, and one for each of the 511 other pairs). On the other hand, if we take some club $c$, we can have one student belonging to no clubs and 512 other students all belonging to $c$ and to the 512 possible subsets of the other 9 clubs, respectively. It is readily checked that this arrangement meets the conditions - for the three-student condition, either all three students are in $c$, or one is the student who belongs to no clubs and we reduce to the two-student condition - so 513 is achievable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo differentiable real functions $f(x)$ and $g(x)$ satisfy\n$$\n\\frac{f^{\\prime}(x)}{g^{\\prime}(x)}=e^{f(x)-g(x)}\n$$\nfor all $x$, and $f(0)=g(2003)=1$. Find the largest constant $c$ such that $f(2003)>c$ for all such functions $f, g$.",
    "options": [],
    "answer": "1 - ln 2",
    "solution": "Solution: $1-\\ln 2$\n\nRearranging the given equation gives $f^{\\prime}(x) e^{-f(x)}=g^{\\prime}(x) e^{-g(x)}$ for all $x$, so\n$$\n\\frac{d}{d x}\\left(e^{-f(x)}-e^{-g(x)}\\right)=-f^{\\prime}(x) e^{-f(x)}+g^{\\prime}(x) e^{-g(x)}=0.\n$$\nThus, $e^{-f(x)}-e^{-g(x)}$ is a constant, and it must be less than $e^{-f(0)}=e^{-1}$. Thus,\n$$\ne^{-f(2003)}<e^{-g(2003)}+e^{-1}=2 e^{-1}=e^{\\ln 2-1} \\Rightarrow f(2003)>1-\\ln 2.\n$$\nOn the other hand, we can find positive-valued functions $e^{-f(x)}, e^{-g(x)}$ that take on the required values at $0$ and $2003$ and have constant difference arbitrarily close to $e^{-1}$. For example, for arbitrarily large $t$, we can set\n$$\ne^{-f(x)}=e^{-(t(2003-x)+1)}+e^{-1}-e^{-(2003 t+1)}\n$$\nand\n$$\ne^{-g(x)}=e^{-(t(2003-x)+1)},\n$$\nand we can check that the resulting functions $f, g$ satisfy the required conditions. Thus, we can make $f(2003)$ arbitrarily close to $1-\\ln 2$, so this is the answer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor how many integers $n$, for $1 \\leq n \\leq 1000$, is the number $\\frac{1}{2}\\binom{2 n}{n}$ even?",
    "options": [],
    "answer": "990",
    "solution": "Solution:\n\n990\n\nIn fact, the expression $\\binom{2 n}{n}$ is always even, and it is not a multiple of four if and only if $n$ is a power of 2, and there are 10 powers of 2 between 1 and 1000.\n\nLet $f(N)$ denote the number of factors of 2 in $N$. Thus,\n\n$$\nf(n!)=\\left\\lfloor\\frac{n}{2}\\right\\rfloor+\\left\\lfloor\\frac{n}{4}\\right\\rfloor+\\left\\lfloor\\frac{n}{8}\\right\\rfloor+\\cdots=\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{2^{k}}\\right\\rfloor .\n$$\n\nAlso, it is clear that $f(a b)=f(a)+f(b)$ and $f\\left(\\frac{a}{b}\\right)=f(a)-f(b)$ for integers $a, b$. Now for any positive integer $n$, let $m$ be the integer such that $2^{m} \\leq n<2^{m+1}$. Then\n$$\n\\begin{aligned}\nf\\left(\\binom{2 n}{n}\\right)=f\\left(\\frac{(2 n)!}{n!n!}\\right) & =\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{2 n}{2^{k}}\\right\\rfloor-2\\left(\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{2^{k}}\\right\\rfloor\\right) \\\\\n& =\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{2^{k-1}}\\right\\rfloor-2\\left(\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{2^{k}}\\right\\rfloor\\right) \\\\\n& =\\lfloor n\\rfloor-\\left(\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{2^{k}}\\right\\rfloor\\right) \\\\\n& =n-\\left(\\sum_{k=1}^{m}\\left\\lfloor\\frac{n}{2^{k}}\\right\\rfloor\\right) \\\\\n& \\geq n-\\left(\\sum_{k=1}^{m} \\frac{n}{2^{k}}\\right) \\\\\n& =n-n\\left(\\frac{2^{m}-1}{2^{m}}\\right)=\\frac{n}{2^{m}} \\geq 1\n\\end{aligned}\n$$\nBoth equalities hold when $n=2^{m}$, and otherwise, $f\\left(\\binom{2 n}{n}\\right)>1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider a $2003$-gon inscribed in a circle and a triangulation of it with diagonals intersecting only at vertices. What is the smallest possible number of obtuse triangles in the triangulation?",
    "options": [],
    "answer": "1999",
    "solution": "Solution:\n\nBy induction, it follows easily that any triangulation of an $n$-gon inscribed in a circle has $n-2$ triangles. A triangle is obtuse unless it contains the center of the circle in its interior (in which case it is acute) or on one of its edges (in which case it is right). It is then clear that there are at most $2$ non-obtuse triangles, and $2$ is achieved when the center of the circle is on one of the diagonals of the triangulation. So the minimum number of obtuse triangles is $2001-2=1999$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least $60^{\\circ}$. Prove that\n$$\n\\frac{1}{3} |AB^3 - AD^3| \\leq |BC^3 - CD^3| \\leq 3|AB^3 - AD^3|.\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds if and only if AB = AD (equivalently BC = CD), i.e., ABCD is a tangential kite.",
    "solution": "By symmetry, we only need to prove the first inequality.\nBecause quadrilateral $ABCD$ has an incircle, we have $AB + CD = BC + AD$, or $AB - AD = BC - CD$. It suffices to prove that\n$$\n\\frac{1}{3}(AB^2 + AB \\cdot AD + AD^2) \\leq BC^2 + BC \\cdot CD + CD^2.\n$$\nBy the given condition, $60^{\\circ} \\leq \\angle A, \\angle C \\leq 120^{\\circ}$, and so $\\frac{1}{2} \\geq \\cos A, \\cos C \\geq -\\frac{1}{2}$. Applying the **Law of Cosines** to triangle $ABD$ yields\n$$\n\\begin{aligned}\nBD^2 &= AB^2 - 2AB \\cdot AD \\cos A + AD^2 \\\\\n&\\geq AB^2 - AB \\cdot AD + AD^2 \\\\\n&\\geq \\frac{1}{3}(AB^2 + AB \\cdot AD + AD^2).\n\\end{aligned}\n$$\nThe last inequality is equivalent to the inequality $3AB^2 - 3AB \\cdot AD + 3AD^2 \\geq AB^2 + AB \\cdot AD + AD^2$, or $AB^2 - 2AB \\cdot AD + AD^2 \\geq 0$, which is evident. The last equality holds if and only if $AB = AD$.\n\nOn the other hand, applying the Law of Cosines to triangle $BCD$ yields\n$$\nBD^2 = BC^2 - 2BC \\cdot CD \\cos C + CD^2 \\leq BC^2 + BC \\cdot CD + CD^2.\n$$\nCombining the last two inequalities gives the desired result.\n\nFor the given inequalities to have equality, we must have $AB = AD$. This condition is also sufficient, because all the entries in the equalities are 0. Thus, equality holds if and only if $ABCD$ is a **kite** with $AB = AD$ and $BC = CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA tiling of the plane with polygons consists of placing the polygons in the plane so that interiors of polygons do not overlap, each vertex of one polygon coincides with a vertex of another polygon, and no point of the plane is left uncovered. A unit polygon is a polygon with all sides of length one.\nIt is quite easy to tile the plane with infinitely many unit squares. Likewise, it is easy to tile the plane with infinitely many unit equilateral triangles.\n\na. Prove that there is a tiling of the plane with infinitely many unit squares and infinitely many unit equilateral triangles in the same tiling.\n\nb. Prove that it is impossible to find a tiling of the plane with infinitely many unit squares and finitely many (and at least one) unit equilateral triangles in the same tiling.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nThis can be done easily with parallel rows of squares and triangles, as shown.\n\n![](attached_image_1.png)\n\nOf course, other tilings are possible.\n\n\nb.\nMany people received only partial credit for part (b), because their arguments were not rigorous. In order to show that a tiling is not possible, you need a completely general argument, that handles all cases. Since there are infinitely many cases, this can be problematical. The way out is a very neat idea known as the Extreme Principle, which essentially says, \"focus on the largest or smallest entity.\" The advantage of this approach is that we now are reduced to studying just one polygon in the infinite plane.\n\nMethod 1:\nSuppose, to the contrary, that there were such a tiling. Since there are only finitely many triangles, there is a vertex that is \"northernmost.\" If there are ties, pick the vertex that lies furthest to the \"east.\" At most one other triangle (located to the west) can share this vertex. This is a contradiction, since the only way that triangles and squares can share a vertex in a tiling is with 3 triangles and two squares (in order to add up to 360 degrees).\n\n\nMethod 2:\nSuppose, to the contrary, that there were such a tiling. Since there are only finitely many triangles, they are contained in a bounded region. Let $N$ be a square that lies to the \"north\" of (i.e., its south side is further north than any point in the region containing the triangles). Likewise, let $E, W, S$ be squares that lie, respectively, east, west, and south of this region.\n\nObserve that the east and west neighbors of $N$ must be squares, since there are no triangles that far north. Thus, there is an infinite east-west chain of connected squares containing $N$. Likewise, there is an infinite east-west chain of squares containing $S$, and there are two infinite north-south chains, one containing $W$, and one containing $E$.\n\nThese four chains meet, forming a rectangular boundary of connected squares that completely encloses the region containing the triangles. Now we have a contradiction: This rectangular region has rational (in fact, integer) area, yet it is tiled with a non-zero number of equilateral triangles, plus, perhaps, some squares. But the area of each triangle is $\\sqrt{3}/4$, and hence the area of the entire collection of triangles and squares inside this rectangular region is irrational.\n\n![](attached_image_2.png)\n\n\nMethod 3:\nSuppose that there were such a tiling. By similar reasoning to method 2, we deduce that the tiling is \"eventually\" all squares; thus it can be formed by starting with an all-squares tiling, and then removing a finite number of squares, and filling in the \"holes\" with a finite number of triangles. The \"holes\" only have 90-degree angles; they can never be filled with triangles, since 90 is not a multiple of 60.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be real numbers such that\n$$\n(a_1^2 + a_2^2 + \\cdots + a_n^2 - 1)(b_1^2 + b_2^2 + \\cdots + b_n^2 - 1) \n> (a_1b_1 + a_2b_2 + \\cdots + a_nb_n - 1)^2.\n$$\nShow that $a_1^2 + a_2^2 + \\cdots + a_n^2 > 1$ and $b_1^2 + b_2^2 + \\cdots + b_n^2 > 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If exactly one of $a_1^2 + a_2^2 + \\dots + a_n^2$ and $b_1^2 + b_2^2 + \\dots + b_n^2$ is no greater than $1$, then the left-hand side of the inequality is no greater than $0$, and inequality $(*)$ is trivial. Now we assume that both are no greater than $1$. Set $a = 1 - a_1^2 - a_2^2 - \\dots - a_n^2$ and $b = 1 - b_1^2 - b_2^2 - \\dots - b_n^2$. Then both $a$ and $b$ are nonnegative real numbers. Multiplying both sides of the inequality $(*)$ by $4$ gives\n$$\n(2 - 2a_1b_1 - 2a_2b_2 - \\dots - 2a_n b_n)^2 \\geq 4ab\n$$\nNote that\n$$\n\\begin{aligned}\n& 2 - 2a_1b_1 - 2a_2b_2 - \\cdots - 2a_n b_n \\\\\n&= (a_1 - b_1)^2 + (a_2 - b_2)^2 + \\cdots + (a_n - b_n)^2 + a + b \\\\\n&\\geq a + b \\geq 0.\n\\end{aligned}\n$$\nIt follows that\n$$\n(2 - 2a_1b_1 - 2a_2b_2 - \\cdots - 2a_n b_n)^2 \\geq (a+b)^2 \\geq 4ab,\n$$\nas desired.\nWe use the following inequality of Aczel: If $x_1, \\dots, x_m, y_1, \\dots, y_m \\in \\mathbb{R}$ such that $x_1^2 > x_2^2 + \\dots + x_m^2$, then\n$$\n(x_1y_1 - x_2y_2 - \\dots - x_m y_m)^2 \\geq (x_1^2 - x_2^2 - \\dots - x_m^2)(y_1^2 - y_2^2 - \\dots - y_m^2).\n$$\nTo prove Aczel's Inequality, we consider the quadratic function\n$$\nf(t) = (x_1 t + y_1)^2 - \\sum_{i=2}^{m} (x_i t + y_i)^2\n$$\nand note that $f(-y_1/x_1) \\leq 0$. It follows that the discriminant is nonnegative, hence proving the desired inequality. Now we proceed to solve the problem at hand. It is clear that $a_1^2 + \\dots + a_n^2 - 1$ and $b_1^2 + \\dots + b_n^2 - 1$ have the same sign. If both are negative, then Aczel's Inequality yields\n$$\n(1 - a_1 b_1 - \\dots - a_n b_n)^2 \\geq (1 - a_1^2 - \\dots - a_n^2)(1 - b_1^2 - \\dots - b_n^2),\n$$\ncontradicting the given inequality.\nDefine the vectors $\\mathbf{u} = [a_1, a_2, \\dots, a_n]$ and $\\mathbf{v} = [b_1, b_2, \\dots, b_n]$, and place $\\mathbf{u}$ and $\\mathbf{v}$ tail by tail at the origin $O$ to form two sides of a triangle in $\\mathbb{R}^n$ by letting $A$ and $B$ be the points with the coordinates of $\\mathbf{u}$ and $\\mathbf{v}$. The given inequality is equivalent to $(\\mathbf{u} \\cdot \\mathbf{u} - 1)(\\mathbf{v} \\cdot \\mathbf{v} - 1) > (\\mathbf{u} \\cdot \\mathbf{v} - 1)^2$, where “$\\cdot$” denotes the **dot product** of two vectors. Expanding both sides and rearranging terms yields\n$$\n(\\mathbf{u} \\cdot \\mathbf{u})(\\mathbf{v} \\cdot \\mathbf{v}) - \\mathbf{u} \\cdot \\mathbf{u} - \\mathbf{v} \\cdot \\mathbf{v} > (\\mathbf{u} \\cdot \\mathbf{v})^2 - 2(\\mathbf{u} \\cdot \\mathbf{v}),\n$$\nor\n$$\n(\\mathbf{u} \\cdot \\mathbf{u})(\\mathbf{v} \\cdot \\mathbf{v}) - (\\mathbf{u} \\cdot \\mathbf{v})^2 > \\mathbf{u} \\cdot \\mathbf{u} + \\mathbf{v} \\cdot \\mathbf{v} - 2(\\mathbf{u} \\cdot \\mathbf{v}).\n$$\nFactoring the right-hand side of the above inequality gives\n$$\n(\\mathbf{u} \\cdot \\mathbf{u})(\\mathbf{v} \\cdot \\mathbf{v}) - (\\mathbf{u} \\cdot \\mathbf{v})^2 > (\\mathbf{u} - \\mathbf{v}) \\cdot (\\mathbf{u} - \\mathbf{v}).\n$$\nBy the vector form of the **Law of Cosines**, we have $(\\mathbf{u} \\cdot \\mathbf{v})^2 = (\\mathbf{u} \\cdot \\mathbf{u})(\\mathbf{v} \\cdot \\mathbf{v}) \\cos^2 \\angle AOB$. Hence\n$$\n(\\mathbf{u} \\cdot \\mathbf{u})(\\mathbf{v} \\cdot \\mathbf{v})(1 - \\cos^2 \\angle AOB) > (\\mathbf{u} - \\mathbf{v}) \\cdot (\\mathbf{u} - \\mathbf{v}),\n$$\nor\n$$\n(\\mathbf{u} \\cdot \\mathbf{u})(\\mathbf{v} \\cdot \\mathbf{v}) \\sin^2 \\angle AOB > (\\mathbf{u} - \\mathbf{v}) \\cdot (\\mathbf{u} - \\mathbf{v});\n$$\nthat is, $OA^2 \\cdot OB^2 \\sin^2 \\angle AOB > AB^2$. Hence\n$$\n[AOB] = \\frac{1}{2} OA \\cdot OB^2 \\sin \\angle AOB > \\frac{1}{2} \\cdot AB.\n$$\nTherefore, the length of the altitude from $O$ to $AB$ must be greater than $1$. Yet the altitude is the shortest segment connecting $O$ to $AB$, and so $1 < OA^2 = \\mathbf{u} \\cdot \\mathbf{u} = \\sum_{i=1}^n a_i^2$ and $1 < OB^2 = \\mathbf{v} \\cdot \\mathbf{v} = \\sum_{i=1}^n b_i^2$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many ordered pairs of integers $(a, b)$ satisfy all of the following inequalities?\n$$\n\\begin{aligned}\na^{2}+b^{2} & <16 \\\\\na^{2}+b^{2} & <8a \\\\\na^{2}+b^{2} & <8b\n\\end{aligned}\n$$",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nThis is easiest to see by simply graphing the inequalities. They correspond to the (strict) interiors of circles of radius $4$ and centers at $(0,0)$, $(4,0)$, $(0,4)$, respectively. So we can see that there are $6$ lattice points in their intersection (circled in the figure).\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are 1000 rooms in a row along a long corridor. Initially the first room contains 1000 people and the remaining rooms are empty. Each minute, the following happens: for each room containing more than one person, someone in that room decides it is too crowded and moves to the next room. All these movements are simultaneous (so nobody moves more than once within a minute). After one hour, how many different rooms will have people in them?",
    "options": [],
    "answer": "31",
    "solution": "Solution:\n\nWe can prove by induction on $n$ that the following pattern holds for $0 \\leq n \\leq 499$: after $2n$ minutes, the first room contains $1000-2n$ people and the next $n$ rooms each contain $2$ people, and after $2n+1$ minutes, the first room contains $1000-(2n+1)$ people, the next $n$ rooms each contain $2$ people, and the next room after that contains $1$ person. So, after $60$ minutes, we have one room with $940$ people and $30$ rooms with $2$ people each.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn trapezoid $ABCD$, $AD$ is parallel to $BC$. $\\angle A = \\angle D = 45^{\\circ}$, while $\\angle B = \\angle C = 135^{\\circ}$. If $AB = 6$ and the area of $ABCD$ is $30$, find $BC$.",
    "options": [],
    "answer": "2√2",
    "solution": "Solution:\n\nDraw altitudes from $B$ and $C$ to $AD$ and label the points of intersection $X$ and $Y$, respectively. Then $ABX$ and $CDY$ are $45^{\\circ}$-$45^{\\circ}$-$90^{\\circ}$ triangles with $BX = CY = 3\\sqrt{2}$. So, the area of $ABX$ and the area of $CDY$ are each $9$, meaning that the area of rectangle $BCYX$ is $12$. Since $BX = 3\\sqrt{2}$, $BC = 12/(3\\sqrt{2}) = 2\\sqrt{2}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = \\sin (\\sin x)$. Evaluate $\\lim_{h \\rightarrow 0} \\frac{f(x+h) - f(h)}{x}$ at $x = \\pi$.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nThe expression $\\frac{f(x+h) - f(h)}{x}$ is continuous at $h = 0$, so the limit is just $\\frac{f(x) - f(0)}{x}$. Letting $x = \\pi$ yields $\\frac{\\sin (\\sin \\pi) - \\sin (\\sin 0)}{\\pi} = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the value of\n$$\n\\binom{6}{1} 2^{1}+\\binom{6}{2} 2^{2}+\\binom{6}{3} 2^{3}+\\binom{6}{4} 2^{4}+\\binom{6}{5} 2^{5}+\\binom{6}{6} 2^{6} .\n$$",
    "options": [],
    "answer": "728",
    "solution": "Solution:\nThis sum is the binomial expansion of $(1+2)^{6}$, except that it is missing the first term, $\\binom{6}{0} 2^{0}=1$. So we get $3^{6}-1=728$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn infinite table of nonnegative integers is constructed as follows: in the top row, some number is $1$ and all other numbers are $0$'s; in each subsequent row, every number is the sum of some two of the three closest numbers in the preceding row. An example of such a table is shown below.\n\n$$\n\\begin{array}{ccccccccccc}\n\\cdots & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \\cdots \\\\\n\\cdots & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & \\cdots \\\\\n\\cdots & 0 & 0 & 0 & 1 & 1 & 2 & 1 & 0 & 0 & \\cdots \\\\\n\\cdots & 0 & 0 & 1 & 1 & 3 & 3 & 2 & 0 & 0 & \\cdots \\\\\n\\cdots & 0 & 1 & 2 & 4 & 4 & 6 & 3 & 2 & 0 & \\cdots \\\\\n. \\cdot & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\ddots\n\\end{array}\n$$\n\nThe top row (with the one $1$) is called row $0$; the next row is row $1$; the next row is row $2$, and so forth.\n\nNote that the following problems require you to prove the statements for every table that can be constructed by the process described above, not just for the example shown.\n\n1. Show that any number in row $n$ (for $n>0$) is at most $2^{n-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe use induction on $n$. It is clear that any number in row $1$ is at most $1 = 2^{0}$. Now, if every number in row $n$ is at most $2^{n-1}$, then every number in row $n+1$ is the sum of two numbers in row $n$ and so is at most $2^{n-1} + 2^{n-1} = 2^{n}$. This gives the induction step, and the result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA floor is tiled with equilateral triangles of side length $1$, as shown. If you drop a needle of length $2$ somewhere on the floor, what is the largest number of triangles it could end up intersecting? (Only count the triangles whose interiors are met by the needle - touching along edges or at corners doesn't qualify.)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "8",
    "solution": "Solution:\nLet $L$ be the union of all the lines of the tiling. Imagine walking from one end of the needle to the other. We enter a new triangle precisely when we cross one of the lines of the tiling. Therefore, the problem is equivalent to maximizing the number of times the needle crosses $L$.\n\nNow, the lines of the tiling each run in one of three directions. It is clear that the needle cannot cross more than three lines in any given direction, since the lines are a distance $\\sqrt{3} / 2$ apart and the needle would therefore have to be of length greater than $3 \\sqrt{3} / 2 > 2$.\n\nMoreover, it cannot cross three lines in each of two different directions. To see this, notice that its endpoints would have to lie in either the two light-shaded regions or the two dark-shaded regions shown, but the closest two points of such opposite regions are at a distance of $2$ (twice the length of a side of a triangle), so the needle cannot penetrate both regions.\n\n![](attached_image_2.png)\n\nTherefore, the needle can cross at most three lines in one direction and two lines in each of the other two directions, making for a maximum of $3+2+2=7$ crossings and $7+1=8$ triangles intersected. The example shows that $8$ is achievable, as long as the needle has length greater than $\\sqrt{3} < 2$.\n\n![](attached_image_3.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA lattice point is a point whose coordinates are both integers. Suppose Johann walks in a line from the point $(0,2004)$ to a random lattice point in the interior (not on the boundary) of the square with vertices $(0,0)$, $(0,99)$, $(99,99)$, $(99,0)$. What is the probability that his path, including the endpoints, contains an even number of lattice points?",
    "options": [],
    "answer": "3/4",
    "solution": "Solution:\nIf Johann picks the point $(a, b)$, the path will contain $\\gcd(a, 2004-b)+1$ points. There will be an odd number of points in the path if $\\gcd(a, 2004-b)$ is even, which is true if and only if $a$ and $b$ are both even. Since there are $49^{2}$ points with $a, b$ both even and $98^{2}$ total points, the probability that the path contains an even number of points is\n\n$$\n\\frac{98^{2}-49^{2}}{98^{2}}=\\frac{49^{2}\\left(2^{2}-1^{2}\\right)}{49^{2}\\left(2^{2}\\right)}=\\frac{3}{4} .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRight triangle $X Y Z$ has right angle at $Y$ and $X Y = 228$, $Y Z = 2004$. Angle $Y$ is trisected, and the angle trisectors intersect $X Z$ at $P$ and $Q$ so that $X, P, Q, Z$ lie on $X Z$ in that order. Find the value of $(P Y + Y Z)(Q Y + X Y)$.",
    "options": [],
    "answer": "1370736",
    "solution": "Solution:\n\nThe triangle's area is $(228 \\cdot 2004) / 2 = 228456$. All the angles at $Y$ are $30$ degrees, so by the sine area formula, the areas of the three small triangles in the diagram are $Q Y \\cdot Y Z / 4$, $P Y \\cdot Q Y / 4$, and $X Y \\cdot P Y / 4$, which sum to the area of the triangle. So expanding $(P Y + Y Z)(Q Y + X Y)$, we see that it equals\n$$\n4 \\cdot 228456 + X Y \\cdot Y Z = 6 \\cdot 228456 = 1370736\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P(x) = x^{3} - \\frac{3}{2} x^{2} + x + \\frac{1}{4}$. Let $P^{[1]}(x) = P(x)$, and for $n \\geq 1$, let $P^{[n+1]}(x) = P^{[n]}(P(x))$. Evaluate $\\int_{0}^{1} P^{[2004]}(x) \\, d x$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n$1 / 2$\n\nNote that $P(1-x) = 1 - P(x)$. It follows easily by induction that $P^{[k]}(1-x) = 1 - P^{[k]}(x)$ for all positive integers $k$. Hence\n$$\n\\begin{aligned}\n\\int_{0}^{1} P^{[2004]}(x) \\, d x & = \\int_{0}^{1} 1 - P^{[2004]}(1-x) \\, d x \\\\\n& = 1 - \\int_{0}^{1} P^{[2004]}(1-x) \\, d x \\\\\n& = 1 - \\int_{0}^{1} P^{[2004]}(u) \\, d u \\quad (u = 1-x)\n\\end{aligned}\n$$\nHence $\\int_{0}^{1} P^{[2004]}(x) \\, d x = 1 / 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a game similar to three card monte, the dealer places three cards on the table: the queen of spades and two red cards. The cards are placed in a row, and the queen starts in the center; the card configuration is thus $RQR$. The dealer proceeds to move. With each move, the dealer randomly switches the center card with one of the two edge cards (so the configuration after the first move is either $RRQ$ or $QRR$). What is the probability that, after $2004$ moves, the center card is the queen?",
    "options": [],
    "answer": "1/3 + 1/(3*2^2003)",
    "solution": "Solution:\n\nIf the probability that the queen is the center card after move $n$ is $p_{n}$, then the probability that the queen is an edge card is $1-p_{n}$, and the probability that the queen is the center card after move $n+1$ is $p_{n+1} = \\frac{1-p_{n}}{2}$. This recursion allows us to calculate the first few values of $p_{n}$. We might then notice in $1, 0, \\frac{1}{2}, \\frac{1}{4}, \\frac{3}{8}, \\frac{5}{16}, \\frac{11}{32}, \\cdots$, that the value of each fraction is close to $1/3$, and getting closer for larger $n$. In fact, subtracting $1/3$ from each fraction yields $\\frac{2}{3}, -\\frac{1}{3}, \\frac{1}{6}, -\\frac{1}{12}, \\frac{1}{24}, -\\frac{1}{48}, \\cdots$. This suggests the formula $p_{n} = \\frac{1}{3} + \\frac{2}{3}\\left(-\\frac{1}{2}\\right)^{n}$, and one can then prove that this formula is in fact correct by induction. Thus, $p(2004) = \\frac{1}{3} + \\frac{2}{3}\\left(-\\frac{1}{2}\\right)^{2004} = \\frac{1}{3} + \\frac{1}{3 \\cdot 2^{2003}}$.\n\nThe recurrence can also be solved without guessing—by generating functions, for example, or by using the fundamental theorem of linear recurrences, which ensures that the solution is of the form $p_{n} = a + b(-1/2)^{n}$ for some constants $a, b$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThere exists a polynomial $P$ of degree $5$ with the following property: if $z$ is a complex number such that $z^{5} + 2004 z = 1$, then $P\\left(z^{2}\\right) = 0$. Calculate the quotient $P(1) / P(-1)$.",
    "options": [],
    "answer": "-2010012/2010013",
    "solution": "Solution:\nLet $z_{1}, \\ldots, z_{5}$ be the roots of $Q(z) = z^{5} + 2004 z - 1$. We can check these are distinct (by using the fact that there's one in a small neighborhood of each root of $z^{5} + 2004 z$, or by noting that $Q(z)$ is relatively prime to its derivative). And certainly none of the roots of $Q$ is the negative of another, since $z^{5} + 2004 z = 1$ implies $(-z)^{5} + 2004(-z) = -1$, so their squares are distinct as well. Then, $z_{1}^{2}, \\ldots, z_{5}^{2}$ are the roots of $P$, so if we write $C$ for the leading coefficient of $P$, we have\n$$\n\\begin{aligned}\n\\frac{P(1)}{P(-1)} & = \\frac{C\\left(1-z_{1}^{2}\\right) \\cdots\\left(1-z_{5}^{2}\\right)}{C\\left(-1-z_{1}^{2}\\right) \\cdots\\left(-1-z_{5}^{2}\\right)} \\\\\n& = \\frac{\\left[\\left(1-z_{1}\\right) \\cdots\\left(1-z_{5}\\right)\\right] \\cdot \\left[\\left(1+z_{1}\\right) \\cdots\\left(1+z_{5}\\right)\\right]}{\\left[\\left(i-z_{1}\\right) \\cdots\\left(i-z_{5}\\right)\\right] \\cdot \\left[\\left(i+z_{1}\\right) \\cdots\\left(i+z_{5}\\right)\\right]} \\\\\n& = \\frac{\\left[\\left(1-z_{1}\\right) \\cdots\\left(1-z_{5}\\right)\\right] \\cdot \\left[\\left(-1-z_{1}\\right) \\cdots\\left(-1-z_{5}\\right)\\right]}{\\left[\\left(i-z_{1}\\right) \\cdots\\left(i-z_{5}\\right)\\right] \\cdot \\left[\\left(-i-z_{1}\\right) \\cdots\\left(-i-z_{5}\\right)\\right]} \\\\\n& = \\frac{\\left(1^{5} + 2004 \\cdot 1 - 1\\right)\\left(-1^{5} + 2004 \\cdot (-1) - 1\\right)}{\\left(i^{5} + 2004 \\cdot i - 1\\right)\\left(-i^{5} + 2004 \\cdot (-i) - 1\\right)} \\\\\n& = \\frac{(2004)(-2006)}{(-1 + 2005 i)(-1 - 2005 i)} \\\\\n& = -\\frac{2005^{2} - 1}{2005^{2} + 1} \\\\\n& = -4020024 / 4020026 = -2010012 / 2010013 .\n\\end{aligned}\n$$\n\n\nAlternative Solution: In fact, we can construct the polynomial $P$ explicitly (up to multiplication by a constant). We write $P\\left(z^{2}\\right)$ as a polynomial in $z$; it must use only even powers of $z$ and be divisible by $z^{5} + 2004 z - 1$, so we are inspired to try a difference of squares,\n$$\nP\\left(z^{2}\\right) = \\left(z^{5} + 2004 z - 1\\right)\\left(z^{5} + 2004 z + 1\\right) = \\left(z^{5} + 2004 z\\right)^{2} - 1^{2} = z^{2}\\left(z^{4} + 2004\\right)^{2} - 1,\n$$\ngiving\n$$\nP(z) = z\\left(z^{2} + 2004\\right)^{2} - 1 .\n$$\nNow plugging in $z = 1$ and $z = -1$ rapidly gives $\\left(2005^{2} - 1\\right) / \\left(-2005^{2} - 1\\right)$ as before.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = x^{2} + x^{4} + x^{6} + x^{8} + \\cdots$, for all real $x$ such that the sum converges. For how many real numbers $x$ does $f(x) = x$?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nClearly $x = 0$ works. Otherwise, we want $x = \\dfrac{x^{2}}{1 - x^{2}}$, or $x^{2} + x - 1 = 0$. Discard the negative root (since the sum doesn't converge there), but $\\dfrac{-1 + \\sqrt{5}}{2}$ works, for a total of 2 values.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all numbers $n$ with the following property: there is exactly one set of 8 different positive integers whose sum is $n$.",
    "options": [],
    "answer": "36, 37",
    "solution": "Solution:\n36, 37\n\nThe sum of 8 different positive integers is at least $1+2+3+\\cdots+8=36$, so we must have $n \\geq 36$. Now $n=36$ satisfies the desired property, since in this case we must have equality - the eight numbers must be $1, \\ldots, 8$.\n\nAnd if $n=37$ the eight numbers must be $1,2, \\ldots, 7,9$: if the highest number is 8 then the sum is $36<n$, while if the highest number is more than 9 the sum is $>1+2+\\cdots+7+9=37=n$. So the highest number must be 9, and then the remaining numbers must be $1,2, \\ldots, 7$. Thus $n=37$ also has the desired property.\n\nHowever, no other values of $n$ work: if $n>37$ then $\\{1,2,3, \\ldots, 7, n-28\\}$ and $\\{1,2, \\ldots, 6,8, n-29\\}$ are both sets of 8 distinct positive integers whose sum is $n$. So $n=36,37$ are the only solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA convex quadrilateral is drawn in the coordinate plane such that each of its vertices $(x, y)$ satisfies the equations $x^{2}+y^{2}=73$ and $x y=24$. What is the area of this quadrilateral?",
    "options": [],
    "answer": "110",
    "solution": "Solution:\nThe vertices all satisfy $(x+y)^{2}=x^{2}+y^{2}+2 x y=73+2 \\cdot 24=121$, so $x+y= \\pm 11$. Similarly, $(x-y)^{2}=x^{2}+y^{2}-2 x y=73-2 \\cdot 24=25$, so $x-y= \\pm 5$. Thus, there are four solutions: $(x, y)=(8,3),(3,8),(-3,-8),(-8,-3)$. All four of these solutions satisfy the original equations. The quadrilateral is therefore a rectangle with side lengths of $5 \\sqrt{2}$ and $11 \\sqrt{2}$, so its area is $110$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive odd integer. Prove that\n$$\n\\left\\lfloor\\log _{2} n\\right\\rfloor+\\left\\lfloor\\log _{2}(n / 3)\\right\\rfloor+\\left\\lfloor\\log _{2}(n / 5)\\right\\rfloor+\\left\\lfloor\\log _{2}(n / 7)\\right\\rfloor+\\cdots+\\left\\lfloor\\log _{2}(n / n)\\right\\rfloor=(n-1) / 2 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that $\\left\\lfloor\\log _{2} k\\right\\rfloor$ is the cardinality of the set $\\left\\{2,4,8, \\ldots, 2^{\\left\\lfloor\\log _{2} k\\right\\rfloor}\\right\\}$, i.e., the number of powers of $2$ that are even and are at most $k$. Then $\\left\\lfloor\\log _{2}(n / k)\\right\\rfloor$ is the number of even powers of $2$ that are at most $n / k$, or equivalently (multiplying each such number by $k$) the number of positive even numbers $\\leq n$ whose greatest odd divisor is $k$. Summing over all odd $k$, we get the number of even numbers $\\leq n$, which is just $(n-1) / 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integer solutions $(m, n)$ to the following equation:\n$$\nm^{2} = 1! + 2! + \\cdots + n!.\n$$",
    "options": [],
    "answer": "[(1, 1), (3, 3)]",
    "solution": "Solution:\n$(1,1), (3,3)$\nA square must end in the digit $0, 1, 4, 5, 6$, or $9$. If $n \\geq 4$, then $1! + 2! + \\cdots + n!$ ends in the digit $3$, so cannot be a square. A simple check for the remaining cases reveals that the only solutions are $(1,1)$ and $(3,3)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\sigma(n)$ denote the sum of the (positive) divisors of $n$, including $1$ and $n$ itself.\nProve that\n$$\n\\sigma(1)+\\sigma(2)+\\sigma(3)+\\cdots+\\sigma(n) \\leq n^{2}\n$$\nfor every positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe $i$th term on the left is the sum of all $d$ dividing $i$. If we write this sum out explicitly, then each term $d=1,2,\\ldots,n$ appears $\\lfloor n/d \\rfloor$ times—once for each multiple of $d$ that is $\\leq n$. Thus, the sum equals\n$$\n\\begin{aligned}\n\\lfloor n/1 \\rfloor + 2\\lfloor n/2 \\rfloor + 3\\lfloor n/3 \\rfloor + \\cdots + n\\lfloor n/n \\rfloor &\\leq n/1 + 2n/2 + 3n/3 + \\cdots + n/n \\\\\n&= n + n + \\cdots + n \\\\\n&= n^{2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA regular decagon $A_{0} A_{1} A_{2} \\cdots A_{9}$ is given in the plane. Compute $\\angle A_{0} A_{3} A_{7}$ in degrees.",
    "options": [],
    "answer": "54°",
    "solution": "Solution:\n$54^{\\circ}$\n\nPut the decagon in a circle. Each side subtends an arc of $360^{\\circ} / 10 = 36^{\\circ}$. The inscribed angle $\\angle A_{0} A_{3} A_{7}$ contains 3 segments, namely $A_{7} A_{8}$, $A_{8} A_{9}$, $A_{9} A_{0}$, so the angle is $108^{\\circ} / 2 = 54^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that\n$$\n\\frac{\\sigma(1)}{1}+\\frac{\\sigma(2)}{2}+\\frac{\\sigma(3)}{3}+\\cdots+\\frac{\\sigma(n)}{n} \\leq 2 n\n$$\nfor every positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThis is similar to the previous solution. If $d$ is a divisor of $i$, then so is $i/d$, and $(i/d)/i = 1/d$. Summing over all $d$, we see that $\\sigma(i)/i$ is the sum of the reciprocals of the divisors of $i$, for each positive integer $i$. So, summing over all $i$ from $1$ to $n$, we get the value $1/d$ appearing $\\lfloor n/d \\rfloor$ times, once for each multiple of $d$ that is at most $n$. In particular, the sum is\n$$\n\\frac{1}{1}\\left\\lfloor\\frac{n}{1}\\right\\rfloor+\\frac{1}{2}\\left\\lfloor\\frac{n}{2}\\right\\rfloor+\\frac{1}{3}\\left\\lfloor\\frac{n}{3}\\right\\rfloor+\\cdots+\\frac{1}{n}\\left\\lfloor\\frac{n}{n}\\right\\rfloor < \\frac{n}{1^{2}}+\\frac{n}{2^{2}}+\\cdots+\\frac{n}{n^{2}}.\n$$\nSo now all we need is $1/1^{2} + 1/2^{2} + \\cdots + 1/n^{2} < 2$. This can be obtained from the classic formula $1/1^{2} + 1/2^{2} + \\cdots = \\pi^{2}/6$, or from the more elementary estimate\n$$\n\\begin{aligned}\n1/2^{2} + 1/3^{2} + \\cdots + 1/n^{2} &< 1/(1 \\cdot 2) + 1/(2 \\cdot 3) + \\cdots + 1/((n-1) \\cdot n) \\\\\n&= (1/1 - 1/2) + (1/2 - 1/3) + \\cdots + (1/(n-1) - 1/n) \\\\\n&= 1 - 1/n \\\\\n&< 1.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAn $n$-string is a string of digits formed by writing the numbers $1, 2, \\ldots, n$ in some order (in base ten). For example, one possible 10-string is\n35728910461\nWhat is the smallest $n > 1$ such that there exists a palindromic $n$-string?",
    "options": [],
    "answer": "19",
    "solution": "Solution:\nThe following is such a string for $n=19$ :\n$$\n9|18|7|16|5|14|3|12|1|10|11|2|13|4|15|6|17|8|19\n$$\nwhere the vertical bars indicate breaks between the numbers. On the other hand, to see that $n=19$ is the minimum, notice that only one digit can occur an odd number of times in a palindromic $n$-string (namely the center digit). If $n \\leq 9$, then (say) the digits $1, 2$ each appear once in any $n$-string, so we cannot have a palindrome. If $10 \\leq n \\leq 18$, then $0, 9$ each appear once, and we again cannot have a palindrome. So $19$ is the smallest possible $n$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKate has four red socks and four blue socks. If she randomly divides these eight socks into four pairs, what is the probability that none of the pairs will be mismatched? That is, what is the probability that each pair will consist either of two red socks or of two blue socks?",
    "options": [],
    "answer": "3/35",
    "solution": "Solution:\n\n$3 / 35$\n\nThe number of ways Kate can divide the four red socks into two pairs is $\\binom{4}{2} / 2 = 3$. The number of ways she can divide the four blue socks into two pairs is also $3$. Therefore, the number of ways she can form two pairs of red socks and two pairs of blue socks is $3 \\cdot 3 = 9$.\n\nThe total number of ways she can divide the eight socks into four pairs is $[8!/(2!\\cdot 2!\\cdot 2!\\cdot 2!)] / 4! = 105$, so the probability that the socks come out paired correctly is $9 / 105 = 3 / 35$.\n\nTo see why $105$ is the correct denominator, we can look at each $2!$ term as representing the double counting of pair $(ab)$ and pair $(ba)$, while the $4!$ term represents the number of different orders in which we can select the same four pairs. Alternatively, we know that there are three ways to select two pairs from four socks. To select three pairs from six socks, there are five different choices for the first sock's partner and then three ways to pair up the remaining four socks, for a total of $5 \\cdot 3 = 15$ pairings. To select four pairs from eight socks, there are seven different choices for the first sock's partner and then fifteen ways to pair up the remaining six socks, for a total of $7 \\cdot 15 = 105$ pairings.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn a spherical planet with diameter $10,000~\\mathrm{km}$, powerful explosives are placed at the north and south poles. The explosives are designed to vaporize all matter within $5,000~\\mathrm{km}$ of ground zero and leave anything beyond $5,000~\\mathrm{km}$ untouched. After the explosives are set off, what is the new surface area of the planet, in square kilometers?",
    "options": [],
    "answer": "100000000π",
    "solution": "Solution:\n$100,000,000 \\pi$\n\nThe explosives have the same radius as the planet, so the surface area of the \"cap\" removed is the same as the new surface area revealed in the resulting \"dimple.\" Thus the area is preserved by the explosion and remains $\\pi \\cdot (10,000)^2$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe Fibonacci numbers are defined by $F_{1} = F_{2} = 1$, and $F_{n} = F_{n-1} + F_{n-2}$ for $n \\geq 3$. If the number\n$$\n\\frac{F_{2003}}{F_{2002}} - \\frac{F_{2004}}{F_{2003}}\n$$\nis written as a fraction in lowest terms, what is the numerator?",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nBefore reducing, the numerator is $F_{2003}^{2} - F_{2002} F_{2004}$. We claim $F_{n}^{2} - F_{n-1} F_{n+1} = (-1)^{n+1}$, which will immediately imply that the answer is $1$ (no reducing required). This claim is straightforward to prove by induction on $n$: it holds for $n=2$, and if it holds for some $n$, then\n$$\nF_{n+1}^{2} - F_{n} F_{n+2} = F_{n+1}(F_{n-1} + F_{n}) - F_{n}(F_{n} + F_{n+1}) = F_{n+1} F_{n-1} - F_{n}^{2} = -(-1)^{n+1} = (-1)^{n+2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA given line passes through the center $O$ of a circle. The line intersects the circle at points $A$ and $B$. Point $P$ lies in the exterior of the circle and does not lie on the line $A B$. Using only an unmarked straightedge, construct a line through $P$, perpendicular to the line $A B$. Give complete instructions for the construction and prove that it works.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. Draw a line from $P$ to $A$, intersecting the circle at $C$.\n2. Draw a line from $P$ to $B$, intersecting the circle at $D$.\n3. Draw lines $A D$ and $B C$, and let $E$ be their point of intersection.\n4. Draw a line from $P$ through $E$; this will be the desired perpendicular line.\n\nThis works because $A D \\perp P B$ and $B C \\perp P A$; hence $A D$ and $B C$ are altitudes of triangle $A P B$. It is well known that the three altitudes of a triangle intersect in a point, so $E$ is the intersection of all three altitudes. It follows that the line through $P E$ is an altitude.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a_1, \\dots, a_n$ are integers whose greatest common divisor is $1$. Let $S$ be a set of integers with the following properties.\n\na. For $i = 1, \\dots, n$, $a_i \\in S$.\n\nb. For $i, j = 1, \\dots, n$ (not necessarily distinct), $a_i - a_j \\in S$.\n\nc. For any integers $x, y \\in S$, if $x + y \\in S$, then $x - y \\in S$.\n\nProve that $S$ must be equal to the set of all integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We may as well assume that none of the $a_i$ is equal to $0$. We start with the following observations.\n\nd. $0 = a_1 - a_1 \\in S$ by (b).\n\ne. $-s = 0 - s \\in S$ whenever $s \\in S$, by (a) and (d).\n\nf. If $x, y \\in S$ and $x - y \\in S$, then $x + y \\in S$ by (c) and (e).\n\nBy (f) plus strong induction on $m$, we have that $ms \\in S$ for any $m \\ge 0$ whenever $s \\in S$. By (d) and (e), the same holds even if $m \\le 0$, and so we have the following.\n\ng. For $i = 1, \\dots, n$, $S$ contains all multiples of $a_i$.\n\nWe next verify that\n\nh. For $i, j \\in \\{1, \\dots, n\\}$ and any integers $c_i, c_j$, $c_i a_i + c_j a_j \\in S$.\n\nWe do this by induction on $|c_i| + |c_j|$. If $|c_i| \\le 1$ and $|c_j| \\le 1$, this follows from (b), (d), (f), so we may assume that $\\max\\{|c_i|, |c_j|\\} \\ge 2$. Suppose without loss of generality (by switching $i$ with $j$ and/or negating both $c_i$ and $c_j$) that $c_i \\ge 2$; then\n$$\nc_i a_i + c_j a_j = a_i + ((c_i - 1)a_i + c_j a_j)\n$$\nand we have $a_i \\in S$, $(c_i - 1)a_i + c_j a_j \\in S$ by the induction hypothesis, and $(c_i - 2)a_i + c_j a_j \\in S$ again by the induction hypothesis. So $c_i a_i + c_j a_j \\in S$ by (f), and (h) is verified.\n\nLet $e_i$ be the largest integer such that $2^{e_i}$ divides $a_i$; without loss of generality we may assume that $e_1 \\ge e_2 \\ge \\dots \\ge e_n$. Let $d_i$ be the greatest common divisor of $a_1, \\dots, a_i$. We prove by induction on $i$ that $S$ contains all multiples of $d_i$ for $i = 1, \\dots, n$; the case $i = n$ is the desired result. Our base cases are $i = 1$ and $i = 2$, which follow from (g) and (h), respectively.\n\nAssume that $S$ contains all multiples of $d_i$, for some $2 \\le i < n$. Let $T$ be the set of integers $m$ such that $m$ is divisible by $d_i$ and $m + r a_{i+1} \\in S$ for all integers $r$. Then $T$ contains nonzero positive and negative numbers, namely any multiple of $a_i$ by (h). By (c), if $t \\in T$ and $s$ divisible by $d_i$ (so in $S$) satisfy $t - s \\in T$, then $t + s \\in T$. By taking $t = s = d_i$, we deduce that $2d_i \\in T$; by induction (as in the proof of (g)), we have $2md_i \\in T$ for any integer $m$ (positive, negative or zero).\n\nFrom the way we ordered the $a_i$, we see that the highest power of $2$ dividing $d_i$ is greater than or equal to the highest power of $2$ dividing $a_{i+1}$. In other words, $a_{i+1}/d_{i+1}$ is odd. We can thus find integers $f, g$ with $f$ even such that $f d_i + g a_{i+1} = d_{i+1}$. (Choose such a pair without any restriction on $f$, and replace $(f, g)$ with $(f - a_{i+1}/d_{i+1}, g + d_i/d_{i+1})$ if needed to get an even $f$.) Then for any integer $r$, we have $r f d_i \\in T$ and so $r d_{i+1} \\in S$. This completes the induction and the proof of the desired result.\n(By Tony Zhang) We present a different way of completing the proof after the observing (d) through (h) of the preceding solution. We proceed to prove the following lemma by induction:\n\n**Lemma** Let $m \\ge 2$. For $i_1, i_2, \\dots, i_m \\in \\{1, \\dots, n\\}$ and $k_{i_1}, \\dots, k_{i_m} \\in \\mathbb{Z}$:\n$$\n(k_{i_1} a_{i_1} + k_{i_2} a_{i_2}) + 2(k_{i_3} a_{i_3} + \\dots + k_{i_m} a_{i_m}) \\in S.\n$$\n*Proof:* Observation (h) proves the $m=2$ case, so now assume that the Lemma is true for all $m$ less than or equal to some $r$. Now by induction hypothesis, the following terms are in $S$:\n$$\n(k_{i_1} a_{i_1} + k_{i_{r+1}} a_{i_{r+1}}) + 2(k_{i_3} a_{i_3} + \\dots + k_{i_r} a_{i_r}) \\in S\n$$\n$$\n(k_{i_2} a_{i_2} + k_{i_{r+1}} a_{i_{r+1}}) \\in S.\n$$\nYet their difference is\n$$\n(k_{i_1} a_{i_1} - k_{i_2} a_{i_2}) + 2(k_{i_3} a_{i_3} + \\dots + k_{i_r} a_{i_r}),\n$$\nwhich is in $S$ by induction hypothesis, so by observation (f), their sum is in $S$:\n$$\n(k_{i_1} a_{i_1} + k_{i_2} a_{i_2}) + 2(k_{i_3} a_{i_3} + \\dots + k_{i_r} a_{i_r} + k_{i_{r+1}} a_{i_{r+1}}),\n$$\nwhich completes the induction, and the proof of the Lemma.\n\nWe apply the Lemma to prove that $1 \\in S$; by the comment following observation (f) in the previous solution, this will prove that $S = \\mathbb{Z}$. Apply the lemma with $m = n$, $i_1 = 1$, $i_2 = 2$, ..., $i_n = n$. Since we are given that the greatest common divisor of the $a_i$ is $1$, there exist integers $k_1, \\dots, k_n$ such that $k_1 a_1 + \\dots + k_n a_n = 1$. Since we can't have all $a_i$ even, without loss of generality, assume that $a_1$ is odd. Now let $i$ iterate from $2$ to $n$, and at each stage perform the following operations if $k_i$ is odd:\n\na. replace $k_i$ with $k_i + a_1$,\n\nb. replace $k_1$ with $k_1 - a_i$.\n\nNote that this preserves the sum $k_1 a_1 + \\dots + k_n a_n = 1$, but it makes all the $k_i$ between $2$ and $n$ even; therefore, the lemma applies, and we find that $1 \\in S$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Consider sequences $a_0, a_1, \\dots, a_n$ such that $a_i \\in \\{1, 2, \\dots, n\\}$ for each $i$ and $a_n = a_0$.\n\na. Call such a sequence *good* if for all $i = 1, 2, \\dots, n$, $a_i - a_{i-1} \\not\\equiv i \\pmod n$. Suppose that $n$ is odd. Find the number of good sequences.\n\nb. Call such a sequence *great* if for all $i = 1, 2, \\dots, n$, $a_i - a_{i-1} \\not\\equiv i, 2i \\pmod n$. Suppose that $n$ is an odd prime. Find the number of great sequences.",
    "options": [],
    "answer": "a: (n−1)^n − (n−1); b: (n−1)((n−2)^{n−1} − 1)",
    "solution": "**First Solution:** The answer is $(n-1)^n - (n-1)$ for part (a) and $(n-1)((n-2)^{n-1} - 1)$ for part (b).\n\na. Observe that the number of good sequences is clearly the same for any choice of $a_0$. For fixed $a_0$, call the condition $a_i - a_{i-1} \\not\\equiv i \\pmod n$ *condition (i)*. Now let $S_i$ be the set of sequences $A = \\{a_i\\}_{i=1}^n, a_i \\in \\{1, 2, \\dots, n\\}$ such that $A$ satisfies conditions (1), (2), \\dots, $(n-i)$ and fails to satisfy conditions $(n-i+2)$, $(n-i+3)$, \\dots, $(n)$. Note that there is no constraint on whether or not $A$ satisfies condition $(n-i+1)$. Finally, let $F$ be the set of sequences that fails all conditions (1), (2), \\dots, $(n)$. Then we claim that the number of good sequences starting with $a_0$ is\n$$\nm = |S_1| - |S_2| + |S_3| - \\dots + |S_n| - |F|.\n$$\nConsider the following table, representing the conditions on sequences in $S_i$.\n$$\n\\begin{array}{c|ccccccc}\nS_1 & (1) & (2) & (3) & \\dots & (n-2) & (n-1) & \\\\\n\\hline\nS_2 & (1) & (2) & (3) & \\dots & (n-2) & \\overline{(n)} & \\\\\n\\hline\nS_3 & (1) & (2) & (3) & \\dots & \\overline{(n-1)} & \\overline{(n)} & \\\\\n\\hline\n\\vdots & \\vdots & \\vdots & \\vdots & \\dots & \\vdots & \\vdots & \\vdots \\\\\n\\hline\nS_n & \\overline{(2)} & \\overline{(3)} & \\dots & \\overline{(n-2)} & \\overline{(n-1)} & \\overline{(n)} & \\\\\n\\hline\nF & (1) & (2) & (3) & \\dots & \\overline{(n-2)} & \\overline{(n-1)} & \\overline{(n)}\n\\end{array}\n$$\nConditions that appear as $(j)$ in row $i$ are satisfied by sequences in $S_i$, and those that appear as $\\overline{(j)}$ are failed.\n\nNow take any sequence $A$. If $A$ is good, then $A$ is counted exactly once in $|S_1|$ and is not counted anywhere else. Otherwise, if $A$ is not good, let $(i)$ be the first condition that $A$ fails. If $i > 1$, then $A$ belongs to exactly the sets $S_{n-1}, S_{n-2}$. As one of $|S_{n-1}|, |S_{n-2}|$ is added and one is subtracted, $A$ has no net contribution to $m$. Finally, if $i = 1$, then $A$ belongs to $S_n$ and $F$, so again $A$ does not contribute to $m$. Therefore, $m$ is indeed the number of good sequences.\n\nWe are left with computing the sizes of the $S_i$'s and $F$. This is easy, as the failure of the final $i-1$ conditions determines the terms $a_{n-1}, a_{n-2}, \\dots, a_{n-i+1}$, whereas there are just $n-1$ ways to satisfy each of the conditions (1), (2), \\dots, $(n-i)$, producing $(n-1)^{n-i}$ subsequences $a_0, a_1, \\dots, a_{n-i}$. As there is no constraint on $a_{n-i+1} - a_{n-i}$, we have found all sequences in $S_i$. Hence, $|S_i| = (n-1)^{n-i}$. As for $F$, there is clearly exactly one way to fail every single condition, because $1+2+\\dots+(n-1) = \\frac{n(n-1)}{2}$ is divisible by $n$. Therefore, $|S_n| - F = 0$ and\n$$\n\\begin{align*}\nm &= |S_1| - |S_2| + \\dots - |S_{n-1}| \\\\\n&= (n-1)^{n-1} - (n-1)^{n-2} + \\dots - (n-1) \\\\\n&= \\frac{(n-1)^n - (n-1)}{n}.\n\\end{align*}\n$$\nAs there were originally $n$ choices for $a_0$, multiplying by $n$ gives the desired answer.\n\nb. We can use the same argument as in part (a), but we must recompute the values of $|S_i|$ and $|F|$. Note that for $1 \\le i \\le n-1$, there are exactly two ways to fail condition (i) (because $n$ is odd) while condition (n) is degenerate and reduces to the single restriction $a_n - a_{n-1} \\not\\equiv 0 \\pmod{n}$. Hence, we have\n$$\n|S_1| = (n-2)^{n-1} \\quad \\text{and} \\quad |S_i| = 2^{i-2}(n-2)^{n-i},\n$$\nfor $2 \\le i \\le n$. Finding $|F|$ is trickier in this case, however. Considering the differences $b_i = a_i - a_{i-1}$, $i = 1, 2, \\dots, n$, we wish to count the number of choices of $\\{b_i\\}_{i=1}^n$ such that\n$$\nb_1 + b_2 + \\dots + b_n \\equiv 0 \\pmod{n},\n$$\nwhere $b_i = i$ or $2i \\pmod{n}$ for each $i$. Note that because of the degenerate condition $(n)$, $b_n \\equiv 0 \\pmod{n}$ is forced; hence, we are actually looking at $2^{n-1}$ sums $b_1 + b_2 + \\dots + b_{n-1}$ rather than $2^n$. Now, as $n \\mid [1 + 2 + \\dots + (n-1)] = \\frac{n(n-1)}{2}$, we can equivalently consider sums of terms $c_i = b_i - i$. That is, we wish to find the number of choices of $c_1, c_2, \\dots, c_{n-1}$ such that\n$$\nc_1 + c_2 + \\dots + c_{n-1} \\equiv 0 \\pmod{n},\n$$\nwhere $c_i = 0$ or $i$ for each $i$. We now introduce an extra term $c_n$ that can be either 0 or $n$. The only effect of this extra term is to double the number of choices of $\\{c_i\\}_{i=1}^n$ compared to the number of choices of $\\{c_i\\}_{i=1}^{n-1}$. That is, letting\n$$\nF' = \\{(c_1, c_2, \\dots, c_n) \\mid \\begin{aligned}[t] & c_i = 0 \\text{ or } i, \\\\ & c_1 + c_2 + \\dots + c_n \\equiv 0 \\pmod{n} \\end{aligned}\\},\n$$\nwe have $|F| = \\frac{1}{2}|F'|$.\n\nObserve that we can identify choices $(c_1, c_2, \\dots, c_n)$ with subsets of $T = \\{1, 2, \\dots, n\\}$; the elements of $F'$ are then simply the subsets of $T$ that have sum divisible by $n$. The empty set and the set $T$ are two sets that clearly satisfy this property. All remaining subsets $U \\subset T$ have from 1 to $n-1$ elements. Write $U+k$ for the set $\\{x+k \\mid x \\in U\\}$ (taken mod $n$); this is the rotation of $U$ by $k$ shifts to the right. Then because $n$ is prime and $0 < |U| < n$, the subsets $U, U+1, U+2, \\dots, U+(n-1)$ form an equivalence class of exactly $n$ distinct subsets. Furthermore, the sums of the elements in $U$ and $U + k$ differ by $k|U|(\\text{mod } n)$, so exactly one element of each class has sum divisible by $n$. It follows that\n$$\n|F'| = 2 + \\frac{2^n - 2}{n}, \\quad |F| = \\frac{1}{2}|F'| = \\frac{2^{n-1} + n - 1}{n}.\n$$\nTherefore,\n$$\n\\begin{align*}\nm &= |S_1| - |S_2| + |S_3| - \\dots + |S_n| - |F| \\\\\n&= (n-2)^{n-1} - (n-2)^{n-2} + 2(n-2)^{n-3} \\\\\n&\\quad - 2^2(n-2)^{n-4} + \\dots + 2^{n-2} - \\frac{2^{n-1} + n-1}{n} \\\\\n&= (n-2)^{n-1} - \\frac{(n-2)^{n-1} - 2^{n-1}}{n} - \\frac{2^{n-1} + n-1}{n} \\\\\n&= \\frac{(n-1)(n-2)^{n-1} - (n-1)}{n}.\n\\end{align*}\n$$\nAs before, multiplying by $n$ gives the desired answer.\n\n\n**Second Solution:** The answer is $(n-1)^n - (n-1)$ for part (a) and $(n-1)((n-2)^{n-1} - 1)$ for part (b). Observe that the number of good sequences is clearly the same for any choice of $a_0$. Fixing $a_0$, we represent the choices for the differences $a_i - a_{i-1} \\pmod n$ with the generating function\n$$\ng_i(x) = 1 + x + x^2 + \\dots + x^{i-1} + x^{i+1} + \\dots + x^{n-1} = p(x) - x^i,\n$$\nwhere $p(x) = 1 + x + x^2 + \\cdots + x^{n-1}$. Then the generating function representing all the possibilities for $a_1 - a_0, a_2 - a_1, \\ldots, a_{n-1} - a_{n-2}$ is\n$$\ng(x) = \\prod_{i=1}^{n-1} g_i(x) = \\prod_{i=1}^{n-1} (p(x) - x^i).\n$$\nThe condition that $a_0 - a_{n-1} \\not\\equiv 0 \\pmod{n}$ means that we wish to sum the coefficients of the $x^k$ terms of $g(x)$ for $n \\not\\mid k$.\n\nNow the sum of all the coefficients is simply $g(1) = (n-1)^{n-1}$. The sum of the coefficients of the $x^k$ terms for $n \\mid k$ can be obtained by taking\n$$\n\\frac{1}{n} \\sum_{j=0}^{n-1} g(\\epsilon^j),\n$$\nwhere $\\epsilon = e^{2\\pi i/n}$ is a primitive $n^{\\text{th}}$ root of unity. As we noted above, $g(1) = (n-1)^{n-1}$. For $n > j > 0$, note that $p(\\epsilon^j) = 0$, so\n$$\ng(\\epsilon^j) = \\prod_{k=1}^{n-1} -\\epsilon^{kj} = (-1)^{n-1} \\epsilon^{n(n-1)/2} j = 1,\n$$\nas $n$ is odd. Therefore, the sum of the coefficients of the terms with powers not dividing $n$ is\n$$\n(n-1)^{n-1} - \\frac{1}{n}((n-1)^{n-1} + (n-1)) = \\frac{1}{n}((n-1)^n - (n-1)).\n$$\nRemembering that we had $n$ choices of $a_0$ to begin with, we obtain $(n-1)^n - (n-1)$ as the answer to part (a).\n\nFor part (b), we change our generating functions to\n$$\nh_i(x) = p(x) - (x^i + x^{2i}), \\quad h(x) = \\prod_{i=1}^{n-1} h_i(x).\n$$\nNote that for $1 \\le i \\le n-1$, the terms $x^i$ and $x^{2i}$ both appear because $i \\ne 2i \\pmod n$ as $n$ is odd. For the final condition, however, $n \\equiv 2n \\equiv 0 \\pmod n$, so we are still counting those terms—and only those terms—with powers not divisible by $n$. Hence, using the same method as before, we have $h(1) = (n-2)^{n-1}$,\n$$\nh(\\epsilon^j) = \\prod_{k=1}^{n-1} -\\epsilon^{kj}(1 + \\epsilon^{kj}) = \\prod_{k=1}^{n-1} (1 + \\epsilon^k),\n$$\nfor $1 \\le j \\le n-1$, as $n$ is prime. Now, $1+\\epsilon^i$, $i=0,1,\\dots,n-1$, are all of the roots of the polynomial $(x-1)^n-1$. Therefore, their product is $(-1)^n = -1$ times the constant coefficient of the polynomial, $-2$. Hence, the product of the roots not including $i=0$ (in which case $1+\\epsilon^0 = 2$) is 1. Therefore, $h(\\epsilon^j) = 1$.\n\nIt follows that the number of sequences for a fixed $a_0$ is\n$$\n(n-2)^{n-1} - \\frac{1}{n}((n-2)^{n-1} + (n-1)) = \\frac{1}{n}((n-1)((n-2)^{n-1} - 1)),\n$$\nso the answer to part (b) is $(n-1)((n-2)^{n-1} - 1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose the function $f(x)-f(2 x)$ has derivative $5$ at $x=1$ and derivative $7$ at $x=2$. Find the derivative of $f(x)-f(4 x)$ at $x=1$.",
    "options": [],
    "answer": "19",
    "solution": "Solution:\nLet $g(x)=f(x)-f(2 x)$. Then we want the derivative of\n$$\nf(x)-f(4 x)=(f(x)-f(2 x))+(f(2 x)-f(4 x))=g(x)+g(2 x)\n$$\nat $x=1$. This is $g'(x)+2 g'(2 x)$ at $x=1$, or $5+2 \\cdot 7=19$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the largest whole number that is equal to the product of its digits?",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nSuppose the number $n$ has $k+1$ digits, the first of which is $d$. Then the number is at least $d \\cdot 10^{k}$. On the other hand, each of the digits after the first is at most $9$, so the product of the digits is at most $d \\cdot 9^{k}$. Thus, if $n$ equals the product of its digits, then\n$$\nd \\cdot 10^{k} \\leq n \\leq d \\cdot 9^{k}\n$$\nwhich forces $k=0$, i.e., the number has only one digit. So $n=9$ is clearly the largest possible value.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest number $n$ such that $(2004!)!$ is divisible by $((n!)!)!$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nFor positive integers $a, b$, we have\n$$\na!\\mid b!\\quad \\Leftrightarrow \\quad a!\\leq b!\\quad \\Leftrightarrow \\quad a \\leq b .\n$$\nThus,\n$$\n((n!)!)!\\mid(2004!)!\\Leftrightarrow(n!)!\\leq 2004!\\Leftrightarrow n!\\leq 2004 \\quad \\Leftrightarrow \\quad n \\leq 6 .\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAndrea flips a fair coin repeatedly, continuing until she either flips two heads in a row (the sequence $H H$) or flips tails followed by heads (the sequence $T H$). What is the probability that she will stop after flipping $H H$?",
    "options": [],
    "answer": "1/4",
    "solution": "Solution:\nThe only way that Andrea can ever flip $H H$ is if she never flips $T$, in which case she must flip two heads immediately at the beginning. This happens with probability $\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways can you mark 8 squares of an $8 \\times 8$ chessboard so that no two marked squares are in the same row or column, and none of the four corner squares is marked? (Rotations and reflections are considered different.)",
    "options": [],
    "answer": "21600",
    "solution": "Solution:\n\nIn the top row, you can mark any of the 6 squares that is not a corner. In the bottom row, you can then mark any of the 5 squares that is not a corner and not in the same column as the square just marked. Then, in the second row, you have 6 choices for a square not in the same column as either of the two squares already marked; then there are 5 choices remaining for the third row, and so on down to 1 for the seventh row, in which you make the last mark. Thus, altogether, there are $6 \\cdot 5 \\cdot (6 \\cdot 5 \\cdots 1) = 30 \\cdot 6! = 30 \\cdot 720 = 21600$ possible sets of squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf the three points\n$$\n\\begin{aligned}\n& (1, a, b) \\\\\n& (a, 2, b) \\\\\n& (a, b, 3)\n\\end{aligned}\n$$\nare collinear (in 3-space), what is the value of $a+b$?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nThe first two points are distinct (otherwise we would have $a=1$ and $a=2$ simultaneously), and they both lie on the plane $z=b$, so the whole line is in this plane and $b=3$. Reasoning similarly with the last two points gives $a=1$, so $a+b=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo positive rational numbers $x$ and $y$, when written in lowest terms, have the property that the sum of their numerators is $9$ and the sum of their denominators is $10$. What is the largest possible value of $x + y$?",
    "options": [],
    "answer": "73/9",
    "solution": "Solution:\n\nFor fixed denominators $a < b$ (with sum $10$), we maximize the sum of the fractions by giving the smaller denominator as large a numerator as possible: $8/a + 1/b$. Then, if $a \\geq 2$, this quantity is at most $8/2 + 1/1 = 5$, which is clearly smaller than the sum we get by setting $a = 1$, namely $8/1 + 1/9 = 73/9$. So this is the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all ordered pairs of integers $(x, y)$ such that $3^{x} 4^{y} = 2^{x+y} + 2^{2(x+y)-1}$.",
    "options": [],
    "answer": "(1,1), (2,2), (0,1)",
    "solution": "Solution:\nThe right side is $2^{x+y} (1 + 2^{x+y-1})$. If the second factor is odd, it needs to be a power of $3$, so the only options are $x+y=2$ and $x+y=4$. This leads to two solutions, namely $(1,1)$ and $(2,2)$. The second factor can also be even, if $x+y-1=0$. Then $x+y=1$ and $3^{x} 4^{y} = 2 + 2$, giving $(0,1)$ as the only other solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nI have written a strictly increasing sequence of six positive integers, such that each number (besides the first) is a multiple of the one before it, and the sum of all six numbers is $79$. What is the largest number in my sequence?",
    "options": [],
    "answer": "48",
    "solution": "Solution:\n\nIf the fourth number is $\\geq 12$, then the last three numbers must sum to at least $12 + 2 \\cdot 12 + 2^{2} \\cdot 12 = 84 > 79$. This is impossible, so the fourth number must be less than $12$. Then the only way we can have the required divisibilities among the first four numbers is if they are $1, 2, 4, 8$. So the last two numbers now sum to $79 - 15 = 64$. If we call these numbers $8a, 8ab$ ($a, b > 1$) then we get $a(1 + b) = a + ab = 8$, which forces $a = 2, b = 3$. So the last two numbers are $16, 48$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest integer $n$ such that $3^{512}-1$ is divisible by $2^{n}$.",
    "options": [],
    "answer": "11",
    "solution": "Solution:\nWrite\n$$\n\\begin{aligned}\n3^{512}-1 & =\\left(3^{256}+1\\right)\\left(3^{256}-1\\right)=\\left(3^{256}+1\\right)\\left(3^{128}+1\\right)\\left(3^{128}-1\\right) \\\\\n& =\\cdots=\\left(3^{256}+1\\right)\\left(3^{128}+1\\right) \\cdots(3+1)(3-1)\n\\end{aligned}\n$$\nNow each factor $3^{2^{k}}+1$, $k \\geq 1$, is divisible by just one factor of $2$, since $3^{2^{k}}+1=\\left(3^{2}\\right)^{2^{k-1}}+1 \\equiv 1^{2^{k-1}}+1=2 \\pmod{4}$. Thus we get $8$ factors of $2$ here, and the remaining terms $(3+1)(3-1)=8$ give us $3$ more factors of $2$, for a total of $11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWe say a point is contained in a square if it is in its interior or on its boundary. Three unit squares are given in the plane such that there is a point contained in all three. Furthermore, three points $A, B, C$, are given, each contained in at least one of the squares. Find the maximum area of triangle $A B C$.",
    "options": [],
    "answer": "3 sqrt(3) / 2",
    "solution": "Solution:\n$3 \\sqrt{3} / 2$\n\nLet $X$ be a point contained in all three squares. The distance from $X$ to any point in any of the three squares is at most $\\sqrt{2}$, the length of the diagonal of the squares. Therefore, triangle $A B C$ is contained in a circle of radius $\\sqrt{2}$, so its circumradius is at most $\\sqrt{2}$. The triangle with greatest area that satisfies this property is the equilateral triangle in a circle of radius $\\sqrt{2}$. (This can be proved, for example, by considering that the maximum altitude to any given side is obtained by putting the opposite vertex at the midpoint of its arc, and it follows that all the vertices are equidistant.) The equilateral triangle is also attainable, since making $X$ the circumcenter and positioning the squares such that $A X, B X$, and $C X$ are diagonals (of the three squares) and $A B C$ is equilateral, leads to such a triangle. This triangle has area $3 \\sqrt{3} / 2$, which may be calculated, for example, using the sine formula for area applied to $A B X, A C X$, and $B C X$, to get $3 / 2 (\\sqrt{2})^{2} \\sin 120^{\\circ}$. (See diagram, next page.)\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $x^{3} - a x^{2} + b x - 48$ is a polynomial with three positive roots $p, q$, and $r$ such that $p < q < r$. What is the minimum possible value of $1/p + 2/q + 3/r$?",
    "options": [],
    "answer": "3/2",
    "solution": "Solution:\nWe know $p q r = 48$ since the product of the roots of a cubic is the constant term. Now,\n$$\n\\frac{1}{p} + \\frac{2}{q} + \\frac{3}{r} \\geq 3 \\sqrt[3]{\\frac{6}{p q r}} = \\frac{3}{2}\n$$\nby AM-GM, with equality when $1/p = 2/q = 3/r$. This occurs when $p = 2$, $q = 4$, $r = 6$, so $3/2$ is in fact the minimum possible value.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many of the integers $1, 2, \\ldots, 2004$ can be represented as $\\frac{mn+1}{m+n}$ for positive integers $m$ and $n$?",
    "options": [],
    "answer": "2004",
    "solution": "Solution:\nFor any positive integer $a$, we can let $m = a^{2} + a - 1$, $n = a + 1$ to see that every positive integer has this property, so the answer is $2004$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA regular hexagon has one side along the diameter of a semicircle, and the two opposite vertices on the semicircle. Find the area of the hexagon if the diameter of the semicircle is $1$.",
    "options": [],
    "answer": "3√3/26",
    "solution": "Solution:\nThe midpoint of the side of the hexagon on the diameter is the center of the circle. Draw the segment from this center to a vertex of the hexagon on the circle. This segment, whose length is $1/2$, is the hypotenuse of a right triangle whose legs have lengths $a/2$ and $a\\sqrt{3}$, where $a$ is a side of the hexagon. So $1/4 = a^{2}(1/4 + 3)$, so $a^{2} = 1/13$. The hexagon consists of $6$ equilateral triangles of side length $a$, so the area of the hexagon is $3a^{2}\\sqrt{3}/2 = 3\\sqrt{3}/26$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$$\n\\binom{2003}{1} + \\binom{2003}{4} + \\binom{2003}{7} + \\cdots + \\binom{2003}{2002}\n$$",
    "options": [],
    "answer": "(2^2003 - 2)/3",
    "solution": "Solution:\n\nLet $\\omega = -1/2 + i \\sqrt{3}/2$ be a complex cube root of unity. Then, by the binomial theorem, we have\n$$\n\\begin{aligned}\n\\omega^{2}(\\omega+1)^{2003} &= \\binom{2003}{0} \\omega^{2} + \\binom{2003}{1} \\omega^{3} + \\binom{2003}{2} \\omega^{4} + \\cdots + \\binom{2003}{2003} \\omega^{2005} \\\\\n2^{2003} &= \\binom{2003}{0} + \\binom{2003}{1} + \\binom{2003}{2} + \\cdots + \\binom{2003}{2003} \\\\\n\\omega^{-2}(\\omega^{-1}+1)^{2003} &= \\binom{2003}{0} \\omega^{-2} + \\binom{2003}{1} \\omega^{-3} + \\binom{2003}{2} \\omega^{-4} + \\cdots + \\binom{2003}{2003} \\omega^{-2005}\n\\end{aligned}\n$$\nIf we add these together, then the terms $\\binom{2003}{n}$ for $n \\equiv 1 \\pmod{3}$ appear with coefficient $3$, while the remaining terms appear with coefficient $1 + \\omega + \\omega^{2} = 0$. Thus the desired sum is just $\\left(\\omega^{2}(\\omega+1)^{2003} + 2^{2003} + \\omega^{-2}(\\omega^{-1}+1)^{2003}\\right)/3$. Simplifying using $\\omega+1 = -\\omega^{2}$ and $\\omega^{-1}+1 = -\\omega$ gives $\\left(-1 + 2^{2003} + -1\\right)/3 = \\left(2^{2003} - 2\\right)/3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA regular dodecahedron is projected orthogonally onto a plane, and its image is an $n$-sided polygon. What is the smallest possible value of $n$?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nWe can achieve $6$ by projecting onto a plane perpendicular to an edge of the dodecahedron. Indeed, if we imagine viewing the dodecahedron in such a direction, then $4$ of the faces are projected to line segments (namely, the two faces adjacent to the edge and the two opposite faces), and of the remaining $8$ faces, $4$ appear on the front of the dodecahedron and the other $4$ are on the back. Thus, the dodecahedron appears as shown.\n\n![](attached_image_1.png)\n\nTo see that we cannot do better, note that, by central symmetry, the number of edges of the projection must be even. So we just need to show that the answer cannot be $4$. But if the projection had $4$ sides, one of the vertices would give a projection forming an acute angle, which is not possible. So $6$ is the answer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46933,
    "subject": "Mathematics (Multi-modal)",
    "question": "For what real values of $k > 0$ is it possible to dissect a $1 \\times k$ rectangle into two similar, but noncongruent, polygons?",
    "options": [],
    "answer": "k ≠ 1",
    "solution": "**First Solution:** We will show that a dissection satisfying the requirements of the problem is possible if and only if $k \\neq 1$.\n\nWe first show by contradiction that such a dissection is not possible when $k=1$. Assume that we have such a dissection. The common boundary of the two dissecting polygons must be a single broken line connecting two points on the boundary of the square (otherwise either the square is subdivided in more than two pieces or one of the polygons is inside the other). The two dissecting polygons must have the same number of vertices. They share all the vertices on the common boundary, so they have to use the same number of corners of the square as their own vertices. Therefore, the common boundary must connect two opposite sides of the square (otherwise one of the polygons will contain at least three corners of the square, while the other at most two). However, this means that each of the dissecting polygons must use an entire side of the square as one of its sides, and thus each polygon has a side of length $1$. A side of longest length in one of the polygons is either a side on the common boundary or, if all those sides have length less than $1$, it is a side of the square. But this is also true of the other polygon, which means that the longest side length in the two polygons is the same. This is impossible since they are similar but not congruent, so we have a contradiction.\n\nWe now construct a dissection satisfying the requirements of the problem when $k \\neq 1$. Notice that we may assume that $k > 1$, because a $1 \\times k$ rectangle is similar to a $1 \\times \\frac{1}{k}$ rectangle.\n\nWe first construct a dissection of an appropriately chosen rectangle (denoted by $ABCD$ below) into two similar incongruent polygons. The construction depends on two parameters ($n$ and $r$ below). By appropriate choice of these parameters we show that the constructed rectangle can be made similar to a $1 \\times k$ rectangle, for any $k > 1$. The construction follows.\n\nLet $r > 1$ be a real number. For any positive integer $n$, consider the following sequence of $2n + 2$ points:\n$$\n\\begin{aligned}\nA_0 &= (0, 0), \\quad A_1 = (1, 0), \\quad A_2 = (1, r), \\quad A_3 = (1 + r^2, r), \\\\\nA_4 &= (1 + r^2, r + r^3), \\quad A_5 = (1 + r^2 + r^4, r + r^3),\n\\end{aligned}\n$$\nand so on, until\n$$\nA_{2n+1} = (1 + r^2 + r^4 + \\dots + r^{2n},\\ r + r^3 + r^5 + \\dots + r^{2n-1}).\n$$\nDefine a rectangle $ABCD$ by $A = A_0, C = A_{2n+1}$,\n$$\nB = (1 + r^2 + \\dots + r^{2n}, 0), \\quad \\text{and} \\quad D = (0, r + r^3 + \\dots + r^{2n-1}).\n$$\n\n![](attached_image_1.png)\n\nThe sides of the $(2n + 2)$-gon $A_1A_2\\dots A_{2n+1}B$ have lengths $r, r^2, r^3, \\dots, r^{2n}, r + r^3 + r^5 + \\dots + r^{2n-1}, r^2 + r^4 + r^6 + \\dots + r^{2n}$, and the sides of the $(2n+2)$-gon $A_0A_1A_2\\dots A_{2n}D$ have lengths $1, r, r^2, \\dots, r^{2n-1}, 1 + r^2 + r^4 + \\dots + r^{2n-2}, r + r^3 + r^5 + \\dots + r^{2n-1}$, respectively. These two polygons dissect the rectangle $ABCD$ and, apart from orientation, it is clear that they are similar but incongruent, with coefficient of similarity $r > 1$. The rectangle $ABCD$ and its dissection are thus constructed.\n\nThe rectangle $ABCD$ is similar to a rectangle of size $1 \\times f_n(r)$, where\n$$\nf_n(r) = \\frac{1 + r^2 + \\dots + r^{2n}}{r + r^3 + \\dots + r^{2n-1}}.\n$$\nIt remains to show that $f_n(r)$ can have any value $k > 1$ for appropriate choices of $n$ and $r$. Choose $n$ sufficiently large so that $1 + \\frac{1}{n} < k$. Since\n$$\nf_n(1) = 1 + \\frac{1}{n} < k < k \\frac{1 + k^2 + k^4 + \\dots + k^{2n}}{k^2 + k^4 + \\dots + k^{2n}} = f_n(k)\n$$\nand $f_n(r)$ is a continuous function for positive $r$, there exists an $r$ such that $1 < r < k$ and $f_n(r) = k$, so we are done.\n\n\n**Second Solution:** (By Oleg Golberg) We present another proof of the fact that $k = 1$ is impossible. Assume for the sake of contradiction that we have a dissection of a unit square into two polygons that are similar but not congruent. As in the first solution, the dissection must be accomplished via a single path connecting opposite sides of the square. Without loss of generality, suppose that the endpoints of the path are $K \\in BC$ and $L \\in AD$, where $K$ is a corner of the square if and only if $L$ is the opposite corner. Also, without loss of generality, assume that the right-hand part is strictly smaller than the left-hand part (they are given to be similar but not congruent).\n\n![](attached_image_2.png)\n\nNow, the right-hand part is supposed to be similar to the left-hand part, so let the function $F$ map the right-hand polygon to the left-hand polygon according to the similarity. Observe that the right-hand polygon has the property that if one draws the two perpendiculars to $CD$ at $C$ and $D$, then these lines completely bound the right-hand polygon. Therefore, after applying $F$, the same property must hold for $F(CD)$; this must be some side of the left-hand polygon, and its perpendiculars at $F(C)$ and $F(D)$ must bound the entire left-hand polygon. In particular, they must bound $A$ and $B$. However, there are only a few ways this can be done:\n\n(a) $F(\\{C, D\\}) = \\{A, B\\}$. Yet the lengths of $CD$ and $AB$ are equal, so this violates the fact that the similarity is not a congruence.\n\n(b) $F(\\{C, D\\}) = \\{K, L\\}$, and $KL \\parallel AB$. This has the same problem as the first case.\n\n(c) $F(\\{C, D\\}) = \\{B, K\\}$. Since the right-hand polygon is strictly smaller than the left-hand one, this forces $BK > 1 \\Rightarrow BC > 1$, contradicting the fact that $ABCD$ is a square.\n\n(d) $F(\\{C, D\\}) = \\{A, L\\}$. This has a similar problem to the previous case.\n\nTherefore, all cases yield contradictions, so we have proven $k \\neq 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNASA has proposed populating Mars with $2,004$ settlements. The only way to get from one settlement to another will be by a connecting tunnel. A bored bureaucrat draws on a map of Mars, randomly placing $N$ tunnels connecting the settlements in such a way that no two settlements have more than one tunnel connecting them. What is the smallest value of $N$ that guarantees that, no matter how the tunnels are drawn, it will be possible to travel between any two settlements?",
    "options": [],
    "answer": "2005004",
    "solution": "Solution:\n\nThe problem is equivalent, in general, to finding the least number of edges required so that a graph on $n$ vertices will be connected, i.e., one can reach any vertex from any other vertex by following the edges of the graph. (We are letting settlements be vertices and tunnels be edges, of course.) This value is $\\binom{n-1}{2} + 1$.\n\nHere $\\binom{m}{2}$ counts the number of all possible pairs in a group of $m$ people, or equivalently, the number of edges in a graph with $m$ vertices where every two vertices are connected with an edge. This latter graph is called a complete graph on $m$ vertices.\n\nTo see that the minimum number of edges must be $\\binom{n-1}{2} + 1$, we first observe that it cannot be less than this, since $n-1$ vertices can be connected to one another with $\\binom{n-1}{2}$ edges, leaving the $n$th vertex isolated.\n\nNext we will show that $\\binom{n-1}{2} + 1$ edges will guarantee that the graph is connected.\n\nMethod 1: Suppose to the contrary, that the graph is not connected. Then it consists of $k$ connected components, each containing $v_{1}, v_{2}, \\ldots, v_{k}$ vertices. Each component has at most $\\binom{v_{i}}{2}$ edges. We claim that\n$$\n\\binom{v_{1}}{2} + \\binom{v_{2}}{2} + \\cdots + \\binom{v_{k}}{2} \\leq \\binom{n-1}{2}\n$$\nwhich establishes the contradiction.\n\nMethod 2: Since there are at most $\\binom{n}{2}$ tunnels possible, there will be at most\n$$\n\\binom{n}{2} - \\left(\\binom{n-1}{2} + 1\\right) = n-2\n$$\ntunnels that are not drawn. Call these \"antitunnels.\" Suppose to the contrary, that the graph is not connected. Then two settlements, $A$ and $B$ will not be connected. Thus, $A$ and $B$ are joined by an antitunnel. Furthermore, for each settlement $X$ that is neither $A$ nor $B$, there can be no path drawn from $A$ to $X$ and then from $X$ to $B$. In other words, at least one of the connections $A X$ or $X B$ must be an antitunnel. However, this would require $n-2$ antitunnels, in addition to the antitunnel joining $A$ and $B$. Thus $n-1$ antitunnels are needed, but at most $n-2$ are available; a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46935,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $2004 \\times 2004$ array of points is drawn. Find the largest integer $n$ such that it is possible to draw a convex $n$-sided polygon whose vertices lie on the points of the array.",
    "options": [],
    "answer": "561",
    "solution": "For a vector $v = (x, y)$, define $\\|v\\| = |x| + |y|$, the so-called taxicab distance (or taxicab norm). Embed the array of points in the plane such that they correspond to the lattice points in $\\{(x, y) : 1 \\le x, y \\le 2004\\}$.\n\nConsider a convex $n$-gon drawn in our square array, and imagine that we walk along the edges in a counterclockwise direction. Then we can orient each edge and obtain a set of $n$ nonzero vectors $S = \\{v_i = (x_i, y_i)\\}$, with integer coordinates, whose sum is $(0,0)$. $S$ has several further properties. First, no two vectors in $S$ are positive multiples of each other by convexity (if $i \\ne j$ and the directed edges $v_i$ and $v_j$ are parallel and pointing in the same direction, then our polygon cannot be strictly convex.) Second, the sum of the $x_i$ which are positive is at most $2003$, and the same is true for the sum of the $y_i > 0$, as well as the sums of the $|x_i|$ and $|y_i|$ for $x_i < 0$ and $y_i < 0$. This is true because all the vectors with, say, $x_i$ positive will correspond to adjacent edges (by convexity), and if one traces these edges in order on the polygon, one must start from a point within $\\{(x, y) : 1 \\le x, y \\le 2004\\}$ and finish at a point in that same region; therefore the total displacement in the $x$-dimension is bounded by $2003$. In particular, this implies that $\\sum \\|v_i\\| \\le 8012$. Third, given $S$ that satisfies the above properties, we can construct a convex polygon that fits within the bounds: since the polygon should be convex, we must place the vectors end to end ordered by the angle (measured counterclockwise) that they make with the positive $x$-axis, and the resulting polygon will fit within the array because of the given inequalities.\n\nWe now show that it is impossible to draw a $562$-gon in the array. We will prove that for every set $S$ of $562$ vectors such that no two are positive multiples of each other, $\\sum \\|v_i\\| > 8012$. We can even ignore the condition that $\\sum v_i = (0,0)$. Let us now try to minimize $\\sum \\|v_i\\|$. Since no two vectors in $S$ are positive multiples of each other, we may assume that for each $i$, $\\gcd(|x_i|, |y_i|) = 1$, or else we might as well scale that vector down by that gcd (it will only reduce $\\sum \\|v_i\\|$).\n\nNow:\n\n**Lemma** For any positive integer $k$, the maximum number of distinct $v_i = (x_i, y_i)$ that satisfy $\\|v_i\\| = k$ and $\\gcd(|x_i|, |y_i|) = 1$ is $4\\phi(k)$. Recall that $\\phi(k)$ is Euler's totient function, which counts the number of positive integers $z \\le k$ that satisfy $\\gcd(z, k) = 1$.\n\n*Proof:* For $k=1$, the lemma is trivial. For $k \\ge 2$, the gcd condition forces that neither of $x_i$ or $y_i$ can ever be $0$. So, it suffices to show that the number of $v_i$ with $x_i > 0$ and $y_i > 0$ is $\\phi(k)$, because there are exactly $4$ ways to choose pairs of signs, for the coordinates of each $v_i$. Well, since $\\|v_i\\| = k$, this means that $|x_i| + |y_i| = k$. Hence $\\gcd(|x_i|, |y_i|) = 1 \\Rightarrow \\gcd(|x_i|, k) = 1$. By definition of $\\phi(k)$, there are precisely $\\phi(k)$ ways to choose positive $x_i$ to satisfy this condition; each way yields exactly one positive choice for $y_i$, so we are done with the proof of this lemma. ■\n\nNow for $n = 562$, since $\\sum_{k=1}^{21} 4\\phi(k) = 560$, we minimize $\\sum \\|v_i\\|$ when we use all $560$ vectors with $\\|v_i\\| \\le 21$, and $2$ more vectors with $\\|v_i\\| = 22$. The value of this sum is $\\sum \\|v_i\\| \\ge \\sum_{k=1}^{21} 4k\\phi(k) + 2 \\cdot 22 = 8032 > 8012$, as was claimed. Thus $n \\le 561$.\n\nIndeed, we now construct a $561$-gon in the array by constructing a suitable set $S$. Start by setting $S$ to be the set of all $560$ integer vectors $(x_i, y_i)$ with norm at most $21$ and $\\gcd(|x_i|, |y_i|) = 1$. By symmetry, the vector sum of these $560$ vectors is $(0, 0)$. Also by symmetry, the sum of all of the positive $x_i$ is exactly $1/4$ of the sum of all of the taxicab norms, so it is $\\sum_{k=1}^{21} k\\phi(k) = 1997$. This is the span of the $560$-gon in the $x$-dimension. Similarly, that is also the span of the $560$-gon in the $y$-dimension.\n\nNext, remove from $S$ the vectors $(20, 1)$ and $(-11, 10)$, and insert the vectors $(5, 17)$, $(-17, -5)$, and $(21, -1)$. Now we have a $561$-gon. Since the vector sum of $(20, 1)$ and $(-11, 10)$ is equal to the vector sum of the $3$ new vectors, this preserves the fact that $\\sum v_i = (0, 0)$, so we can still have a closed, convex polygon. Let us see what has happened to the $x$-span and $y$-span: we removed $20$ from the sum of positive $x_i$, but added back $5$ and $21$. This is a net gain of $6$, so the sum of positive $x_i$ will increase from $1997$ to $2003$—this exactly matches the upper limit imposed by the second observation at the beginning of the proof. Similarly, the sum of positive $y_i$ will lose summands of $1$ and $10$ from the deletion, but it will gain $17$ from the insertion. This is also a net gain of $6$, so we also have $2003$ as the final sum of positive $y_i$. Since the vector sum of all $v_i$ is $(0, 0)$, similar results with $2003$ hold for the negative $x_i$ and $y_i$. By the third observation at the beginning of the proof, we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA class of 10 students took a math test. Each problem was solved by exactly 7 of the students. If the first nine students each solved 4 problems, how many problems did the tenth student solve?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nSuppose the last student solved $n$ problems, and the total number of problems on the test was $p$. Then the total number of correct solutions written was $7p$ (seven per problem), and also equal to $36 + n$ (the sum of the students' scores), so $p = (36 + n)/7$. The smallest $n \\geq 0$ for which this is an integer is $n = 6$. But we also must have $n \\leq p$, so $7n \\leq 36 + n$, and solving gives $n \\leq 6$. Thus $n = 6$ is the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind $\\lim_{x \\rightarrow \\infty}\\left(\\sqrt[3]{x^{3}+x^{2}}-\\sqrt[3]{x^{3}-x^{2}}\\right)$.",
    "options": [],
    "answer": "2/3",
    "solution": "Solution: $\\frac{2}{3}$\nObserve that\n$$\n\\lim_{x \\rightarrow \\infty}\\left[(x+1/3)-\\sqrt[3]{x^{3}+x^{2}}\\right]=\\lim_{x \\rightarrow \\infty} \\frac{x/3+1/27}{\\left(\\sqrt[3]{x^{3}+x^{2}}\\right)^{2}+\\left(\\sqrt[3]{x^{3}+x^{2}}\\right)(x+1/3)+(x+1/3)^{2}},\n$$\nby factoring the numerator as a difference of cubes. The numerator is linear in $x$, while the denominator is at least $3x^{2}$, so the limit as $x \\rightarrow \\infty$ is $0$. By similar arguments,\n$$\n\\lim_{x \\rightarrow \\infty}\\left[(x-1/3)-\\sqrt[3]{x^{3}-x^{2}}\\right]=0.\n$$\nSo, the desired limit equals\n$$\n\\frac{2}{3}+\\lim_{x \\rightarrow \\infty}\\left[(x-1/3)-\\sqrt[3]{x^{3}-x^{2}}\\right]-\\lim_{x \\rightarrow \\infty}\\left[(x+1/3)-\\sqrt[3]{x^{3}+x^{2}}\\right]=\\frac{2}{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA swimming pool is in the shape of a circle with diameter $60$ ft. The depth varies linearly along the east-west direction from $3$ ft at the shallow end in the east to $15$ ft at the diving end in the west (this is so that divers look impressive against the sunset) but does not vary at all along the north-south direction. What is the volume of the pool, in $\\mathrm{ft}^{3}$?",
    "options": [],
    "answer": "8100 π ft^3",
    "solution": "Solution:\nTake another copy of the pool, turn it upside-down, and put the two together to form a cylinder. It has height $18$ ft and radius $30$ ft, so the volume is $\\pi (30\\ \\mathrm{ft})^{2} \\cdot 18\\ \\mathrm{ft} = 16200 \\pi\\ \\mathrm{ft}^{3}$; since our pool is only half of that, the answer is $8100 \\pi\\ \\mathrm{ft}^{3}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet\n$$\nS(n, r)=\\binom{n-1}{r-1}+\\binom{n-1}{r}+\\binom{n-1}{r+1}+\\cdots+\\binom{n-1}{n-1}\n$$\nfor all $n, r>0$, and in particular $S(n, r)=0$ if $r>n>0$. Prove that the number in row $n$ of the table, $r$ columns to the left of the 1 in the top row, is at most $S(n, r)$. (Hint: First prove that $S(n-1, r-1)+S(n-1, r)=S(n, r)$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst, we prove the statement in the hint: adding the $i$th term of the sum for $S(n-1, r-1)$ to the $i$th term for $S(n-1, r)$, for each $i$, we get that $S(n-1, r-1)+S(n-1, r)$ equals\n$$\n\\begin{gathered}\n\\left(\\binom{n-2}{r-2}+\\binom{n-2}{r-1}\\right)+\\left(\\binom{n-2}{r-1}+\\binom{n-2}{r}\\right)+\\cdots \\\\\n\\cdots+\\left(\\binom{n-2}{n-3}+\\binom{n-2}{n-2}\\right)+\\binom{n-2}{n-2} \\\\\n=\\binom{n-1}{r-1}+\\binom{n-1}{r}+\\cdots+\\binom{n-1}{n-1}=S(n, r) .\n\\end{gathered}\n$$\nNow we can prove the main statement by induction on $n$. The base case $n=1$ is clear. If the statement holds for $n-1$, then first suppose $r>1$. Then the number in row $n$, $r$ columns to the left, is the sum of two of the three numbers above it, which, by the induction hypothesis, are at most $S(n-1, r-1), S(n-1, r), S(n-1, r+1)$ respectively. Since the first two of these are greater than the last (because the summation formula gives $S(n-1, r-1)=S(n-1, r)+\\binom{n-1}{r-2}$ and $S(n-1, r)=S(n-1, r+1)+\\binom{n-1}{r-1}$), we have an upper bound of $S(n-1, r-1)+S(n-1, r)=S(n, r)$ by the above. So the result follows by induction. Finally, in the case $r=1$, the quantity in question is just $2^{n-1}$, and the result holds by Problem 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $f$ is a function that assigns to each real number $x$ a value $f(x)$, and suppose the equation\n$$\nf\\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\\right)=f\\left(x_{1}\\right)+f\\left(x_{2}\\right)+f\\left(x_{3}\\right)+f\\left(x_{4}\\right)+f\\left(x_{5}\\right)-8\n$$\nholds for all real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. What is $f(0)$ ?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nPlug in $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$. Then the equation reads $f(0)=5 f(0)-8$, so $4 f(0)=8$, so $f(0)=2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf the system of equations\n$$\n\\begin{aligned}\n& |x+y|=99 \\\\\n& |x-y|=c\n\\end{aligned}\n$$\nhas exactly two real solutions $(x, y)$, find the value of $c$.",
    "options": [],
    "answer": "c = 0",
    "solution": "Solution:\nIf $c<0$, there are no solutions. If $c>0$ then we have four possible systems of linear equations given by $x+y= \\pm 99$, $x-y= \\pm c$, giving four solutions $(x, y)$. So we must have $c=0$, and then we do get two solutions ($x=y$, so they must both equal $\\pm 99 / 2$).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute:\n$$\n\\left\\lfloor\\frac{2005^{3}}{2003 \\cdot 2004}-\\frac{2003^{3}}{2004 \\cdot 2005}\\right\\rfloor .\n$$",
    "options": [],
    "answer": "8",
    "solution": "Solution: 8\nLet $x=2004$. Then the expression inside the floor brackets is\n$$\n\\frac{(x+1)^{3}}{(x-1) x}-\\frac{(x-1)^{3}}{x(x+1)}=\\frac{(x+1)^{4}-(x-1)^{4}}{(x-1) x(x+1)}=\\frac{8 x^{3}+8 x}{x^{3}-x}=8+\\frac{16 x}{x^{3}-x} .\n$$\nSince $x$ is certainly large enough that $0<16 x /(x^{3}-x)<1$, the answer is 8 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe have an $n$-gon, and each of its vertices is labeled with a number from the set $\\{1, \\ldots, 10\\}$. We know that for any pair of distinct numbers from this set there is at least one side of the polygon whose endpoints have these two numbers. Find the smallest possible value of $n$.",
    "options": [],
    "answer": "50",
    "solution": "Solution:\n\nEach number be paired with each of the 9 other numbers, but each vertex can be used in at most 2 different pairs, so each number must occur on at least $\\lceil 9 / 2\\rceil=5$ different vertices. Thus, we need at least $10 \\cdot 5=50$ vertices, so $n \\geq 50$.\n\nTo see that $n=50$ is feasible, let the numbers $1, \\ldots, 10$ be the vertices of a complete graph. Then each vertex has degree $9$, and there are $\\binom{10}{2}=45$ edges. If we attach extra copies of the edges $1-2, 3-4, 5-6, 7-8$, and $9-10$, then every vertex will have degree $10$. In particular, the graph has an Eulerian tour, so we can follow this tour, successively numbering vertices of the $50$-gon according to the vertices of the graph we visit. Then, for each edge of the graph, there will be a corresponding edge of the polygon with the same two vertex labels on its endpoints. It follows that every pair of distinct numbers occurs at the endpoints of some edge of the polygon, and so $n=50$ is the answer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$P$ is a point inside triangle $ABC$, and lines $AP$, $BP$, $CP$ intersect the opposite sides $BC$, $CA$, $AB$ in points $D$, $E$, $F$, respectively. It is given that $\\angle APB = 90^{\\circ}$, and that $AC = BC$ and $AB = BD$. We also know that $BF = 1$, and that $BC = 999$. Find $AF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "499/500",
    "solution": "Solution:\n\n![](attached_image_1.png)\nLet $AC = BC = s$, $AB = BD = t$. Since $BP$ is the altitude in isosceles triangle $ABD$, it bisects angle $B$. So, the Angle Bisector Theorem in triangle $ABC$ gives $AE / EC = AB / BC = t / s$. Meanwhile, $CD / DB = (s - t) / t$. Now Ceva's theorem gives us\n$$\n\\begin{gathered}\n\\frac{AF}{FB} = \\left(\\frac{AE}{EC}\\right) \\cdot \\left(\\frac{CD}{DB}\\right) = \\frac{s - t}{s} \\\\\n\\Rightarrow \\frac{AB}{FB} = 1 + \\frac{s - t}{s} = \\frac{2s - t}{s} \\Rightarrow FB = \\frac{st}{2s - t} .\n\\end{gathered}\n$$\nNow we know $s = 999$, but we need to find $t$ given that $st / (2s - t) = FB = 1$. So $st = 2s - t \\Rightarrow t = 2s / (s + 1)$, and then\n$$\nAF = FB \\cdot \\frac{AF}{FB} = 1 \\cdot \\frac{s - t}{s} = \\frac{(s^2 - s)/(s + 1)}{s} = \\frac{s - 1}{s + 1} = \\frac{499}{500} .\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$$\nb_{i}= \\begin{cases}1 & \\text{ if } i \\text{ is a multiple of } 3 \\\\ 0 & \\text{ otherwise }\\end{cases}\n$$\nLet $\\{a_{i}\\}$ be a sequence of elements of $\\{0,1\\}$ such that\n$$\nb_{n} \\equiv a_{n-1}+a_{n}+a_{n+1} \\quad(\\bmod 2)\n$$\nfor $0 \\leq n \\leq 59$ ($a_{0}=a_{60}$ and $a_{-1}=a_{59}$). Find all possible values of $4 a_{0}+2 a_{1}+a_{2}$.",
    "options": [],
    "answer": "0, 3, 5, 6",
    "solution": "Solution:\n\nTry the four possible combinations of values for $a_{0}$ and $a_{1}$. Since we can write $a_{n} \\equiv b_{n-1}-a_{n-2}-a_{n-1}$, these two numbers completely determine the solution $\\{a_{i}\\}$ beginning with them (if there is one).\n\nFor $a_{0}=a_{1}=0$, we can check that the sequence beginning $0,0,0,0,1,1$ and repeating every 6 indices is a possible solution for $\\{a_{i}\\}$, so one possible value for $4 a_{0}+2 a_{1}+a_{2}$ is $0$.\n\nThe other three combinations for $a_{0}$ and $a_{1}$ similarly lead to valid sequences (produced by repeating the sextuples $0,1,1,1,0,1$; $1,0,1,1,1,0$; $1,1,0,1,0,1$, respectively); we thus obtain the values $3$, $5$, and $6$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 46946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA plane $P$ slices through a cube of volume $1$ with a cross-section in the shape of a regular hexagon. This cube also has an inscribed sphere, whose intersection with $P$ is a circle. What is the area of the region inside the regular hexagon but outside the circle?",
    "options": [],
    "answer": "(3*sqrt(3)-pi)/4",
    "solution": "Solution:\nOne can show that the hexagon must have as its vertices the midpoints of six edges of the cube, as illustrated; for example, this readily follows from the fact that opposite sides of the hexagons and the medians between them are parallel. We then conclude that the side of the hexagon is $\\sqrt{2} / 2$ (since it cuts off an isosceles triangle of leg $1 / 2$ from each face), so the area is $\\left(\\frac{3}{2}\\right)\\left(\\frac{\\sqrt{2}}{2}\\right)^{2}\\left(\\sqrt{3}\\right) = \\frac{3 \\sqrt{3}}{4}$. Also, the plane passes through the center of the sphere by symmetry, so it cuts out a cross section of radius $1 / 2$, whose area (which is contained entirely inside the hexagon) is then $\\pi / 4$. The sought area is thus $\\frac{3 \\sqrt{3} - \\pi}{4}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of 20-tuples of integers $x_{1}, \\ldots, x_{10}, y_{1}, \\ldots, y_{10}$ with the following properties:\n\n- $1 \\leq x_{i} \\leq 10$ and $1 \\leq y_{i} \\leq 10$ for each $i$;\n- $x_{i} \\leq x_{i+1}$ for $i=1, \\ldots, 9$;\n- if $x_{i}=x_{i+1}$, then $y_{i} \\leq y_{i+1}$.",
    "options": [],
    "answer": "C(109,10)",
    "solution": "Solution:\nBy setting $z_{i}=10 x_{i}+y_{i}$, we see that the problem is equivalent to choosing a nondecreasing sequence of numbers $z_{1}, z_{2}, \\ldots, z_{10}$ from the values $11,12, \\ldots, 110$. Making a further substitution by setting $w_{i}=z_{i}-11+i$, we see that the problem is equivalent to choosing a strictly increasing sequence of numbers $w_{1}, \\ldots, w_{10}$ from among the values $1,2, \\ldots, 109$. There are $\\binom{109}{10}$ ways to do this.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are eleven positive integers $n$ such that there exists a convex polygon with $n$ sides whose angles, in degrees, are unequal integers that are in arithmetic progression. Find the sum of these values of $n$.",
    "options": [],
    "answer": "106",
    "solution": "Solution:\n\nThe sum of the angles of an $n$-gon is $(n-2) 180$, so the average angle measure is $(n-2) 180 / n$. The common difference in this arithmetic progression is at least $1$, so the difference between the largest and smallest angles is at least $n-1$. So the largest angle is at least $(n-1)/2 + (n-2) 180 / n$. Since the polygon is convex, this quantity is no larger than $179$: $(n-1)/2 - 360/n \\leq -1$, so that $360/n - n/2 \\geq 1/2$. Multiplying by $2n$ gives $720 - n^{2} \\geq n$. So $n(n+1) \\leq 720$, which forces $n \\leq 26$.\n\nOf course, since the common difference is an integer, and the angle measures are integers, $(n-2) 180 / n$ must be an integer or a half integer, so $(n-2) 360 / n = 360 - 720 / n$ is an integer, and then $720 / n$ must be an integer. This leaves only $n = 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24$ as possibilities.\n\nWhen $n$ is even, $(n-2) 180 / n$ is not an angle of the polygon, but the mean of the two middle angles. So the common difference is at least $2$ when $(n-2) 180 / n$ is an integer. For $n=20$, the middle angle is $162$, so the largest angle is at least $162 + 38/2 = 181$, since $38$ is no larger than the difference between the smallest and largest angles. For $n=24$, the middle angle is $165$, again leading to a contradiction. So no solution exists for $n=20,24$. All of the others possess solutions:\n\n| $n$ | angles |\n| :---: | :--- |\n| 3 | $59, 60, 61$ |\n| 4 | $87, 89, 91, 93$ |\n| 5 | $106, 107, 108, 109, 110$ |\n| 6 | $115, 117, 119, 121, 123, 125$ |\n| 8 | $128, 130, 132, 134, 136, 138, 140, 142$ |\n| 9 | $136, \\ldots, 144$ |\n| 10 | $135, 137, 139, \\ldots, 153$ |\n| 12 | $139, 141, 143, \\ldots, 161$ |\n| 15 | $149, 150, \\ldots, 163$ |\n| 16 | $150, 151, \\ldots, 165$ |\n| 18 | $143, 145, \\ldots, 177$ |\n\n(These solutions are quite easy to construct.) The desired value is then $3 + 4 + 5 + 6 + 8 + 9 + 10 + 12 + 15 + 16 + 18 = 106$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor a string of $P$ 's and $Q$ 's, the value is defined to be the product of the positions of the $P$ 's. For example, the string $P P Q P Q Q$ has value $1 \\cdot 2 \\cdot 4=8$.\n\nAlso, a string is called antipalindromic if writing it backwards, then turning all the $P$ 's into $Q$ 's and vice versa, produces the original string. For example, $P P Q P Q Q$ is antipalindromic.\n\nThere are $2^{1002}$ antipalindromic strings of length $2004$. Find the sum of the reciprocals of their values.",
    "options": [],
    "answer": "2005^1002/2004!",
    "solution": "Solution:\n$2005^{1002} / 2004$ !\nConsider the product\n$$\n\\left(\\frac{1}{1}+\\frac{1}{2004}\\right)\\left(\\frac{1}{2}+\\frac{1}{2003}\\right)\\left(\\frac{1}{3}+\\frac{1}{2002}\\right) \\cdots\\left(\\frac{1}{1002}+\\frac{1}{1003}\\right) .\n$$\nThis product expands to $2^{1002}$ terms, and each term gives the reciprocal of the value of a corresponding antipalindromic string of $P$ 's and $Q$ 's as follows: if we choose the term $1 / n$ for the $n$th factor, then our string has a $P$ in position $n$ and $Q$ in position $2005-n$; if we choose the term $1 /(2005-n)$, then we get a $Q$ in position $n$ and $P$ in position $2005-n$. Conversely, each antipalindromic string has its value represented by exactly one of our $2^{1002}$ terms. So the value of the product is the number we are looking for. But when we simplify this product, the $n$th factor becomes $1 / n+1 /(2005-n)=2005 / n(2005-n)$. Multiplying these together, we get 1002 factors of 2005 in the numerator and each integer from 1 to 2004 exactly once in the denominator, for a total of $2005^{1002}$ / $2004!$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSimplify $\\prod_{k=1}^{2004} \\sin (2 \\pi k / 4009)$.",
    "options": [],
    "answer": "sqrt(4009)/2^2004",
    "solution": "Solution:\n$\\frac{\\sqrt{4009}}{2^{2004}}$\n\nLet $\\zeta = e^{2 \\pi i / 4009}$ so that $\\sin (2 \\pi k / 4009) = \\frac{\\zeta^{k} - \\zeta^{-k}}{2i}$ and $x^{4009} - 1 = \\prod_{k=0}^{4008} (x - \\zeta^{k})$. Hence $1 + x + \\cdots + x^{4008} = \\prod_{k=1}^{4008} (x - \\zeta^{k})$. Comparing constant coefficients gives $\\prod_{k=1}^{4008} \\zeta^{k} = 1$, setting $x = 1$ gives $\\prod_{k=1}^{4008} (1 - \\zeta^{k}) = 4009$, and setting $x = -1$ gives $\\prod_{k=1}^{4008} (1 + \\zeta^{k}) = 1$. Now, note that $\\sin (2 \\pi (4009 - k) / 4009) = -\\sin (2 \\pi k / 4009)$, so\n$$\n\\begin{aligned}\n\\left(\\prod_{k=1}^{2004} \\sin (2 \\pi k / 4009)\\right)^{2} &= (-1)^{2004} \\prod_{k=1}^{4008} \\sin (2 \\pi k / 4009) \\\\\n&= \\prod_{k=1}^{4008} \\frac{\\zeta^{k} - \\zeta^{-k}}{2i} \\\\\n&= \\frac{1}{(2i)^{4008}} \\prod_{k=1}^{4008} \\frac{\\zeta^{2k} - 1}{\\zeta^{k}} \\\\\n&= \\frac{1}{2^{4008}} \\prod_{k=1}^{4008} (\\zeta^{2k} - 1) \\\\\n&= \\frac{1}{2^{4008}} \\prod_{k=1}^{4008} (\\zeta^{k} - 1)(\\zeta^{k} + 1) \\\\\n&= \\frac{4009 \\cdot 1}{2^{4008}} .\n\\end{aligned}\n$$\nHowever, $\\sin(x)$ is nonnegative on the interval $[0, \\pi]$, so our product is positive. Hence it is $\\frac{\\sqrt{4009}}{2^{2004}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S=\\{p_{1} p_{2} \\cdots p_{n} \\mid p_{1}, p_{2}, \\ldots, p_{n}$ are distinct primes and $p_{1}, \\ldots, p_{n}<30\\}$. Assume $1$ is in $S$. Let $a_{1}$ be an element of $S$. We define, for all positive integers $n$:\n$$\n\\begin{gathered}\na_{n+1}=a_{n} /(n+1) \\quad \\text{ if } a_{n} \\text{ is divisible by } n+1 \\\\\na_{n+1}=(n+2) a_{n} \\\\\n\\text{ if } a_{n} \\text{ is not divisible by } n+1\n\\end{gathered}\n$$\nHow many distinct possible values of $a_{1}$ are there such that $a_{j}=a_{1}$ for infinitely many $j$'s?",
    "options": [],
    "answer": "512",
    "solution": "Solution:\nIf $a_{1}$ is odd, then we can see by induction that $a_{j}=(j+1) a_{1}$ when $j$ is even and $a_{j}=a_{1}$ when $j$ is odd (using the fact that no even $j$ can divide $a_{1}$). So we have infinitely many $j$'s for which $a_{j}=a_{1}$.\n\nIf $a_{1}>2$ is even, then $a_{2}$ is odd, since $a_{2}=a_{1} / 2$, and $a_{1}$ may have only one factor of $2$. Now, in general, let $p=\\min (\\{p_{1}, \\ldots, p_{n}\\} \\setminus\\{2\\})$. Suppose $1<j<p$. By induction, we have $a_{j}=(j+1) a_{1} / 2$ when $j$ is odd, and $a_{j}=a_{1} / 2$ when $j$ is even. So $a_{i} \\neq a_{1}$ for all $1<j<p$. It follows that $a_{p}=a_{1} / 2 p$. Then, again using induction, we get for all nonnegative integers $k$ that $a_{p+k}=a_{p}$ if $k$ is even, and $a_{p+k}=(p+k+1) a_{p}$ if $k$ is odd. Clearly, $a_{p} \\neq a_{1}$ and $p+k+1 \\neq 2 p$ when $k$ is odd (the left side is odd, and the right side even). It follows that $a_{j}=a_{1}$ for no $j>1$.\n\nFinally, when $a_{1}=2$, we can check inductively that $a_{j}=j+1$ for $j$ odd and $a_{j}=1$ for $j$ even.\n\nSo our answer is just the number of odd elements in $S$. There are $9$ odd prime numbers smaller than $30$, so the answer is $2^{9}=512$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou want to arrange the numbers $1,2,3, \\ldots, 25$ in a sequence with the following property: if $n$ is divisible by $m$, then the $n$th number is divisible by the $m$th number. How many such sequences are there?",
    "options": [],
    "answer": "24",
    "solution": "Solution:\nLet the rearranged numbers be $a_{1}, \\ldots, a_{25}$. The number of pairs $(n, m)$ with $n \\mid m$ must equal the number of pairs with $a_{n} \\mid a_{m}$, but since each pair of the former type is also of the latter type, the converse must be true as well. Thus, $n \\mid m$ if and only if $a_{n} \\mid a_{m}$. Now for each $n=1,2, \\ldots, 6$, the number of values divisible by $n$ uniquely determines $n$, so $n=a_{n}$. Similarly, $7, 8$ must either be kept fixed by the rearrangement or interchanged, because they are the only values that divide exactly $2$ other numbers in the sequence; since $7$ is prime and $8$ is not, we conclude they are kept fixed. Then we can easily check by induction that $n=a_{n}$ for all larger composite numbers $n \\leq 25$ (by using $m=a_{m}$ for all proper factors $m$ of $n$) and $n=11$ (because it is the only prime that divides exactly $1$ other number). So we have only the primes $n=13,17,19,23$ left to rearrange, and it is easily seen that these can be permuted arbitrarily, leaving $4!$ possible orderings altogether.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 46953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose one is given $n$ real numbers, not all zero, but such that their sum is zero. Prove that one can label these numbers $a_{1}, a_{2}, \\ldots, a_{n}$ in such a manner that\n$$\na_{1} a_{2}+a_{2} a_{3}+\\cdots+a_{n-1} a_{n}+a_{n} a_{1}<0 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet the given numbers (in an arbitrary order) be $b_{1}, b_{2}, \\ldots, b_{n}$. For every possible permutation $\\pi$ of $\\{1,2, \\ldots, n\\}$, consider the sum\n$$\nb_{\\pi(1)} b_{\\pi(2)}+b_{\\pi(2)} b_{\\pi(3)}+\\cdots+b_{\\pi(n-1)} b_{\\pi(n)}+b_{\\pi(n)} b_{\\pi(1)} .\n$$\nWe wish to show that some such sum is negative, so assume otherwise. For every two distinct elements $i, j \\in\\{1, \\ldots, n\\}$, the term $b_{i} b_{j}$ appears $N$ times among these sums, where $N$ does not depend on $i, j$, by symmetry. (In fact, one can show that $N=n(n-2)$ !.) Each such sum is assumed to be nonnegative; adding these inequalities for all permutations $\\pi$, and dividing by $N$, we have\n$$\n\\sum_{i \\neq j} b_{i} b_{j} \\geq 0\n$$\nHowever, we also know that $\\sum_{i} b_{i}^{2}>0$ (strictly, since not all $b_{i}$ are zero). Thus\n$$\n\\left(b_{1}+\\cdots+b_{n}\\right)^{2}=\\sum_{i} b_{i}^{2}+\\sum_{i \\neq j} b_{i} b_{j}>0\n$$\nBut since $b_{1}+\\cdots+b_{n}=0$, we have a contradiction. So our assumption was false, and the needed negative sum does exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle and let $D$ be a point in its interior. Construct a circle $\\omega_1$ passing through $B$ and $D$ and a circle $\\omega_2$ passing through $C$ and $D$ such that the point of intersection of $\\omega_1$ and $\\omega_2$ other than $D$ lies on line $AD$. Denote by $E$ and $F$ the points where $\\omega_1$ and $\\omega_2$ intersect side $BC$, respectively, and by $X$ and $Y$ the intersections of lines $DF$, $AB$ and $DE$, $AC$, respectively. Prove that $XY \\parallel BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution:** Let circles $\\omega_1$ and $\\omega_2$ meet again at $R$ (other than $D$), and let $\\omega_1$ and $\\omega_2$ intersect again with segments $AB$ and $AC$, respectively, at $P$ and $Q$ (other than $B$ and $C$). By the **Power of a Point Theorem**, we have $AP \\cdot AB = AR \\cdot AD = AQ \\cdot AC$; that is, points $A, R$, and $D$ lie on the **radical axis** of the circles $\\omega_1$ and $\\omega_2$. Therefore, points $B, C, Q$, and $P$ lie on a circle. Consequently, $\\angle AQP = \\angle ABC$. To prove that $XY \\parallel BC$, it suffices to show that $\\angle AXY = \\angle ABC$, or $\\angle AQP = \\angle ABC = \\angle AXY$; that is, to prove that $P, Q, Y$, and $X$ lie on a circle.\n\nBecause $BCQP$ is cyclic, $\\angle AQP = \\angle ABC$. Because $BDRP$ is cyclic, $\\angle PDY = \\angle PBE = \\angle ABC$. Hence $\\angle AQP = \\angle PDY$, implying that $P, Q, Y$ and $D$ lie on a circle. Similarly, we can show that $P, Q, D$, and $X$ lie on a circle. We conclude that $P, Q, Y, D$, and $X$ all lie on the circumcircle of triangle $PQD$, completing the proof.\n\n\n**Second Solution:** It suffices to show that\n$$\n\\frac{AX}{XB} = \\frac{AY}{YC}. \\qquad (*)\n$$\nLet line $AD$ and segment $BC$ meet at $G$. Applying **Menelaus's Theorem** to triangle $ABG$ and line $XF$, triangle $ACG$ and line $YE$ gives\n$$\n\\frac{AX \\cdot BF \\cdot GD}{XB \\cdot FG \\cdot DA} = 1 \\text{ and } \\frac{AY \\cdot CE \\cdot GD}{YC \\cdot EG \\cdot DA} = 1.\n$$\nHence\n$$\n\\frac{AX \\cdot BF}{XB \\cdot FG} = \\frac{AY \\cdot CE}{YC \\cdot EG}.\n$$\nBy the equation (*), it suffices to show that\n$$\n\\frac{BF}{FG} = \\frac{CE}{EG},\n$$\nor $EG \\cdot BF = CE \\cdot FG$. By the Power of a point Theorem, we have $EG \\cdot BG = GD \\cdot GR = GF \\cdot GC$, implying that\n$$\n\\begin{aligned}\nEG \\cdot BF &= EG(BG + GF) = EG \\cdot BG + EG \\cdot GF \\\\\n&= GF \\cdot GC + EG \\cdot GF = GF(GC + GE) \\\\\n&= CE \\cdot FG,\n\\end{aligned}\n$$\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice and Bob play a game on a $6$ by $6$ grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players.",
    "options": [],
    "answer": "Bob has a winning strategy.",
    "solution": "**First Solution:** Bob can win as follows.\nAfter each of his moves, Bob can insure that the maximum number in each row is a square in $A \\cup B$, where $A$ and $B$ are the sets of squares marked with A's and B's in the following diagram, respectively.\n![](attached_image_1.png)\n**Proof:** Bob pairs each square of $A \\cup B$ with a square in the same row that is not in $A \\cup B$, so that each square of the grid is in exactly one pair. Whenever Alice plays in one square of a pair, Bob will play in the other square of the pair on his next turn. If Alice moves with $x$ in $A \\cup B$, Bob writes $y$ with $y < x$ in the paired square. If Alice moves with $x$ not in $A \\cup B$, Bob writes $z$ with $z > x$ in the paired square in $A \\cup B$. So after Bob's turn, the maximum of each pair is in $A \\cup B$, and thus the maximum of each row is in $A \\cup B$. ■\nSo when all the numbers are written, the maximum square in row $1$ is in $B$ and the maximum square in row $6$ is in $A$. Since there is no path from $B$ to $A$ that stays in $A \\cup B$, Bob wins.\n\n\n**Second Solution:** (By Tiankai Liu) Let $P$ be the property that the following conditions are met:\n(a) Row $1$, column $c$ is empty if and only if row $2$, column $c+3$ is (taken modulo $6$).\n(b) If any numbers have been written in the first or second rows, let $c_i$ be the index of the column with the largest number in row $i$; we require that $|c_1 - c_2| = 3$.\nThe initial configuration of the grid (empty) clearly satisfies $P$, and Bob can preserve $P$ by executing the following strategy after each of Alice's moves:\n**Case 1: Alice moves somewhere in the lower $4$ rows.**\nBob should pick a random square somewhere in the bottom four rows, and write any rational number that has not yet been chosen. This is always possible because the number of squares in the bottom four rows is $24$, and if Bob follows this strategy, there will always be an odd number of empty squares (i.e. nonzero) in this section every time Alice's move puts him into this case.\n**Case 2: Alice moves somewhere in the top two rows.**\nSuppose that she chose the square in row $r$, column $c$. If the number she wrote is now the largest number in row $r$, then Bob should choose a number larger than every number written on the board, and write it in the square at row $3-r$, column $c+3$ (taken modulo $6$). On the other hand, if Alice's new number is not the largest number in row $r$, then Bob should choose a number smaller than every number written on the board, and write it in row $3-r$, column $c+3$ (taken modulo $6$).\nIt is clear that by the end of Bob's move, $P$ is preserved in both cases. Yet this implies that at the end of the game, $|c_1 - c_2| = 3$, which means that the black squares in the first two rows don't touch anywhere. Therefore, there is no way Alice can connect the top and bottom of the grid with a continuous path, because the blackened region will be disconnected.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 46956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = \\cos (\\cos (\\cos (\\cos (\\cos (\\cos (\\cos (\\cos x)))))))$, and suppose that the number $a$ satisfies the equation $a = \\cos a$. Express $f'(a)$ as a polynomial in $a$.",
    "options": [],
    "answer": "a^8 - 4 a^6 + 6 a^4 - 4 a^2 + 1",
    "solution": "Solution: $a^{8} - 4 a^{6} + 6 a^{4} - 4 a^{2} + 1$\nThis is an exercise using the chain rule. Define $f_{0}(x) = x$ and $f_{n}(x) = \\cos f_{n-1}(x)$ for $n \\geq 0$. We will show by induction that $f_{n}(a) = a$ and $f_{n}'(a) = (-\\sin a)^{n}$ for all $n$. The case $n = 0$ is clear. Then $f_{n}(a) = \\cos f_{n-1}(a) = \\cos a = a$, and\n$$\nf_{n}'(a) = f_{n-1}'(a) \\cdot \\left(-\\sin f_{n-1}(a)\\right) = (-\\sin a)^{n-1} \\cdot (-\\sin a) = (-\\sin a)^{n}\n$$\nby induction. Now, $f(x) = f_{8}(x)$, so $f'(a) = (-\\sin a)^{8} = \\sin^{8} a$. But $\\sin^{2} a = 1 - \\cos^{2} a = 1 - a^{2}$, so $f'(a) = (1 - a^{2})^{4} = a^{8} - 4 a^{6} + 6 a^{4} - 4 a^{2} + 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 46957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways can you mark 8 squares of an $8 \\times 8$ chessboard so that no two marked squares are in the same row or column, and none of the four corner squares is marked? (Rotations and reflections are considered different.)",
    "options": [],
    "answer": "21600",
    "solution": "Solution:\n\nIn the top row, you can mark any of the 6 squares that is not a corner. In the bottom row, you can then mark any of the 5 squares that is not a corner and not in the same column as the square just marked. Then, in the second row, you have 6 choices for a square not in the same column as either of the two squares already marked; then there are 5 choices remaining for the third row, and so on down to 1 for the seventh row, in which you make the last mark. Thus, altogether, there are $6 \\cdot 5 \\cdot (6 \\cdot 5 \\cdots 1) = 30 \\cdot 6! = 30 \\cdot 720 = 21600$ possible sets of squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAndrea flips a fair coin repeatedly, continuing until she either flips two heads in a row (the sequence $H H$) or flips tails followed by heads (the sequence $T H$). What is the probability that she will stop after flipping $H H$?",
    "options": [],
    "answer": "1/4",
    "solution": "Solution:\n\nThe only way that Andrea can ever flip $H H$ is if she never flips $T$, in which case she must flip two heads immediately at the beginning. This happens with probability $\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA tree grows in a rather peculiar manner. Lateral cross-sections of the trunk, leaves, branches, twigs, and so forth are circles. The trunk is $1$ meter in diameter to a height of $1$ meter, at which point it splits into two sections, each with diameter $0.5$ meter. These sections are each one meter long, at which point they each split into two sections, each with diameter $0.25$ meter. This continues indefinitely: every section of tree is $1$ meter long and splits into two smaller sections, each with half the diameter of the previous.\nWhat is the total volume of the tree?",
    "options": [],
    "answer": "pi/2",
    "solution": "Solution:\nIf we count the trunk as level $0$, the two sections emerging from it as level $1$, and so forth, then the $n$th level consists of $2^{n}$ sections each with diameter $1 / 2^{n}$, for a volume of $2^{n}\\left(\\pi / 4 \\cdot 2^{-2 n}\\right) = (\\pi / 4) \\cdot 2^{-n}$. So the total volume is given by a simple infinite sum,\n$$\n0.25 \\pi \\cdot (1 + 1/2 + 1/4 + \\ldots) = 0.25 \\pi \\cdot 2 = \\pi / 2.\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate the sum\n$$\n\\frac{1}{2\\lfloor\\sqrt{1}\\rfloor+1}+\\frac{1}{2\\lfloor\\sqrt{2}\\rfloor+1}+\\frac{1}{2\\lfloor\\sqrt{3}\\rfloor+1}+\\cdots+\\frac{1}{2\\lfloor\\sqrt{100}\\rfloor+1}.\n$$",
    "options": [],
    "answer": "190/21",
    "solution": "Solution:\nThe first three terms all equal $1/3$, then the next five all equal $1/5$; more generally, for each $a=1,2,\\ldots,9$, the terms $1/(2\\lfloor\\sqrt{a^{2}}\\rfloor+1)$ to $1/(2\\lfloor\\sqrt{a^{2}+2a}\\rfloor+1)$ all equal $1/(2a+1)$, and there are $2a+1$ such terms. Thus our terms can be arranged into 9 groups, each with sum $1$, and only the last term $1/(2\\lfloor\\sqrt{100}\\rfloor+1)$ remains, so the answer is $9+1/21=190/21$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA horse stands at the corner of a chessboard, a white square. With each jump, the horse can move either two squares horizontally and one vertically or two vertically and one horizontally (like a knight moves). The horse earns two carrots every time it lands on a black square, but it must pay a carrot in rent to rabbit who owns the chessboard for every move it makes. When the horse reaches the square on which it began, it can leave. What is the maximum number of carrots the horse can earn without touching any square more than twice?\n![](attached_image_1.png)",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nThe horse must alternate white and black squares, and it ends on the same square where it started. Thus it lands on the same number of black squares ($b$) as white squares ($w$). Thus, its net earnings will be $2b - (b + w) = b - w = 0$ carrots, regardless of its path.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$P$ is inside rectangle $ABCD$. $PA = 2$, $PB = 3$, and $PC = 10$. Find $PD$.",
    "options": [],
    "answer": "sqrt(95)",
    "solution": "Solution:\n\nDraw perpendiculars from $P$ to $E$ on $AB$, $F$ on $BC$, $G$ on $CD$, and $H$ on $DA$, and let $AH = BF = w$, $HD = FC = x$, $AE = DG = y$, and $EB = GC = z$. Then $PA^{2} = w^{2} + y^{2}$, $PB^{2} = w^{2} + z^{2}$, $PC^{2} = x^{2} + z^{2}$, and $PD^{2} = x^{2} + y^{2}$. Adding and subtracting, we see that $PD^{2} = PA^{2} - PB^{2} + PC^{2} = 95$, so $PD = \\sqrt{95}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou would like to provide airline service to the 10 cities in the nation of Schizophrenia, by instituting a certain number of two-way routes between cities. Unfortunately, the government is about to divide Schizophrenia into two warring countries of five cities each, and you don't know which cities will be in each new country. All airplane service between the two new countries will be discontinued. However, you want to make sure that you set up your routes so that, for any two cities in the same new country, it will be possible to get from one city to the other (without leaving the country).\nWhat is the minimum number of routes you must set up to be assured of doing this, no matter how the government divides up the country?",
    "options": [],
    "answer": "30",
    "solution": "Solution:\nEach city $C$ must be directly connected to at least 6 other cities, since otherwise the government could put $C$ in one country and all its connecting cities in the other country, and there would be no way out of $C$. This means that we have 6 routes for each of 10 cities, counted twice (since each route has two endpoints) $\\Rightarrow 6 \\cdot 10 / 2=30$ routes. On the other hand, this is enough: picture the cities arranged around a circle, and each city connected to its 3 closest neighbors in either direction. Then if $C$ and $D$ are in the same country but mutually inaccessible, this means that on each arc of the circle between $C$ and $D$, there must be (at least) three consecutive cities in the other country. Then this second country would have 6 cities, which is impossible. So our arrangement achieves the goal with 30 routes.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA tetrahedron has all its faces triangles with sides $13, 14, 15$. What is its volume?",
    "options": [],
    "answer": "42√55",
    "solution": "Solution:\nLet $ABC$ be a triangle with $AB = 13$, $BC = 14$, $CA = 15$. Let $AD$, $BE$ be altitudes. Then $BD = 5$, $CD = 9$. (If you don't already know this, it can be deduced from the Pythagorean Theorem: $CD^{2} - BD^{2} = (CD^{2} + AD^{2}) - (BD^{2} + AD^{2}) = AC^{2} - AB^{2} = 56$, while $CD + BD = BC = 14$, giving $CD - BD = 56 / 14 = 4$, and now solve the linear system.) Also, $AD = \\sqrt{AB^{2} - BD^{2}} = 12$. Similar reasoning gives $AE = 33 / 5$, $EC = 42 / 5$.\n\n![](attached_image_1.png)\n\nNow let $F$ be the point on $BC$ such that $CF = BD = 5$, and let $G$ be on $AC$ such that $CG = AE = 33 / 5$. Imagine placing face $ABC$ flat on the table, and letting $X$ be a point in space with $CX = 13$, $BX = 14$. By mentally rotating triangle $BCX$ about line $BC$, we can see that $X$ lies on the plane perpendicular to $BC$ through $F$. In particular, this holds if $X$ is the fourth vertex of our tetrahedron $ABCX$. Similarly, $X$ lies on the plane perpendicular to $AC$ through $G$. Let the mutual intersection of these two planes and plane $ABC$ be $H$. Then $XH$ is the altitude of the tetrahedron.\n\nTo find $XH$, extend $FH$ to meet $AC$ at $I$. Then $\\triangle CFI \\sim \\triangle CDA$, a 3-4-5 triangle, so $FI = CF \\cdot 4 / 3 = 20 / 3$, and $CI = CF \\cdot 5 / 3 = 25 / 3$. Then $IG = CI - CG = 26 / 15$, and $HI = IG \\cdot 5 / 4 = 13 / 6$. This leads to $HF = FI - HI = 9 / 2$, and finally $XH = \\sqrt{XF^{2} - HF^{2}} = \\sqrt{AD^{2} - HF^{2}} = 3 \\sqrt{55} / 2$.\n\nNow $XABC$ is a tetrahedron whose base $\\triangle ABC$ has area $AD \\cdot BC / 2 = 12 \\cdot 14 / 2 = 84$, and whose height $XH$ is $3 \\sqrt{55} / 2$, so its volume is $(84)(3 \\sqrt{55} / 2) / 3 = 42 \\sqrt{55}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 46965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n$S$ is a set of complex numbers such that if $u, v \\in S$, then $u v \\in S$ and $u^{2}+v^{2} \\in S$. Suppose that the number $N$ of elements of $S$ with absolute value at most $1$ is finite. What is the largest possible value of $N$?",
    "options": [],
    "answer": "13",
    "solution": "Solution:\nFirst, if $S$ contained some $u \\neq 0$ with absolute value $<1$, then (by the first condition) every power of $u$ would be in $S$, and $S$ would contain infinitely many different numbers of absolute value $<1$. This is a contradiction.\n\nNow suppose $S$ contains some number $u$ of absolute value $1$ and argument $\\theta$. If $\\theta$ is not an integer multiple of $\\pi / 6$, then $u$ has some power $v$ whose argument lies strictly between $\\theta+\\pi / 3$ and $\\theta+\\pi / 2$. Then $u^{2}+v^{2}=u^{2}\\left(1+(v / u)^{2}\\right)$ has absolute value between $0$ and $1$, since $(v / u)^{2}$ lies on the unit circle with angle strictly between $2 \\pi / 3$ and $\\pi$. But $u^{2}+v^{2} \\in S$, so this is a contradiction.\n\nThis shows that the only possible elements of $S$ with absolute value $\\leq 1$ are $0$ and the points on the unit circle whose arguments are multiples of $\\pi / 6$, giving $N \\leq 1+12=13$.\n\nTo show that $N=13$ is attainable, we need to show that there exists a possible set $S$ containing all these points. Let $T$ be the set of all numbers of the form $a+b \\omega$, where $a, b$ are integers and $\\omega$ is a complex cube root of $1$. Since $\\omega^{2}=-1-\\omega$, $T$ is closed under multiplication and addition. Then, if we let $S$ be the set of numbers $u$ such that $u^{2} \\in T$, $S$ has the required properties, and it contains the $13$ complex numbers specified, so we're in business.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWrite down an integer from $0$ to $20$ inclusive. This problem will be scored as follows: if $N$ is the second-largest number from among the responses submitted, then each team that submits $N$ gets $N$ points, and everyone else gets zero. (If every team picks the same number then nobody gets any points.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe only Nash equilibria of this game (where each team plays its best possible move given the other teams' choices) are fairly degenerate: every team but one plays $1$, and the remaining team is more likely to choose $2$ than any higher number. Of course, we cannot assume perfectly rational play in reality - nor are the utility functions the same, since the goal is to score higher than other teams, not to maximize one's own expected number of points. It will be interesting to see what the submissions are.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShown on your answer sheet is a $20 \\times 20$ grid. Place as many queens as you can so that each of them attacks at most one other queen. (A queen is a chess piece that can move any number of squares horizontally, vertically, or diagonally.) It's not very hard to get 20 queens, so you get no points for that, but you get 5 points for each further queen beyond 20. You can mark the grid by placing a dot in each square that contains a queen.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAn elementary argument shows there cannot be more than 26 queens: we cannot have more than 2 in a row or column (or else the middle queen would attack the other two), so if we had 27 queens, there would be at least 7 columns with more than one queen and thus at most 13 queens that are alone in their respective columns. Similarly, there would be at most 13 queens that are alone in their respective rows. This leaves $27-13-13=1$ queen who is not alone in her row or column, and she therefore attacks two other queens, contradiction.\n\nOf course, this is not a very strong argument since it makes no use of the diagonals. The best possible number of queens is not known to us; the following construction gives 23:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA binary string of length $n$ is a sequence of $n$ digits, each of which is $0$ or $1$. The distance between two binary strings of the same length is the number of positions in which they disagree; for example, the distance between the strings $01101011$ and $00101110$ is $3$ since they differ in the second, sixth, and eighth positions.\n\nFind as many binary strings of length $8$ as you can, such that the distance between any two of them is at least $3$. You get one point per string.",
    "options": [],
    "answer": "20",
    "solution": "Solution:\nThe maximum possible number of such strings is $20$. An example of a set attaining this bound is\n\n| 00000000 | 00110101 |\n| :--- | :--- |\n| 11001010 | 10011110 |\n| 11100001 | 01101011 |\n| 11010100 | 01100110 |\n| 10111001 | 10010011 |\n| 01111100 | 11001101 |\n| 00111010 | 10101100 |\n| 01010111 | 11110010 |\n| 00001111 | 01011001 |\n| 10100111 | 11111111 |\n\nThis example is taken from page 57 of F. J. MacWilliams and N. J. A. Sloane, *The Theory of Error Correcting Codes* (New York: Elsevier Publishing, 1977). The proof that $20$ is the best possible is elementary but too long to reproduce here; see pages 537-541 of MacWilliams and Sloane for details.\n\nIn general, a set of $M$ strings of length $n$ such that any two have a distance of at least $d$ is called an $(n, M, d)$-code. These objects are of basic importance in coding theory, which studies how to transmit information through a channel with a known error rate. For example, since the code given above has minimum distance $3$, I can transmit to you a message consisting of strings in this code, and even if there is a possible error rate of one digit in each string, you will still be able to determine the intended message uniquely.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind (with proof) all monic polynomials $f(x)$ with integer coefficients that satisfy the following two conditions.\n1. $f(0)=2004$.\n2. If $x$ is irrational, then $f(x)$ is also irrational.\n\n(Notes: A polynomial is monic if its highest degree term has coefficient 1. Thus, $f(x)=x^{4}-5 x^{3}-4 x+7$ is an example of a monic polynomial with integer coefficients.\nA number $x$ is rational if it can be written as a fraction of two integers. A number $x$ is irrational if it is a real number which cannot be written as a fraction of two integers. For example, $2 / 5$ and $-9$ are rational, while $\\sqrt{2}$ and $\\pi$ are well known to be irrational.)",
    "options": [],
    "answer": "f(x) = x + 2004",
    "solution": "Solution:\nThe polynomial $x+2004$ certainly meets the two conditions. In fact, this is the only one. We will prove this using three ingredients: the infinitude of primes, the Rational Roots Theorem for polynomials, and the approximation principle that $x^{n}$ dominates any polynomial of lower degree, for large enough $x$.\n\nNote that the only monic constant polynomial is $f(x)=1$, which fails Condition 1; and that the only monic degree 1 polynomial satisfying Condition 1 is $f(x)=x+2004$. Thus, we need to eliminate all polynomials of degree 2 or more. To this end, it is sufficient to show that, given any monic polynomial $f(x)$ with integer coefficients of degree 2 or more, there exists an integer $a$ such that $f(x)+a=0$ has an irrational solution $x$ (for then $f(x)=-a$ is rational for an irrational number $x$).\n\nLet $f(x)=x^{n}+c_{n-1} x^{n-1}+\\cdots+c_{1} x+c_{0}$ be a polynomial with integer coefficients, with $n \\geq 2$. It may be the case that $f(x)$ has no real roots, for example, if $n$ is even and the graph of $y=f(x)$ lies above the $x$-axis. But certainly, if $a$ is a sufficiently large negative integer, we can guarantee that $f(x)+a=0$ will have at least one real solution. In fact, by further making $a$ a larger negative number, we can ensure that, say, the largest of the solutions of $f(x)+a=0$ has absolute value bigger than 1: $|x|>1$.\n\nMoreover, regardless of how large a negative number $a$ needs to be, we can choose $a$ so that $c_{0}+a=-p$ where $p$ is prime. This is because there are infinitely many prime numbers. Now we can apply the Rational Roots Theorem, according to which all rational solutions $\\frac{r}{s}$ of the monic integer coefficient polynomial $f(x)+a$ must satisfy: $s$ divides the leading coefficient of $f(x)$ and $r$ divides the last (free term) of $f(x)$; in other words, $s$ divides 1 and $r$ divides $p$. Since $p$ is prime, this gives only four possible rational solutions: $x= \\pm 1, \\pm p$. Since we have ensured that $|x|>1$, we are left with $x= \\pm p$.\n\nLet $g(x)=f(x)+a$. From the well known inequalities of absolute values $|y+z| \\geq |y|-|z|$ and $|y+z| \\leq |y|+|z|$, we obtain:\n$$\n|g(x)|=\\left|x^{n}+c_{n-1} x^{n-1}+\\ldots+c_{1} x-p\\right| \\geq \\left|x^{n}\\right|-\\left|c_{n-1} x^{n-1}+\\ldots+c_{1} x-p\\right|\n$$\nand as long as $|x|>1$ and $n \\geq 2$:\n$$\n\\begin{aligned}\n\\left|c_{n-1} x^{n-1}+\\cdots+c_{1} x-p\\right| & \\leq \\left|c_{n-1}\\right||x|^{n-1}+\\cdots+\\left|c_{1}\\right||x|+p \\\\\n& \\leq \\left(\\left|c_{n-1}\\right|+\\cdots+\\left|c_{1}\\right|\\right)|x|^{n-1}+p^{n-1}\n\\end{aligned}\n$$\nIf we let $S=\\left|c_{n-1}\\right|+\\cdots+\\left|c_{1}\\right|$, we can put everything together:\n$$\n|g( \\pm p)| \\geq p^{n}-(S+1) p^{n-1}=p^{n-1}(p-(S+1))\n$$\nSince $S$ is fixed, we can choose the prime $p$ large enough so that $p>S+1$, and hence the quantity $p-(S+1)$ is positive. Therefore, $g( \\pm p) \\neq 0$.\n\nThus $g(x)$ has a real zero $x$, which cannot be rational since the only possibilities for rational zeros $\\pm p$ fail to be zeros by the above. We conclude that $x$ is an irrational root of $g(x)$, whereas $f(x)=-a$ is an integer, hence rational. This contradicts Condition 2, and eliminates all polynomials of degree 2 or more.\n\nFinally, we are left with only one possible solution: $f(x)=x+2004$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 46970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers. Prove that\n$$\n(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \\geq (a + b + c)^3.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution:** For any positive number $x$, the quantities $x^2 - 1$ and $x^3 - 1$ have the same sign. Thus, we have $0 \\le (x^3 - 1)(x^2 - 1) = x^5 - x^3 - x^2 + 1$, or\n$$\nx^5 - x^2 + 3 \\geq x^3 + 2.\n$$\nIt follows that\n$$\n(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \\geq (a^3 + 2)(b^3 + 2)(c^3 + 2).\n$$\nIt suffices to show that\n$$\n(a^3 + 2)(b^3 + 2)(c^3 + 2) \\geq (a + b + c)^3. \\quad (*)\n$$\nWe finish with three approaches.\n\n*   **First approach** Expanding both sides of inequality (*) and cancelling like terms gives\n$$\n\\begin{aligned} & a^3b^3c^3 + 3(a^3 + b^3 + c^3) + 2(a^3b^3 + b^3c^3 + c^3a^3) + 8 \\\\ & \\geq 3(a^2b + b^2a + b^2c + c^2b + c^2a + ac^2) + 6abc. \\end{aligned}\n$$\nBy the **AM-GM Inequality**, we have $a^3 + a^3 b^3 + 1 \\ge 3a^2b$. Combining similar results, the desired inequality reduces to\n$$\na^3 b^3 c^3 + a^3 + b^3 + c^3 + 1 + 1 \\geq 6abc,\n$$\nwhich is evident by the AM-GM Inequality.\n\n\n**Second Solution:** By the AM-GM Inequality,\n$$\n\\frac{a^5 + a^5 + 1 + 1 + 1}{5} \\geq a^2\n$$\nwith equality when $a = 1$. Hence the left hand side of the desired inequality is bounded by\n$$\n\\frac{(3a^5 + 12)(3b^5 + 12)(3c^5 + 12)}{125},\n$$\nso it suffices to prove that\n$$\n(3a^5 + 12)(3b^5 + 12)(3c^5 + 12) \\geq 125(a + b + c)^3.\n$$\nAfter expanding the expressions by brute force, we find that this is equivalent to the following, which we will refer to as (†):\n$$\n\\begin{aligned} & 27a^5b^5c^5 + 108 \\sum_{\\text{cyc}} a^5b^5 + 432 \\sum_{\\text{cyc}} a^5 + 12^3 \\\\ & \\geq 125 \\left( \\sum_{\\text{cyc}} a^3 + 3 \\sum_{\\text{cyc}} a^2b + 3 \\sum_{\\text{cyc}} ab^2 + 6abc \\right). \\end{aligned}\n$$\n---\nNext, we apply AM-GM seven times to derive the following family of inequalities, referred as (‡),\n(1) $a^5 + a^5 + a^5 + 1 + 1 \\ge 5a^3$, and so\n$$\n25 \\left[ 6 + \\sum_{\\text{cyc}} 3a^5 \\ge 5 \\sum_{\\text{cyc}} a^3 \\right]. \\qquad (\\ddagger)\n$$\n(2) $a^5 b^5 c^5 + 1 + 1 + 1 + 1 \\ge 5abc$, and so\n$$\n27 [a^5 b^5 c^5 + 4 \\ge 5abc]. \\qquad (\\ddagger)\n$$\n(3) $a^5 + a^5 b^5 + 1 + 1 + 1 \\ge 5a^2b$, and so\n$$\n54 \\left[ \\sum_{\\text{cyc}} a^5 + \\sum_{\\text{cyc}} a^5 b^5 + 9 \\ge \\sum_{\\text{cyc}} a^2 b \\right]. \\qquad (\\ddagger)\n$$\nLikewise, we have\n$$\n54 \\left[ \\sum_{\\text{cyc}} b^5 + \\sum_{\\text{cyc}} a^5 b^5 + 9 \\ge \\sum_{\\text{cyc}} ab^2 \\right]. \\qquad (\\ddagger)\n$$\n(4) $a^5 + b^5 + c^5 + 1 + 1 \\ge 5abc$, and so\n$$\n123 [a^5 + b^5 + c^5 + 2 \\ge 5abc]. \\qquad (\\ddagger)\n$$\n(5) $a^5 + a^5 + b^5 + 1 + 1 \\ge 5a^2b$, and so\n$$\n21 \\left[ \\sum_{\\text{cyc}} 2a^5 + \\sum_{\\text{cyc}} b^5 + 6 \\ge \\sum_{\\text{cyc}} a^2 b \\right]. \\qquad (\\ddagger)\n$$\nLikewise, we have\n$$\n21 \\left[ \\sum_{\\text{cyc}} 2b^5 + \\sum_{\\text{cyc}} a^5 + 6 \\ge \\sum_{\\text{cyc}} ab^2 \\right]. \\qquad (\\ddagger)\n$$\n---\nMiraculously, the sum of the seven inequalities marked (‡) is precisely inequality (†), which was what we wanted. The equality condition in each application of AM-GM is $a = b = c = 1$, so that is the equality condition for (†) as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = (0,0,0)$ be the origin in the three dimensional coordinate space. The *weight* of a point is the sum of the absolute values of its coordinates. A point is a *primitive lattice point* if all its coordinates are integers with their greatest common divisor equal to 1. A square $ABCD$ is called a *unbalanced primitive integer square* if it has integer side length and the points $B$ and $D$ are primitive lattice points with different weights.\nShow that there are infinitely many unbalanced primitive integer squares $AB_iC_iD_i$ such that the plane containing the squares are not parallel to each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution:** Let $(a, b, c)$ be a **Pythagorean Triple**, that is, $a, b, c$ are positive integers with $a^2 + b^2 = c^2$. The key facts are\n(i) for any positive integers $m$ and $n$ such that $m > n$, $(m^2 - n^2, 2mn, m^2 + n^2)$ is a Pythagorean triple; and\n(ii) the vector $[c-a, c-b, a+b-c]$ has integer length $2c-a-b$.\n\nTo establish (ii), we compute\n$$\n\\begin{align*}\n& (c-a)^2 + (c-b)^2 + (a+b-c)^2 - (2c-a-b)^2 \\\\\n&= 2c^2 + a^2 + b^2 - 2c(a+b) + (a+b)^2 - 2c(a+b) + c^2 \\\\\n& \\quad -4c^2 + 4c(a+b) - (a+b)^2 \\\\\n&= a^2 + b^2 - c^2 = 0.\n\\end{align*}\n$$\nPutting these two facts together shows that the vector $[2n^2, (m-n)^2, 2n(m-n)]$ has integer length. For $i = 1, 2, \\dots$, let $o_i = 2i + 1$. We set $n = 1$ and $m - n = o_i$, and define vectors\n$$\n\\overrightarrow{AB_i} = \\mathbf{u}_i = [2, 2o_i, o_i^2], \\quad \\overrightarrow{AD_i} = \\mathbf{v}_i = [2o_i, o_i^2 - 2, -2o_i],\n$$\nand\n$$\n\\mathbf{w}_i = [-o_i^2, 2o_i, -2].\n$$\nIt is not difficult to check that $2^2 + (2o_i)^2 + (o_i^2)^2 = (o_i^2 + 2)^2$,\nand that\n$$\n\\begin{align*}\n& (2o_i)^2 + (o_i^2 - 2)^2 + (-2o_i)^2 \\\\\n&= 8o_i^2 + o_i^4 - 4o_i^2 + 4 \\\\\n&= o_i^4 + 4o_i^2 + 4 = (o_i^2 + 2)^2,\n\\end{align*}\n$$\nimplying that the vectors $\\mathbf{u}_i$, $\\mathbf{v}_i$, $\\mathbf{w}_i$ have the same integer length. Because $\\gcd(2, o_i) = \\gcd(2o_i, o_i^2 - 2) = 1$, points $B_i = (2, 2o_i, o_i^2)$ and $D_i = (2o_i, o_i^2 - 2, -2o_i)$ are primitive lattice points. It is also routine to check that the weights of $B_i$ and $D_i$ are distinct, because $2 + 2o_i + o_i^2 \\neq 4o_i + o_i^2 - 2$, that is, $2 \\neq o_i$. Finally, we claim that $AB_i \\perp AD_i$, because the **dot product**\n$$\n\\begin{align*}\n\\mathbf{u}_i \\cdot \\mathbf{v}_i &= 4o_i + 2o_i(o_i^2 - 2) - 2o_i^3 \\\\\n&= 4o_i + 2o_i^3 - 4o_i - 2o_i^3 = 0.\n\\end{align*}\n$$\nIt follows that there are infinitely many unbalanced primary integer squares $AB_iC_iD_i$. (We can set $C_i = B_i + v_i$.)\n\n\n**Second Solution:** We maintain the same notation as in the first solution. For $i = 1, 2, \\dots$, let $e_i = 2i$. We set $n = 1$ and $m - n = e_i$. Then $[2n^2, (m-n)^2, 2n(m-n)] = [2, 4i^2, 4i] = 2[1, 2i^2, 2i]$. We define vectors\n$$\n\\overrightarrow{AB_i} = \\mathbf{u}_i' = [1, 2i, 2i^2], \\quad \\overrightarrow{AD_i} = \\mathbf{v}_i' = [-2i, 1 - 2i^2, 2i],\n$$\nand\n$$\n\\mathbf{w}_i' = [-2i^2, 2i, -1].\n$$\nFor each $i$, the ordered triples of vectors $(\\mathbf{u}_i, \\mathbf{v}_i, \\mathbf{w}_i)$ and $(\\mathbf{u}_i', \\mathbf{v}_i', \\mathbf{w}_i')$ are dual to each other. It is straightforward to check that pairs of vectors $(\\mathbf{u}_i', \\mathbf{v}_i')$ provides another family of infinitely many unbalanced primitive integer squares $AB_iC_iD_i$ such that the planes containing the squares are not parallel to each other.\n\n\n**Third Solution:** Let $(a, b, c)$ be a primitive Pythagorean triple, that is, $a, b, c$ are positive integers with $a < b < c$, $a^2 + b^2 = c^2$ and $\\text{gcd}(a, b, c) = 1$. It is not difficult to show that $\\text{gcd}(a, b) = \\text{gcd}(c, b) = \\text{gcd}(c, a) = 1$. In order to find a vector with integer length, we consider $c^4 = c^2(a^2 + b^2) = c^2a^2 + c^2b^2 = (a^2 + b^2)a^2 + c^2b^2 = c^2a^2 + (a^2 + b^2)b^2$. Set\n$$\n\\overrightarrow{AB} = \\mathbf{u} = [-ab, b^2, ac] \\quad \\text{and} \\quad \\overrightarrow{AD} = \\mathbf{v} = [a^2, -ab, bc].\n$$\nHence $C = (a^2 - ab, b^2 - ab, c(a+b))$.\nBy construction, $AB^2 = AD^2 = (a^2 + b^2)c^2 = c^4$ and\n$$\n\\overrightarrow{AB} \\cdot \\overrightarrow{AD} = \\mathbf{u} \\cdot \\mathbf{v} = ab(-a^2 - b^2 + c^2) = 0.\n$$\nWe also note that the difference of the weights of $B$ and $D$ is equal to\n$$\n\\begin{aligned} & (b^2 + ab + ac) - (a^2 + ab + bc) \\\\ &= b^2 - a^2 + ac - bc \\\\ &= (b-a)(a+b) + c(a-b) \\\\ &= (b-a)(a+b-c), \\end{aligned}\n$$\nwhich is nonzero because $\\gcd(a, b) = 1$ and $a, b$, and $c$ are sides of a triangle. Because $\\gcd(a, b) = \\gcd(c, b) = \\gcd(c, a) = 1$, $\\gcd(b^2, ac) = \\gcd(a^2, bc) = 1$. Therefore, $ABCD$ is a unbalanced primitive integer square (associated to this primitive Pythagorean triple).\nWe note that vector $\\mathbf{w} = [b, a, 0]$ is perpendicular to the plane containing the square $ABCD$. Indeed,\n$$\n\\mathbf{u} \\cdot \\mathbf{w} = -ab^2 + ab^2 = 0 = a^2b - a^2b = \\mathbf{v} \\cdot \\mathbf{w}.\n$$\nFinally, it is not difficult to see that for distinct primitive Pythagorean triples $(a, b, c)$ and $(a', b', c')$, the vectors $\\mathbf{w} = [b, a, 0]$ and $\\mathbf{w}' = [b', a', 0]$ are not parallel to each other. Hence, the unbalanced primitive integer squares associated with distinct primitive Pythagorean triples lie in different planes. It is well known that there are infinitely many primitive Pythagorean triples, so our proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA mouse is sitting in a toy car on a negligibly small turntable. The car cannot turn on its own, but the mouse can control when the car is launched and when the car stops (the car has brakes). When the mouse chooses to launch, the car will immediately leave the turntable on a straight trajectory at $1$ meter per second.\n\nSuddenly someone turns on the turntable; it spins at $30$ rpm. Consider the set $S$ of points the mouse can reach in his car within $1$ second after the turntable is set in motion. (For example, the arrows in the figure below represent two possible paths the mouse can take.) What is the area of $S$, in square meters?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "π/6",
    "solution": "Solution:\nThe mouse can wait while the table rotates through some angle $\\theta$ and then spend the remainder of the time moving along that ray at $1~\\mathrm{m}/\\mathrm{s}$. He can reach any point between the starting point and the furthest reachable point along the ray, $(1-\\theta/\\pi)$ meters out. So the area is given by the polar integral\n$$\n\\int_{0}^{\\pi} \\frac{(1-\\theta/\\pi)^{2}}{2} d\\theta = \\frac{1}{2} \\cdot \\frac{1}{\\pi^{2}} \\int_{0}^{\\pi} \\phi^{2} d\\phi = \\pi/6\n$$\n(where we have used the change of variables $\\phi = \\pi - \\theta$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEight strangers are preparing to play bridge. How many ways can they be grouped into two bridge games - that is, into unordered pairs of unordered pairs of people?",
    "options": [],
    "answer": "315",
    "solution": "Solution:\n315\nPutting 8 people into 4 pairs and putting those 4 pairs into 2 pairs of pairs are independent. If the people are numbered from 1 to 8, there are 7 ways to choose the person to pair with person 1. Then there are 5 ways to choose the person to pair with the person who has the lowest remaining number, 3 ways to choose the next, and 1 way to choose the last (because there are only 2 people remaining). Thus, there are $7 \\cdot 5 \\cdot 3 \\cdot 1$ ways to assign 8 people to pairs and similarly there are $3 \\cdot 1$ ways to assign 4 pairs to 2 pairs of pairs, so there are $7 \\cdot 5 \\cdot 3 \\cdot 3=315$ ways.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAugustin has six $1 \\times 2 \\times \\pi$ bricks. He stacks them, one on top of another, to form a tower six bricks high. Each brick can be in any orientation so long as it rests flat on top of the next brick below it (or on the floor). How many distinct heights of towers can he make?",
    "options": [],
    "answer": "28",
    "solution": "Solution:\n\nIf there are $k$ bricks which are placed so that they contribute either $1$ or $2$ height, then the height of these $k$ bricks can be any integer from $k$ to $2k$. Furthermore, towers with different values of $k$ cannot have the same height. Thus, for each $k$ there are $k+1$ possible tower heights, and since $k$ is any integer from $0$ to $6$, there are $1+2+3+4+5+6+7=28$ possible heights.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the area of the region of the $x y$-plane defined by the inequality $|x|+|y|+|x+y| \\leq 1$.",
    "options": [],
    "answer": "3/4",
    "solution": "Solution:\nTo graph this region we divide the $x y$-plane into six sectors depending on which of $x, y, x+y$ are $\\geq 0$, or $\\leq 0$. The inequality simplifies in each case:\n\n| Sector | Inequality | Simplified inequality |\n| :---: | :---: | :---: |\n| $x \\geq 0, y \\geq 0, x+y \\geq 0$ | $x+y+(x+y) \\leq 1$ | $x+y \\leq 1 / 2$ |\n| $x \\geq 0, y \\leq 0, x+y \\geq 0$ | $x-y+(x+y) \\leq 1$ | $x \\leq 1 / 2$ |\n| $x \\geq 0, y \\leq 0, x+y \\leq 0$ | $x-y-(x+y) \\leq 1$ | $y \\geq-1 / 2$ |\n| $x \\leq 0, y \\geq 0, x+y \\geq 0$ | $-x+y+(x+y) \\leq 1$ | $y \\leq 1 / 2$ |\n| $x \\leq 0, y \\geq 0, x+y \\leq 0$ | $-x+y-(x+y) \\leq 1$ | $x \\geq-1 / 2$ |\n| $x \\leq 0, y \\leq 0, x+y \\leq 0$ | $-x-y-(x+y) \\leq 1$ | $x+y \\geq-1 / 2$ |\n\nWe then draw the region; we get a hexagon as shown. The hexagon intersects each region in an isosceles right triangle of area $1 / 8$, so the total area is $6 \\cdot 1 / 8=3 / 4$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA pair of successive numbers in the same row is called a switch pair if one number in the pair is even and the other is odd.\nProve that the number of switch pairs in row $n$ is at most twice the number of odd numbers in row $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEach switch pair contains an odd number, and each odd number can belong to at most two switch pairs (since it has only two neighbors).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA rectangle has perimeter $10$ and diagonal $\\sqrt{15}$. What is its area?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nIf the sides are $x$ and $y$, we have $2x + 2y = 10$, so $x + y = 5$, and $\\sqrt{x^{2} + y^{2}} = \\sqrt{15}$, so $x^{2} + y^{2} = 15$.\n\nSquaring the first equation gives $x^{2} + 2xy + y^{2} = 25$, and subtracting the second equation gives $2xy = 10$, so the area is $xy = 5$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThere exists a positive real number $x$ such that $\\cos \\left(\\tan^{-1}(x)\\right)=x$. Find the value of $x^{2}$.",
    "options": [],
    "answer": "(-1 + sqrt(5)) / 2",
    "solution": "Solution:\nDraw a right triangle with legs $1$, $x$; then the angle $\\theta$ opposite $x$ is $\\tan^{-1} x$, and we can compute $\\cos (\\theta)=1 / \\sqrt{x^{2}+1}$. Thus, we only need to solve $x=1 / \\sqrt{x^{2}+1}$. This is equivalent to $x \\sqrt{x^{2}+1}=1$. Square both sides to get $x^{4}+x^{2}=1 \\Rightarrow x^{4}+x^{2}-1=0$. Use the quadratic formula to get the solution $x^{2}=(-1+\\sqrt{5}) / 2$ (unique since $x^{2}$ must be positive).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA best-of-9 series is to be played between two teams; that is, the first team to win $5$ games is the winner. The Mathletes have a chance of $2/3$ of winning any given game. What is the probability that exactly $7$ games will need to be played to determine a winner?",
    "options": [],
    "answer": "20/81",
    "solution": "Solution:\nIf the Mathletes are to win, they must win exactly $5$ out of the $7$ games. One of the $5$ games they win must be the $7$th game, because otherwise they would win the tournament before $7$ games are completed. Thus, in the first $6$ games, the Mathletes must win $4$ games and lose $2$. The probability of this happening and the Mathletes winning the last game is\n$$\n\\left[\\binom{6}{2} \\cdot\\left(\\frac{2}{3}\\right)^{4} \\cdot\\left(\\frac{1}{3}\\right)^{2}\\right] \\cdot\\left(\\frac{2}{3}\\right) .\n$$\nLikewise, the probability of the other team winning on the $7$th game is\n$$\n\\left[\\binom{6}{2} \\cdot\\left(\\frac{1}{3}\\right)^{4} \\cdot\\left(\\frac{2}{3}\\right)^{2}\\right] \\cdot\\left(\\frac{1}{3}\\right)\n$$\nSumming these values, we obtain $160/729 + 20/729 = 20/81$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N}_0^+$ and $\\mathbb{Q}$ be the set of nonnegative integers and rational numbers, respectively. Define the function $f : \\mathbb{N}_0^+ \\to \\mathbb{Q}$ by $f(0) = 0$ and\n$$f(3n + k) = -\\frac{3f(n)}{2} + k, \\quad \\text{for } k = 0, 1, 2.$$ \nProve that $f$ is one-to-one, and determine its range.",
    "options": [],
    "answer": "Range(f) = { m/2^n : m ∈ ℤ, n ∈ ℕ_0^+ } (the dyadic rationals). Also, f is injective.",
    "solution": "We prove that the range of $f$ is the set $T$ of rational numbers of the form $m/2^n$ for $m \\in \\mathbb{Z}$ and $n \\in \\mathbb{N}_0^+$ (also known as the *dyadic rational numbers*). For $x, y \\in T$, we write $x \\equiv y \\pmod 3$ to mean that the numerator of $x - y$, when written in lowest terms, is divisible by 3. Define the function $g : T \\to \\{0, 1, 2\\}$ by declaring that for $x \\in T$, $g(x)$ is the unique element of $\\{0, 1, 2\\}$ such that $g(x) \\equiv x \\pmod 3$.\n\nWe first check that $f$ is one-to-one. Suppose that $f(a) = f(b)$ for some $a, b \\in \\mathbb{N}_0^+$ with $a < b$; choose such a pair with $b$ minimal. Now note that $f(a) \\equiv a \\pmod 3$ and $f(b) \\equiv b \\pmod 3$ from the definition of $f$. Hence $a \\equiv b \\pmod 3$. Choose the $i \\in \\{0, 1, 2\\}$ that is congruent to $a$ and $b$ modulo 3, and put $a' = (a-i)/3$ and $b' = (b-i)/3$. Then $f(a') = f(b')$, but $a' < b'$ and $b' < b$ since $b > 0$. This contradicts the choice of $a$ and $b$. Hence no such pairs exist.\n\nWe next verify that $T$, which clearly contains the range of $f$, is in fact equal to it. Define the map $h : T \\to T$ by setting $h(x) = \\frac{2}{3}(g(x)-x)$. Then $x$ is in the image of $f$ whenever $h(x)$ is: if $h(x) = f(a)$, then $x = f(3a + g(x))$. Hence it suffices to show that if one starts from $x$ and applies $h$ repeatedly, one eventually ends up with an element of the range of $f$.\n\nFirst note that if $x$ is not an integer, then $g(x) - x$ has the same denominator as $x$, and $h(x)$ has denominator half of that. Hence some $x_i$ is an integer.\n\nNext, by the **Triangle Inequality**,\n$$\n|h(x)| \\leq \\frac{2}{3}(2 + |x|) < |x|\n$$\nwhenever $|x| > 4$. Thus, our process eventually hits an integer of absolute value bounded by 4. Simple computation yields the following table:\n\n| x   | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |\n|-----|----|----|----|----|---|---|---|---|---|\n| h(x)|  4 |  2 |  2 |  2 | 0 | 0 | 0 | -2| -2|\n\n¿From the table, it is clear that starting from any point in $\\{-4, -3, \\ldots, 4\\}$, within 4 applications of $h$, our process hits zero. Since $f(0) = 0$, this value is in the range of $f$, so we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na and b are positive integers. When written in binary, $a$ has 2004 1's, and $b$ has 2005 1's (not necessarily consecutive). What is the smallest number of 1's $a+b$ could possibly have?",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nConsider the following addition:\n$$\n\\begin{array}{r}\n111\\cdots 100\\cdots 01 \\\\\n+ \\\\\n11\\cdots 11\n\\end{array}\n$$\nBy making the blocks of 1's and 0's appropriately long, we can ensure that the addends respectively contain 2004 and 2005 1's. (To be precise, we get $a = 2^{4008} - 2^{2005} + 1$ and $b = 2^{2005} - 1$.) Then the sum has only one 1. And clearly it is not possible to get any less than one 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the ordered quadruple of digits $(A, B, C, D)$, with $A > B > C > D$, such that\n$$\n\\begin{aligned}\n& ABCD \\\\\n- & DCBA \\\\\n= & BDAC .\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(7, 6, 4, 1)",
    "solution": "Solution:\nSince $D < A$, when $A$ is subtracted from $D$ in the ones' column, there will be a borrow from $C$ in the tens' column. Thus, $D + 10 - A = C$.\n\nNext, consider the subtraction in the tens' column, $(C - 1) - B$. Since $C < B$, there will be a borrow from the hundreds' column, so $(C - 1 + 10) - B = A$.\n\nIn the hundreds' column, $B - 1 \\geq C$, so we do not need to borrow from $A$ in the thousands' column. Thus, $(B - 1) - C = D$ and $A - D = B$.\n\nWe thus have a system of four equations in four variables $A, B, C, D$, and solving by standard methods (e.g. substitution) produces $(A, B, C, D) = (7, 6, 4, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn trapezoid $ABCD$ shown, $AD$ is parallel to $BC$, and $AB = 6$, $BC = 7$, $CD = 8$, $AD = 17$. If sides $AB$ and $CD$ are extended to meet at $E$, find the resulting angle at $E$ (in degrees).\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "90",
    "solution": "Solution:\n\nChoose point $F$ on $AD$ so that $BCDF$ is a parallelogram. Then $BF = CD = 8$, and $AF = AD - DF = AD - BC = 10$, so $\\triangle ABF$ is a $6$-$8$-$10$ right triangle. The required angle is equal to $\\angle ABF = 90^{\\circ}$.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA committee of 5 is to be chosen from a group of 9 people. How many ways can it be chosen, if Bill and Karl must serve together or not at all, and Alice and Jane refuse to serve with each other?",
    "options": [],
    "answer": "41",
    "solution": "Solution:\nIf Bill and Karl are on the committee, there are $\\binom{7}{3} = 35$ ways for the other group members to be chosen. However, if Alice and Jane are on the committee with Bill and Karl, there are $\\binom{5}{1} = 5$ ways for the last member to be chosen, yielding 5 unacceptable committees.\n\nIf Bill and Karl are not on the committee, there are $\\binom{7}{5} = 21$ ways for the 5 members to be chosen, but again if Alice and Jane were to be on the committee, there would be $\\binom{5}{3} = 10$ ways to choose the other three members, yielding 10 more unacceptable committees.\n\nSo, we obtain $(35 - 5) + (21 - 10) = 41$ ways the committee can be chosen.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 46985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest integer $n$ such that $\\sqrt{n+99}-\\sqrt{n}<1$.",
    "options": [],
    "answer": "2402",
    "solution": "Solution:\nThis is equivalent to\n$$\n\\begin{aligned}\n\\sqrt{n+99} &< \\sqrt{n} + 1 \\\\\nn+99 &< n + 1 + 2\\sqrt{n} \\\\\n99 &< 1 + 2\\sqrt{n} \\\\\n98 &< 2\\sqrt{n} \\\\\n49 &< \\sqrt{n}\n\\end{aligned}\n$$\nSo the smallest integer $n$ with this property is $49^{2} + 1 = 2402$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor $x > 0$, let $f(x) = x^{x}$. Find all values of $x$ for which $f(x) = f'(x)$.",
    "options": [],
    "answer": "x = 1",
    "solution": "Solution:\n\nLet $g(x) = \\log f(x) = x \\log x$. Then $f'(x)/f(x) = g'(x) = 1 + \\log x$. Therefore $f(x) = f'(x)$ when $1 + \\log x = 1$, that is, when $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real solutions to $x^{4} + (2 - x)^{4} = 34$.",
    "options": [],
    "answer": "x = 1 ± sqrt(2)",
    "solution": "Solution:\n$1 \\pm \\sqrt{2}$\nLet $y = 2 - x$, so $x + y = 2$ and $x^{4} + y^{4} = 34$. We know\n$$\n(x + y)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4x y^{3} + y^{4} = x^{4} + y^{4} + 2 x y (2 x^{2} + 2 y^{2} + 3 x y)\n$$\nMoreover, $x^{2} + y^{2} = (x + y)^{2} - 2 x y$, so the preceding equation becomes $2^{4} = 34 + 2 x y (2 \\cdot 2^{2} - x y )$, or $(x y)^{2} - 8 x y - 9 = 0$. Hence $x y = 9$ or $-1$. Solving $x y = 9$, $x + y = 2$ produces complex solutions, and solving $x y = -1$, $x + y = 2$ produces $(x, y) = (1 + \\sqrt{2}, 1 - \\sqrt{2})$ or $(1 - \\sqrt{2}, 1 + \\sqrt{2})$. Thus, $x = 1 \\pm \\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x$, $y$, $k$ are positive reals such that\n$$\n3 = k^{2}\\left(\\frac{x^{2}}{y^{2}} + \\frac{y^{2}}{x^{2}}\\right) + k\\left(\\frac{x}{y} + \\frac{y}{x}\\right),\n$$\nfind the maximum possible value of $k$.",
    "options": [],
    "answer": "(sqrt(7) - 1)/2",
    "solution": "Solution:\n$$\n\\frac{-1 + \\sqrt{7}}{2}\n$$\nWe have $3 = k^{2}\\left(\\frac{x^{2}}{y^{2}} + \\frac{y^{2}}{x^{2}}\\right) + k\\left(\\frac{x}{y} + \\frac{y}{x}\\right) \\geq 2k^{2} + 2k$, hence $7 \\geq 4k^{2} + 4k + 1 = (2k + 1)^{2}$, hence $k \\leq (\\sqrt{7} - 1)/2$. Obviously $k$ can assume this value, if we let $x = y = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the shortest distance from the line $3x + 4y = 25$ to the circle $x^{2} + y^{2} = 6x - 8y$.",
    "options": [],
    "answer": "7/5",
    "solution": "Solution:\nThe circle is $(x-3)^{2} + (y+4)^{2} = 5^{2}$. The center $(3, -4)$ is a distance of\n$$\n\\frac{|3 \\cdot 3 + 4 \\cdot (-4) - 25|}{\\sqrt{3^{2} + 4^{2}}} = \\frac{32}{5}\n$$\nfrom the line, so we subtract $5$ for the radius of the circle and get $\\frac{7}{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A C E$ be a triangle with a point $B$ on segment $A C$ and a point $D$ on segment $C E$ such that $B D$ is parallel to $A E$. A point $Y$ is chosen on segment $A E$, and segment $C Y$ is drawn. Let $X$ be the intersection of $C Y$ and $B D$. If $C X=5, X Y=3$, what is the ratio of the area of trapezoid $A B D E$ to the area of triangle $B C D$?",
    "options": [],
    "answer": "39/25",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nDraw the altitude from $C$ to $A E$, intersecting line $B D$ at $K$ and line $A E$ at $L$. Then $C K$ is the altitude of triangle $B C D$, so triangles $C K X$ and $C L Y$ are similar. Since $C Y / C X=8 / 5, C L / C K=8 / 5$. Also triangles $C K B$ and $C L A$ are similar, so that $C A / C B=8 / 5$, and triangles $B C D$ and $A C E$ are similar, so that $A E / B D=8 / 5$. The area of $A C E$ is $(1 / 2)(A E)(C L)$, and the area of $B C D$ is $(1 / 2)(B D)(C K)$, so the\nratio of the area of $A C E$ to the area of $B C D$ is $64 / 25$. Therefore, the ratio of the area of $A B D E$ to the area of $B C D$ is $39 / 25$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFarmer John is grazing his cows at the origin. There is a river that runs east to west 50 feet north of the origin. The barn is 100 feet to the south and 80 feet to the east of the origin. Farmer John leads his cows to the river to take a swim, then the cows leave the river from the same place they entered and Farmer John leads them to the barn. He does this using the shortest path possible, and the total distance he travels is $d$ feet. Find the value of $d$.",
    "options": [],
    "answer": "40*sqrt(29)",
    "solution": "Solution:\nSuppose we move the barn to its reflection across the river's edge. Then paths from the origin to the river and then to the old barn location correspond to paths from the origin to the river and then to the new barn location, by reflecting the second part of the path across the river, and corresponding paths have the same length. Now the shortest path from the origin to the river and then to the new barn location is a straight line. The new barn location is 200 feet north and 80 feet east of the origin, so the value of $d$ is $\\sqrt{200^{2}+80^{2}}=40 \\sqrt{29}$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYet another trapezoid $ABCD$ has $AD$ parallel to $BC$. $AC$ and $BD$ intersect at $P$. If $[ADP]/[BCP] = 1/2$, find $[ADP]/[ABCD]$. (Here the notation $[P_1 \\cdots P_n]$ denotes the area of the polygon $P_1 \\cdots P_n$.)",
    "options": [],
    "answer": "3 - 2√2",
    "solution": "Solution:\n\nA homothety (scaling) about $P$ takes triangle $ADP$ into $BCP$, since $AD$, $BC$ are parallel and $A, P, C; B, P, D$ are collinear. The ratio of homothety is thus $\\sqrt{2}$. It follows that, if we rescale to put $[ADP]=1$, then $[ABP]=[CDP]=\\sqrt{2}$, just by the ratios of lengths of bases. So $[ABCD]=3+2\\sqrt{2}$, so $[ADP]/[ABCD]=1/(3+2\\sqrt{2})$. Simplifying this, we get $3-2\\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWe have a polyhedron such that an ant can walk from one vertex to another, traveling only along edges, and traversing every edge exactly once. What is the smallest possible total number of vertices, edges, and faces of this polyhedron?",
    "options": [],
    "answer": "20",
    "solution": "Solution:\nThis is obtainable by construction. Consider two tetrahedrons glued along a face; this gives us 5 vertices, 9 edges, and 6 faces, for a total of 20, and one readily checks that the required Eulerian path exists.\n\nNow, to see that we cannot do better, first notice that the number $v$ of vertices is at least 5, since otherwise we must have a tetrahedron, which does not have an Eulerian path. Each vertex is incident to at least 3 edges, and in fact, since there is an Eulerian path, all except possibly two vertices are incident to an even number of edges. So the number of edges is at least $(3+3+4+4+4)/2$ (since each edge meets two vertices) $=9$.\n\nFinally, if $f=4$ then each face must be a triangle, because there are only 3 other faces for it to share edges with, and we are again in the case of a tetrahedron, which is impossible; therefore $f \\geq 5$. So $f+v+e \\geq 5+5+9=19$. But since $f+v-e=2-2g$ (where $g$ is the number of holes in the polyhedron), $f+v+e$ must be even. This strengthens our bound to 20 as needed.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 46994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x$ and $y$ are real numbers with $(x+y)^{4} = x - y$, what is the maximum possible value of $y$?",
    "options": [],
    "answer": "3 * cube_root(2) / 16",
    "solution": "Solution:\n$$\n3 \\sqrt[3]{2} / 16\n$$\nBy drawing the graph of the curve (as shown), which is just a $135^{\\circ}$ clockwise rotation and scaling of $y = x^{4}$, we see that the maximum is achieved at the unique point where $d y / d x = 0$. Implicit differentiation gives $4(d y / d x + 1)(x + y)^{3} = 1 - d y / d x$, so setting $d y / d x = 0$ gives $4(x + y)^{3} = 1$. So $x + y = 1 / \\sqrt[3]{4} = \\sqrt[3]{2} / 2$, and $x - y = (x + y)^{4} = \\sqrt[3]{2} / 8$. Subtracting and dividing by 2 gives $y = (\\sqrt[3]{2} / 2 - \\sqrt[3]{2} / 8) / 2 = 3 \\sqrt[3]{2} / 16$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA triangle has side lengths $18$, $24$, and $30$. Find the area of the triangle whose vertices are the incenter, circumcenter, and centroid of the original triangle.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nThere are many solutions to this problem, which is straightforward. The given triangle is a right $3$-$4$-$5$ triangle, so the circumcenter is the midpoint of the hypotenuse. Coordinatizing for convenience, put the vertex at $(0,0)$ and the other vertices at $(0,18)$ and $(24,0)$. Then the circumcenter is $(12,9)$. The centroid is at one-third the sum of the three vertices, which is $(8,6)$. Finally, since the area equals the inradius times half the perimeter, we can see that the inradius is $(18 \\cdot 24 / 2) /([18+24+30] / 2)=6$. So the incenter of the triangle is $(6,6)$. So the small triangle has a base of length $2$ and a height of $3$, hence its area is $3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 46996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou have a $10 \\times 10$ grid of squares. You write a number in each square as follows: you write $1,2,3, \\ldots, 10$ from left to right across the top row, then $11,12, \\ldots, 20$ across the second row, and so on, ending with $100$ in the bottom right square. You then write a second number in each square, writing $1,2, \\ldots, 10$ in the first column (from top to bottom), then $11,12, \\ldots, 20$ in the second column, and so forth.\nWhen this process is finished, how many squares will have the property that their two numbers sum to $101$?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nThe number in the $i$th row, $j$th column will receive the numbers $10(i-1)+j$ and $10(j-1)+i$, so the question is how many pairs $(i, j)$ ($1 \\leq i, j \\leq 10$) will have\n$$\n101 = [10(i-1)+j] + [10(j-1)+i] \\quad \\Leftrightarrow \\quad 121 = 11i + 11j = 11(i+j).\n$$\nNow it is clear that this is achieved by the ten pairs $(1,10), (2,9), (3,8), \\ldots, (10,1)$ and no others.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA freight train leaves the town of Jenkinsville at 1:00 PM traveling due east at constant speed. Jim, a hobo, sneaks onto the train and falls asleep. At the same time, Julie leaves Jenkinsville on her bicycle, traveling along a straight road in a northeasterly direction (but not due northeast) at 10 miles per hour. At 1:12 PM, Jim rolls over in his sleep and falls from the train onto the side of the tracks. He wakes up and immediately begins walking at 3.5 miles per hour directly towards the road on which Julie is riding. Jim reaches the road at 2:12 PM, just as Julie is riding by. What is the speed of the train in miles per hour?",
    "options": [],
    "answer": "62.5",
    "solution": "Solution:\nJulie's distance is $(10\\ \\mathrm{mph}) \\cdot (6/5\\ \\mathrm{hrs}) = 12$ miles. Jim's walking distance, after falling off the train, is $(3.5\\ \\mathrm{mph}) \\cdot (1\\ \\mathrm{hr}) = 3.5$ miles at a right angle to the road. Therefore, Jim rode the train for $\\sqrt{12^{2} + 3.5^{2}} = \\frac{1}{2} \\sqrt{24^{2} + 7^{2}} = 25/2$ miles, and its speed is $(25/2\\ \\mathrm{mi}) / (1/5\\ \\mathrm{hr}) = 62.5\\ \\mathrm{mph}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nI have chosen five of the numbers $\\{1,2,3,4,5,6,7\\}$. If I told you what their product was, that would not be enough information for you to figure out whether their sum was even or odd. What is their product?",
    "options": [],
    "answer": "420",
    "solution": "Solution: 420\nGiving you the product of the five numbers is equivalent to telling you the product of the two numbers I didn't choose. The only possible products that are achieved by more than one pair of numbers are $12(\\{3,4\\}$ and $\\{2,6\\})$ and $6(\\{1,6\\}$ and $\\{2,3\\})$. But in the second case, you at least know that the two unchosen numbers have odd sum (and so the five chosen numbers have odd sum also). Therefore, the first case must hold, and the product of the five chosen numbers is\n$$\n1 \\cdot 2 \\cdot 5 \\cdot 6 \\cdot 7=1 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 7=420\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 46999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $S$ is a finite set with a binary operation $\\star$ - that is, for any elements $a, b$ of $S$, there is defined an element $a \\star b$ of $S$. It is given that $(a \\star b) \\star(a \\star b)=b \\star a$ for all $a, b \\in S$.\n\nProve that $a \\star b=b \\star a$ for all $a, b \\in S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe have\n$$\n\\begin{aligned}\na \\star b & =(b \\star a) \\star(b \\star a) \\\\\n& =([a \\star b] \\star[a \\star b]) \\star([a \\star b] \\star[a \\star b]) \\\\\n& =[a \\star b] \\star[a \\star b] \\\\\n& =b \\star a .\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$ be a real number such that $x^{3} + 4x = 8$. Determine the value of $x^{7} + 64x^{2}$.",
    "options": [],
    "answer": "128",
    "solution": "Solution:\nFor any integer $n \\geq 0$, the given implies $x^{n+3} = -4x^{n+1} + 8x^{n}$, so we can rewrite any such power of $x$ in terms of lower powers. Carrying out this process iteratively gives\n$$\n\\begin{aligned}\nx^{7} & = -4x^{5} + 8x^{4} \\\\\n& = 8x^{4} + 16x^{3} - 32x^{2} \\\\\n& = 16x^{3} - 64x^{2} + 64x \\\\\n& = -64x^{2} + 128.\n\\end{aligned}\n$$\nThus, our answer is $128$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA positive integer $n$ is picante if $n!$ ends in the same number of zeroes whether written in base $7$ or in base $8$. How many of the numbers $1, 2, \\ldots, 2004$ are picante?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nThe number of zeroes in base $7$ is the total number of factors of $7$ in $1 \\cdot 2 \\cdots n$, which is\n$$\n\\left\\lfloor \\frac{n}{7} \\right\\rfloor + \\left\\lfloor \\frac{n}{7^2} \\right\\rfloor + \\left\\lfloor \\frac{n}{7^3} \\right\\rfloor + \\cdots.\n$$\nThe number of zeroes in base $8$ is $\\lfloor a \\rfloor$, where\n$$\na = \\left( \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{2^2} \\right\\rfloor + \\left\\lfloor \\frac{n}{2^3} \\right\\rfloor + \\cdots \\right) / 3\n$$\nis one-third the number of factors of $2$ in the product $n!$. Now $\\left\\lfloor \\frac{n}{2^k} \\right\\rfloor / 3 \\geq \\left\\lfloor \\frac{n}{7^k} \\right\\rfloor$ for all $k$, since $\\left( n / 2^k \\right) / 3 \\geq n / 7^k$. But $n$ can only be picante if the two sums differ by at most $2/3$, so in particular this requires $\\left( \\left\\lfloor n / 2^2 \\right\\rfloor \\right) / 3 \\leq \\left\\lfloor n / 7^2 \\right\\rfloor + 2/3 \\Leftrightarrow \\left\\lfloor n / 4 \\right\\rfloor \\leq 3 \\left\\lfloor n / 49 \\right\\rfloor + 2$. This cannot happen for $n \\geq 12$; checking the remaining few cases by hand, we find $n = 1, 2, 3, 7$ are picante, for a total of $4$ values.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA classroom consists of a $5 \\times 5$ array of desks, to be filled by anywhere from $0$ to $25$ students, inclusive. No student will sit at a desk unless either all other desks in its row or all others in its column are filled (or both). Considering only the set of desks that are occupied (and not which student sits at each desk), how many possible arrangements are there?",
    "options": [],
    "answer": "962",
    "solution": "Solution:\nThe set of empty desks must be of the form (non-full rows) $\\times$ (non-full columns): each empty desk is in a non-full column and a non-full row, and the given condition implies that each desk in such a position is empty. So if there are fewer than $25$ students, then both of these sets are nonempty; we have $2^{5}-1=31$ possible sets of non-full rows, and $31$ sets of non-full columns, for $961$ possible arrangements. Alternatively, there may be $25$ students, and then only $1$ arrangement is possible. Thus there are $962$ possibilities altogether.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUrn $A$ contains $4$ white balls and $2$ red balls. Urn $B$ contains $3$ red balls and $3$ black balls. An urn is randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability that the first ball drawn was red, given that the second ball drawn was black?",
    "options": [],
    "answer": "7/15",
    "solution": "Solution:\n\nThis is a case of conditional probability; the answer is the probability that the first ball is red and the second ball is black, divided by the probability that the second ball is black.\n\nFirst, we compute the numerator. If the first ball is drawn from Urn $A$, we have a probability of $2/6$ of getting a red ball, then a probability of $1/2$ of drawing the second ball from Urn $B$, and a further probability of $3/6$ of drawing a black ball. If the first ball is drawn from Urn $B$, we have probability $3/6$ of getting a red ball, then $1/2$ of drawing the second ball from Urn $B$, and $3/5$ of getting a black ball. So our numerator is\n$$\n\\frac{1}{2}\\left(\\frac{2}{6} \\cdot \\frac{1}{2} \\cdot \\frac{3}{6}+\\frac{3}{6} \\cdot \\frac{1}{2} \\cdot \\frac{3}{5}\\right)=\\frac{7}{60} .\n$$\n\nWe similarly compute the denominator: if the first ball is drawn from Urn $A$, we have a probability of $1/2$ of drawing the second ball from Urn $B$, and $3/6$ of drawing a black ball. If the first ball is drawn from Urn $B$, then we have probability $3/6$ that it is red, in which case the second ball will be black with probability $(1/2) \\cdot (3/5)$, and probability $3/6$ that the first ball is black, in which case the second is black with probability $(1/2) \\cdot (2/5)$. So overall, our denominator is\n$$\n\\frac{1}{2}\\left(\\frac{1}{2} \\cdot \\frac{3}{6}+\\frac{3}{6}\\left[\\frac{1}{2} \\cdot \\frac{3}{5}+\\frac{1}{2} \\cdot \\frac{2}{5}\\right]\\right)=\\frac{1}{4}\n$$\n\nThus, the desired conditional probability is $(7/60)/(1/4)=7/15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence of positive integers is defined by $a_{0}=1$ and $a_{n+1}=a_{n}^{2}+1$ for each $n \\geq 0$. Find $\\operatorname{gcd}\\left(a_{999}, a_{2004}\\right)$.",
    "options": [],
    "answer": "677",
    "solution": "Solution:\nIf $d$ is the relevant greatest common divisor, then $a_{1000}=a_{999}^{2}+1 \\equiv 1=a_{0} \\pmod{d}$, which implies (by induction) that the sequence is periodic modulo $d$, with period $1000$. In particular, $a_{4} \\equiv a_{2004} \\equiv 0$. So $d$ must divide $a_{4}$. Conversely, we can see that $a_{5}=a_{4}^{2}+1 \\equiv 1=a_{0}$ modulo $a_{4}$, so (again by induction) the sequence is periodic modulo $a_{4}$ with period $5$, and hence $a_{999}, a_{2004}$ are indeed both divisible by $a_{4}$. So the answer is $a_{4}$, which we can compute directly; it is $677$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest positive integer $n$ such that $1+2+3+\\cdots+n^{2}$ is divisible by $1+2+3+\\cdots+n$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nThe statement is\n$$\n\\frac{n(n+1)}{2} \\mid \\frac{n^{2}(n^{2}+1)}{2} \\Leftrightarrow n+1 \\mid n(n^{2}+1) = n^{3}+n.\n$$\nBut $n+1$ also divides $(n+1)(n^{2}-n+2) = n^{3}+n+2$, so $n+1$ must divide $2$. Hence, $n$ cannot be greater than $1$. And $n=1$ clearly works, so that is the answer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for any two types of dominoes, there exists a rectangle that can be tiled by dominoes of either type.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNote that a type $(a, b)$ domino tiles a $\\max\\{1, 2a\\} \\times 2b$ rectangle (see diagram for $a > 0$). Then both type $(a, b)$ and type $(a', b')$ dominoes tile a $\\left(\\max\\{1, 2a\\}, \\max\\{1, 2a'\\}\\right) \\times (2b \\cdot 2b')$ rectangle.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many real numbers $x$ are solutions to the following equation?\n$$\n|x-1|=|x-2|+|x-3|\n$$",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nIf $x<1$, the equation becomes $(1-x)=(2-x)+(3-x)$ which simplifies to $x=4$, contradicting the assumption $x<1$.\n\nIf $1 \\leq x \\leq 2$, we get $(x-1)=(2-x)+(3-x)$, which gives $x=2$.\n\nIf $2 \\leq x \\leq 3$, we get $(x-1)=(x-2)+(3-x)$, which again gives $x=2$.\n\nIf $x \\geq 3$, we get $(x-1)=(x-2)+(x-3)$, or $x=4$.\n\nSo $2$ and $4$ are the only solutions, and the answer is $2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = x^{3} + a x + b$, with $a \\neq b$, and suppose the tangent lines to the graph of $f$ at $x = a$ and $x = b$ are parallel. Find $f(1)$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nSince $f'(x) = 3x^{2} + a$, we must have $3a^{2} + a = 3b^{2} + a$. Then $a^{2} = b^{2}$, and since $a \\neq b$, $a = -b$. Thus $f(1) = 1 + a + b = 1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 47009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA true-false test has ten questions. If you answer five questions \"true\" and five \"false,\" your score is guaranteed to be at least four. How many answer keys are there for which this is true?",
    "options": [],
    "answer": "22",
    "solution": "Solution:\nSuppose that either nine or ten of the questions have the same answer. Then no matter which five questions we pick to have this answer, we will be right at least four times. Conversely, suppose that there are at least two questions with each answer; we will show that we can get a score less than four. By symmetry, assume there are at least five questions whose answer is true. Then if we label five of these false, not only will we get these five wrong, but we will also have answered all the false questions with true, for a total of at least seven incorrect. There are 2 ways for all the questions to have the same answer, and $2 \\cdot 10 = 20$ ways for one question to have a different answer from the others, for a total of 22 ways.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn an infinite checkerboard, the union of any two distinct unit squares is called a (disconnected) domino. A domino is said to be of type $(a, b)$, with $a \\leq b$ integers not both zero, if the centers of the two squares are separated by a distance of $a$ in one orthogonal direction and $b$ in the other. (For instance, an ordinary connected domino is of type $(0,1)$, and a domino of type $(1,2)$ contains two squares separated by a knight's move.)\n\n![](attached_image_1.png)\n\nEach of the three pairs of squares above forms a domino of type $(1,2)$.\n\nTwo dominoes are said to be congruent if they are of the same type. A rectangle is said to be $(a, b)$-tileable if it can be partitioned into dominoes of type $(a, b)$.\n\nLet $0 < m \\leq n$ be integers. How many different (i.e., noncongruent) dominoes can be formed by choosing two squares of an $m \\times n$ array?",
    "options": [],
    "answer": "mn - m(m-1)/2 - 1",
    "solution": "Solution:\n\nWe must have $0 \\leq a < m$, $0 \\leq b < n$, $a \\leq b$, and $a$ and $b$ not both $0$. The number of pairs $(a, b)$ with $b < a < m$ is $m(m-1)/2$, so the answer is\n$$\nm n - \\frac{m(m-1)}{2} - 1 = m n - \\frac{m^2 - m + 2}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe volume of a cube (in cubic inches) plus three times the total length of its edges (in inches) is equal to twice its surface area (in square inches). How many inches long is its long diagonal?",
    "options": [],
    "answer": "6 sqrt(3)",
    "solution": "Solution:\n\nIf the side length of the cube is $s$ inches, then the condition implies\n$$\ns^{3} + 3 \\cdot 12s = 2 \\cdot 6s^{2},\n$$\nor\n$$\ns(s^{2} - 12s + 36) = s(s-6)^{2} = 0.\n$$\nTherefore $s = 6$, and the long diagonal has length $s \\sqrt{3} = 6 \\sqrt{3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $AB$ be the diameter of a semicircle $\\Gamma$. Two circles, $\\omega_1$ and $\\omega_2$, externally tangent to each other and internally tangent to $\\Gamma$, are tangent to the line $AB$ at $P$ and $Q$, respectively, and to semicircular arc $AB$ at $C$ and $D$, respectively, with $AP < AQ$. Suppose $F$ lies on $\\Gamma$ such that $\\angle FQB = \\angle CQA$ and that $\\angle ABF = 80^\\circ$. Find $\\angle PDQ$ in degrees.",
    "options": [],
    "answer": "35",
    "solution": "Solution:\n\nExtend the semicircle centered at $O$ to an entire circle $\\omega$, and let the reflection of $F$ over $AB$ be $F'$. Then $CQF'$ is a straight line. Also, the homothety centered at $C$ taking $\\omega_1$ into $\\omega$ takes $P$ to a point $X$ on $\\omega$ and $AB$ to the parallel line tangent to $\\omega$ at $X$. Therefore, $X$ is the midpoint of semicircle $AXB$, and $C$, $P$, and $X$ lie on a line. Similarly, $D$, $Q$, and $X$ lie on a line. So,\n$$\n45^\\circ = \\angle XCB = \\angle PCB = \\angle PCQ + \\angle QCB = \\angle PCQ + 10^\\circ,\n$$\nsince $\\angle QCB = \\angle F'CB = \\angle F'AB = \\angle FAB = 90^\\circ - \\angle ABF = 10^\\circ$. Thus $\\angle PCQ = 35^\\circ$. We will show that $\\angle PCQ = \\angle PDQ$ to get that $\\angle PDQ = 35^\\circ$.\n\nNote that $\\angle XPQ$ subtends the sum of arcs $AC$ and $BX$, which is equal to arc $XC$. Therefore $\\angle XPQ = \\angle CDX$, so $CDQP$ is cyclic and $\\angle PCQ = \\angle PDQ$. The conclusion follows.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$, $y$, and $z$ be three roots of unity. Prove that $x + y + z$ is also a root of unity if and only if $x + y = 0$, $y + z = 0$, or $z + x = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAgain, we consider the geometric picture. Arrange the vectors $x$, $y$, $z$, and $-x-y-z$ so as to form a quadrilateral. If they are all roots of unity, they form a quadrilateral all of whose side lengths are $1$. If the quadrilateral is degenerate, then two of the vectors sum to $0$, which implies the result. But even if it is not degenerate, the quadrilateral must be a rhombus, and since opposite sides of a rhombus are parallel, this again implies that two of the four roots of unity sum to $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of the absolute values of the roots of $x^{4}-4 x^{3}-4 x^{2}+16 x-8=0$.",
    "options": [],
    "answer": "2+2√2+2√3",
    "solution": "Solution: $2+2 \\sqrt{2}+2 \\sqrt{3}$\n$$\n\\begin{aligned}\nx^{4}-4 x^{3}-4 x^{2}+16 x-8 & =\\left(x^{4}-4 x^{3}+4 x^{2}\\right)-\\left(8 x^{2}-16 x+8\\right) \\\\\n& =x^{2}(x-2)^{2}-8(x-1)^{2} \\\\\n& =\\left(x^{2}-2 x\\right)^{2}-(2 \\sqrt{2} x-2 \\sqrt{2})^{2} \\\\\n& =\\left(x^{2}-(2+2 \\sqrt{2}) x+2 \\sqrt{2}\\right)\\left(x^{2}-(2-2 \\sqrt{2}) x-2 \\sqrt{2}\\right)\n\\end{aligned}\n$$\nBut noting that $(1+\\sqrt{2})^{2}=3+2 \\sqrt{2}$ and completing the square,\n$$\n\\begin{aligned}\nx^{2}-(2+2 \\sqrt{2}) x+2 \\sqrt{2} & =x^{2}-(2+2 \\sqrt{2}) x+3+2 \\sqrt{2}-3 \\\\\n& =(x-(1+\\sqrt{2}))^{2}-(\\sqrt{3})^{2} \\\\\n& =(x-1-\\sqrt{2}+\\sqrt{3})(x-1-\\sqrt{2}-\\sqrt{3}) .\n\\end{aligned}\n$$\nLikewise,\n$$\nx^{2}-(2-2 \\sqrt{2}) x-2 \\sqrt{2}=(x-1+\\sqrt{2}+\\sqrt{3})(x-1+\\sqrt{2}-\\sqrt{3})\n$$\nso the roots of the quartic are $1 \\pm \\sqrt{2} \\pm \\sqrt{3}$. Only one of these is negative, namely $1-\\sqrt{2}-\\sqrt{3}$, so the sum of the absolute values of the roots is\n$(1+\\sqrt{2}+\\sqrt{3})+(1+\\sqrt{2}-\\sqrt{3})+(1-\\sqrt{2}+\\sqrt{3})-(1-\\sqrt{2}-\\sqrt{3})=2+2 \\sqrt{2}+2 \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou are given a set of cards labeled from $1$ to $100$. You wish to make piles of three cards such that in any pile, the number on one of the cards is the product of the numbers on the other two cards. However, no card can be in more than one pile. What is the maximum number of piles you can form at once?",
    "options": [],
    "answer": "8",
    "solution": "Solution: $8$\nCertainly, the two factors in any pile cannot both be at least $10$, since then the product would be at least $10 \\times 11 > 100$. Also, the number $1$ can not appear in any pile, since then the other two cards in the pile would have to be the same. So each pile must use one of the numbers $2, 3, \\ldots, 9$ as one of the factors, meaning we have at most $8$ piles. Conversely, it is easy to construct a set of $8$ such piles, for example:\n$$\n\\begin{array}{llll}\n\\{9, 11, 99\\} & \\{8, 12, 96\\} & \\{7, 13, 91\\} & \\{6, 14, 84\\} \\\\\n\\{5, 15, 75\\} & \\{4, 16, 64\\} & \\{3, 17, 51\\} & \\{2, 18, 36\\}\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou start out with a big pile of $3^{2004}$ cards, with the numbers $1,2,3, \\ldots, 3^{2004}$ written on them. You arrange the cards into groups of three any way you like; from each group, you keep the card with the largest number and discard the other two. You now again arrange these $3^{2003}$ remaining cards into groups of three any way you like, and in each group, keep the card with the smallest number and discard the other two. You now have $3^{2002}$ cards, and you again arrange these into groups of three and keep the largest number in each group. You proceed in this manner, alternating between keeping the largest number and keeping the smallest number in each group, until you have just one card left.\nHow many different values are possible for the number on this final card?",
    "options": [],
    "answer": "3^{2004} - 2*3^{1002} + 2",
    "solution": "Solution:\n$3^{2004}-2 \\cdot 3^{1002}+2$\nWe claim that if you have cards numbered $1,2, \\ldots, 3^{2 n}$ and perform $2 n$ successive grouping operations, then $c$ is a possible value for your last remaining card if and only if\n$$\n3^{n} \\leq c \\leq 3^{2 n}-3^{n}+1\n$$\nThis gives $3^{2 n}-2 \\cdot 3^{n}+2$ possible values of $c$, for a final answer of $3^{2004}-2 \\cdot 3^{1002}+2$. Indeed, notice that the last remaining card $c$ must have been the largest of some set of three at the $(2 n-1)$th step; each of these was in turn the largest of some set of three (and so $c$ was the largest of some set of 9 cards) remaining at the $(2 n-3)$th step; each of these was in turn the largest of some set of three (and so $c$ was the largest of some set of 27) remaining at the $(2 n-5)$th step; continuing in this manner, we get that $c$ was the largest of some $3^{n}$ cards at the first step, so $c \\geq 3^{n}$. A similar analysis of all of the steps in which we save the smallest card gives that $c$ is the smallest of some set of $3^{n}$ initial cards, so $c \\leq 3^{2 n}-3^{n}+1$.\nTo see that any $c$ in this interval is indeed possible, we will carry out the groupings inductively so that, after $2 i$ steps, the following condition is satisfied: if the numbers remaining are $a_{1}<a_{2}<\\cdots<a_{3^{2(n-i)}}$, then $c$ is one of these, and there are at least $3^{n-i}-1$ numbers smaller than $c$ and at least $3^{n-i}-1$ numbers larger than $c$. This is certainly true when $i=0$, so it suffices to show that if it holds for some $i<n$, we can perform the grouping so that the condition will still hold for $i+1$. Well, we first group the smallest numbers as $\\{a_{1}, a_{2}, a_{3}\\},\\{a_{4}, a_{5}, a_{6}\\}, \\ldots,\\{a_{3^{n-i}-5}, a_{3^{n-i}-4}, a_{3^{n-i}-3}\\}$. We then group the remaining numbers in such a way that $c$ and the largest $3^{n-i}-1$ numbers are each the largest in its respective group; it is easy to see that we can do this. After retaining the largest number in each group, we will then have at least $3^{n-i-1}-1$ numbers smaller than $c$ remaining and at least $3^{n-i}-1$ numbers larger than $c$ remaining. And for the next grouping, we similarly group the largest $3^{n-i}-3$ numbers into $3^{n-i-1}-1$ groups, and arrange the remaining numbers so that the smallest $3^{n-i-1}-1$ numbers and $c$ are all the smallest in their groups. After this round of discarding, then $c$ will be retained, and we will still have at least $3^{n-i-1}-1$ numbers larger than $c$ and $3^{n-i-1}$ numbers smaller than $c$. This proves the induction step, and now the solution is complete.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $\\lfloor x\\rfloor$ denote the greatest integer less than or equal to $x$. How many positive integers less than $2005$ can be expressed in the form $\\lfloor x\\lfloor x\\rfloor\\rfloor$ for some positive real $x$?",
    "options": [],
    "answer": "990",
    "solution": "Solution:\n\nLet $\\{x\\} = x - \\lfloor x\\rfloor$ be the fractional part of $x$. Note that\n$$\n\\lfloor x\\lfloor x\\rfloor\\rfloor = \\lfloor (\\lfloor x\\rfloor + \\{x\\})\\lfloor x\\rfloor \\rfloor = \\lfloor x\\rfloor^{2} + \\lfloor \\{x\\}\\lfloor x\\rfloor \\rfloor.\n$$\nBecause $\\{x\\}$ may take on any value in the half-open interval $[0,1)$, the quantity $\\lfloor \\{x\\}\\lfloor x\\rfloor \\rfloor$ can take on any integer value between $0$ and $\\lfloor x\\rfloor - 1$, inclusive.\n\nIf $\\lfloor x\\rfloor = n$, then $\\lfloor x\\lfloor x\\rfloor\\rfloor$ can be any of the numbers $n^{2}, n^{2} + 1, \\ldots, n^{2} + n - 1$. In other words, there are precisely $n$ possible values that $\\lfloor x\\lfloor x\\rfloor\\rfloor$ can take, and moreover, all of them are less than $(n+1)^{2}$.\n\nBecause $44^{2} + 43 = 1979 < 2005$ and $45^{2} = 2025 > 2005$, $n$ can range between $1$ and $44$, inclusive. Therefore, the answer is\n$$\n\\sum_{n=1}^{44} n = \\frac{44 \\cdot 45}{2} = 990\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P$ be a regular $k$-gon inscribed in a circle of radius $1$. Find the sum of the squares of the lengths of all the sides and diagonals of $P$.",
    "options": [],
    "answer": "k^2",
    "solution": "Solution:\nPlace the vertices of $P$ at the $k$th roots of unity, $1, \\omega, \\omega^{2}, \\ldots, \\omega^{k-1}$. We will first calculate the sum of the squares of the lengths of the sides and diagonals that contain the vertex $1$. This is\n$$\n\\begin{aligned}\n\\sum_{i=0}^{k-1} \\left|1-\\omega^{i}\\right|^{2} & = \\sum_{i=0}^{k-1} \\left(1-\\omega^{i}\\right)\\left(1-\\bar{\\omega}^{i}\\right) \\\\\n& = \\sum_{i=0}^{k-1} \\left(2-\\omega^{i}-\\bar{\\omega}^{i}\\right) \\\\\n& = 2k - 2 \\sum_{i=0}^{k-1} \\omega^{i} \\\\\n& = 2k\n\\end{aligned}\n$$\nusing the fact that $1+\\omega+\\cdots+\\omega^{k-1}=0$. Now, by symmetry, this is the sum of the squares of the lengths of the sides and diagonals emanating from any vertex. Since there are $k$ vertices and each segment has two endpoints, the total sum is $2k \\cdot k / 2 = k^{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the smallest integer $x$ larger than $1$ such that $x^{2}$ ends in the same three digits as $x$ does?",
    "options": [],
    "answer": "376",
    "solution": "Solution:\n\nThe condition is that $1000 \\mid x^{2}-x = x(x-1)$. Since $1000 = 2^{3} \\cdot 5^{3}$, and $2$ cannot divide both $x$ and $x-1$, $2^{3} = 8$ must divide one of them. Similarly, $5^{3} = 125$ must divide either $x$ or $x-1$. We try successive values of $x$ that are congruent to $0$ or $1$ modulo $125$ and see which ones have the property that $x$ or $x-1$ is divisible by $8$. It is easy to check that $125, 126, 250, 251$, and $375$ do not work, but the next value, $376$, does, so this is the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f: \\mathbf{R} \\rightarrow \\mathbf{R}$ be a smooth function such that $f'(x) = f(1-x)$ for all $x$ and $f(0) = 1$. Find $f(1)$.",
    "options": [],
    "answer": "sec 1 + tan 1",
    "solution": "Solution:\nDifferentiating the given equation gives $f''(x) = -f(x)$. This has solutions of the form $A \\cos(x) + B \\sin(x)$. Since $f(0) = 1$, $A = 1$. Then $f'(x) = B \\cos(x) - \\sin(x)$ and\n\n$$\n\\begin{aligned}\nf(1-x) & = \\cos(1-x) + B \\sin(1-x) \\\\\n& = \\cos 1 \\cos x + \\sin 1 \\sin x + B \\sin 1 \\cos x - B \\cos 1 \\sin x \\\\\n& = (\\cos 1 + B \\sin 1) \\cos x + (\\sin 1 - B \\cos 1) \\sin x\n\\end{aligned}\n$$\n\nTherefore, $B = \\cos 1 + B \\sin 1$ and $-1 = \\sin 1 - B \\cos 1$, both of which yield as solutions\n\n$$\nB = \\frac{\\cos 1}{1 - \\sin 1} = \\frac{1 + \\sin 1}{\\cos 1} = \\sec 1 + \\tan 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $P(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{0}$ be a polynomial with real coefficients, $a_{n} \\neq 0$. Suppose every root of $P$ is a root of unity, but $P(1) \\neq 0$. Show that the coefficients of $P$ are symmetric; that is, show that $a_{n} = a_{0}, a_{n-1} = a_{1}, \\ldots$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince the coefficients of $P$ are real, the complex conjugates of the roots of $P$ are also roots of $P$. Now, if $x$ is a root of unity, then $x^{-1} = \\bar{x}$. But the roots of\n$$\nx^{n} P\\left(x^{-1}\\right) = a_{0} x^{n} + a_{1} x^{n-1} + \\cdots + a_{n}\n$$\nare then just the complex conjugates of the roots of $P$, so they are the roots of $P$. Therefore, $P(x)$ and $x^{n} P\\left(x^{-1}\\right)$ differ by a constant multiple $c$. Since $a_{n} = c a_{0}$ and $a_{0} = c a_{n}$, $c$ is either $1$ or $-1$. But if it were $-1$, then\n$$\nP(1) = a_{n} + a_{n-1} + \\cdots + a_{0} = \\frac{1}{2}\\left(\\left(a_{n} + a_{0}\\right) + \\left(a_{n-1} + a_{1}\\right) + \\cdots + \\left(a_{0} + a_{n}\\right)\\right) = 0,\n$$\na contradiction. Therefore $c = 1$, giving the result.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe Dingoberry Farm is a $10$ mile by $10$ mile square, broken up into $1$ mile by $1$ mile patches. Each patch is farmed either by Farmer Keith or by Farmer Ann. Whenever Ann farms a patch, she also farms all the patches due west of it and all the patches due south of it. Ann puts up a scarecrow on each of her patches that is adjacent to exactly two of Keith's patches (and nowhere else). If Ann farms a total of $30$ patches, what is the largest number of scarecrows she could put up?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\nWhenever Ann farms a patch $P$, she also farms all the patches due west of $P$ and due south of $P$. So, the only way she can put a scarecrow on $P$ is if Keith farms the patch immediately north of $P$ and the patch immediately east of $P$, in which case Ann cannot farm any of the patches due north of $P$ or due east of $P$. That is, Ann can only put a scarecrow on $P$ if it is the easternmost patch she farms in its east-west row, and the northernmost in its north-south column. In particular, all of her scarecrow patches are in different rows and columns. Suppose that she puts up $n$ scarecrows. The farthest south of these must be in the $10$th row or above, so she farms at least $1$ patch in that column; the second-farthest south must be in the $9$th row above, so she farms at least $2$ patches in that column; the third-farthest south must be in the $8$th row or above, so she farms at least $3$ patches in that column, and so forth, for a total of at least\n$$\n1+2+\\cdots+n = \\frac{n(n+1)}{2}\n$$\npatches. If Ann farms a total of $30 < 8 \\cdot 9 / 2$ patches, then we have $n < 8$. On the other hand, $n = 7$ scarecrows are possible, as shown:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S=\\{s_{0}, \\ldots, s_{n}\\}$ be a finite set of integers, and define $S+k=\\{s_{0}+k, \\ldots, s_{n}+k\\}$. We say that $S$ and $T$ are equivalent, written $S \\sim T$, if $T=S+k$ for some $k$. Given a (possibly infinite) set of integers $A$, we say that $S$ tiles $A$ if $A$ can be partitioned into subsets equivalent to $S$. Such a partition is called a tiling of $A$ by $S$.\n\nFind all sets $S$ with minimum element $1$ that tile $A=\\{1, \\ldots, 12\\}$.",
    "options": [],
    "answer": "{1}, {1,2}, {1,3}, {1,4}, {1,7}, {1,2,3}, {1,3,5}, {1,5,9}, {1,2,3,4}, {1,2,7,8}, {1,4,7,10}, {1,2,3,4,5,6}, {1,2,3,7,8,9}, {1,2,5,6,9,10}, {1,3,5,7,9,11}, {1,2,3,4,5,6,7,8,9,10,11,12}",
    "solution": "Solution:\nThis can be done by brute force. Alternatively, note that if $P(x)$ and $Q(x)$ are polynomials with coefficients either $0$ or $1$ with $P(x) Q(x)=x+x^{2}+\\cdots+x^{12}$, then the set consisting of the exponents of nonzero terms in $P$ tiles $A$. Either way, we find that $S$ is one of the following: $\\{1\\},\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{1,7\\},\\{1,2,3\\}$, $\\{1,3,5\\},\\{1,5,9\\},\\{1,2,3,4\\},\\{1,2,7,8\\},\\{1,4,7,10\\},\\{1,2,3,4,5,6\\},\\{1,2,3,7,8,9\\}$, $\\{1,2,5,6,9,10\\},\\{1,3,5,7,9,11\\}$, or $A$ itself.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTwo vertices of a cube are given in space. The locus of points that could be a third vertex of the cube is the union of $n$ circles. Find $n$.",
    "options": [],
    "answer": "10",
    "solution": "Solution:\nLet the distance between the two given vertices be $1$. If the two given vertices are adjacent, then the other vertices lie on four circles, two of radius $1$ and two of radius $\\sqrt{2}$. If the two vertices are separated by a diagonal of a face of the cube, then the locus of possible vertices adjacent to both of them is a circle of radius $\\frac{1}{2}$, the locus of possible vertices adjacent to exactly one of them is two circles of radius $\\frac{\\sqrt{2}}{2}$, and the locus of possible vertices adjacent to neither of them is a circle of radius $\\frac{\\sqrt{3}}{2}$. If the two given vertices are separated by a long diagonal, then each of the other vertices lie on one of two circles of radius $\\frac{\\sqrt{2}}{3}$, for a total of $10$ circles.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $B$ be a set of integers either bounded below or bounded above. Then show that if $S$ tiles all other integers $\\mathbf{Z} \\backslash B$, then $S$ tiles all integers $\\mathbf{Z}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume $B$ is bounded above; the other case is analogous. Let $a$ be the difference between the largest and smallest element of $S$. Denote the sets in the partition of $\\mathbf{Z} \\backslash B$ by $S_{k}$, $k \\in \\mathbf{Z}$, such that the minimum element of $S_{k}$, which we will denote $c_{k}$, is strictly increasing as $k$ increases. Since $B$ is bounded above, there exists some $k_{0}$ such that $c_{k_{0}}$ is larger than all the elements of $B$. Let\n$$\nT_{l}=\\bigcup_{k=l}^{\\infty} S_{k}\n$$\nSuppose $l \\geq k_{0}$. Note that any element in $S_{k}$, $k<l$, is at most $c_{l}-1+a$. Therefore, $T_{l}$ contains all integers that are at least $c_{l}+a$. Since the minimum element of $T_{l}$ is $c_{l}$, $T_{l}$ is completely determined by which of the integers $c_{l}+1, c_{l}+2, \\ldots, c_{l}+a-1$ it contains. This implies that there are at most $2^{a-1}$ possible nonequivalent sets $T_{l}$ when $l \\geq k_{0}$ (here we extend the notion of equivalence to infinite sets in the natural way.) By the Pigeonhole Principle, there must then be some $l_{2}>l_{1} \\geq k_{0}$ such that $T_{l_{1}} \\sim T_{l_{2}}$. But then it is easy to see that the set $S_{l_{1}} \\cup S_{l_{1}+1} \\cup \\cdots \\cup S_{l_{2}-1}$ tiles $\\mathbf{Z}$, so $S$ tiles $\\mathbf{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S=\\{s_{0}, \\ldots, s_{n}\\}$ be a finite set of integers, and define $S+k=\\{s_{0}+k, \\ldots, s_{n}+k\\}$. We say that $S$ and $T$ are equivalent, written $S \\sim T$, if $T=S+k$ for some $k$. Given a (possibly infinite) set of integers $A$, we say that $S$ tiles $A$ if $A$ can be partitioned into subsets equivalent to $S$. Such a partition is called a tiling of $A$ by $S$.\n\nExhibit a set $S$ which tiles the integers $\\mathbf{Z}$ but not the natural numbers $\\mathbf{N}$.",
    "options": [],
    "answer": "{1,3,4,6}",
    "solution": "Solution:\n\nOne example is $\\{1,3,4,6\\}$. Since its elements are all distinct modulo $4$, it tiles $\\mathbf{Z}$ by translation by multiples of $4$. On the other hand, it is easy to see that it cannot tile $\\mathbf{N}$: $1$ is contained in $\\{1,3,4,6\\}$, but then there is no possible set for $2$ to be contained in that does not overlap. Another example is $\\{1,2,6\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTriangle $ABC$ has $AB=1$, $BC=\\sqrt{7}$, and $CA=\\sqrt{3}$. Let $\\ell_1$ be the line through $A$ perpendicular to $AB$, $\\ell_2$ the line through $B$ perpendicular to $AC$, and $P$ the point of intersection of $\\ell_1$ and $\\ell_2$. Find $PC$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nBy the Law of Cosines, $\\angle BAC = \\cos^{-1} \\left(\\frac{3+1-7}{2 \\sqrt{3}}\\right) = \\cos^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right) = 150^\\circ$.\n\nIf we let $Q$ be the intersection of $\\ell_2$ and $AC$, we notice that $\\angle QBA = 90^\\circ - \\angle QAB = 90^\\circ - 30^\\circ = 60^\\circ$.\n\nIt follows that triangle $ABP$ is a $30$-$60$-$90$ triangle and thus $PB = 2$ and $PA = \\sqrt{3}$.\n\nFinally, we have $\\angle PAC = 360^\\circ - (90^\\circ + 150^\\circ) = 120^\\circ$, and\n$$\nPC = \\left(PA^2 + AC^2 - 2 PA \\cdot AC \\cos 120^\\circ\\right)^{1/2} = (3 + 3 + 3)^{1/2} = 3.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThree noncollinear points and a line $\\ell$ are given in the plane. Suppose no two of the points lie on a line parallel to $\\ell$ (or $\\ell$ itself). There are exactly $n$ lines perpendicular to $\\ell$ with the following property: the three circles with centers at the given points and tangent to the line all concur at some point. Find all possible values of $n$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nThe condition for the line is that each of the three points lies at an equal distance from the line as from some fixed point; in other words, the line is the directrix of a parabola containing the three points. Three noncollinear points in the coordinate plane determine a quadratic polynomial in $x$ unless two of the points have the same $x$-coordinate. Therefore, given the direction of the directrix, three noncollinear points determine a parabola, unless two of the points lie on a line perpendicular to the directrix. This case is ruled out by the given condition, so the answer is 1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S=\\{s_{0}, \\ldots, s_{n}\\}$ be a finite set of integers, and define $S+k=\\{s_{0}+k, \\ldots, s_{n}+k\\}$. We say that $S$ and $T$ are equivalent, written $S \\sim T$, if $T=S+k$ for some $k$. Given a (possibly infinite) set of integers $A$, we say that $S$ tiles $A$ if $A$ can be partitioned into subsets equivalent to $S$. Such a partition is called a tiling of $A$ by $S$.\n\nSuppose that $S$ tiles the set of all integer cubes. Prove that $S$ has only one element.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the difference between the smallest and largest element of $S$ be $a$. Then the set equivalent to $S$ that contains $b^{3}$ can only contain integers between $b^{3}-a$ and $b^{3}+a$, inclusive. But for sufficiently large $b$, $b^{3}$ is the only cube in this range, so $S$ can only have one element.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $S$ tiles the natural numbers $\\mathbf{N}$. Show that $S$ tiles the set $\\{1,2, \\ldots, k\\}$ for some positive integer $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nUsing the notation from above, we can find $l_{1}<l_{2}$ such that $T_{l_{1}} \\sim T_{l_{2}}$. By the same argument as in problem 12, as long as $T_{l_{1}} \\neq \\mathbf{N}$, there is a unique choice for $S_{l_{1}-1}$ that contains the largest integer not in $T_{l_{1}}$. Since the same can be said for $T_{l_{2}}$, we must have that $T_{l_{1}-1} \\sim T_{l_{2}-1}$. Continuing in this manner, we find that there must exist some $l$ for which $\\mathbf{N} \\sim T_{l}$; then $S$ tiles $\\mathbf{N} \\backslash T_{l}=\\left\\{1,2, \\cdots, c_{l}-1\\right\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S$ be the set of lattice points inside the circle $x^{2}+y^{2}=11$. Let $M$ be the greatest area of any triangle with vertices in $S$. How many triangles with vertices in $S$ have area $M$?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\nThe boundary of the convex hull of $S$ consists of points with $(x, y)$ or $(y, x) = (0, \\pm 3)$, $(\\pm 1, \\pm 3)$, and $(\\pm 2, \\pm 2)$. For any triangle $T$ with vertices in $S$, we can increase its area by moving a vertex not on the boundary to some point on the boundary. Thus, if $T$ has area $M$, its vertices are all on the boundary of $S$.\n\nThe next step is to see (either by inspection or by noting that $T$ has area no larger than that of an equilateral triangle inscribed in a circle of radius $\\sqrt{10}$, which has area less than $13$) that $M = 12$. There are $16$ triangles with area $12$, all congruent to one of the following three: vertices $(2,2), (1,-3)$, and $(-3,1)$; vertices $(3,-1), (-3,-1)$, and $(1,3)$; or vertices $(3,-1), (-3,-1)$, and $(0,3)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $S$ tiles $\\mathbb{N}$. Show that $S$ is symmetric; that is, if $-S=\\{-s_{n}, \\ldots,-s_{0}\\}$, show that $S \\sim -S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume without loss of generality that the minimum element of $S$ is $0$. By the previous problem, $S$ tiles the set $\\{1,2, \\ldots, k\\}$ for some positive integer $k$. Then let $P(x)$ be the polynomial $\\sum_{i=0}^{n} x^{s_{i}}$. To say that the set $\\{1,2, \\ldots, k\\}$, or equivalently the set $\\{0,1, \\ldots, k-1\\}$, is tiled by $S$ is to say that there is some polynomial $Q(x)$ with coefficients $0$ or $1$ such that $P(x) Q(x) = 1 + x + \\cdots + x^{k-1} = \\frac{x^{k}-1}{x-1}$. It follows that all the roots of $P(x)$ are roots of unity, but $P(1) \\neq 0$. By question 11 above, this implies that $P(x)$ is symmetric. Therefore, $s_{0} + s_{n} = s_{1} + s_{n-1} = \\cdots = s_{n} + s_{0}$, so $S$ is symmetric.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S=\\{s_{0}, \\ldots, s_{n}\\}$ be a finite set of integers, and define $S+k=\\{s_{0}+k, \\ldots, s_{n}+k\\}$. We say that $S$ and $T$ are equivalent, written $S \\sim T$, if $T=S+k$ for some $k$. Given a (possibly infinite) set of integers $A$, we say that $S$ tiles $A$ if $A$ can be partitioned into subsets equivalent to $S$. Such a partition is called a tiling of $A$ by $S$.\n\nSuppose that $S$ tiles the set of odd prime numbers. Prove that $S$ has only one element.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider the set $S_{0}$ equivalent to $S$ that contains $3$. If it contains $5$ but not $7$, then the set $S_{1}$ equivalent to $S$ containing $7$ must contain $9$, which is not prime. Likewise, $S_{0}$ cannot contain $7$ but not $5$, because then the set $S_{1}$ containing $5$ must contain $9$. Suppose $S_{0}$ contains $3, 5$, and $7$. Then any other set $S_{1}$ of the tiling contains elements $p, p+2$, and $p+4$. But not all of these can be prime, because one of them is divisible by $3$. Finally, suppose $S_{0}$ contains $3$ and has second-smallest element $p>7$. Then the set $S_{1}$ containing $5$ does not contain $7$ but does contain $p+2$, and the set $S_{2}$ containing $7$ contains $p+4$. But as before, not all of $p, p+2$, and $p+4$ can be prime. Therefore $S$ has no second-smallest element, so it has only one element.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA regular octahedron has a side length of $1$. What is the distance between two opposite faces?",
    "options": [],
    "answer": "√6/3",
    "solution": "Solution: $\\sqrt{6} / 3$\nImagine orienting the octahedron so that the two opposite faces are horizontal. Project onto a horizontal plane; these two faces are congruent equilateral triangles which (when projected) have the same center and opposite orientations. Hence, the vertices of the octahedron project to the vertices of a regular hexagon $A B C D E F$.\n\n![](attached_image_1.png)\n\nLet $O$ be the center of the hexagon and $M$ the midpoint of $A C$. Now $A B M$ is a 30-60-90 triangle, so $A B = A M / (\\sqrt{3} / 2) = (1 / 2) / (\\sqrt{3} / 2) = \\sqrt{3} / 3$. If we let $d$ denote the desired vertical distance between the opposite faces (which project to $A C E$ and $B D F$), then by the Pythagorean Theorem, $A B^{2} + d^{2} = 1^{2}$, so $d = \\sqrt{1 - A B^{2}} = \\sqrt{6} / 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\sqrt[{2 \\sqrt{2 \\sqrt[3]{2 \\sqrt[4]{2 \\sqrt[5]{2 \\cdots}}}}}]{ }\n$$",
    "options": [],
    "answer": "2^(e-1)",
    "solution": "Solution:\nTaking the base $2$ logarithm of the expression gives\n$$\n1+\\frac{1}{2}\\left(1+\\frac{1}{3}\\left(1+\\frac{1}{4}(1+\\cdots)\\right)\\right)=1+\\frac{1}{2!}+\\frac{1}{3!}+\\frac{1}{4!}+\\cdots=e-1.\n$$\nTherefore the expression is just $2^{e-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a$, $b$, and $c$ are random real numbers from $0$ to $1$, independently and uniformly chosen, what is the average (expected) value of the smallest of $a$, $b$, and $c$?",
    "options": [],
    "answer": "1/4",
    "solution": "Solution:\nLet $d$ be a fourth random variable, also chosen uniformly from $[0,1]$. For fixed $a$, $b$, and $c$, the probability that $d < \\min \\{a, b, c\\}$ is evidently equal to $\\min \\{a, b, c\\}$. Hence,\nif we average over all choices of $a$, $b$, $c$, the average value of $\\min \\{a, b, c\\}$ is equal to the probability that, when $a$, $b$, $c$, and $d$ are independently randomly chosen, $d < \\min \\{a, b, c\\}$, i.e., that $d$ is the smallest of the four variables. On the other hand, by symmetry, the probability that $d$ is the smallest of the four is simply equal to $1/4$, so that is our answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRegular tetrahedron $A B C D$ is projected onto a plane sending $A, B, C$, and $D$ to $A', B', C'$, and $D'$ respectively. Suppose $A' B' C' D'$ is a convex quadrilateral with $A' B' = A' D'$ and $C' B' = C' D'$, and suppose that the area of $A' B' C' D' = 4$. Given these conditions, the set of possible lengths of $A B$ consists of all real numbers in the interval $[a, b)$. Compute $b$.",
    "options": [],
    "answer": "2 sqrt[4]{6}",
    "solution": "Solution:\n\n$2 \\sqrt[4]{6}$\n\nThe value of $b$ occurs when the quadrilateral $A' B' C' D'$ degenerates to an isosceles triangle. This occurs when the altitude from $A$ to $B C D$ is parallel to the plane. Let $s = A B$. Then the altitude from $A$ intersects the center $E$ of face $B C D$. Since $E B = \\frac{s}{\\sqrt{3}}$, it follows that $A' C' = A E = \\sqrt{s^2 - \\frac{s^2}{3}} = \\frac{s \\sqrt{6}}{3}$. Then since $B D$ is parallel to the plane, $B' D' = s$. Then the area of $A' B' C' D'$ is $4 = \\frac{1}{2} \\cdot \\frac{s^2 \\sqrt{6}}{3}$, implying $s^2 = 4 \\sqrt{6}$, or $s = 2 \\sqrt[4]{6}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots$ be a sequence of integers with infinitely many positive and infinitely many negative terms. Suppose that for each positive integer $n$, the numbers $a_1, a_2, \\dots, a_n$ leave distinct remainders upon division by $n$. Prove that every integer occurs exactly once in the sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The conditions of the problem can be reformulated by saying that for every positive integer $n$, the numbers $a_1, a_2, \\dots, a_n$ form a complete set of residues modulo $n$. We proceed our proof as the following.\n\n(1) First, we claim that the sequence consists of distinct integers; that is, if $1 \\le i < j$, then $a_i \\ne a_j$. Otherwise the set $\\{a_1, a_2, \\dots, a_j\\}$ would contain at most $j-1$ distinct residues modulo $j$, violating our new formulation of the conditions of the problem.\n\n(2) Second, we show that numbers in the sequence are fairly close to each other. More precisely, we claim that if $1 \\le i < j \\le n$, then $|a_i - a_j| \\le n - 1$. For if $m = |a_i - a_j| \\ge n$, then the set $\\{a_1, a_2, \\dots, a_m\\}$ would contain two numbers congruent modulo $m$, violating our new formulation of the conditions of the problem.\n\n(3) Third, we show that the set $\\{a_1, a_2, \\dots, a_n\\}$ contains a block of consecutive numbers. Indeed, for every positive integer $n$, let $i_n$ and $j_n$ be the indices such that $a_{i_n}$ and $a_{j_n}$ are respectively the smallest and the largest number among $a_1, a_2, \\dots, a_n$. By (2), we conclude that $a_{j_n} - a_{i_n} = |a_{j_n} - a_{i_n}| \\le n - 1$. By (1), we conclude that $\\{a_1, a_2, \\dots, a_n\\}$ consists of all integers between $a_{i_n}$ and $a_{j_n}$ (inclusive).\n\n(4) Finally, we show that every integer appears in the sequence. Let $x$ be an arbitrary integer. Because $a_k < 0$ for infinitely many indices $k$ and the terms of the sequence are distinct, it follows that there exists $i$ such that $a_i < x$. Likewise, there exists $j$ such that $x < a_j$. Let $n$ be an integer with $n \\ge \\max\\{i, j\\}$. By (3), we conclude that every number between $a_i$ and $a_j$, including $x$ in particular, is in $\\{a_1, a_2, \\dots, a_n\\}$. Our proof is thus complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many real numbers $x$ are solutions to the following equation?\n$$\n2003^{x} + 2004^{x} = 2005^{x}\n$$",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nRewrite the equation as $(2003 / 2005)^{x} + (2004 / 2005)^{x} = 1$. The left side is strictly decreasing in $x$, so there cannot be more than one solution. On the other hand, the left side equals $2 > 1$ when $x = 0$ and goes to $0$ when $x$ is very large, so it must equal $1$ somewhere in between. Therefore there is one solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$, $y$, and $z$ be positive real numbers such that $(x \\cdot y) + z = (x + z) \\cdot (y + z)$. What is the maximum possible value of $x y z$?",
    "options": [],
    "answer": "1/27",
    "solution": "Solution:\nThe condition is equivalent to $z^{2} + (x + y - 1) z = 0$. Since $z$ is positive, $z = 1 - x - y$, so $x + y + z = 1$. By the AM-GM inequality,\n$$\nx y z \\leq \\left(\\frac{x + y + z}{3}\\right)^{3} = \\frac{1}{27}\n$$\nwith equality when $x = y = z = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many nonempty subsets of $\\{1,2,3, \\ldots, 12\\}$ have the property that the sum of the largest element and the smallest element is $13$?",
    "options": [],
    "answer": "1365",
    "solution": "Solution:\nIf $a$ is the smallest element of such a set, then $13-a$ is the largest element, and for the remaining elements we may choose any (or none) of the $12-2a$ elements $a+1, a+2, \\ldots, (13-a)-1$. Thus there are $2^{12-2a}$ such sets whose smallest element is $a$. Also, $13-a \\geq a$ clearly implies $a < 7$. Summing over all $a = 1, 2, \\ldots, 6$, we get a total of\n$$\n2^{10} + 2^{8} + 2^{6} + \\cdots + 2^{0} = 4^{5} + 4^{4} + \\cdots + 4^{0} = \\frac{4^{6} - 1}{4 - 1} = \\frac{4095}{3} = 1365\n$$\npossible sets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA plane curve is parameterized by $x(t) = \\int_{t}^{\\infty} \\frac{\\cos u}{u} du$ and $y(t) = \\int_{t}^{\\infty} \\frac{\\sin u}{u} du$ for $1 \\leq t \\leq 2$. What is the length of the curve?",
    "options": [],
    "answer": "ln 2",
    "solution": "Solution: $\\ln 2$\nBy the Second Fundamental Theorem of Calculus, $\\frac{d x}{d t} = -\\frac{\\cos t}{t}$ and $\\frac{d y}{d t} = -\\frac{\\sin t}{t}$. Therefore, the length of the curve is\n$$\n\\int_{1}^{2} \\sqrt{\\left(\\frac{d x}{d t}\\right)^{2} + \\left(\\frac{d y}{d t}\\right)^{2}} d t = \\int_{1}^{2} \\sqrt{\\frac{\\cos^{2} t}{t^{2}} + \\frac{\\sin^{2} t}{t^{2}}} d t = \\int_{1}^{2} \\frac{1}{t} d t = [\\ln t]_{1}^{2} = \\ln 2\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a regular tetrahedron with side length $2$. The plane parallel to edges $AB$ and $CD$ and lying halfway between them cuts $ABCD$ into two pieces. Find the surface area of one of these pieces.",
    "options": [],
    "answer": "1 + 2√3",
    "solution": "Solution:\n\n$1 + 2\\sqrt{3}$\n\nThe plane intersects each face of the tetrahedron in a midline of the face; by symmetry it follows that the intersection of the plane with the tetrahedron is a square of side length $1$. The surface area of each piece is half the total surface area of the tetrahedron plus the area of the square, that is,\n$$\n\\frac{1}{2} \\cdot 4 \\cdot \\frac{2^{2} \\sqrt{3}}{4} + 1 = 1 + 2\\sqrt{3}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $0 < a \\leq b$ and $4 \\nmid m n$. Prove that the number of ways in which an $m \\times n$ rectangle can be partitioned into dominoes of type $(a, b)$ is even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf the rectangle is tileable, it can be partitioned into an odd number of dominoes. Consider the reflection of the partitioned rectangle over one axis. This gives another partition of the rectangle. In fact, it cannot be the same partition, for suppose it were. Then we can pair each domino with its reflected image, but since there are an odd number of dominoes, one must reflect into itself. Since $a > 0$, this is not possible. Therefore, we can pair off partitions and their reflections, and it follows that the total number of partitions is even.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA true-false test has ten questions. If you answer five questions \"true\" and five \"false,\" your score is guaranteed to be at least four. How many answer keys are there for which this is true?",
    "options": [],
    "answer": "22",
    "solution": "Solution:\nSuppose that either nine or ten of the questions have the same answer. Then no matter which five questions we pick to have this answer, we will be right at least four times. Conversely, suppose that there are at least two questions with each answer; we will show that we can get a score less than four. By symmetry, assume there are at least five questions whose answer is true. Then if we label five of these false, not only will we get these five wrong, but we will also have answered all the false questions with true, for a total of at least seven incorrect. There are 2 ways for all the questions to have the same answer, and $2 \\cdot 10=20$ ways for one question to have a different answer from the others, for a total of 22 ways.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind three real numbers $a < b < c$ satisfying:\n$$\n\\begin{aligned}\na + b + c & = 21 / 4 \\\\\n1 / a + 1 / b + 1 / c & = 21 / 4 \\\\\na b c & = 1\n\\end{aligned}\n$$",
    "options": [],
    "answer": "1/4, 1, 4",
    "solution": "Solution:\nBy inspection, one notices that if $b$ is a number such that $b + 1 / b = 17 / 4$, then $a = 1$, $c = 1 / b$ will work. Again by inspection (or by solving the quadratic $b^{2} - 17 b / 4 + 1 = 0$), one finds $b = 1 / 4$ or $4$, so the numbers are $1 / 4$, $1$, and $4$.\n\nAlternative Solution:\nNote that $a b + b c + c a = a b c (1 / a + 1 / b + 1 / c) = 21 / 4$, so $a$, $b$, and $c$ are the roots of the polynomial $x^{3} - 21 x^{2} / 4 + 21 x / 4 - 1$, which factors as $\\frac{1}{4}(x - 1)(x - 4)(4 x - 1)$, giving the same answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $n$ is a positive integer, let $s(n)$ denote the sum of the digits of $n$. We say that $n$ is zesty if there exist positive integers $x$ and $y$ greater than 1 such that $x y = n$ and $s(x) s(y) = s(n)$. How many zesty two-digit numbers are there?",
    "options": [],
    "answer": "34",
    "solution": "Solution:\n\nLet $n$ be a zesty two-digit number, and let $x$ and $y$ be as in the problem statement. Clearly if both $x$ and $y$ are one-digit numbers, then $s(x) s(y) = n \\neq s(n)$. Thus either $x$ is a two-digit number or $y$ is. Assume without loss of generality that it is $x$. If $x = 10a + b$, $1 \\leq a \\leq 9$ and $0 \\leq b \\leq 9$, then $n = 10a y + b y$. If both $a y$ and $b y$ are less than $10$, then $s(n) = a y + b y$, but if either is at least $10$, then $s(n) < a y + b y$. It follows that the two digits of $n$ share a common factor greater than $1$, namely $y$. It is now easy to count the zesty two-digit numbers by first digit starting with $2$; there are a total of $5 + 4 + 5 + 2 + 7 + 2 + 5 + 4 = 34$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $ABC$ with altitude $AD$, $\\angle BAC = 45^{\\circ}$, $DB = 3$, and $CD = 2$. Find the area of triangle $ABC$.",
    "options": [],
    "answer": "15",
    "solution": "Solution:\n\nSuppose first that $D$ lies between $B$ and $C$. Let $ABC$ be inscribed in circle $\\omega$, and extend $AD$ to intersect $\\omega$ again at $E$. Note that $A$ subtends a quarter of the circle, so in particular, the chord through $C$ perpendicular to $BC$ and parallel to $AD$ has length $BC = 5$. Therefore, $AD = 5 + DE$. By power of a point, $6 = BD \\cdot DC = AD \\cdot DE = AD^2 - 5AD$, implying $AD = 6$, so the area of $ABC$ is $\\frac{1}{2} BC \\cdot AD = 15$.\n\nIf $D$ does not lie between $B$ and $C$, then $BC = 1$, so $A$ lies on a circle of radius $\\sqrt{2}/2$ through $B$ and $C$. But then it is easy to check that the perpendicular to $BC$ through $D$ cannot intersect the circle, a contradiction.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind\n$$\n\\{\\ln (1+e)\\}+\\left\\{\\ln \\left(1+e^{2}\\right)\\right\\}+\\left\\{\\ln \\left(1+e^{4}\\right)\\right\\}+\\left\\{\\ln \\left(1+e^{8}\\right)\\right\\}+\\cdots,\n$$\nwhere $\\{x\\}=x-\\lfloor x\\rfloor$ denotes the fractional part of $x$.",
    "options": [],
    "answer": "1 - ln(e - 1)",
    "solution": "Solution:\n$1-\\ln (e-1)$\n\nSince $\\ln \\left(1+e^{2^{k}}\\right)$ is just larger than $2^{k}$, its fractional part is $\\ln \\left(1+e^{2^{k}}\\right)-\\ln e^{2^{k}}=\\ln \\left(1+e^{-2^{k}}\\right)$. But now notice that\n$$\n\\prod_{k=0}^{n}\\left(1+x^{2^{k}}\\right)=1+x+x^{2}+\\cdots+x^{2^{n+1}-1}\n$$\n(This is easily proven by induction or by noting that every nonnegative integer less than $2^{n+1}$ has a unique ( $n+1$ )-bit binary expansion.) If $|x|<1$, this product converges to $\\frac{1}{1-x}$ as $n$ goes to infinity. Therefore,\n$$\n\\sum_{k=0}^{\\infty} \\ln \\left(1+e^{-2^{k}}\\right)=\\ln \\prod_{k=0}^{\\infty}\\left(1+\\left(e^{-1}\\right)^{2^{k}}\\right)=\\ln \\frac{1}{1-e^{-1}}=\\ln \\frac{e}{e-1}=1-\\ln (e-1)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sides of a regular hexagon are trisected, resulting in 18 points, including vertices. These points, starting with a vertex, are numbered clockwise as $A_{1}, A_{2}, \\ldots, A_{18}$. The line segment $A_{k} A_{k+4}$ is drawn for $k=1,4,7,10,13,16$, where indices are taken modulo 18. These segments define a region containing the center of the hexagon. Find the ratio of the area of this region to the area of the large hexagon.",
    "options": [],
    "answer": "9/13",
    "solution": "Solution:\n\nLet us assume all sides are of side length $3$. Consider the triangle $A_{1} A_{4} A_{5}$. Let $P$ be the point of intersection of $A_{1} A_{5}$ with $A_{4} A_{8}$. This is a vertex of the inner hexagon. Then $\\angle A_{4} A_{1} A_{5} = \\angle A_{5} A_{4} P$, by symmetry. It follows that $A_{1} A_{4} A_{5} \\sim A_{4} P A_{5}$. Also, $\\angle A_{1} A_{4} A_{5} = 120^{\\circ}$, so by the Law of Cosines $A_{1} A_{5} = \\sqrt{13}$. It follows that $P A_{5} = \\left(A_{4} A_{5}\\right) \\cdot \\left(A_{4} A_{5}\\right) / \\left(A_{1} A_{5}\\right) = 1 / \\sqrt{13}$. Let $Q$ be the intersection of $A_{1} A_{5}$ and $A_{16} A_{2}$. By similar reasoning, $A_{1} Q = 3 / \\sqrt{13}$, so $P Q = A_{1} A_{5} - A_{1} Q - P A_{5} = 9 / \\sqrt{13}$. By symmetry, the inner region is a regular hexagon with side length $9 / \\sqrt{13}$. Hence the ratio of the area of the smaller to larger hexagon is $(3 / \\sqrt{13})^{2} = 9 / 13$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 47051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the base 10 arithmetic problem $H M M T + G U T S = R O U N D$, each distinct letter represents a different digit, and leading zeroes are not allowed. What is the maximum possible value of $R O U N D$?",
    "options": [],
    "answer": "16352",
    "solution": "Solution:\n\nClearly $R = 1$, and from the hundreds column, $M = 0$ or $9$. Since $H + G = 9 + O$ or $10 + O$, it is easy to see that $O$ can be at most $7$, in which case $H$ and $G$ must be $8$ and $9$, so $M = 0$. But because of the tens column, we must have $S + T \\geq 10$, and in fact since $D$ cannot be $0$ or $1$, $S + T \\geq 12$, which is impossible given the remaining choices. Therefore, $O$ is at most $6$.\n\nSuppose $O = 6$ and $M = 9$. Then we must have $H$ and $G$ be $7$ and $8$. With the remaining digits $0, 2, 3, 4$, and $5$, we must have in the ones column that $T$ and $S$ are $2$ and $3$, which leaves no possibility for $N$. If instead $M = 0$, then $H$ and $G$ are $7$ and $9$. Since again $S + T \\geq 12$ and $N = T + 1$, the only possibility is $S = 8$, $T = 4$, and $N = 5$, giving $R O U N D = 16352 = 7004 + 9348 = 9004 + 7348$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn ant starts at one vertex of a tetrahedron. Each minute it walks along a random edge to an adjacent vertex. What is the probability that after one hour the ant winds up at the same vertex it started at?",
    "options": [],
    "answer": "(3^59 + 1)/(4 * 3^59)",
    "solution": "Solution:\n\nLet $p_n$ be the probability that the ant is at the original vertex after $n$ minutes; then $p_0 = 1$. The chance that the ant is at each of the other three vertices after $n$ minutes is $\\frac{1}{3}(1 - p_n)$. Since the ant can only walk to the original vertex from one of the three others, and at each there is a $\\frac{1}{3}$ probability of doing so, we have that\n$$\np_{n+1} = \\frac{1}{3}(1 - p_n).\n$$\nLet $q_n = p_n - \\frac{1}{4}$. Substituting this into the recurrence, we find that\n$$\nq_{n+1} = \\frac{1}{4} + \\frac{1}{3}\\left(-q_n - \\frac{3}{4}\\right) = -\\frac{1}{3} q_n.\n$$\nSince $q_0 = \\frac{3}{4}$, $q_n = \\frac{3}{4} \\cdot \\left(-\\frac{1}{3}\\right)^n$. In particular, this implies that\n$$\np_{60} = \\frac{1}{4} + q_{60} = \\frac{1}{4} + \\frac{3}{4} \\cdot \\frac{1}{3^{60}} = \\frac{3^{59} + 1}{4 \\cdot 3^{59}}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $ABC$, $AC = 3 AB$. Let $AD$ bisect angle $A$ with $D$ lying on $BC$, and let $E$ be the foot of the perpendicular from $C$ to $AD$. Find $[ABD]/[CDE]$. (Here, $[XYZ]$ denotes the area of triangle $XYZ$.)",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n\n$1/3$\n\nBy the Angle Bisector Theorem, $DC/DB = AC/AB = 3$. We will show that $AD = DE$. Let $CE$ intersect $AB$ at $F$. Then since $AE$ bisects angle $A$, $AF = AC = 3 AB$, and $EF = EC$. Let $G$ be the midpoint of $BF$. Then $BG = GF$, so $GE \\parallel BC$. But then since $B$ is the midpoint of $AG$, $D$ must be the midpoint of $AE$, as desired. Then $[ABD]/[CDE] = (AD \\cdot BD)/(ED \\cdot CD) = 1/3$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a chess-playing club, some of the players take lessons from other players. It is possible (but not necessary) for two players both to take lessons from each other. It so happens that for any three distinct members of the club, $A$, $B$, and $C$, exactly one of the following three statements is true: $A$ takes lessons from $B$; $B$ takes lessons from $C$; $C$ takes lessons from $A$. What is the largest number of players there can be?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nIf $P$, $Q$, $R$, $S$, and $T$ are any five distinct players, then consider all pairs $A$, $B \\in \\{P, Q, R, S, T\\}$ such that $A$ takes lessons from $B$. Each pair contributes to exactly three triples $(A, B, C)$ (one for each of the choices of $C$ distinct from $A$ and $B$); three triples $(C, A, B)$; and three triples $(B, C, A)$. On the other hand, there are $5 \\times 4 \\times 3 = 60$ ordered triples of distinct players among these five, and each includes exactly one of our lesson-taking pairs. That means that there are $60 / 9$ such pairs. But this number isn't an integer, so there cannot be five distinct people in the club.\n\nOn the other hand, there can be four people, $P$, $Q$, $R$, and $S$: let $P$ and $Q$ both take lessons from each other, and let $R$ and $S$ both take lessons from each other; it is easy to check that this meets the conditions. Thus the maximum number of players is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThere are three pairs of real numbers $\\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right)$, and $\\left(x_{3}, y_{3}\\right)$ that satisfy both $x^{3}-3 x y^{2}=2005$ and $y^{3}-3 x^{2} y=2004$. Compute $\\left(1-\\frac{x_{1}}{y_{1}}\\right)\\left(1-\\frac{x_{2}}{y_{2}}\\right)\\left(1-\\frac{x_{3}}{y_{3}}\\right)$.",
    "options": [],
    "answer": "1/1002",
    "solution": "Solution:\nBy the given, $2004 \\left(x^{3}-3 x y^{2}\\right)-2005\\left(y^{3}-3 x^{2} y\\right)=0$. Dividing both sides by $y^{3}$ and setting $t=\\frac{x}{y}$ yields $2004\\left(t^{3}-3 t\\right)-2005\\left(1-3 t^{2}\\right)=0$. A quick check shows that this cubic has three real roots. Since the three roots are precisely $\\frac{x_{1}}{y_{1}}, \\frac{x_{2}}{y_{2}},$ and $\\frac{x_{3}}{y_{3}}$, we must have $2004\\left(t^{3}-3 t\\right)-2005\\left(1-3 t^{2}\\right)=2004\\left(t-\\frac{x_{1}}{y_{1}}\\right)\\left(t-\\frac{x_{2}}{y_{2}}\\right)\\left(t-\\frac{x_{3}}{y_{3}}\\right)$. Therefore,\n$$\n\\left(1-\\frac{x_{1}}{y_{1}}\\right)\\left(1-\\frac{x_{2}}{y_{2}}\\right)\\left(1-\\frac{x_{3}}{y_{3}}\\right)=\\frac{2004\\left(1^{3}-3(1)\\right)-2005\\left(1-3(1)^{2}\\right)}{2004}=\\frac{1}{1002} .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n > 0$ be an integer. Each face of a regular tetrahedron is painted in one of $n$ colors (the faces are not necessarily painted different colors). Suppose there are $n^{3}$ possible colorings, where rotations, but not reflections, of the same coloring are considered the same. Find all possible values of $n$.",
    "options": [],
    "answer": "1 and 11",
    "solution": "Solution:\nWe count the possible number of colorings. If four colors are used, there are two different colorings that are mirror images of each other, for a total of $2\\binom{n}{4}$ colorings. If three colors are used, we choose one color to use twice (which determines the coloring), for a total of $3\\binom{n}{3}$ colorings. If two colors are used, we can either choose one of those colors and color three faces with it, or we can color two faces each color, for a total of $3\\binom{n}{2}$ colorings. Finally, we can also use only one color, for $\\binom{n}{1}$ colorings. This gives a total of\n$$\n2\\binom{n}{4} + 3\\binom{n}{3} + 3\\binom{n}{2} + \\binom{n}{1} = \\frac{1}{12} n^{2}(n^{2} + 11)\n$$\ncolorings. Setting this equal to $n^{3}$, we get the equation $n^{2}(n^{2} + 11) = 12 n^{3}$, or equivalently $n^{2}(n - 1)(n - 11) = 0$, giving the answers $1$ and $11$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47057,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\frac{x^5 - x^2}{x^5 + y^2 + z^2} + \\frac{y^5 - y^2}{y^5 + z^2 + x^2} + \\frac{z^5 - z^2}{z^5 + x^2 + y^2} \\ge 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that\n$$\n\\frac{x^5 - x^2}{x^5 + y^2 + z^2} = 1 - \\frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2}\n$$\nand its cyclic analogous forms. The given inequality is equivalent to\n$$\n\\frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2} + \\frac{x^2 + y^2 + z^2}{y^5 + z^2 + x^2} + \\frac{x^2 + y^2 + z^2}{z^5 + x^2 + y^2} \\le 3. \\quad (\\dagger)\n$$\nIn view of the **Cauchy-Schwarz Inequality** and the condition $xyz \\ge 1$, we have\n$$\n\\begin{aligned} (x^5 + y^2 + z^2)(yz + y^2 + z^2) &\\ge \\left(x^{\\frac{5}{2}}(yz)^{\\frac{1}{2}} + y^2 + z^2\\right)^2 \\\\ &\\ge (x^2 + y^2 + z^2)^2, \\end{aligned}\n$$\nor\n$$\n\\frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2} \\le \\frac{yz + y^2 + z^2}{x^2 + y^2 + z^2}.\n$$\n\nTaking the cyclic sum of the above inequality and analogous forms gives\n$$\n\\frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2} + \\frac{x^2 + y^2 + z^2}{y^5 + z^2 + x^2} + \\frac{x^2 + y^2 + z^2}{z^5 + x^2 + y^2} \\le 2 + \\frac{yz + zx + xy}{x^2 + y^2 + z^2}.\n$$\nIt suffices to show that $xy + yz + zx \\le x^2 + y^2 + z^2$, which is well known (and can be easily shown by the Cauchy-Schwarz Inequality or $(x - y)^2 + (y - z)^2 + (z - x)^2 \\ge 0$).\nGiven a function $f$ of $n$ variables, we define the symmetric sum\n$$\n\\sum_{\\text{sym}} f(x_1, \\dots, x_n) = \\sum_{\\sigma} f(x_{\\sigma(1)}, \\dots, x_{\\sigma(n)})\n$$\nwhere $\\sigma$ runs over all permutations of $1, \\dots, n$ (for a total of $n!$ terms). For example, if $n = 3$, and we write $x, y, z$ for $x_1, x_2, x_3$,\n$$\n\\begin{aligned} \\sum_{\\text{sym}} x^3 &= 2x^3 + 2y^3 + 2z^3 \\\\ \\sum_{\\text{sym}} x^2y &= x^2y + y^2z + z^2x + x^2z + y^2x + z^2y \\\\ \\sum_{\\text{sym}} xyz &= 6xyz. \\end{aligned}\n$$\nIf $xyz = t^3 \\ge 1$, set $x = tx_1, y = ty_1$, and $z = tz_1$. Then $x_1, y_1$, and $z_1$ are real numbers with $x_1y_1z_1 = 1$. Note that\n$$\n\\frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2} = \\frac{x_1^2 + y_1^2 + z_1^2}{x_1^5 t^3 + y_1^2 + z_1^2} \\le \\frac{x_1^2 + y_1^2 + z_1^2}{x_1^5 + y_1^2 + z_1^2}\n$$\nWe may further assume $xyz = 1$ for inequality (†). We establish inequality (†) in a very mechanical way. Multiplying both sides of the inequality by\n$$\n(x^5 + y^2 + z^2)(y^5 + z^2 + x^2)(z^5 + x^2 + y^2)\n$$\nand canceling the like terms reduces the desired inequality to\n$$\n\\sum_{\\text{sym}} (x^9 + x^3 + 4x^7y^5) \\ge \\sum_{\\text{sym}} (x^6 + x^3y^3 + 2x^5y^4 + 2x^4y^2). \\quad (\\dagger')\n$$\nBy the AM-GM Inequality, we have $x^6+y^6 \\ge 2x^3y^3$ and $x^9+x^3 \\ge 2x^6$ and their symmetric analogous forms. Adding them together shows that\n$$\n\\sum_{\\text{sym}} (x^9 + x^3) \\ge \\sum_{\\text{sym}} (x^6 + x^3 y^3). \\quad (*)\n$$\nIt suffices to show that\n$$\n\\sum_{\\text{sym}} x^7 y^5 \\ge \\sum_{\\text{sym}} x^5 y^4 = \\sum_{\\text{sym}} x^6 y^5 z \\quad (**)\n$$\nand\n$$\n\\sum_{\\text{sym}} x^7 y^5 \\ge \\sum_{\\text{sym}} x^4 y^2 = \\sum_{\\text{sym}} x^6 y^4 z^2. \\quad (***)\n$$\n(Based on work by Jiayin Kang from China) We shall prove something more, namely that\n$$\n\\frac{x^5}{x^5 + y^2 + z^2} + \\frac{y^5}{y^5 + z^2 + x^2} + \\frac{z^5}{z^5 + x^2 + y^2} \\geq 1 \\quad (\\ddag)\n$$\nand\n$$\n1 \\geq \\frac{x^2}{x^5 + y^2 + z^2} + \\frac{y^2}{y^5 + z^2 + x^2} + \\frac{z^2}{z^5 + x^2 + y^2}. \\quad (\\ddag')\n$$\nNote that (‡) follows from adding\n$$\n\\frac{x^5}{x^5 + y^2 + z^2} \\geq \\frac{x^4}{x^4 + y^4 + z^4} \\quad (\\text{or } x(y^4 + z^4) \\geq y^2 + z^2)\n$$\nwith its analogous cyclic inequalities. By the AM-GM Inequality, we have\n$$\n2x(y^4 + z^4) \\geq x(y^2 + z^2)^2 \\geq 2xyz(y^2 + z^2) \\geq 2(y^2 + z^2),\n$$\nas desired.\n(Based on work by Hyun Soo Kim) We present a third approach of establishing inequality (†). Because $y^2 + z^2 \\geq 2yz$ and $xyz \\geq 1$, we have\n$$\n\\frac{1}{x^5 + y^2 + z^2} \\leq \\frac{1}{\\frac{x^4}{yz} + y^2 + z^2} \\leq \\frac{1}{\\frac{2x^4}{y^2+z^2} + y^2 + z^2}\n$$\nand the cyclic analogous forms. Thus it suffices to show that\n$$\n\\frac{x^2 + y^2 + z^2}{\\frac{2x^4}{y^2+z^2} + y^2 + z^2} + \\frac{x^2 + y^2 + z^2}{\\frac{2y^4}{z^2+x^2} + z^2 + x^2} + \\frac{x^2 + y^2 + z^2}{\\frac{2z^4}{x^2+y^2} + x^2 + y^2} \\leq 3.\n$$\nHowever, since this is a homogeneous inequality, the condition $xyz \\geq 1$ is not relevant anymore.\nFurthermore, we may assume that $x^2 + y^2 + z^2 = 3$. Then the inequality reduces to\n$$\n\\frac{1}{\\frac{2x^4}{3-x^2} + 3 - x^2} + \\frac{1}{\\frac{2y^4}{3-y^2} + 3 - y^2} + \\frac{1}{\\frac{2z^4}{3-z^2} + 3 - z^2} \\leq 1,\n$$\nor\n$$\n\\frac{3 - x^2}{3x^4 - 6x^2 + 9} + \\frac{3 - y^2}{3y^4 - 6y^2 + 9} + \\frac{3 - z^2}{3z^4 - 6z^2 + 9} \\leq 1,\n$$\nwhere $x, y$, and $z$ are positive real numbers with $x^2 + y^2 + z^2 = 3$.\nBecause $3x^4 - 6x^2 + 9 = 3(x^2 - 1)^2 + 6 \\geq 6$ and $3 - x^2 = y^2 + z^2 \\geq 0$, we obtain\n$$\n\\frac{3 - x^2}{3x^4 - 6x^2 + 9} \\leq \\frac{3 - x^2}{6}.\n$$\nAdding the above inequality and the cyclic analogous forms gives\n$$\n\\frac{3 - x^2}{3x^4 - 6x^2 + 9} + \\frac{3 - y^2}{3y^4 - 6y^2 + 9} + \\frac{3 - z^2}{3z^4 - 6z^2 + 9} \\\\ \\leq \\frac{9 - (x^2 + y^2 + z^2)}{6} = 1,\n$$\nas desired.\n(Based on work by Xuancheng Shao from China) We claim that\n$$\n\\frac{1}{x^5 + y^2 + z^2} \\le \\frac{\\frac{3}{2} \\cdot (y^2 + z^2)}{(x^2 + y^2 + z^2)^2}.\n$$\nAdding the above inequality and the cyclic analogous forms yields the desired inequality (†).\nBecause $xyz \\ge 1$, $x \\ge \\frac{1}{yz}$, and so\n$$\n\\frac{1}{x^5 + y^2 + z^2} \\le \\frac{1}{\\frac{x^4}{yz} + y^2 + z^2},\n$$\nor\n$$\n\\frac{1}{x^5 + y^2 + z^2} \\le \\frac{yz}{x^4 + yz(y^2 + z^2)}.\n$$\nIt suffices to show that\n$$\n\\frac{2yz}{x^4 + yz(y^2 + z^2)} \\le \\frac{3(y^2 + z^2)}{(x^2 + y^2 + z^2)^2},\n$$\nor\n$$\n2yz(x^2 + y^2 + z^2)^2 \\le 3x^4(y^2 + z^2) + 3yz(y^2 + z^2)^2.\n$$\nExpanding the left-hand side of the last inequality in $x^2$ and $y^2 + z^2$ gives\n$$\n2x^4yz + 4x^2yz(y^2 + z^2) \\le 3x^4(y^2 + z^2) + yz(y^2 + z^2)^2.\n$$\nBecause $3x^4(y^2 + z^2) \\ge 6x^4yz$, it suffices to show that\n$$\n4x^2yz(y^2 + z^2) \\le 4x^4yz + yz(y^2 + z^2)^2,\n$$\nwhich is evident by the AM-GM Inequality.\n(Based on work by Iurie Boreico from Moldova) Note that\n$$\n\\frac{x^5 - x^2}{x^5 + y^2 + z^2} \\ge \\frac{x^5 - x^2}{x^3(x^2 + y^2 + z^2)}\n$$\nis equivalent to\n$$\n\\frac{(x^3 - 1)^2(y^2 + z^2)}{x(x^5 + y^2 + z^2)(x^2 + y^2 + z^2)} \\ge 0,\n$$\nwhich is true for all positive $x, y, z$. Hence\n$$\n\\frac{x^5 - x^2}{x^5 + y^2 + z^2} \\ge \\frac{x^2 - \\frac{1}{x}}{x^2 + y^2 + z^2}.\n$$\nSumming the above inequality with its analogous cyclic inequalities, we see that the desired result follows from\n$$\nx^2 + y^2 + z^2 - \\frac{1}{x} - \\frac{1}{y} - \\frac{1}{z} \\ge 0.\n$$\nSince $xyz \\ge 1$,\n$$\n\\begin{aligned}\n& x^2 + y^2 + z^2 - \\frac{1}{x} - \\frac{1}{y} - \\frac{1}{z} \\\\\n= & x^2 + y^2 + z^2 - \\frac{yz + xz + xy}{xyz} \\\\\n\\ge & x^2 + x^2 + z^2 - yz - xz - xy \\\\\n= & \\frac{(x-y)^2 + (y-z)^2 + (z-x)^2}{2} \\ge 0,\n\\end{aligned}\n$$\nso we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum\n$$\n\\frac{2^{1}}{4^{1}-1}+\\frac{2^{2}}{4^{2}-1}+\\frac{2^{4}}{4^{4}-1}+\\frac{2^{8}}{4^{8}-1}+\\cdots .\n$$",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nNotice that\n$$\n\\frac{2^{2^{k}}}{4^{2^{k}}-1}=\\frac{2^{2^{k}}+1}{4^{2^{k}}-1}-\\frac{1}{4^{2^{k}}-1}=\\frac{1}{2^{2^{k}}-1}-\\frac{1}{4^{2^{k}}-1}=\\frac{1}{4^{2^{k-1}}-1}-\\frac{1}{4^{2^{k}}-1} .\n$$\nTherefore, the sum telescopes as\n$$\n\\left(\\frac{1}{4^{2^{-1}}-1}-\\frac{1}{4^{2^{0}}-1}\\right)+\\left(\\frac{1}{4^{2^{0}}-1}-\\frac{1}{4^{2^{1}}-1}\\right)+\\left(\\frac{1}{4^{2^{1}}-1}-\\frac{1}{4^{2^{2}}-1}\\right)+\\cdots\n$$\nand evaluates to $1 /(4^{2^{-1}}-1)=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a regular tetrahedron with side length $2$. The plane parallel to edges $AB$ and $CD$ and lying halfway between them cuts $ABCD$ into two pieces. Find the surface area of one of these pieces.",
    "options": [],
    "answer": "1 + 2*sqrt(3)",
    "solution": "Solution:\n\nThe plane intersects each face of the tetrahedron in a midline of the face; by symmetry it follows that the intersection of the plane with the tetrahedron is a square of side length $1$. The surface area of each piece is half the total surface area of the tetrahedron plus the area of the square, that is,\n$$\n\\frac{1}{2} \\cdot 4 \\cdot \\frac{2^{2} \\sqrt{3}}{4} + 1 = 1 + 2\\sqrt{3}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $f: \\mathbf{R} \\rightarrow \\mathbf{R}$ be a continuous function with $\\int_{0}^{1} f(x) f^{\\prime}(x) d x=0$ and $\\int_{0}^{1} f(x)^{2} f^{\\prime}(x) d x=18$. What is $\\int_{0}^{1} f(x)^{4} f^{\\prime}(x) d x$?",
    "options": [],
    "answer": "486/5",
    "solution": "Solution:\n\n$0=\\int_{0}^{1} f(x) f^{\\prime}(x) d x=\\int_{f(0)}^{f(1)} u d u=\\frac{1}{2}\\left(f(1)^{2}-f(0)^{2}\\right)$, and\n\n$18=\\int_{0}^{1} f(x)^{2} f^{\\prime}(x) d x=\\int_{f(0)}^{f(1)} u^{2} d u=\\frac{1}{3}\\left(f(1)^{3}-f(0)^{3}\\right)$.\n\nThe first equation implies $f(0)= \\pm f(1)$. The second equation shows that $f(0) \\neq f(1)$, and in fact $54=f(1)^{3}-f(0)^{3}=2 f(1)^{3}$, so $f(1)=3$ and $f(0)=-3$. Then\n\n$\\int_{0}^{1} f(x)^{4} f^{\\prime}(x) d x=\\int_{f(0)}^{f(1)} u^{4} d u=\\frac{1}{5}\\left(f(1)^{5}-f(0)^{5}\\right)=\\frac{1}{5}(243+243)=\\frac{486}{5}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWorking together, Jack and Jill can paint a house in 3 days; Jill and Joe can paint the same house in 4 days; or Joe and Jack can paint the house in 6 days. If Jill, Joe, and Jack all work together, how many days will it take them?",
    "options": [],
    "answer": "8/3 days",
    "solution": "Solution:\n\nSuppose that Jack paints $x$ houses per day, Jill paints $y$ houses per day, and Joe paints $z$ houses per day. Together, Jack and Jill paint $1 / 3$ of a house in a day - that is,\n$$\nx+y=1 / 3 .\n$$\nSimilarly,\n$$\ny+z=1 / 4\n$$\nand\n$$\nz+x=1 / 6\n$$\nAdding all three equations and dividing by 2 gives\n$$\nx+y+z=3 / 8\n$$\nSo, working together, the three folks can paint $3 / 8$ houses in a day, or $8 / 3$ days per house.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe Red Sox play the Yankees in a best-of-seven series that ends as soon as one team wins four games. Suppose that the probability that the Red Sox win Game $n$ is $\\frac{n-1}{6}$. What is the probability that the Red Sox will win the series?",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nNote that if we imagine that the series always continues to seven games even after one team has won four, this will never change the winner of the series. Notice also that the probability that the Red Sox will win Game $n$ is precisely the probability that the Yankees will win Game $8-n$. Therefore, the probability that the Yankees win at least four games is the same as the probability that the Red Sox win at least four games, namely $1/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$, $y$, and $z$ be distinct real numbers that sum to $0$. Find the maximum possible value of\n$$\n\\frac{x y + y z + z x}{x^{2} + y^{2} + z^{2}}.\n$$",
    "options": [],
    "answer": "-1/2",
    "solution": "Solution:\n$-1/2$\n\nNote that $0 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2 x y + 2 y z + 2 z x$. Rearranging, we get that $x y + y z + z x = -\\frac{1}{2}(x^2 + y^2 + z^2)$, so that in fact the quantity is always equal to $-1/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a rectangle with area $1$, and let $E$ lie on side $CD$. What is the area of the triangle formed by the centroids of triangles $ABE$, $BCE$, and $ADE$?",
    "options": [],
    "answer": "1/9",
    "solution": "Solution:\nLet the centroids of $ABE$, $BCE$, and $ADE$ be denoted by $X$, $Y$, and $Z$, respectively. Let $d(P, QR)$ denote the distance from $P$ to line $QR$. Since the centroid lies two-thirds of the distance from each vertex to the midpoint of the opposite edge, $d(X, AB) = d(Y, CD) = d(Z, CD) = \\frac{1}{3} BC$, so $YZ$ is parallel to $CD$ and $d(X, YZ) = BC - \\frac{2}{3} BC = \\frac{1}{3} BC$.\n\nLikewise, $d(Z, AD) = \\frac{1}{3} DE$ and $d(Y, BC) = \\frac{1}{3} CE$, so that since $YZ$ is perpendicular to $AD$ and $BC$, we have that $YZ = CD - \\frac{1}{3}(DE + CE) = \\frac{2}{3} CD$. Therefore, the area of $XYZ$ is $\\frac{1}{2}\\left(\\frac{1}{3} BC\\right)\\left(\\frac{2}{3} CD\\right) = \\frac{1}{9} BC \\cdot CD = \\frac{1}{9}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow that no rectangle of the form $1 \\times k$ or $2 \\times n$, where $4 \\nmid n$, is $(1,2)$-tileable.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe claim is obvious for $1 \\times k$ rectangles. For the others, color the first two columns black, the next two white, the next two black, etc. Each $(1,2)$ domino will contain one square of each color, so in order to be tileable, the rectangle must contain the same number of black and white squares. This is the case only when $4 \\mid n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cuboctahedron is a polyhedron whose faces are squares and equilateral triangles such that two squares and two triangles alternate around each vertex, as shown.\n![](attached_image_1.png)\nWhat is the volume of a cuboctahedron of side length $1$?",
    "options": [],
    "answer": "5 sqrt(2) / 3",
    "solution": "Solution: $5 \\sqrt{2} / 3$\nWe can construct a cube such that the vertices of the cuboctahedron are the midpoints of the edges of the cube.\n![](attached_image_2.png)\nLet $s$ be the side length of this cube. Now, the cuboctahedron is obtained from the cube by cutting a tetrahedron from each corner. Each such tetrahedron has a base in the form of an isosceles right triangle of area $(s / 2)^{2} / 2$ and height $s / 2$ for a volume of $(s / 2)^{3} / 6$. The total volume of the cuboctahedron is therefore\n$$\ns^{3}-8 \\cdot(s / 2)^{3} / 6=5 s^{3} / 6\n$$\nNow, the side of the cuboctahedron is the hypotenuse of an isosceles right triangle of leg $s / 2$; thus $1=(s / 2) \\sqrt{2}$, giving $s=\\sqrt{2}$, so the volume of the cuboctahedron is $5 \\sqrt{2} / 3$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe L shape made by adjoining three congruent squares can be subdivided into four smaller L shapes.\n![](attached_image_1.png)\nEach of these can in turn be subdivided, and so forth. If we perform 2005 successive subdivisions, how many of the $4^{2005}$ L's left at the end will be in the same orientation as the original one?",
    "options": [],
    "answer": "4^{2004} + 2^{2004}",
    "solution": "Solution:\n$4^{2004} + 2^{2004}$\n\nAfter $n$ successive subdivisions, let $a_{n}$ be the number of small L's in the same orientation as the original one; let $b_{n}$ be the number of small L's that have this orientation rotated counterclockwise $90^{\\circ}$; let $c_{n}$ be the number of small L's that are rotated $180^{\\circ}$; and let $d_{n}$ be the number of small L's that are rotated $270^{\\circ}$. When an L is subdivided, it produces two smaller L's of the same orientation, one of each of the neighboring orientations, and none of the opposite orientation. Therefore,\n$$(a_{n+1}, b_{n+1}, c_{n+1}, d_{n+1}) = (d_{n} + 2a_{n} + b_{n},\\ a_{n} + 2b_{n} + c_{n},\\ b_{n} + 2c_{n} + d_{n},\\ c_{n} + 2d_{n} + a_{n})$$\nIt is now straightforward to show by induction that\n$$\n(a_{n}, b_{n}, c_{n}, d_{n}) = (4^{n-1} + 2^{n-1},\\ 4^{n-1},\\ 4^{n-1} - 2^{n-1},\\ 4^{n-1})\n$$\nfor each $n \\geq 1$. In particular, our desired answer is $a_{2005} = 4^{2004} + 2^{2004}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1} = 3$, and for $n \\geq 1$, let $a_{n+1} = (n+1) a_{n} - n$. Find the smallest $m \\geq 2005$ such that $a_{m+1} - 1 \\mid a_{m}^{2} - 1$.",
    "options": [],
    "answer": "2010",
    "solution": "Solution:\nWe will show that $a_{n} = 2 \\cdot n! + 1$ by induction. Indeed, the claim is obvious for $n = 1$, and $(n+1)(2 \\cdot n! + 1) - n = 2 \\cdot (n+1)! + 1$.\n\nThen we wish to find $m \\geq 2005$ such that $2(m+1)! \\mid 4(m!)^{2} + 4 m!$, or dividing by $2 \\cdot m!$, we want $m+1 \\mid 2(m! + 1)$.\n\nSuppose $m+1$ is composite. Then it has a proper divisor $d > 2$, and since $d \\mid m!$, we must have $d \\mid 2$, which is impossible. Therefore, $m+1$ must be prime, and if this is the case, then $m+1 \\mid m! + 1$ by Wilson's Theorem.\n\nTherefore, since the smallest prime greater than $2005$ is $2011$, the smallest possible value of $m$ is $2010$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriangle $ABC$ has incircle $\\omega$ which touches $AB$ at $C_1$, $BC$ at $A_1$, and $CA$ at $B_1$. Let $A_2$ be the reflection of $A_1$ over the midpoint of $BC$, and define $B_2$ and $C_2$ similarly. Let $A_3$ be the intersection of $AA_2$ with $\\omega$ that is closer to $A$, and define $B_3$ and $C_3$ similarly. If $AB=9$, $BC=10$, and $CA=13$, find $\\left[A_3 B_3 C_3\\right] / [ABC]$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)",
    "options": [],
    "answer": "14/65",
    "solution": "Solution:\n\nNotice that $A_2$ is the point of tangency of the excircle opposite $A$ to $BC$. Therefore, by considering the homothety centered at $A$ taking the excircle to the incircle, we notice that $A_3$ is the intersection of $\\omega$ and the tangent line parallel to $BC$. It follows that $A_1 B_1 C_1$ is congruent to $A_3 B_3 C_3$ by reflecting through the center of $\\omega$. We therefore need only find $\\left[A_1 B_1 C_1\\right] / [ABC]$. Since\n$$\n\\frac{\\left[A_1 BC_1\\right]}{[ABC]} = \\frac{A_1 B \\cdot BC_1}{AB \\cdot BC} = \\frac{((9+10-13)/2)^2}{9 \\cdot 10} = \\frac{1}{10}\n$$\nand likewise $\\left[A_1 B_1 C\\right] / [ABC] = 49/130$ and $\\left[AB_1 C_1\\right] / [ABC] = 4/13$, we get that\n$$\n\\frac{\\left[A_3 B_3 C_3\\right]}{[ABC]} = 1 - \\frac{1}{10} - \\frac{49}{130} - \\frac{4}{13} = \\frac{14}{65}.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA regular octahedron $A B C D E F$ is given such that $A D$, $B E$, and $C F$ are perpendicular. Let $G$, $H$, and $I$ lie on edges $A B$, $B C$, and $C A$ respectively such that $\\frac{A G}{G B}=\\frac{B H}{H C}=\\frac{C I}{I A}=\\rho$. For some choice of $\\rho>1$, $G H$, $H I$, and $I G$ are three edges of a regular icosahedron, eight of whose faces are inscribed in the faces of $A B C D E F$. Find $\\rho$.",
    "options": [],
    "answer": "(1+sqrt(5))/2",
    "solution": "Solution:\nLet $J$ lie on edge $C E$ such that $\\frac{E J}{J C}=\\rho$. Then we must have that $H I J$ is another face of the icosahedron, so in particular, $H I=H J$. But since $B C$ and $C E$ are perpendicular, $H J=H C \\sqrt{2}$. By the Law of Cosines, $H I^{2}=H C^{2}+C I^{2}-2 H C \\cdot C I \\cos 60^{\\circ}=H C^{2}\\left(1+\\rho^{2}-\\rho\\right)$. Therefore, $2=1+\\rho^{2}-\\rho$, or $\\rho^{2}-\\rho-1=0$, giving $\\rho=\\frac{1+\\sqrt{5}}{2}$.\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p=2^{24036583}-1$, the largest prime currently known. For how many positive integers $c$ do the quadratics $\\pm x^{2} \\pm p x \\pm c$ all have rational roots?",
    "options": [],
    "answer": "0",
    "solution": "Solution: 0\nThis is equivalent to both discriminants $p^{2} \\pm 4 c$ being squares. In other words, $p^{2}$ must be the average of two squares $a^{2}$ and $b^{2}$. Note that $a$ and $b$ must have the same parity, and that $\\left(\\frac{a+b}{2}\\right)^{2}+\\left(\\frac{a-b}{2}\\right)^{2}=\\frac{a^{2}+b^{2}}{2}=p^{2}$. Therefore, $p$ must be the hypotenuse in a Pythagorean triple. Such triples are parametrized by $k\\left(m^{2}-n^{2}, 2 m n, m^{2}+n^{2}\\right)$. But $p \\equiv 3(\\bmod 4)$ and is therefore not the sum of two squares. This implies that $p$ is not the hypotenuse of any Pythagorean triple, so the answer is 0 .",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOne hundred people are in line to see a movie. Each person wants to sit in the front row, which contains one hundred seats, and each has a favorite seat, chosen randomly and independently. They enter the row one at a time from the far right. As they walk, if they reach their favorite seat, they sit, but to avoid stepping over people, if they encounter a person already seated, they sit to that person's right. If the seat furthest to the right is already taken, they sit in a different row. What is the most likely number of people that will get to sit in the first row?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nLet $S(i)$ be the favorite seat of the $i$th person, counting from the right. Let $P(n)$ be the probability that at least $n$ people get to sit. At least $n$ people sit if and only if $S(1) \\geq n, S(2) \\geq n-1, \\ldots, S(n) \\geq 1$. This has probability:\n$$\nP(n) = \\frac{100-(n-1)}{100} \\cdot \\frac{100-(n-2)}{100} \\cdots \\frac{100}{100} = \\frac{100!}{(100-n)! \\cdot 100^{n}}\n$$\nThe probability, $Q(n)$, that exactly $n$ people sit is\n$$\nQ(n) = P(n) - P(n+1) = \\frac{100!}{(100-n)! \\cdot 100^{n}} - \\frac{100!}{(99-n)! \\cdot 100^{n+1}} = \\frac{100! \\cdot n}{(100-n)! \\cdot 100^{n+1}}\n$$\nNow,\n$$\n\\frac{Q(n)}{Q(n-1)} = \\frac{100! \\cdot n}{(100-n)! \\cdot 100^{n+1}} \\cdot \\frac{(101-n)! \\cdot 100^{n}}{100! \\cdot (n-1)} = \\frac{n(101-n)}{100(n-1)} = \\frac{101 n - n^{2}}{100 n - 100},\n$$\nwhich is greater than $1$ exactly when $n^{2} - n - 100 < 0$, that is, for $n \\leq 10$. Therefore, the maximum value of $Q(n)$ occurs for $n = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, \\ldots, a_{2005}$ be real numbers such that\n$$\n\\begin{array}{ccccccccccc}\na_{1} \\cdot 1 & + & a_{2} \\cdot 2 & + & a_{3} \\cdot 3 & + & \\cdots & + & a_{2005} \\cdot 2005 & = & 0 \\\\\na_{1} \\cdot 1^{2} & + & a_{2} \\cdot 2^{2} & + & a_{3} \\cdot 3^{2} & + & \\cdots & + & a_{2005} \\cdot 2005^{2} & = & 0 \\\\\na_{1} \\cdot 1^{3} & + & a_{2} \\cdot 2^{3} & + & a_{3} \\cdot 3^{3} & + & \\cdots & + & a_{2005} \\cdot 2005^{3} & = & 0 \\\\\n\\vdots & & \\vdots & & \\vdots & & & & \\vdots & & \\vdots \\\\\na_{1} \\cdot 1^{2004} & + & a_{2} \\cdot 2^{2004} & + & a_{3} \\cdot 3^{2004} & + & \\cdots & + & a_{2005} \\cdot 2005^{2004} & = & 0\n\\end{array}\n$$\nand\n$$\na_{1} \\cdot 1^{2005}+a_{2} \\cdot 2^{2005}+a_{3} \\cdot 3^{2005}+\\cdots+a_{2005} \\cdot 2005^{2005}=1 .\n$$",
    "options": [],
    "answer": "1/2004!",
    "solution": "Solution:\n$1 / 2004!$\n\nThe polynomial $p(x) = x(x-2)(x-3) \\cdots (x-2005) / 2004!$ has zero constant term, has the numbers $2, 3, \\ldots, 2005$ as roots, and satisfies $p(1) = 1$. Multiplying the $n$th equation by the coefficient of $x^{n}$ in the polynomial $p(x)$ and summing over all $n$ gives\n$$\na_{1} p(1) + a_{2} p(2) + a_{3} p(3) + \\cdots + a_{2005} p(2005) = 1 / 2004!\n$$\n(since the leading coefficient is $1 / 2004!$). The left side just reduces to $a_{1}$, so $1 / 2004!$ is the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn how many ways can the set of ordered pairs of integers be colored red and blue such that for all $a$ and $b$, the points $(a, b)$, $(-1-b, a+1)$, and $(1-b, a-1)$ are all the same color?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\nLet $\\varphi_{1}$ and $\\varphi_{2}$ be $90^{\\circ}$ counterclockwise rotations about $(-1,0)$ and $(1,0)$, respectively. Then $\\varphi_{1}(a, b)=(-1-b, a+1)$, and $\\varphi_{2}(a, b)=(1-b, a-1)$. Therefore, the possible colorings are precisely those preserved under these rotations. Since $\\varphi_{1}(1,0)=(-1,2)$, the colorings must also be preserved under $90^{\\circ}$ rotations about $(-1,2)$. Similarly, one can show that they must be preserved under rotations about any point $(x, y)$, where $x$ is odd and $y$ is even. Decompose the lattice points as follows:\n$$\n\\begin{aligned}\nL_{1} & =\\{(x, y) \\mid x+y \\equiv 0 \\quad(\\bmod 2)\\} \\\\\nL_{2} & =\\{(x, y) \\mid x \\equiv y-1 \\equiv 0 \\quad(\\bmod 2)\\} \\\\\nL_{3} & =\\{(x, y) \\mid x+y-1 \\equiv y-x+1 \\equiv 0 \\quad(\\bmod 4)\\} \\\\\nL_{4} & =\\{(x, y) \\mid x+y+1 \\equiv y-x-1 \\equiv 0 \\quad(\\bmod 4)\\}\n\\end{aligned}\n$$\nWithin any of these sublattices, any point can be brought to any other through appropriate rotations, but no point can be brought to any point in a different sublattice. It follows that every sublattice must be colored in one color, but that different sublattices can be colored differently. Since each of these sublattices can be colored in one of two colors, there are $2^{4}=16$ possible colorings.\n\n| 1 | 2 | 1 | 2 | 1 | 2 | 1 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 4 | 1 | 3 | 1 | 4 | 1 | 3 |\n| 1 | 2 | 1 | 2 | 1 | 2 | 1 |\n| 3 | 1 | 4 | 1 | 3 | 1 | 4 |\n| 1 | 2 | 1 | 2 | 1 | 2 | 1 |\n| 4 | 1 | 3 | 1 | 4 | 1 | 3 |\n| 1 | 2 | 1 | 2 | 1 | 2 | 1 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many regions of the plane are bounded by the graph of\n$$\nx^{6}-x^{5}+3 x^{4} y^{2}+10 x^{3} y^{2}+3 x^{2} y^{4}-5 x y^{4}+y^{6}=0 ?\n$$",
    "options": [],
    "answer": "5",
    "solution": "Solution: 5\nThe left-hand side decomposes as\n$$\n\\left(x^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}+y^{6}\\right)-\\left(x^{5}-10 x^{3} y^{2}+5 x y^{4}\\right)=\\left(x^{2}+y^{2}\\right)^{3}-\\left(x^{5}-10 x^{3} y^{2}+5 x y^{4}\\right) .\n$$\nNow, note that\n$$\n(x+i y)^{5}=x^{5}+5 i x^{4} y-10 x^{3} y^{2}-10 i x^{2} y^{3}+5 x y^{4}+i y^{5}\n$$\nso that our function is just $\\left(x^{2}+y^{2}\\right)^{3}-\\Re\\left((x+i y)^{5}\\right)$. Switching to polar coordinates, this is $r^{6}-Re\\left(r^{5}(\\cos \\theta+i \\sin \\theta)^{5}\\right)=r^{6}-r^{5} \\cos 5 \\theta$ by de Moivre's rule. The graph of our function is then the graph of $r^{6}-r^{5} \\cos 5 \\theta=0$, or, more suitably, of $r=\\cos 5 \\theta$. This is a five-petal rose, so the answer is 5 .\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 47076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence $a_1, a_2, \\dots$ defined by\n$$\na_n = 2^n + 3^n + 6^n - 1.\n$$\nfor all positive integers $n$. Determine all positive integers that are relatively prime to every term of the sequence.",
    "options": [],
    "answer": "1",
    "solution": "The answer is that $1$ is the only such number. It suffices to show that every prime $p$ divides $a_n$ for some positive integer $n$. Note that both $p=2$ and $p=3$ divide $a_2 = 2^2 + 3^2 + 6^2 - 1 = 48$.\n\nNow we assume that $p \\ge 5$. By **Fermat's Little Theorem**, we have $2^{p-1} \\equiv 3^{p-1} \\equiv 6^{p-1} \\equiv 1 \\pmod{p}$. Then\n$$\n3 \\cdot 2^{p-1} + 2 \\cdot 3^{p-1} + 6^{p-1} \\equiv 3 + 2 + 1 \\equiv 6 \\pmod{p},\n$$\nor, $6(2^{p-2} + 3^{p-2} + 6^{p-2} - 1) \\equiv 0 \\pmod{p}$; that is, $6a_{p-2}$ is divisible by $p$. Because $p$ is relatively prime to $6$, $a_{p-2}$ is divisible by $p$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the probability that in a randomly chosen arrangement of the numbers and letters in \"HMMT2005,\" one can read either \"HMMT\" or \"2005\" from left to right? (For example, in \"5HM0M20T,\" one can read \"HMMT.\")",
    "options": [],
    "answer": "23/144",
    "solution": "Solution:\n$\\frac{23}{144}$\nTo read \"HMMT,\" there are $\\binom{8}{4}$ ways to place the letters, and $\\frac{4!}{2}$ ways to place the numbers. Similarly, there are $\\binom{8}{4} \\frac{4!}{2}$ arrangements where one can read \"2005.\" The number of arrangements in which one can read both is just $\\binom{8}{4}$. The total number of arrangements is $\\frac{8!}{4}$, thus the answer is\n$$\n\\frac{\\binom{8}{4} \\frac{4!}{2} + \\binom{8}{4} \\frac{4!}{2} - \\binom{8}{4}}{\\frac{8!}{4}} = \\binom{8}{4} \\frac{4}{8!} \\cdot 23 = \\frac{23}{144}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real solutions to $x^{3} + (x+1)^{3} + (x+2)^{3} = (x+3)^{3}$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nThe equation simplifies to $3x^{3} + 9x^{2} + 15x + 9 = x^{3} + 9x^{2} + 27x + 27$, or equivalently, $2x^{3} - 12x - 18 = 2(x-3)(x^{2} + 3x + 3) = 0$. The discriminant of $x^{2} + 3x + 3$ is $-3 < 0$, so the only real solution is $x = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a, b, c > 0$, what is the smallest possible value of $\\left\\lfloor\\frac{a+b}{c}\\right\\rfloor + \\left\\lfloor\\frac{b+c}{a}\\right\\rfloor + \\left\\lfloor\\frac{c+a}{b}\\right\\rfloor$? (Note that $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.)",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nSince $\\lfloor x \\rfloor > x - 1$ for all $x$, we have that\n$$\n\\begin{aligned}\n\\left\\lfloor\\frac{a+b}{c}\\right\\rfloor + \\left\\lfloor\\frac{b+c}{a}\\right\\rfloor + \\left\\lfloor\\frac{c+a}{b}\\right\\rfloor &> \\frac{a+b}{c} + \\frac{b+c}{a} + \\frac{c+a}{b} - 3 \\\\\n&= \\left(\\frac{a}{b} + \\frac{b}{a}\\right) + \\left(\\frac{b}{c} + \\frac{c}{b}\\right) + \\left(\\frac{c}{a} + \\frac{a}{c}\\right) - 3\n\\end{aligned}\n$$\nBut by the AM-GM inequality, each of the first three terms in the last line is at least $2$. Therefore, the left-hand side is greater than $2 + 2 + 2 - 3 = 3$. Since it is an integer, the smallest value it can be is $4$. This is in fact attainable by letting $(a, b, c) = (6, 8, 9)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f: \\mathbf{R} \\rightarrow \\mathbf{R}$ be a smooth function such that $f'(x)^2 = f(x) f''(x)$ for all $x$. Suppose $f(0) = 1$ and $f^{(4)}(0) = 9$. Find all possible values of $f'(0)$.",
    "options": [],
    "answer": "± sqrt(3)",
    "solution": "Solution: $\\pm \\sqrt{3}$\nLet $f'(0) = a$. Then the equation gives $f''(0) = a^2$. Differentiating the given equation gives\n$$\n2 f'(x) f''(x) = f(x) f'''(x) + f'(x) f''(x)\n$$\nor $f'(x) f''(x) = f(x) f'''(x)$. Differentiating once more gives\n$$\nf'(x) f'''(x) + f''(x)^2 = f(x) f^{(4)}(x) + f'(x) f'''(x)\n$$\nor $f''(x)^2 = f(x) f^{(4)}(x)$, giving $9 = f^{(4)}(0) = a^4$. Thus $a = \\pm \\sqrt{3}$. These are indeed both attainable by $f(x) = e^{\\pm x \\sqrt{3}}$.\n\nAlternative Solution: Rewrite the given equation as $\\frac{f''(x)}{f'(x)} = \\frac{f'(x)}{f(x)}$. Integrating both sides gives $\\ln f'(x) = \\ln f(x) + C_1$, and exponentiating gives $f'(x) = C f(x)$. This has solution $f(x) = A e^{C x}$ for constants $A$ and $C$. Since $f(0) = 1$, $A = 1$, and differentiating we find that $C^4 = f^{(4)}(0) = 9$, yielding $f'(0) = C = \\pm \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $X Y Z$ be a triangle with $\\angle X = 60^{\\circ}$ and $\\angle Y = 45^{\\circ}$. A circle with center $P$ passes through points $A$ and $B$ on side $X Y$, $C$ and $D$ on side $Y Z$, and $E$ and $F$ on side $Z X$. Suppose $A B = C D = E F$. Find $\\angle X P Y$ in degrees.",
    "options": [],
    "answer": "255/2 degrees",
    "solution": "Solution:\n\nSince $P A B$, $P C D$, and $P E F$ are all isosceles triangles with equal legs and equal bases, they are congruent. It follows that the heights of each are the same, so that $P$ is equidistant from the sides of $X Y Z$. Therefore, $P$ is the incenter and therefore lies on the angle bisectors of $X Y Z$. Thus $\\angle Y X P = \\frac{1}{2} \\angle Y X Z = 30^{\\circ}$ and $\\angle P Y X = \\frac{1}{2} \\angle Z Y X = \\frac{45^{\\circ}}{2}$. Therefore $\\angle X P Y = 180^{\\circ} - 30^{\\circ} - \\frac{45^{\\circ}}{2} = \\frac{255^{\\circ}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn how many ways can $8$ people be arranged in a line if Alice and Bob must be next to each other, and Carol must be somewhere behind Dan?",
    "options": [],
    "answer": "5040",
    "solution": "Solution:\n\nLet us place Alice and Bob as a single person; there are then $7! = 5040$ different arrangements. Alice can be in front of Bob or vice versa, multiplying the number of possibilities by $2$, but Carol is behind Dan in exactly half of those, so that the answer is just $5040$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a town of $n$ people, a governing council is elected as follows: each person casts one vote for some person in the town, and anyone that receives at least five votes is elected to council. Let $c(n)$ denote the average number of people elected to council if everyone votes randomly. Find $\\lim_{n \\rightarrow \\infty} c(n) / n$.",
    "options": [],
    "answer": "1 - 65/(24e)",
    "solution": "Solution:\n\n$1 - 65 / 24e$\n\nLet $c_{k}(n)$ denote the expected number of people that will receive exactly $k$ votes. We will show that $\\lim_{n \\rightarrow \\infty} c_{k}(n) / n = 1/(e \\cdot k!)$. The probability that any given person receives exactly $k$ votes, which is the same as the average proportion of people that receive exactly $k$ votes, is\n$$\n\\binom{n}{k} \\cdot \\left(\\frac{1}{n}\\right)^{k} \\cdot \\left(\\frac{n-1}{n}\\right)^{n-k} = \\left(\\frac{n-1}{n}\\right)^{n} \\cdot \\frac{n(n-1) \\cdots (n-k+1)}{k! \\cdot (n-1)^{k}}.\n$$\nTaking the limit as $n \\rightarrow \\infty$ and noting that $\\lim_{n \\rightarrow \\infty} \\left(1-\\frac{1}{n}\\right)^{n} = \\frac{1}{e}$ gives that the limit is $1/(e \\cdot k!)$, as desired. Therefore, the limit of the average proportion of the town that receives at least five votes is\n$$\n1 - \\frac{1}{e}\\left(\\frac{1}{0!} + \\frac{1}{1!} + \\frac{1}{2!} + \\frac{1}{3!} + \\frac{1}{4!}\\right) = 1 - \\frac{65}{24e}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThere are 42 stepping stones in a pond, arranged along a circle. You are standing on one of the stones. You would like to jump among the stones so that you move counterclockwise by either 1 stone or 7 stones at each jump. Moreover, you would like to do this in such a way that you visit each stone (except for the starting spot) exactly once before returning to your initial stone for the first time. In how many ways can you do this?",
    "options": [],
    "answer": "63",
    "solution": "Solution:\nNumber the stones $0, 1, \\ldots, 41$, treating the numbers as values modulo $42$, and let $r_n$ be the length of your jump from stone $n$. If you jump from stone $n$ to $n+7$, then you cannot jump from stone $n+6$ to $n+7$ and so must jump from $n+6$ to $n+13$. That is, if $r_n = 7$, then $r_{n+6} = 7$ also. It follows that the 7 values $r_n, r_{n+6}, r_{n+12}, \\ldots, r_{n+36}$ are all equal: if one of them is $7$, then by the preceding argument applied repeatedly, all of them must be $7$, and otherwise all of them are $1$.\n\nNow, for $n = 0, 1, 2, \\ldots, 42$, let $s_n$ be the stone you are on after $n$ jumps. Then $s_{n+1} = s_n + r_{s_n}$, and we have $s_{n+1} = s_n + r_{s_n} \\equiv s_n + 1 \\pmod{6}$. By induction, $s_{n+i} \\equiv s_n + i \\pmod{6}$; in particular $s_{n+6} \\equiv s_n$, so $r_{s_n+6} = r_{s_n}$. That is, the sequence of jump lengths is periodic with period $6$ and so is uniquely determined by the first $6$ jumps. So this gives us at most $2^6 = 64$ possible sequences of jumps $r_{s_0}, r_{s_1}, \\ldots, r_{s_{41}}$.\n\nNow, the condition that you visit each stone exactly once before returning to the original stone just means that $s_0, s_1, \\ldots, s_{41}$ are distinct and $s_{42} = s_0$. If all jumps are length $7$, then $s_6 = s_0$, so this cannot happen. On the other hand, if the jumps are not all of length $7$, then we claim $s_0, \\ldots, s_{41}$ are indeed all distinct. Indeed, suppose $s_i = s_j$ for some $0 \\leq i < j < 42$. Since $s_j \\equiv s_i + (j-i) \\pmod{6}$, we have $j \\equiv i \\pmod{6}$, so $j-i = 6k$ for some $k$. Moreover, since the sequence of jump lengths has period $6$, we have\n$$\ns_{i+6} - s_i = s_{i+12} - s_{i+6} = \\cdots = s_{i+6k} - s_{i+6(k-1)}\n$$\nCalling this common value $l$, we have $k l \\equiv 0 \\pmod{42}$. But $l$ is divisible by $6$, and $j-i < 42 \\Rightarrow k < 7$ means that $k$ is not divisible by $7$, so $l$ must be. So $l$, the sum of six successive jump lengths, is divisible by $42$. Hence the jumps must all be of length $7$, as claimed.\n\nThis shows that, for the $64-1=63$ sequences of jumps that have period $6$ and are not all of length $7$, you do indeed reach every stone once before returning to the starting point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn how many ways can 6 purple balls and 6 green balls be placed into a $4 \\times 4$ grid of boxes such that every row and column contains two balls of one color and one ball of the other color? Only one ball may be placed in each box, and rotations and reflections of a single configuration are considered different.",
    "options": [],
    "answer": "5184",
    "solution": "Solution:\nIn each row or column, exactly one box is left empty. There are $4! = 24$ ways to choose the empty spots. Once that has been done, there are 6 ways to choose which two rows have 2 purple balls each. Now, assume without loss of generality that boxes $(1,1)$, $(2,2)$, $(3,3)$, and $(4,4)$ are the empty ones, and that rows 1 and 2 have two purple balls each. Let $A, B, C$, and $D$ denote the $2 \\times 2$ squares in the top left, top right, bottom left, and bottom right corners, respectively (so $A$ is formed by the first two rows and first two columns, etc.). Let $a, b, c$, and $d$ denote the number of purple balls in $A, B, C$, and $D$, respectively. Then $0 \\leq a, d \\leq 2$, $a+b=4$, and $b+d \\leq 4$, so $a \\geq d$.\nNow suppose we are given the numbers $a$ and $d$, satisfying $0 \\leq d \\leq a \\leq 2$. Fortunately, the numbers of ways to color the balls in $A, B, C$, and $D$ are independent of each other. For example, given $a=1$ and $d=0$, there are 2 ways to color $A$ and 1 way to color $D$ and, no matter how the coloring of $A$ is done, there are always 2 ways to color $B$ and 3 ways to color $C$. The numbers of ways to choose the colors of all the balls is as follows:\n\n| $a \\backslash d$ | 0 | 1 | 2 |\n| :---: | :---: | :---: | :---: |\n| 0 | $1 \\cdot (1 \\cdot 2) \\cdot 1 = 2$ | 0 | 0 |\n| 1 | $2 \\cdot (2 \\cdot 3) \\cdot 1 = 12$ | $2 \\cdot (1 \\cdot 1) \\cdot 2 = 4$ | 0 |\n| 2 | $1 \\cdot (2 \\cdot 2) \\cdot 1 = 4$ | $1 \\cdot (3 \\cdot 2) \\cdot 2 = 12$ | $1 \\cdot (2 \\cdot 1) \\cdot 1 = 2$ |\n\nIn each square above, the four factors are the number of ways of arranging the balls in $A$, $B$, $C$, and $D$, respectively. Summing this over all pairs $(a, d)$ satisfying $0 \\leq d \\leq a \\leq 2$ gives a total of 36. The answer is therefore $24 \\cdot 6 \\cdot 36 = 5184$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWrite down an integer $N$ between $0$ and $20$ inclusive. If more than $N$ teams write down $N$, your score is $N$; otherwise it is $0$.",
    "options": [],
    "answer": "20",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWrite down a set $S$ of positive integers, all greater than $1$, such that for each $x \\in S$, $x$ is a proper divisor of $(P / x) + 1$, where $P$ is the product of all the elements of $S$. Your score is $2n$, where $n = |S|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind a long binary word containing a small number of square subwords. Specifically, write down a binary word of any length $n \\leq 50$. Your score will be $\\max \\{0, n-s\\}$, where $s$ is the number of occurrences of square subwords. (That is, each different square subword will be counted according to the number of times it appears.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoug and Ryan are competing in the 2005 Wiffle Ball Home Run Derby. In each round, each player takes a series of swings. Each swing results in either a home run or an out, and an out ends the series. When Doug swings, the probability that he will hit a home run is $1/3$. When Ryan swings, the probability that he will hit a home run is $1/2$. In one round, what is the probability that Doug will hit more home runs than Ryan hits?",
    "options": [],
    "answer": "1/5",
    "solution": "Solution:\n\nDenote this probability by $p$. Doug hits more home runs if he hits a home run on his first try when Ryan does not, or if they both hit home runs on their first try and Doug hits more home runs thereafter. The probability of the first case occurring is $\\frac{1}{3} \\cdot \\frac{1}{2} = \\frac{1}{6}$, and the probability of the second case occurring is $\\frac{1}{3} \\cdot \\frac{1}{2} \\cdot p = \\frac{p}{6}$. Therefore $p = \\frac{1}{6} + \\frac{p}{6}$, which we solve to find $p = \\frac{1}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor how many integers $n$ between $1$ and $2005$, inclusive, is $2 \\cdot 6 \\cdot 10 \\cdots (4n-2)$ divisible by $n!$?",
    "options": [],
    "answer": "2005",
    "solution": "Solution:\nNote that\n$$\n\\begin{aligned}\n2 \\cdot 6 \\cdot 10 \\cdots (4n-2) &= 2^n \\cdot 1 \\cdot 3 \\cdot 5 \\cdots (2n-1) \\\\\n&= 2^n \\cdot \\frac{1 \\cdot 2 \\cdot 3 \\cdots 2n}{2 \\cdot 4 \\cdot 6 \\cdots 2n} \\\\\n&= \\frac{1 \\cdot 2 \\cdot 3 \\cdots 2n}{1 \\cdot 2 \\cdot 3 \\cdots n}\n\\end{aligned}\n$$\nthat is, it is just $\\frac{(2n)!}{n!}$. Therefore, since $\\frac{(2n)!}{(n!)^2} = \\binom{2n}{n}$ is always an integer, the answer is $2005$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou and I play the following game on an $8 \\times 8$ square grid of boxes: Initially, every box is empty. On your turn, you choose an empty box and draw an $X$ in it; if any of the four adjacent boxes are empty, you mark them with an $X$ as well. (Two boxes are adjacent if they share an edge.) We alternate turns, with you moving first, and whoever draws the last $X$ wins. How many choices do you have for a first move that will enable you to guarantee a win no matter how I play?",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nI can follow a symmetry strategy: whenever you play in the box $S$, I play in the image of $S$ under the $180^{\\circ}$ rotation about the center of the board. This ensures that the board will always be centrally symmetric at the beginning of your turn. Thus, if you play in an empty box $S$, its symmetric image $S^{\\prime}$ is also empty at the beginning of your turn, and it remains so after your turn, since the even size of the board ensures that $S$ can be neither equal to nor adjacent to $S^{\\prime}$. In particular, I always have a move available. Since the first person without an available move loses, you are guaranteed to lose. So the answer is that you have 0 choices for a first move that will guarantee your win.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 47092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn how many ways can 4 purple balls and 4 green balls be placed into a $4 \\times 4$ grid such that every row and column contains one purple ball and one green ball? Only one ball may be placed in each box, and rotations and reflections of a single configuration are considered different.",
    "options": [],
    "answer": "216",
    "solution": "Solution:\n\nThere are $4! = 24$ ways to place the four purple balls into the grid. Choose any purple ball, and place two green balls, one in its row and the other in its column. There are four boxes that do not yet lie in the same row or column as a green ball, and at least one of these contains a purple ball (otherwise the two rows containing green balls would contain the original purple ball as well as the two in the columns not containing green balls). It is then easy to see that there is a unique way to place the remaining green balls. Therefore, there are a total of $24 \\cdot 9 = 216$ ways.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow that for $b$ even, there exists some $M$ such that for every $n > M$, a $2b \\times n$ rectangle is $(1, b)$-tileable.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRecall from above that we can tile a $2 \\times 2b$ rectangle. Four columns of a $(b+1) \\times 2b$ rectangle can be tiled as shown below, and repeating this $\\frac{b}{2}$ times tiles the entire rectangle. Since any integer at least $b$ can be written as a positive linear combination of $2$ and $b+1$, we can tile any $2b \\times n$ rectangle for $n \\geq b$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTen positive integers are arranged around a circle. Each number is one more than the greatest common divisor of its two neighbors. What is the sum of the ten numbers?",
    "options": [],
    "answer": "28",
    "solution": "Solution:\n\nFirst note that all the integers must be at least $2$, because the greatest common divisor of any two positive integers is at least $1$. Let $n$ be the largest integer in the circle. The greatest common divisor of its two neighbors is $n-1$. Therefore, each of the two neighbors is at least $n-1$ but at most $n$, so since $n-1 \\nmid n$ for $n-1 \\geq 2$, they must both be equal to $n-1$.\n\nLet $m$ be one of the numbers on the other side of $n-1$ from $n$. Then $\\operatorname{gcd}(n, m) = n-2$. Since $n-2 \\geq 0$, $n-2 \\mid n$ only for $n=3$ or $4$.\n\nIf $n=3$, each number must be $2$ or $3$, and it is easy to check that there is no solution.\n\nIf $n=4$, then it is again not hard to find that there is a unique solution up to rotation, namely $4\\ 3\\ 2\\ 2\\ 3\\ 4\\ 3\\ 2\\ 2\\ 3$. The only possible sum is therefore $28$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCalculate\n$$\n\\lim_{x \\rightarrow 0^{+}} \\left(x^{x^{x}} - x^{x}\\right).\n$$",
    "options": [],
    "answer": "-1",
    "solution": "Solution:\nWe first calculate $\\lim_{x \\rightarrow 0^{+}} x^{x}$: it is just $\\exp\\left(\\lim_{x \\rightarrow 0^{+}} x \\ln x\\right)$. But\n$$\n\\lim_{x \\rightarrow 0^{+}} x \\ln x = \\lim_{x \\rightarrow 0^{+}} \\frac{\\ln x}{1/x} = \\lim_{x \\rightarrow 0^{+}} \\frac{1/x}{-1/x^{2}} = \\lim_{x \\rightarrow 0^{+}} -x = 0\n$$\nby L'Hôpital's Rule. Therefore $\\lim_{x \\rightarrow 0^{+}} x^{x} = 1$. Then $\\lim_{x \\rightarrow 0^{+}} x^{x^{x}} = 0^{1} = 0$, so $\\lim_{x \\rightarrow 0^{+}} \\left(x^{x^{x}} - x^{x}\\right) = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that for $b$ even, there exists some $M$ such that for every $m, n > M$ with $m n$ even, an $m \\times n$ rectangle is $(1, b)$-tileable.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nBy the diagram below, it is possible to tile a $(2 b+2) \\times (4 b+1)$ rectangle.\n\n![](attached_image_1.png)\n\nSince we can already tile a $(2 b+2) \\times 2 b$ rectangle by above, and $2 b$ is relatively prime to $4 b+1$, this will allow us to tile any $(2 b+2) \\times n$ rectangle for $n$ sufficiently large. Combining this with the previous problem, this will allow us to tile any $m \\times n$ rectangle for $m$ and $n$ sufficiently large and $m$ even, completing the proof.\n\nTo tile the $(2 b+2) \\times (4 b+1)$ rectangle, we first tile the following piece:\n\n![](attached_image_1.png)\n\nThis is then combined with two $2 \\times 2 b$ rectangles, a $2 b \\times b$ rectangle, and a $2 b \\times (2 b+1)$ rectangle as follows:\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA triangular piece of paper of area $1$ is folded along a line parallel to one of the sides and pressed flat. What is the minimum possible area of the resulting figure?",
    "options": [],
    "answer": "2/3",
    "solution": "Solution:\nLet the triangle be denoted $ABC$, and suppose we fold parallel to $BC$. Let the distance from $A$ to $BC$ be $h$, and suppose we fold along a line at a distance of $c h$ from $A$. We will assume that neither angle $B$ nor $C$ is obtuse, for the area of overlap will only be smaller if either is obtuse.\n\nIf $c \\leq \\frac{1}{2}$, then $A$ does not fold past the edge $BC$, so the overlap is a triangle similar to the original with height $c h$; the area of the figure is then $1 - c^2 \\geq \\frac{3}{4}$.\n\nSuppose $c > \\frac{1}{2}$, so that $A$ does fold past $BC$. Then the overlap is a trapezoid formed by taking a triangle of height $c h$ similar to the original and removing a triangle of height $(2c - 1)h$ similar to the original. The area of the resulting figure is thus $1 - c^2 + (2c - 1)^2 = 3c^2 - 4c + 2$.\n\nThis is minimized when $c = \\frac{2}{3}$, when the area is $\\frac{2}{3} < \\frac{3}{4}$; the minimum possible area is therefore $\\frac{2}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn an election, there are two candidates, $A$ and $B$, who each have $5$ supporters. Each supporter, independent of other supporters, has a $\\frac{1}{2}$ probability of voting for his or her candidate and a $\\frac{1}{2}$ probability of being lazy and not voting. What is the probability of a tie (which includes the case in which no one votes)?",
    "options": [],
    "answer": "63/256",
    "solution": "Solution:\n\nThe probability that exactly $k$ supporters of $A$ vote and exactly $k$ supporters of $B$ vote is $\\binom{5}{k}^2 \\cdot \\frac{1}{2^{10}}$. Summing over $k$ from $0$ to $5$ gives\n$$\n\\left(\\frac{1}{2^{10}}\\right)(1+25+100+100+25+1)=\\frac{252}{1024}=\\frac{63}{256}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $m \\circ n = \\dfrac{m+n}{mn+4}$. Compute $((\\cdots((2005 \\circ 2004) \\circ 2003) \\circ \\cdots \\circ 1) \\circ 0)$.",
    "options": [],
    "answer": "1/12",
    "solution": "Solution:\nNote that $m \\circ 2 = \\dfrac{m+2}{2m+4} = \\dfrac{1}{2}$, so the quantity we wish to find is just $\\left(\\dfrac{1}{2} \\circ 1\\right) \\circ 0 = \\dfrac{1}{3} \\circ 0 = \\dfrac{1}{12}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of the $x$-coordinates of the distinct points of intersection of the plane curves given by $x^{2}=x+y+4$ and $y^{2}=y-15 x+36$.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nSubstituting $y = x^{2} - x - 4$ into the second equation yields\n$$\n\\begin{aligned}\n0 &= \\left(x^{2} - x - 4\\right)^{2} - \\left(x^{2} - x - 4\\right) + 15x - 36 \\\\\n&= x^{4} - 2x^{3} - 7x^{2} + 8x + 16 - x^{2} + x + 4 + 15x - 36 \\\\\n&= x^{4} - 2x^{3} - 8x^{2} + 24x - 16 \\\\\n&= (x-2)\\left(x^{3} - 8x + 8\\right) = (x-2)^{2}\\left(x^{2} + 2x - 4\\right) .\n\\end{aligned}\n$$\nThis quartic has three distinct real roots at $x = 2, -1 \\pm \\sqrt{5}$. Each of these yields a distinct point of intersection, so the answer is their sum, $0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cube with side length $2$ is inscribed in a sphere. A second cube, with faces parallel to the first, is inscribed between the sphere and one face of the first cube. What is the length of a side of the smaller cube?",
    "options": [],
    "answer": "2/3",
    "solution": "Solution:\nFirst note that the long diagonal of the cube has length $2\\sqrt{3}$, so the radius of the sphere is $\\sqrt{3}$. Let $x$ be the side length of the smaller cube. Then the distance from the center of the sphere to the far face of the smaller cube is $1 + x$, while the distance from the center of the far face to a vertex lying on the sphere is $\\frac{x\\sqrt{2}}{2}$. Therefore, the square of the radius is\n$$\n3 = (1 + x)^2 + \\frac{x^2}{2},\n$$\nor\n$$\n3x^2 + 4x - 4 = (3x - 2)(x + 2) = 0,\n$$\nso $x = \\frac{2}{3}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 47102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThree fair six-sided dice, each numbered $1$ through $6$, are rolled. What is the probability that the three numbers that come up can form the sides of a triangle?",
    "options": [],
    "answer": "37/72",
    "solution": "Solution:\n$37 / 72$\n\nDenote this probability by $p$, and let the three numbers that come up be $x$, $y$, and $z$. We will calculate $1-p$ instead: $1-p$ is the probability that $x \\geq y+z$, $y \\geq z+x$, or $z \\geq x+y$. Since these three events are mutually exclusive, $1-p$ is just $3$ times the probability that $x \\geq y+z$. This happens with probability $(0+1+3+6+10+15)/216 = 35/216$, so the answer is $1-3 \\cdot (35/216) = 1-35/72 = 37/72$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe graph of $r=2+\\cos 2\\theta$ and its reflection over the line $y=x$ bound five regions in the plane. Find the area of the region containing the origin.",
    "options": [],
    "answer": "9π/2 − 8",
    "solution": "Solution: $\\frac{9\\pi}{2} - 8$\n![](attached_image_1.png)\nThe original graph is closer to the origin than its reflection for $\\theta \\in \\left(\\frac{\\pi}{4}, \\frac{3\\pi}{4}\\right) \\cup \\left(\\frac{5\\pi}{4}, \\frac{7\\pi}{4}\\right)$, and the region is symmetric about the origin. Therefore the area we wish to find is the polar integral\n$$\n\\begin{aligned}\n4 \\int_{\\frac{\\pi}{4}}^{\\frac{3\\pi}{4}} \\frac{1}{2}(2+\\cos 2\\theta)^2 d\\theta &= 2 \\int_{\\frac{\\pi}{4}}^{\\frac{3\\pi}{4}} \\left(4 + 4\\cos 2\\theta + \\cos^2 2\\theta\\right) d\\theta \\\\\n&= 2 \\int_{\\frac{\\pi}{4}}^{\\frac{3\\pi}{4}} \\left(4 + 4\\cos 2\\theta + \\frac{1}{2}(1 + \\cos 4\\theta)\\right) d\\theta \\\\\n&= \\left[9\\theta + 4\\sin 2\\theta + \\frac{1}{4} \\sin 4\\theta\\right]_{\\frac{\\pi}{4}}^{\\frac{3\\pi}{4}} \\\\\n&= \\left(\\frac{27\\pi}{4} - 4\\right) - \\left(\\frac{9\\pi}{4} + 4\\right) = \\frac{9\\pi}{2} - 8\n\\end{aligned}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 47104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $k$ be an integer such that $k \\mid a$ and $k \\mid b$. Prove that if an $m \\times n$ rectangle is $(a, b)$-tileable, then $2k \\mid m$ or $2k \\mid n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe prove the following lemma.\n\nLemma. Let $k$ be a positive integer such that $k \\mid a$ and $k \\mid b$. Then an $m \\times n$ rectangle is $(a, b)$-tileable if and only if an $m' \\times n'$ rectangle is $\\left(\\frac{a}{k}, \\frac{b}{k}\\right)$-tileable for $\\left\\lfloor\\frac{m}{k}\\right\\rfloor \\leq m' \\leq \\left\\lceil\\frac{m}{k}\\right\\rceil$ and $\\left\\lfloor\\frac{n}{k}\\right\\rfloor \\leq n' \\leq \\left\\lceil\\frac{n}{k}\\right\\rceil$. (Here, $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$, while $\\lceil x\\rceil$ denotes the least integer greater than or equal to $x$.)\n\nProof. Number the rows and columns in order. For each pair $0 \\leq i, j < k$, consider the set of squares in a row congruent to $i$ modulo $k$ and in a column congruent to $j$ modulo $k$. If one square of a type $(a, b)$ domino lies in this set, then so does the other. We can therefore partition the rectangle into these sets and then tile these sets instead. Each such set is a rectangular array of dimensions $m' \\times n'$, with $\\left\\lfloor\\frac{m}{k}\\right\\rfloor \\leq m' \\leq \\left\\lceil\\frac{m}{k}\\right\\rceil$ and $\\left\\lfloor\\frac{n}{k}\\right\\rfloor \\leq n' \\leq \\left\\lceil\\frac{n}{k}\\right\\rceil$, and a type $(a, b)$ domino on the original rectangle is a type $\\left(\\frac{a}{k}, \\frac{b}{k}\\right)$ domino on this new array. Since all possible pairs $(m', n')$ occur, the result follows.\n\nSuppose $2k \\nmid m$ and $2k \\nmid n$. Then at least one of $\\left\\lfloor\\frac{m}{k}\\right\\rfloor$ and $\\left\\lceil\\frac{m}{k}\\right\\rceil$ is odd, so we can choose $m'$ odd. Likewise we can choose $n'$ odd. But then an $m' \\times n'$ rectangle has odd area and so cannot be tileable, implying that the $m \\times n$ rectangle is not tileable.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a tetrahedron such that edges $AB$, $AC$, and $AD$ are mutually perpendicular. Let the areas of triangles $ABC$, $ACD$, and $ADB$ be denoted by $x$, $y$, and $z$, respectively. In terms of $x$, $y$, and $z$, find the area of triangle $BCD$.",
    "options": [],
    "answer": "sqrt(x^2 + y^2 + z^2)",
    "solution": "Solution:\n\n$\\sqrt{x^{2}+y^{2}+z^{2}}$\n\nPlace $A$, $B$, $C$, and $D$ at $(0,0,0)$, $(b, 0,0)$, $(0, c, 0)$, and $(0,0, d)$ in Cartesian coordinate space, with $b$, $c$, and $d$ positive. Then the plane through $B$, $C$, and $D$ is given by the equation $\\frac{x}{b}+\\frac{y}{c}+\\frac{z}{d}=1$. The distance from the origin to this plane is then\n\n$$\n\\frac{1}{\\sqrt{\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}}}=\\frac{b c d}{\\sqrt{b^{2} c^{2}+c^{2} d^{2}+d^{2} b^{2}}}=\\frac{b c d}{2 \\sqrt{x^{2}+y^{2}+z^{2}}}\n$$\n\nThen if the area of $BCD$ is $K$, the volume of the tetrahedron is\n$$\n\\frac{b c d}{6}=\\frac{b c d K}{6 \\sqrt{x^{2}+y^{2}+z^{2}}}\n$$\nimplying $K=\\sqrt{x^{2}+y^{2}+z^{2}}$.\n\n\nAlternative Solution: The area of $BCD$ is also half the length of the cross product of the vectors $\\overrightarrow{BC}=(0,-c, d)$ and $\\overrightarrow{BD}=(-b, 0, d)$. This cross product is $(-c d,-d b,-b c)=-2(y, z, x)$, which has length $2 \\sqrt{x^{2}+y^{2}+z^{2}}$. Thus the area of $BCD$ is $\\sqrt{x^{2}+y^{2}+z^{2}}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTwo ants, one starting at $(-1,1)$, the other at $(1,1)$, walk to the right along the parabola $y = x^{2}$ such that their midpoint moves along the line $y = 1$ with constant speed $1$. When the left ant first hits the line $y = \\frac{1}{2}$, what is its speed?",
    "options": [],
    "answer": "3 sqrt(3) - 3",
    "solution": "Solution: $3 \\sqrt{3} - 3$\nWhen the left ant first hits the line $y = \\frac{1}{2}$, the right ant hits the line $y = \\frac{3}{2}$. The left ant is then at $\\left(-\\frac{\\sqrt{2}}{2}, \\frac{1}{2}\\right)$, and the right ant is at $\\left(\\frac{\\sqrt{6}}{2}, \\frac{3}{2}\\right)$. Let the left ant have velocity with components $v_{x}$ and $v_{y}$, the right ant velocity with components $w_{x}$ and $w_{y}$. Since $\\frac{d y}{d x} = 2x$, $\\frac{v_{y}}{v_{x}} = -\\sqrt{2}$ and $\\frac{w_{y}}{w_{x}} = \\sqrt{6}$. Since the midpoint of the ants moves at speed $1$ along the line $y = 1$, $\\frac{1}{2}(v_{x} + w_{x}) = 1$ and $\\frac{1}{2}(v_{y} + w_{y}) = 0$. Therefore, $\\sqrt{2} v_{x} = -v_{y} = w_{y} = \\sqrt{6} w_{x} = \\sqrt{6}(2 - v_{x})$. Solving for $v_{x}$ gives $\\frac{2 \\sqrt{6}}{\\sqrt{6} + \\sqrt{2}} = 3 - \\sqrt{3}$. Then the speed of the left ant is\n$$\n\\sqrt{v_{x}^{2} + v_{y}^{2}} = \\sqrt{v_{x}^{2} + \\left(-\\sqrt{2} v_{x}\\right)^{2}} = \\sqrt{3} |v_{x}| = 3 \\sqrt{3} - 3\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a, b, c > 0$, what is the smallest possible value of $\\left\\lfloor\\frac{a+b}{c}\\right\\rfloor + \\left\\lfloor\\frac{b+c}{a}\\right\\rfloor + \\left\\lfloor\\frac{c+a}{b}\\right\\rfloor$? (Note that $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.)",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nSince $\\lfloor x \\rfloor > x - 1$ for all $x$, we have that\n$$\n\\begin{aligned}\n\\left\\lfloor\\frac{a+b}{c}\\right\\rfloor + \\left\\lfloor\\frac{b+c}{a}\\right\\rfloor + \\left\\lfloor\\frac{c+a}{b}\\right\\rfloor &> \\frac{a+b}{c} + \\frac{b+c}{a} + \\frac{c+a}{b} - 3 \\\\\n&= \\left(\\frac{a}{b} + \\frac{b}{a}\\right) + \\left(\\frac{b}{c} + \\frac{c}{b}\\right) + \\left(\\frac{c}{a} + \\frac{a}{c}\\right) - 3\n\\end{aligned}\n$$\nBut by the AM-GM inequality, each of the first three terms in the last line is at least $2$. Therefore, the lefthand side is greater than $2 + 2 + 2 - 3 = 3$. Since it is an integer, the smallest value it can be is therefore $4$. This is in fact attainable by letting $(a, b, c) = (6, 8, 9)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat 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": "Solution:\nIf 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. In 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 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. This is indeed attainable as in the diagram below.\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind a real, irreducible quartic polynomial with leading coefficient 1 whose roots are all twelfth roots of unity.",
    "options": [],
    "answer": "x^4 - x^2 + 1",
    "solution": "Solution:\n\nAll twelfth roots of unity are roots of\n$$\n\\begin{aligned}\nx^{12}-1 & =\\left(x^{6}-1\\right)\\left(x^{6}+1\\right) \\\\\n& =\\left(x^{3}-1\\right)\\left(x^{3}+1\\right)\\left(x^{6}+1\\right) \\\\\n& =(x-1)\\left(x^{2}+x+1\\right)(x+1)\\left(x^{2}-x+1\\right)\\left(x^{2}+1\\right)\\left(x^{4}-x^{2}+1\\right)\n\\end{aligned}\n$$\nso the answer is $x^{4}-x^{2}+1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree distinct lines are drawn in the plane. Suppose there exist exactly $n$ circles in the plane tangent to all three lines. Find all possible values of $n$.",
    "options": [],
    "answer": "0, 2, 4",
    "solution": "Solution:\n\nIf the three lines form a triangle, then there are 4 circles, namely the incircle and the three excircles.\n\nIf the three lines concur or are all parallel, then there are 0 circles.\n\nIf two lines are parallel and the third is not, then there are 2 circles lying between the two parallel lines, one on each side of the transverse line.\n\nThese are the only possible configurations, so the answers are $0$, $2$, and $4$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$ be a positive real number. Find the maximum possible value of\n$$\n\\frac{x^{2}+2-\\sqrt{x^{4}+4}}{x} .\n$$",
    "options": [],
    "answer": "2 sqrt(2) - 2",
    "solution": "Solution:\n$2 \\sqrt{2}-2$\n\nRationalizing the numerator, we get\n$$\n\\begin{aligned}\n\\frac{x^{2}+2-\\sqrt{x^{4}+4}}{x} \\cdot \\frac{x^{2}+2+\\sqrt{x^{4}+4}}{x^{2}+2+\\sqrt{x^{4}+4}} & =\\frac{\\left(x^{2}+2\\right)^{2}-\\left(x^{4}+4\\right)}{x\\left(x^{2}+2+\\sqrt{x^{4}+4}\\right)} \\\\\n& =\\frac{4 x^{2}}{x\\left(x^{2}+2+\\sqrt{x^{4}+4}\\right)} \\\\\n& =\\frac{4}{\\frac{1}{x}\\left(x^{2}+2+\\sqrt{x^{4}+4}\\right)} \\\\\n& =\\frac{4}{x+\\frac{2}{x}+\\sqrt{x^{2}+\\frac{4}{x^{2}}}}\n\\end{aligned}\n$$\nSince we wish to maximize this quantity, we wish to minimize the denominator. By AM-GM, $x+\\frac{2}{x} \\geq 2 \\sqrt{2}$ and $x^{2}+\\frac{4}{x^{2}} \\geq 4$, so that the denominator is at least $2 \\sqrt{2}+2$. Therefore,\n$$\n\\frac{x^{2}+2-\\sqrt{x^{4}+4}}{x} \\leq \\frac{4}{2 \\sqrt{2}+2}=2 \\sqrt{2}-2,\n$$\nwith equality when $x=\\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFive people of different heights are standing in line from shortest to tallest. As it happens, the tops of their heads are all collinear; also, for any two successive people, the horizontal distance between them equals the height of the shorter person. If the shortest person is 3 feet tall and the tallest person is 7 feet tall, how tall is the middle person, in feet?",
    "options": [],
    "answer": "sqrt(21)",
    "solution": "Solution:\n\nIf $A$, $B$, and $C$ are the tops of the heads of three successive people and $D$, $E$, and $F$ are their respective feet, let $P$ be the foot of the perpendicular from $A$ to $BE$ and let $Q$ be the foot of the perpendicular from $B$ to $CF$. Then, by equal angles, $\\triangle ABP \\sim \\triangle BCQ$, so\n$$\n\\frac{CF}{BE} = \\frac{CF}{BQ} = \\frac{CQ}{BQ} + 1 = \\frac{BP}{AP} + 1 = \\frac{BE}{AP} = \\frac{BE}{AD}\n$$\nTherefore the heights of successive people are in geometric progression. Hence, the heights of all five people are in geometric progression, so the middle height is $\\sqrt{3 \\cdot 7} = \\sqrt{21}$ feet.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvery second, Andrea writes down a random digit uniformly chosen from the set $\\{1,2,3,4\\}$. She stops when the last two numbers she has written sum to a prime number. What is the probability that the last number she writes down is 1?",
    "options": [],
    "answer": "15/44",
    "solution": "Solution:\n\nLet $p_{n}$ be the probability that the last number she writes down is 1 when the first number she writes down is $n$.\n\nSuppose she starts by writing 2 or 4. Then she can continue writing either 2 or 4, but the first time she writes 1 or 3, she stops. Therefore $p_{2} = p_{4} = \\frac{1}{2}$.\n\nSuppose she starts by writing 1. Then she stops if she writes 1, 2, or 4, but continues if she writes 3. Therefore $p_{1} = \\frac{1}{4}(1 + p_{3})$.\n\nIf she starts by writing 3, then she stops if she writes 2 or 4 and otherwise continues. Therefore $p_{3} = \\frac{1}{4}(p_{1} + p_{3}) = \\frac{1}{16}(1 + 5p_{3})$.\n\nSolving gives $p_{3} = \\frac{1}{11}$ and $p_{1} = \\frac{3}{11}$.\n\nThe probability we want to find is therefore $\\frac{1}{4}(p_{1} + p_{2} + p_{3} + p_{4}) = \\frac{15}{44}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat 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": "Solution:\n\nIf 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. In 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 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. This is indeed attainable as in the diagram below.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $x$ is a fifth root of unity. Find, in radical form, all possible values of\n$$\n2x + \\frac{1}{1+x} + \\frac{x}{1+x^{2}} + \\frac{x^{2}}{1+x^{3}} + \\frac{x^{3}}{1+x^{4}}.\n$$",
    "options": [],
    "answer": "4, -1 + sqrt(5), -1 - sqrt(5)",
    "solution": "Solution:\nNote that\n$$\n\\begin{gathered}\n\\frac{x}{1+x^{2}} + \\frac{x^{2}}{1+x^{3}} = \\frac{x^{6}}{1+x^{2}} + \\frac{x^{4}}{x^{2}+x^{5}} = \\frac{x^{4}+x^{6}}{1+x^{2}} = x^{4} = \\frac{1}{x}, \\text{ and } \\\\\n\\frac{1}{1+x} + \\frac{x^{3}}{1+x^{4}} = \\frac{x^{5}}{1+x} + \\frac{x^{4}}{x+x^{5}} = \\frac{x^{4}+x^{5}}{1+x} = x^{4} = \\frac{1}{x}\n\\end{gathered}\n$$\nTherefore, the sum is just $2x + \\frac{2}{x}$. If $x=1$, this is $4$. Otherwise, let $y = x + \\frac{1}{x}$. Then $x$ satisfies\n$$\n0 = 1 + x + x^{2} + x^{3} + x^{4} = \\left(x^{2} + \\frac{1}{x^{2}} + 2\\right) + \\left(x + \\frac{1}{x}\\right) - 1 = y^{2} + y - 1,\n$$\nso solving this quadratic yields $y = \\frac{-1 \\pm \\sqrt{5}}{2}$, or $2y = -1 \\pm \\sqrt{5}$. Since each value of $y$ can correspond to only 2 possible values of $x$, and there are 4 possible values of $x$ besides $1$, both of these values for $y$ are possible, which yields the answers, $4$ and $-1 \\pm \\sqrt{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $T$ be a triangle with side lengths $26$, $51$, and $73$. Let $S$ be the set of points inside $T$ which do not lie within a distance of $5$ of any side of $T$. Find the area of $S$.",
    "options": [],
    "answer": "135/28",
    "solution": "Solution:\nNote that the sides of $S$ are parallel to the sides of $T$, so $S$ is a triangle similar to $T$.\n\nThe semiperimeter of $T$ is $s = \\frac{1}{2}(26 + 51 + 73) = 75$.\n\nBy Heron's formula, the area of $T$ is $\\sqrt{75 \\cdot 49 \\cdot 24 \\cdot 2} = 420$.\n\nIf $r$ is the inradius of $T$, then the area of $T$ is $r s$, so $r = 420 / 75 = 28 / 5$.\n\nIt follows that the inradius of $S$ is $r - 5 = 3 / 5$, and the ratio of similitude between $S$ and $T$ is $3 / 28$.\n\nTherefore, the area of $S$ is $420 \\cdot (3 / 28)^2 = 135 / 28$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47117,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a convex quadrilateral inscribed in a circle with shortest side $AB$. The ratio $[BCD]/[ABD]$ is an integer (where $[XYZ]$ denotes the area of triangle $XYZ$.) If the lengths of $AB$, $BC$, $CD$, and $DA$ are distinct integers no greater than $10$, find the largest possible value of $AB$.",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nNote that\n$$\n\\frac{[BCD]}{[ABD]} = \\frac{\\frac{1}{2} BC \\cdot CD \\cdot \\sin C}{\\frac{1}{2} DA \\cdot AB \\cdot \\sin A} = \\frac{BC \\cdot CD}{DA \\cdot AB}\n$$\nsince $\\angle A$ and $\\angle C$ are supplementary. If $AB \\geq 6$, it is easy to check that no assignment of lengths to the four sides yields an integer ratio, but if $AB = 5$, we can let $BC = 10$, $CD = 9$, and $DA = 6$ for a ratio of $3$. The maximum value for $AB$ is therefore $5$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $f$ is a continuous real function such that $f(x-1)+f(x+1) \\geq x+f(x)$ for all $x$, what is the minimum possible value of $\\int_{1}^{2005} f(x) d x$?",
    "options": [],
    "answer": "2010012",
    "solution": "Solution:\n\nLet $g(x) = f(x) - x$. Then\n$$\ng(x-1) + x - 1 + g(x+1) + x + 1 \\geq x + g(x) + x\n$$\nor $g(x-1) + g(x+1) \\geq g(x)$. But now,\n$$\ng(x+3) \\geq g(x+2) - g(x+1) \\geq -g(x)\n$$\nTherefore\n$$\n\\begin{aligned}\n\\int_{a}^{a+6} g(x) d x & = \\int_{a}^{a+3} g(x) d x + \\int_{a+3}^{a+6} g(x) d x \\\\\n& = \\int_{a}^{a+3} (g(x) + g(x+3)) d x \\geq 0\n\\end{aligned}\n$$\nIt follows that\n$$\n\\int_{1}^{2005} g(x) = \\sum_{n=0}^{333} \\int_{6n+1}^{6n+7} g(x) d x \\geq 0\n$$\nso that\n$$\n\\int_{1}^{2005} f(x) d x = \\int_{1}^{2005} (g(x) + x) d x \\geq \\int_{1}^{2005} x d x = \\left[\\frac{x^{2}}{2}\\right]_{1}^{2005} = \\frac{2005^{2} - 1}{2} = 2010012\n$$\nEquality holds for $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $k$th root of unity is any complex number $\\omega$ such that $\\omega^{k}=1$.\n\nLet $x$ and $y$ be two $k$th roots of unity. Prove that $(x+y)^{k}$ is real.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that\n$$\n\\begin{aligned}\n(x+y)^{k} & = \\sum_{i=0}^{k} \\binom{k}{i} x^{i} y^{k-i} \\\\\n& = \\frac{1}{2} \\sum_{i=0}^{k} \\binom{k}{i} \\left(x^{i} y^{k-i} + x^{k-i} y^{i}\\right)\n\\end{aligned}\n$$\nby pairing the $i$th and $(k-i)$th terms. But $x^{k-i} y^{i} = \\left(x^{i} y^{k-i}\\right)^{-1}$ since $x$ and $y$ are $k$th roots of unity. Moreover, since $x$ and $y$ have absolute value $1$, so does $x^{i} y^{k-i}$, so $x^{k-i} y^{i}$ is in fact its complex conjugate. It follows that their sum is real, thus so is $(x+y)^{k}$.\n\nThis can also be shown geometrically. The argument of $x$ (the angle between the vector $x$ and the positive $x$-axis) is an integer multiple of $\\frac{2 \\pi}{k}$, as is the argument of $y$. Since $x+y$ bisects the angle between $x$ and $y$, its argument is an integer multiple of $\\frac{\\pi}{k}$. Multiplying this angle by $k$ gives a multiple of $\\pi$, so $(x+y)^{k}$ is real.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\sum_{n=0}^{\\infty} \\frac{n}{n^{4}+n^{2}+1}\n$$",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\nNote that\n$$\nn^{4}+n^{2}+1=\\left(n^{4}+2 n^{2}+1\\right)-n^{2}=\\left(n^{2}+1\\right)^{2}-n^{2}=\\left(n^{2}+n+1\\right)\\left(n^{2}-n+1\\right)\n$$\nDecomposing into partial fractions, we find that\n$$\n\\frac{n}{n^{4}+n^{2}+1}=\\frac{1}{2}\\left(\\frac{1}{n^{2}-n+1}-\\frac{1}{n^{2}+n+1}\\right) .\n$$\nNow, note that if $f(n)=\\frac{1}{n^{2}-n+1}$, then $f(n+1)=\\frac{1}{(n+1)^{2}-(n+1)+1}=\\frac{1}{n^{2}+n+1}$. It follows that\n$$\n\\sum_{n=0}^{\\infty} \\frac{n}{n^{4}+n^{2}+1}=\\frac{1}{2}((f(0)-f(1))+(f(1)-f(2))+(f(2)-f(3))+\\cdots) .\n$$\nSince $f(n)$ tends towards 0 as $n$ gets large, this sum telescopes to $f(0) / 2=1 / 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A_{1} A_{2} \\ldots A_{k}$ be a regular $k$-gon inscribed in a circle of radius $1$, and let $P$ be a point lying on or inside the circumcircle. Find the maximum possible value of $\\left(PA_{1}\\right)\\left(PA_{2}\\right) \\cdots\\left(PA_{k}\\right)$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nPlace the vertices at the $k$th roots of unity, $1, \\omega, \\ldots, \\omega^{k-1}$, and place $P$ at some complex number $p$. Then\n$$\n\\begin{aligned}\n\\left(\\left(PA_{1}\\right)\\left(PA_{2}\\right) \\cdots\\left(PA_{k}\\right)\\right)^{2} & = \\prod_{i=0}^{k-1}\\left|p-\\omega^{i}\\right|^{2} \\\\\n& = \\left|p^{k}-1\\right|^{2},\n\\end{aligned}\n$$\nsince $x^{k}-1=(x-1)(x-\\omega) \\cdots\\left(x-\\omega^{k-1}\\right)$. This is maximized when $p^{k}$ is as far as possible from $1$, which occurs when $p^{k}=-1$. Therefore, the maximum possible value of $\\left(PA_{1}\\right)\\left(PA_{2}\\right) \\cdots\\left(PA_{k}\\right)$ is $2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe number $27,000,001$ has exactly four prime factors. Find their sum.",
    "options": [],
    "answer": "652",
    "solution": "Solution:\nFirst, we factor\n$$\n\\begin{aligned}\n27x^{6} + 1 &= (3x^{2})^{3} + 1 \\\\\n&= (3x^{2} + 1)(9x^{4} - 3x^{2} + 1) \\\\\n&= (3x^{2} + 1)((9x^{4} + 6x^{2} + 1) - 9x^{2}) \\\\\n&= (3x^{2} + 1)((3x^{2} + 1)^{2} - (3x)^{2}) \\\\\n&= (3x^{2} + 1)(3x^{2} + 3x + 1)(3x^{2} - 3x + 1)\n\\end{aligned}\n$$\nLetting $x = 10$, we get that $27,000,001 = 301 \\cdot 331 \\cdot 271$. A quick check shows that $301 = 7 \\cdot 43$, so that $27,000,001 = 7 \\cdot 43 \\cdot 271 \\cdot 331$. Each factor here is prime, and their sum is $652$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn how many ways can the cells of a $4 \\times 4$ table be filled in with the digits $1,2, \\ldots, 9$ so that each of the 4-digit numbers formed by the columns is divisible by each of the 4-digit numbers formed by the rows?",
    "options": [],
    "answer": "9",
    "solution": "Solution:\nIf $a$ and $b$ are 4-digit numbers with the same first digit, and $a$ divides $b$, then since $b < a + 1000 \\leq 2a$, $b$ must equal $a$. In particular, since the number formed by the first row of the table divides the number in the first column (and both have the same first digit), these numbers must be equal; call their common value $n$. Then, for $k = 2, 3$, or $4$, we find that the number in the $k$th column and the number in the $k$th row have the same first digit (namely the $k$th digit of $n$), so by the same reasoning, they are equal. Also, the smallest number $b$ formed by any column is divisible by the largest number $a$ formed by any row, but by the symmetry just proven, $a$ is also the largest number formed by any column, so $a \\geq b$. Since $b$ is divisible by $a$, we must have equality. Then all columns contain the same number - and hence all rows also contain the same number - which is only possible if all 16 cells contain the same digit. Conversely, for each $d = 1, \\ldots, 9$, filling in all 16 cells with the digit $d$ clearly gives a table meeting the required condition, so we have exactly 9 such tables, one for each digit.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\sum_{k=0}^{\\infty} \\frac{4}{(4 k)!}\n$$",
    "options": [],
    "answer": "e + 1/e + 2 cos 1",
    "solution": "Solution: $e + 1 / e + 2 \\cos 1$\nThis is the power series\n$$\n4 + \\frac{4 x^{4}}{4!} + \\frac{4 x^{8}}{8!} + \\cdots\n$$\nevaluated at $x = 1$. But this power series can be written as the sum\n$$\n\\begin{aligned}\n& \\left(1 + \\frac{x}{1!} + \\frac{x^{2}}{2!} + \\frac{x^{3}}{3!} + \\frac{x^{4}}{4!} + \\frac{x^{5}}{5!} + \\frac{x^{6}}{6!} + \\frac{x^{7}}{7!} + \\cdots\\right) \\\\\n+ & \\left(1 - \\frac{x}{1!} + \\frac{x^{2}}{2!} - \\frac{x^{3}}{3!} + \\frac{x^{4}}{4!} - \\frac{x^{5}}{5!} + \\frac{x^{6}}{6!} - \\frac{x^{7}}{7!} + \\cdots\\right) \\\\\n+ & 2\\left(1 - \\frac{x^{2}}{2!} + \\frac{x^{4}}{4!} - \\frac{x^{6}}{6!} + \\cdots\\right) \\\\\n= & e^{x} + e^{-x} + 2 \\cos x .\n\\end{aligned}\n$$\nIt follows that the quantity is $e + 1 / e + 2 \\cos 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEight coins are arranged in a circle heads up. A move consists of flipping over two adjacent coins. How many different sequences of six moves leave the coins alternating heads up and tails up?",
    "options": [],
    "answer": "7680",
    "solution": "Solution:\n\nImagine we flip over two adjacent coins by pushing a button halfway between them. Then the outcome depends only on the parities of the number of times that each button is pushed. To flip any coin, we must push the two buttons adjacent to that coin a total of an odd number of times. To flip every other coin, the parities must then progress around the circle as even, even, odd, odd, even, even, odd, odd. There are 4 ways to assign these parities. If we assume each button is pressed either once or not at all, this accounts for only four presses, so some button is also pressed twice more. Suppose this button was already pushed once. There are 4 of these, and the number of possible sequences of presses is then $6!/ 3! = 120$. Suppose it has not already been pressed. There are 4 of these as well, and the number of possible sequences is $6!/ 2! = 360$. The total number of sequences is then $4(4 \\cdot 120 + 4 \\cdot 360) = 7680$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA triangular piece of paper of area $1$ is folded along a line parallel to one of the sides and pressed flat. What is the minimum possible area of the resulting figure?",
    "options": [],
    "answer": "2/3",
    "solution": "Solution:\n\nLet the triangle be denoted $ABC$, and suppose we fold parallel to $BC$. Let the distance from $A$ to $BC$ be $h$, and suppose we fold along a line at a distance of $c h$ from $A$. We will assume that neither angle $B$ nor $C$ is obtuse, for the area of overlap will only be smaller if either is obtuse.\n\nIf $c \\leq \\frac{1}{2}$, then $A$ does not fold past the edge $BC$, so the overlap is a triangle similar to the original with height $c h$; the area of the figure is then $1 - c^2 \\geq \\frac{3}{4}$.\n\nSuppose $c > \\frac{1}{2}$, so that $A$ does fold past $BC$. Then the overlap is a trapezoid formed by taking a triangle of height $c h$ similar to the original and removing a triangle of height $(2c - 1) h$ similar to the original. The area of the resulting figure is thus $1 - c^2 + (2c - 1)^2 = 3c^2 - 4c + 2$.\n\nThis is minimized when $c = \\frac{2}{3}$, when the area is $\\frac{2}{3} < \\frac{3}{4}$; the minimum possible area is therefore $\\frac{2}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFarmer Bill's 1000 animals - ducks, cows, and rabbits - are standing in a circle. In order to feel safe, every duck must either be standing next to at least one cow or between two rabbits. If there are 600 ducks, what is the least number of cows there can be for this to be possible?",
    "options": [],
    "answer": "201",
    "solution": "Solution:\nSuppose Bill has $r$ rabbits and $c$ cows. At most $r-1$ ducks can be between two rabbits: each rabbit can serve up to two such ducks, so at most $2r/2 = r$ ducks will each be served by two rabbits, but we cannot have equality, since this would require alternating between rabbits and ducks all the way around the circle, contradicting the fact that more than half the animals are ducks. Also, at most $2c$ ducks can each be adjacent to a cow. So we need $600 \\leq r-1 + 2c = (400-c)-1 + 2c$, giving $c \\geq 201$. Conversely, an arrangement with 201 cows is possible:\n$$\n\\underbrace{R D R D R \\cdots D R}_{199_{R},\\ 198\\ D}\\ \\underbrace{D C D D C D D C D \\cdots D C D}_{201\\ C,\\ 402\\ D}.\n$$\nSo 201 is the answer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $AC$ be a diameter of a circle $\\omega$ of radius $1$, and let $D$ be the point on $AC$ such that $CD=1/5$. Let $B$ be the point on $\\omega$ such that $DB$ is perpendicular to $AC$, and let $E$ be the midpoint of $DB$. The line tangent to $\\omega$ at $B$ intersects line $CE$ at the point $X$. Compute $AX$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nWe first show that $AX$ is perpendicular to $AC$. Let the tangent to $\\omega$ at $A$ intersect $CB$ at $Z$ and $CE$ at $X'$. Since $ZA$ is parallel to $BD$ and $BE=ED$, $ZX' = X'A$. Therefore, $X'$ is the midpoint of the hypotenuse of the right triangle $ABZ$, so it is also its circumcenter. Thus $X'A = X'B$, and since $X'A$ is tangent to $\\omega$ and $B$ lies on $\\omega$, we must have that $X'B$ is tangent to $\\omega$, so $X = X'$.\n\nLet $O$ be the center of $\\omega$. Then $OD = \\frac{4}{5}$, so $BD = \\frac{3}{5}$ and $DE = \\frac{3}{10}$. Then $AX = DE \\cdot \\frac{AC}{DC} = \\frac{3}{10} \\cdot \\frac{2}{1/5} = 3$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$ and $y$ be two distinct roots of unity. Prove that $x+y$ is also a root of unity if and only if $\\frac{y}{x}$ is a cube root of unity.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThis is easiest to see geometrically. The vectors corresponding to $x$, $y$, and $-x-y$ sum to $0$, so they form a triangle. In order for them all to be roots of unity, they must all have length one, so the triangle must be equilateral. Therefore the angle between $x$ and $y$ is $\\pm \\frac{2 \\pi}{3}$, that is, $\\frac{y}{x}$ is a cube root of unity.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAll the chairs in a classroom are arranged in a square $n \\times n$ array (in other words, $n$ columns and $n$ rows), and every chair is occupied by a student. The teacher decides to rearrange the students according to the following two rules:\n\na. Every student must move to a new chair.\n\nb. A student can only move to an adjacent chair in the same row or to an adjacent chair in the same column. In other words, each student can move only one chair horizontally or vertically.\n\n(Note that the rules above allow two students in adjacent chairs to exchange places.)\n\nShow that this procedure can be done if $n$ is even, and cannot be done if $n$ is odd.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf $n$ is even, there are many ways to do this. One is simply to exchange adjacent students in each row: the student in an odd-numbered chair $k$ exchanges places with the student in chair $k+1$. In other words, exchange students in chairs 1 and 2, those in chairs 3 and 4, and so on. Since there are an even number of students in each row, every student in every row will move and the teacher's two conditions can be satisfied.\n\nIf $n$ is odd, imagine that the chairs are colored alternately black and white as on a chessboard with a black chair in one corner (and hence all four corner chairs are black). It is easy to see that $\\left(n^{2}+1\\right) / 2$ of the chairs are colored black and $\\left(n^{2}-1\\right) / 2$ are white, so there is one more black than white chair. Any valid rearrangement must move each student to a chair of the opposite color. The conditions cannot be satisfied since there is one more black chair than white chair, so some student seated in a black chair will have nowhere to go.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47131,
    "subject": "Mathematics (Multi-modal)",
    "question": "A communications network consisting of some terminals is called a 3-connector if among any three terminals, some two of them can directly communicate with each other. A communications network contains a windmill with $n$ blades if there exist $n$ pairs of terminals $\\{x_1, y_1\\}, \\dots, \\{x_n, y_n\\}$ such that each $x_i$ can directly communicate with the corresponding $y_i$ and there is a hub terminal that can directly communicate with each of the $2n$ terminals $x_1, y_1, \\dots, x_n, y_n$. Determine the minimum value of $f(n)$, in terms of $n$, such that a 3-connector with $f(n)$ terminals always contains a windmill with $n$ blades.",
    "options": [],
    "answer": "f(1) = 6; for n ≥ 2, f(n) = 4n + 1",
    "solution": "The answer is\n$$\nf(n) = \\begin{cases} 6 & \\text{if } n = 1; \\\\ 4n + 1 & \\text{if } n \\ge 2. \\end{cases}\n$$\nWe will use *connected* as a synonym for directly communicating, call a set of $k$ terminals for which each of the $\\binom{k}{2}$ pairs of terminals is connected *complete* and call a set of $2k$ terminals forming $k$ disjoint connected pairs a *k-matching*.\n\nWe first show that $f(n) = 4n + 1$ for $n > 1$. The $4n$-terminal network consisting of two disconnected complete sets of $2n$ terminals clearly does not contain an $n$-bladed windmill (henceforth called an $n$-mill), since such a windmill requires a set of $2n+1$ connected terminals. So we need only demonstrate that $f(n) = 4n + 1$ is sufficient.\n\nNote that we can inductively create a $k$-matching in any subnetwork of $2k + 1$ elements, as there is a connected pair in any set of three or more terminals. Also, the set of terminals that are not connected to a given terminal $x$ must be complete, as otherwise there would be a set of three mutually disconnected terminals. We now proceed by contradiction and assume that there is a $(4n + 1)$-terminal network without an $n$-mill. Any terminal $x$ must then be connected to at least $2n$ terminals, for otherwise there would be a complete set of size at least $2n + 1$, which includes an $n$-mill. In addition, $x$ cannot be directly connected to more than $2n$ terminals, for otherwise we could construct an $n$-matching among these, and therefore an $n$-mill. Therefore every terminal is connected to precisely $2n$ others.\n\nIf we take two terminals $u$ and $v$ that are not connected we can then note that at least one must be connected to the $4n - 1$ remaining terminals, and therefore there must be exactly one, $w$, to which both are connected. The rest of the network now consists of two complete sets of terminals $A$ and $B$ of size $2n - 1$, where every terminal in $A$ is connected to $u$ and not connected to $v$, and every terminal in $B$ is not connected to $u$ and connected to $v$. If $w$ were connected to any terminal in $A$ or $B$, it would form a blade with this element and hub $u$ or $v$ respectively, and we could fill out the rest of an $n$-mill with terminals in $A$ or $B$ respectively. Hence $w$ is only connected to two terminals, and therefore $n = 1$.\n\n![](attached_image_1.png)\n\nExamining the preceding proof, we can find the only 5-terminal network with no 1-mill: With terminals labeled A, B, C, D, and E, the connected pairs are (A, B), (B, C), (C, D), (D, E), and (E, A). (As indicated in the figure above, a pair of terminals are connected if and only if the edge connecting them are darkened.) To show that any 6-terminal network has a 1-mill, we note that any complete set of three terminals is a 1-mill. We again work by contradiction. Any terminal *a* would have to be connected to at least three others, *b*, *c*, and *d*, or the terminals not connected to *a* would form a 1-mill. But then one of the pairs (*b*, *c*), (*c*, *d*), and (*b*, *d*) must be connected, and this creates a 1-mill with that pair and *a*.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and let $s$ be an integer with $0 < s < p$. Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and\n$$\n\\{\\frac{sm}{p}\\} < \\{\\frac{sn}{p}\\} < \\frac{s}{p}\n$$\nif and only if $s$ is not a divisor of $p-1$.\n(For $x$ a real number, let $\\lfloor x \\rfloor$ denote the greatest integer less than or equal to $x$, and let $\\{x\\} = x - \\lfloor x \\rfloor$ denote the fractional part of $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First suppose that $s$ is a divisor of $p-1$; write $d = (p-1)/s$. As $x$ varies among $1, 2, \\dots, p-1$, $\\{sx/p\\}$ takes the values $1/p, 2/p, \\dots, (p-1)/p$ once each in some order. The possible values with $\\{sx/p\\} < s/p$ are precisely $1/p, \\dots, (s-1)/p$. From the fact that $\\{sd/p\\} = (p-1)/p$, we realize that the values $\\{sx/p\\} = (p-1)/p, (p-2)/p, \\dots, (p-s+1)/p$ occur for\n$$\nx = d, 2d, \\dots, (s-1)d\n$$\n(which are all between 0 and $p$), and so the values $\\{sx/p\\} = 1/p, 2/p, \\dots, (s-1)/p$ occur for\n$$\nx = p - d, p - 2d, \\dots, p - (s-1)d,\n$$\nrespectively. From this it is clear that $m$ and $n$ cannot exist as requested.\n\nConversely, suppose that $s$ is not a divisor of $p-1$. Put $m = \\lfloor p/s \\rfloor$; then $m$ is the smallest positive integer such that $\\{ms/p\\} < s/p$, and in fact $\\{ms/p\\} = (ms - p)/p$. However, we cannot have $\\{ms/p\\} = (s-1)/p$ or else $(m-1)s = p-1$, contradicting our hypothesis that $s$ does not divide $p-1$. Hence the unique $n \\in \\{1, \\dots, p-1\\}$ for which $\\{nx/p\\} = (s-1)/p$ has the desired properties (since the fact that $\\{nx/p\\} < s/p$ forces $n \\ge m$, but $m \\ne n$).\nWe prove the contrapositive statement:\nLet $p$ be a prime number and let $s$ be an integer with $0 < s < p$. Prove that the following statements are equivalent:\n(a) $s$ is a divisor of $p-1$;\n(b) if integers $m$ and $n$ are such that $0 < m < p$, $0 < n < p$, and\n$$\n\\{\\frac{sm}{p}\\} < \\{\\frac{sn}{p}\\} < \\frac{s}{p},\n$$\nthen $0 < n < m < p$.\nSince $p$ is prime and $0 < s < p$, $s$ is relatively prime to $p$ and\n$$\nm_1s + a_1p = 1,\\ m_2s + a_2p = 2, \\dots, m_s s + a_s p = s.\n$$\nHence $\\{m_k s/p\\} = k/p$ for $1 \\le k \\le s$.\nStatement (b) holds if and only $0 < m_s < m_{s-1} < \\dots < m_1 < p$. For $1 \\le k \\le s-1$, $m_k s - m_{k+1} s = (a_{k+1} - a_k) p - 1$, or $(m_k - m_{k+1}) s \\equiv -1 \\pmod p$. Since $0 < m_{k+1} < m_k < p$, by (1), we have $m_k - m_{k+1} = d$. We conclude that (b) holds if and only if $m_s, m_{s-1}, \\dots, m_1$ form an arithmetic progression with common difference $-d$. Clearly $m_s = 1$, so $m_1 = 1 + (s-1)d = jp - d + 1$ for some $j$. Then $j=1$ because $m_1$ and $d$ are both positive and less than $p$, so $sd = p-1$. This proves (a).\n\nConversely, if (a) holds, then $sd = p-1$ and $m_k \\equiv -d s m_k \\equiv -d k \\pmod p$. Hence $m_k = p-dk$ for $1 \\le k \\le s$. Thus $m_s, m_{s-1}, \\dots, m_1$ form an arithmetic progression with common difference $-d$. Hence (b) holds.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOctagon $A B C D E F G H$ is equiangular. Given that $A B=1$, $B C=2$, $C D=3$, $D E=4$, and $E F=F G=2$, compute the perimeter of the octagon.",
    "options": [],
    "answer": "20 + √2",
    "solution": "Solution:\n\nExtend sides $A B$, $C D$, $E F$, $G H$ to form a rectangle: let $X$ be the intersection of lines $G H$ and $A B$; $Y$ that of $A B$ and $C D$; $Z$ that of $C D$ and $E F$; and $W$ that of $E F$ and $G H$.\n\nAs $B C=2$, we have $B Y=Y C=\\sqrt{2}$. As $D E=4$, we have $D Z=Z E=2\\sqrt{2}$. As $F G=2$, we have $F W=W G=\\sqrt{2}$.\n\nWe can compute the dimensions of the rectangle: $W X=Y Z=Y C+C D+D Z=3+3\\sqrt{2}$, and $X Y=Z W=Z E+E F+F W=2+3\\sqrt{2}$. Thus, $H X=X A=X Y-A B-B Y=1+2\\sqrt{2}$, and so $A H=\\sqrt{2} H X=4+\\sqrt{2}$, and $G H=W X-W G-H X=2$. The perimeter of the octagon can now be computed by adding up all its sides.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many positive integers $x$ are there such that $3x$ has 3 digits and $4x$ has four digits?",
    "options": [],
    "answer": "84",
    "solution": "Solution:\n\nNote that $x$ must be between $250$ and $333$, inclusive. There are $84$ integers in that interval.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA bear walks one mile south, one mile east, and one mile north, only to find itself where it started. Another bear, more energetic than the first, walks two miles south, two miles east, and two miles north, only to find itself where it started. However, the bears are not white and did not start at the north pole. At most how many miles apart, to the nearest .001 mile, are the two bears' starting points?",
    "options": [],
    "answer": "3.477",
    "solution": "Solution:\nSay the first bear walks a mile south, an integer $n > 0$ times around the south pole, and then a mile north. The middle leg of the first bear's journey is a circle of circumference $1 / n$ around the south pole, and therefore about $\\frac{1}{2 n \\pi}$ miles north of the south pole. (This is not exact even if we assume the Earth is perfectly spherical, but it is correct to about a micron.) Adding this to the mile that the bear walked south/north, we find that it started about $1 + \\frac{1}{2 n \\pi}$ miles from the south pole. Similarly, the second bear started about $2 + \\frac{2}{2 m \\pi}$ miles from the south pole for some integer $m > 0$, so they must have started at most\n$$\n3 + \\frac{1}{2 n \\pi} + \\frac{2}{2 m \\pi} \\leq 3 + \\frac{3}{2 \\pi} \\approx 3.477\n$$\nmiles apart.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDo there exist positive integers $x$, $y$, and $z$ such that $x^{2006} + y^{2006} = z^{2007}$? Explain your answer.",
    "options": [],
    "answer": "Yes; for example, x = y = z = 2.",
    "solution": "Solution:\n\nYes, for example, $x = y = z = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSpring is finally here in Cambridge, and it's time to mow our lawn. For the purpose of these problems, our lawn consists of little clumps of grass arranged in an $m \\times n$ rectangular grid, that is, with $m$ rows running east-west and $n$ columns running north-south. To be even more explicit, we might say our clumps are at the lattice points\n$$\n\\left\\{(x, y) \\in \\mathbb{Z}^2 \\mid 0 \\leq x < n \\text{ and } 0 \\leq y < m \\right\\}.\n$$\nOur machinery consists of a fleet of identical mowbots (or \"mobots\" for short). A mobot is a lawn-mowing machine. To mow our lawn, we begin by choosing a formation: we place as many mobots as we want at various clumps of grass and orient each mobot's head in a certain direction, either north or east (not south or west). At the blow of a whistle, each mobot starts moving in the direction we've chosen, mowing every clump of grass in its path (including the clump it starts on) until it goes off the lawn.\nBecause the spring is so young, our lawn is rather delicate. Consequently, we want to make sure that every clump of grass is mowed once and only once. We will not consider formations that do not meet this criterion.\nOne more thing: two formations are considered \"different\" if there exists a clump of grass for which either (1) for exactly one of the formations does a mobot starts on that clump, or (2) there are mobots starting on this clump for both the formations, but they're oriented in different directions.\n\nProve that the maximum number of mobots you need to mow your lawn is $m+n-1$.",
    "options": [],
    "answer": "m + n − 1",
    "solution": "Solution:\n\nThis is attainable if we place one mobot at each clump in the first row, oriented north, and one mobot in the first column of each row except the first, oriented east.\n\nTo show that at most $m+n-1$ mobots can be used, note that each mobot must mow at least one of the $m+n-1$ clumps in union of the first row and last column.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLarry can swim from Harvard to MIT (with the current of the Charles River) in 40 minutes, or back (against the current) in 45 minutes. How long does it take him to row from Harvard to MIT, if he rows the return trip in 15 minutes? (Assume that the speed of the current and Larry's swimming and rowing speeds relative to the current are all constant.) Express your answer in the format mm:ss.",
    "options": [],
    "answer": "14:24",
    "solution": "Solution:\n\nLet the distance between Harvard and MIT be $1$, and let $c$, $s$, $r$ denote the speeds of the current and Larry's swimming and rowing, respectively. Then we are given\n$$\ns + c = \\frac{1}{40} = \\frac{9}{360}, \\quad s - c = \\frac{1}{45} = \\frac{8}{360}, \\quad r - c = \\frac{1}{15} = \\frac{24}{360},\n$$\nso\n$$\nr + c = (s + c) - (s - c) + (r - c) = \\frac{9 - 8 + 24}{360} = \\frac{25}{360}\n$$\nand it takes Larry $360 / 25 = 14.4$ minutes, or $14:24$, to row from Harvard to MIT.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVernonia High School has 85 seniors, each of whom plays on at least one of the school's three varsity sports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 are on the baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball and football teams; and 13 are on both the baseball and lacrosse teams. Compute the number of seniors playing all three sports, given that twice this number are members of the lacrosse team.",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nSuppose that $n$ seniors play all three sports and that $2n$ are on the lacrosse team. Then, by the principle of inclusion-exclusion,\n$$\n85 = (74 + 26 + 2n) - (17 + 18 + 13) + n = 100 + 2n - 48 + n = 52 + 3n.\n$$\nIt is easily seen that $n = 11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA nonzero polynomial $f(x)$ with real coefficients has the property that $f(x) = f'(x) f''(x)$. What is the leading coefficient of $f(x)$?",
    "options": [],
    "answer": "1/18",
    "solution": "Solution:\nSuppose that the leading term of $f(x)$ is $c x^{n}$, where $c \\neq 0$. Then the leading terms of $f'(x)$ and of $f''(x)$ are $c n x^{n-1}$ and $c n(n-1) x^{n-2}$, respectively, so $c x^{n} = c n x^{n-1} \\cdot c n(n-1) x^{n-2}$, which implies that $n = (n-1) + (n-2)$, or $n = 3$, and $c = c n \\cdot c n(n-1) = 18 c^{2}$, or $c = \\frac{1}{18}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose we have an $n$-gon such that each interior angle, measured in degrees, is a positive integer. Suppose further that all angles are less than $180^{\\circ}$, and that all angles are different sizes. What is the maximum possible value of $n$? Prove your answer.",
    "options": [],
    "answer": "26",
    "solution": "Solution:\n\nLet's work with the exterior angles (each is $180$ minus the interior angle). Then the conditions on the exterior angles are identical to the conditions on the interior angles: each is a positive integer between $1$ and $179$ inclusive. The sum of the exterior angles is exactly $360$.\n\nHowever, the sum of $1$ through $27$ is $27 \\cdot 28 / 2 = 378$, which is too large. We can get $26$ using angles of $1$ through $25$ (sum $325$) and an angle of $35$. The previous problem shows that this is actually possible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSomewhere in the universe, $n$ students are taking a 10-question math competition. Their collective performance is called laughable if, for some pair of questions, there exist 57 students such that either all of them answered both questions correctly or none of them answered both questions correctly. Compute the smallest $n$ such that the performance is necessarily laughable.",
    "options": [],
    "answer": "253",
    "solution": "Solution:\n\nLet $c_{i, j}$ denote the number of students correctly answering questions $i$ and $j$ ($1 \\leq i < j \\leq 10$), and let $w_{i, j}$ denote the number of students getting both questions wrong. An individual student answers $k$ questions correctly and $10-k$ questions incorrectly. This student answers $\\binom{k}{2}$ pairs of questions correctly and $\\binom{10-k}{2}$ pairs of questions incorrectly. Now observe that\n$$\n\\binom{k}{2} + \\binom{10-k}{2} = k^2 - 10k + 45 = (k-5)^2 + 20 \\geq 20\n$$\nTherefore,\n$$\n\\sum_{1 \\leq i < j \\leq 10} c_{i, j} + w_{i, j} \\geq 20n\n$$\nNow if the performance is not laughable, then $c_{i, j} \\leq 56$ and $w_{i, j} \\leq 56$ for all $1 \\leq i < j \\leq 10$. Observe that there are $2 \\binom{10}{2} = 90$ of these variables. Hence, in a boring performance,\n$$\n20n \\leq \\sum_{1 \\leq i < j \\leq 10} c_{i, j} + w_{i, j} \\leq 90 \\cdot 56 = 5040\n$$\nor $n \\leq 252$. In particular this implies that if $n \\geq 253$, the performance is laughable. This is the best bound because $\\binom{10}{5} = 252$, and if each of 252 students correctly answers a different 5-element subset of the 10 questions, then $c_{i, j} = w_{i, j} = 56$ for all $1 \\leq i < j \\leq 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFifteen freshmen are sitting in a circle around a table, but the course assistant (who remains standing) has made only six copies of today's handout. No freshman should get more than one handout, and any freshman who does not get one should be able to read a neighbor's. If the freshmen are distinguishable but the handouts are not, how many ways are there to distribute the six handouts subject to the above conditions?",
    "options": [],
    "answer": "125",
    "solution": "Solution:\n\nSuppose that you are one of the freshmen; then there's a $6 / 15$ chance that you'll get one of the handouts. We may ask, given that you do get a handout, how many ways are there to distribute the rest? We need only multiply the answer to that question by $15 / 6$ to answer the original question.\n\nGoing clockwise around the table from you, one might write down the sizes of the gaps between people with handouts. There are six such gaps, each of size $0$–$2$, and the sum of their sizes must be $15-6=11$. So the gap sizes are either $1,1,1,2,2,2$ in some order, or $0,1,2,2,2,2$ in some order. In the former case, $\\frac{6!}{3!3!}=20$ orders are possible; in the latter, $\\frac{6!}{1!1!4!}=30$ are. Altogether, then, there are $20+30=50$ possibilities.\n\nMultiplying this by $15 / 6$, or $5 / 2$, gives $125$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $f$ and $g$ are differentiable functions such that\n$$\nx g(f(x)) f'(g(x)) g'(x) = f(g(x)) g'(f(x)) f'(x)\n$$\nfor all real $x$. Moreover, $f$ is nonnegative and $g$ is positive. Furthermore,\n$$\n\\int_{0}^{a} f(g(x)) dx = 1 - \\frac{e^{-2a}}{2}\n$$\nfor all reals $a$. Given that $g(f(0)) = 1$, compute the value of $g(f(4))$.",
    "options": [],
    "answer": "e^{-16}",
    "solution": "Solution:\nDifferentiating the given integral with respect to $a$ gives $f(g(a)) = e^{-2a}$. Now\n$$\nx \\frac{d[\\ln (f(g(x)))]}{dx} = x \\frac{f'(g(x)) g'(x)}{f(g(x))} = \\frac{g'(f(x)) f'(x)}{g(f(x))} = \\frac{d[\\ln (g(f(x)))]}{dx}\n$$\nwhere the second equals sign follows from the given. Since $\\ln (f(g(x))) = -2x$, we have $-x^2 + C = \\ln (g(f(x)))$, so $g(f(x)) = K e^{-x^2}$. It follows that $K = 1$ and $g(f(4)) = e^{-16}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA positive integer $n$ is called \"flippant\" if $n$ does not end in $0$ (when written in decimal notation) and, moreover, $n$ and the number obtained by reversing the digits of $n$ are both divisible by $7$. How many flippant integers are there between $10$ and $1000$?",
    "options": [],
    "answer": "17",
    "solution": "Solution:\nWe use the notation \"$|$\" to mean \"divides.\"\n\nThere is only one flippant 2-digit number, namely $77$. Indeed, if $10a + b$ is flippant (where $a, b$ are integers $1$-$9$), then $7 \\mid 10a + b$ and $7 \\mid 10b + a$. Thus,\n$$\n7 \\mid 3(10a + b) - (10b + a) = 29a - 7b = a + 7(4a - b)\n$$\nso that $7 \\mid a$, and similarly $7 \\mid b$, so we'd better have $a = b = 7$.\n\nThere are $16$ flippant 3-digit numbers. First consider the $12$ palindromic ones (ones where the hundreds and units digits are the same): $161, 252, 343, 434, 525, 595, 616, 686, 707, 777, 868$, and $959$.\n\nNow consider the general case: suppose $100a + 10b + c$ is flippant, where $a, b, c$ are integers $1$-$9$. Then $7 \\mid 100a + 10b + c$ and $7 \\mid 100c + 10b + a$, so $7 \\mid (100a + 10b + c) - (100c + 10b + a) = 99(a - c)$, and so $7 \\mid a - c$. In order for this not to result in a palindromic integer, we must have $a - c = \\pm 7$ and, moreover, both $100a + 10b + a$ and $100c + 10b + c$ must be palindromic flippant integers. Consulting our list above, we find $4$ more flippant integers: $168, 259, 861$, and $952$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the maximum value attained by\n$$\n\\frac{x^{4}-x^{2}}{x^{6}+2 x^{3}-1}\n$$\nover real numbers $x>1$.",
    "options": [],
    "answer": "1/6",
    "solution": "Solution:\nWe have the following algebra:\n$$\n\\begin{aligned}\n\\frac{x^{4}-x^{2}}{x^{6}+2 x^{3}-1} & = \\frac{x-\\frac{1}{x}}{x^{3}+2-\\frac{1}{x^{3}}} \\\\\n& = \\frac{x-\\frac{1}{x}}{\\left(x-\\frac{1}{x}\\right)^{3}+2+3\\left(x-\\frac{1}{x}\\right)} \\\\\n& \\leq \\frac{x-\\frac{1}{x}}{3\\left(x-\\frac{1}{x}\\right)+3\\left(x-\\frac{1}{x}\\right)} = \\frac{1}{6}\n\\end{aligned}\n$$\nwhere $\\left(x-\\frac{1}{x}\\right)^{3}+1+1 \\geq 3\\left(x-\\frac{1}{x}\\right)$ in the denominator was deduced by the AM-GM inequality. As a quick check, equality holds where $x-\\frac{1}{x}=1$ or when $x=\\frac{1+\\sqrt{5}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = x^{2} - 2x$. How many distinct real numbers $c$ satisfy $f(f(f(f(c)))) = 3$?",
    "options": [],
    "answer": "9",
    "solution": "Solution:\nWe see the size of the set $f^{-1}\\left(f^{-1}\\left(f^{-1}\\left(f^{-1}(3)\\right)\\right)\\right)$. Note that $f(x) = (x-1)^{2} - 1 = 3$ has two solutions: $x = 3$ and $x = -1$, and that the fixed points $f(x) = x$ are $x = 3$ and $x = 0$. Therefore, the number of real solutions is equal to the number of distinct real numbers $c$ such that $c = 3$, $c = -1$, $f(c) = -1$ or $f(f(c)) = -1$, or $f(f(f(c))) = -1$. The equation $f(x) = -1$ has exactly one root $x = 1$. Thus, the last three equations are equivalent to $c = 1$, $f(c) = 1$, and $f(f(c)) = 1$. $f(c) = 1$ has two solutions, $c = 1 \\pm \\sqrt{2}$, and for each of these two values $c$ there are two preimages. It follows that the answer is $1 + 1 + 1 + 2 + 4 = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriangle $ABC$ has side lengths $AB = 65$, $BC = 33$, and $AC = 56$. Find the radius of the circle tangent to sides $AC$ and $BC$ and to the circumcircle of triangle $ABC$.",
    "options": [],
    "answer": "24",
    "solution": "Solution:\n\nLet $\\Gamma$ be the circumcircle of triangle $ABC$, and let $E$ be the center of the circle tangent to $\\Gamma$ and the sides $AC$ and $BC$. Notice that $\\angle C = 90^{\\circ}$ because $33^{2} + 56^{2} = 65^{2}$. Let $D$ be the second intersection of line $CE$ with $\\Gamma$, so that $D$ is the midpoint of the arc $AB$ away from $C$. Because $\\angle BCD = 45^{\\circ}$, one can easily calculate $CD = 89 \\sqrt{2} / 2$. The power of $E$ with respect to $\\Gamma$ is both $r(65 - r)$ and $r \\sqrt{2} \\cdot (89 \\sqrt{2} / 2 - r \\sqrt{2}) = r(89 - 2r)$, so $r = 89 - 65 = 24$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all positive integers $n > 1$ for which $\\frac{n^{2} + 7n + 136}{n-1}$ is the square of a positive integer.",
    "options": [],
    "answer": "n = 5, 37",
    "solution": "Solution:\n\nWrite $\\frac{n^{2} + 7n + 136}{n-1} = n + \\frac{8n + 136}{n-1} = n + 8 + \\frac{144}{n-1} = 9 + (n-1) + \\frac{144}{n-1}$.\n\nWe seek to find $p$ and $q$ such that $pq = 144$ and $p + q + 9 = k^{2}$.\n\nThe possibilities are seen to be $1 + 144 + 9 = 154$, $2 + 72 + 9 = 83$, $3 + 48 + 9 = 60$, $4 + 36 + 9 = 49$, $6 + 24 + 9 = 39$, $8 + 18 + 9 = 35$, $9 + 16 + 9 = 34$, and $12 + 12 + 9 = 33$.\n\nOf these, $\\{p, q\\} = \\{4, 36\\}$ is the only solution to both equations. Hence $n-1 = 4, 36$ and $n = 5, 37$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the largest positive integer $n$ such that $1! + 2! + 3! + \\cdots + n!$ is a perfect square. Prove that your answer is correct.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nClearly $1! + 2! + 3! = 9$ works. For $n \\geq 4$, we have\n$$\n1! + 2! + 3! + \\cdots + n! \\equiv 1! + 2! + 3! + 4! \\equiv 3 \\pmod{5}\n$$\nbut there are no squares congruent to $3$ modulo $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe lottery cards of a certain lottery contain all nine-digit numbers that can be formed with the digits $1$, $2$ and $3$. There is exactly one number on each lottery card. There are only red, yellow and blue lottery cards. Two lottery numbers that differ from each other in all nine digits always appear on cards of different color. Someone draws a red card and a yellow card. The red card has the number $122222222$ and the yellow card has the number $222222222$. The first prize goes to the lottery card with the number $123123123$. What color(s) can it possibly have? Prove your answer.",
    "options": [],
    "answer": "red",
    "solution": "Solution:\n\nFirst, it can in fact be red, if, say, cards are colored based on the first digit only ($1=$ red, $2=$ yellow, $3=$ blue$)$. We now endeavor to show it must be red.\n\nConsider the cards $333133133$ and $331331331$: they each differ in all their digits from $122222222$ and from $222222222$, so they must both be blue. Now $211311311$ differs in all its digits from both $122222222$ and $333133133$, so it must be yellow. Finally, $123123123$ differs in all its digits from both $331331331$ and $211311311$, so it must be red.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA $3 \\times 3 \\times 3$ cube is built from 27 unit cubes. Suddenly five of those cubes mysteriously teleport away. What is the minimum possible surface area of the remaining solid? Prove your answer.",
    "options": [],
    "answer": "50",
    "solution": "Solution:\nOrient the cube so that its edges are parallel to the $x$-, $y$-, and $z$-axes. A set of three unit cubes whose centers differ only in their $x$-coordinate will be termed an \"$x$-row\"; there are thus nine $x$-rows. Define \"$y$-row\" and \"$z$-row\" similarly.\nTo achieve 50, simply take away one $x$-row and one $y$-row (their union consists of precisely five unit cubes).\n\nTo show that 50 is the minimum: Note that there cannot be two $x$-rows that are both completely removed, as that would imply removing six unit cubes. (Similar statements apply for $y$- and $z$-rows, of course.) It is also impossible for there to be one $x$-row, one $y$-row, and one $z$-row that are all removed, as that would imply removing seven unit cubes. Every $x$-, $y$-, or $z$-row that is not completely removed contributes at least 2 square units to the surface area. Thus, the total surface area is at least $9 \\cdot 2 + 8 \\cdot 2 + 8 \\cdot 2 = 50$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 47153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor each positive integer $n$ let $S_{n}$ denote the set $\\{1,2,3, \\ldots, n\\}$. Compute the number of triples of subsets $A, B, C$ of $S_{2006}$ (not necessarily nonempty or proper) such that $A$ is a subset of $B$ and $S_{2006}-A$ is a subset of $C$.",
    "options": [],
    "answer": "4^{2006}",
    "solution": "Solution:\n\nLet $A_{o}, B_{o}, C_{o}$ be sets satisfying the said conditions. Note that $1 \\in A_{o}$ implies that $1 \\in B_{o}$ and $1 \\notin S_{2006}-A_{o}$ so that 1 may or may not be in $C_{o}$. Also,\n$1 \\notin A_{o}$ implies that $1 \\in S_{2006}-A_{o} \\subset C_{o}$ while 1 may or may not be in $B_{o}$. Thus there are four possibilities for the distribution of 1, and since the same argument holds independently for $2,3, \\ldots, 2006$, the answer is $4^{2006}$ or $2^{4012}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all ordered triples $(x, y, z)$ of positive reals such that $x + y + z = 27$ and $x^{2} + y^{2} + z^{2} - x y - y z - z x = 0$. Prove that your answer is correct.",
    "options": [],
    "answer": "(9, 9, 9)",
    "solution": "Solution:\nWe have $x^{2} + y^{2} + z^{2} - x y - y z - z x = \\frac{(x - y)^{2} + (y - z)^{2} + (z - x)^{2}}{2} = 0$, which implies $x = y = z = \\frac{27}{3} = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHaving lost a game of checkers and my temper, I dash all the pieces to the ground but one. This last checker, which is perfectly circular in shape, remains completely on the board, and happens to cover equal areas of red and black squares. Prove that the center of this piece must lie on a boundary between two squares (or at a junction of four).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose, for the sake of contradiction, that the problem is false. Evidently, at least one boundary between adjacent squares must lie within our checker, or else the checker would exist entirely within one square, meaning it would cover only one color. Note also that a checker's diameter is smaller than the side of any square of the board, so there are at most two such boundaries within our checker (one in each direction). Let $\\ell$ be this, or one of these, boundaries. Draw a diameter $d$ of the checker parallel to $\\ell$. Presumably, the strip of the checker between $\\ell$ and $d$ is part red, part black. These red and black areas are unequal, however, because the center of the checker does not lie on any boundary between squares. But, if we were to swap colors within this strip, then the checker would have equal red and black areas, because then it would be colored in a way such that flipping it across $d$ swaps the colors. This shows that, the way it is currently colored, the checker does not have equal red and black areas. This gives us the desired contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFour circles with radii $1, 2, 3$, and $r$ are externally tangent to one another. Compute $r$. (No proof is necessary.)",
    "options": [],
    "answer": "6/23",
    "solution": "Solution:\n\nLet $A, B, C, P$ be the centers of the circles with radii $1, 2, 3$, and $r$, respectively. Then, $ABC$ is a $3$-$4$-$5$ right triangle. Using the law of cosines in $\\triangle PAB$ yields\n$$\n\\cos \\angle PAB = \\frac{3^2 + (1 + r)^2 - (2 + r)^2}{2 \\cdot 3 \\cdot (1 + r)} = \\frac{3 - r}{3(1 + r)}\n$$\nSimilarly,\n$$\n\\cos \\angle PAC = \\frac{4^2 + (1 + r)^2 - (3 + r)^2}{2 \\cdot 4 \\cdot (1 + r)} = \\frac{2 - r}{2(1 + r)}\n$$\nWe can now use the equation $(\\cos \\angle PAB)^2 + (\\cos \\angle PAC)^2 = 1$, which yields $0 = 23r^2 + 132r - 36 = (23r - 6)(r + 6)$, or $r = 6/23$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA number $n$ is called bummed out if there is exactly one ordered pair of positive integers $(x, y)$ such that\n$$\n\\left\\lfloor x^{2} / y\\right\\rfloor+\\left\\lfloor y^{2} / x\\right\\rfloor=n .\n$$\nFind all bummed out numbers.",
    "options": [],
    "answer": "2, 6, 8, 10",
    "solution": "Solution:\nSuppose $n$ is bummed out. If $(a, b)$ is one solution for $(x, y)$ to the given equation $\\left\\lfloor x^{2} / y\\right\\rfloor+\\left\\lfloor y^{2} / x\\right\\rfloor=n$, then $(b, a)$ is another, so the unique solution $(a, b)$ better have the property that $a=b$ and $n=2 a \\geq 2$. In particular, $n$ is an even positive integer.\n\nNow, if $n=2 a \\geq 12$, then setting $x=a-1 \\geq 5, y=a+1 \\geq 7$, we have\n$$\n\\left\\lfloor\\frac{x^{2}}{y}\\right\\rfloor+\\left\\lfloor\\frac{y^{2}}{x}\\right\\rfloor=\\left\\lfloor a-3+\\frac{4}{a+1}\\right\\rfloor+\\left\\lfloor a+3+\\frac{4}{a-1}\\right\\rfloor=2 a=n\n$$\nso $n$ cannot be bummed out.\n\nMoreover, $\\left\\lfloor 1^{2} / 2\\right\\rfloor+\\left\\lfloor 2^{2} / 1\\right\\rfloor=4$, so 4 is not bummed out. The only possibilities left are $2,6,8$, and 10.\n\nTo check these, note that\n$$\nn=\\left\\lfloor\\frac{x^{2}}{y}\\right\\rfloor+\\left\\lfloor\\frac{y^{2}}{x}\\right\\rfloor>-2+\\frac{x^{2}}{2 y}+\\frac{x^{2}}{2 y}+\\frac{y^{2}}{x} \\geq-2+\\frac{3 x}{\\sqrt[3]{4}}\n$$\nso\n$$\nx<\\frac{\\sqrt[3]{4}}{3}(n+2)<.53(n+2)\n$$\nand similarly for $y$. So we only have to check $x, y \\leq\\lfloor .53(10+2)\\rfloor=6$:\n\n| $x \\backslash y$ | 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | 2 | 4 | 9 | 16 | 25 | 36 |\n| 2 | 4 | 4 | 5 | 9 | 12 | 18 |\n| 3 | 9 | 5 | 6 | 7 | 9 | 13 |\n| 4 | 16 | 9 | 7 | 8 | 9 | 11 |\n| 5 | 25 | 12 | 9 | 9 | 10 | 11 |\n| 6 | 36 | 18 | 13 | 11 | 11 | 12 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the prime factorization of\n$$\n2006^{2} \\cdot 2262 - 669^{2} \\cdot 3599 + 1593^{2} \\cdot 1337\n$$\n(No proof is necessary.)",
    "options": [],
    "answer": "2 * 3 * 7 * 13 * 29 * 59 * 61 * 191",
    "solution": "Solution:\nUpon observing that $2262 = 669 + 1593$, $3599 = 1593 + 2006$, and $1337 = 2006 - 669$, we are inspired to write $a = 2006$, $b = 669$, $c = -1593$. The expression in question then rewrites as $a^{2}(b-c) + b^{2}(c-a) + c^{2}(a-b)$. But, by experimenting in the general case (e.g. setting $a = b$), we find that this polynomial is zero when two of $a, b, c$ are equal. Immediately we see that it factors as $(b-a)(c-b)(a-c)$, so the original expression is a way of writing $(-1337) \\cdot (-2262) \\cdot (3599)$. Now, $1337 = 7 \\cdot 191$, $2262 = 2 \\cdot 3 \\cdot 13 \\cdot 29$, and $3599 = 60^{2} - 1^{2} = 59 \\cdot 61$.\n\n**Answer:** $2 \\cdot 3 \\cdot 7 \\cdot 13 \\cdot 29 \\cdot 59 \\cdot 61 \\cdot 191$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c, d$ be real numbers so that $c, d$ are not both $0$. Define the function\n$$\nm(x) = \\frac{a x + b}{c x + d}\n$$\non all real numbers $x$ except possibly $-d / c$, in the event that $c \\neq 0$. Suppose that the equation $x = m(m(x))$ has at least one solution that is not a solution of $x = m(x)$. Find all possible values of $a + d$. Prove that your answer is correct.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nThat $0$ is a possible value of $a + d$ can be seen by taking $m(x) = -x$, i.e., $a = -d = 1$, $b = c = 0$. We will now show that $0$ is the only possible value of $a + d$.\n\nThe equation $x = m(m(x))$ implies\n$$\nx = \\frac{\\left(a^{2} + b c\\right) x + (a + d) b}{(a + d) c x + \\left(b c + d^{2}\\right)}\n$$\nwhich in turn implies\n$$\n(a + d)\\left[c x^{2} + (-a + d) x - b\\right] = 0\n$$\nSuppose for the sake of contradiction that $a + d \\neq 0$. Then the above equation would further imply\n$$\nc x^{2} + (-a + d) x - b = 0, \\quad x(c x + d) = a x + b,\n$$\nwhich would imply $x = m(x)$ for any $x$ except possibly $-d / c$. But of course $-d / c$ is not a root of $x = m(m(x))$ anyway, so in this case, all solutions of $x = m(m(x))$ are also solutions of $x = m(x)$, a contradiction. So our assumption was wrong, and in fact $a + d = 0$, as claimed.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind, with proof, all positive integer palindromes whose square is also a palindrome.",
    "options": [],
    "answer": "Exactly the palindromes whose digits satisfy that the sum of their squares is less than ten; equivalently, a palindrome has a palindromic square if and only if the sum of squares of its decimal digits is less than ten.",
    "solution": "Solution:\nLet $n := \\sum_{i=0}^{d} a_{i} \\cdot 10^{i}$ be a palindrome, where the $a_{i}$ are digits with $a_{i} = a_{d-i}$ and $a_{d} \\neq 0$. Then, if we let\n$$\nb_{k} := \\sum_{i+j=k} a_{i} a_{j}\n$$\nfor all $0 \\leq k \\leq 2d$, then\n$$\nn^{2} = \\sum_{k=0}^{2d} b_{k} \\cdot 10^{k}\n$$\n(this is not necessarily the decimal expansion of $n^{2}$, however). We have to show that $\\sum_{i=0}^{d} a_{i}^{2} < 10$ if and only if $n^{2}$ is a palindrome.\n\nSuppose $\\sum_{i=0}^{d} a_{i}^{2} < 10$. Then, by the AM-GM inequality, we have\n$$\nb_{k} = \\sum_{i+j=k} a_{i} a_{j} \\leq \\sum_{i+j=k} \\frac{a_{i}^{2} + a_{j}^{2}}{2} \\leq \\sum_{i=0}^{d} \\frac{a_{i}^{2}}{2} + \\sum_{j=0}^{d} \\frac{a_{j}^{2}}{2} < \\frac{10}{2} + \\frac{10}{2} = 10\n$$\nThus, loosely speaking, no carrying is ever done in computing $n \\times n$ by long multiplication, so the digit in the $10^{k}$ place in $n^{2}$ is precisely $b_{k}$, and it's easy to see that $b_{k} = b_{2d-k}$ and that $b_{2d} = a_{d}^{2} \\neq 0$. So $n^{2}$ is indeed a palindrome, as desired.\n\nNow suppose $\\sum_{i=0}^{d} a_{i}^{2} \\geq 10$. Here note that\n$$\nb_{d} = \\sum_{i+j=d} a_{i} a_{j} = \\sum_{i=0}^{d} a_{i} a_{d-i} = \\sum_{i=0}^{d} a_{i}^{2} \\geq 10\n$$\nThus, it cannot be true that, for all $k$, $b_{k}$ represents the $10^{k}$ digit of $n^{2}$, because no digit can be greater than or equal to 10. Let $\\ell$ be the greatest such that $b_{\\ell}$ does not represent the $10^{\\ell}$ digit of $n^{2}$. We are trying to prove that $n^{2}$ cannot be a palindrome. Consider three cases:\n\n- $a_{d} = a_{0} \\geq 4$. In this case we must have $\\ell \\geq 2d$, because $b_{2d} = a_{d}^{2} > 10$.\nIf $a_{0} = 4$, then $n^{2}$ ends in the digit 6, but lies in the interval $[16 \\cdot 10^{2d}, 25 \\cdot 10^{2d})$, and so starts with either a 1 or a 2; thus, $n^{2}$ cannot be a palindrome. Similarly, if $a_{0} = 5$, then $n^{2}$ ends in 5 but starts with 2 or 3; if $a_{0} = 6$, then $n^{2}$ ends in 6 but starts with 3 or 4; if $a_{0} = 7$, then $n^{2}$ ends in 9 but starts with 4, 5, or 6; if $a_{0} = 8$, then $n^{2}$ ends in 4 but starts with 6, 7 or 8; if $a_{0} = 9$, then $n^{2}$ ends in 1 but starts with 8 or 9.\n\n- $\\ell \\geq 2d$ and $a_{d} = a_{0} \\leq 3$.\nHere we do something similar, but with a slight twist. The units digit of $n^{2}$ is $a_{0}^{2}$. Because $\\ell \\geq 2d$, $n^{2}$ must be in the interval $[(a_{0}^{2} + 1) \\cdot 10^{2d}, (a_{0} + 1)^{2} \\cdot 10^{2d})$, which is certainly a subset of the interval $[(a_{0}^{2} + 1) \\cdot 10^{2d}, a_{0}^{2} \\cdot 10^{2d+1})$. No integer in even this larger interval manages to start with the digit $a_{0}^{2}$, so $n^{2}$ cannot be palindromic.\n\n- $\\ell < 2d$.\nHere we can rest assured that $n^{2}$ does have $(2d+1)$ digits - that is, the first digit is in the $10^{2d}$ place. In order for $n^{2}$ to be a palindrome, the digits in the $10^{k}$ and $10^{2d-k}$ places must always be the same.\nNow $b_{\\ell}, b_{\\ell+1}, \\ldots, b_{2d}$ had all better be less than 10, or else $\\ell$ would be greater than what it is. Thus, the numbers just listed do appear as the lowest digits of $n^{2}$ in left-to-right order, although they don't appear as the highest $(2d+1-\\ell)$ digits of $n^{2}$ in right-to-left order. Thus, $n^{2}$ cannot be a palindrome.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence $a_{1}, a_{2}, a_{3}, \\ldots$ of positive reals satisfies $a_{n+1}=\\sqrt{\\frac{1+a_{n}}{2}}$. Determine all $a_{1}$ such that $a_{i}=\\frac{\\sqrt{6}+\\sqrt{2}}{4}$ for some positive integer $i$.",
    "options": [],
    "answer": "a1 ∈ { (sqrt(6)+sqrt(2))/4, sqrt(3)/2, 1/2 }",
    "solution": "Solution:\nClearly $a_{1}<1$, or else $1 \\leq a_{1} \\leq a_{2} \\leq a_{3} \\leq \\ldots$\nWe can therefore write $a_{1}=\\cos \\theta$ for some $0<\\theta<90^{\\circ}$.\nNote that $\\cos \\frac{\\theta}{2}=\\sqrt{\\frac{1+\\cos \\theta}{2}}$, and $\\cos 15^{\\circ}=\\frac{\\sqrt{6}+\\sqrt{2}}{4}$.\nHence, the possibilities for $a_{1}$ are $\\cos 15^{\\circ}, \\cos 30^{\\circ}$, and $\\cos 60^{\\circ}$, which are $\\frac{\\sqrt{2}+\\sqrt{6}}{2}, \\frac{\\sqrt{3}}{2}$, and $\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBegining at a vertex, an ant crawls between the vertices of a regular octahedron. After reaching a vertex, it randomly picks a neighboring vertex (sharing an edge) and walks to that vertex along the adjoining edge (with all possibilities equally likely.) What is the probability that after walking along $2006$ edges, the ant returns to the vertex where it began?",
    "options": [],
    "answer": "(2^2005 + 1) / (3 * 2^2006)",
    "solution": "Solution:\n\nFor each nonnegative integer $n$, let $a_{n}, b_{n}$, and $c_{n}$ denote the respective probabilities that the ant is where it began, at a neighbor of where it began, or is opposite where it began after moving along $n$ edges. We seek $a_{2006}$. We have $a_{0}=1$ and $b_{0}=c_{0}=0$. We also have the recursive system\n$$\n\\begin{aligned}\n& a_{n}=\\frac{b_{n-1}}{4} \\\\\n& b_{n}=a_{n-1}+\\frac{b_{n-1}}{2}+c_{n-1} \\\\\n& c_{n}=\\frac{b_{n-1}}{4}\n\\end{aligned}\n$$\nfor integers $n \\geq 1$. Substituting into the second equation we have $b_{n}=\\frac{b_{n-1}}{2}+\\frac{b_{n-2}}{2}$ for $n \\geq 2$. Solving the characteristic equation $x^{2}-\\frac{x}{2}-\\frac{1}{2}=0$ for $x=1, \\frac{-1}{2}$, we write $b_{n}=a \\cdot 1^{n}+b(-1 / 2)^{n}$. Using $b_{0}=0, b_{1}=1$, we compute\n$$\nb_{n}=\\frac{2}{3} \\cdot\\left(1-(-1 / 2)^{n}\\right)\n$$\nFrom which we find $a_{2006}=\\frac{b_{2005}}{4}=\\frac{1}{6}\\left(1+\\frac{1}{2^{2005}}\\right)=\\frac{2^{2005}+1}{3 \\cdot 2^{2006}}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCyclic quadrilateral $ABCD$ has side lengths $AB = 1$, $BC = 2$, $CD = 3$ and $DA = 4$. Points $P$ and $Q$ are the midpoints of $\\overline{BC}$ and $\\overline{DA}$. Compute $PQ^{2}$.",
    "options": [],
    "answer": "116/35",
    "solution": "Solution:\n\nConstruct $\\overline{AC}$, $\\overline{AQ}$, $\\overline{BQ}$, $\\overline{BD}$, and let $R$ denote the intersection of $\\overline{AC}$ and $\\overline{BD}$. Because $ABCD$ is cyclic, we have that $\\triangle ABR \\sim \\triangle DCR$ and $\\triangle ADR \\sim \\triangle BCR$. Thus, we may write $AR = 4x$, $BR = 2x$, $CR = 6x$, $DR = 12x$. Now, Ptolemy applied to $ABCD$ yields $140x^{2} = 1 \\cdot 3 + 2 \\cdot 4 = 11$.\n\nNow $\\overline{BQ}$ is a median in triangle $ABD$. Hence,\n$$\nBQ^{2} = \\frac{2BA^{2} + 2BD^{2} - AD^{2}}{4}.\n$$\nLikewise,\n$$\nCQ^{2} = \\frac{2CA^{2} + 2CD^{2} - DA^{2}}{4}.\n$$\nBut $PQ$ is a median in triangle $BQC$, so\n$$\nPQ^{2} = \\frac{2BQ^{2} + 2CQ^{2} - BC^{2}}{4} = \\frac{AB^{2} + BD^{2} + CD^{2} + CA^{2} - BC^{2} - AD^{2}}{4} = \\frac{(196 + 100)x^{2} + 1^{2} + 3^{2} - 2^{2} - 4^{2}}{4} = \\frac{148x^{2} - 5}{2} = \\frac{148 \\cdot \\frac{11}{140} - 5}{2} = \\frac{116}{35}.\n$$\n\nAnother solution is possible. Extend $\\overline{AD}$ and $\\overline{BC}$ past $A$ and $B$ to their intersection $S$. Use similar triangles $SAB$ and $SCD$, and similar triangles $SAC$ and $SBD$ to compute $SA$ and $SB$, then apply the Law of Cosines twice, first to compute the cosine of $\\angle A$ and then to compute $PQ^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $AB = 2$, $CA = 3$, $BC = 4$. Let $D$ be the point diametrically opposite $A$ on the circumcircle of $ABC$, and let $E$ lie on line $AD$ such that $D$ is the midpoint of $\\overline{AE}$. Line $l$ passes through $E$ perpendicular to $\\overline{AE}$, and $F$ and $G$ are the intersections of the extensions of $\\overline{AB}$ and $\\overline{AC}$ with $l$. Compute $FG$.",
    "options": [],
    "answer": "1024/45",
    "solution": "Solution:\n\nUsing Heron's formula we arrive at $[ABC] = \\frac{3 \\sqrt{15}}{4}$. Now invoking the relation $[ABC] = \\frac{abc}{4R}$ where $R$ is the circumradius of $ABC$, we compute $R^2 = \\left(\\frac{2 \\cdot 3}{[ABC]^2}\\right) = \\frac{64}{15}$. Now observe that $\\angle ABD$ is right, so that $BDEF$ is a cyclic quadrilateral. Hence $AB \\cdot AF = AD \\cdot AE = 2R \\cdot 4R = \\frac{512}{15}$. Similarly, $AC \\cdot AG = \\frac{512}{15}$. It follows that $BCGF$ is a cyclic quadrilateral, so that triangles $ABC$ and $AGF$ are similar. Then $FG = BC \\cdot \\frac{AF}{AC} = 4 \\cdot \\frac{512}{2 \\cdot 15 \\cdot 3} = \\frac{1024}{45}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSince $24=3+5+7+9$, the number $24$ can be written as the sum of at least two consecutive odd positive integers.\n\na. Can $2005$ be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide a proof why not.\n\nb. Can $2006$ be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide a proof why not.",
    "options": [],
    "answer": "2005 = 397 + 399 + 401 + 403 + 405; 2006 cannot be written as a sum of at least two consecutive odd positive integers.",
    "solution": "Solution:\n\nLet $N = (2k+1) + (2k+3) + \\cdots + (2k+2n-1)$ where $n$ and $k$ are integers, $n \\geq 2$, $k \\geq 0$. Then\n$$\nN = [(2k+1) + (2k+2n-1)] \\frac{n}{2} = (2k+n)n,\n$$\nwhich is a product of two integers with the same parity, since adding the even number $2k$ to the integer $n$ does not change its parity.\n\n1. Since $2005 = 401 \\cdot 5$, taking $n = 5$ gives $2k+5 = 401$, i.e. $2k+1 = 397$. Checking, we see that $397 + 399 + 401 + 403 + 405 = 401 \\cdot 5 = 2005$, so $2005$ can be represented as the sum of at least two consecutive odd positive integers.\n\n2. Since $2006 = 2 \\cdot 1003 = 2 \\cdot 17 \\cdot 59$ is divisible by $2$ but not by $4$, every pair of integers with a product of $2006$ consists of two integers with different parity (i.e. one odd and one even). Therefore $2006$ cannot be represented as the sum of at least two consecutive odd positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47166,
    "subject": "Mathematics (Multi-modal)",
    "question": "In acute triangle $ABC$, segments $AD$, $BE$, and $CF$ are its altitudes, and $H$ is its orthocenter. Circle $\\omega$, centered at $O$, passes through $A$ and $H$ and intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. The circumcircle of triangle $OPQ$ is tangent to segment $BC$ at $R$. Prove that $CR/BR = ED/FD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution:** (Based on work by Ryan Ko) Let $M$ be the midpoint of segment $AH$. Since $\\angle AEH = \\angle AFH = 90^\\circ$, quadrilateral $AEHF$ is cyclic with $M$ as its circumcenter. Hence triangle $EFM$ is isosceles with vertex angle $\\angle EMF = 2\\angle CAB = 2x$. Likewise, triangle $PQO$ is also an isosceles angle with vertex angle $\\angle POQ = 2x$. Therefore, triangles $EFM$ and $PQO$ are similar.\n\n![](attached_image_1.png)\n\nSince $AEHF$ and $APHQ$ are cyclic, we have $\\angle EFH = \\angle EAH = \\angle PQH$ and $\\angle FEH = \\angle FAH = \\angle QPH$. Consequently, triangles $HEF$ and $HPQ$ are similar. It is not difficult to see that quadrilaterals $EHFM$ and $PHQO$ are similar. More precisely, if $\\angle QHF = \\theta$, there is a spiral similarity $S$, centered at $H$ with clockwise rotation angle $\\theta$ and ratio $QH/FH$, that sends $FMEH$ to $QOPH$. Let $R_1$ be the point in between $B$ and $D$ such that $\\angle R_1HD = \\theta$. Then triangles $QHF$ and $R_1HD$ are similar. Hence $S(D) = R_1$. It follows that\n$$\nS(DFME) = R_1QOP.\n$$\nIt is well known that points $D$, $E$, $F$, and $M$ lie on a circle (the **nine-point circle** of triangle $ABC$). (This fact can be established easily by noting that $ABDE$ and $ACDF$ are cyclic, implying that\n$\\angle FDB = \\angle CAF = x$, $\\angle EDC = \\angle BAE = x$, and $\\angle EDF = 180^\\circ - 2x = 180^\\circ - \\angle EMF$.) Since $DFME$ is cyclic, $R_1QOP$ must also be cyclic. By the given conditions of the problem, we conclude that $R_1 = R$, implying that\n$$\nS(DEF) = RPQ,\n$$\nor triangles $DEF$ and $RPQ$ are similar. It follows that\n$$\n\\frac{ED}{FD} = \\frac{PR}{QR}.\n$$\n![](attached_image_2.png)\n\nNow we are ready to finish our proof. Since $ACDF$ and $ABDE$ are cyclic, $\\angle BFD = \\angle AFE = \\angle ACB = z$. Thus $\\angle DFE = 180^\\circ - 2z$. Since triangles $DEF$ and $RPQ$ are similar, $\\angle RQP = 180^\\circ - 2z$. Because $CR$ is tangent to the circumcircle of triangle $PQR$, $\\angle CRP = \\angle RQP = 180^\\circ - 2z$. Thus, in triangle $CPR$, $\\angle CPR = z$, and so it is isosceles with $CR = PR$. Likewise, we have $BR = QR$. Therefore, we have\n$$\n\\frac{ED}{FD} = \\frac{PR}{QR} = \\frac{CR}{BR}.\n$$\n\n\n**Second Solution:** (Based on work by Zarathustra Brady) Let the circumcircle of triangle $BQH$ meet line $BC$ at $R_3$ (other than $B$).\n![](attached_image_3.png)\n\nSince $APHQ$ and $BQHR_3$ are cyclic, $\\angle PHQ = 180^\\circ - \\angle PAQ$ and $\\angle QHR_3 = 180^\\circ - \\angle QBR_3$, implying that $\\angle PHR_3 = 360^\\circ - \\angle PHQ - \\angle QHR_3 = 180^\\circ - \\angle ACB$. Hence $CPHR_3$ is also cyclic.\n\n(We just established a special case of **Miquel's Theorem**.) Because $BQHR_3$ and $CR_3HP$ are cyclic, we have $\\angle QR_3H = \\angle QBH = 90^\\circ - \\angle BAC$ and $\\angle HR_3P = \\angle HCP = 90^\\circ - \\angle BAC$. Hence $\\angle QR_3P = 180^\\circ - 2\\angle BAC = 180^\\circ - 2x$. Likewise, we have $\\angle PQR = 180^\\circ - 2z$ and $\\angle R_3PQ = 180^\\circ - 2y$. As we have shown in the first solution, triangle $DEF$ have the same angles. Hence triangle $R_3PQ$ is similar to triangle $DEF$. Also note that $\\angle POQ + \\angle PR_3Q = 2x + 180^\\circ - 2x = 180^\\circ$, implying that $R_3$ lies on the circumcircle of triangle $OPQ$. By the given condition, have $R_3 = R$. We can then finish our proof as we did in the first solution.\n\n$$\n\\frac{ED}{FD} = \\frac{PR}{QR} = \\frac{CR}{BR}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47167,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k+1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $N/2$.",
    "options": [],
    "answer": "N = 2k^3 + 3k^2 + 3k",
    "solution": "The minimum is $N = 2k^3 + 3k^2 + 3k$. The set\n$$\n\\{k^2+1, k^2+2, \\dots, k^2+2k+1\\}\n$$\nhas sum $2k^3 + 3k^2 + 3k + 1 = N + 1$ which exceeds $N$, but the sum of the $k$ largest elements is only $(2k^3 + 3k^2 + 3k)/2 = N/2$. Thus this $N$ is such a value.\n\nSuppose $N < 2k^3 + 3k^2 + 3k$ and there are positive integers $a_1 < a_2 < \\dots < a_{2k+1}$ with $a_1 + a_2 + \\dots + a_{2k+1} > N$ and $a_{k+2} + \\dots + a_{2k+1} \\le N/2$. Then\n$$\n(a_{k+1} + 1) + (a_{k+1} + 2) + \\dots + (a_{k+1} + k) \\le a_{k+2} + \\dots + a_{2k+1} \\le N/2 < \\frac{2k^3 + 3k^2 + 3k}{2}.\n$$\nThis rearranges to give $2k a_{k+1} \\le N - k^2 - k$ and $a_{k+1} < k^2 + k + 1$. Hence $a_{k+1} \\le k^2 + k$. Combining these we get\n$$\n2(k+1)a_{k+1} \\le N + k^2 + k.\n$$\nWe also have\n$$\n(a_{k+1} - k) + \\dots + (a_{k+1} - 1) + a_{k+1} \\ge a_1 + \\dots + a_{k+1} > N/2\n$$\nor $2(k+1)a_{k+1} > N + k^2 + k$. This contradicts the previous inequality, hence no such set exists for $N < 2k^3 + 3k^2 + 3k$ and the stated value is the minimum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{P}$ be a regular 2006-gon. A diagonal of $\\mathcal{P}$ is called *good segment* if its endpoints divide the boundary of $\\mathcal{P}$ into two parts, each composed of an odd number of sides of $\\mathcal{P}$. The sides of $\\mathcal{P}$ are also called *good segment*.\nSuppose $\\mathcal{P}$ has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of $\\mathcal{P}$. Find the maximum number of isosceles triangles having two *good segments* that could appear in such a configuration.",
    "options": [],
    "answer": "1003",
    "solution": "**First Solution:** We start with the following lemma.\n**Lemma** Let $P_iP_j$ is a diagonal used in $\\mathcal{T}$, and $\\widehat{P_iP_j}$ is non-major and contains $n$ segments of $\\mathcal{P}$, then there are at most $\\lfloor \\frac{n}{2} \\rfloor$ good triangles with vertices on $\\widehat{P_iP_j}$. More precise there are at most\n$$\n\\left\\{ \\begin{array}{ll} \\lfloor \\frac{j-i}{2} \\rfloor, & \\text{if } i < j \\\\ \\lfloor \\frac{j-i+2006}{2} \\rfloor, & \\text{if } i > j \\end{array} \\right.\n$$\ngood triangles with vertices on $\\widehat{P_iP_j}$\n*Proof:* Without loss of generality, we may assume that $i < j$. We induct on $n$.\nThe bases cases for $n=1$ and $n=2$ are trivial. Assume the statement is true for $n$ with $n \\le k$ and $2 \\le k < 1003$. We consider the case $n=k+1$.\nLet $P_iP_aP_j$ be a triangle in $\\mathcal{T}$ with $P$ on $\\widehat{P_iP_j}$. (Note that $P_i, P_a$, and $P_j$ lie on non-major arc $\\widehat{P_iP_j}$ on $\\omega$ in clockwise order. By the induction hypothesis, there are at most\n$$\n\\lfloor \\frac{a-i}{2} \\rfloor \\le \\frac{a-i}{2}\n$$\ngood triangles with vertices on $\\widehat{P_iP_a}$. Similar result holds for $\\widehat{P_aP_j}$.\nBecause $P_iP_aP_j$ is a triangle in $\\mathcal{T}$, we conclude that if a good triangles has its vertices on $\\widehat{P_iP_j}$ then either it is $P_iP_aP_j$, or all its vertices are on exactly one of $\\widehat{P_iP_a}$ or $\\widehat{P_aP_j}$. We can now apply the induction hypothesis $\\widehat{P_iP_a}$ and $\\widehat{P_aP_j}$. We conclude that there are at most\n$$\n1 + \\frac{a-i}{2} + \\frac{j-a}{2} = \\frac{j-i}{2} + 1 \\qquad (\\ddag)\n$$\ngood triangles with vertices on $\\widehat{P_iP_j}$.\nTo finish our proof, we need to reduce the value of the right-hand side of (‡) by 1. We consider the following two cases.\nIn the first case, we assume that $P_iP_aP_j$ is not good. The summand 1 on the right-hand of (†) should be taken out, and we are done.\nIn the second case, we assume that $P_iP_aP_j$ is good. Since $\\widehat{P_iP_j}$ is non-major, $P_iP_j > P_iP_a$ and $P_iP_j > P_aP_j$. We must have $P_iP_a$ and $P_aP_j$ must be the two equal good sides, and both must be the good sides. Hence both $a-i$ and $j-a$ are odd, and so we can improve (†) to\n$$\n\\left\\lfloor \\frac{a-i}{2} \\right\\rfloor \\le \\frac{a-i}{2} - \\frac{1}{2},\n$$\nand similar result hold for $\\widehat{P_aP_j}$. Then (‡) can be improved to\n$$\n1 + \\frac{a-i}{2} - \\frac{1}{2} + \\frac{j-a}{2} - \\frac{1}{2} = \\frac{j-i}{2},\n$$\ncompleting our induction.\n■\nSince $\\widehat{P_iP_j}$ is non-major, $P_aP_b < P_aP_c$ and $P_bP_c < P_aP_c$. Since $P_aP_bP_c$ is good, we must have $P_aP_b$ and $P_bP_c$ be the good segments (with equal lengths). Thus $b-a$ and $c-b$ are both odd. By the induction hypothesis, there are at most\n$$\n\\left\\lfloor \\frac{b-a}{2} \\right\\rfloor = \\frac{b-a}{2} - \\frac{1}{2}\n$$\ngood triangles with vertices on $\\widehat{P_aP_b}$. Similar result holds for $\\widehat{P_bP_c}$.\nNow we prove our main result. Let $P_iP_k$ be the longest diagonal used in $\\mathcal{T}$. Let $P_iP_jP_k$ be a non-obtuse triangle in $\\mathcal{T}$. Without loss of generality, we may assume that $i < j < k$. Since $P_iP_jP_k$ is non-obtuse, $\\widehat{P_iP_j}$, $\\widehat{P_jP_k}$, and $\\widehat{P_kP_i}$ are all non-major. By the lemma, there are at most\n$$\n\\left\\lfloor \\frac{j-i}{2} \\right\\rfloor + \\left\\lfloor \\frac{k-j}{2} \\right\\rfloor + \\left\\lfloor \\frac{i-k+2006}{2} \\right\\rfloor \\\\ \\le \\frac{j-i}{2} + \\frac{k-j}{2} + \\frac{i-k+2006}{2} = 1003\n$$\ngood triangles besides $P_iP_jP_k$.\nIf $P_iP_jP_k$ is not good, we are done. If it is, then exactly two of $j-i$, $k-j$, and $i-k$ are odd, and so (*) is strict inequality. We still have at most $1002+1=1003$ good triangles in this case, completing our proof.\n\n\n**Second Solution:** Let $P_iP_jP_k$ ($i < j < k$) be a good triangle, with $P_iP_j$ and $P_jP_k$ being good segments. This means that there are an odd number of sides of $\\mathcal{P}$ between $P_i$ and $P_j$ and also between $P_j$ and $P_k$. We say $\\widehat{P_iP_j}$ and $\\widehat{P_jP_k}$ belong to triangle ABC.\nAt least one side in each of these groups does not belong to any other good triangle. This is so because any odd triangle whose vertices are among the points between $P_i$ and $P_j$ has two sides of equal length and therefore has an even number of sides belonging to it in total. Eliminating all sides belonging to any other good triangle in $\\widehat{P_iP_j}$ must therefore leave at least one side that belongs to no other good triangle. Same argument applies to $\\widehat{P_jP_k}$. Let us assign these two sides (one in $\\widehat{P_iP_j}$ and one in $\\widehat{P_jP_k}$) to triangle $P_iP_jP_k$.\nTo each good triangle we have thus assigned a pair of sides, with no two good triangles sharing an assigned side. It follows that at most 1003 good triangles can appear in the triangulation; that is, $M \\le 1003$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\sum_{n_{60}=0}^{2} \\sum_{n_{59}=0}^{n_{60}} \\cdots \\sum_{n_{2}=0}^{n_{3}} \\sum_{n_{1}=0}^{n_{2}} \\sum_{n_{0}=0}^{n_{1}} 1\n$$",
    "options": [],
    "answer": "1953",
    "solution": "Solution:\nThe given sum counts the number of non-decreasing 61-tuples of integers $\\left(n_{0}, \\ldots, n_{60}\\right)$ from the set $\\{0,1,2\\}$. Such 61-tuples are in one-to-one correspondence with strictly increasing 61-tuples of integers $\\left(m_{0}, \\ldots, m_{60}\\right)$ from the set $\\{0,1,2, \\ldots, 62\\}$: simply let $m_{k}=n_{k}+k$. But the number of such $\\left(m_{0}, \\ldots, m_{60}\\right)$ is almost by definition $\\binom{63}{61}=\\binom{63}{2}=1953$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $ABC$ is a scalene right triangle, and $P$ is the point on hypotenuse $\\overline{AC}$ such that $\\angle ABP = 45^{\\circ}$. Given that $AP = 1$ and $CP = 2$, compute the area of $ABC$.",
    "options": [],
    "answer": "9/5",
    "solution": "Solution:\n\nNotice that $\\overline{BP}$ bisects the right angle at $B$. Thus, we write $AB = 2x$, $BC = x$. By the Pythagorean theorem, $5x^{2} = 9$, from which the area $\\frac{1}{2}(x)(2x) = x^{2} = \\frac{9}{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor this problem, our lawn is an $m \\times n$ rectangular grid of clumps, that is, with $m$ rows running east-west and $n$ columns running north-south. To be even more explicit, we might say our clumps are at the lattice points\n$$\n\\left\\{(x, y) \\in \\mathbb{Z}^2 \\mid 0 \\leq x < n \\text{ and } 0 \\leq y < m \\right\\}\n$$\nHowever, mobots are now allowed to be oriented to go either north or east only. So one allowable formation for $m=2, n=3$ might be as follows:\n$$\n\\begin{array}{lll}\n\\cdot & \\rightarrow \\\\\n\\uparrow & \\rightarrow & .\n\\end{array}\n$$\nProve that the number of allowable formations for given $m$ and $n$ is $\\frac{(m+n)!}{m!n!}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThere is a one-to-one correspondence between allowable formations and paths from $(0,0)$ to $(n, m)$ made up of $n$ moves 1 unit to the right and $m$ moves 1 unit up. The correspondence works as follows: There must be a mobot at $(0,0)$, so start the path there. If that mobot is oriented to move up, then move to the right; if that mobot is oriented to move to the right, then move up. In doing so, you will meet another mobot, upon which you can repeat the above process, until you leave the lawn. Once you leave the lawn, there will be only one way to proceed to $(n, m)$: either keep going right, or keep going up. (For the example in the problem, our path would be $(0,0)-(1,0)-(1,1)-(1,2)-(2,2)-(3,2)$.)\n\nConversely, given any path from $(0,0)$ to $(n, m)$, we can derive back a mobot formation: place a mobot at every lattice point of the lawn that the path touches, and don't orient that mobot in the same direction as the path takes when leaving that point.\n\nIt is easy to check that this is indeed a one-to-one correspondence as claimed. Every path from $(0,0)$ to $(n, m)$ consists of $n$ moves to the right and $m$ moves up done in an arbitrary order, and there are precisely $\\binom{m+n}{m}=\\frac{(m+n)!}{m!n!}$ orders.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute the positive integer less than $1000$ which has exactly $29$ positive proper divisors. (Here we refer to positive integer divisors other than the number itself.)",
    "options": [],
    "answer": "720",
    "solution": "Solution:\n\nRecall that the number $N = p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{k}^{e_{k}}$ (where the $p_{i}$ are distinct primes) has exactly $(e_{1}+1)(e_{2}+1) \\cdots (e_{k}+1)$ positive integer divisors including itself. We seek $N < 1000$ such that this expression is $30$. Since $30 = 2 \\cdot 3 \\cdot 5$, we take $e_{1} = 1$, $e_{2} = 2$, $e_{3} = 4$. Then we see that $N = 5^{1} 3^{2} 2^{4} = 720$ is satisfactory.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute $\\lim_{x \\rightarrow 0} \\frac{e^{x \\cos x}-1-x}{\\sin \\left(x^{2}\\right)}$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nLet's compute all the relevant Maclaurin series expansions, up to the quadratic terms:\n$$\nx \\cos x = x + \\ldots, \\quad e^{x \\cos x} = 1 + x + \\frac{1}{2} x^{2} + \\ldots, \\quad \\sin \\left(x^{2}\\right) = x^{2} + \\ldots\n$$\nso\n$$\n\\lim_{x \\rightarrow 0} \\frac{e^{x \\cos x}-1-x}{\\sin \\left(x^{2}\\right)} = \\lim_{x \\rightarrow 0} \\frac{\\frac{1}{2} x^{2} + \\ldots}{x^{2} + \\ldots} = \\frac{1}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $O$ be the intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let $S_{1}, S_{2}, S_{3}$, and $S_{4}$ denote the areas of the triangles $ABO$, $BCO$, $CDO$, and $DAO$.\n\na) Prove that $S_{1} \\cdot S_{3} = S_{2} \\cdot S_{4}$.\n\nb) Does there exist a quadrilateral $ABCD$ such that $S_{1}, S_{2}, S_{3}$, and $S_{4}$ are consecutive positive integers in some order?",
    "options": [],
    "answer": "no",
    "solution": "Solution:\n\na. Let $M$ and $N$ be feet of perpendiculars from $B$ and $D$ to $AC$. Then $S_{1} = AO \\cdot BM / 2$, $S_{2} = CO \\cdot BM / 2$, $S_{3} = CO \\cdot DN / 2$, and $S_{4} = AO \\cdot DN / 2$. Now the desired statement follows immediately from the previous four relations.\n\nb. We will prove that the answer to the question is no. Assume the opposite, i.e. that $n, n+1, n+2, n+3$ are the given areas, where $n$ is a positive integer. Since $S_{1} \\cdot S_{3} = S_{2} \\cdot S_{4}$ and $n < n+1 < n+2 < n+3$ we must have $n(n+3) = (n+1)(n+2)$ which is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real solutions $(x, y)$ of the system $x^{2}+y=12=y^{2}+x$.",
    "options": [],
    "answer": "(3, 3), (-4, -4), ((1 + 3*sqrt(5))/2, (1 - 3*sqrt(5))/2), ((1 - 3*sqrt(5))/2, (1 + 3*sqrt(5))/2)",
    "solution": "Solution:\nWe have $x^{2}+y = y^{2}+x$ which can be written as $(x-y)(x+y-1)=0$.\n\nThe case $x = y$ yields $x^{2}+x-12=0$, hence $(x, y) = (3, 3)$ or $(-4, -4)$.\n\nThe case $y = 1-x$ yields $x^{2}+1-x-12 = x^{2}-x-11=0$ which has solutions $x = \\frac{1 \\pm \\sqrt{1+44}}{2} = \\frac{1 \\pm 3\\sqrt{5}}{2}$. The other two solutions follow.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the probability that two cards randomly selected (without replacement) from a standard 52-card deck are neither of the same value nor the same suit?",
    "options": [],
    "answer": "12/17",
    "solution": "Solution:\n\nAfter choosing a first card, the second needs to be in one of the other three suits and of a different value. Hence, the answer is $\\frac{3 \\cdot 12}{52-1} = \\frac{12}{17}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind\n$$\n\\frac{2^{2}}{2^{2}-1} \\cdot \\frac{3^{2}}{3^{2}-1} \\cdot \\frac{4^{2}}{4^{2}-1} \\cdots \\frac{2006^{2}}{2006^{2}-1}\n$$",
    "options": [],
    "answer": "4012/2007",
    "solution": "Solution:\n\n$$\n\\prod_{k=2}^{2006} \\frac{k^{2}}{k^{2}-1} = \\prod_{k=2}^{2006} \\frac{k^{2}}{(k-1)(k+1)} = \\prod_{k=2}^{2006} \\frac{k}{k-1} \\prod_{k=2}^{2006} \\frac{k}{k+1} = \\frac{2006}{1} \\cdot \\frac{2}{2007} = \\frac{4012}{2007}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute the number of real solutions $(x, y, z, w)$ to the system of equations:\n$$\n\\begin{array}{rlrl}\nx & = z + w + z w x & z & = x + y + x y z \\\\\ny & = w + x + w x y & w & = y + z + y z w\n\\end{array}\n$$",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nThe first equation rewrites as $x = \\frac{w + z}{1 - w z}$, which is a fairly strong reason to consider trigonometric substitution. Let $x = \\tan(a)$, $y = \\tan(b)$, $z = \\tan(c)$, and $w = \\tan(d)$, where $-90^{\\circ} < a, b, c, d < 90^{\\circ}$. Under modulo $180^{\\circ}$, we find $a \\equiv c + d$, $b \\equiv d + a$, $c \\equiv a + b$, $d \\equiv b + c$. Adding all of these together yields $a + b + c + d \\equiv 0$. Then $a \\equiv c + d \\equiv -a - b$ so $b \\equiv -2a$. Similarly, $c \\equiv -2b$, $d \\equiv -2c$, $d \\equiv -2a$. Hence, $c \\equiv -2b \\equiv 4a$, $d \\equiv -2c \\equiv -8a$, and $a \\equiv -2d \\equiv 16a$, so the only possible solutions are $(a, b, c, d) \\equiv (t, -2t, 4t, -8t)$ where $15t \\equiv 0$. Checking these, we see that actually $5t \\equiv 0$, which yields 5 solutions. Our division by $1 - y z$ is valid since $1 - y z = 0$ iff $y z = 1$, but $x = y + z + x y z$ so $y = -z$, which implies that $y z \\leq 0 < 1$, which is impossible. (The solutions we have computed are in fact $(0, 0, 0, 0)$ and the cyclic permutations of $\\left(\\tan(36^{\\circ}), \\tan(-72^{\\circ}), \\tan(-36^{\\circ}), \\tan(72^{\\circ})\\right)$.)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest positive integer $k$ such that $z^{10}+z^{9}+z^{6}+z^{5}+z^{4}+z+1$ divides $z^{k}-1$.",
    "options": [],
    "answer": "84",
    "solution": "Solution:\nLet $Q(z)$ denote the polynomial divisor. We need that the roots of $Q$ are $k$-th roots of unity. With this in mind, we might observe that solutions to $z^{7}=1$ and $z \\neq 1$ are roots of $Q$, which leads to its factorization. Alternatively, we note that\n$$\n(z-1) Q(z)=z^{11}-z^{9}+z^{7}-z^{4}+z^{2}-1=\\left(z^{4}-z^{2}+1\\right)\\left(z^{7}-1\\right)\n$$\nSolving for the roots of the first factor, $z^{2}=\\frac{1+i \\sqrt{3}}{2}= \\pm \\operatorname{cis} \\pi / 3$ (we use the notation $\\operatorname{cis}(x)=\\cos (x)+i \\sin (x)$) so that $z= \\pm \\operatorname{cis}( \\pm \\pi / 6)$. These are primitive 12-th roots of unity. The other roots of $Q(z)$ are the primitive 7-th roots of unity (we introduced $z=1$ by multiplication.) It follows that the answer is $\\operatorname{lcm}[12,7]=84$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x)$ be a degree $2006$ polynomial with complex roots $c_{1}, c_{2}, \\ldots, c_{2006}$, such that the set\n$$\n\\left\\{\\left|c_{1}\\right|,\\left|c_{2}\\right|, \\ldots,\\left|c_{2006}\\right|\\right\\}\n$$\nconsists of exactly $1006$ distinct values. What is the minimum number of real roots of $f(x)$?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nThe complex roots of the polynomial must come in pairs, $c_{i}$ and $\\overline{c_{i}}$, both of which have the same absolute value. If $n$ is the number of distinct absolute values $\\left|c_{i}\\right|$ corresponding to those of non-real roots, then there are at least $2n$ non-real roots of $f(x)$. Thus $f(x)$ can have at most $2006-2n$ real roots. However, it must have at least $1006-n$ real roots, as $\\left|c_{i}\\right|$ takes on $1006-n$ more values. By definition of $n$, these all correspond to real roots. Therefore $1006-n \\leq$ number of real roots $\\leq 2006-2n$, so $n \\leq 1000$, and number of real roots $\\geq 1006-n \\geq 6$. It is easy to see that equality is attainable.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{0}, a_{1}, a_{2}, \\ldots$ be a sequence of real numbers defined by $a_{0}=21$, $a_{1}=35$, and $a_{n+2}=4 a_{n+1}-4 a_{n}+n^{2}$ for $n \\geq 2$. Compute the remainder obtained when $a_{2006}$ is divided by $100$.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nNo pattern is evident in the first few terms, so we look for a formula for $a_{n}$. If we write $a_{n}=A n^{2}+B n+C+b_{n}$ and put $b_{n+2}=4 b_{n+1}-4 b_{n}$. Rewriting the original recurrence, we find\n$$\n\\begin{aligned}\n& A n^{2}+(4 A+B) n+(4 A+2 B+C)+b_{n+2} \\\\\n& =4\\left(A n^{2}+(2 A+B) n+(A+B+C)+b_{n+1}\\right)-4\\left(A n^{2}+B n+C+b_{n}\\right)+n^{2} \\\\\n& \\quad=n^{2}+8 A n+(4 A+4 B)+4 b_{n+1}-4 b_{n}\n\\end{aligned}\n$$\nSolving, $A=1$, $B=4$, $C=8$. With this information, we can solve for $b_{0}=1$ and $b_{1}=6$. Since the characteristic equation of the recurrence of the $b_{i}$ is $x^{2}-4 x+4=(x-2)^{2}=0$, we have $b_{n}=(D n+E) \\cdot 2^{n}$ for some constants $D$ and $E$. Using the known values $b_{0}$ and $b_{1}$, we compute $D=2$ and $E=1$, and finally\n$$\na_{n}=n^{2}+4 n+8+(2 n+1) \\cdot 2^{n}\n$$\nNow, taking modulo $100$, we have $a_{2006} \\equiv 6^{2}+4 \\cdot 6+8+13 \\cdot 2^{2006} \\pmod{100}$. Evidently $2^{2006} \\equiv 0 \\pmod{4}$, but by Euler's theorem $2^{\\phi(25)} \\equiv 2^{20} \\equiv 1 \\pmod{25}$, and so $2^{2006} \\equiv 2^{6} \\equiv 14 \\pmod{25}$. Now the Chinese remainder theorem yields $2^{2006} \\equiv 64 \\pmod{100}$, and we compute $a_{2006} \\equiv 36+24+8+13 \\cdot 64 \\equiv 0 \\pmod{100}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo $18$-$24$-$30$ triangles in the plane share the same circumcircle as well as the same incircle. What's the area of the region common to both the triangles?",
    "options": [],
    "answer": "132",
    "solution": "Solution:\n\nNotice, first of all, that $18$-$24$-$30$ is $6$ times $3$-$4$-$5$, so the triangles are right. Thus, the midpoint of the hypotenuse of each is the center of their common circumcircle, and the inradius is $\\frac{1}{2}(18+24-30)=6$. Let one of the triangles be $ABC$, where $\\angle A < \\angle B < \\angle C = 90^{\\circ}$. Now the line $\\ell$ joining the midpoints of sides $AB$ and $AC$ is tangent to the incircle, because it is the right distance ($12$) from line $BC$. So, the hypotenuse of the other triangle lies along $\\ell$. We may formulate this thus: The hypotenuse of each triangle is parallel to the shorter leg, and therefore perpendicular to the longer leg, of the other. Now it is not hard to see, as a result of these parallel and perpendicularisms, that the other triangle \"cuts off\" at each vertex of $\\triangle ABC$ a smaller, similar right triangle. If we compute the dimensions of these smaller triangles, we find that they are as follows: $9$-$12$-$15$ at $A$, $6$-$8$-$10$ at $B$, and $3$-$4$-$5$ at $C$. The total area chopped off of $\\triangle ABC$ is thus\n$$\n\\frac{9 \\cdot 12}{2} + \\frac{6 \\cdot 8}{2} + \\frac{3 \\cdot 4}{2} = 54 + 24 + 6 = 84.\n$$\nThe area of $\\triangle ABC$ is $18 \\cdot 24 / 2 = 216$. The area of the region common to both the original triangles is thus $216 - 84 = 132$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoints $A$, $C$, and $B$ lie on a line in that order such that $AC = 4$ and $BC = 2$. Circles $\\omega_{1}$, $\\omega_{2}$, and $\\omega_{3}$ have $\\overline{BC}$, $\\overline{AC}$, and $\\overline{AB}$ as diameters. Circle $\\Gamma$ is externally tangent to $\\omega_{1}$ and $\\omega_{2}$ at $D$ and $E$ respectively, and is internally tangent to $\\omega_{3}$. Compute the circumradius of triangle $CDE$.",
    "options": [],
    "answer": "2/3",
    "solution": "Solution:\n\nLet the center of $\\omega_{i}$ be $O_{i}$ for $i=1,2,3$ and let $O$ denote the center of $\\Gamma$. Then $O$, $D$, and $O_{1}$ are collinear, as are $O$, $E$, and $O_{2}$. Denote by $F$ the point of tangency between $\\Gamma$ and $\\omega_{3}$; then $F$, $O$, and $O_{3}$ are collinear. Writing $r$ for the radius of $\\Gamma$ we have $OO_{1} = r + 2$, $OO_{2} = r + 1$, $OO_{3} = 3 - r$. Now since $O_{1}O_{3} = 1$ and $O_{3}O_{2} = 2$, we apply Stewart's theorem:\n$$\n\\begin{aligned}\nOO_{1}^{2} \\cdot O_{2}O_{3} + OO_{2}^{2} \\cdot O_{1}O_{3} & = OO_{3}^{2} \\cdot O_{1}O_{2} + O_{1}O_{3} \\cdot O_{3}O_{2} \\cdot O_{1}O_{2} \\\\\n2(r+2)^{2} + (r+1)^{2} & = 3(3-r)^{2} + 1 \\cdot 2 \\cdot 3\n\\end{aligned}\n$$\nWe find $r = \\frac{6}{7}$. Now the key observation is that the circumcircle of triangle $CDE$ is the incircle of triangle $OO_{1}O_{2}$. We easily compute the sides of $OO_{1}O_{2}$ to be $\\frac{13}{7}$, $\\frac{20}{7}$, and $3$. By Heron's formula, the area of $OO_{1}O_{2}$ is $\\frac{18}{7}$, but the semiperimeter is $\\frac{27}{7}$, so the desired radius is $\\frac{2}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a \\geq b \\geq c$ be real numbers such that\n$$\n\\begin{aligned}\na^{2} b c + a b^{2} c + a b c^{2} + 8 &= a + b + c \\\\\na^{2} b + a^{2} c + b^{2} c + b^{2} a + c^{2} a + c^{2} b + 3 a b c &= -4 \\\\\na^{2} b^{2} c + a b^{2} c^{2} + a^{2} b c^{2} &= 2 + a b + b c + c a\n\\end{aligned}\n$$\nIf $a + b + c > 0$, then compute the integer nearest to $a^{5}$.",
    "options": [],
    "answer": "1279",
    "solution": "Solution:\nWe factor the first and third givens, obtaining the system\n$$\n\\begin{aligned}\na^{2} b c + a b^{2} c + a b c^{2} - a - b - c = (a b c - 1)(a + b + c) &= -8 \\\\\na^{2} b + a^{2} c + b^{2} c + b^{2} a + c^{2} a + c^{2} b + 3 a b c = (a b + b c + c a)(a + b + c) &= -4 \\\\\na^{2} b^{2} c + a b^{2} c^{2} + a^{2} b c^{2} - a b - b c - c a = (a b c - 1)(a b + b c + c a) &= 2\n\\end{aligned}\n$$\nWriting $X = a + b + c$, $Y = a b + b c + c a$, $Z = a b c - 1$, we have $X Z = -8$, $X Y = -4$, $Y Z = 2$. Multiplying the three yields $(X Y Z)^{2} = 64$ from which $X Y Z = \\pm 8$. Since we are given $X > 0$, multiplying the last equation by $X$ we have $2 X = X Y Z = \\pm 8$. Evidently $X Y Z = 8$ from which $X = 4$, $Y = -1$, $Z = -2$. We conclude that $a, b, c$ are the roots of the polynomial $P(t) = t^{3} - 4 t^{2} - t + 1$. Thus, $P(a) = a^{3} - 4 a^{2} - a + 1 = 0$, and also $P(b) = P(c) = 0$. Now since $P(1/2) = -\\frac{3}{8}$, $P(0) = 1$ and $P(-2/3) = -\\frac{11}{27}$, we conclude that $-2/3 < c < 0 < b < 1/2 < a$. It follows that $|b^{5} + c^{5}| < \\frac{1}{2}$. Thus, we compute $a^{5} + b^{5} + c^{5}$.\n\nDefining $S_{n} = a^{n} + b^{n} + c^{n}$, we have $S_{n+3} = 4 S_{n+2} + S_{n+1} - S_{n}$ for $n \\geq 0$. Evidently $S_{0} = 3$, $S_{1} = 4$, $S_{2} = (a + b + c)^{2} - 2(a b + b c + c a) = 18$. Then $S_{3} = 4 \\cdot 18 + 4 - 3 = 73$, $S_{4} = 4 \\cdot 73 + 18 - 4 = 306$, and $S_{5} = 4 \\cdot 306 + 73 - 18 = 1279$. Since $|b^{5} + c^{5}| < \\frac{1}{2}$, we conclude that $|S_{5} - a^{5}| < \\frac{1}{2}$, or that 1279 is the integer nearest to $a^{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $N$ denote the number of subsets of $\\{1,2,3, \\ldots, 100\\}$ that contain more prime numbers than multiples of $4$. Compute the largest integer $k$ such that $2^{k}$ divides $N$.",
    "options": [],
    "answer": "52",
    "solution": "Solution:\nLet $S$ denote a subset with the said property. Note that there are $25$ multiples of $4$ and $25$ primes in the set $\\{1,2,3, \\ldots, 100\\}$, with no overlap between the two. Let $T$ denote the subset of $50$ numbers that are neither prime nor a multiple of $4$, and let $U$ denote the $50$ other numbers. Elements of $T$ can be arbitrarily included in or excluded by $S$.\n\nConsider $S \\cap U = S_{1}$ and $U - S = S_{2}$ (the set difference is defined to be all elements of $U$ that are not in $S$). $S_{1}$ and $S_{2}$ are two disjoint sets such that $U = S_{1} \\cap S_{2}$. If $S_{1}$ contains more multiples of $4$ than primes, then $S_{2}$ contains more primes than multiples of $4$, and conversely. Furthermore, $S_{1}$ contains an equal number of primes and multiples of $4$ if and only if $S_{2}$ contains equal numbers as well. Let $V$ denote an arbitrary subset of $T$. It follows from examining pairs of sets $V \\cup S_{1}$ and $V \\cup S_{2}$ that\n$$\n\\begin{aligned}\nN & = 2^{50} \\cdot \\frac{1}{2}\\left(2^{50} - \\sum_{k=0}^{25} \\binom{25}{k}^{2}\\right) \\\\\n& = 2^{49} \\cdot \\left(2^{50} - \\binom{50}{25}\\right)\n\\end{aligned}\n$$\nSince $50!$ is divisible by $2$ exactly $25+12+6+3+1=47$ times while $25!$ is divisible by $2$ exactly $12+6+3+1=22$ times, it follows that $\\binom{50}{25}$ is divisible by $2$ exactly $3$ times, so the answer is $49+3=52$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA pebble is shaped as the intersection of a cube of side length $1$ with the solid sphere tangent to all of the cube's edges. What is the surface area of this pebble?",
    "options": [],
    "answer": "((6*sqrt(2)-5)/2) * pi",
    "solution": "Solution:\nImagine drawing the sphere and the cube. Take a cross section, with a plane parallel to two of the cube's faces, passing through the sphere's center. In this cross section, the sphere looks like a circle, and the cube looks like a square (of side length $1$) inscribed in that circle. We can now calculate that the sphere has diameter\n$$\nd := \\sqrt{2}\n$$\nand surface area\n$$\nS := \\pi d^{2} = 2 \\pi,\n$$\nand that the sphere protrudes a distance of\n$$\nx := \\frac{\\sqrt{2} - 1}{2}\n$$\nout from any given face of the cube.\n\nIt is known that the surface area chopped off from a sphere by any plane is proportional to the perpendicular distance thus chopped off. Thus, each face of the cube chops off a fraction\n$$\n\\frac{x}{d}\n$$\nof the sphere's surface. The surface area of the pebble contributed by the sphere is thus\n$$\nS \\cdot \\left(1 - 6 \\cdot \\frac{x}{d}\\right),\n$$\nwhereas the cube contributes $6$ circles of radius $\\frac{1}{2}$, with total area\n$$\n6 \\cdot \\pi \\left(\\frac{1}{2}\\right)^{2} = \\frac{3}{2} \\pi.\n$$\nThe pebble's surface area is therefore\n$$\nS \\cdot \\left(1 - 6 \\cdot \\frac{x}{d}\\right) + \\frac{3}{2} \\pi = 2 \\pi \\cdot \\left(1 - 6 \\cdot \\frac{\\sqrt{2} - 1}{2 \\sqrt{2}}\\right) + \\frac{3}{2} \\pi = \\frac{6 \\sqrt{2} - 5}{2} \\pi.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the area in the first quadrant bounded by the hyperbola $x^{2}-y^{2}=1$, the $x$-axis, and the line $3x=4y$.",
    "options": [],
    "answer": "ln(7)/4",
    "solution": "Solution:\n\nConvert to polar coordinates: the hyperbola becomes\n$$\n1 = r^{2} (\\cos^{2} \\theta - \\sin^{2} \\theta) = r^{2} \\cos(2\\theta)\n$$\nso, letting $\\alpha := \\arctan(3/4)$, the area is\n$$\nS := \\int_{0}^{\\alpha} \\frac{r^{2}}{2} d\\theta = \\frac{1}{2} \\int_{0}^{\\alpha} \\sec(2\\theta) d\\theta = \\left. \\frac{1}{4} \\ln |\\sec(2\\theta) + \\tan(2\\theta)| \\right|_{0}^{\\alpha}.\n$$\nNow\n$$\n\\begin{gathered}\n\\tan(2\\alpha) = \\frac{2 \\tan \\alpha}{1 - \\tan^{2} \\alpha} = \\frac{2 \\cdot (3/4)}{1 - (3/4)^{2}} = \\frac{3/2}{7/16} = \\frac{24}{7} \\\\\n\\sec(2\\alpha) = \\sqrt{1 + \\tan^{2}(2\\alpha)} = \\frac{25}{7}\n\\end{gathered}\n$$\nso\n$$\nS = \\frac{1}{4} \\left( \\ln \\left| \\frac{25}{7} + \\frac{24}{7} \\right| - \\ln |1 + 0| \\right) = \\frac{\\ln 7}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least real number $k$ with the following property: if the real numbers $x$, $y$, and $z$ are not all positive, then\n$$\nk(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) \\geq (xyz)^2 - xyz + 1.\n$$",
    "options": [],
    "answer": "16/9",
    "solution": "The answer is $k = \\frac{16}{9}$.\nWe start with a lemma.\n\nLemma 1. If real numbers $s$ and $t$ are not all positive, then\n$$\n\\frac{4}{3}(s^2 - s + 1)(t^2 - t + 1) \\geq (st)^2 - st + 1. \\quad (*)\n$$\nProof: Without loss of generality, we assume that $s \\geq t$.\nWe first assume that $s \\geq 0 \\geq t$. Setting $u = -t$, $(*)$ reads\n$$\n\\frac{4}{3}(s^2 - s + 1)(u^2 + u + 1) \\geq (su)^2 + su + 1,\n$$\nor\n$$\n4(s^2 - s + 1)(u^2 + u + 1) \\geq 3s^2u^2 + 3su + 3.\n$$\nExpanding the left-hand side gives\n$$\n4s^2u^2 + 4s^2u - 4su^2 - 4su + 4s^2 + 4u^2 - 4s + 4u + 4 \\geq 3s^2u^2 + 3su + 3,\n$$\nor\n$$\ns^2u^2 + 4u^2 + 4s^2 + 1 + 4s^2u + 4u \\geq 4su^2 + 4s + 7su\n$$\nwhich is evident as $s^2u^2 + 4u^2 \\geq 4su^2$, $4s^2 + 1 \\geq 4s$, and $4s^2u + 4u \\geq 8su \\geq 7su$.\n\nWe second assume that $0 \\geq s \\geq t$. Let $v = -s$. By our previous argument, we have\n$$\n\\frac{4}{3}(v^2 - v + 1)(t^2 - t + 1) \\geq (vt)^2 - vt + 1.\n$$\nIt is clear that $t^2 - t + 1 > 0$, $s^2 - s + 1 \\geq v^2 - v + 1$, and $(vt)^2 - vt + 1 \\geq (st)^2 - st + 1$. Combining the last four inequalities gives $(*)$, and this completes the proof of the lemma.\n\nNow we show that if $x, y, z$ are not all positive real numbers, then\n$$\n\\frac{16}{9}(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) \\geq (xyz)^2 - xyz + 1. \\quad (**)\n$$\nWe consider three cases.\n\na. We assume that $y \\geq 0$. Setting $(s, t) = (y, z)$ and then $(s, t) = (x, yz)$ in the lemma gives the desired result.\n\nb. We assume that $0 \\geq y$. Setting $(s, t) = (x, y)$ and then $(s, t) = (xy, z)$ in the lemma gives the desired result.\n\nFinally, we confirm that the minimum value of $k$ is $\\frac{16}{9}$ by noting that the equality holds in $(**)$ when $(x, y, z) = (\\frac{1}{2}, \\frac{1}{2}, 0)$.\nWe establish $(**)$ by showing\n$$\ng(z) = \\frac{16}{9}(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) - (xyz)^2 + xyz - 1 \\geq 0.\n$$\nNote that $g(z)$ is a quadratic in $z$ whose axis of symmetry (found by comparing the linear and quadratic terms) is at\n$$\n\\begin{aligned}\nz &= \\frac{1}{2} - \\frac{9}{32} \\cdot \\frac{xy}{(x^2 - x + 1)(y^2 - y + 1)} \\\\\n  &= \\frac{1}{2} - \\frac{9}{32} \\cdot \\frac{1}{\\left(x + \\frac{1}{x} - 1\\right)\\left(y + \\frac{1}{y} - 1\\right)}.\n\\end{aligned}\n$$\nFor any $t$, we have $|x + \\frac{1}{x} - 1| \\ge 1$, so the absolute value of the second quantity on the right-hand side of the above equation is at most $\\frac{9}{32}$, which is less than $\\frac{1}{2}$. That is, the axis of symmetry occurs to the right side of the $y$-axis, so we only decrease the difference between the sides by replacing $z$ by $0$. But when $z = 0$, we only need to show\n$$\ng(0) = \\frac{16}{9}(x^2 - x + 1)(y^2 - y + 1) - 1 \\geq 0,\n$$\nwhich is evident as $t^2 - t + 1 = (t - \\frac{1}{2})^2 + \\frac{3}{4} \\ge \\frac{3}{4}$.\nThis is the Calculus version of the second solution. We maintain the same notation as in the second solution. We have\n$$\n\\frac{dg}{dz} = \\frac{16}{9}(2z - 1)(x^2 - x + 1)(y^2 - y + 1) - 2zx^2y^2 + xy\n$$\nor\n$$\n\\frac{dg}{dz} = 2z \\left[ \\frac{4}{3}(x^2 - x + 1)\\frac{4}{3}(y^2 - y + 1) - x^2y^2 \\right] + \\left[ xy - \\frac{4}{3}(x^2 - x + 1)\\frac{4}{3}(y^2 - y + 1) \\right]. \\quad (\\dagger)\n$$\nIt is evident that\n$$\n\\frac{4}{3}(t^2 - t + 1) \\geq t^2 \\geq 0\n$$\nas it is equivalent to $t^2 - 4t + 4 = (t - 2)^2 \\ge 0$. It follows that\n$$\n2z \\left[ \\frac{4}{3}(x^2 - x + 1)\\frac{4}{3}(y^2 - y + 1) - x^2y^2 \\right] \\le 0;\n$$\nthat is, the first summand on the right-hand side of $(\\dagger)$ is not positive. It is also evident that\n$$\n\\frac{4}{3}(t^2 - t + 1) \\geq t\n$$\nas it is equivalent to $4t^2 - 7t + 4 = 4\\left(t - \\frac{7}{8}\\right)^2 + \\frac{15}{16} > 0$. If $y \\ge 0$, then multiplying the inequalities\n$$\n\\frac{4}{3}(x^2 - x + 1) \\ge x \\ge 0 \\quad \\text{and} \\quad \\frac{4}{3}(y^2 - y + 1) \\ge y \\ge 0\n$$\ngives\n$$\n\\frac{4}{3}(x^2 - x + 1)\\frac{4}{3}(y^2 - y + 1) - xy \\ge 0.\n$$\nIf $y < 0$, then $xy < 0$, and so\n$$\n\\frac{4}{3}(x^2 - x + 1)\\frac{4}{3}(y^2 - y + 1) \\ge 0 \\ge xy.\n$$\nIn either case, we have shown that the second summand in $(\\dagger)$ is also negative. We conclude that $\\frac{dg}{dz} \\le 0$ for $z \\le 0$. Hence $g(z)$ reaches minimum when $z = 0$, and we can finish as we did in the second solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47189,
    "subject": "Mathematics (Multi-modal)",
    "question": "For integral $m$, let $p(m)$ be the greatest prime divisor of $m$. By convention, we set $p(\\pm 1) = 1$ and $p(0) = \\infty$. Find all polynomials $f$ with integer coefficients such that the sequence $\\{p(f(n^2)) - 2n\\}_{n \\ge 0}$ is bounded above. (In particular, this requires $f(n^2) \\ne 0$ for $n \\ge 0$.)",
    "options": [],
    "answer": "All such polynomials are exactly f(x) = c ∏_{i=1}^k (4x − a_i^2), where c is a nonzero integer and a_i are odd positive integers.",
    "solution": "The polynomial $f$ has the required properties if and only if\n$$\nf(x) = c(4x - a_1^2)(4x - a_2^2)\\cdots(4x - a_k^2), \\quad (*)\n$$\nwhere $a_1, a_2, \\dots, a_k$ are odd positive integers and $c$ is a nonzero integer. It is straightforward to verify that polynomials given by $(*)$ have the required property. If $p$ is a prime divisor of $f(n^2)$ but not of $c$, then $p|(2n-a_j)$ or $p|(2n+a_j)$ for some $j \\le k$. Hence $p-2n \\le \\max\\{a_1, a_2, \\dots, a_k\\}$. The prime divisors of $c$ form a finite set and do affect whether or not the given sequence is bounded above. The rest of the proof is devoted to showing that any $f$ for which $\\{p(f(n^2)) - 2n\\}_{n \\ge 0}$ is bounded above is given by $(*)$.\n\nLet $\\mathbb{Z}[x]$ denote the set of all polynomials with integral coefficients. Given $f \\in \\mathbb{Z}[x]$, let $\\mathcal{P}(f)$ denote the set of those primes that divide at least one of the numbers in the sequence $\\{f(n)\\}_{n \\ge 0}$. The solution is based on the following lemma.\n\n**Lemma** If $f \\in \\mathbb{Z}[x]$ is a nonconstant polynomial then $\\mathcal{P}(f)$ is infinite.\n\n*Proof:* Repeated use will be made of the following basic fact: if $a$ and $b$ are distinct integers and $f \\in \\mathbb{Z}[x]$, then $a-b$ divides $f(a)-f(b)$. If $f(0)=0$, then $p$ divides $f(p)$ for every prime $p$, so $\\mathcal{P}(f)$ is infinite. If $f(0)=1$, then every prime divisor $p$ of $f(n!)$ satisfies $p>n$. Otherwise $p$ divides $n!$, which in turn divides $f(n!)-f(0)=f(n!)-1$. This yields $p|1$, which is false. Hence $f(0)=1$ implies that $\\mathcal{P}(f)$ is infinite. To complete the proof, set $g(x) = f(f(0)x)/f(0)$ and observe that $g \\in \\mathbb{Z}[x]$ and $g(0)=1$. The preceding argument shows that $\\mathcal{P}(g)$ is infinite, and it follows that $\\mathcal{P}(f)$ is infinite. $\\blacksquare$\n\nSuppose $f \\in \\mathbb{Z}[x]$ is nonconstant and there exists a number $M$ such that $p(f(n^2)) - 2n \\le M$ for all $n \\ge 0$. Application of the lemma to $f(x^2)$ shows that there is an infinite sequence of distinct primes $\\{p_j\\}$ and a corresponding infinite sequence of nonnegative integers $\\{k_j\\}$ such that $p_j|f(k_j^2)$ for all $j \\ge 1$. Consider the sequence $\\{r_j\\}$ where $r_j = \\min\\{k_j \\pmod{p_j}, p_j - k_j \\pmod{p_j}\\}$. Then $0 \\le r_j \\le (p_j-1)/2$ and $p_j|f(r_j^2)$. Hence $2r_j+1 \\le p_j \\le p(f(r_j^2)) \\le M+2r_j$, so $1 \\le p_j - 2r_j \\le M$ for all $j \\ge 1$. It follows that there is an integer $a_1$ such that $1 \\le a_1 \\le M$ and $a_1 = p_j - 2r_j$ for infinitely many $j$. Let $m = \\deg f$. Then $p_j|4^m f((p_j-a_1)/2)^2)$ and $4^m f((x-a_1)/2)^2) \\in \\mathbb{Z}[x]$. Consequently, $p_j|f((a_1/2)^2)$ for infinitely many $j$, which shows that $(a_1/2)^2$ is a zero of $f$. Since $f(n^2) \\ne 0$ for $n \\ge 0$, $a_1$ must be odd. Then $f(x) = (4x-a_1^2)g(x)$ where $g \\in \\mathbb{Z}[x]$. (See the note below.) Observe that $\\{p(g(n^2)) - 2n\\}_{n \\ge 0}$ must be bounded above. If $g$ is constant, we are done. If $g$ is nonconstant, the argument can be repeated to show that $f$ is given by $(*)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $A B C$, choose point $A_{1}$ on side $B C$, point $B_{1}$ on side $C A$, and point $C_{1}$ on side $A B$ in such a way that the three segments $A A_{1}$, $B B_{1}$, and $C C_{1}$ intersect in one point $P$. Prove that $P$ is the centroid of triangle $A B C$ if and only if $P$ is the centroid of triangle $A_{1} B_{1} C_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume first that $P$ is the centroid of $\\triangle A B C$. This means that $A A_{1}$, $B B_{1}$ and $C C_{1}$ are the medians of $\\triangle A B C$, and hence $A_{1}$, $B_{1}$ and $C_{1}$ are the midpoints of $B C$, $C A$ and $A B$, respectively. As a midsegment of $\\triangle A B C$, segment $B_{1} A_{1}$ is parallel to $A B$, and similarly, $C_{1} A_{1}$ is parallel to $A C$. In the parallelogram $A C_{1} A_{1} B_{1}$, the diagonals $A A_{1}$ and $B_{1} C_{1}$ bisect each other. In particular, $A A_{1}$ contains the median of $\\triangle A_{1} B_{1} C_{1}$ passing through vertex $A_{1}$. Similarly, $B B_{1}$, respectively $C C_{1}$, contains the median of $\\triangle A_{1} B_{1} C_{1}$ passing through vertex $B_{1}$, respectively $C_{1}$. Since $A A_{1}$, $B B_{1}$ and $C C_{1}$ intersect in $P$ and they are extensions of the three medians of $\\triangle A B C$, by definition, $P$ is also the centroid of $\\triangle A_{1} B_{1} C_{1}$, and we are done with this part.\n\n![](attached_image_1.png)\n\nFor the other part of the problem, assume that $P$ is the centroid of $\\triangle A_{1} B_{1} C_{1}$. We construct $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ in the following way. Draw a line through $A_{1}$ parallel to $B_{1} C_{1}$, another line through $B_{1}$ parallel to $C_{1} A_{1}$, and a third line through $C_{1}$ parallel to $A_{1} B_{1}$. Let the three lines intersect pairwise in points $A^{\\prime}$, $B^{\\prime}$ and $C^{\\prime}$ so that $A_{1}$ lies on $B^{\\prime} C^{\\prime}$, $B_{1}$ lies on $C^{\\prime} A^{\\prime}$, and $C_{1}$ lies on $A^{\\prime} B^{\\prime}$. From parallelograms $B^{\\prime} A_{1} B_{1} C_{1}$ and $C^{\\prime} B_{1} C_{1} A_{1}$, it follows that $C^{\\prime} A_{1}=B_{1} C_{1}=B^{\\prime} A_{1}$, i.e. $A_{1}$ is the midpoint of $B^{\\prime} C^{\\prime}$. Similarly, $B_{1}$ and $C_{1}$ are the midpoints of $C^{\\prime} A^{\\prime}$ and $A^{\\prime} B^{\\prime}$, respectively. Further, from the same parallelograms follows, for instance, that $C^{\\prime} C_{1}$ bisects $A_{1} B_{1}$, and hence $C^{\\prime} C_{1}$ contains the median of $\\triangle A_{1} B_{1} C_{1}$ and therefore it passes through point $P$. Similarly, $B^{\\prime} B_{1}$ and $A^{\\prime} A_{1}$ pass through $P$. Thus, $A^{\\prime} A_{1}$, $B^{\\prime} B_{1}$ and $C^{\\prime} C_{1}$ intersect in point $P$, and are the medians in $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$. Hence $P$ is the centroid of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ too. (Note that $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ could be alternatively constructed as the image of $\\triangle A_{1} B_{1} C_{1}$ via a homothety with center $P$ and ratio 1:2, following the famous fact that every centroid divides the medians in ratio 2:1 counted from the vertices.)\n\nIt remains to show that $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ coincides with $\\triangle A B C$. By contradiction, suppose that they are not the same triangle. Without loss of generality, we may assume that point $A$ is different from point $A^{\\prime}$. By hypothesis and construction, both $A$ and $A^{\\prime}$ lie on line $A_{1} P$, and similarly both $B$ and $B^{\\prime}$ lie on line $B_{1} P$, and both $C$ and $C^{\\prime}$ lie on line $C_{1} P$. Since $A$ and $A^{\\prime}$ are different, it is not hard to see that there are only two possibilities for their relative positions: $A^{\\prime}$ is between $A$ and $P$, or $A$ is between $A^{\\prime}$ and $P$. In the first case, $A$ is outside of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$, while in the second case, $A$ is inside $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$. If we are in the first case, drawing line $A C_{1}$ to intersect $B^{\\prime} B_{1}$ in $B$ places $B$ inside $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$; then drawing line $B A_{1}$ to intersect $C^{\\prime} C_{1}$ in $C$ places $C$ outside $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$. Thus, $\\triangle A P C$ properly contains $\\triangle A^{\\prime} P C^{\\prime}$, and hence $B_{1}$ is inside $\\triangle A P C$; well, this certainly doesn't make sense since $B_{1}$ is supposed to lie on side $A C$ by hypothesis. This contradiction eliminates the first case mentioned above.\nThe second case ($A$ between $A^{\\prime}$ and $P$) is treated similarly and leads to point $B_{1}$ being outside $\\triangle A P C$, while it is supposed to lie on side $A C$: again a contradiction.\n\nThis exhausts all possible situations and implies that the initial assumption of $A$ and $A^{\\prime}$ being different is false. But if $A=A^{\\prime}$, one quickly discovers that $B$ and $B^{\\prime}$ must also coincide, as well as $C$ and $C^{\\prime}$. In conclusion, $\\triangle A B C$ is $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ constructed earlier, and hence $\\triangle A B C$'s centroid is already shown to be point $P$. This concludes the proof.\nSolution:\n\nThe first part of the problem: \"if $P$ is the centroid of $\\triangle A B C$ then $P$ is also the centroid of $\\triangle A_{1} B_{1} C_{1}$\", is shown the way same as in Solution 1.\n\nFor the second part, suppose that $P$ is the centroid of $\\triangle A_{1} B_{1} C_{1}$. Denote by $A_{2}$, $B_{2}$ and $C_{2}$ the midpoints of the corresponding sides of $\\triangle A_{1} B_{1} C_{1}$ so that $A$, $A_{2}$, $P$, $A_{1}$ are collinear, etc. We shall use the following:\n\n[Part of] Menelaus Theorem. Given $\\triangle X Y Z$, let line $l$ intersect line $X Y$ in point $L$, line $Y Z$ in point $M$, and line $Z X$ in point $N$. ($L$, $M$ and $N$ can each be inside or outside of the corresponding side of $\\triangle X Y Z$.) Then\n$$\n\\frac{X L}{L Y} \\cdot \\frac{Y M}{M Z} \\cdot \\frac{Z N}{N X}=1\n$$\n\n![](attached_image_2.png)\n\nApply Menelaus Theorem to $\\triangle B_{1} C_{1} B$ and line $A A_{2} P$:\n$$\n\\frac{C_{1} A}{A B} \\cdot \\frac{B P}{P B_{1}} \\cdot \\frac{B_{1} A_{2}}{A_{2} C_{1}}=1\n$$\n\nSince $B_{1} A_{2}=A_{2} C_{1}$, we obtain\n$$\n\\frac{C_{1} A}{A B}=\\frac{P B_{1}}{B P}\n$$\n\nUsing now $\\triangle B_{1} A_{1} B$ and line $C C_{2} P$, we similarly obtain\n$$\n\\frac{A_{1} C}{C B}=\\frac{P B_{1}}{B P}\n$$\n\nEquating (1) and (2) yields:\n$$\n\\frac{C_{1} A}{A B}=\\frac{A_{1} C}{C B} \\Rightarrow 1-\\frac{C_{1} A}{A B}=1-\\frac{A_{1} C}{C B} \\Rightarrow \\frac{A B-C_{1} A}{A B}=\\frac{C B-A_{1} C}{C B} \\Rightarrow \\frac{C_{1} B}{A B}=\\frac{A_{1} B}{C B}.\n$$\n\nThe last equality, and the fact that $\\triangle C_{1} B A_{1}$ and $\\triangle A B C$ share the same angle $\\angle B$, shows that these two triangles are similar, from which we can conclude that $C_{1} A_{1}$ is parallel to $A C$. Applying similar arguments, one shows that $A_{1} B_{1}$ is parallel to $B A$, and $B_{1} C_{1}$ is parallel to $C B$. Now we are in a situation identical to the construction of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ in Solution 1, where we easily showed that $P$ is the centroid of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$, and hence in our present situation, $P$ is the centroid of $\\triangle A B C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A$, $B$, $C$, and $D$ be points on a circle such that $AB = 11$ and $CD = 19$. Point $P$ is on segment $AB$ with $AP = 6$, and $Q$ is on segment $CD$ with $CQ = 7$. The line through $P$ and $Q$ intersects the circle at $X$ and $Y$. If $PQ = 27$, find $XY$.",
    "options": [],
    "answer": "31",
    "solution": "Solution:\n\nSuppose $X$, $P$, $Q$, $Y$ lie in that order. Let $PX = x$ and $QY = y$. By power of a point from $P$, $x \\cdot (27 + y) = 30$, and by power of a point from $Q$, $y \\cdot (27 + x) = 84$. Subtracting the first from the second, $27 \\cdot (y - x) = 54$, so $y = x + 2$. Now, $x \\cdot (29 + x) = 30$, and we find $x = 1, -30$. Since $-30$ makes no sense, we take $x = 1$ and obtain $XY = 1 + 27 + 3 = 31$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA moth starts at vertex $A$ of a certain cube and is trying to get to vertex $B$, which is opposite $A$, in five or fewer \"steps,\" where a step consists in traveling along an edge from one vertex to another. The moth will stop as soon as it reaches $B$. How many ways can the moth achieve its objective?",
    "options": [],
    "answer": "48",
    "solution": "Solution:\n\nLet $X$, $Y$, $Z$ be the three directions in which the moth can initially go. We can symbolize the trajectory of the moth by a sequence of $X$'s, $Y$'s, and $Z$'s in the obvious way: whenever the moth takes a step in a direction parallel or opposite to $X$, we write down $X$, and so on.\n\nThe moth can reach $B$ in either exactly $3$ or exactly $5$ steps. A path of length $3$ must be symbolized by $X Y Z$ in some order. There are $3! = 6$ such orders.\n\nA trajectory of length $5$ must be symbolized by $X Y Z X X$, $X Y Z Y Y$, or $X Y Z Z Z$, in some order. There are $3 \\cdot \\frac{5!}{3!1!1!} = 3 \\cdot 20 = 60$ possibilities here. However, we must remember to subtract out those trajectories that already arrive at $B$ by the $3$rd step: there are $3 \\cdot 6 = 18$ of those.\n\nThe answer is thus $60 - 18 + 6 = 48$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAt a nursery, $2006$ babies sit in a circle. Suddenly each baby pokes the baby immediately to either its left or its right, with equal probability. What is the expected number of unpoked babies?",
    "options": [],
    "answer": "1003/2",
    "solution": "Solution:\n\nThe probability that any given baby goes unpoked is $1/4$. So the answer is $2006/4 = 1003/2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAt time $0$, an ant is at $(1,0)$ and a spider is at $(-1,0)$. The ant starts walking counterclockwise along the unit circle, and the spider starts creeping to the right along the $x$-axis. It so happens that the ant's horizontal speed is always half the spider's. What will the shortest distance ever between the ant and the spider be?",
    "options": [],
    "answer": "sqrt(7/8)",
    "solution": "Solution:\n\nPicture an instant in time where the ant and spider have $x$-coordinates $a$ and $s$, respectively. If $1 \\leq s \\leq 3$, then $a \\leq 0$, and the distance between the bugs is at least $1$. If $s > 3$, then, needless to say, the distance between the bugs is at least $2$. If $-1 \\leq s \\leq 1$, then $s = 1 - 2a$, and the distance between the bugs is\n$$\n\\sqrt{(a - (1 - 2a))^2 + (1 - a^2)} = \\sqrt{8a^2 - 6a + 2} = \\sqrt{\\frac{(8a - 3)^2 + 7}{8}}\n$$\nwhich attains the minimum value of $\\sqrt{7/8}$ when $a = 3/8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $C$ be the unit circle. Four distinct, smaller congruent circles $C_{1}, C_{2}, C_{3}, C_{4}$ are internally tangent to $C$ such that $C_{i}$ is externally tangent to $C_{i-1}$ and $C_{i+1}$ for $i=1, \\ldots, 4$ where $C_{5}$ denotes $C_{1}$ and $C_{0}$ represents $C_{4}$. Compute the radius of $C_{1}$.",
    "options": [],
    "answer": "sqrt(2) - 1",
    "solution": "Solution:\nLet $O$ and $O'$ be the centers of $C$ and $C_{1}$ respectively, and let $C_{1}$ be tangent to $C, C_{2}, C_{4}$ at $P, Q$, and $R$ respectively. Observe that $Q O R O'$ is a square and that $P, O'$, and $O$ are collinear. Thus, if $r$ is the desired radius, $1 = r + O O' = r + r \\sqrt{2}$, so that $r = \\frac{1}{\\sqrt{2}+1} = \\sqrt{2} - 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2006 vertices of a regular 2007-gon are red. The remaining vertex is green. Let $G$ be the total number of polygons whose one vertex is green and the others are red. Denote by $R$ the number of polygons whose all vertices are red. Which number is bigger, $R$ or $G$? Explain your answer.",
    "options": [],
    "answer": "G > R",
    "solution": "Solution:\n\nWe will prove that $G \\geq R$. For each polygon $\\mathcal{P}$ with all red vertices we can correspond a polygon with one green vertex (namely we can add the green vertex to the set of vertices of $\\mathcal{P}$). Thus $G \\geq R$. However, $G > R$ since the triangles with one green vertex can't be corresponded to some polygon whose all vertices are red.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$A B C$ is an acute triangle with incircle $\\omega$. $\\omega$ is tangent to sides $\\overline{B C}$, $\\overline{C A}$, and $\\overline{A B}$ at $D$, $E$, and $F$ respectively. $P$ is a point on the altitude from $A$ such that $\\Gamma$, the circle with diameter $\\overline{A P}$, is tangent to $\\omega$. $\\Gamma$ intersects $\\overline{A C}$ and $\\overline{A B}$ at $X$ and $Y$ respectively. Given $X Y = 8$, $A E = 15$, and that the radius of $\\Gamma$ is $5$, compute $B D \\cdot D C$.",
    "options": [],
    "answer": "675/4",
    "solution": "Solution:\n\nBy the Law of Sines we have $\\sin \\angle A = \\frac{X Y}{A P} = \\frac{4}{5}$. Let $I$, $T$, and $Q$ denote the center of $\\omega$, the point of tangency between $\\omega$ and $\\Gamma$, and the center of $\\Gamma$ respectively. Since we are told $A B C$ is acute, we can compute $\\tan \\frac{\\angle A}{2} = \\frac{1}{2}$. Since $\\angle E A I = \\frac{\\angle A}{2}$ and $\\overline{A E}$ is tangent to $\\omega$, we find $r = \\frac{A E}{2} = \\frac{15}{2}$.\n\nLet $H$ be the foot of the altitude from $A$ to $\\overline{B C}$. Define $h_{T}$ to be the homothety about $T$ which sends $\\Gamma$ to $\\omega$. We have $h_{T}(\\overline{A Q}) = \\overline{D I}$, and conclude that $A$, $T$, and $D$ are collinear. Now since $\\overline{A P}$ is a diameter of $\\Gamma$, $\\angle P A T$ is right, implying that $D T H P$ is cyclic. Invoking Power of a Point twice, we have $225 = A E^{2} = A T \\cdot A D = A P \\cdot A H$. Because we are given radius of $\\Gamma$ we can find $A P = 10$ and $A H = \\frac{45}{2} = h_{a}$.\n\nIf we write $a$, $b$, $c$, $s$ in the usual manner with respect to triangle $A B C$, we seek $B D \\cdot D C = (s-b)(s-c)$. But recall that Heron's formula gives us\n$$\n\\sqrt{s(s-a)(s-b)(s-c)} = K\n$$\nwhere $K$ is the area of triangle $A B C$. Writing $K = r s$, we have $(s-b)(s-c) = \\frac{r^{2} s}{s-a}$. Knowing $r = \\frac{15}{2}$, we need only compute the ratio $\\frac{s}{a}$. By writing $K = \\frac{1}{2} a h_{a} = r s$, we find $\\frac{s}{a} = \\frac{h_{a}}{2 r} = \\frac{3}{2}$.\n\nNow we compute our answer,\n$$\n\\frac{r^{2} s}{s-a} = \\left(\\frac{15}{2}\\right)^{2} \\cdot \\frac{\\frac{s}{a}}{\\frac{s}{a} - 1} = \\frac{675}{4}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn bridge, a standard 52-card deck is dealt in the usual way to 4 players. By convention, each hand is assigned a number of \"points\" based on the formula\n$$\n4 \\times (\\# \\{ A \\text{'s} \\}) + 3 \\times (\\# \\{ K \\text{'s} \\}) + 2 \\times (\\# \\{ Q \\text{'s} \\}) + 1 \\times (\\# \\{ J \\text{'s} \\})\n$$\nGiven that a particular hand has exactly 4 cards that are A, K, Q, or J, find the probability that its point value is 13 or higher.",
    "options": [],
    "answer": "197/1820",
    "solution": "Solution:\nObviously, we can ignore the cards lower than $J$. Simply enumerate the ways to get at least 13 points: $AAAA$ (1), $AAAK$ (16), $AAAQ$ (16), $AAAJ$ (16), $AAKK$ (36), $AAKQ$ (96), $AKKK$ (16). The numbers in parentheses represent the number of ways to choose the suits, given the choices for the values. We see that there are a total of $1 + 16 + 16 + 16 + 36 + 96 + 16 = 197$ ways to get at least 13. There are a total of $\\binom{16}{4} = 1820$ possible ways to choose 4 cards from the 16 total $A$'s, $K$'s, $Q$'s, and $J$'s. Hence the answer is $\\frac{197}{1820}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence is defined by $A_{0}=0$, $A_{1}=1$, $A_{2}=2$, and, for integers $n \\geq 3$,\n$$\nA_{n}=\\frac{A_{n-1}+A_{n-2}+A_{n-3}}{3}+\\frac{1}{n^{4}-n^{2}}\n$$\n\nCompute $\\lim _{N \\rightarrow \\infty} A_{N}$.",
    "options": [],
    "answer": "13/6 - π^2/12",
    "solution": "Solution:\nIf we sum the given equation for $n=3,4,5, \\ldots, N$, we obtain\n$$\n\\sum_{n=3}^{N} A_{n}=\\sum_{n=3}^{N} \\frac{A_{n-1}+A_{n-2}+A_{n-3}}{3}+\\frac{1}{n^{4}-n^{2}}\n$$\nThis reduces dramatically to\n$$\nA_{N}+\\frac{2 A_{N-1}}{3}+\\frac{A_{N-2}}{3}=A_{2}+\\frac{2 A_{1}}{3}+\\frac{A_{0}}{3}+\\sum_{n=3}^{N} \\frac{1}{n^{4}-n^{2}}\n$$\nLet $\\lim _{N \\rightarrow \\infty} A_{N}=L$. Under this limit, the left hand side of $\\left(^{*}\\right)$ is simply $2 L$. We compute the sum on the right with the help of partial fractions\n$$\n\\begin{aligned}\n\\lim _{N \\rightarrow \\infty} \\sum_{n=3}^{N} \\frac{1}{n^{4}-n^{2}} & =\\sum_{n=3}^{\\infty} \\frac{1}{n^{2}-1}-\\frac{1}{n^{2}} \\\\\n& =\\left(\\sum_{n=3}^{\\infty} \\frac{1}{2}\\left(\\frac{1}{n-1}-\\frac{1}{n+1}\\right)\\right)+\\frac{1}{1^{2}}+\\frac{1}{2^{2}}-\\sum_{n=1}^{\\infty} \\frac{1}{n^{2}} \\\\\n& =\\frac{1}{2}\\left(\\frac{1}{2}+\\frac{1}{3}\\right)+\\frac{5}{4}-\\frac{\\pi^{2}}{6} \\\\\n& =\\frac{5}{3}-\\frac{\\pi^{2}}{6}\n\\end{aligned}\n$$\nWith this we easily find $L=\\frac{1}{2} \\cdot\\left(2+\\frac{2}{3} \\cdot 1+\\frac{1}{3} \\cdot 0+\\frac{5}{3}-\\frac{\\pi^{2}}{6}\\right)=\\frac{13}{6}-\\frac{\\pi^{2}}{12}$, and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFour points are independently chosen uniformly at random from the interior of a regular dodecahedron. What is the probability that they form a tetrahedron whose interior contains the dodecahedron's center?",
    "options": [],
    "answer": "1/8",
    "solution": "Solution:\nTo any tetrahedron $P_{1} P_{2} P_{3} P_{4}$ we can associate a quadruple $\\left(\\epsilon_{(i j k)}\\right)$, where $(i j k)$ ranges over all conjugates of the cycle $(123)$ in the alternating group $A_{4}$: $\\epsilon_{i j k}$ is the sign of the directed volume $\\left[O P_{i} P_{j} P_{k}\\right]$. Assume that, for a given tetrahedron $P_{1} P_{2} P_{3} P_{4}$, all members of its quadruple are nonzero (this happens with probability $1$). For $1 \\leq i \\leq 4$, if we replace $P_{i}$ with its reflection through the origin, the three members of the tetrahedron's quadruple that involve $P_{i}$ all flip sign, because each $\\left[O P_{i} P_{j} P_{k}\\right]$ is a linear function of the vector $\\overrightarrow{O P}$. Thus, if we consider the $16$ sister tetrahedra obtained by choosing independently whether to flip each $P_{i}$ through the origin, the quadruples range through all $16$ possibilities (namely, all the quadruples consisting of $\\pm 1$'s). Two of these $16$ tetrahedra, namely those with quadruples $(1,1,1,1)$ and $(-1,-1,-1,-1)$, will contain the origin.\nSo the answer is $2 / 16 = 1 / 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\sum_{n=1}^{\\infty} \\frac{2 n+5}{2^{n} \\cdot\\left(n^{3}+7 n^{2}+14 n+8\\right)}\n$$",
    "options": [],
    "answer": "137/24 - 8 ln 2",
    "solution": "Solution:\nFirst, we manipulate using partial fractions and telescoping:\n$$\n\\begin{aligned}\n\\sum_{n=1}^{\\infty} \\frac{2 n+5}{2^{n} \\cdot\\left(n^{3}+7 n^{2}+14 n+8\\right)} & =\\frac{1}{2} \\cdot \\sum_{n=1}^{\\infty} \\frac{1}{2^{n}}\\left(\\frac{2}{n+1}-\\frac{1}{n+2}-\\frac{1}{n+4}\\right) \\\\\n& =\\frac{1}{4}-\\frac{1}{2} \\sum_{n=1}^{\\infty} \\frac{1}{2^{n} \\cdot(n+4)}\n\\end{aligned}\n$$\nNow, consider the function $f(r, k):=\\sum_{n=1}^{\\infty} \\frac{r^{n}}{n^{k}}$. We have\n$$\n\\begin{aligned}\n\\frac{\\partial f(r, k)}{\\partial r} & =\\frac{\\partial}{\\partial r} \\sum_{n=1}^{\\infty} \\frac{r^{n}}{n^{k}}=\\sum_{n=1}^{\\infty} \\frac{\\partial}{\\partial r}\\left[\\frac{r^{n}}{n^{k}}\\right]=\\sum_{n=1}^{\\infty} \\frac{r^{n-1}}{n^{k-1}}=\\frac{1}{r} f(r, k-1) \\\\\n\\frac{d f(r, 1)}{d r} & =\\frac{1}{r} \\sum_{n=1}^{\\infty} \\frac{r^{n}}{n^{0}}=\\frac{1}{r} \\cdot \\frac{r}{1-r}=\\frac{1}{1-r} \\\\\nf(r, 1) & =\\int \\frac{d r}{1-r}=-\\ln (1-r)+f(0,1)\n\\end{aligned}\n$$\nBy inspection, $f(0,1)=0$, so $f\\left(\\frac{1}{2}, 1\\right)=\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot 2^{n}}=\\ln (2)$. It is easy to compute the desired sum in terms of $f\\left(\\frac{1}{2}, 1\\right)$, and we find $\\sum_{n=1}^{\\infty} \\frac{1}{2^{n}(n+4)}=16 \\ln (2)-\\frac{131}{12}$. Hence, our final answer is $\\frac{137}{24}-8 \\ln (2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $ABC$ is a triangle with incircle $\\omega$, and $\\omega$ is tangent to $\\overline{BC}$ and $\\overline{CA}$ at $D$ and $E$ respectively. The bisectors of $\\angle A$ and $\\angle B$ intersect line $DE$ at $F$ and $G$ respectively, such that $BF=1$ and $FG=GA=6$. Compute the radius of $\\omega$.",
    "options": [],
    "answer": "2√5/5",
    "solution": "Solution:\nLet $\\alpha, \\beta, \\gamma$ denote the measures of $\\frac{1}{2} \\angle A, \\frac{1}{2} \\angle B, \\frac{1}{2} \\angle C$, respectively. We have $m \\angle CEF = 90^\\circ - \\gamma$, $m \\angle FEA = 90^\\circ + \\gamma$, $m \\angle AFG = m \\angle AFE = 180^\\circ - \\alpha - (90^\\circ + \\gamma) = \\beta = m \\angle ABG$, so $ABFG$ is cyclic.\n\nNow $AG = GF$ implies that $\\overline{BG}$ bisects $\\angle ABF$. Since $\\overline{BG}$ by definition bisects $\\angle ABC$, we see that $F$ must lie on $\\overline{BC}$. Hence, $F = D$.\n\nIf $I$ denotes the incenter of triangle $ABC$, then $\\overline{ID}$ is perpendicular to $\\overline{BC}$, but since $A, I, F$ are collinear, we have that $\\overline{AD} \\perp \\overline{BC}$. Hence, $ABC$ is isosceles with $AB = AC$. Furthermore, $BC = 2BF = 2$.\n\nMoreover, since $ABFG$ is cyclic, $\\angle BGA$ is a right angle. Construct $F'$ on minor $\\operatorname{arc} GF$ such that $BF' = 6$ and $F'G = 1$, and let $AB = x$. By the Pythagorean theorem, $AF' = BG = \\sqrt{x^2 - 36}$, so that Ptolemy applied to $ABF'G$ yields $x^2 - 36 = x + 36$. We have $(x - 9)(x + 8) = 0$. Since $x$ is a length we find $x = 9$. Now we have $AB = AC = 9$.\n\nPythagoras applied to triangle $ABD$ now yields $AD = \\sqrt{9^2 - 1^2} = 4\\sqrt{5}$, which enables us to compute $[ABC] = \\frac{1}{2} \\cdot 2 \\cdot 4\\sqrt{5} = 4\\sqrt{5}$. Since the area of a triangle is also equal to its semiperimeter times its inradius, we have $4\\sqrt{5} = 10r$ or $r = \\frac{2\\sqrt{5}}{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA fat coin is one which, when tossed, has a $\\frac{2}{5}$ probability of being heads, $\\frac{2}{5}$ of being tails, and $\\frac{1}{5}$ of landing on its edge. Mr. Fat starts at $0$ on the real line. Every minute, he tosses a fat coin. If it's heads, he moves left, decreasing his coordinate by $1$; if it's tails, he moves right, increasing his coordinate by $1$. If the coin lands on its edge, he moves back to $0$. If Mr. Fat does this ad infinitum, what fraction of his time will he spend at $0$?",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n\nFor $n \\in \\mathbb{Z}$, let $a_n$ be the fraction of the time Mr. Fat spends at $n$. By symmetry, $a_n = a_{-n}$ for all $n$.\n\nFor $n > 0$, we have\n$$\na_n = \\frac{2}{5} a_{n-1} + \\frac{2}{5} a_{n+1},\n$$\nor\n$$\na_{n+1} = \\frac{5}{2} a_n - a_{n-1}.\n$$\nThis Fibonacci-like recurrence can be solved explicitly to obtain\n$$\na_n = \\alpha \\cdot 2^{|n|} + \\beta \\cdot 2^{-|n|}\n$$\nfor all $n \\in \\mathbb{Z}$. Now we also have\n$$\n\\sum_{n \\in \\mathbb{Z}} a_n = 1,\n$$\nso we better have $\\alpha = 0$, so that $a_0 = \\beta$ and $a_{\\pm 1} = \\frac{\\beta}{2}$.\n\nNow we also have\n$$\na_0 = \\frac{2}{5} a_{-1} + \\frac{2}{5} a_1 + \\frac{1}{5},\n$$\nso $\\beta = \\frac{1}{3}$.\n\nThis matches perfectly with $\\sum_{n \\in \\mathbb{Z}} a_n = 1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that there are $k \\ge 2$ positive rational numbers $a_1, a_2, \\dots, a_k$ satisfying $a_1 + a_2 + \\dots + a_k = a_1 \\cdot a_2 \\dots a_k = n$.",
    "options": [],
    "answer": "n = 4 or n ≥ 6",
    "solution": "The answer is $n = 4$ or $n \\ge 6$.\n\nI. First, we prove that each $n \\in \\{4, 6, 7, 8, 9, \\dots\\}$ satisfies the condition.\n\n(1). If $n = 2k \\ge 4$ is even, we set $(a_1, a_2, \\dots, a_k) = (k, 2, 1, \\dots, 1)$:\n$$\na_1 + a_2 + \\dots + a_k = k + 2 + 1 \\cdot (k-2) = 2k = n,\n$$\nand\n$$\na_1 \\cdot a_2 \\cdot \\dots \\cdot a_k = 2k = n.\n$$\n\n(2). If $n = 2k + 3 \\ge 9$ is odd, we set $(a_1, a_2, \\dots, a_k) = (k + \\frac{3}{2}, \\frac{1}{2}, 4, 1, \\dots, 1)$:\n$$\na_1 + a_2 + \\dots + a_k = k + \\frac{3}{2} + \\frac{1}{2} + 4 + (k-3) = 2k + 3 = n,\n$$\nand\n$$\na_1 \\cdot a_2 \\cdot \\dots \\cdot a_k = \\left(k + \\frac{3}{2}\\right) \\cdot \\frac{1}{2} \\cdot 4 = 2k + 3 = n.\n$$\n\n(3). A very special case is $n = 7$, in which we set $(a_1, a_2, a_3) = (\\frac{4}{3}, \\frac{7}{6}, \\frac{9}{2})$. It is also easy to check that\n$$a_1 + a_2 + a_3 = a_1 \\cdot a_2 \\cdot a_3 = 7 = n.$$\n\nII. Second, we prove by contradiction that each $n \\in \\{1, 2, 3, 5\\}$ fails to satisfy the condition.\nSuppose, on the contrary, that there is a set of $k \\ge 2$ positive rational numbers whose sum and product are both $n \\in \\{1, 2, 3, 5\\}$. By the Arithmetic-Geometric Mean inequality, we have\n$$\nn^{1/k} = \\sqrt[k]{a_1 \\cdot a_2 \\cdot \\dots \\cdot a_k} \\le \\frac{a_1 + a_2 + \\dots + a_k}{k} = \\frac{n}{k},\n$$\nwhich gives\n$$\nn \\ge k^{\\frac{k}{k-1}} = k^{1+\\frac{1}{k-1}}.\n$$\nNote that $n > 5$ whenever $k = 3, 4$, or $k \\ge 5$:\n$$\nk = 3 \\Rightarrow n \\ge 3\\sqrt{3} = 5.196\\dots > 5;\n$$\n$$\nk = 4 \\Rightarrow n \\ge 4\\sqrt[3]{4} = 6.349\\dots > 5;\n$$\n$$\nk \\ge 5 \\Rightarrow n \\ge 5^{1+\\frac{1}{k-1}} > 5.\n$$\nThis proves that none of the integers 1, 2, 3, or 5 can be represented as the sum and, at the same time, as the product of three or more positive numbers $a_1, a_2, \\dots, a_k$, rational or irrational.\nThe remaining case $k = 2$ also goes to a contradiction. Indeed, $a_1 + a_2 = a_1a_2 = n$ implies that $n = a_1^2/(a_1 - 1)$ and thus $a_1$ satisfies the quadratic\n$$\na_1^2 - n a_1 + n = 0.\n$$\nSince $a_1$ is supposed to be rational, the discriminant $n^2 - 4n$ must be a perfect square. However, it can be easily checked that this is not the case for any $n \\in \\{1, 2, 3, 5\\}$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $n$ squares of an infinite square grid are colored grey, and the rest are colored white. At each step, a new grid of squares is obtained based on the previous one, as follows. For each location in the grid, examine that square, the square immediately above, and the square immediately to the right. If there are two or three grey squares among these three, then in the next grid, color that location grey; otherwise, color it white. Prove that after at most $n$ steps all the squares in the grid will be white.\n\nBelow is an example with $n=4$. The first grid shows the initial configuration, and the second grid shows the configuration after one step.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n(Sketch) Use strong induction. Consider the smallest rectangle $R$ that contains all the black squares. Suppose this rectangle contains $r$ squares. Assume any rectangle that contains $k < r$ squares will convert to all white squares after $k$ steps. Since $R$ is the smallest rectangle, its left-most column and its bottom row must contain some black squares. So the rectangle with the left-most column of $R$ removed and the rectangle with the bottom row removed must both have less than $r$ black squares. So both of those rectangles must be converted to all white after $r-1$ steps. (We are using the fact that squares below and to the left of a rectangle cannot affect the evolution of the squares within that rectangle.) So, after $r-1$ steps, $R$ would have been converted to all white squares except for possibly the lower left corner square. But even if that square is black, the $r$th conversion will convert that square into white square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find, with proof, the least positive integer $d_n$ which cannot be expressed in the form\n$$\n\\sum_{i=1}^{n} (-1)^{a_i} 2^{b_i},\n$$\nwhere $a_i$ and $b_i$ are nonnegative integers for each $i$.",
    "options": [],
    "answer": "(2^{2n+1} + 1) / 3",
    "solution": "The answer is $d_n = \\dfrac{2^{2n+1} + 1}{3}$. We first show that $d_n$ cannot be obtained. For any $p$ let $t(p)$ be the minimum $n$ required to express $p$ in the desired form and call any realization of this minimum a minimal representation. If $p$ is even, any sequence of $b_i$ that can produce $p$ must contain an even number of zeros. If this number is nonzero, then canceling one against another or replacing two with a $b_i = 1$ term would reduce the number of terms in the sum. Thus a minimal representation cannot contain a $b_i = 0$ term, and by dividing each term by two we see that $t(2m) = t(m)$. If $p$ is odd, there must be at least one $b_i = 0$ and removing it gives a sequence that produces either $p-1$ or $p+1$. Hence\n$$\nt(2m-1) = 1 + \\min(t(2m-2), t(2m)) = 1 + \\min(t(m-1), t(m)).\n$$\nWith $d_n$ as defined above and $c_n = \\dfrac{2^{2n} - 1}{3}$, we have $d_0 = c_1 = 1$, so $t(d_0) = t(c_1) = 1$ and\n$$\nt(d_n) = 1 + \\min(t(d_{n-1}), t(c_n)) \\quad \\text{and} \\quad t(c_n) = 1 + \\min(t(d_{n-1}), t(c_{n-1})).\n$$\nHence, by induction, $t(c_n) = n$ and $t(d_n) = n+1$ and $d_n$ cannot be obtained by a sum with $n$ terms.\n\nNext we show by induction on $n$ that any positive integer less than $d_n$ can be obtained with $n$ terms. By the inductive hypothesis and symmetry about zero, it suffices to show that by adding one summand we can reach every $p$ in the range $d_{n-1} \\le p < d_n$ from an integer $q$ in the range $-d_{n-1} < q < d_{n-1}$. Suppose that $c_n + 1 \\le p \\le d_n - 1$. By using a term $2^{2n-1}$, we see that $t(p) \\le 1 + t(|p - 2^{2n-1}|)$. Since $d_n - 1 - 2^{2n-1} = 2^{2n-1} - (c_n + 1) = d_{n-1} - 1$, it follows from the inductive hypothesis that $t(p) \\le n$. Now suppose that $d_{n-1} \\le p \\le c_n$. By using a term $2^{2n-2}$, we see that $t(p) \\le 1 + t(|p - 2^{2n-2}|)$. Since $c_n - 2^{2n-2} = 2^{2n-2} - d_{n-1} = c_{n-1} < d_{n-1}$, it again follows that $t(p) \\le n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA sequence of numbers $\\{a_n\\}$ is given by $a_1 = 1$, $a_{n+1} = 2a_n + \\sqrt{3a_n^2 + 1}$ for $n \\geq 1$. Prove that each term of the sequence is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFrom the definition of the sequence we see that $\\left(a_{n+1} - 2a_n\\right)^2 = 3a_n^2 + 1$. After simplification we get\n$$\na_{n+1}^2 + a_n^2 - 4a_n a_{n+1} = 1\n$$\nAdding $3a_{n+1}^2$ to both sides of the last equation gives us $\\left(2a_{n+1} - a_n\\right)^2 = 3a_{n+1}^2 + 1$ and after taking the square root of both sides we get\n$$\n\\sqrt{3a_{n+1}^2 + 1} = 2a_{n+1} - a_n\n$$\nOn the other hand, from the definition of the sequence we see that\n$$\n\\sqrt{3a_{n+1}^2 + 1} = a_{n+2} - 2a_{n+1}.\n$$\nFrom (1) and (2) we conclude that $a_{n+2} = 4a_{n+1} - a_n$ which together with $a_1 = 1$, $a_2 = 4$ implies that all terms are integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute $\\sum_{k=1}^{\\infty} \\frac{k^{4}}{k!}$.",
    "options": [],
    "answer": "15e",
    "solution": "Solution:\n\nDefine, for non-negative integers $n$,\n$$\nS_{n}:=\\sum_{k=0}^{\\infty} \\frac{k^{n}}{k!}\n$$\nwhere $0^{0}=1$ when it occurs. Then $S_{0}=e$, and, for $n \\geq 1$,\n$$\nS_{n}=\\sum_{k=0}^{\\infty} \\frac{k^{n}}{k!}=\\sum_{k=1}^{\\infty} \\frac{k^{n}}{k!}=\\sum_{k=0}^{\\infty} \\frac{(k+1)^{n}}{(k+1)!}=\\sum_{k=0}^{\\infty} \\frac{(k+1)^{n-1}}{k!}=\\sum_{i=0}^{n-1}\\binom{n-1}{i} S_{i},\n$$\nso we can compute inductively that $S_{1}=e, S_{2}=2e, S_{3}=5e$, and $S_{4}=15e$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA dot is marked at each vertex of a triangle $A B C$. Then, $2$, $3$, and $7$ more dots are marked on the sides $A B$, $B C$, and $C A$, respectively. How many triangles have their vertices at these dots?",
    "options": [],
    "answer": "357",
    "solution": "Solution:\nAltogether there are $3+2+3+7=15$ dots, and thus $\\binom{15}{3}=455$ combinations of $3$ dots. Of these combinations, $\\binom{2+2}{3}+\\binom{2+3}{3}+\\binom{2+7}{3}=4+10+84=98$ do not give triangles because they are collinear (the rest do give triangles). Thus $455-98=357$ different triangles can be formed.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAnn and Anne are in bumper cars starting 50 meters apart. Each one approaches the other at a constant ground speed of $10~\\mathrm{km}/\\mathrm{hr}$. A fly starts at Ann, flies to Anne, then back to Ann, and so on, back and forth until it gets crushed when the two bumper cars collide. When going from Ann to Anne, the fly flies at $20~\\mathrm{km}/\\mathrm{hr}$; when going in the opposite direction the fly flies at $30~\\mathrm{km}/\\mathrm{hr}$ (thanks to a breeze). How many meters does the fly fly?",
    "options": [],
    "answer": "55",
    "solution": "Solution:\n\nSuppose that at a given instant the fly is at Ann and the two cars are $12d$ apart. Then, while each of the cars travels $4d$, the fly travels $8d$ and meets Anne. Then the fly turns around, and while each of the cars travels $d$, the fly travels $3d$ and meets Ann again. So, in this process described, each car travels a total of $5d$ while the fly travels $11d$. So the fly will travel $\\frac{11}{5}$ times the distance traveled by each bumper car: $\\frac{11}{5} \\cdot \\frac{50}{2} = 55$ meters.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind\n$$\n\\frac{\\sqrt{31+\\sqrt{31+\\sqrt{31+\\ldots}}}}{\\sqrt{1+\\sqrt{1+\\sqrt{1+\\ldots}}}}\n$$",
    "options": [],
    "answer": "6 - sqrt(5)",
    "solution": "Solution:\nLet the numerator be $x$ and the denominator $y$. Then $x^{2} = 31 + x$, so, as $x > 0$, we have\n$$\nx = \\frac{1 + \\sqrt{1 + 4 \\cdot 31}}{2} = \\frac{1 + 5 \\sqrt{5}}{2}.\n$$\nSimilarly we compute that\n$$\ny = \\frac{1 + \\sqrt{1 + 4 \\cdot 1}}{2} = \\frac{1 + \\sqrt{5}}{2}\n$$\nso that\n$$\n\\frac{x}{y} = \\frac{1 + 5 \\sqrt{5}}{1 + \\sqrt{5}} = \\frac{1 + 5 \\sqrt{5}}{1 + \\sqrt{5}} \\cdot \\frac{1 - \\sqrt{5}}{1 - \\sqrt{5}} = \\frac{-24 + 4 \\sqrt{5}}{-4} = 6 - \\sqrt{5}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVernonia High School has 85 seniors, each of whom plays on at least one of the school's three varsity sports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 are on the baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball and football teams; and 13 are on both the baseball and lacrosse teams. Compute the number of seniors playing all three sports, given that twice this number are members of the lacrosse team.",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nSuppose that $n$ seniors play all three sports and that $2n$ are on the lacrosse team. Then, by the principle of inclusion-exclusion,\n$$\n85 = (74 + 26 + 2n) - (17 + 18 + 13) + n = 100 + 2n - 48 + n = 52 + 3n.\n$$\nIt is easily seen that $n = 11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle such that $AB=2$, $CA=3$, and $BC=4$. A semicircle with its diameter on $\\overline{BC}$ is tangent to $\\overline{AB}$ and $\\overline{AC}$. Compute the area of the semicircle.",
    "options": [],
    "answer": "27π/40",
    "solution": "Solution:\nLet $O$, $D$, and $E$ be the midpoint of the diameter and the points of tangency with $\\overline{AB}$ and $\\overline{AC}$ respectively. Then $[ABC]=[AOB]+[AOC]=\\frac{1}{2}(AB+AC) r$, where $r$ is the radius of the semicircle. Now by Heron's formula, $[ABC]=\\sqrt{\\frac{9}{2} \\cdot \\frac{1}{2} \\cdot \\frac{3}{2} \\cdot \\frac{5}{2}}=\\frac{3 \\sqrt{15}}{4}$. We solve for $r=\\frac{3 \\sqrt{15}}{10}$ and compute $\\frac{1}{2} \\pi r^{2}=\\frac{27 \\pi}{40}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, \\ldots$ be a sequence defined by $a_{1}=a_{2}=1$ and $a_{n+2}=a_{n+1}+a_{n}$ for $n \\geq 1$. Find\n$$\n\\sum_{n=1}^{\\infty} \\frac{a_{n}}{4^{n+1}}\n$$",
    "options": [],
    "answer": "1/11",
    "solution": "Solution:\nLet $X$ denote the desired sum. Note that\n$$\n\\begin{array}{rl}\nX & =  \\frac{1}{4^{2}}+\\frac{1}{4^{3}}+\\frac{2}{4^{4}}+\\frac{3}{4^{5}}+\\frac{5}{4^{6}}+\\ldots \\\\\n4 X & =\\quad \\frac{1}{4^{1}}+\\frac{1}{4^{2}}+\\frac{2}{4^{3}}+\\frac{3}{4^{4}}+\\frac{5}{4^{5}}+\\frac{8}{4^{6}}+\\ldots \\\\\n16 X &=\\frac{1}{4^{0}}+\\frac{1}{4^{1}}+\\frac{2}{4^{2}}+\\frac{3}{4^{3}}+\\frac{5}{4^{4}}+\\frac{8}{4^{5}}+\\frac{13}{4^{6}}+\\ldots\n\\end{array}\n$$\nso that $X+4 X=16 X-1$, and $X=1 / 11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\sum_{k=1}^{\\infty} \\frac{3k+1}{2k^{3}+k^{2}} \\cdot (-1)^{k+1}\n$$",
    "options": [],
    "answer": "ln(2) + π/2 + π^2/12 − 2",
    "solution": "Solution:\nVia partial fraction decomposition we write the sum as\n$$\n\\sum_{k=1}^{\\infty} \\left(\\frac{1}{k} - \\frac{2}{1+2k} + \\frac{1}{k^{2}}\\right)(-1)^{k+1}\n$$\nNow recall that\n$$\n\\begin{aligned}\n& \\sum_{k=1}^{\\infty} \\frac{1}{k^{2}} = \\frac{\\pi^{2}}{6} = S_{1} \\\\\n& \\sum_{k=1}^{\\infty} \\frac{(-1)^{k+1}}{k} = \\ln(2) = S_{2} \\\\\n& \\sum_{k=1}^{\\infty} \\frac{(-1)^{k+1}}{2k-1} = \\frac{\\pi}{4} = S_{3}\n\\end{aligned}\n$$\nManipulating (1), we deduce\n$$\n\\begin{aligned}\n\\sum_{k=1}^{\\infty} \\frac{(-1)^{k+1}}{k^{2}} &= \\left(\\sum_{k=1}^{\\infty} \\frac{1}{k^{2}}\\right) - 2 \\cdot \\sum_{k=1}^{\\infty} \\frac{1}{(2k)^{2}} \\\\\n= \\frac{\\pi^{2}}{6} - 2/4 \\cdot \\frac{\\pi^{2}}{6} = \\frac{\\pi^{2}}{12} &= S_{4}\n\\end{aligned}\n$$\nIt is then easily seen that the answer is equal to $S_{2} + 2 \\cdot S_{3} - 2 + S_{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\Gamma$ denote the circumcircle of triangle $A B C$. Point $D$ is on $\\overline{A B}$ such that $\\overline{C D}$ bisects $\\angle A C B$. Points $P$ and $Q$ are on $\\Gamma$ such that $\\overline{P Q}$ passes through $D$ and is perpendicular to $\\overline{C D}$. Compute $P Q$, given that $B C=20$, $C A=80$, $A B=65$.",
    "options": [],
    "answer": "4√745",
    "solution": "Solution:\nSuppose that $P$ lies between $A$ and $B$ and $Q$ lies between $A$ and $C$, and let line $P Q$ intersect lines $A C$ and $B C$ at $E$ and $F$ respectively. As usual, we write $a$, $b$, $c$ for the lengths of $B C$, $C A$, $A B$.\n\nBy the angle bisector theorem, $A D / D B = A C / C B$ so that $A D = \\frac{b c}{a+b}$ and $B D = \\frac{a c}{a+b}$.\n\nNow by Stewart's theorem,\n$$\nc \\cdot C D^{2} + \\left(\\frac{a c}{a+b}\\right)\\left(\\frac{b c}{a+b}\\right) c = \\frac{a^{2} b c}{a+b} + \\frac{a b^{2} c}{a+b}\n$$\nfrom which\n$$\nC D^{2} = \\frac{a b \\left((a+b)^{2} - c^{2}\\right)}{(a+b)^{2}}\n$$\n\nNow observe that triangles $C D E$ and $C D F$ are congruent, so $E D = D F$.\n\nBy Menelaus' theorem,\n$$\n\\frac{C A}{A E} \\cdot \\frac{E D}{D F} \\cdot \\frac{F B}{B C} = 1\n$$\nso that\n$$\n\\frac{C A}{B C} = \\frac{A E}{F B}\n$$\nSince $C F = C E$ while $b > a$, it follows that $A E = \\frac{b(b-a)}{a+b}$ so that $E C = \\frac{2 a b}{a+b}$.\n\nFinally,\n$$\nD E = \\sqrt{C E^{2} - C D^{2}} = \\frac{\\sqrt{a b \\left(c^{2} - (a-b)^{2}\\right)}}{a+b}\n$$\n\nPlugging in $a=20$, $b=80$, $c=65$, we see that $A E = 48$, $E C = 32$, $D E = 10$ as well as $A D = 52$, $B D = 13$.\n\nNow let $P D = x$, $Q E = y$. By power of a point about $D$ and $E$, we have $x(y+10) = 676$ and $y(x+10) = 1536$. Subtracting one from the other, we see that $y = x + 86$. Therefore,\n$$\nx^{2} + 96 x - 676 = 0\n$$\nfrom which\n$$\nx = -48 + 2 \\sqrt{745}\n$$\nFinally,\n$$\nP Q = x + y + 10 = 4 \\sqrt{745}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose hypothetically that a certain, very corrupt political entity in a universe holds an election with two candidates, say $A$ and $B$. A total of $5,825,043$ votes are cast, but, in a sudden rainstorm, all the ballots get soaked. Undaunted, the election officials decide to guess what the ballots say. Each ballot has a $51\\%$ chance of being deemed a vote for $A$, and a $49\\%$ chance of being deemed a vote for $B$. The probability that $B$ will win is $10^{-X}$. What is $X$ rounded to the nearest 10 ?",
    "options": [],
    "answer": "510",
    "solution": "Solution:\n\nLet $N=2912521$, so that the number of ballots cast is $2N+1$. Let $P$ be the probability that $B$ wins, and let $\\alpha=51\\%$ and $\\beta=49\\%$ and $\\gamma=\\beta/\\alpha<1$. We have\n$$\n10^{-X}=P=\\sum_{i=0}^{N}\\binom{2N+1}{N-i} \\alpha^{N-i} \\beta^{N+1+i}=\\alpha^{N} \\beta^{N+1} \\sum_{i=0}^{N}\\binom{2N+1}{N-i} \\gamma^{i}\n$$\n(think of $2i+1$ as representing $B$'s margin of victory). Now\n$$\n\\frac{2^{2N+1}}{2N+1}<\\binom{2N+1}{N}<\\sum_{i=0}^{N}\\binom{2N+1}{N-i} \\gamma^{i}<2^{2N+1}\n$$\nSo\n$-X=\\log P=N \\log \\alpha+(N+1) \\log \\beta+(2N+1) \\log 2-\\epsilon=N \\log (2\\alpha)+(N+1) \\log (2\\beta)-\\epsilon$, where $0<\\epsilon<\\log (2N+1)<7$. With a calculator, we find that\n$$\n-X \\approx 25048.2-25554.2-\\epsilon=-506.0-\\epsilon\n$$\nso $X \\approx 510$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWrite down at least one, and up to ten, different 3-digit prime numbers. If you somehow fail to do this, we will ignore your submission for this problem. Otherwise, you're entered into a game with other teams. In this game, you start with 10 points, and each number you write down is like a bet: if no one else writes that number, you gain 1 point, but if anyone else writes that number, you lose 1 point. Thus, your score on this problem can be anything from 0 to 20.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThere are 143 three-digit primes. None of the following necessarily applies to the actual contest, but it might be useful to think about. Suppose that you're trying to maximize your expected score on this problem. Then you should write down a number if you think the probability that someone else is writing it is less than $1 / 2$ (of course, limit yourself to 10 numbers). You should avoid writing down any number if you think the probability that someone else is writing it is more than $1 / 2$ (of course, write down at least 1 number). Suppose that you expect a total of $M$ different numbers are going to be written down, but have no idea what numbers they might be. If you think $M \\geq 72$, you should write down 10 numbers at random; if $M \\leq 71$, you should write just 1 number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$$\n\\begin{array}{rlrr}\nA=(0,428) & B=(9,85) & C=(42,865) & D=(192,875) \\\\\nE=(193,219) & F=(204,108) & G=(292,219) & H=(316,378) \\\\\nI=(375,688) & J=(597,498) & K=(679,766) & L=(739,641) \\\\\n& M=(772,307) & & N=(793,0) \\\\\n&\n\\end{array}\n$$\nA fly starts at $A$, visits all the other points, and comes back to $A$ in such a way as to minimize the total distance covered. What path did the fly take? Give the names of the points it visits in order. Your score will be\n$$\n20+\\lfloor\\text{ the optimal distance }\\rfloor-\\lfloor\\text{ your distance }\\rfloor\n$$\nor $0$, whichever is greater.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThis problem is an instance of the Traveling Salesman Problem, which is NP-hard. There is an obvious algorithm in $O(n!)$ time (where $n$ is the number of points), but faster algorithms exist. Nonetheless, the best strategy for solving this problem is probably to draw the points and exercise your geometric intuition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn your answer sheet, clearly mark at least seven points, as long as\n(i) No three are collinear.\n(ii) No seven form a convex heptagon.\nPlease do not cross out any points; erase if you can do so neatly. If the graders deem that your paper is too messy, or if they determine that you violated one of those conditions, your submission for this problem will be disqualified. Otherwise, your score will be the number of points you marked minus $6$, even if you actually violated one of the conditions but were able to fool the graders.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThis is the heptagon case of what is known as the \"Happy Ending\" or \"Erdős-Szekeres\" problem, which in general asks, For any integer $n \\geq 3$, what is the smallest $N(n)$, such that any $N(n)$ points in the plane in general position determine a convex $n$-gon? It is known that such an $N(n)$ always exists and is finite (in fact a specific upper bound has been found). The best known lower bound is $N(n) \\geq 2^{n-2}+1$; Erdős and Szekeres conjectured that this bound is tight. The $n \\leq 5$ cases have been known for some time. According to the Wikipedia, the $n=6$ case is solved but unpublished, and for $n \\geq 7$, the problem remains open.\nFor a discussion, see\nW. Morris and V. Soltan. The Erdős-Szekeres Problem on Points in Convex Position-A Survey, Bulletin of the American Math Monthly. 37 (2000), 437-458.\nThis article is available at\nhttp://www.ams.org/bull/2000-37-04/S0273-0979-00-00877-6/home.html.\nIf $N(7)=33$, the highest sure score on this problem would be $32-6=26$. It is not known whether there exist arbitrarily large sets of points that will fool the graders.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a polynomial of degree $n > 1$ with integer coefficients and let $k$ be a positive integer. Consider the polynomial\n$$\nQ(x) = \\underbrace{P(P(\\dots(P(x)\\dots)))}_{k \\text{ P's}}\n$$\nProve that there are at most $n$ integers $t$ such that $Q(t) = t$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\mathbb{Z}$ denote the set of integers. We define\n$$\nS_P = \\{t \\mid t \\in \\mathbb{Z} \\text{ and } P(t) = t\\} \\quad \\text{and} \\quad S_Q = \\{t \\mid t \\in \\mathbb{Z} \\text{ and } Q(t) = t\\}.\n$$\nClearly, $S_P$ is a subset of $S_Q$. Also note that there are at most $n$ elements in $S_P$. This is so because that $t \\in S_P$ if and only if $t$ is a root of polynomial $P(x) - x = 0$ of degree $n$, which has at most $n$ roots. If $S_Q = S_P$, we have nothing to prove. We assume that $S_P$ is a proper subset of $S_Q$, and that $t \\in S_Q$ but $t \\notin S_P$.\nConsider the sequence $\\{t_i\\}_{i=0}^{\\infty}$ with $t_0 = t$, $t_{i+1} = P(t_i)$ for every nonnegative integer $i$. Since $t \\in S_Q$, $t_k = Q(t_0) = Q(t) = t = t_0$.\nSince the polynomial $a - b$ divides the polynomial $a^m - b^m$ (where $m$ is a nonnegative integer), it is not difficult to see that the polynomial $a - b$ divides the polynomial $P(a) - P(b)$, where $P(x)$ is a polynomial with integer coefficients. In our current problem, then, we can conclude that the integer sequence $\\{t_i\\}_{i=0}^{\\infty}$ satisfies the following sequence of divisibility relations\n$$\n(t_{i+1} - t_i) \\mid (P(t_{i+1}) - P(t_i)) = t_{i+2} - t_{i+1}\n$$\nfor every nonnegative integer $i$. Since $t_{k+1} - t_k = t_1 - t_0 = P(t) - t \\ne 0$, each term in the chain of differences\n$t_1 - t_0, t_2 - t_1, \\dots, t_k - t_{k-1}, t_{k+1} - t_k$\nis a nonzero divisor of the next one, and since $t_{k+1} - t_k = t_1 - t_0$, all these differences have equal absolute values. Let $t_i = \\max\\{t_0, t_1, \\dots, t_k\\}$. Then $t_{i-1} - t_i = -(t_i - t_{i+1})$, or $t_{i-1} = t_{i+1}$. It is then not difficult to see that $t_{i+2} = t_i$ for every $i$; that is,\n$$\nt_1 = P(t_0) \\quad \\text{and} \\quad t_0 = P(t_1) \\quad \\text{or} \\quad P(P(t_0)) = t_0.\n$$\n\nTherefore,\n$$\nS_Q = \\{t \\mid t \\in \\mathbb{Z} \\text{ and } P(P(t)) = t\\}.\n$$\nWithout loss of generality, we may assume that $t_0 < t_1$. If $s_0$ is another element in $S_Q$, let $s_1 = P(s_0)$. (It is possible that $s_0 \\in S_P$; that is, $s_1 = s_0$.) We further assume without loss of generality that $s_0 < s_1$ and $t_0 < s_0$; that is, $t_0 < s_0 \\le s_1$ and $t_0 < t_1$. Note that $s_1 - t_0$ divides $P(s_1) - P(t_0) = s_0 - t_1$. We must have $t_0 < s_0 < s_1 < t_1$. Note that $s_0 - t_1$ also divides $P(s_0) - P(t_1) = s_1 - t_0$, it follows that $s_0 - t_1 = -(s_1 - t_0)$; that is,\n$$\nt_0 + t_1 = s_0 + s_1 = s_0 + P(s_0).\n$$\nIn other words, $s_0$ is a root of the polynomial $P(x) + x = t_0 + t_1$. Since $P(x) + x$ has degree $n$, there are at most $n$ (integer) roots (including $t_0$) of $P(x) + x$. Hence there are at most $n$ elements in $S_Q$, completing our proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given integer with $n$ greater than $7$, and let $\\mathcal{P}$ be a convex polygon with $n$ sides. Any set of $n-3$ diagonals of $\\mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $\\mathcal{P}$ into $n-2$ triangles. A triangle in the triangulation of $\\mathcal{P}$ is an interior triangle if all of its sides are diagonals of $\\mathcal{P}$.\n\nExpress, in terms of $n$, the number of triangulations of $\\mathcal{P}$ with exactly two interior triangles, in closed form.",
    "options": [],
    "answer": "n * 2^{n-9} * binom(n-4, 4)",
    "solution": "The answer is\n$$\nn2^{n-9} \\binom{n-4}{4}\n$$\nDenote the vertices of $\\mathcal{P}$ counter-clockwise by $A_0, A_1, \\dots, A_{n-1}$. We will count first the number of triangulations of $\\mathcal{P}$ with two interior triangles positioned as in the following figure. We say that such a triangulation starts at $A_0$.\n\n![](attached_image_1.png)\n\nThe numbers $m_1, m_2, n_1, n_2, n_3, n_4$ in the figure denote the number of sides of $P$ determining the regions $N_1, N_2, N_3, N_4$ and $M$ that consist of exterior triangles (triangles that are not interior). The two interior triangles are\n$$\nA_0 A_{n_1} A_{n_1+n_2} \\text{ and } A_{n_1+n_2+m_1} A_{n_1+n_2+m_1+n_3} A_{n_1+n_2+m_1+n_3+n_4},\n$$\nrespectively.\n\nWe will show that triangulations starting at $A_0$ are in bijective correspondence to 7-tuples\n$$\n(m, n_1, n_2, n_3, n_4, w_M, w_N),\n$$\nwhere $m \\ge 0$, $n_1, n_2, n_3, n_4 \\ge 2$ are integers,\n$$\nm + n_1 + n_2 + n_3 + n_4 = n, \\qquad (\\dagger)\n$$\n$w_M$ is a binary sequence (sequence of 0's and 1's) of length $m$ and $w_N$ is a binary sequence of length $n - m - 8$.\n\nIndeed, given a triangulation as in the figure, the numbers $m = m_1 + m_2$ and $n_1, n_2, n_3, n_4$ satisfy $(\\dagger)$ and the associated constraints.\n\nFurther, the triangulation of the outside region $N_1$ determines a binary sequence of length $n_1 - 2$ as follows. Denote the exterior triangle in $N_1$ using the diagonal $A_0 A_{n_1}$ by $T_1$. If $n_1 \\ge 3$, $T_1$ has a unique neighboring exterior triangle in $N_1$, denoted $T_2$. If $n_1 \\ge 4$, the triangle $T_2$ has another neighbor in $N_1$ denoted $T_3$, etc. Thus we have a sequence of $n_1 - 1$ exterior triangles in $N_1$. We encode this sequence as follows. If $T_1$ uses the vertex $A_1$ as its third vertex we encode this by $0$ and if it uses $A_{n_1-1}$ we encode this by $1$. In each case there are two possible choices for the third vertex in $T_2$. If the one with smaller index is used we encode this by $0$ and if the one with larger index is used we encode this by $1$. Eventually, a sequence of $n_1 - 2$ $0$'s and $1$'s is constructed describing the choice of the third vertex in the triangles $T_1, \\dots, T_{n_1-2}$. Finally, there is only one choice for the third vertex in the triangle $T_{n_1-1}$ (this triangle is uniquely determined by the previous one), so we get $2^{n_1-2}$ possible triangulations of $N_1$ encoded in a binary sequence of length $n_1 - 2$. Similarly, there are $2^{n_1-2}$ triangulations of the region $N_i$, $i = 1, 2, 3, 4$, encoded by binary sequences of length $n_i - 2$. Thus a binary sequence $w_N$ of length $n_1 - 2 + n_2 - 2 + n_3 - 2 + n_4 - 2 = n - m - 8$, uniquely determines the triangulations of the regions $N_1, N_2, N_3, N_4$ (once the regions are precisely determined within $P$, which is done once $m_1, m_2, n_1, n_2, n_3$ and $n_4$ are known).\n\nIt remains to uniquely encode the triangulation of the middle region $M$. Denote by $M_1$ the unique exterior triangle in $M$ using the diagonal $A_0A_{n_1+n_2}$. If $m \\ge 2$, $M_1$ has a unique neighboring exterior triangle $M_2$ in $M$. If $m \\ge 3$, the triangle $M_2$ has another neighbor in $M$ denoted $M_3$, etc. Thus we have a sequence of $m$ exterior triangles in $M$. We encode this sequence as follows. If $M_1$ uses the vertex $A_{n_1+n_2+1}$ as its third vertex we encode this by $0$ and if it uses $A_{n-1}$ we encode this by $1$. In each case there are two possible choices for the third vertex in $M_2$. If the one with smaller index is used we encode this by $0$ and if the one with larger index is used we encode this by $1$. Eventually, a sequence of $m$ $0$'s and $1$'s is constructed describing the choice of the third vertex in the triangles $M_1, \\dots, M_m$. Thus a binary sequence $w_M$ of length $m$ uniquely determines the triangulation of the region $M$. In addition such a sequence $w_M$ uniquely determines $m_1$ and $m_2$ as the number of $0$'s and $1$'s respectively in $w_M$ and therefore also the exact position of the middle region $M$ within $P$ (once $n_1$ and $n_2$ are known), which in turn then exactly determines the position of all the regions considered in the figure.\n\nThe number of solutions of the equation $(\\dagger)$ subject to the given constraints is equal to the number of positive integer solutions to the equation\n$$\nx_1 + x_2 + x_3 + x_4 + x_5 = n - 3,\n$$\nwhich is $\\binom{n-4}{4}$ (a sequence of $n-3$ objects is split into $5$ nonempty groups by placing $4$ separators in the $n-4$ available positions between the objects). Thus the number of $7$-tuples $(m, n_1, n_2, n_3, n_4, w_M, w_N)$ describing triangulations as in the figure is\n$$\n2^m \\cdot 2^{n-m-8} \\binom{n-4}{4} = 2^{n-8} \\binom{n-4}{4}.\n$$\n\nFinally, in order to get the total number of triangulations we multiply the above number by $n$ (since we could start building the triangulation at any vertex rather than at $A_0$) and divide by $2$ (since every triangulation is now counted twice, once as starting at one of the interior triangles and once as starting at the other).\n\nThus, the answer is\n$$\nn2^{n-9} \\binom{n-4}{4}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47223,
    "subject": "Mathematics (Multi-modal)",
    "question": "A mathematical frog jumps along the number line. The frog starts at $1$, and jumps according to the following rule: if the frog is at integer $n$, then it can jump either to $n + 1$ or to $n + 2^{m_n+1}$ where $2^{m_n}$ is the largest power of $2$ that is a factor of $n$. Show that if $k \\ge 2$ is a positive integer and $i$ is a nonnegative integer, then the minimum number of jumps needed to reach $2^i k$ is greater than the minimum number of jumps needed to reach $2^i$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $i \\ge 0$ and $k \\ge 1$, let $x_{i,k}$ denote the minimum number of jumps needed to reach the integer $n_{i,k} = 2^i k$. We must prove that\n$$\nx_{i,k} > x_{i,1} \\qquad (*)\n$$\nfor all $i \\ge 0$ and $k \\ge 2$. We prove this using the method of descent.\n\nFirst note that $(*)$ holds for $i = 0$ and all $k \\ge 2$, because it takes $0$ jumps to reach the starting value $n_{0,1} = 1$, and at least one jump to reach $n_{0,k} = k \\ge 2$. Now assume that $(*)$ is not true for all choices of $i$ and $k$. Let $i_0$ be the minimal value of $i$ for which $(*)$ fails for some $k$, let $k_0$ be the minimal value of $k > 1$ for which $x_{i_0,k} \\le x_{i_0,1}$. Then it must be the case that $i_0 \\ge 1$ and $k_0 \\ge 2$.\n\nLet $J_{i_0,k_0}$ be a shortest sequence of $x_{i_0,k_0} + 1$ integers that the frog occupies in jumping from $1$ to $2^{i_0} k_0$. The length of each jump, that is, the difference between consecutive integers in $J_{i_0,k_0}$, is either $1$ or a positive integer power of $2$. The sequence $J_{i_0,k_0}$ cannot contain $2^{i_0}$ because it takes more jumps to reach $2^{i_0} k_0$ than it does to reach $2^{i_0}$. Let $2^{M+1}$, $M \\ge 0$ be the length of the longest jump made in generating $J_{i_0,k_0}$. Such a jump can only be made from a number that is divisible by $2^M$ (and by no higher power of $2$). Thus we must have $M < i_0$, since otherwise a number divisible by $2^{i_0}$ is visited before $2^{i_0} k_0$ is reached, contradicting the definition of $k_0$.\n\nLet $2^{m+1}$ be the length of the jump when the frog jumps over $2^{i_0}$. If this jump starts at $2^m(2t - 1)$ for some positive integer $t$, then it will end at $2^m(2t - 1) + 2^{m+1} = 2^m(2t + 1)$. Since it goes over $2^{i_0}$ we see $2^m(2t - 1) < 2^{i_0} < 2^m(2t + 1)$ or $(2^{i_0-m} - 1)/2 < t < (2^{i_0-m} + 1)/2$. Thus $t = 2^{i_0-m-1}$ and the jump over $2^{i_0}$ is from $2^m(2^{i_0-m} - 1) = 2^{i_0} - 2^m$ to $2^m(2^{i_0-m} + 1) = 2^{i_0} + 2^m$.\n\nConsidering the jumps that generate $J_{i_0,k_0}$, let $N_1$ be the number of jumps from $1$ to $2^{i_0} + 2^m$, and let $N_2$ be the number of jumps from $2^{i_0} + 2^m$ to $2^{i_0} k_0$. By definition of $i_0$, it follows that $2^m$ can be reached from $1$ in less than $N_1$ jumps. On the other hand, because $m < i_0$, the number $2^{i_0}(k_0 - 1)$ can be reached from $2^m$ in exactly $N_2$ jumps by using the same jump length sequence as in jumping from $2^m + 2^{i_0}$ to $2^{i_0} k_0 = 2^{i_0}(k_0 - 1) + 2^i$. The key point here is that the shift by $2^{i_0}$ does not affect any of divisibility conditions needed to make jumps of the same length. In particular, with the exception of the last entry, $2^{i_0} k_0$, all of the elements of $J_{i_0,k_0}$ are of the form $2^p(2t + 1)$ with $p < i_0$, again because of the definition of $k_0$. Because $2^p(2t + 1) - 2^{i_0} = 2^p(2t - 2^{i_0-p} + 1)$ and the number $2t + 2^{i_0-p} + 1$ is odd, a jump of size $2^{p+1}$ can be made from $2^p(2t + 1) - 2^{i_0}$ just as it can be made from $2^p(2t + 1)$.\n\nThus the frog can reach $2^m$ from $1$ in less than $N_1$ jumps, and can then reach $2^{i_0}(k_0 - 1)$ from $2^m$ in $N_2$ jumps. Hence the frog can reach $2^{i_0}(k_0 - 1)$ from $1$ in less than $N_1 + N_2$ jumps, that is, in fewer jumps than needed to get to $2^{i_0} k_0$ and hence in fewer jumps than required to get to $2^{i_0}$. This contradicts the definition of $k_0$.\nSuppose $x_0 = 1, x_1, \\dots, x_t = 2^i k$ are the integers visited by the frog on his trip from $1$ to $2^i k$, $k \\ge 2$. Let $s_j = x_j - x_{j-1}$ be the jump sizes. Define a reduced path $y_j$ inductively by\n$$\ny_j = \\begin{cases} y_{j-1} + s_j & \\text{if } y_{j-1} + s_j \\le 2^i, \\\\ y_{j-1} & \\text{otherwise.} \\end{cases}\n$$\nSay a jump $s_j$ is deleted in the second case. We will show that the distinct integers among the $y_j$ give a shorter path from $1$ to $2^i$. Clearly $y_j \\le 2^i$ for all $j$. Suppose $2^i - 2^{r+1} < y_j \\le 2^i - 2^r$ for some $0 \\le r \\le i - 1$. Then every deleted jump before $y_j$ must have length greater than $2^r$, hence must be a multiple of $2^{r+1}$. Thus $y_j \\equiv x_j \\pmod{2^{r+1}}$. If $y_{j+1} > y_j$, then either $s_{j+1} = 1$ (in which case this is a valid jump) or $s_{j+1}/2 = 2^m$ is the exact power of $2$ dividing $x_j$. In the second case, since $2^r \\ge s_{j+1} > 2^m$, the congruence says $2^m$ is also the exact power of $2$ dividing $y_j$, thus again this is a valid jump. Thus the distinct $y_j$ form a valid path for the frog. If $j = t$ the congruence gives $y_t \\equiv x_t \\equiv 0 \\pmod{2^{r+1}}$, but this is impossible for $2^i - 2^{r+1} < y_t \\le 2^i - 2^r$. Hence we see $y_t = 2^i$, that is, the reduced path ends at $2^i$. Finally since the reduced path ends at $2^i < 2^i k$ at least one jump must have been deleted and it is strictly shorter than the original path.\n(By Brian Lawrence) Suppose $2^i k$ can be reached in $m$ jumps.\n\nOur approach will be to consider the frog's life as a sequence of leaps of certain lengths. We will prove that by removing the longest leaps from the sequence, we generate a valid sequence of leaps that ends at $2^i$. Clearly this sequence will be shorter, since it was obtained by removing leaps. The result will follow.\n\n**Lemma** If we remove the longest leap in the frog's life (or one of the longest, in case of a tie) the sequence of leaps will still be legitimate.\n\n*Proof:* By definition, a leap from $n$ to $n + \\nu$ is legitimate if and only if either (a) $\\nu = 1$, or (b) $\\nu = 2^{m_n+1}$. If all leaps are of length $1$, then clearly removing one leap does not make any others illegitimate; suppose the longest leap has length $2^s$.\n\nThen we remove this leap and consider the effect on all the other leaps. Take an arbitrary leap starting (originally) at $n$, with length $\\nu$. Then $\\nu \\le 2^s$. If $\\nu = 1$ the new leap is legitimate no matter where it starts. Say $\\nu > 1$. Then $\\nu = 2^{m_n+1}$. Now if the leap is before the removed leap, its position is not changed, so $\\nu = 2^{m_n+1}$ and it remains legitimate. If it is after the removed leap, its starting point is moved back to $n - 2^s$. Now since $2^{m_n+1} = \\nu \\le 2^s$, we have $m_n \\le s - 1$; that is, $2^s$ does not divide $n$. Therefore, $2^{m_n}$ is the highest power of $2$ dividing $n - 2^s$, so $\\nu = 2^{m_n-2^s+1}$ and the leap is still legitimate. This proves the Lemma. ■\n\nWe now remove leaps from the frog's sequence of leaps in decreasing order of length. The frog's path has initial length $2^i k - 1$; we claim that at some point its length is $2^i - 1$.\n\nLet the frog's $m$ leaps have lengths\n$$\na_1 \\ge a_2 \\ge a_3 \\ge \\dots \\ge a_m.\n$$\nDefine a function $f$ by\n$$\nf(0) = 2^i k \\\\\nf(i) = f(i-1) - a_i, \\quad 1 \\le i \\le m.\n$$\nClearly $f(i)$ is the frog's final position if we remove the $i$ longest leaps. Note that $f(m) = 1$ – if we remove all leaps, the frog ends up at $1$. Let $f(j)$ be the last value of $f$ that is at least $2^i$. That is, suppose $f(j) \\ge 2^i$, $f(j+1) < 2^i$. Now we have $a_{j+1}|a_k$ for all $k \\le j$ since $\\{a_k\\}$ is a decreasing sequence of powers of $2$. If $a_{j+1} > 2^i$, we have $2^i|a_p$ for $p \\le j$, so $2^i|f(j+1)$. But $0 < f(j+1) < 2^i$, contradiction. Thus $a_{j+1} \\le 2^i$, so, since $a_{j+1}$ is a power of two, $a_{j+1}|2^i$. Since $a_{j+1}|2^i k$ and $a_1, \\dots, a_j$, we know that $a_{j+1}|f(j)$, and $a_{j+1}|f(j+1)$. So $f(j+1)$, $f(j)$ are two consecutive multiples of $a_{j+1}$, and $2^i$ (another such multiple) satisfies $f(j+1) < 2^i \\le f(j)$. Thus we have $2^i = f(j)$, so by removing $j$ leaps we make a path for the frog that is legitimate by the Lemma, and ends on $2^i$.\n\nNow let $m$ be the minimum number of leaps needed to reach $2^i k$. Applying the Lemma and the argument above the frog can reach $2^i$ in only $m-j$ leaps. Since $j > 0$ trivially ($j = 0$ implies $2^i = f(j) = f(0) = 2^i k$) we have $m-j < m$ as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of solutions in positive integers $\\left(k ; a_{1}, a_{2}, \\ldots, a_{k} ; b_{1}, b_{2}, \\ldots, b_{k}\\right)$ to the equation\n$$\na_{1}\\left(b_{1}\\right)+a_{2}\\left(b_{1}+b_{2}\\right)+\\cdots+a_{k}\\left(b_{1}+b_{2}+\\cdots+b_{k}\\right)=7 .\n$$",
    "options": [],
    "answer": "15",
    "solution": "Solution:\nLet $k, a_{1}, \\ldots, a_{k}, b_{1}, \\ldots, b_{k}$ be a solution. Then $b_{1}, b_{1}+b_{2}, \\ldots, b_{1}+\\cdots+b_{k}$ is just some increasing sequence of positive integers. Considering the $a_{i}$ as multiplicities, the $a_{i}$'s and $b_{i}$'s uniquely determine a partition of $7$. Likewise, we can determine $a_{i}$'s and $b_{i}$'s from any partition of $7$, so the number of solutions is $p(7)=15$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the plane, what is the length of the shortest path from $(-2,0)$ to $(2,0)$ that avoids the interior of the unit circle (i.e., circle of radius $1$) centered at the origin?",
    "options": [],
    "answer": "2*sqrt(3) + pi/3",
    "solution": "Solution:\n\nThe path goes in a line segment tangent to the circle, then an arc of the circle, then another line segment tangent to the circle. Since one of these tangent lines and a radius of the circle give two legs of a right triangle with hypotenuse the line from $(0,0)$ to $(-2,0)$ or $(2,0)$, the length of each tangent line is $\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$. Also, because these are $30^{\\circ}-60^{\\circ}-90^{\\circ}$ right triangles, the angle of the arc is $60^{\\circ}$ and has length $\\pi / 3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriangle $A B C$ has side lengths $A B=2 \\sqrt{5}$, $B C=1$, and $C A=5$. Point $D$ is on side $A C$ such that $C D=1$, and $F$ is a point such that $B F=2$ and $C F=3$. Let $E$ be the intersection of lines $A B$ and $D F$. Find the area of $C D E B$.",
    "options": [],
    "answer": "22/35",
    "solution": "Solution:\n\nDraw segment $A F$. Then notice $A F=4$, and we have a right triangle. Now draw line $C E$, let it intersect $A F$ at $G$. By Ceva, $F G=\\frac{4}{3}$ and $A G=\\frac{8}{3}$. Using mass points we find that $\\frac{A E}{E B}=6$ so $\\frac{[A E F]}{[B E F]}=6$, and since $[A B F]=4$, $[B E F]=\\frac{4}{7}$.\n\nIt's easy to see that $[C D F]=\\frac{1}{5}[A C F]=\\frac{6}{5}$, so\n$$\n[B C D E]=[C D F]-[B E F]=\\frac{6}{5}-\\frac{4}{7}=\\frac{22}{35}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a$, $b$ are nonzero real numbers such that $a^{2} + b^{2} = 8 a b$, find the value of $\\left|\\frac{a+b}{a-b}\\right|$.",
    "options": [],
    "answer": "√15/3",
    "solution": "Solution:\nNote that\n$$\n\\left|\\frac{a+b}{a-b}\\right| = \\sqrt{\\frac{(a+b)^2}{(a-b)^2}} = \\sqrt{\\frac{a^2 + b^2 + 2 a b}{a^2 + b^2 - 2 a b}} = \\sqrt{\\frac{10 a b}{6 a b}} = \\frac{\\sqrt{15}}{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTim has a working analog 12-hour clock with two hands that run continuously (instead of, say, jumping on the minute). He also has a clock that runs really slow—at half the correct rate, to be exact. At noon one day, both clocks happen to show the exact time. At any given instant, the hands on each clock form an angle between $0^{\\circ}$ and $180^{\\circ}$ inclusive. At how many times during that day are the angles on the two clocks equal?",
    "options": [],
    "answer": "33",
    "solution": "Solution:\n\nA tricky thing about this problem may be that the angles on the two clocks might be reversed and would still count as being the same (for example, both angles could be $90^{\\circ}$, but the hour hand may be ahead of the minute hand on one clock and behind on the other).\n\nLet $x$, $-12 \\leq x < 12$, denote the number of hours since noon. If we take $0^{\\circ}$ to mean upwards to the \"XII\" and count angles clockwise, then the hour and minute hands of the correct clock are at $30x^{\\circ}$ and $360x^{\\circ}$, and those of the slow clock are at $15x^{\\circ}$ and $180x^{\\circ}$. The two angles are thus $330x^{\\circ}$ and $165x^{\\circ}$, of course after removing multiples of $360^{\\circ}$ and possibly flipping sign; we are looking for solutions to\n$$\n330x^{\\circ} \\equiv 165x^{\\circ} \\pmod{360^{\\circ}} \\text{ or } 330x^{\\circ} \\equiv -165x^{\\circ} \\pmod{360^{\\circ}}\n$$\nIn other words,\n$$\n360 \\mid 165x \\text{ or } 360 \\mid 495x.\n$$\nOr, better yet,\n$$\n\\frac{165}{360}x = \\frac{11}{24}x \\text{ and/or } \\frac{495}{360}x = \\frac{11}{8}x\n$$\nmust be an integer. Now $x$ is any real number in the range $[-12, 12)$, so $11x/8$ ranges in $[-16.5, 16.5)$, an interval that contains 33 integers. For any value of $x$ such that $11x/24$ is an integer, of course $11x/8 = 3 \\times (11x/24)$ is also an integer, so the answer is just 33.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA finite set of circles in the plane is called nice if it satisfies the following three conditions:\n(i) No two circles intersect in more than one point;\n(ii) For every point $A$ of the plane there are at most two circles passing through $A$;\n(iii) Each circle from the set is tangent to exactly 5 other circles from the set.\nDoes there exist a nice set consisting of exactly\n(a) 2006 circles?\n(b) 2007 circles?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "Solution:\n(a) The answer is yes. The following picture shows that there is a nice set consisting of exactly 12 circles. It is also possible to construct a nice set with 22 circles (the picture can be found below). Since $2006 = 158 \\cdot 12 + 5 \\cdot 22$ we can make a nice set of 2006 by making a union of 158 disjoint nice sets of 12 circles each, and 5 disjoint nice sets each of which contains 22 circles.\n![](attached_image_1.png)\n\n(b) We will prove that a nice set can't contain an odd number of circles. Let $n$ be the total number of circles. We will count the number of pairs $(k, P)$ where $k$ is a circle in the set and $P$ a point at which $k$ touches another circle. For each circle $k$ there are exactly 5 such pairs, and hence the total number of pairs is $5n$. For each point $P$ there are exactly 2 pairs corresponding to it. Hence the total number of pairs has to be even, but $5n$ can't be even if $n$ is odd.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\int_{0}^{1} \\frac{d x}{\\sqrt{x}+\\sqrt[3]{x}}\n$$",
    "options": [],
    "answer": "5 - 6 ln(2)",
    "solution": "Solution:\nWriting $x = u^{6}$ so that $d x = 6 u^{5} d u$, we have\n$$\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d x}{\\sqrt{x}+\\sqrt[3]{x}} & = \\int_{0}^{1} \\frac{6 u^{5} d u}{u^{3}+u^{2}} \\\\\n& = 6 \\int_{0}^{1} \\frac{u^{3} d u}{u+1} \\\\\n& = 6 \\int_{0}^{1}\\left(u^{2}-u+1-\\frac{1}{u+1}\\right) d u \\\\\n& = 6\\left(\\frac{u^{3}}{3}-\\frac{u^{2}}{2}+u-\\left.\\ln |u+1|\\right|_{0} ^{1}\\right)=5-6 \\ln (2)\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ such that $AP = AB$ and $AQ = AC$ and $\\angle BAP = \\angle CAQ$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \\perp PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Note:** We present five different approaches. The first three synthetic solutions are all based on the following simple observation.\nWe first note that $APBR$ and $AQCR$ are cyclic quadrilaterals. It is easy to see that triangles $APC$ and $ABQ$ are congruent to each other, implying that $\\angle APR = \\angle APC = \\angle ABQ = \\angle ABR$. Thus, $APBR$ is a cyclic quadrilateral. Likewise, we can show that $AQCR$ is also cyclic.\n\n![](attached_image_1.png)\n\nLet $\\angle PAB = 2x$. Then in isosceles triangle $APB$, $\\angle APB = 90^\\circ - x$. In cyclic quadrilateral $APBR$, $\\angle ARB = 180^\\circ - \\angle APB = 90^\\circ + x$. Likewise, $\\angle ARC = 90^\\circ + x$. Hence $\\angle BRC = 360^\\circ - \\angle ARB - \\angle ARC = 180^\\circ - 2x$. It follows that $\\angle BOC = 4x$.\n\n**First Solution:** Reflect $C$ across line $AQ$ to $D$. Then $\\angle BAD = 4x + \\angle BAC = \\angle BAQ$. It is easy to see that triangles $BAD$ and $PAQ$ are congruent, implying that $\\angle ADB = \\angle AQP = y$.\n\n![](attached_image_2.png)\n\nNote also that $CAD$ and $COB$ are two isosceles triangles with the same vertex angle, and so they are similar to each other. It follows that triangle $CAO$ and $CBD$ are similar by SAS (side-angle-side), implying that $\\angle CAO = \\angle CDB = z$.\n\nThe angle formed by lines $AO$ and $PQ$ is equal to\n$$\n180^\\circ - \\angle OAQ - \\angle AQP = 180^\\circ - \\angle OAC - \\angle CAQ - \\angle AQP = 180^\\circ - z - 2x - y.\n$$\n\nSince $AQ$ is perpendicular to the base $CD$ in isosceles triangle $ACD$, we have\n$$\n90^\\circ = \\angle QAD + \\angle CDA = \\angle QAD + \\angle ADB + \\angle BDC = 2x + y + z.\n$$\nCombining the last two equations yields that fact the angle formed by lines $AO$ and $PQ$ is equal to $90^\\circ$; that is, $AO \\perp PQ$.\n\n\n**Second Solution:** We maintain the same notations as in the first solution. Let $M$ be the midpoint of arc $\\widehat{BC}$ on the circumcircle of triangle $BOC$. Then $BM = CM$. Since triangles $APC$ and $ABQ$ are congruent, $PC = BQ$. Since $BRMC$ is cyclic, $\\angle PCM = \\angle RCM = \\angle RBM = \\angle QBM$. Hence triangles $BMQ$ and $CMP$ are congruent by SAS. It follows triangles $MPQ$ and $MBC$ are similar. Since $\\angle BOC = 4x$, $\\angle MBC = \\angle MCB = x$, and so $\\angle MPQ = x$.\n\n![](attached_image_3.png)\n\nNote that both triangles $PAB$ and $MOB$ are isosceles triangles with vertex angle $2x$; that is, they are similar to each other. Hence triangles $BMP$ and $BOA$ are also similar by SAS, implying that $\\angle OAB = MPB = s$. We also note that in isosceles triangle $APB$,\n$$\n90^\\circ = \\angle APB + \\angle PAB/2 = \\angle APQ + \\angle QPM + \\angle MPB + \\angle PAB/2 = \\angle APQ + 2x + s.\n$$\nPutting the above together, we conclude that\n$$\n\\angle PAO + \\angle APQ = \\angle PAB + \\angle BAO + \\angle APQ = 2x + s + \\angle APQ = 90^\\circ,\n$$\nthat is $AO \\perp PQ$.\n\n\n**Third Solution:** We consider two rotations:\n$\\mathbf{R}_1$: a counterclockwise $2x$ (degree) rotation centered at $A$,\n$\\mathbf{R}_2$: a clockwise $4x$ (degree) rotation centered at $O$.\nLet $\\mathbf{T}$ denote the composition $\\mathbf{R}_1\\mathbf{R}_2\\mathbf{R}_1$. Then $\\mathbf{T}$ is a counterclockwise $2x - 4x + 2x = 0^\\circ$ rotation; that is, $\\mathbf{T}$ is translation. Note that\n$$\n\\mathbf{T}(P) = \\mathbf{R}_1(\\mathbf{R}_2(\\mathbf{R}_1(P))) = \\mathbf{R}_1(\\mathbf{R}_2(B)) = \\mathbf{R}_1(C) = Q,\n$$\nor, $\\mathbf{T}$ is the vector translation $\\overrightarrow{PQ}$.\nLet $A_1 = \\mathbf{R}_2(A)$ and $A_2 = \\mathbf{R}_1(A_1)$. Then $\\mathbf{T}(A) = A_2$; that is, $\\overrightarrow{AA_2} = \\overrightarrow{PQ}$, or $AA_2 \\parallel PQ$.\nBy the definitions of $\\mathbf{R}_2$ and $\\mathbf{R}_1$, we know that triangles $OAA_1$ and $A_1AA_2$ are isosceles triangles with respect vertex angles $\\angle AOA_1 = 4x$ and $\\angle A_1AA_2 = 2x^\\circ$. It is routine to compute that $\\angle OAA_2 = 90^\\circ$; that $AO \\perp AA_2$, or $AO \\perp PQ$.\n\n\n**Fourth Solution:** (By Ian Le) In this solutions, let each lowercase letter denote the number assigned to the point labeled with the corresponding uppercase letter. We further assume that $A$ is origin; that is, let $a = 0$. Let $\\omega = e^{2xi}$ (or $\\omega = \\cos(2x) + i\\sin(2x)$, and $\\omega^{-1} = \\cos(2x) - i\\sin(2x)$). Then because $O$ lies on the perpendicular bisector of $BC$ and $\\angle BOC = 4x$,\n$$\no = c + \\frac{(b-c)i}{2\\omega \\sin(2x)} = c + \\frac{bi}{2\\omega \\sin(2x)} - \\frac{ci}{2\\omega \\sin(2x)}.\n$$\nNote that\n$$\nc - \\frac{ci}{2\\omega \\sin(2x)} = c + \\frac{c\\omega^{-1}}{2i \\sin(2x)} = \\frac{c(\\omega^{-1} + 2i \\sin(2x))}{2i \\sin(2x)} = \\frac{c\\omega}{2i \\sin(2x)},\n$$\nCombining the last two equations gives\n$$\no = \\frac{bi}{2\\omega \\sin(2x)} + \\frac{c\\omega}{2i \\sin(2x)} = -\\frac{b}{2i\\omega \\sin(2x)} + \\frac{c\\omega}{2i \\sin(2x)} = \\frac{1}{2i \\sin(2x)} \\left( c\\omega - \\frac{b}{\\omega} \\right).\n$$\nNow we note that $p = \\frac{b}{\\omega}$ and $q = c\\omega$. Consequently, we obtain\n$$\n\\frac{q-p}{o-a} = 2i \\sin(2x),\n$$\nwhich is clearly a pure imaginary number; that is, $OA \\perp PQ$.\n\n\n**Fifth Solution:** (By Lan Le) In this solutions, we set $BC = a$, $AB = c$, $CA = b$, $A = \\angle BAC$, $B = \\angle ABC$, and $C = \\angle BCA$. We use the fact that\n$$\nOA \\perp PQ \\text{ if and only if } AP^2 - AQ^2 = OP^2 - OQ^2.\n$$\nClearly $AP^2 - AQ^2 = c^2 - b^2$. It remains to show that\n$$\nOP^2 - OQ^2 = c^2 - b^2. \\qquad (*)\n$$\nIn isosceles triangles APB and BOC, $BP = 2c \\sin x$ and $BO = \\frac{a}{2 \\sin(2x)}$. Note that $\\angle PBA + \\angle ABC + \\angle CBO = 90^\\circ - x + B + 90^\\circ - 2x = 180^\\circ + B - 3x$. Applying the **law of cosines** to triangle PBO yields\n$$\nOP^2 = 4c^2 \\sin^2 x + \\frac{a^2}{4 \\sin^2(2x)} + \\frac{ac \\cos(B - 3x)}{\\cos x}.\n$$\nIn exactly the same way, we can show that\n$$\nOQ^2 = 4b^2 \\sin^2 x + \\frac{a^2}{4 \\sin^2(2x)} + \\frac{ab \\cos(C - 3x)}{\\cos x}.\n$$\nHence\n$$\nOP^2 - OQ^2 = 4(c^2 - b^2) \\sin^2 x + \\frac{a}{\\cos x} (c \\cos(B - 3x) - b \\cos(C - 3x)). \\quad (\\dagger)\n$$\nUsing **Addition and Subtraction formulas** and the **law of sines** (more precisely, $c \\sin B = b \\sin C$), we have\n$$\n\\begin{aligned} & c \\cos(B - 3x) - b \\cos(C - 3x) \\\\ = & c \\cos(3x) \\cos B + c \\sin(3x) \\sin B - b \\cos(3x) \\cos C - b \\sin(3x) \\sin C \\\\ = & \\cos(3x)(c \\cos B - b \\cos C). \\end{aligned}\n$$\nSubstituting the last equation into $(\\dagger)$ gives\n$$\nOP^2 - OQ^2 = 4(c^2 - b^2) \\sin^2 x + \\frac{\\cos 3x}{\\cos x} (ac \\cos B - ab \\cos C).\n$$\nNote that\n$$\nac \\cos B - ab \\cos C = c(a \\cos B + b \\cos A) - b(a \\cos C + c \\cos A) = c^2 - b^2.\n$$\nCombining the last equations gives\n$$\nOP^2 - OQ^2 = (c^2 - b^2) \\left( 4 \\sin^2 x + \\frac{\\cos 3x}{\\cos x} \\right).\n$$\nBy the **Triple-angle formulas**, we have $\\cos 3x = 4 \\cos^3 x - 3 \\cos x$, and so\n$$\nOP^2 - OQ^2 = (c^2 - b^2)(4 \\sin^2 x + 4 \\cos^2 x - 3) = c^2 - b^2,\n$$\nwhich is $(*)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC$, respectively, such that $AE/ED = BF/FC$. Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$, respectively. Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$ pass through a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the second intersection of the circumcircles of triangles $TCF$ and $TDE$. Because the quadrilateral $PEDT$ is cyclic, $\\angle PET = \\angle PDT$, or\n$$\n\\angle PEF = \\angle PDC. \\quad (*)\n$$\nBecause the quadrilateral $PFCT$ is cyclic,\n$$\n\\angle PFE = \\angle PFT = \\angle PCT = \\angle PCD. \\quad (**)\n$$\nBy equations (*) and (**), it follows that triangle $PEF$ is similar to triangle $PDC$. Hence $\\angle FPE = \\angle CPD$ and $PF/PE = PC/PD$. Note also that $\\angle FPC = \\angle FPE + \\angle EPC = \\angle CPD + \\angle EPC = \\angle EPD$. Thus, triangle $EPD$ is similar to triangle $FPC$. Another way to say this is that there is a spiral similarity centered at $P$ that sends triangle $PFE$ to triangle $PCD$, which implies that there is also a spiral similarity, centered at $P$, that sends triangle $PFC$ to triangle $PED$, and vice versa. In terms of complex numbers, this amounts to saying that\n$$\n\\frac{D-P}{E-P} = \\frac{C-P}{F-P} \\implies \\frac{E-P}{F-P} = \\frac{D-P}{C-P}.\n$$\n![](attached_image_1.png)\nBecause $AE/ED = BF/FC$, points $A$ and $B$ are obtained by extending corresponding segments of two similar triangles $PED$ and $PFC$, namely, $DE$ and $CF$, by the identical proportion. We conclude that triangle $PDA$ is similar to triangle $PCB$, implying that triangle $PAE$ is similar to triangle $PBF$. Therefore, as shown before, we can establish the similarity between triangles $PBA$ and $PFE$, implying that\n$$\n\\angle PBS = \\angle PBA = \\angle PFE = \\angle PFS \\quad \\text{and} \\quad \\angle PAB = \\angle PEF.\n$$\nWe will give a solution using complex coordinates. The first step is the following lemma.\n\n**Lemma** Suppose $s$ and $t$ are real numbers and $x$, $y$ and $z$ are complex. The circle in the complex plane passing through $x$, $x + ty$ and $x + (s + t)z$ also passes through the point $x + syz/(y - z)$, independent of $t$.\n\n*Proof:* Four points $z_1, z_2, z_3$ and $z_4$ in the complex plane lie on a circle if and only if the cross-ratio\n$$\ncr(z_1, z_2, z_3, z_4) = \\frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}\n$$\nis real. Since we compute\n$$\ncr(x, x + ty, x + (s + t)z, x + syz/(y - z)) = \\frac{s + t}{s}\n$$\nthe given points are on a circle.\n\nLay down complex coordinates with $S = 0$ and $E$ and $F$ on the positive real axis. Then there are real $r_1, r_2$ and $R$ with $B = r_1A$, $F = r_2E$ and $D = E + R(A - E)$ and hence $AE/ED = BF/FC$ gives\n$$\nC = F + R(B - F) = r_2(1 - R)E + r_1RA.\n$$\nThe line $CD$ consists of all points of the form $sC + (1-s)D$ for real $s$. Since $T$ lies on this line and has zero imaginary part, we see from $\\text{Im}(sC + (1-s)D) = (sr_1R + (1-s)R)\\text{Im}(A)$ that it corresponds to $s = -1/(r_1 - 1)$. Thus\n$$\nT = \\frac{r_1D - C}{r_1 - 1} = \\frac{(r_2 - r_1)(R - 1)E}{r_1 - 1}.\n$$\nApply the lemma with $x = E$, $y = A - E$, $z = (r_2 - r_1)E/(r_1 - 1)$, and $s = (r_2 - 1)(r_1 - r_2)$. Setting $t = 1$ gives\n$$\n(x, x + y, x + (s + 1)z) = (E, A, S = 0)\n$$\nand setting $t = R$ gives\n$$\n(x, x + Ry, x + (s + R)z) = (E, D, T).\n$$\nTherefore the circumcircles to $SAE$ and $TDE$ meet at\n$$\nx + \\frac{syz}{y - z} = \\frac{AE(r_1 - r_2)}{(1 - r_1)E - (1 - r_2)A} = \\frac{AF - BE}{A + F - B - E}.\n$$\nThis last expression is invariant under simultaneously interchanging $A$ and $B$ and interchanging $E$ and $F$. Therefore it is also the intersection of the circumcircles of $SBF$ and $TCF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assign to each side $b$ of a convex polygon $\\mathcal{P}$ the maximum area of a triangle that has $b$ as a side and is contained in $\\mathcal{P}$. Show that the sum of the areas assigned to the sides of $\\mathcal{P}$ is at least twice the area of $\\mathcal{P}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define the *weight* of a side $XY$ to be the area assigned to it, and define an *antipoint* of a side of a polygon to be one of the points in the polygon farthest from that side (and consequently forming the triangle with greatest area).\n\n**Lemma** For any side $XY$, $Z$ is an antipoint if and only if the line $l$ through $Z$ parallel to $XY$ does not go through the interior of the polygon. (Note that this means we can assume $Z$ is a vertex, as we shall do henceforth).\n\n*proof:* Clearly, if $Z$ is an antipoint $l$ must not go through the interior of the polygon. Now if $l$ does not go through the interior of the polygon, assume there is a point $Z'$ farther away from $XY$ than $Z$. Since the polygon is convex, the point $XZ' \\cap l$ is in the interior of the polygon, which is a contradiction. ■\n\nSuppose for the sake of contradiction that the sum of the weights of the sides is less than twice the area of some polygon. Then let $S$ be the non-empty set of all convex polygons for which the sum of the weights is strictly less than twice the area. It is easy to check that no polygon in $S$ can be a triangle, so we may assume all polygons in $S$ have at least 4 sides.\n\nWe first prove by contradiction that there is some polygon in $S$ such that all of its sides are parallel to some other side. Suppose the contrary; then consider one of the polygons in $S$ which has the minimal number of sides not parallel to any other side (this exists by the well-ordering principle). Call this polygon $P = A_1A_2\\cdots A_n$, and WLOG let $A_nA_1$ be a side which is not parallel to any other side of $P$.\n\nThen let $A_i$ be the unique antipoint of $A_nA_1$, and let $A_u$ and $A_v$ be respective antipoints of $A_{i-1}A_i$ and $A_iA_{i+1}$. Define $X$ to be the point such that $A_uX\\parallel A_{i-1}A_i$, $A_vX\\parallel A_iA_{i+1}$.\n\nNow consider the set $T \\subset P$ of points that are strictly on the same side of $A_uA_v$ as $A_nA_1$. First of all, for any side in $T$, $A_i$ must be its antipoint, since the line through $A_i$ parallel to $A_jA_{j+1}$ does not go through the interior of $P$. Similarly, any vertex in $T$ is not the antipoint of any side.\n\nWe now look at the polygon $P' = A_vA_{v+1}\\cdots A_{u-1}A_uX$. First of all, it is clear that $P'$ has fewer sides which are not parallel to any other side than $P$. Using $[\\cdot]$ to denote area, we have\n$$\n[P'] - [P] = [A_1A_2\\cdots A_{v-1}A_vXA_uA_{u+1}\\cdots A_n].\n$$\nThe weights of the side $A_jA_{j+1}$ is the same in both $P'$ and $P$ for $v \\le j < u$, but for $P'$, the sum of the weights of the remaining two sides is $[XA_uA_iA_v]$, as $A_i$ is an antipoint of both $A_uX$ and $A_vX$. Meanwhile, the sum of the weights of remaining sides for $P$ is $[A_1A_2\\cdots A_{v-1}A_vA_iA_uA_{u+1}\\cdots A_n]$. Hence the difference in the sums of weights of $P'$ and $P$ is\n$$\n[XA_uA_iA_v] - [A_1A_2\\cdots A_{v-1}A_vA_iA_uA_{u+1}\\cdots A_n] = [A_1A_2\\cdots A_{v-1}A_vXA_uA_{u+1}\\cdots A_n],\n$$\nthe same as the difference in area (and both differences were positive). Therefore, if the sum of weights of $P$ was less than $2[P]$, then certainly the sum of weights of $P'$ must be less than $2[P']$, so that $P' \\in S$. However, this contradicts the minimality of the number of non-parallel sides in $P$, so there exists a polygon in $S$ with opposite sides parallel.\n\nNow, we will let $R$ be the non-empty set of all polygons in $S$ with all sides parallel to the opposite side. Note that all polygons in $R$ must have an even number of sides. We will show that there is a parallelogram in $R$.\n\nSuppose not, and that $Q = B_1B_2\\cdots B_{2m}$ is one of the polygons in $R$ with the minimal number of sides, and $m \\ge 3$. Let $X = B_1B_2 \\cap B_{2m-1}B_{2m}$ and $Y = B_{m-1}B_m \\cap B_{m+2}B_{m+1}$. Set $Q' = XB_2B_3\\cdots B_{m-1}YB_{m+2}\\cdots B_{2m}$. We propose that the increase in the sum of weights going from $Q$ to $Q'$ is at most twice the increase in area, so that $Q' \\in R$.\n\nTo aid us, we will let $h_X$ and $h_Y$ be the respective distances of $X$ and $Y$ from $B_{2m}B_1$ and $B_mB_{m+1}$.\n\nThe increase in weight is\n$$\n\\begin{align*}\n& [XB_{m+1}B_1] + [XB_{2m}B_m] + [YB_mB_{2m}] + [YB_{m+1}B_1] - [B_1B_{2m}B_m] - [B_{2m}B_mB_{m+1}] \\\\\n&= [XB_1Y] + [XB_{2m}Y] - [B_1B_{2m}B_m] + [YB_mX] + [YB_{m+1}X] - [B_{2m}B_mB_{m+1}] \\\\\n&= [XB_1B_{2m}] + \\frac{h_Y \\cdot B_1B_{2m}}{2} + [YB_mB_{m+1}] + \\frac{h_X \\cdot B_mB_{m+1}}{2}\n\\end{align*}\n$$\nwhile the increase in area is $[XB_1B_{2m}] + [YB_mB_{m+1}]$. It remains to show that the first expression is at most twice the second, or in other words, to show that\n$$\n\\frac{h_Y \\cdot B_1 B_{2m}}{2} + \\frac{h_X \\cdot B_m B_{m+1}}{2} \\le [XB_1 B_{2m}] + [Y B_m B_{m+1}] = \\frac{h_X \\cdot B_1 B_{2m}}{2} + \\frac{h_Y \\cdot B_m B_{m+1}}{2},\n$$\nwhich is equivalent to\n$$\n(h_X - h_Y)(B_1 B_{2m} - B_m B_{m+1}) \\geq 0\n$$\nNoting that triangles $B_1B_{2m}X$ and $B_{m+1}B_mY$ are similar, we have $h_X/h_Y = B_1B_{2m}/B_mB_{m+1}$, so the above inequality holds.\n\nWith the inequality proven, we now know that $Q' \\in R$, and yet $Q'$ has fewer sides than $Q$. This contradicts the minimality of the number of sides of $Q$, so there exists a parallelogram in $R$. However, the sum of the weights of a parallelogram clearly equals twice its area, so this contradicts the entire existence of $S$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOctagon $A B C D E F G H$ is equiangular. Given that $A B=1$, $B C=2$, $C D=3$, $D E=4$, and $E F=F G=2$, compute the perimeter of the octagon.",
    "options": [],
    "answer": "20 + sqrt(2)",
    "solution": "Solution:\n\nExtend sides $A B$, $C D$, $E F$, $G H$ to form a rectangle: let $X$ be the intersection of lines $G H$ and $A B$; $Y$ that of $A B$ and $C D$; $Z$ that of $C D$ and $E F$; and $W$ that of $E F$ and $G H$.\n\nAs $B C=2$, we have $B Y=Y C=\\sqrt{2}$. As $D E=4$, we have $D Z=Z E=2\\sqrt{2}$. As $F G=2$, we have $F W=W G=\\sqrt{2}$.\n\nWe can compute the dimensions of the rectangle: $W X=Y Z=Y C+C D+D Z=3+3\\sqrt{2}$, and $X Y=Z W=Z E+E F+F W=2+3\\sqrt{2}$.\n\nThus, $H X=X A=X Y-A B-B Y=1+2\\sqrt{2}$, and so $A H=\\sqrt{2} H X=4+\\sqrt{2}$, and $G H=W X-W G-H X=2$.\n\nThe perimeter of the octagon can now be computed by adding up all its sides.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be an integer at least $5$. At most how many diagonals of a regular $n$-gon can be simultaneously drawn so that no two are parallel? Prove your answer.",
    "options": [],
    "answer": "n",
    "solution": "Solution:\n\nLet $O$ be the center of the $n$-gon. Let us consider two cases, based on the parity of $n$:\n\n- $n$ is odd. In this case, for each diagonal $d$, there is exactly one vertex $D$ of the $n$-gon, such that $d$ is perpendicular to line $O D$; and of course, for each vertex $D$, there is at least one diagonal $d$ perpendicular to $O D$, because $n \\geq 5$. The problem of picking a bunch of $d$'s so that no two are parallel is thus transmuted into one of picking a bunch of $d$'s so that none of the corresponding $D$'s are the same. Well, go figure.\n\n- $n$ is even. What can I say? For each diagonal $d$, the perpendicular dropped from $O$ to $d$ either passes through two opposite vertices of the $n$-gon, or else bisects two opposite sides. Conversely, for each line joining opposite vertices or bisecting opposite sides, there is at least one diagonal perpendicular to it, because $n \\geq 6$. By reasoning similar to the odd case, we find the answer to be $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor how many ordered triplets $(a, b, c)$ of positive integers less than $10$ is the product $a \\times b \\times c$ divisible by $20$?",
    "options": [],
    "answer": "102",
    "solution": "Solution:\n\nOne number must be $5$. The other two must have a product divisible by $4$. Either both are even, or one is divisible by $4$ and the other is odd. In the former case, there are $48 = 3 \\times 4 \\times 4$ possibilities: $3$ positions for the $5$, and any of $4$ even numbers to fill the other two. In the latter case, there are $54 = 3 \\times 2 \\times 9$ possibilities: $3$ positions and $2$ choices for the multiple of $4$, and $9$ ways to fill the other two positions using at least one $5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $ABC$ is a triangle such that $AB = 13$, $BC = 15$, and $CA = 14$. Say $D$ is the midpoint of $\\overline{BC}$, $E$ is the midpoint of $\\overline{AD}$, $F$ is the midpoint of $\\overline{BE}$, and $G$ is the midpoint of $\\overline{DF}$. Compute the area of triangle $EFG$.",
    "options": [],
    "answer": "21/4",
    "solution": "Solution:\n\nBy Heron's formula, $[ABC] = \\sqrt{21(21-15)(21-14)(21-13)} = 84$. Now, unwinding the midpoint conditions yields $[EFG] = \\frac{[DEF]}{2} = \\frac{[BDE]}{4} = \\frac{[ABD]}{8} = \\frac{[ABC]}{16} = \\frac{84}{16} = \\frac{21}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA triangle with vertices at $(1003,0)$, $(1004,3)$, and $(1005,1)$ in the $xy$-plane is revolved all the way around the $y$-axis. Find the volume of the solid thus obtained.",
    "options": [],
    "answer": "5020π",
    "solution": "Solution:\nLet $T \\subset \\mathbb{R}^2$ denote the triangle, including its interior. Then $T$'s area is $5/2$, and its centroid is $(1004, 4/3)$, so\n$$\n\\int_{(x, y) \\in T} x \\, dx \\, dy = \\frac{5}{2} \\cdot 1004 = 2510\n$$\nWe are interested in the volume\n$$\n\\int_{(x, y) \\in T} 2\\pi x \\, dx \\, dy\n$$\nbut this is just $2\\pi \\cdot 2510 = 5020\\pi$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 47239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$, $b$, $c$ be the roots of $x^{3}-9 x^{2}+11 x-1=0$, and let $s=\\sqrt{a}+\\sqrt{b}+\\sqrt{c}$. Find $s^{4}-18 s^{2}-8 s$.",
    "options": [],
    "answer": "-37",
    "solution": "Solution:\n\nFirst of all, as the left side of the first given equation takes values $-1, 2, -7$, and $32$ when $x=0, 1, 2$, and $3$, respectively, we know that $a$, $b$, and $c$ are distinct positive reals. Let $t=\\sqrt{a b}+\\sqrt{b c}+\\sqrt{c a}$, and note that\n$$\n\\begin{aligned}\ns^{2} & = a+b+c+2 t = 9+2 t \\\\\nt^{2} & = a b+b c+c a+2 \\sqrt{a b c} s = 11+2 s \\\\\ns^{4} & = (9+2 t)^{2} = 81+36 t+4 t^{2} = 81+36 t+44+8 s = 125+36 t+8 s, \\\\\n18 s^{2} & = 162+36 t\n\\end{aligned}\n$$\nso that $s^{4}-18 s^{2}-8 s = -37$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose we have a regular hexagon and draw all its sides and diagonals. Into how many regions do the segments divide the hexagon? (No proof is necessary.)",
    "options": [],
    "answer": "24",
    "solution": "Solution:\nAn accurate diagram and a careful count yields the answer.\n\n**Answer:** $24$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle of radius $t$ is tangent to the hypotenuse, the incircle, and one leg of an isosceles right triangle with inradius $r=1+\\sin \\frac{\\pi}{8}$. Find $r t$.",
    "options": [],
    "answer": "(2+\\sqrt{2})/4",
    "solution": "Solution:\nThe distance between the point of tangency of the two circles and the nearest vertex of the triangle is seen to be both $r\\left(\\csc \\frac{\\pi}{8}-1\\right)$ and $t\\left(\\csc \\frac{\\pi}{8}+1\\right)$, so\n$$\n\\begin{aligned}\nr t & =\\frac{r^{2}\\left(\\csc \\frac{\\pi}{8}-1\\right)}{\\csc \\frac{\\pi}{8}+1}=\\frac{\\left(1+\\sin \\frac{\\pi}{8}\\right)^{2}\\left(1-\\sin \\frac{\\pi}{8}\\right)}{1+\\sin \\frac{\\pi}{8}}=1-\\sin ^{2} \\frac{\\pi}{8} \\\\\n& =\\frac{1}{2}+\\frac{1-2 \\sin ^{2} \\frac{\\pi}{8}}{2}=\\frac{1}{2}+\\frac{\\cos \\frac{\\pi}{4}}{2}=\\frac{1}{2}+\\frac{\\sqrt{2}}{4}=\\frac{2+\\sqrt{2}}{4} .\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer, and let Pushover be a game played by two players, standing squarely facing each other, pushing each other, where the first person to lose balance loses. At the HMPT, $2^{n+1}$ competitors, numbered 1 through $2^{n+1}$ clockwise, stand in a circle. They are equals in Pushover: whenever two of them face off, each has a $50\\%$ probability of victory. The tournament unfolds in $n+1$ rounds. In each round, the referee randomly chooses one of the surviving players, and the players pair off going clockwise, starting from the chosen one. Each pair faces off in Pushover, and the losers leave the circle. What is the probability that players 1 and $2^{n}$ face each other in the last round? Express your answer in terms of $n$.",
    "options": [],
    "answer": "(2^n - 1)/8^n",
    "solution": "Solution:\n\nAt any point during this competition, we shall say that the situation is living if both players 1 and $2^{n}$ are still in the running. A living situation is far if those two players are diametrically opposite each other, and near otherwise, in which case (as one can check inductively) they must be just one person shy of that maximal separation. At the start of the tournament, the situation is living and near. In each of rounds 1 to $n$, a far situation can never become near, and a near situation can stay near or become far with equal likelihood.\n\nIn each of rounds 1 to $n-1$, a living situation has a $1/4$ probability of staying living. Therefore, at the end of round $k$, where $1 \\leq k \\leq n-1$, the situation is near with probability $1/8^{k}$, and far with probability $1/4^{k} - 1/8^{k}$. In round $n$, a far situation has a $1/4$ probability of staying living, whereas a near situation has only a $1/8$ probability of staying living. But if the situation is living at the beginning of the last round, it can only be far, so we can say with complete generality that, at the end of round $k$, where $1 \\leq k \\leq n$, the situation is living and far with probability $1/4^{k} - 1/8^{k}$.\n\nWe are interested in finding the probability that the situation is living at the end of round $n$ (and hence far); that probability is thus\n$$\n\\frac{1}{4^{n}} - \\frac{1}{8^{n}} = \\frac{2^{n} - 1}{8^{n}}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive real numbers $c$ such that the graph of $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ given by $f(x) = x^{3} - c x$ has the property that the circle of curvature at any local extremum is centered at a point on the $x$-axis.",
    "options": [],
    "answer": "sqrt(3)/2",
    "solution": "Solution:\nThe equation $0 = f'(x) = 3x^{2} - c$ has two real roots: $\\pm \\sqrt{c / 3}$. Let $a := \\sqrt{c / 3}$. As $f''(-a) = -6 \\sqrt{c / 3} < 0$, $f$ has a unique local maximum at $x = -a$.\n\nBecause $f$ has half-turn symmetry about the origin, it suffices to consider this local extremum. The radius of curvature at any local extremum is\n$$\nr(x) = \\frac{1}{|f''(x)|} = \\frac{1}{6|x|},\n$$\nso the condition in the problem is equivalent to\n$$\n\\begin{gathered}\nr(-a) = f(-a) \\\\\n\\frac{1}{6a} = -a(a^{2} - c) \\\\\n1 = 6a^{2}(c - a^{2}) = 2c(2c / 3) \\\\\nc = \\sqrt{3} / 2\n\\end{gathered}",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 47244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a convex $n$-gon, $n \\geq 4$, at most how many diagonals can be drawn such that each drawn diagonal intersects every other drawn diagonal either in the interior of the $n$-gon or at a vertex? Prove your answer.",
    "options": [],
    "answer": "Maximum equals n for all n ≥ 5, and equals 2 for n = 4.",
    "solution": "Solution:\n\nFirst of all, assume without loss of generality that the $n$-gon is regular (this has no effect as far as diagonal intersection is concerned). Also, treat $n=4$ as a special case; obviously the answer is 2 here.\n\nIf $n$ is odd, simply draw $n$ diagonals, connecting each vertex to the ones $(n-1) / 2$ vertices away (in either direction).\n\nIf $n$ is even, first draw the $n / 2$ diagonals connecting pairs of vertices $n / 2$ vertices apart. Then, there are $n$ diagonals connecting pairs of vertices $n / 2-1$ vertices apart; they come in $n / 2$ pairs of parallel diagonals; from each such pair, randomly pick one diagonal and draw it.\n\nTo see that these constructions work, note that two diagonals, each connecting pairs of vertices at least $n / 2-1$ vertices apart, can fail to intersect or share a vertex only if they are parallel.\n\nThe previous problem shows that these constructions are optimal.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe train schedule in Hummut is hopelessly unreliable. Train $A$ will enter Intersection $X$ from the west at a random time between 9:00 am and 2:30 pm; each moment in that interval is equally likely. Train $B$ will enter the same intersection from the north at a random time between 9:30 am and 12:30 pm, independent of Train $A$; again, each moment in the interval is equally likely. If each train takes 45 minutes to clear the intersection, what is the probability of a collision today?",
    "options": [],
    "answer": "13/48",
    "solution": "Solution:\n\nSuppose we fix the time at which Train $B$ arrives at Intersection $X$; then call the interval during which Train $A$ could arrive (given its schedule) and collide with Train $B$ the \"disaster window.\"\n\nWe consider two cases:\n\ni. Train $B$ enters Intersection $X$ between 9:30 and 9:45. If Train $B$ arrives at 9:30, the disaster window is from 9:00 to 10:15, an interval of $1 \\frac{1}{4}$ hours. If Train $B$ arrives at 9:45, the disaster window is $1 \\frac{1}{2}$ hours long. Thus, the disaster window has an average length of $\\left(1 \\frac{1}{4}+1 \\frac{1}{2}\\right) \\div 2=\\frac{11}{8}$. From 9:00 to 2:30 is $5 \\frac{1}{2}$ hours. The probability of a collision is thus $\\frac{11}{8} \\div 5 \\frac{1}{2}=\\frac{1}{4}$.\n\nii. Train $B$ enters Intersection $X$ between 9:45 and 12:30. Here the disaster window is always $1 \\frac{1}{2}$ hours long, so the probability of a collision is $1 \\frac{1}{2} \\div 5 \\frac{1}{2}=\\frac{3}{11}$.\n\nFrom 9:30 to 12:30 is 3 hours. Now case (i) occurs with probability $\\frac{1}{4} \\div 3=\\frac{1}{12}$, and case (ii) occurs with probability $\\frac{11}{12}$. The overall probability of a collision is therefore $\\frac{1}{12} \\cdot \\frac{1}{4}+\\frac{11}{12} \\cdot \\frac{3}{11}=\\frac{1}{48}+\\frac{1}{4}=\\frac{13}{48}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $x$ such that\n$$\nx^{2} + \\left\\lfloor \\frac{x}{2} \\right\\rfloor + \\left\\lfloor \\frac{x}{3} \\right\\rfloor = 10\n$$",
    "options": [],
    "answer": "-sqrt(14)",
    "solution": "Solution:\nEvidently $x^{2}$ must be an integer. Well, there aren't that many things to check, are there? Among positive $x$, $\\sqrt{8}$ is too small and $\\sqrt{9}$ is too big; among negative $x$, $-\\sqrt{15}$ is too small and $-\\sqrt{13}$ is too big.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the smallest positive integer $n$ such that $n^{2}$ and $(n+1)^{2}$ both contain the digit 7 but $(n+2)^{2}$ does not?",
    "options": [],
    "answer": "27",
    "solution": "Solution:\nThe last digit of a square is never 7. No two-digit squares begin with 7. There are no 3-digit squares beginning with the digits $17, 27, 37$, or $47$. In fact, the smallest square containing the digit 7 is $576 = 24^{2}$. Checking the next few numbers, we see that $25^{2} = 625$, $26^{2} = 676$, $27^{2} = 729$, $28^{2} = 784$, and $29^{2} = 841$, so the answer is 27.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $ABCD$ is an isosceles trapezoid in which $\\overline{AB} \\parallel \\overline{CD}$. Two mutually externally tangent circles $\\omega_1$ and $\\omega_2$ are inscribed in $ABCD$ such that $\\omega_1$ is tangent to $\\overline{AB}$, $\\overline{BC}$, and $\\overline{CD}$ while $\\omega_2$ is tangent to $\\overline{AB}$, $\\overline{DA}$, and $\\overline{CD}$. Given that $AB=1$, $CD=6$, compute the radius of either circle.",
    "options": [],
    "answer": "3/7",
    "solution": "Solution:\n\nLet the radius of both circles be $r$, and let $\\omega_1$ be centered at $O_1$. Let $\\omega_1$ be tangent to $\\overline{AB}$, $\\overline{BC}$, and $\\overline{CD}$ at $P$, $Q$, and $R$ respectively. Then, by symmetry, $PB = \\frac{1}{2} - r$ and $RC = 3 - r$. By equal tangents from $B$ and $C$, $BQ = \\frac{1}{2} - r$ and $QC = 3 - r$.\n\nNow, $\\angle BO_1C$ is right because $m \\angle O_1BC + m \\angle BCO_1 = \\frac{1}{2}(m \\angle PBC + m \\angle BCR) = 90^\\circ$.\n\nSince $\\overline{O_1Q} \\perp \\overline{BC}$, $r^2 = O_1Q^2 = BQ \\cdot QC = \\left(\\frac{1}{2} - r\\right)(3 - r) = r^2 - \\frac{7}{2}r + \\frac{3}{2}$.\n\nSolving, we find $r = \\frac{3}{7}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet\n$$\nf(x) = x^{4} - 6x^{3} + 26x^{2} - 46x + 65.\n$$\nLet the roots of $f(x)$ be $a_{k} + i b_{k}$ for $k = 1, 2, 3, 4$. Given that the $a_{k}, b_{k}$ are all integers, find $|b_{1}| + |b_{2}| + |b_{3}| + |b_{4}|$.",
    "options": [],
    "answer": "10",
    "solution": "Solution:\nThe roots of $f(x)$ must come in complex-conjugate pairs. We can then say that $a_{1} = a_{2}$ and $b_{1} = -b_{2}$; $a_{3} = a_{4}$ and $b_{3} = -b_{4}$. The constant term of $f(x)$ is the product of these, so $5 \\cdot 13 = (a_{1}^{2} + b_{1}^{2})(a_{3}^{2} + b_{3}^{2})$. Since $a_{k}$ and $b_{k}$ are integers for all $k$, and it is simple to check that $1$ and $i$ are not roots of $f(x)$, we must have $a_{1}^{2} + b_{1}^{2} = 5$ and $a_{3}^{2} + b_{3}^{2} = 13$. The only possible ways to write these sums with positive integers is $1^{2} + 2^{2} = 5$ and $2^{2} + 3^{2} = 13$, so the values of $a_{1}$ and $b_{1}$ up to sign are $1$ and $2$; and $a_{3}$ and $b_{3}$ up to sign are $2$ and $3$. From the $x^{3}$ coefficient of $f(x)$, we get that $a_{1} + a_{2} + a_{3} + a_{4} = 6$, so $a_{1} + a_{3} = 3$. From the limits we already have, this tells us that $a_{1} = 1$ and $a_{3} = 2$. Therefore $b_{1}, b_{2} = \\pm 2$ and $b_{3}, b_{4} = \\pm 3$, so the required sum is $2 + 2 + 3 + 3 = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose we have an octagon with all angles of $135^{\\circ}$, and consecutive sides of alternating length $1$ and $\\sqrt{2}$. We draw all its sides and diagonals. Into how many regions do the segments divide the octagon? (No proof is necessary.)",
    "options": [],
    "answer": "84",
    "solution": "Solution:\n\nThe easiest way to see the answer is to view the octagon as five unit squares in a cross arrangement, with four half-squares wedged at the corners. The center square is divided into $8$ regions. The other $4$ squares are each divided into $15$ regions. The $4$ half-squares are each divided into $4$ regions. The answer is thus $8 + 4 \\times 15 + 4 \\times 4 = 84$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriangle $A B C$ has a right angle at $B$. Point $D$ lies on side $B C$ such that $3 \\angle B A D = \\angle B A C$. Given $A C = 2$ and $C D = 1$, compute $B D$.",
    "options": [],
    "answer": "3/8",
    "solution": "Solution:\n\nLet $B D = x$. We reflect $D$ over $A B$ to $D'$. Then $D D' = 2x$, but $A D$ bisects $C A D'$, so $4x = A D' = A D$. Also, $A D = \\sqrt{x^2 + A B^2} = \\sqrt{x^2 + A C^2 - B C^2} = \\sqrt{x^2 + 4 - (x + 1)^2} = \\sqrt{3 - 2x}$. We have the quadratic $16x^2 = 3 - 2x$ which gives $x = 3/8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways are there to label the faces of a regular octahedron with the integers $1$–$8$, using each exactly once, so that any two faces that share an edge have numbers that are relatively prime? Physically realizable rotations are considered indistinguishable, but physically unrealizable reflections are considered different.",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nWell, instead of labeling the faces of a regular octahedron, we may label the vertices of a cube. Then, as no two even numbers may be adjacent, the even numbers better form a regular tetrahedron, which can be done in $2$ ways (because rotations are indistinguishable but reflections are different). Then $3$ must be opposite $6$, and the remaining numbers – $1, 5, 7$ – may be filled in at will, in $3! = 6$ ways. The answer is thus $2 \\times 6 = 12$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn how many ways can we enter numbers from the set $\\{1,2,3,4\\}$ into a $4 \\times 4$ array so that all of the following conditions hold?\n\n(a) Each row contains all four numbers.\n\n(b) Each column contains all four numbers.\n\n(c) Each \"quadrant\" contains all four numbers. (The quadrants are the four corner $2 \\times 2$ squares.)",
    "options": [],
    "answer": "288",
    "solution": "Solution:\n\nCall a filled $4 \\times 4$ array satisfying the given conditions cool. There are $4!$ possibilities for the first row; WLOG, let it be $1234$. Since each quadrant has to contain all four numbers, we have exactly four possibilities for the second row, namely:\n(i) $3412$\n(ii) $3421$\n(iii) $4312$\n(iv) $4321$\n\nI claim that the number of cool arrays with (i) is equal to those with (iv), and that the number of cool arrays with (ii) is equal to those with (iii). Let's first consider (i) and (iv). Now, (i) is\n1234\n3412\nwhile (iv) is\n4321\nIn (iv), switch $3$ and $4$ (relabeling doesn't affect the coolness of the array); then, it becomes\n3421\nNow, interchange the last two columns, which also does not affect the coolness. This gives us (i). Hence, the cool arrays with (i) and the cool arrays with (iv) have a $1:1$ correspondence. Using the exact same argument, we can show that the number of cool arrays with (ii) equals those with (iii).\n\nSo we only need consider cases (i) and (ii). It is easy to verify that there are four cool arrays with (i), determined precisely by, say, the first two entries of the third row; and two with (ii), determined precisely by, say the first entry of the third row. Hence, the answer is $4! \\times (4+2) \\times 2 = 288$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA regular 12-sided polygon is inscribed in a circle of radius $1$. How many chords of the circle that join two of the vertices of the 12-gon have lengths whose squares are rational? (No proof is necessary.)",
    "options": [],
    "answer": "42",
    "solution": "Solution:\n\nThe chords joining vertices subtend minor arcs of $30^{\\circ}, 60^{\\circ}, 90^{\\circ}, 120^{\\circ}, 150^{\\circ}$, or $180^{\\circ}$. There are $12$ chords of each of the first five kinds and $6$ diameters. For a chord with central angle $\\theta$, we can draw radii from the two endpoints of the chord to the center of the circle. By the law of cosines, the square of the length of the chord is $1+1-2 \\cos \\theta$, which is rational when $\\theta$ is $60^{\\circ}, 90^{\\circ}, 120^{\\circ}$, or $180^{\\circ}$. The answer is thus $12+12+12+6=42$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a regular $n$-gon with sides of length $1$, what is the smallest radius $r$ such that there is a non-empty intersection of $n$ circles of radius $r$ centered at the vertices of the $n$-gon? Give $r$ as a formula in terms of $n$. Be sure to prove your answer.",
    "options": [],
    "answer": "r = 1 / (2 sin(pi/n))",
    "solution": "Solution:\n\nIt is easy to see that, with this $r$, all the circles pass through the center of the $n$-gon. The following proves that this $r$ is necessary even if the word \"circle\" is replaced by the word \"disk.\"\n\nFor $n$ even, it is easy to see using symmetry that containing the center point is necessary and sufficient. For $n$ odd, there is more work to do. Again, containing the center point is sufficient. To see its necessity, consider three circles: a circle at a vertex $A$, and the two circles on the segment $BC$ opposite $A$. Circles $B$ and $C$ intersect in a region $R$ symmetric about the perpendicular bisector of $BC$, with the closest point of $R$ to $A$ being on this line. Hence, circle $A$ must intersect $R$ at some point on the perpendicular bisector of $BC$; and thus we see the entire perpendicular bisector of $BC$ inside of the $n$-gon is contained in the circles. Now, this bisector contains the center of the $n$-gon, so some circle must contain the center. But by symmetry, if one circle contains the center, all do. Thus, in any case, it is necessary and sufficient for $r$ to be large enough so the center is contained in a circle. Basic trigonometry gives the answer, which equals the distance between a vertex and the center.\n\nAnswer: $\\quad r=\\frac{1}{2} \\csc \\frac{180^{\\circ}}{n}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSix people, all of different weights, are trying to build a human pyramid: that is, they get into the formation\n\n> A\n> B C\n> D E F\n\nWe say that someone not in the bottom row is \"supported by\" each of the two closest people beneath her or him. How many different pyramids are possible, if nobody can be supported by anybody of lower weight?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\nWithout loss of generality, let the weights of the people be $1, 2, 3, 4, 5,$ and $6$. Clearly we must have $A=1$. Then, equally clearly, either $B$ or $C$ must be $2$.\n\nSuppose $B=2$: Then either $C$ or $D$ must be $3$. If $C=3$, we have $3!=6$ possibilities to fill the bottom row. If $D=3$, then $C=4$ and we have $2!=2$ possibilities to fill $E$ and $F$. Altogether there are $6+2=8$ possibilities in this case.\n\nSuppose $C=2$: then, similarly, there are $8$ possibilities here.\n\nAltogether there are $8+8=16$ possibilities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\int_{0}^{\\pi / 3} x \\tan^{2}(x) \\, dx\n$$",
    "options": [],
    "answer": "-π^2/18 + π√3/3 - ln(2)",
    "solution": "Solution:\n$$\n\\begin{aligned}\n\\int_{0}^{\\pi / 3} x \\tan^{2}(x) \\, dx & = \\int_{0}^{\\pi / 3} x\\left(-1+\\frac{1}{\\cos^{2}(x)}\\right) dx \\\\\n& = -\\left.\\frac{x^{2}}{2}\\right|_{0}^{\\pi / 3} + \\int_{0}^{\\pi / 3} \\frac{x \\, dx}{\\cos^{2}(x)} \\\\\n& = -\\left.\\frac{x^{2}}{2}\\right|_{0}^{\\pi / 3} + \\left(\\left. x \\tan(x) \\right|_{0}^{\\pi / 3} - \\int_{0}^{\\pi / 3} \\tan(x) dx \\right) \\quad (u = x; \\ dv = \\frac{dx}{\\cos^{2}(x)}) \\\\\n& = -\\frac{x^{2}}{2} + x \\tan(x) + \\left. \\ln |\\cos(x)| \\right|_{0}^{\\pi / 3} = -\\frac{\\pi^{2}}{18} + \\frac{\\pi \\sqrt{3}}{3} - \\ln(2)\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSolve for all complex numbers $z$ such that $z^{4} + 4z^{2} + 6 = z$.",
    "options": [],
    "answer": "z = (1 ± i√7)/2, z = (-1 ± i√11)/2",
    "solution": "Solution:\n\nRewrite the given equation as $\\left(z^{2} + 2\\right)^{2} + 2 = z$. Observe that a solution to $z^{2} + 2 = z$ is a solution of the quartic by substitution of the left hand side into itself. This gives $z = \\frac{1 \\pm i \\sqrt{7}}{2}$. But now, we know that $z^{2} - z + 2$ divides into $\\left(z^{2} + 2\\right)^{2} - z + 2 = z^{4} + 4z^{2} - z + 6$. Factoring it out, we obtain $\\left(z^{2} - z + 2\\right)\\left(z^{2} + z + 3\\right) = z^{4} + 4z^{2} - z + 6$. Finally, the second term leads to the solutions $z = \\frac{-1 \\pm i \\sqrt{11}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow a way to construct an equiangular hexagon with side lengths $1,2,3,4,5$, and $6$ (not necessarily in that order).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe trick is to view an equiangular hexagon as an equilateral triangle with its corners cut off. Consider an equilateral triangle with side length $9$, and cut off equilateral triangles of side length $1$, $2$, and $3$ from its corners. This yields an equiangular hexagon with sides of length $1,6,2,4,3,5$ in that order.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTake a unit sphere $S$, i.e., a sphere with radius $1$. Circumscribe a cube $C$ about $S$, and inscribe a cube $D$ in $S$, so that every edge of cube $C$ is parallel to some edge of cube $D$. What is the shortest possible distance from a point on a face of $C$ to a point on a face of $D$?",
    "options": [],
    "answer": "1 - 1/√3",
    "solution": "Solution:\n\nUsing the Pythagorean theorem, we know that the length of a diagonal of a cube of edge length $s$ is $s \\sqrt{3}$. Since $D$ is inscribed in a sphere that has diameter $2$, this means that its side length is $2 / \\sqrt{3}$.\n\nThe distance from a face of $D$ to a face of $C$ will be the distance between them along any line perpendicular to both of them; take such a line passing through the center of $S$. The distance from the center to any face of $D$ along this line will be half the side length of $D$, or $1 / \\sqrt{3}$. The distance from the center to the edge of $C$ is the radius of $S$, which is $1$. Therefore the desired distance is $1 - 1 / \\sqrt{3} = 1 - \\frac{\\sqrt{3}}{3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute the sum of all real numbers $x$ such that\n$$\n2 x^{6}-3 x^{5}+3 x^{4}+x^{3}-3 x^{2}+3 x-1=0\n$$",
    "options": [],
    "answer": "-1/2",
    "solution": "Solution:\nThe carefully worded problem statement suggests that repeated roots might be involved (not to be double counted), as well as complex roots (not to be counted). Let $P(x)=2 x^{6}-3 x^{5}+3 x^{4}+x^{3}-3 x^{2}+3 x-1$. Now, $a$ is a double root of the polynomial $P(x)$ if and only if $P(a)=P^{\\prime}(a)=0$. Hence, we consider the system\n$$\n\\begin{array}{r}\nP(a)=2 a^{6}-3 a^{5}+3 a^{4}+a^{3}-3 a^{2}+3 a-1=0 \\\\\nP^{\\prime}(a)=12 a^{5}-15 a^{4}+12 a^{3}+3 a^{2}-6 a+3=0 \\\\\n\\Longrightarrow 3 a^{4}+8 a^{3}-15 a^{2}+18 a-7=0 \\\\\n37 a^{3}-57 a^{2}+57 a-20=0 \\\\\na^{2}-a+1=0\n\\end{array}\n$$\nWe have used polynomial long division to deduce that any double root must be a root of $a^{2}-a+1$! With this information, we can see that $P(x)=\\left(x^{2}-x+1\\right)^{2}\\left(2 x^{2}+x-1\\right)$. The real roots are easily computed via the quadratic formula, leading to an answer of $-\\frac{1}{2}$. In fact the repeated roots were complex.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFour spheres, each of radius $r$, lie inside a regular tetrahedron with side length $1$ such that each sphere is tangent to three faces of the tetrahedron and to the other three spheres. Find $r$.",
    "options": [],
    "answer": "(sqrt(6) - 1)/10",
    "solution": "Solution:\n\nLet $O$ be the center of the sphere that is tangent to the faces $ABC$, $ABD$, and $BCD$. Let $P, Q$ be the feet of the perpendiculars from $O$ to $ABC$ and $ABD$ respectively. Let $R$ be the foot of the perpendicular from $P$ to $AB$. Then, $OPRQ$ is a quadrilateral such that $\\angle P$, $\\angle Q$ are right angles and $OP = OQ = r$. Also, $\\angle R$ is the dihedral angle between faces $ABC$ and $ABD$, so $\\cos \\angle R = 1/3$. We can then compute $QR = \\sqrt{2} r$, so $BR = \\sqrt{6} r$. Hence, $1 = AB = 2(\\sqrt{6} r) + 2r = 2r(\\sqrt{6} + 1)$, so $r = (\\sqrt{6} - 1)/10$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 47263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEight celebrities meet at a party. It so happens that each celebrity shakes hands with exactly two others. A fan makes a list of all unordered pairs of celebrities who shook hands with each other. If order does not matter, how many different lists are possible?",
    "options": [],
    "answer": "3507",
    "solution": "Solution:\n\nLet the celebrities get into one or more circles so that each circle has at least three celebrities, and each celebrity shook hands precisely with his or her neighbors in the circle.\n\nLet's consider the possible circle sizes:\n\n- There's one big circle with all $8$ celebrities. Depending on the ordering of the people in the circle, the fan's list can still vary. Literally speaking, there are $7!$ different circles $8$ people can make: fix one of the people, and then there are $7$ choices for the person to the right, $6$ for the person after that, and so on. But this would be double-counting because, as far as the fan's list goes, it makes no difference if we \"reverse\" the order of all the people. Thus, there are $7!/2 = 2520$ different possible lists here.\n\n- $5+3$. In this case there are $\\binom{8}{5}$ ways to split into the two circles, $\\frac{4!}{2}$ essentially different ways of ordering the $5$-circle, and $\\frac{2!}{2}$ ways for the $3$-circle, giving a total count of $56 \\cdot 12 \\cdot 1 = 672$.\n\n- $4+4$. In this case there are $\\binom{8}{4} / 2 = 35$ ways to split into the two circles (we divide by $2$ because here, unlike in the $5+3$ case, it does not matter which circle is which), and $\\frac{3!}{2} = 3$ ways of ordering each, giving a total count of $35 \\cdot 3 \\cdot 3 = 315$.\n\nAdding them up, we get $2520 + 672 + 315 = 3507$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute the value of the infinite series\n$$\n\\sum_{n=2}^{\\infty} \\frac{n^{4}+3 n^{2}+10 n+10}{2^{n} \\cdot\\left(n^{4}+4\\right)}\n$$",
    "options": [],
    "answer": "11/10",
    "solution": "Solution:\nWe employ the difference of squares identity, uncovering the factorization of the denominator: $n^{4}+4=\\left(n^{2}+2\\right)^{2}-(2 n)^{2}=\\left(n^{2}-2 n+2\\right)\\left(n^{2}+2 n+2\\right)$. Now,\n$$\n\\begin{aligned}\n\\frac{n^{4}+3 n^{2}+10 n+10}{n^{4}+4} & =1+\\frac{3 n^{2}+10 n+6}{n^{4}+4} \\\\\n& =1+\\frac{4}{n^{2}-2 n+2}-\\frac{1}{n^{2}+2 n+2} \\\\\n\\Longrightarrow \\sum_{n=2}^{\\infty} \\frac{n^{4}+3 n^{2}+10 n+10}{2^{n} \\cdot\\left(n^{4}+4\\right)} & =\\sum_{n=2}^{\\infty} \\frac{1}{2^{n}}+\\frac{4}{2^{n} \\cdot\\left(n^{2}-2 n+2\\right)}-\\frac{1}{2^{n} \\cdot\\left(n^{2}+2 n+2\\right)} \\\\\n& =\\frac{1}{2}+\\sum_{n=2}^{\\infty} \\frac{1}{2^{n-2} \\cdot\\left((n-1)^{2}+1\\right)}-\\frac{1}{2^{n} \\cdot\\left((n+1)^{2}+1\\right)}\n\\end{aligned}\n$$\nThe last series telescopes to $\\frac{1}{2}+\\frac{1}{10}$, which leads to an answer of $\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{10}=\\frac{11}{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFour unit circles are centered at the vertices of a unit square, one circle at each vertex. What is the area of the region common to all four circles?",
    "options": [],
    "answer": "pi/3 + 1 - sqrt(3)",
    "solution": "Solution:\n\nThe desired region consists of a small square and four \"circle segments,\" i.e. regions of a circle bounded by a chord and an arc. The side of this small square is just the chord of a unit circle that cuts off an angle of $30^{\\circ}$, and the circle segments are bounded by that chord and the circle. Using the law of cosines (in an isosceles triangle with unit leg length and vertex angle $30^{\\circ}$), we find that the square of the length of the chord is equal to $2-\\sqrt{3}$. We can also compute the area of each circle segment, namely $\\frac{\\pi}{12}-\\frac{1}{2}(1)(1) \\sin 30^{\\circ}=\\frac{\\pi}{12}-\\frac{1}{4}$. Hence, the desired region has area $2-\\sqrt{3}+4\\left(\\frac{\\pi}{12}-\\frac{1}{4}\\right)=\\frac{\\pi}{3}+1-\\sqrt{3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n \\geq 3$ be a positive integer. Prove that given any $n$ angles $0 < \\theta_{1}, \\theta_{2}, \\ldots, \\theta_{n} < 180^{\\circ}$, such that their sum is $180(n-2)$ degrees, there exists a convex $n$-gon having exactly those angles, in that order.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe induct on $n$. The statement holds trivially for $n=3$, as all triangles are convex.\n\nNow, suppose that the statement is true for $n-1$, where $n \\geq 4$. Let $\\theta_{1}, \\theta_{2}, \\ldots, \\theta_{n}$ be $n$ angles less than $180^{\\circ}$ whose sum equals $180(n-2)$ degrees. The statement is clearly true if $n=4$ and $\\theta_{1}=\\theta_{3}=180^{\\circ}-\\theta_{2}=180^{\\circ}-\\theta_{4}$ since we can easily form a parallelogram, so assume otherwise.\n\nI claim that there exist two adjacent angles whose sum is greater than $180^{\\circ}$. Assume otherwise. Then, we have $\\theta_{i}+\\theta_{i+1} \\leq 180$ for $i=1,2, \\ldots, n$, where $\\theta_{n+1}=\\theta_{1}$. Summing these inequalities over all $i$ yields $2 \\cdot 180(n-2) \\leq 180 n$, which is equivalent to $n \\leq 4$. Of course, we can have $n=4$ if and only if we have equality in each of the above inequalities, forcing us to have a parallelogram contrary to our assumption.\n\nHence, we have two adjacent angles with sum greater than $180^{\\circ}$. Without loss of generality, let these angles be $\\theta_{n-1}$ and $\\theta_{n}$, relabeling if necessary. By the inductive hypothesis, we may construct an $(n-1)$-gon with angles $\\theta_{1}, \\theta_{2}, \\ldots, \\theta_{n-2}, \\theta_{n-1}+\\theta_{n}-180^{\\circ}$, as these angles are each less than $180^{\\circ}$ and their sum equals $180(n-3)$ degrees. Consider the vertex with angle $\\theta_{n-1}+\\theta_{n}-180^{\\circ}$. Note that we can \"clip off\" a triangle with angles $\\theta_{n-1}+\\theta_{n}-180^{\\circ}, 180^{\\circ}-\\theta_{n-1}$, and $180^{\\circ}-\\theta_{n}$ at this vertex, yielding an $n$-gon with the desired angles, completing the inductive step.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTim has a working analog 12-hour clock with two hands that run continuously (instead of, say, jumping on the minute). He also has a clock that runs really slow—at half the correct rate, to be exact. At noon one day, both clocks happen to show the exact time. At any given instant, the hands on each clock form an angle between $0^{\\circ}$ and $180^{\\circ}$ inclusive. At how many times during that day are the angles on the two clocks equal?",
    "options": [],
    "answer": "33",
    "solution": "Solution:\n\nA tricky thing about this problem may be that the angles on the two clocks might be reversed and would still count as being the same (for example, both angles could be $90^{\\circ}$, but the hour hand may be ahead of the minute hand on one clock and behind on the other).\n\nLet $x$, $-12 \\leq x < 12$, denote the number of hours since noon. If we take $0^{\\circ}$ to mean upwards to the \"XII\" and count angles clockwise, then the hour and minute hands of the correct clock are at $30x^{\\circ}$ and $360x^{\\circ}$, and those of the slow clock are at $15x^{\\circ}$ and $180x^{\\circ}$. The two angles are thus $330x^{\\circ}$ and $165x^{\\circ}$, of course after removing multiples of $360^{\\circ}$ and possibly flipping sign; we are looking for solutions to\n$$\n330x^{\\circ} \\equiv 165x^{\\circ} \\pmod{360^{\\circ}} \\text{ or } 330x^{\\circ} \\equiv -165x^{\\circ} \\pmod{360^{\\circ}}\n$$\nIn other words,\n$360 \\mid 165x$ or $360 \\mid 495x$.\nOr, better yet,\n$$\n\\frac{165}{360}x = \\frac{11}{24}x \\text{ and/or } \\frac{495}{360}x = \\frac{11}{8}x\n$$\nmust be an integer. Now $x$ is any real number in the range $[-12, 12)$, so $11x/8$ ranges in $[-16.5, 16.5)$, an interval that contains 33 integers. For any value of $x$ such that $11x/24$ is an integer, of course $11x/8 = 3 \\times (11x/24)$ is also an integer, so the answer is just 33.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute $(x-a)(x-b)\\cdots(x-z)$",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Define a sequence by setting $a_1 = n$ and, for each $k > 1$, letting $a_k$ be the unique integer in the range $0 \\le a_k \\le k - 1$ for which $a_1 + a_2 + \\cdots + a_k$ is divisible by $k$. For instance, when $n = 9$ the obtained sequence is $9, 1, 2, 0, 3, 3, 3, \\dots$. Prove that for any $n$ the sequence $a_1, a_2, a_3, \\dots$ eventually becomes constant.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $k \\ge 1$, let\n$$\ns_k = a_1 + a_2 + \\cdots + a_k.\n$$\nWe have\n$$\n\\frac{s_{k+1}}{k+1} < \\frac{s_{k+1}}{k} = \\frac{s_k + a_{k+1}}{k} \\le \\frac{s_k + k}{k} = \\frac{s_k}{k} + 1.\n$$\nOn the other hand, for each $k$, $s_k/k$ is a positive integer. Therefore\n$$\n\\frac{s_{k+1}}{k+1} \\le \\frac{s_k}{k},\n$$\nand the sequence of quotients $s_k/k$ is eventually constant. If $s_{k+1}/(k+1) = s_k/k$, then\n$$\na_{k+1} = s_{k+1} - s_k = \\frac{(k+1)s_k}{k} - s_k = \\frac{s_k}{k},\n$$\nshowing that the sequence $a_k$ is eventually constant as well.\nFor $k \\ge 1$, let\n$$\ns_k = a_1 + a_2 + \\cdots + a_k \\quad \\text{and} \\quad \\frac{s_k}{k} = q_k.\n$$\nSince $a_k \\le k - 1$, for $k \\ge 2$, we have\n$$\ns_k = a_1 + a_2 + a_3 + \\cdots + a_k \\le n + 1 + 2 + \\cdots + (k-1) = n + \\frac{k(k-1)}{2}.\n$$\nLet $m$ be a positive integer such that $n \\le \\frac{m(m+1)}{2}$ (such an integer clearly exists). Then\n$$\nq_m = \\frac{s_m}{m} \\le \\frac{n}{m} + \\frac{m-1}{2} \\le \\frac{m+1}{2} + \\frac{m-1}{2} = m.\n$$\nWe claim that\n$$\nq_m = a_{m+1} = a_{m+2} = a_{m+3} = a_{m+4} = \\dots\n$$\nThis follows from the fact that the sequence $a_1, a_2, a_3, \\dots$ is uniquely determined and choosing $a_{m+i} = q_m$, for $i \\ge 1$, satisfies the range condition\n$$\n0 \\le a_{m+i} = q_m \\le m \\le m+i-1,\n$$\nand yields\n$$\ns_{m+i} = s_m + iq_m = mq_m + iq_m = (m+i)q_m.\n$$\nFor $k \\ge 1$, let\n$$\ns_k = a_1 + a_2 + \\cdots + a_k.\n$$\nWe claim that for some $m$ we have $s_m = m(m-1)$. To this end, consider the sequence which computes the differences between $s_k$ and $k(k-1)$, i.e., whose $k$-th term is $s_k - k(k-1)$. Note that the first term of this sequence is positive (it is equal to $n$) and that its terms are strictly decreasing since\n$$\n(s_k - k(k-1)) - (s_{k+1} - (k+1)k) = 2k - a_{k+1} \\ge 2k - k = k \\ge 1.\n$$\nFurther, a negative term cannot immediately follow a positive term. Suppose otherwise, namely that $s_k > k(k-1)$ and $s_{k+1} < (k+1)k$. Since $s_k$ and $s_{k+1}$ are divisible by $k$ and $k+1$, respectively, we can tighten the above inequalities to $s_k \\ge k^2$ and $s_{k+1} \\le (k+1)(k-1) = k^2 - 1$. But this would imply that $s_k > s_{k+1}$, a contradiction. We conclude that the sequence of differences must eventually include a term equal to zero.\nLet $m$ be a positive integer such that $s_m = m(m-1)$. We claim that\n$$\nm - 1 = a_{m+1} = a_{m+2} = a_{m+3} = a_{m+4} = \\dots\n$$\nThis follows from the fact that the sequence $a_1, a_2, a_3, \\dots$ is uniquely determined and choosing $a_{m+i} = m-1$, for $i \\ge 1$, satisfies the range condition\n$$\n0 \\le a_{m+i} = m-1 \\le m+i-1,\n$$\nand yields\n$$\ns_{m+i} = s_m + i(m-1) = m(m-1) + i(m-1) = (m+i)(m-1).\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA 15-inch-long stick has four marks on it, dividing it into five segments of length $1$, $2$, $3$, $4$, and $5$ inches (although not necessarily in that order) to make a \"ruler.\" Here is an example.\n\n![](attached_image_1.png)\n\nUsing this ruler, you could measure $8$ inches (between the marks $B$ and $D$) and $11$ inches (between the end of the ruler at $A$ and the mark at $E$), but there's no way you could measure $12$ inches.\n\nProve that it is impossible to place the four marks on the stick such that the five segments have length $1$, $2$, $3$, $4$, and $5$ inches, and such that every integer distance from $1$ inch through $15$ inches could be measured.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIn order to measure $14$ inches, one mark must be $1$ inch from an end of the ruler. Likewise, in order to measure $13$ inches, there must be another mark that is $2$ inches from an end of the ruler. Without loss of generality, suppose the leftmost mark is $1$ inch from the end, and the rightmost mark is $2$ inches from the other end.\n\nNext, we observe that the second mark from the left must be $5$ inches from the first, or else it would be impossible to measure $6$ inches.\n\nAt this point, there are only two cases to consider: either the distances between marks are, in order,\n$$\n1, 5, 4, 3, 2\n$$\nor the distances are\n$$\n1, 5, 3, 4, 2\n$$\nIn the first case, we cannot measure $8$ inches, and in the second case, we cannot measure $10$ inches. We conclude that it is impossible.\nSolution:\n\nIn order to make a measurement, you must choose a pair of marks, where the two endpoints are included. Since there are $6$ marks, that means that we can measure at most $15$ different segments ($6 \\times 5 / 2$). Therefore a ruler that can measure every length from $1$'' through $15$'' must have exactly one way of measuring each length.\n\nIf the $1$'' segment is next to the $2$'', $3$'', or $4$'' segments, then that would make a second way of measuring $3$'', $4$'', or $5$''. Therefore, the $1$'' segment must be next to the $5$'' segment only: the $1$'' is on the end, next to the $5$''. Now, if the $2$'' segment is next to the $3$'' or $4$'' segments, then that would make a second way of measuring $5$'' or $6$''. So the $2$'' segment is also next to the $5$'' segment only. But that leaves nowhere to put the $3$'' and $4$'' segments.\nSolution:\n\nLet $a, b, c, d, e$ be the segments in order (equalling $1, 2, 3, 4, 5$, but not necessarily in that order). As above, there are only $15$ possible segments that can be measured. The sum of all $15$ segments (in inches) is\n$$\n1 + 2 + \\cdots + 15 = 15 \\times 16 / 2.\n$$\nBut this is also equal to $5a + 8b + 9c + 8d + 5e$.\nHence\n$$\n5a + 8b + 9c + 8d + 5e = 15 \\times 16 / 2 = 15 \\times 8.\n$$\nNow note that $15 = a + b + c + d + e$, so\n$$\n5a + 8b + 9c + 8d + 5e = 8a + 8b + 8c + 8d + 8e.\n$$\nSubtract $5a + 8b + 8c + 8d + 5e$ from both sides to get $c = 3a + 3e$.\nBut $a + e$ is at least $3$ and $c$ is at most $5$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate the functions $\\phi(n)$, $\\sigma(n)$, and $\\tau(n)$ for $n=12$, $n=2007$, and $n=2^{2007}$.",
    "options": [],
    "answer": "For 12: φ=4, σ=28, τ=6. For 2007: φ=1332, σ=2912, τ=6. For 2^2007: φ=2^2006, σ=2^2008−1, τ=2008.",
    "solution": "Solution:\n\nFor $n=12=2^{2} \\cdot 3^{1}$,\n$$\n\\phi(12)=2(2-1)(3-1)=4, \\quad \\sigma(12)=(1+2+4)(1+3)=28, \\quad \\tau(12)=(2+1)(1+1)=6\n$$\n\nFor $n=2007=3^{2} \\cdot 223$,\n$$\n\\phi(2007)=3(3-1)(223-1)=1332, \\quad \\sigma(2007)=(1+3+9)(1+223)=2912, \\quad \\tau(2007)=(2+1)(1+1)=6\n$$\n\nFor $n=2^{2007}$,\n$$\n\\phi\\left(2^{2007}\\right)=2^{2006}, \\quad \\sigma\\left(2^{2007}\\right)=\\left(1+2+\\cdots+2^{2007}\\right)=2^{2008}-1, \\quad \\tau\\left(2^{2007}\\right)=2007+1=2008\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the sum of the positive integer divisors of $2^{2007}$.",
    "options": [],
    "answer": "2^{2008} - 1",
    "solution": "Solution:\n\n$2^{2007}$ has divisors $1, 2, 2^2, \\ldots, 2^{2007}$. The sum is\n$$\n1 + 2 + 2^2 + \\cdots + 2^{2007} = 2^{2008} - 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute\n$$\n\\left\\lfloor\\frac{2007!+2004!}{2006!+2005!}\\right\\rfloor\n$$\n(Note that $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$.)",
    "options": [],
    "answer": "2006",
    "solution": "Solution:\n\nAnswer: $2006$. We have\n$$\n\\left\\lfloor\\frac{2007!+2004!}{2006!+2005!}\\right\\rfloor = \\left\\lfloor\\frac{\\left(2007 \\cdot 2006 + \\frac{1}{2005}\\right) \\cdot 2005!}{(2006+1) \\cdot 2005!}\\right\\rfloor = \\left\\lfloor\\frac{2007 \\cdot 2006 + \\frac{1}{2005}}{2007}\\right\\rfloor = \\left\\lfloor 2006 + \\frac{1}{2005 \\cdot 2007}\\right\\rfloor\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMichael has $16$ white socks, $3$ blue socks, and $6$ red socks in a drawer. Ever the lazy college student, he has overslept and is late for his favorite team's season-opener. Because he is now in such a rush to get from Harvard to Foxborough, he randomly takes socks from the drawer (one at a time) until he has a pair of the same color. What is the largest number of socks he could possibly withdraw in this fashion?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nAnswer: $4$. It is possible for him to begin with three socks of different colors, but an instance of the Pigeon Hole Principle is that among any four objects of three types some two are the same type.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cube of edge length $s > 0$ has the property that its surface area is equal to the sum of its volume and five times its edge length. Compute all possible values of $s$.",
    "options": [],
    "answer": "1, 5",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute:\n$$\n\\lim_{x \\rightarrow 0} \\frac{x^{2}}{1-\\cos(x)}\n$$",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nSince $\\sin^{2}(x) = 1 - \\cos^{2}(x)$, we multiply the numerator and denominator by $1 + \\cos(x)$ and use the fact that $x / \\sin(x) \\rightarrow 1$, obtaining\n$$\n\\lim_{x \\rightarrow 0} \\frac{x^{2}}{1-\\cos(x)} = \\lim_{x \\rightarrow 0} \\frac{x^{2}(1+\\cos(x))}{1-\\cos^{2}(x)} = \\lim_{x \\rightarrow 0}\\left(\\frac{x}{\\sin(x)}\\right)^{2} \\cdot 2 = 2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cube of edge length $s > 0$ has the property that its surface area is equal to the sum of its volume and five times its edge length. Compute all possible values of $s$.",
    "options": [],
    "answer": "1 and 5",
    "solution": "Solution:\nThe volume of the cube is $s^{3}$ and its surface area is $6s^{2}$, so we have\n$$\n6s^{2} = s^{3} + 5s\n$$\nor\n$$\n0 = s^{3} - 6s^{2} + 5s = s(s-1)(s-5)\n$$\nThus, the possible values of $s$ are $1$ and $5$ (since $s > 0$).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine the sequence of positive integers $a_{n}$ recursively by $a_{1}=7$ and $a_{n}=7^{a_{n-1}}$ for all $n \\geq 2$. Determine the last two digits of $a_{2007}$.",
    "options": [],
    "answer": "43",
    "solution": "Solution:\n\nAnswer: 43. Note that the last two digits of $7^{4}$ are 01. Also, $a_{2006}=7^{a_{2005}}=(-1)^{a_{2005}}=-1=3$ $(\\bmod 4)$ since $a_{2005}$ is odd. Therefore, $a_{2007}=7^{a_{2006}}=7^{3}=43\\ (\\bmod\\ 100)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA positive real number $x$ is such that\n$$\n\\sqrt[3]{1-x^{3}}+\\sqrt[3]{1+x^{3}}=1\n$$\nFind $x^{2}$.",
    "options": [],
    "answer": "∛28 / 3",
    "solution": "Solution:\nAnswer: $\\frac{\\sqrt[3]{28}}{3}$. Cubing the given equation yields\n$$\n1=\\left(1-x^{3}\\right)+3 \\sqrt[3]{\\left(1-x^{3}\\right)\\left(1+x^{3}\\right)}\\left(\\sqrt[3]{1-x^{3}}+\\sqrt[3]{1+x^{3}}\\right)+\\left(1+x^{3}\\right)=2+3 \\sqrt[3]{1-x^{6}}\n$$\nThen $\\frac{-1}{3}=\\sqrt[3]{1-x^{6}}$, so $\\frac{-1}{27}=1-x^{6}$ and $x^{6}=\\frac{28}{27}$ and $x^{2}=\\frac{\\sqrt[3]{28}}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe polynomial $f(x) = x^{2007} + 17 x^{2006} + 1$ has distinct zeroes $r_1, \\ldots, r_{2007}$. A polynomial $P$ of degree $2007$ has the property that $P\\left(r_j + \\frac{1}{r_j}\\right) = 0$ for $j = 1, \\ldots, 2007$. Determine the value of $\\frac{P(1)}{P(-1)}$.",
    "options": [],
    "answer": "289/259",
    "solution": "Solution:\n\nAnswer: $\\mathbf{289}$.\n\nFor some constant $k$, we have\n$$\nP(z) = k \\prod_{j=1}^{2007} \\left(z - \\left(r_j + \\frac{1}{r_j}\\right)\\right)\n$$\nNow writing $\\omega^3 = 1$ with $\\omega \\neq 1$, we have $\\omega^2 + \\omega = -1$. Then\n$$\n\\begin{gathered}\n\\frac{P(1)}{P(-1)} = \\frac{k \\prod_{j=1}^{2007} \\left(1 - \\left(r_j + \\frac{1}{r_j}\\right)\\right)}{k \\prod_{j=1}^{2007} \\left(-1 - \\left(r_j + \\frac{1}{r_j}\\right)\\right)} = \\prod_{j=1}^{2007} \\frac{r_j^2 - r_j + 1}{r_j^2 + r_j + 1} = \\prod_{j=1}^{2007} \\frac{\\left(-\\omega - r_j\\right)\\left(-\\omega^2 - r_j\\right)}{\\left(\\omega - r_j\\right)\\left(\\omega^2 - r_j\\right)} \\\\\n= \\frac{f(-\\omega) f\\left(-\\omega^2\\right)}{f(\\omega) f\\left(\\omega^2\\right)} = \\frac{\\left(-\\omega^{2007} + 17 \\omega^{2006} + 1\\right)\\left(-\\left(\\omega^2\\right)^{2007} + 17\\left(\\omega^2\\right)^{2006} + 1\\right)}{\\left(\\omega^{2007} + 17 \\omega^{2006} + 1\\right)\\left(\\left(\\omega^2\\right)^{2007} + 17\\left(\\omega^2\\right)^{2006} + 1\\right)} = \\frac{\\left(17 \\omega^2\\right)(17 \\omega)}{\\left(2 + 17 \\omega^2\\right)(2 + 17 \\omega\\right)} \\\\\n= \\frac{289}{4 + 34\\left(\\omega + \\omega^2\\right) + 289} = \\frac{289}{259}.\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$\\alpha_{1}, \\alpha_{2}, \\alpha_{3}$, and $\\alpha_{4}$ are the complex roots of the equation $x^{4}+2 x^{3}+2=0$. Determine the unordered set\n$$\n\\left\\{\\alpha_{1} \\alpha_{2}+\\alpha_{3} \\alpha_{4}, \\alpha_{1} \\alpha_{3}+\\alpha_{2} \\alpha_{4}, \\alpha_{1} \\alpha_{4}+\\alpha_{2} \\alpha_{3}\\right\\}\n$$",
    "options": [],
    "answer": "{-2, 1 - sqrt(5), 1 + sqrt(5)}",
    "solution": "Solution:\n\n$\\{1 \\pm \\sqrt{5},\\ -2\\}$. Same as Algebra $\\# 9$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA subset $S$ of the nonnegative integers is called supported if it contains $0$, and $k+8, k+9 \\in S$ for all $k \\in S$. How many supported sets are there?",
    "options": [],
    "answer": "1430",
    "solution": "Solution:\nAnswer: $1430$. Note that every supported set $S$ contains $0,8,9,16,17,18,24$-$27,32$-$36,40$-$45,48$-$54$, and all $n \\geq 55$. Now define $\\bar{S} := \\mathbb{Z}^{+} \\setminus S$, which is a subset of $\\{1$-$7,10$-$15,19$-$23,28$-31,37,38,39,46,47,55\\}$ satisfying the opposite property that $k \\in \\bar{S} \\Longrightarrow k-8, k-9 \\in \\bar{S}$.\n\n![](attached_image_1.png)\n\nConsider the above arrangement after removing the numbers not in $\\bar{S}$. The condition that $S$ be supported ensures that sets $\\bar{S}$ are in bijective correspondence with paths from $(0,0)$ to $(16,0)$ consisting of discrete steps of $\\langle 1,1\\rangle$ and $\\langle 1,-1\\rangle$ and lying above the $x$-axis: from the modified version of the above diagram, a unique path passes through the top items left in each column. The number of such paths is the 8th Catalan number, so the answer is $C_{8} = \\frac{1}{8+1} \\binom{16}{8} = \\frac{12870}{9} = 1430$. (Incidentally, $16$ choose $8$ was computed in an earlier problem.) Without the explicit formula for Catalan numbers, the answer can be computed recursively by filling in the number of ways a path can reach $(16,0)$ from each position in the figure. One works right to left, obtaining the following:\n\n|   |   |   |   |   |   |   |   | 1 |   |   |   |   |   |   |   |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|   |   |   |   |   |   |   | 8 |   | 1 |   |   |   |   |   |   |\n|   |   |   |   |   |   | 35 |   | 7 |   | 1 |   |   |   |   |   |\n|   |   |   |   |   | 110 |   | 27 |   | 6 |   | 1 |   |   |   |   |\n|   |   |   |   | 275 |   | 75 |   | 20 |   | 5 |   | 1 |   |   |   |\n|   |   |   | 572 |   | 165 |   | 48 |   | 14 |   | 4 |   | 1 |   |   |\n|   |   | 1001 |   | 297 |   | 90 |   | 28 |   | 9 |   | 3 |   | 1 |   |\n|   | 1430 |   | 429 |   | 132 |   | 42 |   | 14 |   | 5 |   | 2 |   | 1 |\n| 1430 |   | 429 |   | 132 |   | 42 |   | 14 |   | 5 |   | 2 |   | 1 |   |\n\nOne can exploit symmetry and, having determined the middle column, sum the squares: $1^{2} + 7^{2} + 20^{2} + 28^{2} + 14^{2} = 1430$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all pairs $(n, k)$ of positive integers such that\n$$\nsigma(n) \\phi(n) = \\frac{n^{2}}{k}\n$$",
    "options": [],
    "answer": "(1, 1)",
    "solution": "Solution:\nAnswer: $(1, 1)$.\n\nIt is clear that for a given integer $n$, there is at most one integer $k$ for which the equation holds. For $n=1$ this is $k=1$. But, for $n>1$, problem 1 asserts that $\\sigma(n) \\phi(n) \\leq n^{2}-1 < n^{2}$, so that $k \\geq 2$. We now claim that $2 > \\frac{n^{2}}{\\sigma(n) \\phi(n)}$. Write $n = p_{1}^{e_{1}} \\cdots p_{k}^{e_{k}}$, where the $p_{i}$ are distinct primes and $e_{i} \\geq 1$ for all $i$, and let $q_{1} < q_{2} < \\cdots$ be the primes in ascending order. Then\n$$\n\\begin{aligned}\n& \\frac{n^{2}}{\\sigma(n) \\phi(n)} = \\prod_{i=1}^{k} \\frac{p_{i}^{2 e_{i}}}{\\frac{p_{i}^{e_{i}+1}-1}{p_{i}-1} \\cdot (p_{i}-1) p_{i}^{e_{i}-1}} = \\prod_{i=1}^{k} \\frac{p_{i}^{2 e_{i}}}{p_{i}^{2 e_{i}} - p_{i}^{e_{i}-1}} \\\\\n& \\quad = \\prod_{i=1}^{k} \\frac{1}{1 - p_{i}^{-1-e_{i}}} \\leq \\prod_{i=1}^{k} \\frac{1}{1 - p_{i}^{-2}} < \\prod_{i=1}^{\\infty} \\frac{1}{1 - q_{i}^{-2}} \\\\\n& \\quad = \\prod_{i=1}^{\\infty} \\left( \\sum_{j=0}^{\\infty} \\frac{1}{q_{i}^{2j}} \\right ) = \\sum_{n=1}^{\\infty} \\frac{1}{n^{2}} \\\\\n& \\quad < 1 + \\sum_{n=2}^{\\infty} \\frac{1}{n^{2}-1} = 1 + \\frac{1}{2} \\left( \\left( \\frac{1}{1} - \\frac{1}{3} \\right ) + \\left( \\frac{1}{2} - \\frac{1}{4} \\right ) + \\left( \\frac{1}{3} - \\frac{1}{5} \\right ) + \\cdots \\right ) = \\frac{7}{4} < 2.\n\\end{aligned}\n$$\nIt follows that there can be no solutions to $k = \\frac{n^{2}}{\\sigma(n) \\phi(n)}$ other than $n = k = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is a convex quadrilateral such that $AB = 2$, $BC = 3$, $CD = 7$, and $AD = 6$. It also has an incircle. Given that $\\angle ABC$ is right, determine the radius of this incircle.",
    "options": [],
    "answer": "(1 + sqrt(13))/3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\int_{0}^{\\infty} \\frac{e^{-x} \\sin (x)}{x} d x\n$$",
    "options": [],
    "answer": "pi/4",
    "solution": "Solution:\nAnswer: $\\frac{\\pi}{4}$. We can compute the integral by introducing a parameter and exchanging the order of integration:\n$$\n\\begin{aligned}\n\\int_{0}^{\\infty} e^{-x}\\left(\\frac{\\sin (x)}{x}\\right) \\mathrm{d} x & =\\int_{0}^{\\infty} e^{-x}\\left(\\int_{0}^{1} \\cos (a x) \\mathrm{d} a\\right) \\mathrm{d} x=\\int_{0}^{1}\\left(\\int_{0}^{\\infty} e^{-x} \\cos (a x) \\mathrm{d} x\\right) \\mathrm{d} a \\\\\n& =\\int_{0}^{1} \\operatorname{Re}\\left[\\int_{0}^{\\infty} e^{(-1+a i) x} \\mathrm{~d} x\\right] \\mathrm{d} a=\\int_{0}^{1} \\operatorname{Re}\\left[\\left.\\frac{e^{(-1+a i) x}}{-1+a i}\\right|_{x=0} ^{\\infty}\\right] \\mathrm{d} a \\\\\n& =\\int_{0}^{1} \\operatorname{Re}\\left[\\frac{1}{1-a i}\\right] \\mathrm{d} a=\\int_{0}^{1} \\operatorname{Re}\\left[\\frac{1+a i}{1+a^{2}}\\right] \\mathrm{d} a \\\\\n& =\\int_{0}^{1} \\frac{1}{1+a^{2}} \\mathrm{~d} a=\\left.\\tan ^{-1}(a)\\right|_{a=0} ^{1}=\\frac{\\pi}{4}\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a cyclic quadrilateral, and let $P$ be the intersection of its two diagonals. Points $R$, $S$, $T$, and $U$ are feet of the perpendiculars from $P$ to sides $AB$, $BC$, $CD$, and $AD$, respectively. Show that quadrilateral $RSTU$ is bicentric if and only if $AC \\perp BD$. (Note that a quadrilateral is called inscriptible if it has an incircle; a quadrilateral is called bicentric if it is both cyclic and inscriptible.)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst we show that $RSTU$ is always inscriptible. Note that in addition to $ABCD$, we have cyclic quadrilaterals $ARPU$ and $BSPR$. Thus,\n$$\n\\angle PRU = \\angle PAU = \\angle CAD = \\angle CBD = \\angle SBP = \\angle SRP\n$$\nand it follows that $P$ lies on the bisector of $\\angle SRU$. Analogously, $P$ lies on the bisectors of $\\angle TSR$ and $\\angle UTS$, so is equidistant from lines $UR$, $RS$, $ST$, and $TU$, and $RSTU$ is inscriptible having incenter $P$.\n\nNow we show that $RSTU$ is cyclic if and only if the diagonals of $ABCD$ are orthogonal. We have\n$$\n\\begin{aligned}\n& \\angle APB = \\pi - \\angle BAP - \\angle PBA = \\pi - \\angle RAP - \\angle PBR = \\pi - \\angle RUP - \\angle PSR \\\\\n& \\quad = \\pi - \\frac{1}{2}(\\angle RUT + \\angle TSR)\n\\end{aligned}\n$$\nIt follows that $\\angle APB = \\frac{\\pi}{2}$ if and only if $\\angle RUT + \\angle TSR = \\pi$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all nonconstant polynomials $P(x)$, with real coefficients and having only real zeros, such that $P(x+1) P\\left(x^{2}-x+1\\right)=P\\left(x^{3}+1\\right)$ for all real numbers $x$.",
    "options": [],
    "answer": "P(x) = x^k for any positive integer k",
    "solution": "Solution:\nAnswer: $\\left\\{P(x)=x^{k} \\mid k \\in \\mathbb{Z}^{+}\\right\\}$.\n\nNote that if $P(\\alpha)=0$, then by setting $x=\\alpha-1$ in the given equation, we find $0=P\\left(x^{3}+1\\right)=P\\left(\\alpha^{3}-3 \\alpha^{2}+3 \\alpha\\right)$. Because $P$ is nonconstant, it has at least one zero. Because $P$ has finite degree, there exist minimal and maximal roots of $P$. Writing $\\alpha^{3}-3 \\alpha^{2}+3 \\alpha \\geq \\alpha \\Longleftrightarrow \\alpha(\\alpha-1)(\\alpha-2) \\geq 0$, we see that the largest zero of $P$ cannot exceed $2$. Likewise, the smallest zero cannot be negative, so all of the zeroes of $P$ lie in $[0,2]$. Moreover, if $\\alpha \\notin\\{0,1,2\\}$ is a zero of $P$, then $\\alpha' = \\alpha^{3}-3 \\alpha^{2}+3 \\alpha$ is another zero of $P$ that lies strictly between $\\alpha$ and $1$. Because $P$ has only finitely many zeroes, all of its zeroes must lie in $\\{0,1,2\\}$.\n\nNow write $P(x)=k x^{p}(x-1)^{q}(x-2)^{r}$ for nonnegative integers $p, q$ and $r$ having a positive sum. The given equation becomes\n$$\n\\begin{aligned}\n& k^{2}(x+1)^{p} x^{q}(x-1)^{r}\\left(x^{2}-x+1\\right)^{p}\\left(x^{2}-x\\right)^{q}\\left(x^{2}-x-1\\right)^{r} = P(x+1) P\\left(x^{2}-x+1\\right) \\\\\n& \\quad = P\\left(x^{3}+1\\right) = k\\left(x^{3}+1\\right)^{p} x^{3q}\\left(x^{3}-1\\right)^{r}\n\\end{aligned}\n$$\nFor the leading coefficients to agree, we require $k=k^{2}$. Because the leading coefficient is nonzero, $P$ must be monic. In $(*), r$ must be zero lest $P\\left(x^{3}+1\\right)=0$ have complex roots. Then $q$ must be zero as well. For, if $q$ is positive, then $P\\left(x^{2}-x+1\\right)=0$ has $1$ as a root while $P\\left(x^{3}+1\\right)$ does not. Finally, the remaining possibilities are $P(x)=x^{p}$ for $p$ an arbitrary positive integer. It is easily seen that these polynomials are satisfactory.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the largest integer $n$ such that $7^{2048}-1$ is divisible by $2^{n}$.",
    "options": [],
    "answer": "14",
    "solution": "Solution:\nAnswer: 14. We have\n$$\n7^{2048}-1=(7-1)(7+1)\\left(7^{2}+1\\right)\\left(7^{4}+1\\right) \\cdots\\left(7^{1024}+1\\right)\n$$\nIn the expansion, the eleven terms other than $7+1$ are divisible by $2$ exactly once, as can be checked easily with modulo $4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe are given some similar triangles. Their areas are $1^{2}, 3^{2}, 5^{2}, \\ldots$, and $49^{2}$. If the smallest triangle has a perimeter of $4$, what is the sum of all the triangles' perimeters?",
    "options": [],
    "answer": "2500",
    "solution": "Solution:\n\nBecause the triangles are all similar, they all have the same ratio of perimeter squared to area, or, equivalently, the same ratio of perimeter to the square root of area. Because the latter ratio is $4$ for the smallest triangle, it is $4$ for all the triangles, and thus their perimeters are $4 \\cdot 1, 4 \\cdot 3, 4 \\cdot 5, \\ldots, 4 \\cdot 49$, and the sum of these numbers is $4(1+3+5+\\cdots+49)=4\\left(25^{2}\\right)=2500$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind an explicit, closed form formula for\n$$\n\\sum_{k=1}^{n} \\frac{k \\cdot(-1)^{k} \\cdot\\binom{n}{k}}{n+k+1}\n$$",
    "options": [],
    "answer": "-n!(n+1)!/(2n+1)!",
    "solution": "Solution:\nAnswer: $\\frac{-1}{\\binom{2 n+1}{n}}$ or $-\\frac{n!(n+1)!}{(2 n+1)!}$ or obvious equivalent.\n\nConsider the interpolation of the polynomial $P(x)=x \\cdot n!$ at $x=0,1, \\ldots, n$. We obtain the identity\n$$\n\\begin{aligned}\nP(x) & =x \\cdot n!=\\sum_{k=0}^{n} k \\cdot n!\\prod_{j \\neq k} \\frac{x-j}{k-j} \\\\\n& =\\sum_{k=0}^{n} k \\cdot n!\\cdot \\frac{x(x-1) \\cdots(x-k+1)(x-k-1) \\cdots(x-n)}{k!(n-k)!(-1)^{n-k}} \\\\\n& =\\sum_{k=1}^{n} k \\cdot(-1)^{n-k} \\cdot\\binom{n}{k} \\cdot x(x-1) \\cdots(x-k+1)(x-k-1) \\cdots(x-n)\n\\end{aligned}\n$$\nThis identity is valid for all complex numbers $x$, but, to extract a factor $\\frac{1}{n+k+1}$ from the valid product of each summand, we set $x=-n-1$, so that\n\n$-(n+1)!=\\sum_{k=1}^{n} k(-1)^{n-k}\\binom{n}{k}(-n-1) \\cdots(-n-k)(-n-k-2) \\cdots(-2 n-1)=\\sum_{k=1}^{n} \\frac{k(-1)^{k}\\binom{n}{k}(2 n+1)!}{n!(n+k+1)}$.\n\nFinally,\n$$\n\\sum_{k=1}^{n} \\frac{k \\cdot(-1)^{k} \\cdot\\binom{n}{k}}{n+k+1}=\\frac{-n!(n+1)!}{(2 n+1)!}=\\frac{-1}{\\binom{2 n+1}{n}} \\cdot \\square\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoints $A$, $B$, and $C$ lie in that order on line $\\ell$, such that $AB = 3$ and $BC = 2$. Point $H$ is such that $CH$ is perpendicular to $\\ell$. Determine the length $CH$ such that $\\angle AHB$ is as large as possible.",
    "options": [],
    "answer": "sqrt(10)",
    "solution": "Solution:\n\nLet $\\omega$ denote the circumcircle of triangle $ABH$. Since $AB$ is fixed, the smaller the radius of $\\omega$, the bigger the angle $AHB$. If $\\omega$ crosses the line $CH$ in more than one point, then there exists a smaller circle that goes through $A$ and $B$ that crosses $CH$ at a point $H'$. But angle $AH'B$ is greater than $AHB$, contradicting our assumption that $H$ is the optimal spot. Thus the circle $\\omega$ crosses the line $CH$ at exactly one spot: i.e., $\\omega$ is tangent to $CH$ at $H$.\n\nBy Power of a Point, $CH^2 = CA \\cdot CB = 5 \\cdot 2 = 10$, so $CH = \\sqrt{10}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $AB = 7$, $BC = 9$, and $CA = 4$. Let $D$ be the point such that $AB \\parallel CD$ and $CA \\parallel BD$. Let $R$ be a point within triangle $BCD$. Lines $\\ell$ and $m$ going through $R$ are parallel to $CA$ and $AB$ respectively. Line $\\ell$ meets $AB$ and $BC$ at $P$ and $P'$, respectively, and $m$ meets $CA$ and $BC$ at $Q$ and $Q'$, respectively. If $S$ denotes the largest possible sum of the areas of triangles $BPP'$, $RP'Q'$, and $CQQ'$, determine the value of $S^2$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "180",
    "solution": "Solution:\n\nAnswer: $180$. Let $R'$ denote the intersection of the lines through $Q'$ and $P'$ parallel to $\\ell$ and $m$ respectively. Then $[RP'Q'] = [R'P'Q']$. Triangles $BPP'$, $R'P'Q'$, and $CQQ'$ lie in $ABC$ without overlap, so that on the one hand, $S \\leq ABC$. On the other, this bound is realizable by taking $R$ to be a vertex of triangle $BCD$. We compute the square of the area of $ABC$ to be $10 \\cdot (10-9) \\cdot (10-7) \\cdot (10-4) = 180$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConvex quadrilateral $ABCD$ has right angles $\\angle A$ and $\\angle C$ and is such that $AB = BC$ and $AD = CD$. The diagonals $AC$ and $BD$ intersect at point $M$. Points $P$ and $Q$ lie on the circumcircle of triangle $AMB$ and segment $CD$, respectively, such that points $P$, $M$, and $Q$ are collinear. Suppose that $m \\angle ABC = 160^\\circ$ and $m \\angle QMC = 40^\\circ$. Find $MP \\cdot MQ$, given that $MC = 6$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "36",
    "solution": "Solution:\n\nAnswer: $36$. Note that $m \\angle QPB = m \\angle MPB = m \\angle MAB = m \\angle CAB = \\angle BCA = \\angle CDB$. Thus, $MP \\cdot MQ = MB \\cdot MD$. On the other hand, segment $CM$ is an altitude of right triangle $BCD$, so $MB \\cdot MD = MC^{2} = 36$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine $x \\star y = \\frac{\\sqrt{x^{2} + 3 x y + y^{2} - 2 x - 2 y + 4}}{x y + 4}$. Compute\n$$\n((\\cdots((2007 \\star 2006) \\star 2005) \\star \\cdots) \\star 1).\n$$",
    "options": [],
    "answer": "sqrt(15)/9",
    "solution": "Solution:\n\nAnswer: $\\frac{\\sqrt{15}}{9}$.\n\nNote that\n$$\nx \\star 2 = \\frac{\\sqrt{x^{2} + 6x + 4 - 2x - 4 + 4}}{2x + 4} = \\frac{\\sqrt{(x+2)^{2}}}{2(x+2)} = \\frac{1}{2}\n$$\nfor $x > -2$.\n\nBecause $x \\star y > 0$ if $x, y > 0$, we need only compute\n$$\n\\frac{1}{2} \\star 1 = \\frac{\\sqrt{\\frac{1}{4} + \\frac{3}{2} + 1 - 3 + 4}}{\\frac{1}{2} + 4} = \\frac{\\sqrt{15}}{9}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47295,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square grid on the Euclidean plane consists of all points $(m, n)$, where $m$ and $n$ are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least $5$?",
    "options": [],
    "answer": "No",
    "solution": "It is not possible. The proof is by contradiction. Suppose that such a covering family $\\mathcal{F}$ exists. Let $D(P, \\rho)$ denote the disc with center $P$ and radius $\\rho$. Start with an arbitrary disc $D(O, r)$ that does not overlap any member of $\\mathcal{F}$. Then $D(O, r)$ covers no grid point. Take the disc $D(O, r)$ to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then $D(O, r)$ is tangent to at least three discs in $\\mathcal{F}$. Observe that there must be two of the three tangent discs, say $D(A, a)$ and $D(B, b)$, such that $\\angle AOB \\le 120^\\circ$. By the Law of Cosines applied to triangle $ABO$,\n$$\n(a+b)^2 \\le (a+r)^2 + (b+r)^2 + (a+r)(b+r),\n$$\nwhich yields\n$$\nab \\le 3(a+b)r + 3r^2, \\quad \\text{and thus}\\quad 12r^2 \\ge (a-3r)(b-3r).\n$$\nNote that $r < 1/\\sqrt{2}$ because $D(O, r)$ covers no grid point, and $(a-3r)(b-3r) \\ge (5-3r)^2$ because each disc in $\\mathcal{F}$ has radius at least $5$. Hence $2\\sqrt{3}r \\ge (5-3r)$, which gives $5 \\le (3+2\\sqrt{3})r < (3+2\\sqrt{3})/\\sqrt{2}$ and thus $5\\sqrt{2} < 3+2\\sqrt{3}$. Squaring both sides of this inequality yields $50 < 21+12\\sqrt{3} < 21+12 \\cdot 2 = 45$. This contradiction completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe points of the plane are colored in black and white so that whenever three vertices of a parallelogram are the same color, the fourth vertex is that color, too. Prove that all the points of the plane are the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose not. Let $A$ be a white point and $B$ a black point. Their midpoint $C$ is one of the two colors; without loss of generality suppose $C$ is black. Now pick any point $D$ not collinear with $A, B, C$, and construct $E$ so that $C A D E$ is a parallelogram. If $D, E$ are both white, then $C A D E$ has three white vertices and one black vertex, impossible; if they are both black, then $C A D E$ has three black vertices and one white vertex, impossible. So $D$ and $E$ are opposite colors.\n\nBut $B C D E$ is also a parallelogram, since $B C=A C=D E$ and lines $B C, D E$ are parallel. However, $B C D E$ it has three black vertices and one white vertex. Thus we have a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $a_1 \\le a_2 \\le \\dots \\le a_n$ and $b_1 \\le b_2 \\le \\dots \\le b_n$ be two nondecreasing sequences of real numbers such that\n$$\na_1 + \\dots + a_i \\le b_1 + \\dots + b_i \\quad \\text{for every } i = 1, \\dots, n-1\n$$\nand\n$$\na_1 + \\dots + a_n = b_1 + \\dots + b_n.\n$$\n\nSuppose that for any real number $m$, the number of pairs $(i, j)$ with $a_i - a_j = m$ equals the number of pairs $(k, \\ell)$ with $b_k - b_\\ell = m$. Prove that $a_i = b_i$ for $i = 1, \\dots, n$.\n\n**Note:** It is important to interpret the condition that for any real number $m$, the number of pairs $(i, j)$ with $a_i - a_j = m$ equals the number of pairs $(k, \\ell)$ with $b_k - b_\\ell = m$. It means that we have two identical multi-sets (a multi-set is a set that allows repeated elements)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First Solution:** Put $s_n = a_1 + \\cdots + a_n = b_1 + \\cdots + b_n$. Then\n$$\n\\begin{aligned}\n2 \\sum_{i=1}^{n-1} (a_1 + \\cdots + a_i) &= 2(n-1)a_1 + 2(n-2)a_2 + \\cdots + 2(1)a_{n-1} \\\\\n&= (n-1)a_1 + (n-3)a_2 + \\cdots + (1-n)a_n + (n-1)s_n \\\\\n&= (n-1)s_n + \\sum_{1 \\le i < j \\le n} (a_i - a_j)\n\\end{aligned}\n$$\nand similarly\n$$\n2 \\sum_{i=1}^{n-1} (b_1 + \\cdots + b_i) = (n-1)s_n + \\sum_{1 \\le k < \\ell \\le n} (b_k - b_\\ell).\n$$\nBy (*), these two quantities are equal, so\n$$\n2 \\sum_{i=1}^{n-1} (a_1 + \\cdots + a_i) = 2 \\sum_{i=1}^{n-1} (b_1 + \\cdots + b_i).\n$$\nConsequently, each of the inequalities $a_1 + \\cdots + a_i \\le b_1 + \\cdots + b_i$ for $i = 1, \\dots, n-1$ must be an equality. Since we also have equality for $i = n$ by assumption, we deduce that $a_i = b_i$ for $i = 1, \\dots, n$, as desired.\n\n\n**Second Solution:** Expanding both sides of (**) yields\n$$\n(n-1) \\sum_{i=1}^{n} a_i^2 + 2 \\sum_{1 \\le i < j \\le n} a_i a_j = (n-1) \\sum_{i=1}^{n} b_i^2 + 2 \\sum_{1 \\le k < \\ell \\le n} b_k b_\\ell.\n$$\nSquaring both sides of the given equation $a_1 + \\cdots + a_n = b_1 + \\cdots + b_n$ gives\n$$\n\\sum_{i=1}^{n} a_i^2 + 2 \\sum_{1 \\le i < j \\le n} a_i a_j = \\sum_{i=1}^{n} b_i^2 + 2 \\sum_{1 \\le k < \\ell \\le n} b_k b_\\ell.\n$$\nFrom the above relations we easily deduce that\n$$\n\\sum_{i=1}^{n} a_i^2 = \\sum_{i=1}^{n} b_i^2.\n$$\nBy the Cauchy-Schwarz inequality, we obtain that\n$$\n\\left(\\sum_{i=1}^{n} b_i^2\\right)^2 = \\left(\\sum_{i=1}^{n} a_i^2\\right) \\left(\\sum_{i=1}^{n} b_i^2\\right) \\ge \\left(\\sum_{i=1}^{n} a_i b_i\\right)^2\n$$\nor\n$$\n\\sum_{i=1}^{n} b_i^2 \\geq \\left| \\sum_{i=1}^{n} a_i b_i \\right| \\geq \\sum_{i=1}^{n} a_i b_i. \\qquad (\\dagger)\n$$\nWe set $s_i = a_1 + \\dots + a_i$ and $t_i = b_1 + \\dots + b_i$ for every $1 \\le i \\le n$. By Abel's summation formula, we have\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} a_i b_i &= s_1 b_1 + [s_2 - s_1] b_2 + [s_3 - s_2] b_3 + \\dots + [s_n - s_{n-1}] b_n \\\\\n&= s_1(b_1 - b_2) + s_2(b_2 - b_3) + \\dots + s_{n-1}(b_{n-1} - b_n) + s_n b_n.\n\\end{aligned}\n$$\nBy the given conditions, $s_i \\le t_i$ and $b_i - b_{i+1} \\le 0$ for every $1 \\le i \\le n-1$ and $s_n = t_n$. It follows that\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} a_i b_i &\\ge t_1(b_1 - b_2) + t_2(b_2 - b_3) + \\dots + t_{n-1}(b_{n-1} - b_n) + t_n b_n \\\\\n&= t_1 b_1 + [t_2 - t_1] b_2 + [t_3 - t_2] b_3 + \\dots + [t_n - t_{n-1}] b_n = \\sum_{i=1}^{n} b_i^2.\n\\end{aligned}\n$$\nCombining the last inequality and (\\dagger), we conclude that the equality case holds for every inequality we discussed above. In particular, $s_i = t_i$ for $i = 1, \\dots, n$. These inequalities immediately give us $a_n = b_n$, $a_{n-1} = b_{n-1}$, $\\dots$, $a_1 = b_1$ and the problem is solved.\n\n\n**Third Solution:** If $u = (u_1, u_2, \\dots, u_n)$ and $v = (v_1, v_2, \\dots, v_n)$ are two nonincreasing sequences, we say that $u$ majorizes $v$ if $u_1 + \\dots + u_n = v_1 + \\dots + v_n$ and $u_1 + \\dots + u_i \\ge v_1 + \\dots + v_i$ for $i = 1, 2, \\dots, n-1$. It is not difficult to see that $(a_n, \\dots, a_1)$ majorizes $(b_n, \\dots, b_1)$. By a theorem of Birkhoff, it follows that there are constants $c_\\sigma \\in (0, 1]$, where $\\sigma$ runs over some set $S$ of permutations of $\\{1, \\dots, n\\}$, with $\\sum_{\\sigma \\in S} c_\\sigma = 1$ and\n$$\n\\sum_{\\sigma \\in S} c_{\\sigma} a_{\\sigma(i)} = b_i \\quad \\text{for } i = 1, 2, \\dots, n.\n$$\nWe will prove the inequality\n$$\n\\sum_{i,j=1}^{n} |a_i - a_j| \\ge \\sum_{i,j=1}^{n} |b_i - b_j|,\n$$\nwith equality if and only if $a_i = b_i$ for $i = 1, \\dots, n$. With this result, we complete our proof by noting (***).\nWe have\n$$\n\\begin{aligned}\n\\sum_{i,j=1}^{n} |a_i - a_j| &= \\sum_{\\sigma \\in S} c_{\\sigma} \\sum_{i,j=1}^{n} |a_i - a_j| = \\sum_{\\sigma \\in S} c_{\\sigma} \\sum_{i,j=1}^{n} |a_{\\sigma(i)} - a_{\\sigma(j)}| \\\\\n&= \\sum_{i,j=1}^{n} \\sum_{\\sigma \\in S} c_{\\sigma} |a_{\\sigma(i)} - a_{\\sigma(j)}| \\ge \\sum_{i,j=1}^{n} \\left| \\sum_{\\sigma \\in S} c_{\\sigma} (a_{\\sigma(i)} - a_{\\sigma(j)}) \\right| = \\sum_{i,j=1}^{n} |b_i - b_j|,\n\\end{aligned}\n$$\nusing the fact that $|x_1| + \\dots + |x_m| \\ge |x_1| + \\dots + |x_m|$ for all real numbers $x_1, \\dots, x_m$.\nThis establishes the desired inequality; it remains to check the equality condition. For this, we must have\n$$\n\\sum_{\\sigma \\in S} c_{\\sigma} |a_{\\sigma(i)} - a_{\\sigma(j)}| = \\left| \\sum_{\\sigma \\in S} c_{\\sigma} (a_{\\sigma(i)} - a_{\\sigma(j)}) \\right|\n$$\nfor each pair $i$ and $j$; in particular, for each pair $i$ and $j$, the sign of $a_{\\sigma(i)} - a_{\\sigma(j)}$ must be the same for all $\\sigma \\in S$ for which $a_{\\sigma(i)} \\neq a_{\\sigma(j)}$. It follows by the lemma below that the sequence $a_{\\sigma(1)}, \\dots, a_{\\sigma(n)}$ itself must be the same for all $\\sigma \\in S$, yielding $a_i = b_i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many 5-digit numbers $\\overline{a b c d e}$ exist such that digits $b$ and $d$ are each the sum of the digits to their immediate left and right? (That is, $b = a + c$ and $d = c + e$.)",
    "options": [],
    "answer": "330",
    "solution": "Solution:\n\nAnswer: 330. Note that $a > 0$, so that $b > c$, and $e \\geq 0$ so that $d \\geq c$. Conversely, for each choice of $(b, c, d)$ with $b > c$ and $d \\geq c$, there exists a unique pair $(a, e)$ such that $\\overline{a b c d e}$ is a number having the desired property. Thus, we compute\n$$\n\\sum_{c=0}^{9} (9-c)(10-c) = \\sum_{c=0}^{9} c^{2} - 19c + 90 = 330\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA parallelogram has 3 of its vertices at $(1,2)$, $(3,8)$, and $(4,1)$. Compute the sum of all possible $x$ coordinates of the 4th vertex.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\nAnswer: $8$. There are three possibilities: the 4th vertex must be opposite one of the three given vertices. These three possibilities have as a medial triangle the three given vertices, so the sum of their $x$ coordinates is the same as the sum of the $x$ coordinates of the given triangle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA candy company makes 5 colors of jellybeans, which come in equal proportions. If I grab a random sample of 5 jellybeans, what is the probability that I get exactly 2 distinct colors?",
    "options": [],
    "answer": "12/125",
    "solution": "Solution:\nThere are $\\binom{5}{2} = 10$ possible pairs of colors. Each pair of colors contributes $2^{5} - 2 = 30$ sequences of beans that use both colors. Thus, the answer is $10 \\cdot 30 / 5^{5} = 12 / 125$.\n\nAnswer: $\\dfrac{12}{125}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSolve for the positive integer(s) $n$ such that $\\phi\\left(n^{2}\\right)=1000 \\phi(n)$.",
    "options": [],
    "answer": "1000",
    "solution": "Solution:\n\nAnswer: $1000$.\n\nThe unique solution is $n=1000$. For, $\\phi(p n)=p \\phi(n)$ for every prime $p$ dividing $n$, so that $\\phi\\left(n^{2}\\right)=n \\phi(n)$ for all positive integers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe four sides of quadrilateral $ABCD$ are equal in length. Determine the perimeter of $ABCD$ given that it has area $120$ and $AC = 10$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "52",
    "solution": "Solution:\n\nLet $M$ be the midpoint of $AC$. Then triangles $AMB$, $BMC$, $CMD$, and $DMA$ are all right triangles having legs $5$ and $h$ for some $h$.\n\nThe area of $ABCD$ is $120$, but also $4 \\cdot \\left(\\frac{1}{2} \\cdot 5 \\cdot h\\right) = 10h$, so $h = 12$.\n\nThen $AB = BC = CD = DA = \\sqrt{12^2 + 5^2} = 13$, and the perimeter of $ABCD$ is $52$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo reals $x$ and $y$ are such that $x - y = 4$ and $x^{3} - y^{3} = 28$. Compute $x y$.",
    "options": [],
    "answer": "-3",
    "solution": "Solution:\n\nAnswer: $-3$. We have $28 = x^{3} - y^{3} = (x - y)\\left(x^{2} + x y + y^{2}\\right) = (x - y)\\left((x - y)^{2} + 3 x y\\right) = 4 \\cdot (16 + 3 x y)$, from which $x y = -3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the real number $a$ having the property that $f(a)=a$ is a relative minimum of $f(x)=x^{4}-x^{3}-x^{2}+a x+1$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nBeing a relative minimum, we have $0=f'(a)=4 a^{3}-3 a^{2}-2 a+a=a(4 a+1)(a-1)$. Then $a=0, 1, -1/4$ are the only possibilities. However, it is easily seen that $a=1$ is the only value satisfying $f(a)=a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor $a$ a positive real number, let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3}-a x^{2}+a x-a=0$. Determine the smallest possible value of $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}$.",
    "options": [],
    "answer": "-4",
    "solution": "Solution:\n\nAnswer: $-4$. Note that $x_{1}+x_{2}+x_{3}=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a$. Then\n$$\n\\begin{aligned}\n& x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}=\\left(x_{1}+x_{2}+x_{3}\\right)\\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-\\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\\right)\\right) \\\\\n& \\quad=\\left(x_{1}+x_{2}+x_{3}\\right)\\left(\\left(x_{1}+x_{2}+x_{3}\\right)^{2}-3\\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\\right)\\right)=a \\cdot\\left(a^{2}-3 a\\right)=a^{3}-3 a^{2}\n\\end{aligned}\n$$\nThe expression is negative only where $0<a<3$, so we need only consider these values of $a$. Finally, AM-GM gives $\\sqrt[3]{(6-2 a)(a)(a)} \\leq \\frac{(6-2 a)+a+a}{3}=2$, with equality where $a=2$, and this rewrites as $(a-3) a^{2} \\geq -4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBob the bomb-defuser has stumbled upon an active bomb. He opens it up, and finds the red and green wires conveniently located for him to cut. Being a seasoned member of the bomb-squad, Bob quickly determines that it is the green wire that he should cut, and puts his wirecutters on the green wire. But just before he starts to cut, the bomb starts to count down, ticking every second. Each time the bomb ticks, starting at time $t=15$ seconds, Bob panics and has a certain chance to move his wirecutters to the other wire. However, he is a rational man even when panicking, and has a $\\frac{1}{2 t^{2}}$ chance of switching wires at time $t$, regardless of which wire he is about to cut. When the bomb ticks at $t=1$, Bob cuts whatever wire his wirecutters are on, without switching wires. What is the probability that Bob cuts the green wire?",
    "options": [],
    "answer": "23/30",
    "solution": "Solution:\n\nAnswer: $\\frac{23}{30}$. Suppose Bob makes $n$ independent decisions, with probabilities of switching $p_{1}, p_{2}, \\ldots, p_{n}$. Then in the expansion of the product\n$$\nP(x)=\\left(p_{1}+\\left(1-p_{1}\\right) x\\right)\\left(p_{2}+\\left(1-p_{2}\\right) x\\right) \\cdots\\left(p_{n}+\\left(1-p_{n}\\right) x\\right)\n$$\nthe sum of the coefficients of even powers of $x$ gives the probability that Bob makes his original decision. This is just $(P(1)+P(-1)) / 2$, so the probability is just\n$$\n\\frac{1+\\left(1-\\frac{1}{15^{2}}\\right)\\left(1-\\frac{1}{14^{2}}\\right) \\cdots\\left(1-\\frac{1}{2^{2}}\\right)}{2}=\\frac{1+\\frac{14}{15}\\frac{13}{14}\\cdots\\frac{3}{4}}{2}=\\frac{1+\\frac{8}{15}}{2}=\\frac{23}{30}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequence $\\{a_n\\}_{n \\geq 1}$ is defined by $a_{n+2} = 7 a_{n+1} - a_n$ for positive integers $n$ with initial values $a_1 = 1$ and $a_2 = 8$. Another sequence, $\\{b_n\\}$, is defined by the rule $b_{n+2} = 3 b_{n+1} - b_n$ for positive integers $n$ together with the values $b_1 = 1$ and $b_2 = 2$. Find $\\operatorname{gcd}(a_{5000}, b_{501})$.",
    "options": [],
    "answer": "89",
    "solution": "Solution:\n\nAnswer: 89. We show by induction that $a_n = F_{4n-2}$ and $b_n = F_{2n-1}$, where $F_k$ is the $k$\\text{th}$ Fibonacci number. The base cases are clear. As for the inductive steps, note that\n$$\nF_{k+2} = F_{k+1} + F_k = 2 F_k + F_{k-1} = 3 F_k - F_{k-2}\n$$\nand\n$$\nF_{k+4} = 3 F_{k+2} - F_k = 8 F_k + 3 F_{k-2} = 7 F_k - F_{k-4}\n$$\nWe wish to compute the greatest common denominator of $F_{19998}$ and $F_{1001}$. The Fibonacci numbers satisfy the property that $\\operatorname{gcd}(F_m, F_n) = F_{\\operatorname{gcd}(m, n)}$, which can be proven by noting that they are periodic modulo any positive integer. So since $\\operatorname{gcd}(19998, 1001) = 11$, the answer is $F_{11} = 89$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $ABC$, $\\angle ABC$ is obtuse. Point $D$ lies on side $AC$ such that $\\angle ABD$ is right, and point $E$ lies on side $AC$ between $A$ and $D$ such that $BD$ bisects $\\angle EBC$. Find $CE$, given that $AC = 35$, $BC = 7$, and $BE = 5$.",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nAnswer: 10. Reflect $A$ and $E$ over $BD$ to $A'$ and $E'$ respectively. Note that the angle conditions show that $A'$ and $E'$ lie on $AB$ and $BC$ respectively. $B$ is the midpoint of segment $AA'$, and $CE' = BC - BE' = 2$. Menelaus' theorem now gives\n$$\n\\frac{CD}{DA} \\cdot \\frac{AA'}{A'B} \\cdot \\frac{BE'}{E'C} = 1\n$$\nfrom which $DA = 5 CD$ or $CD = AC / 6$. By the angle bisector theorem, $DE = 5 CD / 7$, so that $CE = 12 CD / 7 = 10$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$, $y$, $n$ be positive integers with $n > 1$. How many ordered triples $(x, y, n)$ of solutions are there to the equation $x^{n} - y^{n} = 2^{100}$?",
    "options": [],
    "answer": "49",
    "solution": "Solution:\n\nAnswer: 49. Break all possible values of $n$ into the four cases: $n = 2$, $n = 4$, $n > 4$ and $n$ odd. By Fermat's theorem, no solutions exist for the $n = 4$ case because we may write $y^{4} + (2^{25})^{4} = x^{4}$.\n\nWe show that for $n$ odd, no solutions exist to the more general equation $x^{n} - y^{n} = 2^{k}$ where $k$ is a positive integer. Assume otherwise for contradiction's sake, and suppose on the grounds of well ordering that $k$ is the least exponent for which a solution exists. Clearly $x$ and $y$ must both be even or both odd. If both are odd, we have $(x - y)(x^{n-1} + \\ldots + y^{n-1})$. The right factor of this expression contains an odd number of odd terms whose sum is an odd number greater than 1, impossible. Similarly if $x$ and $y$ are even, write $x = 2u$ and $y = 2v$. The equation becomes $u^{n} - v^{n} = 2^{k-n}$. If $k-n$ is greater than 0, then our choice $k$ could not have been minimal. Otherwise, $k-n = 0$, so that two consecutive positive integers are perfect $n$th powers, which is also absurd.\n\nFor the case that $n$ is even and greater than 4, consider the same generalization and hypotheses. Writing $n = 2m$, we find $(x^{m} - y^{m})(x^{m} + y^{m}) = 2^{k}$. Then $x^{m} - y^{m} = 2^{a} < 2^{k}$. By our previous work, we see that $m$ cannot be an odd integer greater than 1. But then $m$ must also be even, contrary to the minimality of $k$.\n\nFinally, for $n = 2$ we get $x^{2} - y^{2} = 2^{100}$. Factoring the left hand side gives $x - y = 2^{a}$ and $x + y = 2^{b}$, where implicit is $a < b$. Solving, we get $x = 2^{b-1} + 2^{a-1}$ and $y = 2^{b-1} - 2^{a-1}$, for a total of 49 solutions. Namely, those corresponding to $(a, b) = (1, 99), (2, 98), \\cdots, (49, 51)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo real numbers $x$ and $y$ are such that $8 y^{4} + 4 x^{2} y^{2} + 4 x y^{2} + 2 x^{3} + 2 y^{2} + 2 x = x^{2} + 1$. Find all possible values of $x + 2 y^{2}$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nAnswer: $\\frac{1}{2}$.\n\nWriting $a = x + 2 y^{2}$, the given quickly becomes $4 y^{2} a + 2 x^{2} a + a + x = x^{2} + 1$. We can rewrite $4 y^{2} a$ for further reduction to $a(2 a - 2 x) + 2 x^{2} a + a + x = x^{2} + 1$, or\n$$\n2 a^{2} + (2 x^{2} - 2 x + 1) a + (-x^{2} + x - 1) = 0\n$$\nThe quadratic formula produces the discriminant\n$$\n(2 x^{2} - 2 x + 1)^{2} + 8(x^{2} - x + 1) = (2 x^{2} - 2 x + 3)^{2},\n$$\nan identity that can be treated with the difference of squares, so that $a = \\frac{-2 x^{2} + 2 x - 1 \\pm (2 x^{2} - 2 x + 3)}{4} = \\frac{1}{2}, -x^{2} + x - 1$. Now $a$ was constructed from $x$ and $y$, so is not free. Indeed, the second expression flies in the face of the trivial inequality: $a = -x^{2} + x - 1 < -x^{2} + x \\leq x + 2 y^{2} = a$. On the other hand, $a = 1 / 2$ is a bona fide solution to $\\left(^*\\right)$, which is identical to the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is a cyclic quadrilateral in which $AB = 4$, $BC = 3$, $CD = 2$, and $AD = 5$. Diagonals $AC$ and $BD$ intersect at $X$. A circle $\\omega$ passes through $A$ and is tangent to $BD$ at $X$. $\\omega$ intersects $AB$ and $AD$ at $Y$ and $Z$ respectively. Compute $YZ / BD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "115/143",
    "solution": "Solution:\n\nAnswer: $\\frac{115}{143}$. Denote the lengths $AB, BC, CD$, and $DA$ by $a, b, c$, and $d$ respectively. Because $ABCD$ is cyclic, $\\triangle ABX \\sim \\triangle DCX$ and $\\triangle ADX \\sim \\triangle BCX$. It follows that $\\frac{AX}{DX} = \\frac{BX}{CX} = \\frac{a}{c}$ and $\\frac{AX}{BX} = \\frac{DX}{CX} = \\frac{d}{b}$. Therefore we may write $AX = adk$, $BX = abk$, $CX = bck$, and $DX = cdk$ for some $k$.\n\nNow, $\\angle XDC = \\angle BAX = \\angle YXB$ and $\\angle DCX = \\angle XBY$, so $\\triangle BXY \\sim \\triangle CDX$. Thus, $XY = DX \\cdot \\frac{BX}{CD} = cdk \\cdot \\frac{abk}{c} = abd k^2$. Analogously, $XZ = acd k^2$. Note that $XY / XZ = CB / CD$. Since $\\angle YXZ = \\pi - \\angle ZAY = \\angle BCD$, we have that $\\triangle XYZ \\sim \\triangle CBD$. Thus, $YZ / BD = XY / CB = ad k^2$.\n\nFinally, Ptolemy's theorem applied to $ABCD$ gives\n$$\n(ad + bc)k \\cdot (ab + cd)k = ac + bd\n$$\nIt follows that the answer is\n$$\n\\frac{ad(ac + bd)}{(ab + cd)(ad + bc)} = \\frac{20 \\cdot 23}{22 \\cdot 26} = \\frac{115}{143}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of 7-tuples $\\left(n_{1}, \\ldots, n_{7}\\right)$ of integers such that\n$$\n\\sum_{i=1}^{7} n_{i}^{6}=96957\n$$",
    "options": [],
    "answer": "2688",
    "solution": "Solution:\nAnswer: 2688. Consider the equation in modulo 9. All perfect 6th powers are either 0 or 1. Since 9 divides 96957, it must be that each $n_{i}$ is a multiple of 3. Writing $3 a_{i}=n_{i}$ and dividing both sides by $3^{6}$, we have $a_{1}^{6}+\\cdots+a_{7}^{6}=133$. Since sixth powers are nonnegative, $\\left|a_{i}\\right| \\leq 2$. Again considering modulo 9, we see that $a_{i} \\neq 0$. Thus, $a_{i}^{6} \\in\\{1,64\\}$. The only possibility is $133=64+64+1+1+1+1+1$, so $\\left|a_{1}\\right|, \\ldots,\\left|a_{7}\\right|$ consists of two 2's and five 1's. It follows that the answer is $\\binom{7}{2} \\cdot 2^{7}=2688$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute the circumradius of cyclic hexagon $A B C D E F$, which has side lengths $A B = B C = 2$, $C D = D E = 9$, and $E F = F A = 12$.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nAnswer: 8. Construct point $E'$ on the circumcircle of $A B C D E F$ such that $D E' = E F = 12$ and $E' F = D E = 9$; then $\\overline{B E'}$ is a diameter. Let $B E' = d$. Then $C E' = \\sqrt{B E'^2 - B C^2} = \\sqrt{d^2 - 4}$ and $B D = \\sqrt{B E'^2 - D E'^2} = \\sqrt{d^2 - 144}$. Applying Ptolemy's theorem to $B C D E'$ now yields\n$$\n9 \\cdot d + 2 \\cdot 12 = \\sqrt{(d^2 - 4)(d^2 - 144)}\n$$\nSquaring and rearranging, we find $0 = d^4 - 229 d^2 - 432 d = d(d-16)(d^2 + 16 d + 27)$. Since $d$ is a positive real number, $d = 16$, and the circumradius is $8$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence $\\{a_{n}\\}_{n \\geq 1}$ of positive reals is defined by the rule $a_{n+1} a_{n-1}^{5} = a_{n}^{4} a_{n-2}^{2}$ for integers $n > 2$ together with the initial values $a_{1} = 8$, $a_{2} = 64$, and $a_{3} = 1024$. Compute\n$$\n\\sqrt{a_{1} + \\sqrt{a_{2} + \\sqrt{a_{3} + \\cdots}}}\n$$",
    "options": [],
    "answer": "3 sqrt(2)",
    "solution": "Solution:\nAnswer: $3 \\sqrt{2}$.\n\nTaking the base-$2$ log of the sequence $\\{a_{n}\\}$ converts the multiplicative rule to a more familiar additive rule:\n$$\n\\log_{2}(a_{n+1}) - 4 \\log_{2}(a_{n}) + 5 \\log_{2}(a_{n-1}) - 2 \\log_{2}(a_{n-2}) = 0.\n$$\nThe characteristic equation is $0 = x^{3} - 4x^{2} + 5x - 2 = (x-1)^{2}(x-2)$, so $\\log_{2}(a_{n})$ is of the form $a \\cdot n + b + c \\cdot 2^{n}$ and we find $a_{n} = 2^{2n + 2^{n-1}}$.\n\nNow,\n$$\n\\sqrt{a_{1} + \\sqrt{a_{2} + \\sqrt{a_{3} + \\cdots}}} = \\sqrt{2} \\cdot \\sqrt{4 + \\sqrt{16 + \\sqrt{64 + \\cdots}}}\n$$\nWe can estimate the new nested radical expression as $3$, which expands thus\n$$\n3 = \\sqrt{4 + 5} = \\sqrt{4 + \\sqrt{16 + 9}} = \\sqrt{4 + \\sqrt{16 + \\sqrt{64 + 17}}} = \\cdots\n$$\nAs a rigorous confirmation, we have $2^{k} + 1 = \\sqrt{4^{k} + (2^{k+1} + 1)}$, as desired. It follows that the answer is $3 \\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\theta$ be an angle in the interval $(0, \\pi/2)$. Given that $\\cos \\theta$ is irrational, and that $\\cos k\\theta$ and $\\cos[(k+1)\\theta]$ are both rational for some positive integer $k$, show that $\\theta = \\pi/6$.",
    "options": [],
    "answer": "π/6",
    "solution": "Thus both $\\cos(k^2\\theta) = \\cos[k(k\\theta)]$ and $\\cos[(k^2-1)\\theta] = \\cos[(k-1)(k+1)\\theta]$ are rational. By the Addition and subtraction formulas, we have\n$$\n\\cos[(k+1)\\theta] = \\cos k\\theta \\cos \\theta - \\sin k\\theta \\sin \\theta \\quad \\text{and} \\quad \\cos(k^2\\theta) = \\cos[(k^2-1)\\theta] \\cos \\theta - \\sin[(k^2-1)\\theta] \\sin \\theta.\n$$\nSetting $r_1 = \\cos k\\theta$, $r_2 = \\cos[(k+1)\\theta]$, $r_3 = \\cos[(k^2-1)\\theta]$, $r_4 = \\cos(k^2\\theta)$, and $x = \\cos \\theta$ in the above equations yields\n$$\nr_2 = r_1 x \\pm \\sqrt{(1 - r_1^2)(1 - x^2)} \\quad \\text{and} \\quad r_4 = r_3 x \\pm \\sqrt{(1 - r_3^2)(1 - x^2)},\n$$\nor\n$$\n\\pm\\sqrt{(1 - r_1^2)(1 - x^2)} = r_2 - r_1 x \\quad \\text{and} \\quad \\pm\\sqrt{(1 - r_3^2)(1 - x^2)} = r_4 - r_3 x.\n$$\nSquaring these two equations and subtracting the resulting equations gives\n$$\n2(r_1 r_2 - r_3 r_4)x = r_1^2 + r_2^2 - (r_3^2 + r_4^2).\n$$\nSince $r_1, r_2, r_3, r_4$ are rational and $x$ is irrational, we must have $r_1r_2 - r_3r_4 = 0$ or\n$$\n\\cos k\\theta \\cos[(k+1)\\theta] = \\cos(k^2\\theta) \\cos[(k^2-1)\\theta].\n$$\nBy the product-to-sum formulas, we derive\n$$\n\\frac{\\cos[(2k + 1)\\theta] - \\cos \\theta}{2} = \\frac{\\cos[(2k^2 - 1)\\theta] - \\cos \\theta}{2}\n$$\nor $\\cos[(2k + 1)\\theta] - \\cos[(2k^2 - 1)\\theta] = 0$. By the sum-to-product formulas, we obtain\n$$\n2 \\sin[(k - k^2 + 1)\\theta] \\sin[(k^2 + k)\\theta] = 0,\n$$\nimplying that either $(k - k^2 + 1)\\theta$ or $(k^2 + k)\\theta$ is a integral multiple of $\\pi$. Since $k$ is an integer, we conclude that $\\theta = r\\pi$ for some rational number $r$.\nConsidering Lemma 2 for $\\alpha = k\\theta$ and $\\alpha = (k+1)\\theta$, the possible values of $\\cos k\\theta$ and $\\cos[(k+1)\\theta]$ are $0, \\pm 1, \\pm \\frac{1}{2}$. Consequently, both $k\\theta$ and $(k+1)\\theta$ is a integral multiple of $\\frac{\\pi}{6}$. Since $0 < \\theta = k\\theta - (k-1)\\theta < \\frac{\\pi}{2}$, the only possible values of $\\theta$ are $\\frac{\\pi}{3}$ and $\\frac{\\pi}{6}$. Since $\\cos \\theta$ is irrational, $\\theta = \\frac{\\pi}{6}$.\n(Based on the work by Kiran Kedlaya) We maintain the notations used in the first proof. Then $s = 2 \\cos \\theta$ is a root of $S_k(x) - 2r_1$ and $S_{k+1}(x) - 2r_2$ by the definition of $S_n$. Define\n$$\nQ(x) = \\gcd(S_k(x) - 2r_1, S_{k+1}(x) - 2r_2)\n$$\nwhere the gcd is taken over the field of rational numbers. Then $Q(x)$ is a polynomial with rational coefficients, so the sum of its roots (with multiplicities) is rational. Since $s$ is assumed not to be rational, there must be at least one other distinct root $t$ of $Q(x)$.\nNote that the $k$ distinct reals $2\\cos(\\theta + 2\\pi a/k)$ for $a = 0, 1, \\dots, k-1$ form $k$ roots of the degree $k$ polynomial $S_k(x) - 2r_1$, so they compose all of its roots. Similarly, all of the roots of $S_{k+1}(x) - 2r_2$ have the form $2\\cos(\\theta + 2\\pi b/(k+1))$ for $b = 0, 1, \\dots, k$. Note that $s$ and $t$ are roots of $Q(x)$. Therefore roots of both $S_k(x) - 2r_1$ and $S_{k+1}(x) - 2r_2$, and so they must have at least two distinct common roots. Each root $r$ of $Q(x)$ must thus satisfy\n$$\nr = 2 \\cos(\\theta + 2\\pi a/k) = 2 \\cos(\\theta + 2\\pi b/(k+1))\n$$\nfor some $a$ and $b$. We either have $\\theta + 2\\pi a/k = \\theta + 2\\pi b/(k+1)$ and thus $r = 2 \\cos \\theta$ or $\\theta + 2\\pi a/k = -\\theta - 2\\pi b/(k+1)$ and thus\n$$\n\\theta = - \\frac{\\pi[(a+b)k+a]}{k(k+1)}.\n$$\nIn the first case, we obtain $s$, so $t$ must lead to the second value of $\\theta$, as $s \\neq t$.\nTherefore, we can write $\\theta = \\frac{\\pi c}{k(k+1)}$ for some integer $c$. By Lemma 2, $c/k$ and $c/(k+1)$ must both be multiples of $1/6$, since $\\cos k\\theta = \\cos \\frac{c\\pi}{k+1}$ and $\\cos(k+1)\\theta = \\cos \\frac{c\\pi}{k}$ are rational. Therefore, $\\theta = \\frac{c\\pi}{k} - \\frac{c\\pi}{k+1}$ is a multiple of $\\pi/6$. Since $t$ is not rational, $\\theta$ can only be $\\pi/6$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn $\\triangle ABC$, $D$ and $E$ are two points on segment $\\overline{BC}$ such that $BD = CE$ and $\\angle BAD = \\angle CAE$. Prove that $\\triangle ABC$ is isosceles.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTranslate $\\triangle BDA$ horizontally until its side $BD$ coincides with side $EC$, and label the image of point $A$ by $A'$. We now have two triangles $\\triangle ECA$ and $\\triangle ECA'$ which share the same base $EC$, have the same height (equal to the height $AH$ of the original $\\triangle ABC$), and equal angles $\\angle EAC$ and $\\angle EA'C$. The last implies that there is a circle $k$ passing through points $E, C, A'$ and $A$. The equal heights condition implies that $BC$ is parallel to $AA'$; yet the only trapezoids inscribed in a circle are isosceles. Therefore, $ECA'A$ is isosceles with $AE = A'C$. Since the diagonals of an isosceles trapezoid are equal, $EA' = CA$. Translating back to the original $\\triangle ABC$, this means that $BA = CA$ and $\\triangle ABC$ is isosceles.\n\n![](attached_image_2.png)\nSolution:\n\nWithout loss of generality, assume that the points $B, D, E$ and $C$ are arranged in this order on the line $\\overleftrightarrow{BC}$; otherwise, switch points $D$ and $E$ in the remainder of the solution.\n\nSuppose that $\\triangle ABC$ is not isosceles. Without loss of generality, let $BA < CA$. Let $AL$ denote the angle bisector of $\\angle BAC$, where $L$ lies on side $BC$. Reflect $\\triangle ABL$ across $AL$ and denote by $B'$ and $D'$ the images of $B$ and $D$, respectively. Since $B'A = BA < CA$, $B'$ is inside side $CA$, and this forces the whole segment $LB'$ to be inside $\\triangle LCA$; in particular, $D'$ is an interior point of segment $AE$.\n\nBecause of the reflection, $\\triangle AD'B'$ has equal area as $\\triangle ADB$. In its turn, $\\triangle ADB$ has the equal area as $\\triangle AEC$ (they have equal bases and heights). This implies that $\\triangle AD'B'$ and $\\triangle AEC$ have equal area, which contradicts the fact that one triangle is properly included in the other.\n\nWe conclude that our supposition is false. Therefore, $AB = AC$ and our triangle is isosceles as desired.\n\n![](attached_image_3.png)\nSolution:\n\nLet $\\angle BAD = \\angle CAE = \\theta$. In the end triangles, $\\triangle ABD$ and $\\triangle ACE$, and $\\triangle ABC$, by the Law of Sines\n$$\n\\frac{a}{\\sin \\theta} = \\frac{m}{\\sin B} = \\frac{n}{\\sin C} \\quad \\text{and} \\quad \\frac{q}{\\sin B} = \\frac{p}{\\sin C}\n$$\nThen $mp = nq$. Suppose $p \\neq q$. Then, without loss of generality, $p < q$, which implies $np < nq = mp$ and $n < m$. But if $p < q$ then $\\angle C < \\angle B$ which implies $\\angle AED = \\angle C + \\theta < \\angle B + \\theta = \\angle ADE$ and $m < n$.\n\nThis is a contradiction, so $p = q$ and $\\triangle ABC$ is isosceles.\nSolution:\n\nLet $\\angle BAD = \\angle CAE = \\theta$. $\\triangle ABD$ and $\\triangle ACE$ have equal areas since they have equal bases $(BD = EC)$ and the same altitude from $A$. Then\n$$\n\\frac{1}{2} p m \\sin \\theta = \\frac{1}{2} q n \\sin \\theta\n$$\nwhich implies $mp = nq$. Now use the same contradiction as in the above solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a set containing $n^2 + n - 1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In order to apply induction, we generalize the result to be proved so that it reads as follows:\n\n**Proposition.** If the $n$-element subsets of a set $S$ with $(n+1)m - 1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.\n\n*Proof:* Fix $n$ and proceed by induction on $m$. The case of $m=1$ is trivial. Assume $m>1$ and that the proposition is true for $m-1$. Let $\\mathcal{P}$ be the partition of the $n$-element subsets into two classes. If all the $n$-element subsets belong to the same class, the result is obvious. Otherwise select two $n$-element subsets $A$ and $B$ from different classes so that their intersection has maximal size. It is easy to see that $|A \\cap B| = n-1$. (If $|A \\cap B| = k < n-1$, then build $C$ from $B$ by replacing some element not in $A \\cap B$ with an element of $A$ not already in $B$. Then $|A \\cap C| = k+1$ and $|B \\cap C| = n-1$ and either $A$ and $C$ or $B$ and $C$ are in different classes.) Removing $A \\cup B$ from $S$, there are $(n+1)(m-1)-1$ elements left. On this set the partition induced by $\\mathcal{P}$ has, by the inductive hypothesis, $m-1$ pairwise disjoint sets in the same class. Adding either $A$ or $B$ as appropriate gives $m$ pairwise disjoint sets in the same class. $\\blacksquare$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a set containing $n^2 + n - 1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In order to apply induction, we generalize the result to be proved so that it reads as follows:\n\n**Proposition.** If the $n$-element subsets of a set $S$ with $(n+1)m-1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.\n\n*Proof.* Fix $n$ and proceed by induction on $m$. The case of $m=1$ is trivial. Assume $m>1$ and that the proposition is true for $m-1$. Let $\\mathcal{P}$ be the partition of the $n$-element subsets into two classes. If all the $n$-element subsets belong to the same class, the result is obvious. Otherwise select two $n$-element subsets $A$ and $B$ from different classes so that their intersection has maximal size. It is easy to see that $|A \\cap B| = n-1$. (If $|A \\cap B| = k < n-1$, then build $C$ from $B$ by replacing some element not in $A \\cap B$ with an element of $A$ not already in $B$. Then $|A \\cap C| = k+1$ and $|B \\cap C| = n-1$ and either $A$ and $C$ or $B$ and $C$ are in different classes.) Removing $A \\cup B$ from $S$, there are $(n+1)(m-1)-1$ elements left. On this set the partition induced by $\\mathcal{P}$ has, by the inductive hypothesis, $m-1$ pairwise disjoint sets in the same class. Adding either $A$ or $B$ as appropriate gives $m$ pairwise disjoint sets in the same class. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe equation $x^{2}+2x=i$ has two complex solutions. Determine the product of their real parts.",
    "options": [],
    "answer": "(1-\\sqrt{2})/2",
    "solution": "Solution:\n\nAnswer: $\\frac{1-\\sqrt{2}}{2}$. Complete the square by adding $1$ to each side. Then $(x+1)^{2}=1+i=e^{\\frac{i \\pi}{4}} \\sqrt{2}$, so $x+1= \\pm e^{\\frac{i \\pi}{8}} \\sqrt[4]{2}$. The desired product is then\n$$\n\\left(-1+\\cos \\left(\\frac{\\pi}{8}\\right) \\sqrt[4]{2}\\right)\\left(-1-\\cos \\left(\\frac{\\pi}{8}\\right) \\sqrt[4]{2}\\right)=1-\\cos ^{2}\\left(\\frac{\\pi}{8}\\right) \\sqrt{2}=1-\\frac{\\left(1+\\cos \\left(\\frac{\\pi}{4}\\right)\\right)}{2} \\sqrt{2}=\\frac{1-\\sqrt{2}}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCircles $\\omega_{1}$, $\\omega_{2}$, and $\\omega_{3}$ are centered at $M$, $N$, and $O$, respectively. The points of tangency between $\\omega_{2}$ and $\\omega_{3}$, $\\omega_{3}$ and $\\omega_{1}$, and $\\omega_{1}$ and $\\omega_{2}$ are tangent at $A$, $B$, and $C$, respectively. Line $MO$ intersects $\\omega_{3}$ and $\\omega_{1}$ again at $P$ and $Q$ respectively, and line $AP$ intersects $\\omega_{2}$ again at $R$. Given that $ABC$ is an equilateral triangle of side length $1$, compute the area of $PQR$.",
    "options": [],
    "answer": "2√3",
    "solution": "Solution:\n\n$\\boxed{2 \\sqrt{3}}$. Note that $ONM$ is an equilateral triangle of side length $2$, so $m \\angle BPA = m \\angle BOA / 2 = \\pi / 6$. Now $BPA$ is a $30$-$60$-$90$ triangle with short side length $1$, so $AP = \\sqrt{3}$. Now $A$ and $B$ are the midpoints of segments $PR$ and $PQ$, so\n\n$$\n[PQR] = \\frac{PR}{PA} \\cdot \\frac{PQ}{PB} [PBA] = 2 \\cdot 2 [PBA] = 2 \\sqrt{3}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for every integer $n$ greater than $1$,\n$$\n\\sigma(n) \\phi(n) \\leq n^{2}-1\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds if and only if n is prime.",
    "solution": "Solution:\nNote that\n$$\n\\sigma(m n) \\phi(m n) = \\sigma(m) \\phi(m) \\sigma(n) \\phi(n) \\leq (m^{2}-1)(n^{2}-1) = (m n)^{2} - (m^{2} + n^{2} - 1) < (m n)^{2} - 1\n$$\nfor any pair of relatively prime positive integers $(m, n)$ other than $(1,1)$. Now, for $p$ a prime and $k$ a positive integer, $\\sigma\\left(p^{k}\\right) = 1 + p + \\cdots + p^{k} = \\frac{p^{k+1}-1}{p-1}$ and $\\phi\\left(p^{k}\\right) = p^{k} - \\frac{1}{p} \\cdot p^{k} = (p-1) p^{k-1}$. Thus,\n$$\n\\sigma\\left(p^{k}\\right) \\phi\\left(p^{k}\\right) = \\frac{p^{k+1}-1}{p-1} \\cdot (p-1) p^{k-1} = (p^{k+1}-1) p^{k-1} = p^{2k} - p^{k-1} \\leq p^{2k} - 1\n$$\nwith equality where $k=1$. It follows that equality holds in the given inequality if and only if $n$ is prime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$A$, $B$, $C$, and $D$ are points on a circle, and segments $\\overline{AC}$ and $\\overline{BD}$ intersect at $P$, such that $AP = 8$, $PC = 1$, and $BD = 6$. Find $BP$, given that $BP < DP$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nLet $BP = x$ and $DP = 6 - x$ (since $BP < DP$ and $BD = 6$).\n\nBy the Power of a Point theorem (intersecting chords),\n$$\nAP \\cdot PC = BP \\cdot PD\n$$\nSubstitute the given values:\n$$\n8 \\cdot 1 = x \\cdot (6 - x)\n$$\n$$\n8 = 6x - x^2\n$$\n$$\nx^2 - 6x + 8 = 0\n$$\n$$\n(x - 4)(x - 2) = 0\n$$\nSo $x = 2$ or $x = 4$.\n\nSince $BP < DP$, $BP = 2$ and $DP = 4$.\n\n**Answer:** $BP = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFive people are crowding into a booth against a wall at a noisy restaurant. If at most three can fit on one side, how many seating arrangements accommodate them all?",
    "options": [],
    "answer": "240",
    "solution": "Solution:\n\nAnswer: $240$. Three people will sit on one side and two sit on the other, giving a factor of two. Then there are $5!$ ways to permute the people.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$ be a positive real number. Find the value of $a$ such that the definite integral\n$$\n\\int_{a}^{a^{2}} \\frac{\\mathrm{d} x}{x+\\sqrt{x}}\n$$\nachieves its smallest possible value.",
    "options": [],
    "answer": "3 - 2√2",
    "solution": "Solution:\nAnswer: $\\sqrt{3-2 \\sqrt{2}}$\nLet $F(a)$ denote the given definite integral. Then\n$$\nF'(a) = \\frac{\\mathrm{d}}{\\mathrm{d} a} \\int_{a}^{a^{2}} \\frac{\\mathrm{d} x}{x+\\sqrt{x}} = 2a \\cdot \\frac{1}{a^{2}+\\sqrt{a^{2}}} - \\frac{1}{a+\\sqrt{a}}.\n$$\nSetting $F'(a) = 0$, we find that $2a + 2\\sqrt{a} = a + 1$ or $(\\sqrt{a} + 1)^{2} = 2$. We find $\\sqrt{a} = \\pm \\sqrt{2} - 1$, and because $\\sqrt{a} > 0$, $a = (\\sqrt{2} - 1)^{2} = 3 - 2\\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree real numbers $x$, $y$, and $z$ are such that $(x+4)/2 = (y+9)/(z-3) = (x+5)/(z-5)$. Determine the value of $x/y$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nAnswer: $1/2$. Because the first and third fractions are equal, adding their numerators and denominators produces another fraction equal to the others: $((x+4)+(x+5))/(2+(z-5)) = (2x+9)/(z-3)$. Then $y+9 = 2x+9$, etc.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is a cyclic quadrilateral in which $AB = 3$, $BC = 5$, $CD = 6$, and $AD = 10$. $M$, $I$, and $T$ are the feet of the perpendiculars from $D$ to lines $AB$, $AC$, and $BC$ respectively. Determine the value of $MI / IT$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "25/9",
    "solution": "Solution:\n\nAnswer: $\\frac{25}{9}$. Quadrilaterals $AMID$ and $DICT$ are cyclic, having right angles $\\angle AMD$, $\\angle AID$, and $\\angle CID$, $\\angle CTD$ respectively. We see that $M$, $I$, and $T$ are collinear. For, $m \\angle MID = \\pi - m \\angle DAM = \\pi - m \\angle DAB = m \\angle BCD = \\pi - m \\angle DCT = \\pi - m \\angle DIT$. Therefore, Menelaus' theorem applied to triangle $MTB$ and line $ICA$ gives\n$$\n\\frac{MI}{IT} \\cdot \\frac{TC}{CB} \\cdot \\frac{BA}{AM} = 1\n$$\nOn the other hand, triangle $ADM$ is similar to triangle $CDT$ since $\\angle AMD \\cong \\angle CTD$ and $\\angle DAM \\cong \\angle DCT$ and thus $AM / CT = AD / CD$. It follows that\n$$\n\\frac{MI}{IT} = \\frac{BC \\cdot AM}{AB \\cdot CT} = \\frac{BC \\cdot AD}{AB \\cdot CD} = \\frac{5 \\cdot 10}{3 \\cdot 6} = \\frac{25}{9}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence $\\{a_{n}\\}_{n \\geq 0}$ of real numbers satisfies the recursion $a_{n+1} = a_{n}^{3} - 3 a_{n}^{2} + 3$ for all positive integers $n$. For how many values of $a_{0}$ does $a_{2007} = a_{0}$?",
    "options": [],
    "answer": "3^{2007}",
    "solution": "Solution:\nAnswer: $\\mathbf{3}^{\\mathbf{2007}}$. If $x$ appears in the sequence, the next term $x^{3} - 3x^{2} + 3$ is the same if and only if $0 = x^{3} - 3x^{2} - x + 3 = (x-3)(x-1)(x+1)$. Moreover, that next term is strictly larger if $x > 3$ and strictly smaller if $x < -1$. It follows that no values of $a_{0}$ with $|a_{0} - 1| > 2$ yield $a_{0} = a_{2007}$.\n\nNow suppose $a_{0} = a_{2007}$ and write $a_{0} = 1 + e^{\\alpha i} + e^{-\\alpha i}$; the values $a_{0}$ we seek will be in bijective correspondence with solutions $\\alpha$ where $0 \\leq \\alpha \\leq \\pi$. Then\n$$\na_{1} = (a_{0} - 1)^{3} - 3a_{0} + 4 = e^{3\\alpha i} + 3e^{\\alpha i} + 3e^{-\\alpha i} + e^{-3\\alpha i} - 3e^{\\alpha i} - 3e^{-\\alpha i} - 3 + 4 = e^{3\\alpha i} + e^{-3\\alpha i} + 1\n$$\nand an easy inductive argument gives $a_{2007} = e^{3^{2007} \\alpha i} + e^{-3^{2007} \\alpha i} + 1$. It follows that $a_{0} = a_{2007}$ is equivalent to $\\cos(\\alpha) = \\cos(3^{2007} \\alpha)$. Now,\n$$\n\\cos(3^{2007} \\alpha) - \\cos(\\alpha) = 2 \\sin\\left(\\frac{3^{2007} + 1}{2} \\alpha\\right) \\sin\\left(\\frac{3^{2007} - 1}{2} \\alpha\\right)\n$$\nso since $\\sin(kx) = 0$ for a positive integer $k$ if and only if $x$ is a multiple of $\\frac{\\pi}{k}$, the solutions $\\alpha$ are $\\{0, \\frac{2\\pi}{3^{2007} - 1}, \\frac{4\\pi}{3^{2007} - 1}, \\ldots, \\pi\\} \\cup \\{0, \\frac{2\\pi}{3^{2007} + 1}, \\ldots, \\pi\\}$. Because our values $k$ are consecutive, these sets overlap only at $0$ and $\\pi$, so there are $3^{2007}$ distinct $\\alpha$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriangle $A B C$ has $A B=4$, $B C=6$, and $A C=5$. Let $O$ denote the circumcenter of $A B C$. The circle $\\Gamma$ is tangent to and surrounds the circumcircles of triangles $A O B$, $B O C$, and $A O C$. Determine the diameter of $\\Gamma$.",
    "options": [],
    "answer": "256√7/17",
    "solution": "Solution:\n\nDenote by $\\omega$, $\\Gamma_{1}$, $\\Gamma_{2}$, and $\\Gamma_{3}$ the circumcenters of triangles $A B C$, $B O C$, $C O A$, and $A O B$, respectively. An inversion about $\\omega$ interchanges $\\Gamma_{1}$ and line $B C$, $\\Gamma_{2}$ and line $C A$, and $\\Gamma_{3}$ and line $A B$. This inversion also preserves tangency between generalized circles, so the image of $\\Gamma$ is a circle tangent to $A B$, $B C$, and $C A$. It is the incircle of $A B C$ because it is closer to $O$ than these lines and $A B C$ is acute.\n\nNow we run a few standard calculations. Where $s$, $r$, and $R$ denote the semiperimeter, inradius, and circumradius of $A B C$, respectively, we have the following:\n$$\n\\begin{aligned}\n& [A B C]=\\sqrt{s(s-a)(s-b)(s-c)}=\\frac{15 \\sqrt{7}}{4} \\\\\n& r=[A B C] / s=\\sqrt{7} / 2 \\\\\n& R=\\frac{a b c}{4[A B C]}=\\frac{8}{\\sqrt{7}} \\\\\n& O I^{2}=R(R-2 r)=\\frac{8}{7}\n\\end{aligned}\n$$\nLet $O I$ intersect the incircle of $A B C$ at $P$ and $Q$, with $I$ between $P$ and $O$. Then $O P=r+O I$ and $O Q=r-O I$, and $\\overline{P Q}$ is a diameter. Under the inversion, $P$ and $Q$ map to $P^{\\prime}$ and $Q^{\\prime}$ respectively. Because $P$, $I$, $O$, and $Q$ are collinear in that order, $P^{\\prime}$ and $Q^{\\prime}$ are diametrically opposed on $\\Gamma$. It follows that the diameter of $\\Gamma$ is\n$$\nP^{\\prime} Q^{\\prime}=O P^{\\prime}+O Q^{\\prime}=\\frac{R^{2}}{O P}+\\frac{R^{2}}{O Q}=R^{2}\\left(\\frac{1}{r+O I}+\\frac{1}{r-O I}\\right)=\\frac{2 r R^{2}}{r^{2}-O I^{2}}.\n$$\nWe plug in the values found above to arrive at $\\frac{256 \\sqrt{7}}{17}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\int_{1}^{2} \\frac{9x+4}{x^{5}+3x^{2}+x} \\, dx\n$$\n(No, your TI-89 doesn't know how to do this one. Yes, the end is near.)",
    "options": [],
    "answer": "ln(80/23)",
    "solution": "Solution:\nAnswer: $\\ln \\frac{80}{23}$. We break the given integral into two pieces:\n$$\n\\int_{1}^{2} \\frac{9x+4}{x^{5}+3x^{2}+x} \\, dx = 5 \\int_{1}^{2} \\frac{x^{4}+3x+1}{x^{5}+3x^{2}+x} \\, dx - \\int_{1}^{2} \\frac{5x^{4}+6x+1}{x^{5}+3x^{2}+x} \\, dx\n$$\nThese two new integrals are easily computed; for, the first integrand reduces to $1/x$ and the second is of the form $f'(x)/f(x)$. We obtain\n$$\n\\left[5 \\ln |x| - \\ln \\left|x^{5}+3x^{2}+x\\right|\\right]_{1}^{2} = \\ln 32 - \\ln 46 + \\ln 5 = \\ln \\frac{80}{23}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEric and Greg are watching their new favorite TV show, The Price is Right. Bob Barker recently raised the intellectual level of his program, and he begins the latest installment with bidding on the following question: How many Carmichael numbers are there less than 100,000?\n\nEach team is to list one nonnegative integer not greater than 100,000. Let $X$ denote the answer to Bob's question. The teams listing $N$, a maximal bid (of those submitted) not greater than $X$, will receive $N$ points, and all other teams will neither receive nor lose points. (A Carmichael number is an odd composite integer $n$ such that $n$ divides $a^{n-1}-1$ for all integers $a$ relatively prime to $n$ with $1<a<n$.)",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\n16? There are 16 such numbers: 561, 1105, 1729, 2465, 2821, 6601, 8911, 10585, 15841, 29341, 41041, 46657, 52633, 62745, 63973, and 75361. The next, 101101, is too large to be counted. Their distribution is considerably more subtle than that of the primes, and it was only recently proven that there are infinitely many Carmichael numbers. For sufficiently large $n$, $C(n) > n^{2/7}$, although this bound has been subsequently improved. (For details, see Alford, W. R.; Granville, A.; and Pomerance, C. \"There are Infinitely Many Carmichael Numbers.\" Annals of Mathematics. 139 (1994), 703-722.) The expectation is that teams are unable to determine that $X=16$; otherwise, the obvious dominant play is listing 16. The Problem Czar, anticipating that teams will attempt to deduce $X$ by considering the point values of the other problems in the triplet, has chosen a value $X$ that is considerably lower. Teams may of course speculate whether this action has been taken, and to what extent, etc... On the actual TV show, many contestants win by guessing prices of 1, or other numbers dramatically lower than the actual price. This strategy is enhanced because of the show's ordered bidding, and will be more difficult here. It will be interesting to see the submissions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are thirteen broken computers situated at the following set $S$ of thirteen points in the plane:\n$$\n\\begin{array}{lll}\nA=(1,10) & B=(976,9) & C=(666,87) \\\\\nD=(377,422) & E=(535,488) & F=(775,488) \\\\\nG=(941,500) & H=(225,583) & I=(388,696) \\\\\nJ=(3,713) & K=(504,872) & L=(560,934) \\\\\n& M=(22,997) &\n\\end{array}\n$$\nAt time $t=0$, a repairman begins moving from one computer to the next, traveling continuously in straight lines at unit speed. Assuming the repairman begins at $A$ and fixes computers instantly, what path does he take to minimize the total downtime of the computers? List the points he visits in order. Your score will be $\\left\\lfloor\\frac{N}{40}\\right\\rfloor$, where\n$$\nN=1000+\\lfloor\\text{ the optimal downtime }\\rfloor-\\lfloor\\text{ your downtime }\\rfloor,\n$$\nor $0$, whichever is greater. By total downtime we mean the sum\n$$\n\\sum_{P \\in S} t_{P}\n$$\nwhere $t_{P}$ is the time at which the repairman reaches $P$.",
    "options": [],
    "answer": "ADHIKLEFGBCJM",
    "solution": "Solution:\n\nAnswer: ADHIKLEFGBCJM. This is an instance of the minimum-latency problem, which is at least NP-hard. There is an easy $O(n!)$ algorithm, but this is unavailable to teams on computational grounds (100 MHz calculators used to seem fast...) The best strategy may be drawing an accurate picture and exercising geometric intuition. The distribution of the points somewhat resembles a short, four-pronged fork with its outermost prongs bent apart; it is plausible to assume that the optimal order respects this shape. The optimal downtime is $24113.147907$, realized by ADHIKLEFGBCJM, though a number of others also receive positive marks.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $\\omega$ denote the incircle of triangle $ABC$. The segments $BC$, $CA$, and $AB$ are tangent to $\\omega$ at $D$, $E$, and $F$, respectively. Point $P$ lies on $EF$ such that segment $PD$ is perpendicular to $BC$. The line $AP$ intersects $BC$ at $Q$. The circles $\\omega_1$ and $\\omega_2$ pass through $B$ and $C$, respectively, and are tangent to $AQ$ at $Q$; the former meets $AB$ again at $X$, and the latter meets $AC$ again at $Y$. The line $XY$ intersects $BC$ at $Z$. Given that $AB=15$, $BC=14$, and $CA=13$, find $\\lfloor XZ \\cdot YZ\\rfloor$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "101",
    "solution": "Solution:\n\nAnswer: 101. Construct $D'$ diametrically opposed to $D$, so that $\\angle DFD'$ and $\\angle DED'$ are right, and note that $P$ lies on $DD'$. By standard angle chasing, $m \\angle FDD' = \\beta$ (half angle $B$) and $m \\angle D'DE = \\gamma$. Thus, $m \\angle DD'F = 90^\\circ - \\beta$ and $m \\angle ED'D = 90^\\circ - \\gamma$. Then by the law of sines, $DE : ED' : D'F : FD = \\cos(\\gamma) : \\sin(\\gamma) : \\sin(\\beta) : \\sin(\\gamma)$. Now using $\\triangle DEP \\sim \\triangle FD'P$ and $\\triangle DFP \\sim \\triangle ED'P$, we have\n$$\n\\frac{EP}{PF} = \\frac{ED \\cdot ED'}{FD \\cdot FD'} = \\frac{\\sin(\\gamma) \\cos(\\gamma)}{\\sin(\\beta) \\sin(\\beta)} = \\frac{c}{b}\n$$\nLet the dilation centered at $A$ sending $E$ to $C$ map $P$ and $F$ to $P'$ and $F'$, respectively. Note that $AF' = AC$ as $AE$ and $AF$ are equal tangents, and $CP' : P'F' = EP : PF = c : b$ by similarity. Then by Menelaus' theorem,\n$$\n1 = \\frac{BQ}{QC} \\cdot \\frac{CP'}{P'F'} \\cdot \\frac{F'A}{AB} = \\frac{BQ}{QC} \\cdot \\frac{c}{b} \\cdot \\frac{b}{c}\n$$\nso that $BQ = QC$ and $AQ$ is actually a median. So, $AQ^2 = \\frac{1}{4}(2b^2 + 2c^2 - a^2) = 148$. Now by Power of a Point, $AB \\cdot AX = AQ^2 = AC \\cdot AY$, so $AX = 148/15$ and $AY = 148/13$. Moreover, $B X C Y$ is cyclic as $\\triangle ABC \\sim \\triangle AYX$. Thus, $XZ \\cdot YZ = BZ \\cdot CZ$, and it suffices to compute $BZ / CZ$. Menelaus once more gives\n$$\n1 = \\frac{BZ}{ZC} \\cdot \\frac{CY}{YA} \\cdot \\frac{AX}{XB}\n$$\nwhence, $BZ / CZ = (AY / AX)(BX / CY) = (15/13)((77 \\cdot 13)/(21 \\cdot 15)) = 11/3$. We write $CZ = 3d$ and $BZ = 11d$. Because $AX < AB$ and $AY < AC$, $Z$ does not lie on segment $BC$. Given the configuration information, $BC = 8d = 14$, so $d = 7/4$, and finally $\\lfloor BZ \\cdot CZ \\rfloor = \\left\\lfloor 33d^2 \\right\\rfloor = \\left\\lfloor \\frac{1617}{16} \\right\\rfloor = 101$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether or not there exist positive integers $a$ and $b$ such that $a$ does not divide $b^n - n$ for all positive integers $n$.",
    "options": [],
    "answer": "No. For any positive integers a and b, there exist infinitely many positive integers n such that a divides b^n − n.",
    "solution": "**Solution 1.** For all pairs of positive integers $a$ and $b$, we claim that there exist infinitely many $n$ such that $a$ divides $b^n - n$.\nWe establish our claim by strong induction on $a$. The base case of $a = 1$ holds trivially. Now, suppose that the claim holds for all $a < a_0$. Since $\\varphi(a) < a$, by the induction hypothesis and by Lemma 1, there are infinitely many $n$ such that\n$$\n\\varphi(a) \\mid (b^n - n) \\quad \\text{and} \\quad b^{n+\\varphi(a)} \\equiv b^n \\pmod{a}.\n$$\nFor each such $n$, set\n$$\nt = \\frac{b^n - n}{\\varphi(a)} \\quad \\text{and} \\quad n_1 = b^n = n + t\\varphi(a).\n$$\n\nIt follows that\n$$\nb^{n_1} - n_1 \\equiv b^{n+t\\varphi(a)} - (n+t\\varphi(a)) \\equiv b^n - n - t\\varphi(a) \\equiv 0 \\pmod{a}.\n$$\nThen, we see that $n_1$ satisfies the desired property. By the induction hypothesis, there are infinitely many $n_1 = b^n$ satisfying the conditions of the claim for $a$, completing the induction.\n\n\n**Solution 2.** We prove that no such $a, b$ exist by proving the following: for any $a, b$, there is an arithmetic progression $n \\equiv h \\pmod m$, with $m$ divisible only by primes less than or equal to the greatest prime factor of $a$, such that $b^n \\equiv n \\pmod a$ for all sufficiently large $n$ satisfying $n \\equiv h \\pmod m$.\nLet us induct on highest prime divisor of $a$. The result is trivial for $a = 1$. Let $p$ be a prime, and suppose that the result is true whenever all the prime divisors of $a$ are less than $p$. Now, suppose that $p$ is the greatest prime divisor of some $a$, and write $a = p^e a_1$, where $a_1$ has all prime factors less than $p$. By the induction hypothesis, there is an arithmetic progression $n \\equiv h_1 \\pmod{m_1}$, with $m_1$ divisible only by primes strictly less than $p$, such that for $n \\equiv h_1 \\pmod{m_1}$ sufficiently large, $b^n \\equiv n \\pmod{a_1}$. There is no harm in assuming that $p-1$ divides $m_1$. In this case, in this arithmetic progression, $b^n$ is eventually constant modulo $p$ due to the lemma. We can thus choose a congruence modulo $p$ so that for $n$ an appropriate residue class modulo $m_1p$, $b^n \\equiv n \\pmod{p}$. In this progression, $b^n$ is constant modulo $p^2$, so we can refine our choice of $n$ modulo $m_1p$ to a choice of $n$ modulo $m_1p^2$ to force $b^n \\equiv n \\pmod{p^2}$. We can then repeat the above process until we obtain $b^n \\equiv n \\pmod{p^e}$. Since we originally had $b^n \\equiv n \\pmod{a_1}$, combining the two congruences using the Chinese Remainder Theorem gives us $b^n \\equiv n \\pmod a$ for all sufficiently large $n$ in the congruence class generated at the last step. This completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $N$ be the number of ordered pairs $(x, y)$ of integers such that\n$$\nx^2 + x y + y^2 \\leq 2007\n$$\nRemember, integers may be positive, negative, or zero!\n\na. Prove that $N$ is odd.\n\nb. Prove that $N$ is not divisible by $3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nIf $(x, y)$ is a pair of integers that satisfies the inequality, then $(-x, -y)$ is also such a pair, since\n$$\n(-x)^2 + (-x)(-y) + (-y)^2 = x^2 + x y + y^2.\n$$\nSo we can match up pairs of solutions to the inequality, $(x, y) \\leftrightarrow (-x, -y)$. Every solution will be paired with a different solution, except for the one remaining solution $(0, 0)$ which is paired with itself. This shows that the number of solutions is odd.\n\nb.\nThis is similar to the previous part, except now that we have to arrange the nonzero solutions into triples instead of pairs. If $(x, y)$ is a solution to the inequality, then so is $(-x-y, x)$, since\n$$\n(-x-y)^2 + (-x-y)x + x^2 = x^2 + x y + y^2\n$$\nApplying this transformation three times in succession gives the cycle\n$$\n(x, y) \\rightarrow (-x-y, x) \\rightarrow (y, -x-y) \\rightarrow (x, y)\n$$\nso we can unambiguously arrange the solutions into cycles of three, of the form $\\{(x, y), (-x-y, x), (y, -x-y)\\}$. Now, if any two solutions in the same cycle are equal, then the third is also equal to them, so every cycle contains either three distinct solutions or just one solution. If a cycle contains just one solution $(x, y)$, then $x = y$ and $y = -x - y$ gives $x = y = 0$. Therefore, the solution $(0, 0)$ forms a cycle by itself, and every other cycle consists of three different solutions, which means that the total number of solutions has remainder $1$ when divided by $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47335,
    "subject": "Mathematics (Multi-modal)",
    "question": "An *animal* with $n$ cells is a connected figure consisting of $n$ equal-sized square cells. Animals are also called *polyominoes*. They can be defined inductively. Two cells are *adjacent* if they share a complete edge. A single cell is an animal, and given an animal with $n$-cells, one with $n+1$ cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.\n\nA dinosaur is an animal with at least $2007$ cells. It is said to be primitive if its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "8025",
    "solution": "**First Solution:** Let $s$ denote the minimum number of cells in a dinosaur; the number this year is $s = 2007$.\n\n**Claim:** The maximum number of cells in a primitive dinosaur is $4(s - 1) + 1 = 8025$.\n\nFirst, a primitive dinosaur can contain up to $4(s - 1) + 1$ cells. To see this, consider a dinosaur in the form of a cross consisting of a central cell and four arms with $s - 1$ cells apiece. No connected figure with at least $s$ cells can be removed without disconnecting the dinosaur.\n\nThe proof that no dinosaur with at least $4(s - 1) + 2$ cells is primitive relies on the following result.\n\n**Lemma** Let $D$ be a dinosaur having at least $4(s - 1) + 2$ cells, and let $R$ (red) and $B$ (black) be two complementary animals in $D$, i.e., $R \\cap B = \\emptyset$ and $R \\cup B = D$. Suppose $|R| \\le s - 1$. Then $R$ can be augmented to produce animals $\\tilde{R} \\supset R$ and $\\tilde{B} = D \\setminus \\tilde{R}$ such that at least one of the following holds:\n\n(i) $|\\tilde{R}| \\ge s$ and $|\\tilde{B}| \\ge s$,\n\n(ii) $|\\tilde{R}| = |R| + 1$,\n\n(iii) $|R| < |\\tilde{R}| \\le s - 1$.\n\n**Proof:** If there is a black cell adjacent to $R$ that can be made red without disconnecting $B$, then (ii) holds. Otherwise, there is a black cell $c$ adjacent to $R$ whose removal disconnects $B$. Of the squares adjacent to $c$, at least one is red, and at least one is black, otherwise $B$ would be disconnected. Then there are at most three resulting components $C_1, C_2, C_3$ of $B$ after the removal of $c$. Without loss of generality, $C_3$ is the largest of the remaining components. (Note that $C_1$ or $C_2$ may be empty.) Now $C_3$ has at least $\\lceil(3s - 2)/3\\rceil = s$ cells. Let $\\tilde{B} = C_3$. Then $|\\tilde{R}| = |R| + |C_1| + |C_2| + 1$. If $|\\tilde{B}| \\le 3s - 2$, then $|\\tilde{R}| \\ge s$ and (i) holds. If $|\\tilde{B}| \\ge 3s - 1$ then either (ii) or (iii) holds, depending on whether $|\\tilde{R}| \\ge s$ or not. \n\nStarting with $|R| = 1$, repeatedly apply the Lemma. Because in alternatives (ii) and (iii) $|R|$ increases but remains less than $s$, alternative (i) eventually must occur. This shows that no dinosaur with at least $4(s - 1) + 2$ cells is primitive.\n\n\n**Second Solution:** (Based on Andrew Geng's solution) Let $s = 2007$. We claim that the answer is $4s - 3 = 8025$.\n\nConsider a graph with the cells as the vertices and whose edges connect adjacent cells. Let $T$ be a spanning tree in this graph. By removing any vertex of $T$, we obtain at most four connected components, which we call the *limbs* of the vertex. Limbs with at least $s$ vertices are called *big*. Suppose that every vertex of $T$ contains a big limb, then consider a walk on $T$ starting from an arbitrary vertex and always moving along the edge towards a big limb. Since $T$ is a finite tree, this walk must traverse back on some edge at some point. Then the two connected components of $T$ made by deleting this edge are both big, so they both contain at least $s$ vertices, which means that the dinosaur is not primitive. It follows that a primitive dinosaur contains some vertex with no big limbs. By removing this vertex, we get at most four connected components with at most $s - 1$ vertices each. This not only shows that a primitive dinosaur has at most $4s - 3$ cells, but also shows that any such dinosaur consists of four limbs of $s-1$ cells each connected to a central cell. It is easy to see that such a dinosaur indeed exists.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThomas and Michael are just two people in a large pool of well qualified candidates for appointment to a problem writing committee for a prestigious college math contest. It is 40 times more likely that both will serve if the size of the committee is increased from its traditional 3 members to a whopping $n$ members. Determine $n$. (Each person in the pool is equally likely to be chosen.)",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\nAnswer: 16. Suppose there are $k$ candidates. Then the probability that both serve on a 3 membered committee is $(k-2)/\\binom{k}{3}$, and the odds that both serve on an $n$ membered committee are $\\binom{k-2}{n-2}/\\binom{k}{n}$. The ratio of the latter to the former is\n$$\n\\frac{\\binom{k}{3}\\binom{k-2}{n-2}}{(k-2)\\binom{k}{n}} = \\frac{k!(k-2)!1!(k-3)!n!(k-n)!}{k!(k-2)!(n-2)!(k-n)!3!(k-3)!} = \\frac{n \\cdot (n-1)}{3!}\n$$\nSolving $n \\cdot (n-1) = 240$ produces $n = 16, -15$, and we discard the latter.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn the Cartesian grid, Johnny wants to travel from $(0,0)$ to $(5,1)$, and he wants to pass through all twelve points in the set $S=\\{(i, j) \\mid 0 \\leq i \\leq 1, 0 \\leq j \\leq 5, i, j \\in \\mathbb{Z}\\}$. Each step, Johnny may go from one point in $S$ to another point in $S$ by a line segment connecting the two points. How many ways are there for Johnny to start at $(0,0)$ and end at $(5,1)$ so that he never crosses his own path?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "252",
    "solution": "Solution:\n\nAnswer: 252. Observe that Johnny needs to pass through the points $(0,0), (1,0), (2,0), \\ldots, (5,0)$ in that order, and he needs to pass through $(0,1), (1,1), (2,1), \\ldots, (5,1)$ in that order, or else he will intersect his own path. Then, the problem is equivalent to interlacing those two sequences together, so that the first term is $(0,0)$ and the final term is $(5,1)$. To do this, we need to select 5 positions out of 10 to have points with $x$-coordinate 0. Hence the answer is $\\binom{10}{5} = 252$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\frac{2^{3}-1}{2^{3}+1} \\cdot \\frac{3^{3}-1}{3^{3}+1} \\cdot \\frac{4^{3}-1}{4^{3}+1} \\cdot \\frac{5^{3}-1}{5^{3}+1} \\cdot \\frac{6^{3}-1}{6^{3}+1}\n$$",
    "options": [],
    "answer": "43/63",
    "solution": "Solution:\nAnswer: $\\frac{43}{63}$\n\nUse the factorizations $n^{3}-1=(n-1)\\left(n^{2}+n+1\\right)$ and $n^{3}+1=(n+1)\\left(n^{2}-n+1\\right)$ to write\n$$\n\\frac{1 \\cdot 7}{3 \\cdot 3} \\cdot \\frac{2 \\cdot 13}{4 \\cdot 7} \\cdot \\frac{3 \\cdot 21}{5 \\cdot 13} \\cdot \\frac{4 \\cdot 31}{6 \\cdot 21} \\cdot \\frac{5 \\cdot 43}{7 \\cdot 31} = \\frac{1 \\cdot 2 \\cdot 43}{3 \\cdot 6 \\cdot 7} = \\frac{43}{63}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$ and $b$ be integer solutions to $17 a + 6 b = 13$. What is the smallest possible positive value for $a - b$?",
    "options": [],
    "answer": "17",
    "solution": "Solution:\n\nFirst group as $17(a-b) + 23b = 13$. Taking this equation modulo $23$, we get $-6(a-b) \\equiv -10 \\pmod{23}$. Since $-4$ is an inverse of $-6$ modulo $23$, then we multiply to get $(a-b) \\equiv 17 \\pmod{23}$. Therefore, the smallest possible positive value for $(a-b)$ is $17$. This can be satisfied by $a = 5$, $b = -12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the real number $\\alpha$ such that the curve $f(x)=e^{x}$ is tangent to the curve $g(x)=\\alpha x^{2}$.",
    "options": [],
    "answer": "e^2/4",
    "solution": "Solution:\n$\\boxed{e^{2}/4}$. Suppose tangency occurs at $x=x_{0}$. Then $e^{x_{0}}=\\alpha x_{0}^{2}$ and $f'(x_{0})=2 \\alpha x_{0}$. On the other hand, $f'(x)=f(x)$, so $\\alpha x_{0}^{2}=2 \\alpha x_{0}$. Clearly, $\\alpha=0$ and $x_{0}=0$ are impossible, so it must be that $x_{0}=2$. Then $\\alpha=e^{x_{0}}/(x_{0}^{2})=e^{2}/4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence consists of the digits $122333444455555 \\ldots$ such that each positive integer $n$ is repeated $n$ times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence.",
    "options": [],
    "answer": "13",
    "solution": "Solution:\nAnswer: 13. Note that $n$ contributes $n \\cdot d(n)$ digits, where $d(n)$ is the number of digits of $n$. Then because $1+\\cdots+99=4950$, we know that the digits of interest appear amongst copies of two digit numbers. Now for $10 \\leq n \\leq 99$, the number of digits in the subsequence up to the last copy of $n$ is\n$$\n1+2+3+\\cdots+9+2 \\cdot(10+\\cdots+n)=2 \\cdot(1+\\cdots+n)-45=n^{2}+n-45\n$$\nSince $67^{2}+67-45=4511$, the two digits are 6 and 7 in some order, so have sum 13.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree brothers Abel, Banach, and Gauss each have portable music players that can share music with each other. Initially, Abel has $9$ songs, Banach has $6$ songs, and Gauss has $3$ songs, and none of these songs are the same. One day, Abel flips a coin to randomly choose one of his brothers and he adds all of that brother's songs to his collection. The next day, Banach flips a coin to randomly choose one of his brothers and he adds all of that brother's collection of songs to his collection. Finally, each brother randomly plays a song from his collection with each song in his collection being equally likely to be chosen. What is the probability that they all play the same song?",
    "options": [],
    "answer": "1/288",
    "solution": "Solution:\n\nIf Abel copies Banach's songs, this can never happen. Therefore, we consider only the cases where Abel copies Gauss's songs. Since all brothers have Gauss's set of songs, the probability that they play the same song is equivalent to the probability that they independently match whichever song Gauss chooses.\n\nCase 1: Abel copies Gauss and Banach copies Gauss ($1/4$ chance) - The probability of songs matching is then $1/12 \\cdot 1/9$.\n\nCase 2: Abel copies Gauss and Banach copies Abel ($1/4$ probability) - The probability of songs matching is then $1/12 \\cdot 1/18$.\n\nWe add the two probabilities together to get $1/4 \\cdot 1/12 \\cdot (1/9 + 1/18) = 1/288$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo sequences of positive integers, $x_{1}, x_{2}, x_{3}, \\ldots$ and $y_{1}, y_{2}, y_{3}, \\ldots$, are given, such that\n$$\ny_{n+1} / x_{n+1}>y_{n} / x_{n}\n$$\nfor each $n \\geq 1$. Prove that there are infinitely many values of $n$ such that $y_{n}>\\sqrt{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose the statement is false. So there are only finitely many values for which $y_{n}>\\sqrt{n}$; suppose there are $R_{1}$ such values. Let $m$ be the largest integer such that $x_{n} / y_{n} \\geq m$ for all $n$ (this is possible since every $x_{n} / y_{n}>0$; note that it may be the case that $m=0$ ). We have $x_{n} / y_{n}<m+1$ for some $n$, say $n=R_{2}$, and since the sequence $\\left(x_{n} / y_{n}\\right)$ is decreasing, we then get $x_{n} / y_{n}<m+1$ for all $n \\geq R_{2}$. Letting $R=R_{1}+R_{2}$, we obtain\n$$\ny_{n} \\leq \\sqrt{n} \\quad \\text{ and } \\quad m \\leq x_{n} / y_{n}<m+1\n$$\nfor all $n$, with at most $R$ possible exceptions.\n\nNow fix any positive integer $N$. The pairs of positive integers $\\left(x_{n}, y_{n}\\right)$ for $n<N^{2}$ are all distinct, since the corresponding values of $y_{n} / x_{n}$ are strictly increasing; and except for at most $R$ of them, the remaining ones all satisfy\n$$\ny_{n} \\leq \\sqrt{n}<N\n$$\nand\n$$\nm y_{n} \\leq x_{n}<m y_{n}+y_{n} .\n$$\nSo we have $N-1$ possible values for $y_{n}$ (namely $1,2, \\ldots, N-1$ ), and for each such value, we have $y_{n}$ choices for $x_{n}$ (namely $m y_{n}, m y_{n}+1, \\ldots, m y_{n}+y_{n}-1$ ), giving $1+2+\\cdots+(N-1)=$ $N(N-1) / 2$ possible pairs obtained in this way. Hence, counting all the pairs $\\left(x_{n}, y_{n}\\right)$ for $n<N^{2}$, we have\n$$\nN^{2}-1-R \\leq N(N-1) / 2 .\n$$\nBut since $R$ is fixed, clearly this inequality will become false for large enough $N$. At this point we have a contradiction, and the problem is solved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ is inscribed in circle $\\omega$. The tangent lines to $\\omega$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS \\perp AT$. Points $B_1$ and $C_1$ lie on ray $ST$ (with $C_1$ in between $B_1$ and $S$) such that $B_1T = BT = C_1T$. Prove that triangles $ABC$ and $AB_1C_1$ are similar to each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution 1.** (Based on work by Oleg Golberg) We start with an important geometric observation.\n\n**Lemma.** Triangle $ABC$ inscribed in circle $\\omega$. Lines $BT$ and $CT$ are tangent to $\\omega$. Let $M$ be the midpoint of side $BC$. Then $\\angle BAT = \\angle CAM$. (Line $AT$ is a symmedian of triangle.)\n\n*Proof.* We consider the above configuration. (If $\\angle BAC$ is obtuse, our proof can be modified slightly.) Let $D$ denote the second intersection (other than $A$) of line $AT$ and circle $\\omega$. Because $BT$ is tangent to $\\omega$ at $B$, $\\angle TBD = \\angle TAB$. Hence triangles $TBD$ and $TAB$ are similar, implying that $BD/AB = TB/TA$. Likewise, triangles $TCD$ and $TAC$ are similar and $CD/AC = TC/TA$. By equal tangents, $TB = TC$. Consequently, we have $BD/AB = TB/TA = TC/TA = CD/AC$, implying that\n$$\nBD \\cdot AC = CD \\cdot AB.\n$$\nBy the **Ptolemy's theorem** to cyclic quadrilateral $ABDC$, we have $BD \\cdot AC + AB \\cdot CD = AD \\cdot BC$.\nCombining the last two equations, we obtain that $2 BD \\cdot AC = AD \\cdot BC$ or\n$$\n\\frac{AC}{AD} = \\frac{BC}{2BD} = \\frac{MC}{BD}.\n$$\nFurther considering that $\\angle ACM = \\angle ACB = \\angle ADB$ (since $ABDC$ is cyclic), we conclude that\ntriangle $ABD$ is similar to triangle $AMC$, implying that $\\angle BAT = \\angle BAD = \\angle CAM$. $\\square$\n\nBecause $BT$ is tangent to $\\omega$, $\\angle CBT = \\angle CAB$, and so\n$$\n\\angle TBA = \\angle ABC + \\angle CBT = \\angle ABC + \\angle CAB = 180^\\circ - \\angle BCA.\n$$\nBy the lemma, we have $\\angle BAT = \\angle CAM$. Applying the Law of Sines to triangles $BAT$ and $CAM$, we obtain\n$$\n\\frac{BT}{AT} = \\frac{\\sin \\angle BAT}{\\sin \\angle TBA} = \\frac{\\sin \\angle CAM}{\\sin \\angle BCA} = \\frac{MC}{AM}.\n$$\nNote that $TB = TC_1$. Thus, $TC_1/TA = MC/MA$. By equal tangents, $TB = TC$. In isosceles triangle $BTC$, $M$ is the midpoint of base $BC$. Consequently, $\\angle TMS = \\angle TAC = \\angle TAS = 90^\\circ$, implying that $TMAP$ is cyclic. Hence $\\angle AMC = \\angle ATC_1$. Because\n$$\n\\frac{AM}{AT} = \\frac{MC}{TC_1} \\quad (*)\n$$\nand $\\angle AMC = \\angle ATC_1$, triangles $MAC$ and $TAC_1$ are similar. Because $BC/BM = B_1C_1/TC_1 = 2$, triangles $ABC$ and $AB_1C_1$ are similar.\n\n\n**Solution 2.** (By Alex Zhai) We maintain the notations in the first proof. As shown at the end of the first proof, it suffices to show that $(*)$.\n\n![](attached_image_1.png)\n\nLet $O$ be the circumcenter of $ABC$. Note that triangles $OMC$, $OCT$ are similar to each other, implying that $OM/OC = OC/OT$ or $OM \\cdot OT = OC^2 = OA^2$. Thus triangles $OAM$ and $OTA$ are also similar to each other. Further note that triangles $OMC$ and $CMT$ are also similar to each other. These similarities (which amount to the circumcircle of $ABC$ being a circle of Apollonius) give\n$$\n\\frac{AM}{AT} = \\frac{OM}{OA} = \\frac{OM}{OC} = \\frac{MC}{CT} = \\frac{MC}{TB} = \\frac{MC}{TC_1},\n$$\nwhich is $(*)$.\n\n\n**Solution 3.** (Based on work by Sherry Gong) We maintain the notations of the previous solutions. Let $\\omega$ intersect lines $AT$ and $AS$ again at $X$ and $Y$ (other than $A$), respectively. Let lines $YB$ and $CX$ meet at $B_2$, and let $YC$ and $BX$ meet at $C_2$. Applying Pascal's theorem to cyclic (degenerated) hexagon $BBYAXC$ shows that intersections of three pairs of lines $BB$ and $AX$, $BY$ and $XC$, and $YA$ and $CB$ are collinear; that is, $B_2, C_2, S$ are collinear. Likewise, applying Pascal's theorem to cyclic (degenerated) hexagon $CCYAXB$ shows that $B_2, C_2, T$ are collinear. We conclude that $B_2, C_2, S, T$ are collinear.\n\n![](attached_image_2.png)\n\nSince $ACXY$ is cyclic, $\\angle YCX = \\angle YAX = 180^\\circ - \\angle XAB = 90^\\circ$. Thus $\\angle B_2CC_2 = \\angle XCC_2 = 180^\\circ - \\angle YCX = 90^\\circ$. Likewise, $\\angle C_2BB_2 = 90^\\circ$. It follows that $BCC_2B_2$ is inscribed in a circle with $B_2C_2$ as its diameter. Thus the circumcenter of this circle is the intersection of lines $ST$ and the perpendicular of segment $BC$. This circumcenter must be $T$, and consequently, $B_2 = B_1$ and $C_2 = C_1$.\n\n![](attached_image_3.png)\n\nBecause $AYBC$ and $B_1C_1CB$ are cyclic, by Miquel's theorem, $SACC_1$. (Indeed, $\\angle ACB = 180^\\circ - \\angle B_1YS$ and $\\angle BCC_1 = 180^\\circ - \\angle YB_1S$ lead to $\\angle ACC_1 = 360^\\circ - \\angle ACB - \\angle BCC_1 = 180^\\circ - YSB_1$.) Also, by Miquel's theorem, $YAC_1B_1$ is cyclic. (Indeed, $\\angle C_1AX = \\angle C_1CS = \\angle SB_1Y$.) By these cyclic quadrilaterals, it is not difficult to obtain $\\angle ACS = \\angle AC_1S$ (or $\\angle ACB = \\angle AC_1B_1$) and $\\angle ABC = \\angle AYC = \\angle AYC_1 = \\angle AB_1C_1$. Consequently, triangles $ABC$ and $AB_1C_1$ are similar to each other.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers. Show that if $4ab - 1$ divides $(4a^2 - 1)^2$, then $a = b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Call a pair $(a, b)$ of positive integers *bad* if $4ab - 1$ divides $(4a^2 - 1)^2$ but $a \\neq b$. In order to prove that bad pairs do not exist, we present two properties of them which provide an infinite descent.\n\n*   **Property (i):** If $(a, b)$ is a bad pair then $(b, a)$ is also bad.\nIndeed, we consider congruence modulo $4ab - 1$. Since $1 = 1^2 \\equiv (4ab)^2 \\pmod{4ab - 1}$, we have\n$$\n(4b^2 - 1)^2 \\equiv (4b^2 - (4ab)^2)^2 = 16b^4(4a^2 - 1)^2 \\equiv 0 \\pmod{4ab - 1}.\n$$\nHence, the number $4ab - 1$ divides $(4b^2 - 1)^2$ as well.\n\n*   **Property (ii):** If $(a, b)$ is a bad pair and $a < b$, then there exists a positive integer $c < a$ such that $(a, c)$ is also bad.\nIndeed, let\n$$\nr = \\frac{(4a^2 - 1)^2}{4ab - 1}.\n$$\nThen $r$ is an integer. Considering congruence modulo $4a$ yields\n$$\nr = -r \\cdot (-1) \\equiv -r(4ab - 1) = -(4a^2 - 1)^2 \\equiv -1 \\pmod{4a}\n$$\nthat is, $r = 4ac - 1$ for some positive integer $c$. Since $a < b$ we obtain that\n$$\n4ac - 1 = \\frac{(4a^2 - 1)^2}{4ab - 1} < 4a^2 - 1\n$$\nand therefore $c < a$. By the construction, the number $r = 4ac - 1$ is a divisor of $(4a^2 - 1)^2$, so $(a, c)$ is a bad pair.\n\nNow suppose that there exists at least one bad pair. Take a bad pair $(a, b)$ such that $2a + b$ attains its smallest possible value. If $a < b$ then Property (ii) provides a bad pair $(a, c)$ with $c < b$ and thus $2a + c < 2a + b$, violating the minimality assumption on $(a, b)$. Otherwise, if $b < a$, Property (i) yields that pair $(b, a)$ is also bad while $2b + a < 2a + b$, again violating the minimality assumption on $(a, b)$. In either case, we reach a contradiction. Hence our assumption was wrong and there are no bad pairs.\nDefine *bad* pair as in the first solution. We present two new properties of bad pairs. Assume that $(a, b)$ is a bad pair.\n\n*   **Property (iii):** $(a, b)$ is a bad pair if and only if $4ab - 1$ divides $(a - b)^2$.\nConsidering modulo $4ab - 1$, we observe that\n$$\n(4a^2 - 1)^2 \\equiv (4a^2 - 4ab)^2 \\equiv 16a^2(a - b)^2 \\pmod{4ab - 1}.\n$$\nSince $\\gcd(4a, 4ab - 1) = 1$, it follows that $4ab - 1$ divides $(4a^2 - 1)^2$ if and only if $4ab - 1$ divides $(a - b)^2$ (which is a symmetric expression in $a$ and $b$). Note that this implies that $(a, b)$ is a bad pair if and only if $(b, a)$ is bad.\n\n* Property (iv): If $(a, b)$ is a bad pair, then there exists a positive integer $c \\neq b$ such that $(b, c)$ is also bad.\nBy Property (iii), we can write $(a-b)^2 = m(4ab-1)$ for some positive integer $m$. We consider the quadratic\n$$\nf(x) = (x-b)^2 - m(4bx-1) = x^2 - 2(2m+1)bx + b^2 + m.\n$$\nLet $x_1$ and $x_2$ be zeros of $f$. Since $f(a) = 0$, we write $x_1 = a$. Note that $x_1x_2 = b^2 + m$, which is positive. Hence $x_2$ is positive. Note also that $x_1 + x_2 = 2(2m+1)b$, which is an integer. Thus $x_2 = c$ is also a positive integer. Finally, we note that $c \\neq b$ since $f(b) = -m(4b^2-1) < 0$. Thus $c$ and $b$ are distinct positive integers such that $(c-b)^2 = m(4cb-1)$. By Property (iii), $(c, b)$ is a bad pair.\n\nBy Property (iii), we may assume that $a > b$. We further assume that $a$ is minimal. By properties (iii) and (iv), we obtain another bad pair $(b, c)$. Note that for the quadratic $f$ with positive leading coefficient, we have the following facts: $f(a) = f(c) = 0$, $f(b) < 0$, and $a > b$. We conclude that $c < b < a$. For bad pair $(b, c)$, we have $b > c$. But $b < a$ violates the minimality assumption on $a$. Hence our assumption was wrong and there is no bad pair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every nonnegative integer $n$, the number $7^{7^n} + 1$ is the product of at least $2n+3$ (not necessarily distinct) primes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The proof is by induction. The base is provided by the $n=0$ case, where $7^{7^0} + 1 = 7^1 + 1 = 2^3$.\n\nTo prove the inductive step, it suffices to show that if $x = 7^{2^{m-1}}$ for some positive integer $m$ then $(x^7+1)/(x+1)$ is composite. As a consequence, $x^7+1$ has at least two more prime factors than does $x+1$.\n\nTo confirm that $(x^7+1)/(x+1)$ is composite, observe that\n$$\n\\begin{aligned}\n\\frac{x^7+1}{x+1} &= \\frac{(x+1)^7 - ((x+1)^7 - (x^7+1))}{x+1} \\\\\n&= (x+1)^6 - \\frac{7x(x^5 + 3x^4 + 5x^3 + 5x^2 + 3x + 1)}{x+1} \\\\\n&= (x+1)^6 - 7x(x^4 + 2x^3 + 3x^2 + 2x + 1) \\\\\n&= (x+1)^6 - 7^{2m}(x^2 + x + 1)^2 \\\\\n&= \\left[(x+1)^3 - 7^m(x^2 + x + 1)\\right]\\left[(x+1)^3 + 7^m(x^2 + x + 1)\\right]\n\\end{aligned}\n$$\nAlso each factor exceeds 1. It suffices to check the smaller one; $\\sqrt{7x} \\le x$ gives\n$$\n\\begin{aligned}\n(x+1)^3 - 7^m(x^2+x+1) &= (x+1)^3 - \\sqrt{7x}(x^2+x+1) \\\\\n&\\geq x^3 + 3x^2 + 3x + 1 - x(x^2+x+1) \\\\\n&= 2x^2 + 2x + 1 \\geq 113 > 1.\n\\end{aligned}\n$$\nHence $(x^7 + 1)/(x + 1)$ is composite and the proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFive marbles of various sizes are placed in a conical funnel. Each marble is in contact with the adjacent marble(s). Also, each marble is in contact all around the funnel wall. The smallest marble has a radius of $8$, and the largest marble has a radius of $18$. What is the radius of the middle marble?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nAnswer: $12$. One can either go through all of the algebra, find the slope of the funnel wall and go from there to figure out the radius of the middle marble. Or one can notice that the answer will just be the geometric mean of $18$ and $8$ which is $12$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute the largest positive integer such that $\\frac{2007!}{2007^{n}}$ is an integer.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nAnswer: 9. Note that $2007 = 3^{2} \\cdot 223$. Using the fact that the number of times a prime $p$ divides $n!$ is given by\n$$\n\\left\\lfloor\\frac{n}{p}\\right\\rfloor + \\left\\lfloor\\frac{n}{p^{2}}\\right\\rfloor + \\left\\lfloor\\frac{n}{p^{3}}\\right\\rfloor + \\cdots\n$$\nit follows that the answer is 9.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the number of ways to select a positive number of squares on an $8 \\times 8$ chessboard such that no two lie in the same row or the same column and no chosen square lies to the left of and below another chosen square.",
    "options": [],
    "answer": "12869",
    "solution": "Solution:\n\nAnswer: $12869 = \\binom{16}{8} - 1$. If $k$ is the number of squares chosen, then there are $\\binom{8}{k}$ ways to choose $k$ columns, and $\\binom{8}{k}$ ways to choose $k$ rows, and this would uniquely determine the set of squares selected. Thus the answer is\n$$\n\\sum_{k=1}^{8} \\binom{8}{k} \\binom{8}{k} = -1 + \\sum_{k=0}^{8} \\binom{8}{k} \\binom{8}{k} = -1 + \\binom{16}{8} = 12869\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the smallest positive integer that is twice a perfect square and three times a perfect cube.",
    "options": [],
    "answer": "648",
    "solution": "Solution:\n\nLet $n$ be such a number. If $n$ is divisible by $2$ and $3$ exactly $e_{2}$ and $e_{3}$ times, then $e_{2}$ is odd and a multiple of three, and $e_{3}$ is even and one more than a multiple of three. The smallest possible exponents are $n_{2}=3$ and $n_{3}=4$. The answer is then $2^{3} \\cdot 3^{4}=648$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfies $f\\left(x^{2}\\right) f^{\\prime\\prime}(x) = f^{\\prime}(x) f^{\\prime}\\left(x^{2}\\right)$ for all real $x$. Given that $f(1) = 1$ and $f^{\\prime\\prime\\prime}(1) = 8$, determine $f^{\\prime}(1) + f^{\\prime\\prime}(1)$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nLet $f^{\\prime}(1) = a$ and $f^{\\prime\\prime}(1) = b$. Then setting $x = 1$ in the given equation, $b = a^{2}$.\n\nDifferentiating the given yields\n$$\n2x f^{\\prime}\\left(x^{2}\\right) f^{\\prime\\prime}(x) + f\\left(x^{2}\\right) f^{\\prime\\prime\\prime}(x) = f^{\\prime\\prime}(x) f^{\\prime}\\left(x^{2}\\right) + 2x f^{\\prime}(x) f^{\\prime\\prime}\\left(x^{2}\\right)\n$$\nPlugging $x = 1$ into this equation gives $2ab + 8 = ab + 2ab$, or $ab = 8$. Then because $a$ and $b$ are real, we obtain the solution $(a, b) = (2, 4)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe curves $y = x^{2}(x-3)^{2}$ and $y = (x^{2}-1)(x-2)$ intersect at a number of points in the real plane. Determine the sum of the $x$-coordinates of these points of intersection.",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nAnswer: 7. Because the first curve touches the $x$-axis at $x=0$ and $x=3$ while the second curve crosses the $x$-axis at $x= \\pm 1$ and $x=2$, there are four points of intersection. In particular, the points of intersection have $x$-coordinates determined by the difference of the two curves:\n$$\n0 = x^{2}(x-3)^{2} - (x^{2}-1)(x-2) = (x^{4} - 6x^{3} + \\cdots) - (x^{3} + \\cdots) = x^{4} - 7x^{3} + \\cdots\n$$\nWe need only the first two coefficients to determine $x_{1} + x_{2} + x_{3} + x_{4} = -\\left(\\frac{-7}{1}\\right) = 7$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA convex quadrilateral is determined by the points of intersection of the curves $x^{4}+y^{4}=100$ and $x y=4$; determine its area.",
    "options": [],
    "answer": "4 sqrt(17)",
    "solution": "Solution:\nAnswer: $4 \\sqrt{17}$. By symmetry, the quadrilateral is a rectangle having $x=y$ and $x=-y$ as axes of symmetry. Let $(a, b)$ with $a>b>0$ be one of the vertices. Then the desired area is\n$$\n(\\sqrt{2}(a-b)) \\cdot (\\sqrt{2}(a+b)) = 2\\left(a^{2}-b^{2}\\right) = 2 \\sqrt{a^{4}-2 a^{2} b^{2}+b^{4}} = 2 \\sqrt{100-2 \\cdot 4^{2}} = 4 \\sqrt{17}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve the identity\n$$\n\\sum_{d \\mid n} \\tau(d)^{3}=\\left(\\sum_{d \\mid n} \\tau(d)\\right)^{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNote that $\\tau^{3}$ is multiplicative; in light of the convolution property just shown, it follows that both sides of the posed equality are multiplicative. Thus, it would suffice to prove the claim for $n$ a power of a prime. So, write $n=p^{k}$ where $p$ is a prime and $k$ is a nonnegative integer. Then\n$$\n\\begin{aligned}\n& \\sum_{d \\mid n} \\tau(d)^{3}=\\sum_{i=0}^{k} \\tau\\left(p^{i}\\right)^{3}=\\sum_{i=0}^{k}(i+1)^{3} \\\\\n& \\quad=1^{3}+\\cdots+(k+1)^{3}=\\frac{(k+1)^{2}(k+2)^{2}}{4}=\\left(\\frac{(k+1)(k+2)}{2}\\right)^{2} \\\\\n& \\quad=\\left(\\sum_{i=0}^{k} \\tau\\left(p^{i}\\right)\\right)^{2}=\\left(\\sum_{d \\mid n} \\tau(d)\\right)^{2}\n\\end{aligned}\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $\\omega, \\Omega$, and $R$ being its incircle, circumcircle, and circumradius, respectively. Circle $\\omega_A$ is tangent internally to $\\Omega$ at $A$ and tangent externally to $\\omega$. Circle $\\Omega_A$ is tangent internally to $\\Omega$ at $A$ and tangent internally to $\\omega$. Let $P_A$ and $Q_A$ denote the centers of $\\omega_A$ and $\\Omega_A$, respectively. Define points $P_B, Q_B, P_C, Q_C$ analogously. Prove that\n$$\n8P_AQ_A \\cdot P_BQ_B \\cdot P_CQ_C \\le R^3,\n$$\nwith equality if and only if triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the incircle touch the sides $AB, BC$, and $CA$ at $C_1, A_1$, and $B_1$, respectively. Set $AB = c, BC = a, CA = b$. By equal tangents, we may assume that $AB_1 = AC_1 = x$, $BC_1 = BA_1 = y$, and $CA_1 = CB_1 = z$. Then $a = y + z, b = z + x, c = x + y$. By the AM-GM inequality, we have $a \\ge 2\\sqrt{yz}$, $b \\ge 2\\sqrt{zx}$, and $c \\ge 2\\sqrt{xy}$. Multiplying the last three inequalities yields\n$$\nabc \\ge 8xyz, \\tag{†}\n$$\nwith equality if and only if $x = y = z$; that is, triangle $ABC$ is equilateral.\nLet $k$ denote the area of triangle $ABC$. By the Extended Law of Sines, $c = 2R \\sin \\angle C$. Hence\n$$\nk = \\frac{ab \\sin \\angle C}{2} = \\frac{abc}{4R} \\quad \\text{or} \\quad R = \\frac{abc}{4k}. \\tag{‡}\n$$\nWe are going to show that\n$$\nP_A Q_A = \\frac{xa^2}{4k}. \\tag{*}\n$$\nIn exactly the same way, we can also establish its cyclic analogous forms\n$$\nP_B Q_B = \\frac{yb^2}{4k} \\quad \\text{and} \\quad P_C Q_C = \\frac{zc^2}{4k}.\n$$\nMultiplying the last three equations together gives\n$$\nP_A Q_A \\cdot P_B Q_B \\cdot P_C Q_C = \\frac{xyz a^2 b^2 c^2}{64k^3}.\n$$\nFurther considering (†) and (‡), we have\n$$\n8P_A Q_A \\cdot P_B Q_B \\cdot P_C Q_C = \\frac{8xyz a^2 b^2 c^2}{64k^3} \\le \\frac{a^3 b^3 c^3}{64k^3} = R^3,\n$$\nwith equality if and only if triangle $ABC$ is equilateral.\nHence it suffices to show (*). Let $r, r_A, r'_A$ denote the radii of $\\omega, \\omega_A, \\Omega_A$, respectively. We consider the inversion $I$ with center $A$ and radius $x$. Clearly, $I(B_1) = B_1$, $I(C_1) = C_1$, and $I(\\omega) = \\omega$. Let ray $AO$ intersect $\\omega_A$ and $\\Omega_A$ at $S$ and $T$, respectively. It is not difficult to see that $AT > AS$, because $\\omega$ is tangent to $\\omega_A$ and $\\Omega_A$ externally and internally, respectively. Set $S_1 = I(S)$ and $T_1 = I(T)$. Let $\\ell$ denote the line tangent to $\\Omega$ at $A$. Then the image of $\\omega_A$ (under the inversion) is the line (denoted by $\\ell_1$) passing through $S_1$ and parallel to $\\ell$, and the image of $\\Omega_A$ is the line (denoted by $\\ell_2$) passing through $T_1$ and parallel to $\\ell$. Furthermore, since $\\omega$ is tangent to both $\\omega_A$ and $\\Omega_A$, $\\ell_1$ and $\\ell_2$ are also tangent to the image of $\\omega$, which is $\\omega$ itself. Thus the distance between these two lines is $2r$; that is, $S_1T_1 = 2r$. Hence we can consider the following configuration. (The darkened circle is $\\omega_A$, and its image is the darkened line $\\ell_1$.)\n\n![](attached_image_1.png)\n\nBy the definition of inversion, we have $AS_1 \\cdot AS = AT_1 \\cdot AT = x^2$. Note that $AS = 2r_A$, $AT = 2r'_A$, and $S_1T_1 = 2r$. We have\n$$\nr_A = \\frac{x^2}{2AS_1}. \\quad \\text{and} \\quad r'_A = \\frac{x^2}{2AT_1} = \\frac{x^2}{2(AS_1 - 2r)}.\n$$\nHence\n$$\nP_A Q_A = AQ_A - AP_A = r'_A - r_A = \\frac{x^2}{2} \\left( \\frac{1}{AS_1 - 2r} + \\frac{1}{AS_1} \\right).\n$$\nLet $H_A$ be the foot of the perpendicular from $A$ to side $BC$. It is well known that $\\angle BAS_1 = \\angle BAO = 90^\\circ - \\angle C = \\angle CAH_A$. Since ray $AI$ bisects $\\angle BAC$, it follows that rays $AS_1$ and $AH_A$ are symmetric with respect to ray $AI$. Further note that both line $l_1$ (passing through $S_1$) and line $BC$ (passing through $H_A$) are tangent to $\\omega$. We conclude that $AS_1 = AH_A$. In light of this observation and using the fact $2k = AH_A \\cdot BC = (AB + BC + CA)r$, we can compute $P_A Q_A$ as follows:\n$$\n\\begin{aligned}\nP_A Q_A &= \\frac{x^2}{2} \\left( \\frac{1}{AH_A - 2r} - \\frac{1}{AH_A} \\right) = \\frac{x^2}{4k} \\left( \\frac{2k}{AH_A - 2r} - \\frac{2k}{AH_A} \\right) \\\\\n&= \\frac{x^2}{4k} \\left( \\frac{1}{\\frac{1}{BC} - \\frac{2}{AB+BC+CA}} - BC \\right) = \\frac{x^2}{4k} \\left( \\frac{1}{\\frac{1}{y+z} - \\frac{1}{x+y+z}} - (y+z) \\right) \\\\\n&= \\frac{x^2}{4k} \\left( \\frac{(y+z)(x+y+z)}{x} - (y+z) \\right) \\\\\n&= \\frac{x(y+z)^2}{4k} = \\frac{xa^2}{4k},\n\\end{aligned}\n$$\nestablishing (*). Our proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Consider\n$$\nS = \\{(x, y, z) : x, y, z \\in \\{0, 1, \\dots, n\\}, x + y + z > 0\\}.\n$$\nas a set of $(n+1)^3 - 1$ points in three-dimensional space. Determine the smallest possible number of planes, the union of which contains $S$ but does not include $(0, 0, 0)$.\n(This problem was suggested by the Netherlands.)",
    "options": [],
    "answer": "3n",
    "solution": "We establish the following key lemma.\n\n**Lemma 1.** Consider a nonzero polynomial $P(x_1, \\dots, x_k)$ in $k$ variables. Suppose that $P$ vanishes at all points $(x_1, \\dots, x_k)$ such that $x_1, \\dots, x_k \\in \\{0, 1, \\dots, n\\}$ and $x_1 + \\dots + x_k > 0$, while $P(0, 0, \\dots, 0) \\ne 0$. Then $\\deg P \\ge kn$.\n\n*Proof.* We induct on $k$. The base case $k=0$ is clear since $P \\ne 0$. We assume that the statement is true for $k = \\ell - 1$ for some positive integer $\\ell$. Now we consider the case $k = \\ell$. Denote for clarity $y = x_k = x_\\ell$.\n\nLet $R(x_1, \\dots, x_{k-1}, y)$ be the residue of $P$ modulo $Q(y) = y(y-1)\\dots(y-n)$. Polynomial $Q(y)$ vanishes at each $y=0, 1, \\dots, n$, hence $P(x_1, \\dots, x_{k-1}, y) = R(x_1, \\dots, x_{k-1}, y)$ for all $x_1, \\dots, x_{k-1}, y \\in \\{0, 1, \\dots, n\\}$. Therefore, $R$ also satisfies the condition of the Lemma; moreover, $\\deg_y R \\le n$. Clearly, $\\deg R \\le \\deg P$, so it suffices to prove that $\\deg R \\ge nk$.\n\nNow, expand polynomial $R$ in the powers of $y$:\n$$\nR(x_1, \\dots, x_{k-1}, y) = R_n(x_1, \\dots, x_{k-1})y^n + R_{n-1}(x_1, \\dots, x_{k-1})y^{n-1} + \\dots + R_0(x_1, \\dots, x_{k-1}).\n$$\nWe show that the polynomial $R_n(x_1, \\dots, x_{k-1})$ satisfies the condition of the induction hypothesis.\n\nConsider the polynomial $T(y) = R(0, \\dots, 0, y)$ of degree $\\le n$. This polynomial has $n$ roots $y = 1, \\dots, n$; on the other hand, $T(y) \\ne 0$ since $T(0) \\ne 0$. Hence $\\deg T = n$, and its leading coefficient is $R_n(0, 0, \\dots, 0) \\ne 0$. (For example, in the case $k=1$ we obtain that coefficient $R_n$ is nonzero.)\n\nSimilarly, take any numbers $a_1, \\dots, a_{k-1} \\in \\{0, 1, \\dots, n\\}$ with $a_1 + \\dots + a_{k-1} > 0$. Substituting $x_i = a_i$ into $R(x_1, \\dots, x_{k-1}, y)$, we get a polynomial in $y$ which vanishes at all points $y = 0, \\dots, n$ and has degree $\\le n$. Therefore, this polynomial is null, hence $R_i(a_1, \\dots, a_{k-1}) = 0$ for all $i = 0, 1, \\dots, n$. In particular, $R_n(a_1, \\dots, a_{k-1}) = 0$.\n\nThus, the polynomial $R_n(x_1, \\dots, x_{k-1})$ satisfies the condition of the induction hypothesis. So, we have $\\deg R_n \\ge (k-1)n$ and $\\deg P \\ge \\deg R \\ge \\deg R_n + n \\ge kn$. $\\square$\n\nNow we can finish the solution. Suppose that there are $N$ planes covering all the points of $S$ but not containing the origin. Let their equations be $a_i x + b_i y + c_i z + d_i = 0$. Consider the polynomial\n$$\nP(x, y, z) = \\prod_{i=1}^{N} (a_i x + b_i y + c_i z + d_i).\n$$\nIt has total degree $N$. This polynomial has the property that $P(x_0, y_0, z_0) = 0$ for any $(x_0, y_0, z_0) \\in S$, while $P(0, 0, 0) \\ne 0$. Hence by Lemma 1 we get $N = \\deg P \\ge 3n$, as desired.\nSuppose $r$ planes are given. As in the first solution, let $A_1, \\dots, A_r$ be (nonzero) linear functions over $\\mathbb{R}^3$ such that the equations of the planes are $A_i(x, y, z) = 0$. Define the polynomial $P_0 = A_1A_2\\cdots A_r$, whose degree is $r$. If a lattice point $(x, y, z)$ satisfies $0 \\le x \\le a$, $0 \\le y \\le b$, $0 \\le z \\le c$, and $x + y + z > 0$, then $P_0(x, y, z) = 0$; however, $P_0(0, 0, 0) \\ne 0$.\n\nPolynomials $P_1, \\dots, P_a$ are defined recursively by $P_{i+1}(x, y, z) = P_i(x+1, y, z) - P_i(x, y, z)$. By induction, we see that $P_i(x, y, z) = 0$ if $0 \\le x \\le a-i$, $0 \\le y \\le b$, $0 \\le z \\le c$, and $x+y+z > 0$, while $P_i(0, 0, 0) \\ne 0$. Furthermore, if $P_i$ is a nonzero polynomial, then its degree is $r-i$.\n\nLet $Q_0 = P_a$, and construct polynomials $Q_1, \\dots, Q_b$ as above. Specifically, define $Q_{i+1}(x, y, z) = Q_i(x, y+1, z) - Q_i(x, y, z)$. Again, we have $Q_i(x, y, z) = 0$ provided $x = 0, 0 \\le y \\le b-i, 0 \\le z \\le c$, and $x+y+z > 0$, while $Q_i(0, 0, 0) \\ne 0$. If $Q_i$ is nonzero, then its degree is $r-a-i$.\n\nFinally, let $R_0 = Q_b$, and define $R_1, \\dots, R_c$ by $R_{i+1}(x, y, z) = R_i(x, y, z+1) - R_i(x, y, z)$. As above, $R_i(x, y, z) = 0$ if $x = y = 0, 0 \\le z \\le c-i$, and $x+y+z > 0$. If $R_i$ is nonzero then its degree is $r-a-b-i$.\n\nConsider the polynomial $R_c(x, y, z)$. Its value at $(0, 0, 0)$ is nonzero, so it is a nonzero polynomial. Its degree, which must be nonnegative, is $r-a-b-c$. Therefore, we have $r \\ge a+b+c$, as desired.\n\nTherefore, $a+b+c$ planes are necessary, and it is possible to cover all the points with $a+b+c$ planes. In the original problem, $a=b=c=n$, so the answer is $3n$.\n\nReal numbers $x_0, \\dots, x_a, y_0, \\dots, y_b, z_0, \\dots, z_c$ are given such that the $x_i$ are distinct, the $y_i$ are distinct, and the $z_i$ are distinct. If a collection of planes covers all points $(x_i, y_j, z_k)$ except $(x_0, y_0, z_0)$, the collection contains at least $a+b+c$ points.\n\nTo prove this we need a more general form of the method of finite differences used above.\n\n**Lemma 2.** Suppose distinct reals $t_0, \\dots, t_n$ are given. Then there exist weights $w_0, \\dots, w_n$, with $w_0 \\neq 0$, such that for each nonnegative integer $i < n$, we have $w_0 t_0^i + \\dots + w_n t_n^i = 0$, while $w_0 t_0^n + \\dots + w_n t_n^n \\neq 0$.\n\n*Proof.* The $n+1$ vectors $(1, t_j, t_j^2, \\dots, t_j^{n-1})$ are linearly dependent over $\\mathbb{R}^n$, so there exist weights $w_j$ such that $w_0 t_0^i + \\dots + w_n t_n^i = 0$ for each $i < n$. However, since the $t_j$ are distinct, the vectors $(1, t_j, t_j^2, \\dots, t_j^n)$ are linearly independent over $\\mathbb{R}^{n+1}$. (This follows from the fact that their determinant, the Vandermonde determinant, is nonzero.) Therefore, we must have $w_0 t_0^n + \\dots + w_n t_n^n \\neq 0$. Also, the vectors $(1, t_j, \\dots, t_j^{n-1})$, for $j = 1, \\dots, n$, are independent. Thus we cannot have $w_0 = 0$. $\\square$\n\nNext we prove an essential result about the weights found above.\n\n**Lemma 3.** Let reals $t_0, \\dots, t_n$ be given, and weights $w_0, \\dots, w_n$ be defined as above. Let $P(t)$ be any polynomial, and define $Q(t) = w_0 P(t+t_0) + \\dots + w_n P(t+t_n)$. If the degree of $P(t)$ is less than $n$, then $Q(t) = 0$; otherwise, the degree of $Q$ is $n$ less than the degree of $P$.\n\n*Proof.* First we prove the result for $P(t) = t^s$. The $t^r$ coefficient of $Q$ is $\\frac{s!}{r!(s-r)!} (w_0 t_0^{s-r} + \\dots + w_n t_n^{s-r})$. This is zero if $r > s-n$ and nonzero if $r \\le s-n$, provided $s \\ge n$. If $s < n$, then all coefficients of $Q$ are zero.\n\nNow suppose $P$ is a general polynomial of degree $s$. Then the $t^s$ term of $P$ is nonzero. Also, the $t^{s-n}$ term of $Q$ depends only on the $t^s$ term of $P$, so it is also nonzero. Thus the degree of $Q$ is $s-n$. $\\square$\n\nFinally, we solve the general form of IMO Problem 6. Suppose there are $r$ planes in the collection, so there exists a polynomial $P$ of degree $r$ such that $P(x_i, y_j, z_k) = 0$ unless $i = j = k = 0$. Apply Lemma 3 to $P$ (regarded as a polynomial in its first variable) with $t_0 = x_0, \\dots, t_a = x_a$, to obtain a polynomial $P_1$ of degree $r-a$ such that $P_1(0, y_j, z_k) = 0$ unless $j = k = 0$. Apply Lemma 3 again (this time on the second variable) to produce a polynomial $P_2$ of degree $r-a-b$ such that $P_2(0, 0, z_k) = 0$ unless $k = 0$. Apply Lemma 3 one more time (on the third variable) to produce a polynomial $P_3$ of degree $r-a-b-c$ such that $P_3(0, 0, 0) \\neq 0$. Since $P_3$ is a nonzero polynomial, we have $r \\ge a+b+c$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47357,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a polynomial $P(x)$ with integer coefficients, $r(2i - 1)$ (for $i = 1, 2, 3, \\dots, 512$) is the remainder obtained when $P(2i - 1)$ is divided by $1024$. The sequence\n$$\n(r(1), r(3), \\dots, r(1023))\n$$\nis called the *remainder sequence* of $P(x)$. A remainder sequence is called *complete* if it is a permutation of $(1, 3, 5, \\dots, 1023)$. Prove that there are no more than $2^{35}$ different complete remainder sequences.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define the polynomials\n$$\n\\begin{align*}\nQ_0(x) &= b_0, \\\\\nQ_1(x) &= b_1 (x + 1), \\\\\nQ_2(x) &= b_2 (x + 1)(x + 3), \\\\\nQ_3(x) &= b_3 (x + 1)(x + 3)(x + 5), \\\\\nQ_4(x) &= b_4 (x + 1)(x + 3)(x + 5)(x + 7), \\\\\nQ_5(x) &= b_5 (x + 1)(x + 3)(x + 5)(x + 7)(x + 9), \\\\\nQ_6(x) &= b_6 (x + 1)(x + 3)(x + 5)(x + 7)(x + 9)(x + 11),\n\\end{align*}\n$$\nwhere\n$$\nb_0 = 2^{10}, \\quad b_1 = 2^9, \\quad \\dots, b_2 = 2^7, \\quad b_3 = 2^6, \\quad b_4 = 2^3, \\quad b_5 = 2^2, \\quad b_6 = 2^0.\n$$\nThe product of $i$ consecutive even integers is divisible by $2^i \\cdot i!$. Therefore, for $i = 0, 1, 2, 3, 4, 5, 6$, we obtain that the product of $i$ consecutive even integers is divisible by $2^0, 2^1, 2^3, 2^4, 2^7, 2^8, 2^{10}$, respectively. This implies that, for any odd integer $x$ and $i = 0, \\dots, 6$, $Q_i(x)$ is divisible by $2^{10}$.\nA polynomial $P(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$ with integer coefficients is called *reduced* if, for $i = 0, \\dots, 5$,\n$$\n0 \\le a_i < b_i. \\qquad (\\dagger)\n$$\nClearly, there are exactly $b_0b_1\\dots b_5 = 2^{10+9+7+6+3+2} = 2^{37}$ distinct reduced polynomials.\nWe show that, for every polynomial $P(x)$ with integer coefficients, there exists a reduced polynomial $\\bar{P}(x)$ such that $P(x)$ and $\\bar{P}(x)$ have the same remainder sequence.\nFirst note that, for $i = 0, \\dots, 6$, and any polynomial $R(x)$ with integer coefficients $P(x)$ and $P(x) - R(x)Q_i(x)$ have the same remainder sequence. This follows from the fact that $Q_i(x)$ is divisible by $2^{10}$, for any odd integer $x$.\n\nIf the degree $d$ of $P(x) = a_0 + \\cdots + a_d x^d$ is higher than $5$ we may replace $P(x)$ by $P(x) - a_d x^{d-6} Q_6(x)$. Indeed, the polynomial $P(x) - a_d x^{d-6} Q_6(x)$ has smaller degree than $P(x)$ and has the same remainder sequence as $P(x)$. We may continue this until we obtain a polynomial that of degree at most $5$ that has the same remainder sequence as $P(x)$.\nWe assume now that $P(x)$ has degree no higher than $5$. If $P(x)$ is reduced we are done. Otherwise, let $i$ be the highest degree of a coefficient $a_i$ of $x^i$ that does not satisfy the range condition $(\\dagger)$. If $q$ is the quotient obtained by dividing $a_i$ by $b_i$ then $P(x)$ and $P(x) - qQ_i(x)$ have the same remainder sequence and the coefficient at degree $i$ in $P(x) - qQ_i(x)$ is in the correct range $0, \\dots, b_i - 1$.\nWe repeat this procedure with the next highest degree that has a coefficient out of range until we reach a reduced polynomial that has the same remainder sequence as $P(x)$.\n\nWe now consider the $2^{37}$ reduced polynomials.\nLet $a = 2^9+1$ and $b = 1$. Then $P(a)-P(b) = (a-b)(a_1+a_2A_2+a_3A_3+a_4A_4+a_5A_5)$, where $A_2 = a+b$, $A_3 = a^2+ab+b^2$, $A_4 = a^3+a^2b+ab^2+b^3$ and $A_5 = a^4+a^3b+a^2b^2+ab^3+b^4$. Since both $a$ and $b$ are odd, $A_2$ and $A_4$ are even, $A_3$ and $A_5$ are odd, and the parity of $a_1+a_2A_2+a_3A_3+a_4A_4+a_5A_5$ is the same as the parity of $a_1+a_3+a_5$. Therefore, if $a_1+a_3+a_5$ is even $P(a)-P(b)$ is divisible by $2^{10}$ and the sequence of remainders of $P(x)$ is not a permutation.\nFor an odd integer $x$, the parity of $P(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$ is the same as the parity of the sum $a_0 + a_1 + \\cdots + a_5$. Thus, only polynomials with odd sum of coefficients have odd remainders.\nTherefore, there are no more remainder sequences that are permutations of $1, 3, \\dots, 1023$ than there are reduced polynomials $P(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$ for which both $a_1 + a_3 + a_5$ and $a_0 + a_1 + a_2 + a_3 + a_4 + a_5$ are odd.\nThere are exactly $2^{36}$ reduced polynomials for which $a_1+a_3+a_5$ is odd. This can be seen by pairing up every reduced polynomial $P(x)$ in which $a_1$ is even with the polynomial $P(x)+x$. Exactly one of the two polynomials in each such pair has odd sum $a_1+a_3+a_5$.\nThere are exactly $2^{35}$ reduced polynomials for which both $a_1+a_3+a_5$ and $a_0+a_1+a_2+a_3+a_4+a_5$ are odd. This can be seen by pairing up every reduced polynomial $P(x)$ in which $a_1+a_3+a_5$ is odd and $a_0$ is even with the polynomial $P(x)+1$. Exactly one of the two polynomials in each such pair has odd sum $a_0+a_1+a_2+a_3+a_4+a_5$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 47358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCircle $\\omega$ has radius $5$ and is centered at $O$. Point $A$ lies outside $\\omega$ such that $OA = 13$. The two tangents to $\\omega$ passing through $A$ are drawn, and points $B$ and $C$ are chosen on them (one on each tangent), such that line $BC$ is tangent to $\\omega$ and $\\omega$ lies outside triangle $ABC$. Compute $AB + AC$ given that $BC = 7$.",
    "options": [],
    "answer": "31",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe positive integer $n$ is such that the numbers $2^{n}$ and $5^{n}$ start with the same digit when written in decimal notation; determine this common leading digit.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nAnswer: $3$. Note $1 = 1^{2} < 2^{2} < 3^{2} < 10 < 4^{2} < \\cdots < 9^{2} < 10^{2} = 100$. Divide $2^{n}$ and $5^{n}$ by $10$ repeatedly until each is reduced to a decimal number less than $10$ but at least $1$; call the resulting numbers $x$ and $y$. Since $(5^{n})(2^{n}) = 10^{n}$, either $x y = 1$ or $x y = 10$. Because $2^{n}$ and $5^{n}$ begin with the same digit, $x$ and $y$ are bounded by the same pair of adjacent integers. It follows that either $x = y = 1$ or $3 \\leq x, y < 4$. Because $n$ is positive, neither $2^{n}$ nor $5^{n}$ is a perfect power of $10$, so the former is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKevin has four red marbles and eight blue marbles. He arranges these twelve marbles randomly, in a ring. Determine the probability that no two red marbles are adjacent.",
    "options": [],
    "answer": "7/33",
    "solution": "Solution:\n\nAnswer: $\\frac{7}{33}$.\n\nSelect any blue marble and consider the remaining eleven marbles, arranged in a line. The proportion of arrangements for which no two red marbles are adjacent will be the same as for the original twelve marbles, arranged in a ring. The total number of ways of arranging $4$ red marbles out of $11$ is $\\binom{11}{4} = 330$.\n\nTo count the number of arrangements such that no two red marbles are adjacent, there must be one blue marble between each two would-be adjacent red marbles. Having fixed the positions of three blue marbles, we have four blue marbles to play with. So the number of ways we can arrange the remaining four marbles is $\\binom{8}{4} = 70$ ways.\n\nThis yields a probability of $70 / 330 = 7 / 33$ as our final answer.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are three video game systems: the Paystation, the WHAT, and the ZBoz2 $\\pi$, and none of these systems will play games for the other systems. Uncle Riemann has three nephews: Bernoulli, Galois, and Dirac. Bernoulli owns a Paystation and a WHAT, Galois owns a WHAT and a ZBoz2 $\\pi$, and Dirac owns a ZBoz2 $\\pi$ and a Paystation. A store sells 4 different games for the Paystation, 6 different games for the WHAT, and 10 different games for the ZBoz2 $\\pi$. Uncle Riemann does not understand the difference between the systems, so he walks into the store and buys 3 random games (not necessarily distinct) and randomly hands them to his nephews. What is the probability that each nephew receives a game he can play?",
    "options": [],
    "answer": "7/25",
    "solution": "Solution:\n\nAnswer: $\\frac{7}{25}$. Since the games are not necessarily distinct, probabilities are independent. Multiplying the odds that each nephew receives a game he can play, we get $10/20 \\cdot 14/20 \\cdot 16/20 = 7/25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe elliptic curve $y^{2}=x^{3}+1$ is tangent to a circle centered at $(4,0)$ at the point $(x_{0}, y_{0})$. Determine the sum of all possible values of $x_{0}$.",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n\nAnswer: $\\frac{1}{3}$.\n\nNote that $y^{2} \\geq 0$, so $x^{3} \\geq -1$ and $x \\geq -1$.\n\nLet the circle be defined by $(x-4)^{2}+y^{2}=c$ for some $c \\geq 0$.\n\nNow differentiate the equations with respect to $x$, obtaining $2y \\frac{\\mathrm{d}y}{\\mathrm{d}x} = 3x^{2}$ from the given and $2y \\frac{\\mathrm{d}y}{\\mathrm{d}x} = -2x + 8$ from the circle.\n\nFor tangency, the two expressions $\\frac{\\mathrm{d}y}{\\mathrm{d}x}$ must be equal if they are well-defined, and this is almost always the case. Thus, $-2x_{0} + 8 = 3x_{0}^{2}$ so $x_{0} = -2$ or $x_{0} = 4/3$, but only the latter corresponds to a point on $y^{2} = x^{3} + 1$.\n\nOtherwise, $y_{0} = 0$, and this gives the trivial solution $x_{0} = -1$.",
    "topic": "Geometry",
    "subtopic": "Non-Euclidean Geometry"
  },
  {
    "id": 47363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the polynomial $P(x) = x^{3} + x^{2} - x + 2$. Determine all real numbers $r$ for which there exists a complex number $z$ not in the reals such that $P(z) = r$.",
    "options": [],
    "answer": "All real r with r < 49/27 or r > 3",
    "solution": "Solution:\n\nAnswer: $r > 3$, $r < \\frac{49}{27}$. Because such roots to polynomial equations come in conjugate pairs, we seek the values $r$ such that $P(x) = r$ has just one real root $x$. Considering the shape of a cubic, we are interested in the boundary values $r$ such that $P(x) - r$ has a repeated zero. Thus, we write\n$$\nP(x) - r = x^{3} + x^{2} - x + (2 - r) = (x - p)^{2}(x - q) = x^{3} - (2p + q)x^{2} + p(p + 2q)x - p^{2}q\n$$\nThen $q = -2p - 1$ and $1 = p(p + 2q) = p(-3p - 2)$ so that $p = 1/3$ or $p = -1$. It follows that the graph of $P(x)$ is horizontal at $x = 1/3$ (a maximum) and $x = -1$ (a minimum), so the desired values $r$ are $r > P(-1) = 3$ and $r < P(1/3) = 1/27 + 1/9 - 1/3 + 2 = 49/27$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriangle $ABC$ has $\\angle A = 90^{\\circ}$, side $BC = 25$, $AB > AC$, and area $150$. Circle $\\omega$ is inscribed in $ABC$, with $M$ its point of tangency on $AC$. Line $BM$ meets $\\omega$ a second time at point $L$. Find the length of segment $BL$.",
    "options": [],
    "answer": "45*sqrt(17)/17",
    "solution": "Solution:\n\nLet $D$ be the foot of the altitude from $A$ to side $BC$. The length of $AD$ is $2 \\cdot 150 / 25 = 12$.\n\nTriangles $ADC$ and $BDA$ are similar, so $CD \\cdot DB = AD^2 = 144 \\Rightarrow BD = 16$ and $CD = 9 \\Rightarrow AB = 20$ and $AC = 15$.\n\nUsing equal tangents or the formula inradius as area divided by semiperimeter, we can find the radius of $\\omega$ to be $5$.\n\nNow, let $N$ be the tangency point of $\\omega$ on $AB$. By power of a point, we have $BL \\cdot BM = BN^2$. Since the center of $\\omega$ together with $M$, $A$, and $N$ determines a square, $BN = 15$ and $BM = 5 \\sqrt{17}$, and we have $BL = 45 \\sqrt{17} / 17$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAndrew has a fair six sided die labeled with $1$ through $6$ as usual. He tosses it repeatedly, and on every third roll writes down the number facing up as long as it is not the $6$. He stops as soon as the last two numbers he has written down are squares or one is a prime and the other is a square. What is the probability that he stops after writing squares consecutively?",
    "options": [],
    "answer": "4/25",
    "solution": "Solution:\n\nAnswer: $\\frac{4}{25}$. We can safely ignore all of the rolls he doesn't record. The probability that he stops after writing two squares consecutively is the same as the probability that he never rolls a prime. For, as soon as the first prime is written, either it must have been preceded by a square or it will be followed by a nonnegative number of additional primes and then a square. So we want the probability that two numbers chosen uniformly with replacement from $\\{1,2,3,4,5\\}$ are both squares, which is $\\left(\\frac{2}{5}\\right)^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that for positive integers $n$,\n$$\n\\sum_{d \\mid n} \\phi(d) = n\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nBoth sides are multiplicative functions of $n$, the right side trivially and the left because for relatively prime positive integers $n$ and $n'$, \n$$\n\\left(\\sum_{d \\mid n} \\phi(d)\\right)\\left(\\sum_{d' \\mid n'} \\phi(d')\\right) = \\sum_{d \\mid n,\\ d' \\mid n'} \\phi(d) \\phi(d'),\n$$\nand $\\phi(d) \\phi(d') = \\phi(dd')$. The identity is then easy to check; since $\\phi(p^k) = p^{k-1}(p-1)$ for positive integers $k$ and $\\phi(1) = 1$, we have\n$$\n\\phi(1) + \\phi(p) + \\cdots + \\phi(p^k) = 1 + (p-1) + (p^2 - p) + \\cdots + (p^k - p^{k-1}) = p^k,\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet the incircle of $ABCD$ be tangent to sides $AB$, $BC$, $CD$, and $AD$ at points $P$, $Q$, $R$, and $S$, respectively. Show that $ABCD$ is cyclic if and only if $PR \\perp QS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the diagonals of $PQRS$ intersect at $T$. Because $\\overline{AP}$ and $\\overline{AS}$ are tangent to $\\omega$ at $P$ and $S$, we may write $\\alpha = \\angle ASP = \\angle SPA = \\angle SQP$ and $\\beta = \\angle CQR = \\angle QRC = \\angle QPR$. Then $\\angle PTQ = \\pi - \\alpha - \\beta$. On the other hand, $\\angle PAS = \\pi - 2\\alpha$ and $\\angle RCQ = \\pi - 2\\beta$, so that $ABCD$ is cyclic if and only if\n$$\n\\pi = \\angle BAD + \\angle DCB = 2\\pi - 2\\alpha - 2\\beta\n$$\nor simply\n$$\n\\pi/2 = \\pi - \\alpha - \\beta = \\angle PTQ\n$$\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn infinite sequence of positive real numbers is defined by $a_{0}=1$ and $a_{n+2}=6 a_{n}-a_{n+1}$ for $n=0,1,2, \\cdots$. Find the possible value(s) of $a_{2007}$.",
    "options": [],
    "answer": "2^{2007}",
    "solution": "Solution:\n\nAnswer: $2^{2007}$. The characteristic equation of the linear homogeneous equation is $m^{2}+m-6=(m+3)(m-2)=0$ with solutions $m=-3$ and $m=2$. Hence the general solution is given by $a_{n}=A(2)^{n}+B(-3)^{n}$ where $A$ and $B$ are constants to be determined. Then we have $a_{n}>0$ for $n \\geq 0$, so necessarily $B=0$, and $a_{0}=1 \\Rightarrow A=1$. Therefore, the unique solution to the recurrence is $a_{n}=2^{n}$ for all $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConvex quadrilateral $A B C D$ has sides $A B = B C = 7$, $C D = 5$, and $A D = 3$. Given additionally that $m \\angle A B C = 60^{\\circ}$, find $B D$.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nAnswer: $8$. Triangle $A B C$ is equilateral, so $A C = 7$ as well. Now the law of cosines shows that $m \\angle C D A = 120^{\\circ}$; i.e., $A B C D$ is cyclic. Ptolemy's theorem now gives $A C \\cdot B D = A B \\cdot C D + A D \\cdot B C$, or simply $B D = C D + A D = 8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCompute\n$$\n\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot (n+1) \\cdot (n+1)!}\n$$",
    "options": [],
    "answer": "3 - e",
    "solution": "Solution:\nAnswer: $3-e$. We write\n$$\n\\begin{gathered}\n\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot (n+1) \\cdot (n+1)!} = \\sum_{n=1}^{\\infty} \\left(\\frac{1}{n} - \\frac{1}{n+1}\\right) \\frac{1}{(n+1)!} = \\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot (n+1)!} - \\sum_{n=1}^{\\infty} \\frac{1}{(n+1) \\cdot (n+1)!} \\\\\n\\frac{1}{2} + \\sum_{n=2}^{\\infty} \\frac{1}{n \\cdot (n+1)!} - \\sum_{n=1}^{\\infty} \\frac{1}{(n+1) \\cdot (n+1)!} = \\frac{1}{2} + \\sum_{n=1}^{\\infty} \\frac{1}{(n+1) \\cdot (n+2)!} - \\frac{1}{(n+1) \\cdot (n+1)!} \\\\\n\\frac{1}{2} + \\sum_{n=1}^{\\infty} \\frac{1-(n+2)}{(n+1) \\cdot (n+2)!} = \\frac{1}{2} - \\left(\\frac{1}{3!} + \\frac{1}{4!} + \\cdots\\right) = 3 - \\left(\\frac{1}{0!} + \\frac{1}{1!} + \\frac{1}{2!} + \\cdots\\right) = 3-e.\n\\end{gathered}\n$$\n\nAlternatively, but with considerably less motivation, we can induce telescoping by adding and subtracting $e-2 = 1/2! + 1/3! + \\cdots$, obtaining\n$$\n\\begin{aligned}\n2-e & + \\sum_{n=1}^{\\infty} \\frac{n(n+1)+1}{n \\cdot (n+1) \\cdot (n+1)!} = 2-e + \\sum_{n=1}^{\\infty} \\frac{(n+1)^2 - n}{n \\cdot (n+1) \\cdot (n+1)!} \\\\\n2 & - e + \\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot n!} - \\frac{1}{(n+1) \\cdot (n+1)!} = 3-e\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMy friend and I are playing a game with the following rules: If one of us says an integer $n$, the opponent then says an integer of their choice between $2 n$ and $3 n$, inclusive. Whoever first says $2007$ or greater loses the game, and their opponent wins. I must begin the game by saying a positive integer less than $10$. With how many of them can I guarantee a win?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nAnswer: $6$. We assume optimal play and begin working backward. I win if I say any number between $1004$ and $2006$. Thus, by saying such a number, my friend can force a win for himself if I ever say a number between $335$ and $1003$. Then I win if I say any number between $168$ and $334$, because my friend must then say one of the losing numbers just considered. Similarly, I lose by saying $56$ through $167$, win by saying $28$ through $55$, lose with $10$ through $17$, win with $5$ through $9$, lose with $2$ through $4$, and win with $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that for positive integers $n$,\n$$\n\\sum_{d \\mid n} \\frac{\\mu(d)}{d} = \\frac{\\phi(n)}{n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOn the grounds of the previous problem, Möbius inversion with $f(k) = \\phi(k)$ and $g(k) = k$ gives:\n$$\n\\phi(n) = f(n) = \\sum_{d \\mid n} g(d) \\mu\\left(\\frac{n}{d}\\right) = \\sum_{d' \\mid n} g\\left(\\frac{n}{d'}\\right) \\mu(d') = \\sum_{d' \\mid n} \\frac{n}{d'} \\mu(d')\n$$\nDividing both sides by $n$ yields:\n$$\n\\frac{\\phi(n)}{n} = \\sum_{d' \\mid n} \\frac{\\mu(d')}{d'}\n$$\nAlternatively, one uses the convolution of the functions $f(k) = n$ and $g(k) = \\frac{\\mu(d)}{d}$. The strategy is the same as the previous convolution proof. For $n = p^{k}$ with $k$ a positive integer, we have $\\phi(n) = p^{k} - p^{k-1}$, while the series reduces to $p^{k} \\cdot \\mu(1) + p^{k} \\cdot \\mu(p) / p = p^{k} - p^{k-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nForty two cards are labeled with the natural numbers $1$ through $42$ and randomly shuffled into a stack. One by one, cards are taken off of the top of the stack until a card labeled with a prime number is removed. How many cards are removed on average?",
    "options": [],
    "answer": "43/14",
    "solution": "Solution:\n\nAnswer: $\\frac{43}{14}$. Note that there are $13$ prime numbers amongst the cards. We may view these as separating the remaining $29$ cards into $14$ groups of nonprimes - those appearing before the first prime, between the first and second, etc. Each of these groups is equally likely to appear first, so $29/14$ nonprimes are removed on average. We are done since exactly one prime is always drawn.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree positive reals $x$, $y$, and $z$ are such that\n$$\n\\begin{aligned}\nx^{2}+2(y-1)(z-1) & =85 \\\\\ny^{2}+2(z-1)(x-1) & =84 \\\\\nz^{2}+2(x-1)(y-1) & =89\n\\end{aligned}\n$$\nCompute $x+y+z$.",
    "options": [],
    "answer": "18",
    "solution": "Solution:\n\nAdd the three equations to obtain\n$$\nx^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x-4 x-4 y-4 z+6=258\n$$\nwhich rewrites as $(x+y+z-2)^{2}=256$. Evidently, $x+y+z=2 \\pm 16$. Since $x$, $y$, and $z$ are positive, $x+y+z>0$ so $x+y+z=2+16=18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJack, Jill, and John play a game in which each randomly picks and then replaces a card from a standard 52 card deck, until a spades card is drawn. What is the probability that Jill draws the spade? (Jack, Jill, and John draw in that order, and the game repeats if no spade is drawn.)",
    "options": [],
    "answer": "12/37",
    "solution": "Solution:\n\n$\\boxed{\\dfrac{12}{37}}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA student at Harvard named Kevin\nWas counting his stones by $11$\nHe messed up $n$ times\nAnd instead counted $9$s\nAnd wound up at $2007$.\nHow many values of $n$ could make this limerick true?",
    "options": [],
    "answer": "21",
    "solution": "Solution:\nAnswer: $21$. The mathematical content is that $9n + 11k = 2007$, for some nonnegative integers $n$ and $k$. As $2007 = 9 \\cdot 223$, $k$ must be divisible by $9$. Using modulo $11$, we see that $n$ is $3$ more than a multiple of $11$. Thus, the possibilities are $n = 223, 212, 201, \\ldots, 3$, which are $21$ in number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that $\\omega$ is a primitive $2007^{\\text{th}}$ root of unity. Find $\\left(2^{2007}-1\\right) \\sum_{j=1}^{2006} \\frac{1}{2-\\omega^{j}}$.\n\nFor this problem only, you may express your answer in the form $m \\cdot n^{k}+p$, where $m, n, k$, and $p$ are positive integers. Note that a number $z$ is a primitive $n^{\\text{th}}$ root of unity if $z^{n}=1$ and $n$ is the smallest number amongst $k=1,2, \\ldots, n$ such that $z^{k}=1$.",
    "options": [],
    "answer": "2005 · 2^{2006} + 1",
    "solution": "Solution:\nAnswer: $2005 \\cdot 2^{2006}+1$. Note that\n$$\n\\begin{aligned}\n& \\frac{1}{z-\\omega}+\\cdots+\\frac{1}{z-\\omega^{2006}}=\\frac{\\sum_{j=1}^{2006} \\prod_{i \\neq j}\\left(z-\\omega^{i}\\right)}{(z-\\omega) \\cdots\\left(z-\\omega^{2006}\\right)} \\\\\n& \\quad=\\frac{\\frac{\\mathrm{d}}{\\mathrm{~d} z}\\left[z^{2006}+z^{2005}+\\cdots+1\\right]}{z^{2006}+z^{2005}+\\cdots+1}=\\frac{2006 z^{2005}+2005 z^{2004}+\\cdots+1}{z^{2006}+z^{2005}+\\cdots+1} \\cdot \\frac{z-1}{z-1} \\\\\n& \\quad=\\frac{2006 z^{2006}-z^{2005}-z^{2004}-\\cdots-1}{z^{2007}-1} \\cdot \\frac{z-1}{z-1}=\\frac{2006 z^{2007}-2007 z^{2006}+1}{\\left(z^{2007}-1\\right)(z-1)} .\n\\end{aligned}\n$$\nPlugging in $z=2$ gives $\\frac{2005 \\cdot 2^{2006}+1}{2^{2007}-1}$; whence the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the positive real number(s) $x$ such that\n$$\n\\frac{1}{2}\\left(3 x^{2}-1\\right)=\\left(x^{2}-50 x-10\\right)\\left(x^{2}+25 x+5\\right).\n$$",
    "options": [],
    "answer": "25 + 2 sqrt(159)",
    "solution": "Solution:\n\nAnswer: $25+2 \\sqrt{159}$.\n\nWrite $a = x^{2} - 50x - 10$ and $b = x^{2} + 25x + 5$; the given becomes\n$$\n\\frac{a + 2b - 1}{2} = ab,\n$$\nso\n$$\n0 = 2ab - a - 2b + 1 = (a - 1)(2b - 1).\n$$\nThen $a - 1 = x^{2} - 50x - 11 = 0$ or $2b - 1 = 2x^{2} + 50x + 9 = 0$.\n\nThe former has a positive root, $x = 25 + 2\\sqrt{159}$, while the latter cannot, for obvious reasons.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine with proof, a simple closed form expression for\n$$\n\\sum_{d \\mid n} \\phi(d) \\tau\\left(\\frac{n}{d}\\right)\n$$",
    "options": [],
    "answer": "σ(n)",
    "solution": "Solution:\nWe claim the series reduces to $\\sigma(n)$. The series counts the ordered triples $(d, x, y)$ with $d \\mid n;\\ x \\mid d;\\ 0 < y \\leq n/d;$ and $(y, n/d) = 1$. To see this, write\n$$\n\\sum_{d \\mid n} \\phi(d) \\tau\\left(\\frac{n}{d}\\right) = \\sum_{d' \\mid n} \\phi\\left(\\frac{n}{d'}\\right) \\tau\\left(d'\\right)\n$$\nso that for a given $d' \\mid n$ we may choose $x$ and $y$ as described above. On the other hand, we can count these triples by groups sharing a given $x$. Fixing $x$ as a divisor of $n$ fixes an integer $\\frac{n}{x}$. Then $d$ varies such that $\\frac{n}{d}$ is a divisor of $\\frac{n}{x}$. For each divisor $\\frac{n}{d}$ of $\\frac{n}{x}$ there are precisely $\\phi\\left(\\frac{n}{d}\\right)$ choices $y$, so that by the lemma from the previous problem, there are $\\frac{n}{x}$ triples $(d, x, y)$ for a given $x$. It follows that there are precisely $\\sigma(n)$ such triples $(d, x, y)$.\n\nAgain, an alternative is to use the multiplicativity of the convolution, although it is now a little more difficult. Write $n = p^{k}$ so that\n$$\n\\begin{aligned}\n\\sum_{d \\mid n} & \\phi(d) \\tau\\left(\\frac{n}{d}\\right) = \\sum_{m=0}^{k} \\phi\\left(p^{m}\\right) \\tau\\left(p^{k-m}\\right) = k+1 + \\sum_{m=1}^{k} p^{m-1}(p-1)(k-m+1) \\\\\n& = k+1 + \\left(\\sum_{m=1}^{k} p^{m}(k-m+1)\\right) - \\left(\\sum_{m=1}^{k} p^{m-1}(k-m+1)\\right) = k+1 + p^{k} - k + \\sum_{m'=1}^{k-1} p^{m'} \\\\\n& = 1 + p + \\cdots + p^{k} = \\sigma\\left(p^{k}\\right)\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A := \\mathbb{Q} \\setminus \\{0, 1\\}$ denote the set of all rationals other than $0$ and $1$. A function $f: A \\rightarrow \\mathbb{R}$ has the property that for all $x \\in A$,\n$$\nf(x) + f\\left(1 - \\frac{1}{x}\\right) = \\log |x|\n$$\nCompute the value of $f(2007)$.",
    "options": [],
    "answer": "log(2007/2006)",
    "solution": "Solution:\nAnswer: $\\log (2007 / 2006)$.\nLet $g: A \\rightarrow A$ be defined by $g(x) := 1 - 1 / x$; the key property is that\n$$\ng(g(g(x))) = 1 - \\frac{1}{1 - \\frac{1}{1 - \\frac{1}{x}}} = x\n$$\nThe given equation rewrites as $f(x) + f(g(x)) = \\log |x|$. Substituting $x = g(y)$ and $x = g(g(z))$ gives the further equations $f(g(y)) + f(g(g(y))) = \\log |g(x)|$ and $f(g(g(z))) + f(z) = \\log |g(g(x))|$. Setting $y$ and $z$ to $x$ and solving the system of three equations for $f(x)$ gives\n$$\nf(x) = \\frac{1}{2} \\cdot (\\log |x| - \\log |g(x)| + \\log |g(g(x))|)\n$$\nFor $x = 2007$, we have $g(x) = \\frac{2006}{2007}$ and $g(g(x)) = \\frac{-1}{2006}$, so that\n$$\nf(2007) = \\frac{\\log |2007| - \\log \\left|\\frac{2006}{2007}\\right| + \\log \\left|\\frac{-1}{2006}\\right|}{2} = \\log (2007 / 2006)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle inscribed in a square,\nHas two chords as shown in a pair.\nIt has radius $2$,\nAnd $P$ bisects $TU$.\nThe chords' intersection is where?\n\n![](attached_image_1.png)\n\nAnswer the question by giving the distance of the point of intersection from the center of the circle.",
    "options": [],
    "answer": "2√2 - 2",
    "solution": "Solution:\nAnswer: $2\\sqrt{2} - 2$.\nLet $OB$ intersect the circle at $X$ and $Y$, and the chord $PM$ at $Q$, such that $O$ lies between $X$ and $Q$. Then $MNXQ$ is a parallelogram. For, $OB \\parallel NM$ by homothety at $C$ and $PM \\parallel NX$ because $MNXP$ is an isosceles trapezoid. It follows that $QX = MN$. Considering that the center of the circle together with points $M$, $C$, and $N$ determines a square of side length $2$, it follows that $MN = 2\\sqrt{2}$, so the answer is $2\\sqrt{2} - 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the largest positive integer $n$ such that there exist positive integers $x, y, z$ so that\n$$\nn^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x+3 x+3 y+3 z-6\n$$",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nThe given equation rewrites as $n^{2}=(x+y+z+1)^{2}+(x+y+z+1)-8$. Writing $r=x+y+z+1$, we have $n^{2}=r^{2}+r-8$. Clearly, one possibility is $n=r=8$, which is realized by $x=y=1, z=6$. On the other hand, for $r>8$, we have $r^{2}<r^{2}+r-8<(r+1)^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA set of six edges of a regular octahedron is called Hamiltonian cycle if the edges in some order constitute a single continuous loop that visits each vertex exactly once. How many ways are there to partition the twelve edges into two Hamiltonian cycles?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nAnswer: 6. Call the octahedron $A B C D E F$, where $A, B$, and $C$ are opposite $D, E$, and $F$, respectively. Note that each Hamiltonian cycle can be described in terms of the order it visits vertices in exactly 12 different ways. Conversely, listing the six vertices in some order determines a Hamiltonian cycle precisely when no pair of opposite vertices are listed consecutively or first-and-last. Suppose we begin with $A B$. If $D$ is listed third, then the final three letters are $C E F$ or $F E C$. Otherwise, $C$ or $F$ is listed next, and each gives three possibilities for the final three. For example $A B C$ is be followed by $D E F, D F E$, or $E D F$. Thus, there are $6 \\cdot 4 \\cdot(2+3+3)=192$ listings. These correspond to $192 / 12=16$ Hamiltonian cycles. Finally, the complement of all but four Hamiltonian cycles is a Hamiltonian cycle. For, each vertex has degree four, so is an endpoint of two edges in the complement of a Hamiltonian cycle, so is also a Hamiltonian cycle unless it describes two opposite faces. It follows that there are six pairs of disjoint Hamiltonian cycles.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is a convex quadrilateral such that $AB < AD$. The diagonal $\\overline{AC}$ bisects $\\angle BAD$, and $m \\angle ABD = 130^{\\circ}$. Let $E$ be a point on the interior of $\\overline{AD}$, and $m \\angle BAD = 40^{\\circ}$. Given that $BC = CD = DE$, determine $m \\angle ACE$ in degrees.",
    "options": [],
    "answer": "55",
    "solution": "Solution:\n\nFirst, we check that $ABCD$ is cyclic. Reflect $B$ over $\\overline{AC}$ to $B'$ on $\\overline{AD}$, and note that $B'C = CD$. Therefore, $m \\angle ADC = m \\angle B'DC = m \\angle CB'D = 180^{\\circ} - m \\angle AB'C = 180^{\\circ} - m \\angle CBA$.\n\nNow $m \\angle CBD = m \\angle CAD = 20^{\\circ}$ and $m \\angle ADC = 180^{\\circ} - m \\angle CBA = 30^{\\circ}$. Triangle $CDE$ is isosceles, so $m \\angle CED = 75^{\\circ}$ and $m \\angle AEC = 105^{\\circ}$. It follows that $m \\angle ECA = 180^{\\circ} - m \\angle AEC - m \\angle CAE = 55^{\\circ}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCompute the number of sequences of numbers $a_{1}, a_{2}, \\ldots, a_{10}$ such that\n\n$$\n\\begin{aligned}\n& \\text{ I. } a_{i}=0 \\text{ or } 1 \\text{ for all } i \\\\\n& \\text{ II. } a_{i} \\cdot a_{i+1}=0 \\text{ for } i=1,2, \\ldots, 9 \\\\\n& \\text{ III. } a_{i} \\cdot a_{i+2}=0 \\text{ for } i=1,2, \\ldots, 8\n\\end{aligned}\n$$",
    "options": [],
    "answer": "60",
    "solution": "Solution:\n\nAnswer: 60. Call such a sequence \"good,\" and let $A_{n}$ be the number of good sequences of length $n$. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be a good sequence. If $a_{1}=0$, then $a_{1}, a_{2}, \\ldots, a_{n}$ is a good sequence if and only if $a_{2}, \\ldots, a_{n}$ is a good sequence, so there are $A_{n-1}$ of them. If $a_{1}=1$, then we must have $a_{2}=a_{3}=0$, and in this case, $a_{1}, a_{2}, \\ldots, a_{n}$ is a good sequence if and only if $a_{4}, a_{5}, \\ldots, a_{n}$ is a good sequence, so there are $A_{n-3}$ of them. We thus obtain the recursive relation $A_{n}=A_{n-1}+A_{n-3}$. Note that $A_{1}=2, A_{2}=3, A_{3}=4$. Plugging these into the recursion eventually yields $A_{10}=60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nI have four distinct rings that I want to wear on my right hand (five distinct fingers). One of these rings is a Canadian ring that must be worn on a finger by itself, the rest I can arrange however I want. If I have two or more rings on the same finger, then I consider different orders of rings along the same finger to be different arrangements. How many different ways can I wear the rings on my fingers?",
    "options": [],
    "answer": "600",
    "solution": "Solution:\n\nAnswer: $600$. First we pick the finger for the Canadian ring. This gives a multiplicative factor of $5$. For distributing the remaining $3$ rings among $4$ fingers, they can either be all on the same finger ($4 \\cdot 3!$ ways), all on different fingers ($\\binom{4}{3} \\cdot 3!$ ways), or two on one finger and one on another ($4 \\cdot \\binom{3}{2} \\cdot 2! \\cdot 3$ ways). Therefore, I have $5 \\cdot (24 + 24 + 72) = 600$ choices.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$\\triangle ABC$ is right angled at $A$. $D$ is a point on $AB$ such that $CD=1$. $AE$ is the altitude from $A$ to $BC$. If $BD=BE=1$, what is the length of $AD$?",
    "options": [],
    "answer": "∛2 − 1",
    "solution": "Solution:\n\nAnswer: $\\sqrt[3]{2}-1$. Let $AD = x$, angle $ABC = t$. We also have $\\angle BCA = 90 - t$ and $\\angle DCA = 90 - 2t$ so that $\\angle ADC = 2t$. Considering triangles $ABE$ and $ADC$, we obtain, respectively,\n$\\cos(t) = 1/(1 + x)$ and $\\cos(2t) = x$. By the double angle formula we get, $(1 + x)^3 = 2$.\n\nAlternatively, construct $M$, the midpoint of segment $BC$, and note that triangles $ABC$, $EBA$, and $MBD$ are similar. Thus, $AB^2 = BC \\cdot BE = BC$. In particular,\n$$\nAB = \\frac{BC}{AB} = \\frac{AB}{BE} = \\frac{BD}{BM} = \\frac{2BD}{BC} = \\frac{2}{AB^2}\n$$\nfrom which $AB = \\sqrt[3]{2}$ and $AD = \\sqrt[3]{2} - 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCyclic quadrilateral $ABCD$ has side lengths $AB = 1$, $BC = 2$, $CD = 3$, and $AD = 4$. Determine $AC / BD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "5/7",
    "solution": "Solution:\n\nAnswer: $\\dfrac{5}{7}$. Let the diagonals intersect at $P$. Note that triangles $ABP$ and $DCP$ are similar, so that $3AP = DP$ and $3BP = CP$. Additionally, triangles $BCP$ and $ADP$ are similar, so that $2BP = AP$. It follows that\n$$\n\\frac{AC}{BD} = \\frac{AP + PC}{BP + PD} = \\frac{2BP + 3BP}{BP + 6BP} = \\frac{5}{7}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S$ denote the set of all triples $(i, j, k)$ of positive integers where $i + j + k = 17$. Compute\n$$\n\\sum_{(i, j, k) \\in S} i j k.\n$$",
    "options": [],
    "answer": "11628",
    "solution": "Solution:\nAnswer: $11628 = \\binom{19}{5}$. We view choosing five objects from a row of $19$ objects in an unusual way. First, remove two of the chosen objects, the second and fourth, which are not adjacent nor at either end, forming three nonempty groups of consecutive objects. We then have $i, j$, and $k$ choices for the first, third, and fifth objects. Because this is a reversible process taking a triple $(i, j, k)$ to $i j k$ choices, the answer is $\\binom{19}{5} = 11628$.\n\nA simple generating functions argument is also possible. Let $s_n = \\sum_{i + j + k = n} i j k$. Then\n$$\n\\sum_{n \\geq 0} s_n x^n = \\left(\\sum_{n \\geq 0} n x^n\\right)^3 = \\left(\\frac{x}{(1-x)^2}\\right)^3 = \\frac{x^3}{(1-x)^6}\n$$\nand so $s_n = \\binom{n+2}{5}$, yielding $s_{17} = \\binom{19}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $n$ such that\n$$\n\\sum_{k=1}^{n} \\phi(k)=\\frac{3 n^{2}+5}{8}\n$$",
    "options": [],
    "answer": "1, 3, 5",
    "solution": "Solution:\nAnswer: $1$, $3$, $5$.\n\nWe contend that the proper relation is\n$$\n\\sum_{k=1}^{n} \\phi(k) \\leq \\frac{3 n^{2}+5}{8}\n$$\nLet $\\Phi(k)$ denote the left hand side of $(*)$. It is trivial to see that for $n \\leq 7$ the posed inequality holds, has equality where $n=1,3,5$, and holds strictly for $n=7$. Note that $\\phi(2 k) \\leq k$ and $\\phi(2 k+1) \\leq 2 k$, the former because $2,4, \\ldots, 2 k$ share a common divisor. It follows that $\\phi(2 k)+\\phi(2 k+1) \\leq 3 k$. Suppose for the sake of induction that $\\Phi(2 k-1)<\\frac{3(2 k-1)^{2}+5}{8}$. Then\n$$\n\\Phi(2 k+1)=\\Phi(2 k-1)+\\phi(2 k)+\\phi(2 k+1)<\\frac{3(2 k-1)^{2}+5}{8}+3 k=\\frac{3(2 k+1)^{2}+5}{8}\n$$\nTo complete the proof, it is enough to note that for a positive integer $k$,\n$$\n\\frac{3(2 k-1)^{2}+5}{8}+k<\\frac{3(2 k)^{2}+5}{8}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\ng is a twice differentiable function over the positive reals such that\n$$\n\\begin{aligned}\ng(x)+2 x^{3} g^{\\prime}(x)+x^{4} g^{\\prime \\prime}(x) & =0 \\quad \\text{ for all positive reals } x . \\\\\n\\lim _{x \\rightarrow \\infty} x g(x) & =1\n\\end{aligned}\n$$\nFind the real number $\\alpha>1$ such that $g(\\alpha)=1 / 2$.",
    "options": [],
    "answer": "6/pi",
    "solution": "Solution:\n\nAnswer: $\\frac{6}{\\pi}$.\n\nIn the first equation, we can convert the expression $2 x^{3} g^{\\prime}(x)+x^{4} g^{\\prime \\prime}(x)$ into the derivative of a product, and in fact a second derivative, by writing $y=1 / x$. Specifically,\n$$\n\\begin{aligned}\n0=g(x)+2 x^{3} g^{\\prime}(x)+x^{4} g^{\\prime \\prime}(x) & =g\\left(\\frac{1}{y}\\right)+2 y^{-3} g^{\\prime}\\left(\\frac{1}{y}\\right)+y^{-4} g^{\\prime \\prime}\\left(\\frac{1}{y}\\right) \\\\\n& =g\\left(\\frac{1}{y}\\right)+\\frac{\\mathrm{d}}{\\mathrm{~d} y}\\left[-y^{-2} g^{\\prime}\\left(\\frac{1}{y}\\right)\\right] \\\\\n& =g\\left(\\frac{1}{y}\\right)+\\frac{\\mathrm{d}^{2}}{\\mathrm{~d} y^{2}}\\left[g\\left(\\frac{1}{y}\\right)\\right]\n\\end{aligned}\n$$\nThus $g\\left(\\frac{1}{y}\\right)=c_{1} \\cos (y)+c_{2} \\sin (y)$ or $g(x)=c_{1} \\cos (1 / x)+c_{2} \\sin (1 / x)$.\n\nNow the second condition gives\n$$\n1=\\lim _{x \\rightarrow \\infty} c_{1} x+c_{2} \\cdot \\frac{\\sin (1 / x)}{1 / x}=c_{2}+\\lim _{x \\rightarrow \\infty} c_{1} x\n$$\nIt must be that $c_{1}=0, c_{2}=1$.\n\nNow since $0<1 / \\alpha<1$, the value of $\\alpha$ such that $g(\\alpha)=\\sin (1 / \\alpha)=1 / 2$ is given by $1 / \\alpha=\\pi / 6$ and so $\\alpha=6 / \\pi$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47392,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a set of $n$ coins with distinct integer weights $w_1, w_2, \\dots, w_n$. It is known that if any coin with weight $w_k$, $1 \\le k \\le n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $n$ for which such a set of coins exists.\n(This problem was suggested by Gregory Galperin.)",
    "options": [],
    "answer": "n = 1 or n is odd and at least 7",
    "solution": "The only such $n$ are $1$ and odd $n$ at least $7$. We divide into cases: $n$ even, $n = 1$, $n = 3$, $n = 5$, and $n \\ge 7$ odd.\n\n**Case 1:** $n$ even. Suppose for contradiction that such a set exists, and choose one with minimal total weight. Let $s$ be the sum of the weights, and note that\n$$\ns - w_1 \\equiv \\dots \\equiv s - w_n \\equiv 0 \\pmod{2}.\n$$\nHence\n$$\nw_1 \\equiv \\dots \\equiv w_n \\pmod{2}.\n$$\nSince $n$ is even, it follows that $s$ is even, and so also that $w_i$ is even for all $i$. But then the set $\\{w_i/2\\}$ is another set of $n$ weights with the desired property, contradicting minimality. Hence $n$ odd fails.\n\n**Case 2:** $n = 1$. Trivially yes.\n\n**Case 3:** $n = 3$. Removing any weight leaves two unequal weights.\n\n**Case 4:** $n = 5$. Suppose for contradiction that such a set exists: order the weights such that $w_1 < w_2 < w_3 < w_4 < w_5$. Let $t = w_2 + w_3 + w_4$. For each $i \\in \\{2, 3, 4\\}$, the only possible ways to split the weights other than $w_i$ are as $w_5 = t - w_i + w_1$ or $w_5 + w_1 = t - w_i$ (in all other combinations, the side with $w_5$ is strictly heavier than the other). By pigeonhole, one of the equations is satisfied for two values of $i$, and the corresponding weights are equal, contradiction.\n\n**Case 5:** $n \\ge 7$ odd. Checking the cases $n = 7$, $n = 9$, and $n = 11$ is straightforward casework: the sets $\\{1, 3, 5, \\dots, 2n - 1\\}$ suffice. For example, if $n = 7$, the set is $\\{1, 3, 5, \\dots, 13\\}$ and\n$$\n\\begin{aligned}\n3 + 5 + 7 + 9 &= 11 + 13 \\\\\n1 + 9 + 13 &= 5 + 7 + 11 \\\\\n1 + 3 + 7 + 11 &= 9 + 13 \\\\\n1 + 9 + 11 &= 3 + 5 + 13 \\\\\n1 + 3 + 5 + 11 &= 7 + 13 \\\\\n1 + 5 + 13 &= 3 + 7 + 9 \\\\\n1 + 3 + 5 + 9 &= 7 + 11.\n\\end{aligned}\n$$\nSuppose there exists such a set for $n = k$ and $n = j$, let $w_1 < w_2 < \\dots < w_k$ and $v_1 < \\dots < v_j$ be the corresponding set of weights and let $s = v_1 + \\dots + v_j$. Consider the set of weights $w_1v_1, w_1v_2, \\dots, w_1v_j, w_2s, w_3s, \\dots, w_ks$. These are distinct (each is less than the next in the list by assumption). If weight $w_1v_i$ is removed, then by the inductive hypothesis the other $v$'s can be sorted into two groups of equal weight, as can the $w$'s other than $w_1$, so the remaining weights can be divided into two groups of equal weight. If weight $w_is$ is removed, then put all the weights $w_1v_1, \\dots, w_1v_j$ together to get weights $w_1s, w_2s, \\dots, w_{i-1}s, w_{i+1}s, \\dots, w_ks$. By the inductive hypothesis these can be divided into two groups of equal weight. Hence if such a set exists for $n = k$ and $n = j$, it exists for $k + j - 1$.\n\nTaking $j = 7$, we see that if a set exists for $n = k$, one also exists for $n = k + 6$. By induction, it follows that a set exists for all $n$ of the form $6m + 7$, $6m + 9$, and $6m + 11$, where $m$ is a non-negative integer. Since this covers all odd numbers greater than $7$, we are done.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for each positive integer $n$, there are pairwise relatively prime integers $k_0, k_1, \\dots, k_n$, all strictly greater than 1, such that $k_0 k_1 \\cdots k_n - 1$ is the product of two consecutive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We proceed by induction. The case $n=1$ is clear, since we may pick $k_0 = 3$ and $k_1 = 7$. Let us assume now that for a certain $n$ there are pairwise relatively prime integers $1 < k_0 < k_1 < \\cdots < k_n$ such that $k_0 k_1 \\cdots k_n - 1 = a_n(a_n - 1)$, for some positive integer $a_n$. Then choosing $k_{n+1} = a_n^2 + a_n + 1$ yields\n$$\nk_0 k_1 \\cdots k_{n+1} = (a_n^2 - a_n + 1)(a_n^2 + a_n + 1) = a_n^4 + a_n^2 + 1,\n$$\nso $k_0 k_1 \\cdots k_{n+1} - 1$ is the product of the two consecutive integers $a_n^2$ and $a_n^2 + 1$. Moreover,\n$$\n\\gcd(k_0 k_1 \\cdots k_n, k_{n+1}) = \\gcd(a_n^2 - a_n + 1, a_n^2 + a_n + 1) = 1,\n$$\nhence $k_0, k_1, \\dots, k_{n+1}$ are pairwise relatively prime. This completes the proof.\nWrite the relation to be proved as\n$$\n4k_0 k_1 \\cdots k_n = 4a(a+1) + 4 = (2a+1)^2 + 3.\n$$\nThere are infinitely many primes for which $-3$ is a quadratic residue. Let $2 < p_0 < p_1 < \\dots < p_n$ be such primes. Using the Chinese Remainder Theorem to specify $a$ modulo $p_n$, we can find an integer $a$ such that $(2a+1)^2 + 3 = 4p_0 p_1 \\cdots p_n m$ for some positive integer $m$. Grouping the factors of $m$ appropriately with the $p_i$'s, we obtain $(2a+1)^2 + 3 = 4k_0 k_1 \\cdots k_n$ with $k_i$ pairwise relatively prime. We then have $k_0 k_1 \\cdots k_n - 1 = a(a+1)$, as desired.\nWe are supposed to show that for every positive integer $n$, there is a positive integer $x$ such that $x(x+1)+1 = x^2+x+1$ has at least $n$ distinct prime divisors. We can actually prove a more general statement.\n\n**Claim.** Let $P(x) = a_d x^d + \\cdots + a_1 x + 1$ be a polynomial of degree $d \\ge 1$ with integer coefficients. Then for every positive integer $n$, there is a positive integer $x$ such that $P(x)$ has at least $n$ distinct prime divisors.\n\nThe proof follows from the following two lemmas.\n\n**Lemma 1.** The following set is infinite:\n$$\nQ = \\{p \\mid p \\text{ a prime for which there is an integer } x \\text{ such that } p \\text{ divides } P(x)\\}.\n$$\n*Proof.* The proof is analogous to Euclid's proof that there are infinitely many primes. Namely, if we assume that there are only finitely many primes $p_1, p_2, \\dots, p_k$ in $Q$, then for each integer $m$, $P(m p_1 p_2 \\cdots p_k)$ is an integer with no prime factors, which must equal 1 or $-1$. However, since $P$ has degree $d \\ge 1$, it takes each of the values 1 and $-1$ at most $d$ times, a contradiction. $\\square$\n\n**Lemma 2.** Let $p_1, p_2, \\dots, p_n$, $n \\ge 1$ be primes in $Q$. Then there is a positive integer $x$ such that $P(x)$ is divisible by $p_1 p_2 \\cdots p_n$.\n*Proof.* For $i = 1, 2, \\dots, n$, since $p_i \\in Q$ we can find an integer $c_i$ such that $P(x)$ is divisible by $p_i$ whenever $x \\equiv c_i(\\text{mod } p_i)$. By the Chinese Remainder Theorem, the system of $n$ congruences $x \\equiv c_i(\\text{mod } p_i)$, $i = 1, 2, \\dots, n$ has positive integer solutions. For every positive integer $x$ that solves this system, $P(x)$ is divisible by $p_1 p_2 \\cdots p_n$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a year ultra-even if all of its digits are even. Thus $2000$, $2002$, $2004$, $2006$, and $2008$ are all ultra-even years. They are all $2$ years apart, which is the shortest possible gap. $2009$ is not an ultra-even year because of the $9$, and $2010$ is not an ultra-even year because of the $1$.\n\na. In the years between the years $1$ and $10000$, what is the longest possible gap between two ultra-even years? Give an example of two ultra-even years that far apart with no ultra-even years between them. Justify your answer.\n\nb. What is the second-shortest possible gap (that is, the shortest gap longer than $2$ years) between two ultra-even years? Again, give an example, and justify your answer.",
    "options": [],
    "answer": "a) 1112; for example, 888 to 2000. b) 12; for example, 8 to 20.",
    "solution": "Solution:\n\na. The longest possible gap has length $1112$. There are four valid examples of two ultra-even years that far apart with no ultra-even years between them: $888$ to $2000$, $2888$ to $4000$, $4888$ to $6000$, or $6888$ to $8000$. ($8888$ to $10000$ is not acceptable as $10000$ is not ultra-even).\n\nTo see that no longer gap is possible, note that if $A$ and $B$ are ultra-even years between $1$ and $10000$ with no ultra-even year between them, and $B-A \\geq 1000$, then $A$ and $B$ must have different digits in their \"thousands place\". Since both of these digits must be even, that means there must be at least one odd digit $d$ in between them, and so all the years between $d000$ and $d999$ must be in the gap between $A$ and $B$. But the smallest ultra-even year greater than $d999$ is $e000$ (where $e = d+1$, of course), and the largest ultra-even year less than $d000$ is $c888$ (where $c = d-1$), so this gap must be of length exactly $1112$.\n\nAnother approach is to consider the increasing sequence of ultra-even years: $2$ (the smallest ultra-even year), $888$, $2000$, $2888$, $4000$, $6000$, $6888$, $8000$, $8888$ (the largest ultra-even year less than $10000$). Every ultra-even year is either in this sequence or lies between two elements of this sequence, so any two ultra-even years with no ultra-even years between them must have a difference that is less than or equal to a difference between two consecutive elements of this set; but the largest such difference is $1112$.\n\nb. The second-shortest possible gap is $12$. Examples include $8$ to $20$, $4268$ to $4280$, etc.\n\nTo see that this really is the second-shortest possible gap, note that every gap must be even, since the difference between any two ultra-even years is even. There are certainly many gaps of length $2$ (from $4022$ to $4024$, for example). We must show that there are no gaps with lengths $4, 6, 8$, or $10$. Any ultra-even year whose units digit is $0, 2, 4$, or $6$ is $2$ less than another ultra-even year (since adding $2$ to this year won't change any other digit). So if an ultra-even year is the lower year of a gap longer than $2$ years, its units digit must be $8$. However, if year $A$ is ultra-even, and has a units digit of $8$, adding $2, 4, 6, 8$, or $10$ to it will result in a year whose tens digit is one greater than the tens digit of year $A$, which will mean the new number is not ultra-even. So $12$ is the smallest possible value for the gap between such years $A$ and the next largest ultra-even number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPositive real numbers $x$, $y$ satisfy the equations $x^{2} + y^{2} = 1$ and $x^{4} + y^{4} = \\frac{17}{18}$. Find $x y$.",
    "options": [],
    "answer": "1/6",
    "solution": "Solution:\n\nWe have\n$$\n2 x^{2} y^{2} = (x^{2} + y^{2})^{2} - (x^{4} + y^{4}) = 1^{2} - \\frac{17}{18} = 1 - \\frac{17}{18} = \\frac{1}{18}.\n$$\nSo\n$$\nx^{2} y^{2} = \\frac{1}{36} \\implies x y = \\frac{1}{6}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of all solutions for $x$ :\n$$\n\\begin{aligned}\nx y & = 1 \\\\\nx + y & = 3\n\\end{aligned}\n$$",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nSubstitute $3 - x$ in for $y$ into the first equation:\n$$\nx(3 - x) = 1 \\Leftrightarrow x^{2} - 3x + 1 = 0\n$$\nThis equation has two distinct roots, each of which corresponds to a possible solution $x$. The sum of the roots of the quadratic equation $a x^{2} + b x + c = 0$ is $\\frac{-b}{a}$, which in this case is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA triangle has sides of length $9$, $40$, and $41$. What is its area?",
    "options": [],
    "answer": "180",
    "solution": "Solution:\nObserve that $9^{2} + 40^{2} = 41^{2}$, so this triangle is right and therefore has area $\\frac{1}{2} \\cdot 9 \\cdot 40 = 180$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = 1 + x + x^{2} + \\cdots + x^{100}$. Find $f'(1)$.",
    "options": [],
    "answer": "5050",
    "solution": "Solution:\nNote that $f'(x) = 1 + 2x + 3x^{2} + \\cdots + 100x^{99}$, so $f'(1) = 1 + 2 + \\cdots + 100 = \\frac{100 \\cdot 101}{2} = 5050$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoe B. first places the black king in one corner of the board. In how many of the 35 remaining squares can he place a white bishop so that it does not check the black king?",
    "options": [],
    "answer": "30",
    "solution": "Solution:\n\nAny square not on the diagonal containing the corner is a possible location, and there are $36 - 6 = 30$ such squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the number of ways of walking from $(0,0)$ to $(5,5)$ using only up and right unit steps such that the path does not pass through any of the following points: $(1,1)$, $(1,4)$, $(4,1)$, $(4,4)$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "34",
    "solution": "Solution:\nIn the following figure, each lattice point (with the bottom-left-most point $(0,0)$) is labeled with the number of ways of reaching there from $(0,0)$. With the exception of the forbidden points, the labels satisfy the recursion formula $f(x, y) = f(x-1, y) + f(x, y-1)$. We see from the diagram that there are 34 ways to reach $(5,5)$.\n\n| 1 | 1 | 5 | 17 | 17 | 34 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 1 | 0 | 4 | 12 | 0 | 17 |\n| 1 | 2 | 4 | 8 | 12 | 17 |\n| 1 | 1 | 2 | 4 | 4 | 5 |\n| 1 | 0 | 1 | 2 | 0 | 1 |\n| 1 | 1 | 1 | 1 | 1 | 1 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all positive integers $p$ such that $p$, $p+4$, and $p+8$ are all prime.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nIf $p=3$, then $p+4=7$ and $p+8=11$, both prime.\n\nIf $p \\neq 3$, then $p$ is not a multiple of $3$ and is therefore of one of the forms $3k+1$ or $3k+2$ ($k \\geq 0$).\n\nIf $p=3k+1$, then $p+8=3k+9=3(k+3)$, which is not prime since $k+3>1$.\n\nIf $p=3k+2$, then $p+4=3k+6=3(k+2)$, which is not prime since $k+2>1$.\n\nThus $p=3$ is the only solution with all three numbers prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the minimum of $x^{2}-2x$ over all real numbers $x$.",
    "options": [],
    "answer": "-1",
    "solution": "Solution:\nWrite $x^{2}-2x = x^{2}-2x+1-1 = (x-1)^{2}-1$. Since $(x-1)^{2} \\geq 0$, it is clear that the minimum is $-1$.\n\nAlternate method: The graph of $y = x^{2}-2x$ is a parabola that opens up. Therefore, the minimum occurs at its vertex, which is at $\\frac{-b}{2a} = \\frac{-(-2)}{2} = 1$. But $1^{2}-2 \\cdot 1 = -1$, so the minimum is $-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEach square of a $100 \\times 100$ grid is colored black or white so that there is at least one square of each color. Prove that there is a point which is a vertex of exactly one black square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLocate the uppermost row that has at least one black square. Then, within that row, find the leftmost black square. By construction, all squares above and/or to the left of that square are white. Therefore, the upper left corner of that square will solve the problem.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $3 \\times 3 \\times 3$ cube composed of 27 unit cubes rests on a horizontal plane. Determine the number of ways of selecting two distinct unit cubes from a $3 \\times 3 \\times 1$ block (the order is irrelevant) with the property that the line joining the centers of the two cubes makes a $45^{\\circ}$ angle with the horizontal plane.",
    "options": [],
    "answer": "60",
    "solution": "Solution:\n\nThere are 6 such slices, and each slice gives 10 valid pairs (with no overcounting). Therefore, there are 60 such pairs.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a unit square (that is, the labels $A$, $B$, $C$, $D$ appear in that order around the square). Let $X$ be a point outside of the square such that the distance from $X$ to $AC$ is equal to the distance from $X$ to $BD$, and also that $AX = \\frac{\\sqrt{2}}{2}$. Determine the value of $CX^{2}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "5/2",
    "solution": "Solution:\n\nSince $X$ is equidistant from $AC$ and $BD$, it must lie on either the perpendicular bisector of $AB$ or the perpendicular bisector of $AD$. It turns that the two cases yield the same answer, so we will just assume the first case. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $CD$. Then, $XM$ is perpendicular to $AB$, so $XM = \\frac{1}{2}$ and thus $XN = \\frac{3}{2}$, $NC = \\frac{1}{2}$. By the Pythagorean Theorem we find $XC = \\frac{\\sqrt{10}}{2}$ and the answer follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFour students from Harvard, one of them named Jack, and five students from MIT, one of them named Jill, are going to see a Boston Celtics game. However, they found out that only 5 tickets remain, so 4 of them must go back. Suppose that at least one student from each school must go see the game, and at least one of Jack and Jill must go see the game, how many ways are there of choosing which 5 people can see the game?",
    "options": [],
    "answer": "104",
    "solution": "Solution:\n\nLet us count the number of ways of distributing the tickets so that one of the conditions is violated. There is $1$ way to give all the tickets to MIT students, and $\\binom{7}{5}$ ways to give all the tickets to the $7$ students other than Jack and Jill. Therefore, the total number of valid ways is $\\binom{9}{5} - 1 - \\binom{7}{5} = 104$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine all pairs $(a, b)$ of real numbers such that $10, a, b, a b$ is an arithmetic progression.",
    "options": [],
    "answer": "(4, -2) and (5/2, -5)",
    "solution": "Solution:\n\nThe answer is $(4, -2)$ and $\\left(\\frac{5}{2}, -5\\right)$.\n\nSince $10, a, b$ is an arithmetic progression, we have\n$$\na = \\frac{1}{2}(10 + b).\n$$\nAlso, we have $a + a b = 2 b$, and so\n$$\na(1 + b) = 2b.\n$$\nSubstituting the expression for $a$ gives\n$$\n(10 + b)(1 + b) = 4b.\n$$\nSolving this quadratic equation gives the solutions $b = -2$ and $b = -5$. The corresponding values for $a$ can be found by $a = \\frac{1}{2}(10 + b)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many different values can $\\angle ABC$ take, where $A, B, C$ are distinct vertices of a cube?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nAnswer: 5. In a unit cube, there are 3 types of triangles, with side lengths $(1, 1, \\sqrt{2})$, $(1, \\sqrt{2}, \\sqrt{3})$ and $(\\sqrt{2}, \\sqrt{2}, \\sqrt{2})$. Together they generate 5 different angle values.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest positive integer $n$ such that $n^{3}+4 n^{2}-15 n-18$ is the cube of an integer.",
    "options": [],
    "answer": "19",
    "solution": "Solution:\nAnswer: 19\n\nNote that the next cube after $n^{3}$ is $(n+1)^{3} = n^{3} + 3 n^{2} + 3 n + 1$. After that, it is $(n+2)^{3} = n^{3} + 6 n^{2} + 12 n + 8$. $n^{3} + 6 n^{2} + 12 n + 8$ is definitely bigger than $n^{3} + 4 n^{2} - 15 n - 18$, so the largest cube that $n^{3} + 4 n^{2} - 15 n - 18$ could be is $(n+1)^{3}$. On the other hand, for $n \\geq 4$, $n^{3} + 4 n^{2} - 15 n - 18$ is larger than $(n-2)^{3} = n^{3} - 6 n^{2} + 12 n - 8$ (as $4 n^{2} - 15 n - 18 > -6 n^{2} + 12 n - 8 \\Longleftrightarrow 10 n^{2} - 27 n - 10 > 0$, which is true for $n \\geq 4$).\n\nSo, we will check for all solutions to $n^{3} + 4 n^{2} - 15 n - 18 = (n-1)^{3}, n^{3}, (n+1)^{3}$.\n\nThe first case yields\n$$\nn^{3} + 4 n^{2} - 15 n - 18 = n^{3} - 3 n^{2} + 3 n - 1 \\Longleftrightarrow 7 n^{2} - 18 n - 17 = 0\n$$\nwhich has no integer solutions.\n\nThe second case yields\n$$\nn^{3} + 4 n^{2} - 15 n - 18 = n^{3} \\Longleftrightarrow 4 n^{2} - 15 n - 18 = 0\n$$\nwhich also has no integer solutions.\n\nThe final case yields\n$$\nn^{3} + 4 n^{2} - 15 n - 18 = n^{3} + 3 n^{2} + 3 n + 1 \\Longleftrightarrow n^{2} - 18 n - 19 = 0\n$$\nwhich has integer solutions $n = -1, 19$. So, the largest possible $n$ is $19$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the number of 8-tuples of nonnegative integers $\\left(a_{1}, a_{2}, a_{3}, a_{4}, b_{1}, b_{2}, b_{3}, b_{4}\\right)$ satisfying $0 \\leq a_{k} \\leq k$, for each $k=1,2,3,4$, and\n$$\na_{1} + a_{2} + a_{3} + a_{4} + 2 b_{1} + 3 b_{2} + 4 b_{3} + 5 b_{4} = 19.\n$$",
    "options": [],
    "answer": "1540",
    "solution": "Answer: 1540 Same as Combinatorics Test problem 10.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate the infinite sum\n$$\n\\sum_{n=0}^{\\infty} \\binom{2 n}{n} \\frac{1}{5^{n}}\n$$",
    "options": [],
    "answer": "sqrt(5)",
    "solution": "Solution:\nAnswer: $\\sqrt{5}$\n\nFirst Solution: Note that\n$$\n\\begin{aligned}\n\\binom{2 n}{n} & =\\frac{(2 n)!}{n!\\cdot n!}=\\frac{(2 n)(2 n-2)(2 n-4) \\cdots(2)}{n!} \\cdot \\frac{(2 n-1)(2 n-3)(2 n-5) \\cdots(1)}{n!} \\\\\n& =2^{n} \\cdot \\frac{(-2)^{n}}{n!}\\left(-\\frac{1}{2}\\right)\\left(-\\frac{1}{2}-1\\right)\\left(-\\frac{1}{2}-2\\right) \\cdots\\left(-\\frac{1}{2}-n+1\\right) \\\\\n& =(-4)^{n}\\binom{-\\frac{1}{2}}{n}\n\\end{aligned}\n$$\nThen, by the binomial theorem, for any real $x$ with $|x|<\\frac{1}{4}$, we have\n$$\n(1-4 x)^{-1 / 2}=\\sum_{n=0}^{\\infty}\\binom{-\\frac{1}{2}}{n}(-4 x)^{n}=\\sum_{n=0}^{\\infty}\\binom{2 n}{n} x^{n} .\n$$\nTherefore,\n$$\n\\sum_{n=0}^{\\infty}\\binom{2 n}{n}\\left(\\frac{1}{5}\\right)^{n}=\\frac{1}{\\sqrt{1-\\frac{4}{5}}}=\\sqrt{5}\n$$\n\n\nSecond Solution: Consider the generating function\n$$\nf(x)=\\sum_{n=0}^{\\infty}\\binom{2 n}{n} x^{n}\n$$\nIt has formal integral given by\n$$\ng(x)=I(f(x))=\\sum_{n=0}^{\\infty} \\frac{1}{n+1}\\binom{2 n}{n} x^{n+1}=\\sum_{n=0}^{\\infty} C_{n} x^{n+1}=x \\sum_{n=0}^{\\infty} C_{n} x^{n}\n$$\nwhere $C_{n}=\\frac{1}{n+1}\\binom{2 n}{n}$ is the $n$th Catalan number. Let $h(x)=\\sum_{n=0}^{\\infty} C_{n} x^{n}$; it suffices to compute this generating function. Note that\n$$\n1+x h(x)^{2}=1+x \\sum_{i, j \\geq 0} C_{i} C_{j} x^{i+j}=1+x \\sum_{k \\geq 0}\\left(\\sum_{i=0}^{k} C_{i} C_{k-i}\\right) x^{k}=1+\\sum_{k \\geq 0} C_{k+1} x^{k+1}=h(x)\n$$\nwhere we've used the recurrence relation for the Catalan numbers. We now solve for $h(x)$ with the quadratic equation to obtain\n$$\nh(x)=\\frac{1 / x \\pm \\sqrt{1 / x^{2}-4 / x}}{2}=\\frac{1 \\pm \\sqrt{1-4 x}}{2 x}\n$$\nNote that we must choose the - sign in the $\\pm$, since the + would lead to a leading term of $\\frac{1}{x}$ for $h$ (by expanding $\\sqrt{1-4 x}$ into a power series). Therefore, we see that\n$$\nf(x)=D(g(x))=D(x h(x))=D\\left(\\frac{1-\\sqrt{1-4 x}}{2}\\right)=\\frac{1}{\\sqrt{1-4 x}}\n$$\nand our answer is hence $f(1 / 5)=\\sqrt{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn the circumcircle of $A B C$, let $A'$ be the midpoint of arc $B C$ (not containing $A$).\n\na. Show that $A, I, A'$ are collinear.\n\nb. Show that $A'$ is the circumcenter of $BIC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\na. Since $A'$ bisects the arc $B C$, the two arcs $A' B$ and $A' C$ are equal, and so $\\angle B A A' = \\angle C A A'$. Thus, $A'$ lies on the angle bisector of $B A C$. Since $I$ also lies on the angle bisector of $B A C$, we see that $A, I, A'$ are collinear.\n\nb. We have\n\n$$\n\\angle C I A' = \\angle A' A C + \\angle I C A = \\angle A' A B + \\angle I C B = \\angle A' C B + \\angle I C B = \\angle I C A'.\n$$\n\nTherefore, $A' I = A' C$. By similar arguments, $A' I = A' B$. So, $A'$ is equidistant from $B, I, C$, and thus is its circumcenter.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an equilateral triangle with side length $2$, and let $\\Gamma$ be a circle with radius $\\frac{1}{2}$ centered at the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on $\\Gamma$, visits all three sides of $ABC$, and ends somewhere on $\\Gamma$ (not necessarily at the starting point). Express your answer in the form of $\\sqrt{p}-q$, where $p$ and $q$ are rational numbers written as reduced fractions.",
    "options": [],
    "answer": "sqrt(4/3) - 1",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate the integral $\\int_{0}^{1} \\ln x \\ln (1-x) d x$.",
    "options": [],
    "answer": "2 - π^2/6",
    "solution": "Solution:\nAnswer: $2-\\frac{\\pi^{2}}{6}$\n\nWe have the MacLaurin expansion $\\ln (1-x) = -x - \\frac{x^{2}}{2} - \\frac{x^{3}}{3} - \\cdots$. So\n$$\n\\int_{0}^{1} \\ln x \\ln (1-x) d x = -\\int_{0}^{1} \\ln x \\sum_{n=1}^{\\infty} \\frac{x^{n}}{n} d x = -\\sum_{n=1}^{\\infty} \\frac{1}{n} \\int_{0}^{1} x^{n} \\ln x d x\n$$\nUsing integration by parts, we get\n$$\n\\int_{0}^{1} x^{n} \\ln x d x = \\left.\\frac{x^{n+1} \\ln x}{n+1}\\right|_{0}^{1} - \\int_{0}^{1} \\frac{x^{n}}{n+1} d x = -\\frac{1}{(n+1)^{2}}\n$$\n(We used the fact that $\\lim_{x \\rightarrow 0} x^{n} \\ln x = 0$ for $n > 0$, which can be proven using l'Hôpital's rule.) Therefore, the original integral equals to\n$$\n\\sum_{n=1}^{\\infty} \\frac{1}{n(n+1)^{2}} = \\sum_{n=1}^{\\infty}\\left(\\frac{1}{n} - \\frac{1}{n+1} - \\frac{1}{(n+1)^{2}}\\right)\n$$\nTelescoping the sum and using the well-known identity $\\sum_{n=1}^{\\infty} \\frac{1}{n^{2}} = \\frac{\\pi^{2}}{6}$, we see that the above sum is equal to $2-\\frac{\\pi^{2}}{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of subsets $S$ of $\\{1,2, \\ldots, 63\\}$ the sum of whose elements is $2008$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nNote that $1+2+\\cdots+63=2016$. So the problem is equivalent to finding the number of subsets of $\\{1,2, \\cdots, 63\\}$ whose sum of elements is $8$.\n\nWe can count this by hand:\n$\\{8\\}$,\n$\\{1,7\\}$,\n$\\{2,6\\}$,\n$\\{3,5\\}$,\n$\\{1,2,5\\}$,\n$\\{1,3,4\\}$.\n\nSo the answer is $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of all positive integers $n$ such that $n$ divides $n^{2}+n+2$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nSince $n$ always divides $n^{2}+n$, the only $n$ that work are divisors of $2$, because if $n$ divides $a$ and $n$ divides $b$, then $n$ divides $a+b$. So the solutions are $1$ and $2$ which sum to $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $BC = 2007$, $CA = 2008$, $AB = 2009$. Let $\\omega$ be an excircle of $ABC$ that touches the line segment $BC$ at $D$, and touches extensions of lines $AC$ and $AB$ at $E$ and $F$, respectively (so that $C$ lies on segment $AE$ and $B$ lies on segment $AF$). Let $O$ be the center of $\\omega$. Let $\\ell$ be the line through $O$ perpendicular to $AD$. Let $\\ell$ meet line $EF$ at $G$. Compute the length $DG$.",
    "options": [],
    "answer": "2014024",
    "solution": "Solution:\n\nAnswer: $2014024$\n\nLet line $AD$ meet $\\omega$ again at $H$. Since $AF$ and $AE$ are tangents to $\\omega$ and $ADH$ is a secant, we see that $DEHF$ is a harmonic quadrilateral. This implies that the pole of $AD$ with respect to $\\omega$ lies on $EF$. Since $\\ell \\perp AD$, the pole of $AD$ lies on $\\ell$. It follows that the pole of $AD$ is $G$.\n\n![](attached_image_1.png)\n\nThus, $G$ must lie on the tangent to $\\omega$ at $D$, so $C, D, B, G$ are collinear. Furthermore, since the pencil of lines $(AE, AF ; AD, AG)$ is harmonic, by intersecting it with the line $BC$, we see that $(C, B ; D, G)$ is harmonic as well. This means that\n$$\n\\frac{BD}{DC} \\cdot \\frac{CG}{GB} = -1\n$$\n(where the lengths are directed.) The semiperimeter of $ABC$ is $s = \\frac{1}{2}(2007 + 2008 + 2009) = 3012$. So $BD = s - 2009 = 1003$ and $CD = s - 2008 = 1004$. Let $x = DG$, then the above equation gives\n$$\n\\frac{1003}{1004} \\cdot \\frac{x + 1004}{x - 1003} = 1\n$$\nSolving gives $x = 2014024$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(r) = \\sum_{j=2}^{2008} \\frac{1}{j^{r}} = \\frac{1}{2^{r}} + \\frac{1}{3^{r}} + \\cdots + \\frac{1}{2008^{r}}$. Find $\\sum_{k=2}^{\\infty} f(k)$.",
    "options": [],
    "answer": "2007/2008",
    "solution": "Solution:\nAnswer: $\\frac{2007}{2008}$\nWe change the order of summation:\n$$\n\\sum_{k=2}^{\\infty} \\sum_{j=2}^{2008} \\frac{1}{j^{k}} = \\sum_{j=2}^{2008} \\sum_{k=2}^{\\infty} \\frac{1}{j^{k}} = \\sum_{j=2}^{2008} \\frac{1}{j^{2}\\left(1-\\frac{1}{j}\\right)} = \\sum_{j=2}^{2008} \\frac{1}{j(j-1)} = \\sum_{j=2}^{2008}\\left(\\frac{1}{j-1}-\\frac{1}{j}\\right) = 1-\\frac{1}{2008} = \\frac{2007}{2008}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAl has a rectangle of integer side lengths $a$ and $b$, and area $1000$. What is the smallest perimeter it could have?",
    "options": [],
    "answer": "130",
    "solution": "Solution:\n\nAnswer: $130$\n\nTo minimize the sum of the side lengths, we need to keep the height and width as close as possible, because the square has the smallest perimeter of all rectangles with a fixed area. So, $40$ and $25$ multiply to $1000$ and are as close as possible—the $40 \\times 25$ rectangle has perimeter $130$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the following system of equations for $w$.\n$$\n\\begin{aligned}\n& 2w + x + y + z = 1 \\\\\n& w + 2x + y + z = 2 \\\\\n& w + x + 2y + z = 2 \\\\\n& w + x + y + 2z = 1\n\\end{aligned}\n$$",
    "options": [],
    "answer": "-1/5",
    "solution": "Solution:\nAdd all the equations together to find that $5x + 5y + 5z + 5w = 6$, or $x + y + z + w = \\frac{6}{5}$. We can now subtract this equation from the first equation to see that $w = \\frac{-1}{5}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose we have an (infinite) cone $\\mathcal{C}$ with apex $A$ and a plane $\\pi$. The intersection of $\\pi$ and $\\mathcal{C}$ is an ellipse $\\mathcal{E}$ with major axis $BC$, such that $B$ is closer to $A$ than $C$, and $BC = 4$, $AC = 5$, $AB = 3$. Suppose we inscribe a sphere in each part of $\\mathcal{C}$ cut up by $\\mathcal{E}$ with both spheres tangent to $\\mathcal{E}$. What is the ratio of the radii of the spheres (smaller to larger)?",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\nAnswer: $\\sqrt{\\frac{1}{3}}$\n\nIt can be seen that the points of tangency of the spheres with $\\mathcal{E}$ must lie on its major axis due to symmetry. Hence, we consider the two-dimensional cross-section with plane $ABC$. Then the two spheres become the incentre and the excentre of the triangle $ABC$, and we are looking for the ratio of the inradius to the exradius. Let $s$, $r$, $r_{a}$ denote the semiperimeter, inradius, and exradius (opposite to $A$) of the triangle $ABC$. We know that the area of $ABC$ can be expressed as both $rs$ and $r_{a}(s - |BC|)$, and so $\\frac{r}{r_{a}} = \\frac{s - |BC|}{s}$. For the given triangle, $s = 6$ and $a = 4$, so the required ratio is $\\frac{1}{3}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 47422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of distinct primes dividing $1 \\cdot 2 \\cdot 3 \\cdots 9 \\cdot 10$.",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nA prime divides this product if and only if it divides one of the multiplicands, so prime divisors of this product must be less than $10$. There are $4$ primes less than $10$, namely, $2, 3, 5$, and $7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P(x)$ be a polynomial with degree $2008$ and leading coefficient $1$ such that\n$$\nP(0)=2007,\\ P(1)=2006,\\ P(2)=2005,\\ \\ldots,\\ P(2007)=0.\n$$\nDetermine the value of $P(2008)$. You may use factorials in your answer.",
    "options": [],
    "answer": "2008! - 1",
    "solution": "Solution:\nAnswer: $2008! - 1$\n\nConsider the polynomial $Q(x) = P(x) + x - 2007$. The given conditions imply that $Q(x)$ is also a polynomial with degree $2008$ and leading coefficient $1$. It follows that\n$$\nQ(x) = x(x-1)(x-2)\\cdots(x-2007).\n$$\nThus,\n$$\nP(x) = x(x-1)(x-2)\\cdots(x-2007) - x + 2007.\n$$\nSetting $x = 2008$ gives the answer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $X$ be the point on side $BC$ such that $BX = CD$. Show that the excircle $ABC$ opposite of vertex $A$ touches segment $BC$ at $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet the excircle touch lines $BC$, $AC$ and $AB$ at $X'$, $Y$ and $Z$, respectively. Using the equal tangent property repeatedly, we have\n$$\nBX' - X'C = BZ - CY = (EY - CY) - (FZ - BZ) = CE - BF = CD - BD.\n$$\nIt follows that $BX' = CD$, and thus $X' = X$. So the excircle touches $BC$ at $X$.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEvaluate the infinite sum $\\sum_{n=1}^{\\infty} \\frac{n}{n^{4}+4}$.",
    "options": [],
    "answer": "3/8",
    "solution": "Solution:\n\n$\\displaystyle\n\\begin{aligned}\n\\sum_{n=1}^{\\infty} \\frac{n}{n^{4}+4} & =\\sum_{n=1}^{\\infty} \\frac{n}{\\left(n^{2}+2 n+2\\right)\\left(n^{2}-2 n+2\\right)} \\\\\n& =\\frac{1}{4} \\sum_{n=1}^{\\infty}\\left(\\frac{1}{n^{2}-2 n+2}-\\frac{1}{n^{2}+2 n+2}\\right) \\\\\n& =\\frac{1}{4} \\sum_{n=1}^{\\infty}\\left(\\frac{1}{(n-1)^{2}+1}-\\frac{1}{(n+1)^{2}+1}\\right)\n\\end{aligned}$\n\nObserve that the sum telescopes. From this we find that the answer is $\\frac{1}{4}\\left(\\frac{1}{0^{2}+1}+\\frac{1}{1^{2}+1}\\right)=\\frac{3}{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou have a $2 \\times 3$ grid filled with integers between $1$ and $9$. The numbers in each row and column are distinct, the first row sums to $23$, and the columns sum to $14$, $16$, and $17$ respectively.\n\n|   | $14$ | $16$ | $17$ |\n| :---: | :---: | :---: | :---: |\n| $23$ | $a$ | $b$ | $c$ |\n|   | $x$ | $y$ | $z$ |\n\nWhat is $x+2y+3z$?",
    "options": [],
    "answer": "49",
    "solution": "Solution:\n\nThe sum of all $6$ numbers is $14+16+17=47$, so $x+y+z=47-23=24$. If three distinct digits sum to $24$, they must be $7$, $8$, and $9$, because any other triple of digits would have a smaller sum. So, we try placing these digits in for $x$, $y$, and $z$, and the only arrangement that does not force equal digits in any row or column is $x=8$, $y=7$, $z=9$. In this case, $x+2y+3z=8+2\\times7+3\\times9=8+14+27=49$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a game show, Bob is faced with 7 doors, 2 of which hide prizes. After he chooses a door, the host opens three other doors, of which one is hiding a prize. Bob chooses to switch to another door. What is the probability that his new door is hiding a prize?",
    "options": [],
    "answer": "5/21",
    "solution": "Solution:\n\nIf Bob initially chooses a door with a prize, then he will not find a prize by switching.\n\nWith probability $\\frac{5}{7}$ his original door does not hide the prize. After the host opens the three doors, the remaining three doors have equal probability of hiding the prize. Therefore, the probability that Bob finds the prize is $\\frac{5}{7} \\times \\frac{1}{3} = \\frac{5}{21}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cat is going up a stairwell with ten stairs. However, instead of walking up the stairs one at a time, the cat jumps, going either two or three stairs up at each step (though if necessary, it will just walk the last step). How many different ways can the cat go from the bottom to the top?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\nThe number of ways for the cat to get to the $i$th step is the number of ways for the cat to get to step $i-2$ plus the number of ways to get to step $i-3$, because for each way to get to step $i$, we can undo the last move the cat made to go back to one of these two steps. The cat can get to step $1$ in $0$ ways, to step $2$ in $1$ way, and to step $3$ in $1$ way. Now we repeatedly use our formula for calculating the number of ways to get to the $i$th step to see that the cat gets to:\n\n| Step | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Number of ways | 0 | 1 | 1 | 1 | 2 | 2 | 3 | 4 | 5 | 7 |\n\nSo our answer is $5+7=12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $p$ and $q$ are positive integers and $\\frac{2008}{2009} < \\frac{p}{q} < \\frac{2009}{2010}$, what is the minimum value of $p$?",
    "options": [],
    "answer": "4017",
    "solution": "Solution:\nAnswer: $4017$\n\nBy multiplying out the fraction inequalities, we find that $2008q + 1 \\leq 2009p$ and $2010p + \\leq 2009q$. Adding $2009$ times the first inequality to $2008$ times the second, we find that $2008 \\cdot 2009q + 4017 \\leq 2008 \\cdot 2009q + p$, or $p \\geq 4017$. This minimum is attained when $\\frac{p}{q} = \\frac{4017}{4019}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoint $A$ lies at $(0,4)$ and point $B$ lies at $(3,8)$. Find the $x$-coordinate of the point $X$ on the $x$-axis maximizing $\\angle A X B$.",
    "options": [],
    "answer": "5 sqrt(2)-3",
    "solution": "Solution:\n\nAnswer: $5 \\sqrt{2}-3$\n\nLet $X$ be a point on the $x$-axis and let $\\theta=\\angle A X B$. We can easily see that the circle with diameter $A B$ does not meet the $x$-axis, so $\\theta \\leq \\pi$. Thus, maximizing $\\theta$ is equivalent to maximizing $\\sin \\theta$. By the Law of Sines, this in turn is equivalent to minimizing the circumradius of triangle $A B X$. This will occur when the circumcircle of $A B X$ is the smaller of the two circles through $A$ and $B$ tangent to the $x$-axis. So let $X$ now be this point of tangency. Extend line $A B$ to meet the $x$-axis at $C=(-3,0)$; by Power of a Point $C X^{2}=C A \\cdot C B=50$ so $C X=5 \\sqrt{2}$. Clearly $X$ has larger $x$-coordinate than $C$, so the $x$-coordinate of $X$ is $5 \\sqrt{2}-3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the last two digits of $17^{17}$, written in base 10.",
    "options": [],
    "answer": "77",
    "solution": "Solution:\nWe are asked to find the remainder when $17^{17}$ is divided by $100$. Write the power as $(7+10)^{17}$ and expand with the binomial theorem:\n$$\n(7+10)^{17} = 7^{17} + 17 \\cdot 7^{16} \\cdot 10 + \\ldots\n$$\nWe can ignore terms with more than one factor of $10$ because these terms are divisible by $100$, so adding them does not change the last two digits. Now, $7^{4} = 2401$, which has remainder $1 \\bmod 100$, so $7^{17}$ has last two digits $07$ and $7^{16} \\cdot 10$ has last two digits $70$. We add these together.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the equation\n$$\n\\sqrt{x+\\sqrt{4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{2008} x+3}}}}}-\\sqrt{x}=1\n$$\nExpress your answer as a reduced fraction with the numerator and denominator written in their prime factorization.",
    "options": [],
    "answer": "1/2^{4016}",
    "solution": "Solution:\nAnswer: $\\frac{1}{2^{4016}}$\n\nRewrite the equation to get\n$$\n\\sqrt{x+\\sqrt{4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{2008} x+3}}}}}=\\sqrt{x}+1\n$$\nSquaring both sides yields\n$$\n\\sqrt{4 x+\\sqrt{\\ldots+\\sqrt{4^{2008} x+3}}}=2 \\sqrt{x}+1\n$$\nSquaring again yields\n$$\n\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{2008} x+3}}}=4 \\sqrt{x}+1\n$$\nOne can see that by continuing this process one gets\n$$\n\\sqrt{4^{2008} x+3}=2^{2008} \\sqrt{x}+1\n$$\nso that $2 \\cdot 2^{2008} \\sqrt{x}=2$. Hence $x=4^{-2008}$.\nIt is also easy to check that this is indeed a solution to the original equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a right triangle with $\\angle A = 90^{\\circ}$. Let $D$ be the midpoint of $AB$ and let $E$ be a point on segment $AC$ such that $AD = AE$. Let $BE$ meet $CD$ at $F$. If $\\angle BFC = 135^{\\circ}$, determine $BC / AB$.",
    "options": [],
    "answer": "sqrt(13)/2",
    "solution": "Solution:\nAnswer: $\\frac{\\sqrt{13}}{2}$\n\nLet $\\alpha = \\angle ADC$ and $\\beta = \\angle ABE$. By the exterior angle theorem, $\\alpha = \\angle BFD + \\beta = 45^{\\circ} + \\beta$. Also, note that $\\tan \\beta = AE / AB = AD / AB = 1/2$. Thus,\n$$\n1 = \\tan 45^{\\circ} = \\tan (\\alpha - \\beta) = \\frac{\\tan \\alpha - \\tan \\beta}{1 + \\tan \\alpha \\tan \\beta} = \\frac{\\tan \\alpha - \\frac{1}{2}}{1 + \\frac{1}{2} \\tan \\alpha}\n$$\nSolving for $\\tan \\alpha$ gives $\\tan \\alpha = 3$. Therefore, $AC = 3 AD = \\frac{3}{2} AB$. Using the Pythagorean Theorem, we find that $BC = \\frac{\\sqrt{13}}{2} AB$. So the answer is $\\frac{\\sqrt{13}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the coefficient of $x^{6}$ in the expansion of\n$$\n(x+1)^{6} \\cdot \\sum_{i=0}^{6} x^{i}\n$$",
    "options": [],
    "answer": "64",
    "solution": "Solution:\nAnswer: 64\nEach term of $(x+1)^{6}$ can be multiplied by a unique power $x^{i}$, $0 \\leq i \\leq 6$, to get a sixth degree term. So the answer is the sum of the coefficients of the terms of $(x+1)^{6}$, which is the same as substituting $x=1$ into this power to get $2^{6}=64$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47435,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P$ be a polynomial with $P(1)=P(2)=\\cdots=P(2007)=0$ and $P(0)=2009!$. $P(x)$ has leading coefficient $1$ and degree $2008$. Find the largest root of $P(x)$.",
    "options": [],
    "answer": "4034072",
    "solution": "Solution:\n$P(0)$ is the constant term of $P(x)$, which is the product of all the roots of the polynomial, because its degree is even. So the product of all $2008$ roots is $2009!$ and the product of the first $2007$ is $2007!$, which means the last root is $\\frac{2009!}{2007!}=2009 \\cdot 2008=4034072$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a regular tetrahedron, and let $O$ be the centroid of triangle $BCD$. Consider the point $P$ on $AO$ such that $P$ minimizes $PA + 2(PB + PC + PD)$. Find $\\sin \\angle PBO$.",
    "options": [],
    "answer": "1/6",
    "solution": "Solution:\n\n$\\boxed{\\dfrac{1}{6}}$\n\nWe translate the problem into one about 2-D geometry. Consider the right triangle $ABO$, and $P$ is some point on $AO$. Then, the choice of $P$ minimizes $PA + 6PB$. Construct the line $\\ell$ through $A$ but outside the triangle $ABO$ so that $\\sin \\angle(AO, \\ell) = \\dfrac{1}{6}$. For whichever $P$ chosen, let $Q$ be the projection of $P$ onto $\\ell$, then $PQ = \\dfrac{1}{6} AP$. Then, since $PA + 6PB = 6(PQ + PB)$, it is equivalent to minimize $PQ + PB$. Observe that this sum is minimized when $B, P, Q$ are collinear and the line through them is perpendicular to $\\ell$ (so that $PQ + PB$ is simply the distance from $B$ to $\\ell$). Then, $\\angle AQB = 90^\\circ$, and since $\\angle AOB = 90^\\circ$ as well, we see that $A, Q, P, B$ are concyclic. Therefore, $\\angle PBO = \\angle OPA = \\angle(AO, \\ell)$, and the sine of this angle is therefore $\\dfrac{1}{6}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider a $7 \\times 7$ chessboard that starts out with all the squares white. We start painting squares black, one at a time, according to the rule that after painting the first square, each newly painted square must be adjacent along a side to only the square just previously painted. The final figure painted will be a connected \"snake\" of squares.\n\na. Show that it is possible to paint 31 squares.\n\nb. Show that it is possible to paint 32 squares.\n\nc. Show that it is possible to paint 33 squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\n![](attached_image_1.png)\n\nb.\n![](attached_image_2.png)\nFigure 1: Solutions to (a), (b) and (c)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, scalene triangle, and let $M$, $N$, and $P$ be the midpoints of $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively. Let the perpendicular bisectors of $\\overline{AB}$ and $\\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $CE$ intersect in point $F$, inside of triangle $ABC$. Prove that points $A$, $N$, $F$, and $P$ all lie on one circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume without loss of generality that $AB > AC$, and we consider the configuration shown below. Our proof can be modified for other configurations.\nLet $O$ be the circumcenter of triangle $ABC$. Point $D$ lies on $PO$, which is the perpendicular bisector of segment $AB$. Thus, $ABO$ is an isosceles triangle with $AD = BD$. Likewise, $AEC$ is isosceles with $AE = EC$. Set $x = \\angle ABD = \\angle BAD$ and $y = \\angle CAE = \\angle ACE$. (Hence $x + y = \\angle BAC$.)\nApplying the law of sines to triangles $ABM$ and $ACM$ gives\n$$\n\\frac{BM}{\\sin x} = \\frac{AB}{\\sin \\angle BMA} \\quad \\text{and} \\quad \\frac{CM}{\\sin y} = \\frac{AC}{\\sin \\angle CMA}.\n$$\nDividing the two equations yields\n$$\n\\frac{BM}{CM} \\cdot \\frac{\\sin y}{\\sin x} = \\frac{AB}{AC} \\cdot \\frac{\\sin \\angle BMA}{\\sin \\angle CMA} = \\frac{AC}{AB},\n$$\nsince $\\sin \\angle BMA = \\sin \\angle CMA$ (as they are supplementary angles). Therefore,\n$$\nBM = MC \\quad \\text{if and only if} \\quad \\frac{\\sin x}{\\sin y} = \\frac{AC}{AB}. \\qquad (*)\n$$\nApplying the law of sines to triangles $ABF$ and $ACF$ gives\n$$\n\\frac{AF}{\\sin x} = \\frac{AB}{\\sin \\angle AFB} \\quad \\text{and} \\quad \\frac{AF}{\\sin y} = \\frac{AC}{\\sin \\angle AFC}.\n$$\nDividing these two equations and noting (*) yields\n$$\n\\frac{\\sin y}{\\sin x} = \\frac{AB}{AC} \\cdot \\frac{\\sin \\angle AFB}{\\sin \\angle AFC} \\quad \\text{or} \\quad \\sin \\angle AFB = \\sin \\angle AFC.\n$$\nSince $\\angle ADF$ is an exterior angle of triangle $ABO$, $\\angle EDF = 2x$. Likewise, $\\angle DEF = 2y$. Thus $\\angle EFD = 180^\\circ - 2x - 2y = 180^\\circ - 2\\angle BAC$. Hence $\\angle BFC = 2\\angle BAC$ (and so $BOFC$ is cyclic), implying that $\\angle AFB + \\angle AFC = 360^\\circ - 2\\angle BAC > 180^\\circ$ (since $\\angle BAC < 90^\\circ$). We must have $\\angle AFB = \\angle AFC = 180^\\circ - \\angle BAC$. Since $BOFC$ is cyclic and $BOC$ is isosceles with vertex angle $BOC = 2\\angle BAC$, $\\angle OFB = \\angle OCB = 90^\\circ - \\angle BAC$. Therefore,\n$$\n\\angle AFO = \\angle AFB - \\angle OFB = 180^\\circ - \\angle BAC - (90^\\circ - \\angle BAC) = 90^\\circ.\n$$\nSince $\\angle APO = \\angle AFO = \\angle ANO = 90^\\circ$, points $A$, $P$, $O$, $F$, $N$ lie on a circle.\n\n\nSolution 2:\n\nWe maintain the notations in the first proof. Let $R$ denote the intersection of lines $MP$ and $CE$. Let $\\omega$ denote the circumcircle of $APON$. We claim that $R$ lies on $\\omega$. Note that $AEC$ is isosceles, so that $MER$ is also isosceles, and so $ARMC$ is an isosceles trapezoid, hence cyclic. Since $PN \\parallel MC$, $ARPN$ is also cyclic.\nAs in the above proof, we can show that $BFOC$ is cyclic. Then $\\angle RFO = \\angle OBC = 90^\\circ - \\angle A$. Moreover, $\\angle RAO = \\angle RAC - \\angle OAC = \\angle C - (90^\\circ - \\angle B) = 90^\\circ - \\angle A$. Thus $AROF$ is cyclic, so that $ARPOFN$ is cyclic.\n\n\nSolution 3:\n\nWe maintain the notations in the first proof. Assume without loss of generality that $AB > AC$, and we consider the configuration shown below. Our proof can be modified for other configurations.\nSince $\\angle OPB = \\angle OMB = \\angle OMC = \\angle ONC = 90^\\circ$, $BPOM$ and $CNOM$ are cyclic. Let ray $AM$ meet the circumcircles of $BPOM$ and $CNOM$ at $X$ and $Y$ (different from $M$), respectively. We have\n$$\n\\angle OXY = \\angle OXM = \\angle OBM = \\angle OBC = \\angle BCO = \\angle MCO = \\angle MYO = \\angle XYO;\n$$\nthat is, triangle $OXY$ is similar to triangle $OBC$. and the rotation centered at $O$ that takes one to the other. In particular, $\\angle XOY = \\angle BAC$. Note also that $APON$ is cyclic, implying that $\\angle NOD = \\angle PAN = \\angle BAC$. Hence, $\\angle XOY = 2\\angle NOD$. Combining the facts that $OX = OY$ and $\\angle XOY = 2\\angle NOD$, we conclude that the image of $X$ under the reflection through line $OD$ (denoted by $R_{OP}$) is the same as that of $Y$ under the reflection through line $ON$ (denoted by $R_{ON}$). Let $F_1$ denote this common image. Note that the image of line $AYE$ under $R_{ON}$ is line $CF_1E$, and the image of line $ADX$ under $R_{OP}$ is line $BDF_1$. Therefore, $F_1$ lies on lines $CE$ and $BD$; that is, $F_1 = F$.\nOn the other hand, since $X$ lies on the circumcircle of $MBPO$, $F = F_1$ must lie on its image under $R_{OP}$, which is the circumcircle of $NAPO$.\n\n\nSolution 4:\n\nInvert the figure about a circle centered at $A$, and let $I_X$ denote the image of the point $X$ under this inversion. Find point $G$ so that $AI_B I_G I_C$ is a parallelogram and let $I_Z$ denote the center of this parallelogram. Note that triangle $BAC$ is similar to triangle $I_C AI_B$ and triangles $BAD$ and $I_D AI_B$. Because $M$ is the midpoint of $AB$ and $I_Z$ is the midpoint of $I_B I_C$, we also have the similarity between the triangles $BAM$ and $I_C AI_Z$. Thus\n$$\n\\angle AI_G I_B = \\angle I_G AI_C = \\angle ZAI_C = \\angle MAB = \\angle DAB = \\angle DBA = \\angle AI_D I_B.\n$$\nHence quadrilateral $AI_BI_DI_G$ is cyclic and, by a similar argument, quadrilateral $AI_CI_EI_G$ is also cyclic. Because the images under the inversion of lines $BDF$ and $CFE$ are circles that intersect in $A$ and $I_F$, it follows that $I_G = I_F$; that is, $G = F$.\nNext note that $I_B, I_Z$, and $I_C$ are collinear and are the images of $I_P, I_F$ (or $I_G$), and $I_N$, respectively, under a homothety centered at $A$ and with ratio $1/2$. It follows that $I_P, I_F$ and $I_N$ are collinear, and then that the points $A, P, F$ and $N$ lie on a circle.\n\n\nSolution 5:\n\nLet $A$ be the origin and denote the complex number of each point by the corresponding lowercase letter. Note that $2n = c$ and $2p = b$. Define point $F_2$ such that\n$$\nf_2 = \\frac{2np}{m} = \\frac{cp}{m} = \\frac{bn}{m};\n$$\nthat is\n$$\n\\frac{f_2}{c} = \\frac{p}{m} \\quad \\text{and} \\quad \\frac{f_2}{b} = \\frac{n}{m}. \\qquad (**)\n$$\nThe first equation in $(**)$ implies that triangles $AF_2C$ and $APM$ are similar (and have the same orientation). Hence\n$$\n\\angle ACF_2 = \\angle AMP = \\angle MAC = \\angle EAC = \\angle ECA;\n$$\nthat is, $F_2$ lies on line $EC$. Similarly, the second equation in $(**)$ shows that $F_2$ lies on line $BD$. Thus, $F_2$ lies on both $BD$ and $CE$; that is, $F_2 = F$.\nWe have established the similarity between triangle $AFC$ (which is $AF_2C$) and triangle $APM$. Thus, triangle $ACM$ is similar to triangle $AFP$ (because of spiral similarities about $A$). Hence $\\angle AFP = \\angle ACM = \\angle ANP$, implying that $ANFP$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$, $Q$, and $R$ be the points on sides $BC$, $CA$, and $AB$ of an acute triangle $ABC$ such that triangle $PQR$ is equilateral and has minimal area among all such equilateral triangles. Prove that the perpendiculars from $A$ to line $QR$, from $B$ to line $RP$, and from $C$ to line $PQ$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(By Zuming Feng) By Miquel's theorem (which can be shown by simple angle chasing), the circumcircles of triangles $AQR$, $BRP$, and $CPQ$ meet at a common point $X$. The key observation is that $XR \\perp AB$, $XP \\perp BC$, and $XQ \\perp CA$. Indeed, if $P_1Q_1R_1$ is an inscribed equilateral triangle, and the circumcircles of triangles $AQ_1R_1$, $BR_1P_1$, and $CP_1Q_1$ meet at a common point $X$. Let $P$, $Q$, and $R$ be the feet of the perpendiculars from $X$ to the sides of the triangle. Quick angle chasing ($\\angle RR_1X = \\angle PP_1X = QQ_1X$) shows that right triangles $XPP_1$, $XQQ_1$, and $XRR_1$ are similar, and so triangles $PQR$ and $P_1Q_1R_1$ are similar. Clearly, $PQR$ is a smaller triangle, and this establishes our observation.\n\n![](attached_image_1.png)\n\nIt is then straightforward to check that the perpendiculars from $A$ to $QR$, $B$ to $RP$, and $C$ to $PQ$ meet at the isogonal conjugate of $X$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nExplain how any unit fraction $\\frac{1}{n}$ can be decomposed into other unit fractions.",
    "options": [],
    "answer": "1/n = 1/(2n) + 1/(3n) + 1/(6n)",
    "solution": "Solution:\n$\\frac{1}{n} = \\frac{1}{2n} + \\frac{1}{3n} + \\frac{1}{6n}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S=\\{1,2, \\ldots, 2008\\}$. For any nonempty subset $A \\subset S$, define $m(A)$ to be the median of $A$ (when $A$ has an even number of elements, $m(A)$ is the average of the middle two elements). Determine the average of $m(A)$, when $A$ is taken over all nonempty subsets of $S$.",
    "options": [],
    "answer": "2009/2",
    "solution": "Solution:\n\nFor any subset $A$, we can define the \"reflected subset\" $A' = \\{i \\mid 2009 - i \\in A\\}$. Then $m(A) = 2009 - m(A')$. Note that as $A$ is taken over all nonempty subsets of $S$, $A'$ goes through all the nonempty subsets of $S$ as well. Thus, the average of $m(A)$ is equal to the average of $\\frac{m(A) + m(A')}{2}$, which is the constant $\\frac{2009}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(n)$ be the number of times you have to hit the $\\sqrt{\\ }$ key on a calculator to get a number less than $2$ starting from $n$. For instance, $f(2)=1$, $f(5)=2$. For how many $1 < m < 2008$ is $f(m)$ odd?",
    "options": [],
    "answer": "242",
    "solution": "Solution:\nAnswer: $242$ This is $[2^{1}, 2^{2}) \\cup [2^{4}, 2^{8}) \\cup [2^{16}, 2^{32}) \\ldots$, and $2^{8} < 2008 < 2^{16}$ so we have exactly the first two intervals.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $\\ell$ be the line through $(0,0)$ and tangent to the curve $y = x^{3} + x + 16$. Find the slope of $\\ell$.",
    "options": [],
    "answer": "13",
    "solution": "Solution:\n\nLet the point of tangency be $(t, t^{3} + t + 16)$, then the slope of $\\ell$ is $\\dfrac{t^{3} + t + 16}{t}$. On the other hand, since $\\dfrac{dy}{dx} = 3t^{2} + 1$, the slope of $\\ell$ is $3t^{2} + 1$. Therefore,\n$$\n\\frac{t^{3} + t + 16}{t} = 3t^{2} + 1\n$$\nSimplifying, we get $t^{3} = 8$, so $t = 2$. It follows that the slope is $3(2)^{2} + 1 = 13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate the sum\n$$\n1 - 2 + 3 - 4 + \\cdots + 2007 - 2008\n$$",
    "options": [],
    "answer": "-1004",
    "solution": "Solution:\nEvery odd integer term can be paired with the next even integer, and this pair sums to $-1$. There are $1004$ such pairs, so the total sum is $-1004$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEach vertex of a regular heptagon is colored either red or blue. Prove that there is an isosceles triangle with all its vertices the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenote the vertices of the heptagon by $A$, $B$, $C$, $D$, $E$, $F$, $G$. Since an alternating arrangement cannot be continued all the way around the heptagon, two adjacent vertices must be the same color, say $A$ and $B$. If any of $C$, $E$, $G$ shares this color, we are done since triangles $A B C$, $A B E$, and $A B G$ are all isosceles. On the other hand, if $C$, $E$, and $G$ are all of the opposite color, we are also done because triangle $C E G$ is isosceles. Thus in all cases we can find an isosceles triangle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven right triangle $ABC$, with $AB = 4$, $BC = 3$, and $CA = 5$. Circle $\\omega$ passes through $A$ and is tangent to $BC$ at $C$. What is the radius of $\\omega$?",
    "options": [],
    "answer": "25/8",
    "solution": "Solution:\n\nThe radius is $\\frac{25}{8}$.\n\nLet $O$ be the center of $\\omega$, and let $M$ be the midpoint of $AC$. Since $OA = OC$, $OM \\perp AC$. Also, $\\angle OCM = \\angle BAC$, and so triangles $ABC$ and $CMO$ are similar. Then, $CO / CM = AC / AB$, from which we obtain that the radius of $\\omega$ is $CO = \\frac{25}{8}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest positive integer $n$ such that $107 n$ has the same last two digits as $n$.",
    "options": [],
    "answer": "50",
    "solution": "Solution:\nThe two numbers have the same last two digits if and only if $100$ divides their difference $106 n$, which happens if and only if $50$ divides $n$.\n\nAnswer: $50$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$ and $b$ be nonzero real numbers. Prove that at least one of the following inequalities is true:\n$$\n\\begin{aligned}\n& \\left|\\frac{a+\\sqrt{a^{2}+2 b^{2}}}{2 b}\\right|<1 \\\\\n& \\left|\\frac{a-\\sqrt{a^{2}+2 b^{2}}}{2 b}\\right|<1\n\\end{aligned}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume for the sake of contradiction that both inequalities are false, that is,\n$$\n1 \\leq\\left|\\frac{a+\\sqrt{a^{2}+2 b^{2}}}{2 b}\\right| \\text{ and } 1 \\leq\\left|\\frac{a-\\sqrt{a^{2}+2 b^{2}}}{2 b}\\right|.\n$$\nMultiplying these inequalities together yields\n$$\n1 \\leq\\left|\\frac{a+\\sqrt{a^{2}+2 b^{2}}}{2 b} \\cdot \\frac{a-\\sqrt{a^{2}+2 b^{2}}}{2 b}\\right|=\\left|\\frac{a^{2}-(a^{2}+2 b^{2})}{4 b^{2}}\\right|=\\left|\\frac{-2 b^{2}}{4 b^{2}}\\right|=\\frac{1}{2},\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an equilateral triangle. Let $\\Omega$ be a circle inscribed in $ABC$ and let $\\omega$ be a circle tangent externally to $\\Omega$ as well as to sides $AB$ and $AC$. Determine the ratio of the radius of $\\Omega$ to the radius of $\\omega$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nAnswer: $3$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the units digit of $7^{2009}$?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\nAnswer: 7\nNote that the units digits of $7^{1}, 7^{2}, 7^{3}, 7^{4}, 7^{5}, 7^{6}, \\ldots$ follows the pattern $7, 9, 3, 1, 7, 9, 3, 1, \\ldots$. The 2009th term in this sequence should be 7.\n\n\nAlternate method:\nNote that the units digit of $7^{4}$ is equal to 1, so the units digit of $(7^{4})^{502}$ is also 1. But $(7^{4})^{502} = 7^{2008}$, so the units digit of $7^{2008}$ is 1, and therefore the units digit of $7^{2009}$ is 7.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle, and let $M$ be the midpoint of side $AB$. If $AB$ is $17$ units long and $CM$ is $8$ units long, find the maximum possible value of the area of $ABC$.",
    "options": [],
    "answer": "68",
    "solution": "Solution:\n\nAnswer: $68$\n\nLet $h$ be the length of the altitude from $C$ to $AB$. Observe that $K = \\frac{1}{2} \\cdot h \\cdot AB \\leq \\frac{1}{2} \\cdot CM \\cdot AB = 68$ and that equality is achieved when $CM \\perp AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n>2$ be a positive integer. Prove that there are $\\frac{1}{2}(n-2)(n+1)$ ways to walk from $(0,0)$ to $(n, 2)$ using only up and right unit steps such that the walk never visits the line $y=x$ after it leaves the origin.",
    "options": [],
    "answer": "(1/2)(n-2)(n+1)",
    "solution": "Solution:\n\nThe first two steps can only go to the right. Then we need to compute the number of ways of walking from $(2,0)$ to $(n, 2)$ which does not pass through the point $(2,2)$. There are $\\binom{n}{2}$ ways to walk from $(2,0)$ to $(n, 2)$, and exactly one of those paths passes through the point $(2,2)$. So the number of valid paths is $\\binom{n}{2}-1=\\frac{1}{2} n(n-1)-1=\\frac{1}{2}(n-2)(n+1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor how many ordered triples $(a, b, c)$ of positive integers are the equations $a b c + 9 = a b + b c + c a$ and $a + b + c = 10$ satisfied?",
    "options": [],
    "answer": "21",
    "solution": "Solution:\n\nSubtracting the first equation from the second, we obtain\n$$\n1 - a - b - c + a b + b c + c a - a b c = (1 - a)(1 - b)(1 - c) = 0.\n$$\nSince $a$, $b$, and $c$ are positive integers, at least one must equal $1$. Note that $a = b = c = 1$ is not a valid triple, so it suffices to consider the cases where exactly two or one of $a, b, c$ are equal to $1$.\n\nIf $a = b = 1$, we obtain $c = 8$ and similarly for the other two cases, so this gives $3$ ordered triples.\n\nIf $a = 1$, then we need $b + c = 9$, which has $6$ solutions for $b, c \\neq 1$; a similar argument for $b$ and $c$ gives a total of $18$ such solutions.\n\nIt is easy to check that all the solutions we found are actually solutions to the original equations. Adding, we find $18 + 3 = 21$ total triples.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nYou have a die with faces labelled $1$ through $6$. On each face, you draw an arrow to an adjacent face, such that if you start on a face and follow the arrows, after $6$ steps you will have passed through every face once and will be back on your starting face. How many ways are there to draw the arrows so that this is true?",
    "options": [],
    "answer": "32",
    "solution": "Solution:\n\nAnswer: $32$\n\nThere are $4$ choices for where to go from face $1$. Consider the $4$ faces adjacent to $1$. We can visit either $1$, $2$, or $3$ of them before visiting the face opposite $1$. If we only visit one of these adjacent faces, we have $4$ choices for which one, then we visit face $6$, opposite face $1$, then we visit the remaining $3$ faces in one of two orders—for a total of $8$ ways. If we visit $2$ adjacent faces first, there are $8$ choices for these two faces, then $2$ choices for the path back from face $6$ to face $1$. Lastly, there are $8$ ways to visit three of the adjacent faces before visiting the opposite face. These choices give $32$ total.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a number overweight if it has at least three positive integer divisors (including $1$ and the number), and call a number obese if it has at least four positive integer divisors (including $1$ and the number). How many positive integers between $1$ and $200$ are overweight, but not obese?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nA positive integer is overweight, but not obese, if it has exactly $3$ factors. This can only happen if that integer is the square of a prime. (If two primes, $p$ and $q$, divide the number, then $p$, $q$, $pq$, and $1$ all divide it, making it at least obese.)\n\nSo, the integers less than $200$ which are squares of a prime are the squares of $2, 3, 5, 7, 11$, and $13$.\n\nThus, there are $6$ such numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $AB = 5$, $BC = 4$ and $AC = 3$. Let $\\mathcal{P}$ and $\\mathcal{Q}$ be squares inside $ABC$ with disjoint interiors such that they both have one side lying on $AB$. Also, the two squares each have an edge lying on a common line perpendicular to $AB$, and $\\mathcal{P}$ has one vertex on $AC$ and $\\mathcal{Q}$ has one vertex on $BC$. Determine the minimum value of the sum of the areas of the two squares.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "144/49",
    "solution": "Solution:\n\n$\\boxed{\\dfrac{144}{49}}$\n\nLet the side lengths of $\\mathcal{P}$ and $\\mathcal{Q}$ be $a$ and $b$, respectively. Label two of the vertices of $\\mathcal{P}$ as $D$ and $E$ so that $D$ lies on $AB$ and $E$ lies on $AC$, and so that $DE$ is perpendicular to $AB$. The triangle $ADE$ is similar to $ACB$. So $AD = \\dfrac{3}{4} a$. Using similar arguments, we find that\n$$\n\\frac{3a}{4} + a + b + \\frac{4b}{3} = AB = 5\n$$\nso\n$$\n\\frac{a}{4} + \\frac{b}{3} = \\frac{5}{7}\n$$\nUsing Cauchy-Schwarz inequality, we get\n$$\n\\left(a^2 + b^2\\right)\\left(\\frac{1}{4^2} + \\frac{1}{3^2}\\right) \\geq \\left(\\frac{a}{4} + \\frac{b}{3}\\right)^2 = \\frac{25}{49}\n$$\nIt follows that\n$$\na^2 + b^2 \\geq \\frac{144}{49}\n$$\nEquality occurs at $a = \\frac{36}{35}$ and $b = \\frac{48}{35}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSandra the Maverick has 5 pairs of shoes in a drawer, each pair a different color. Every day for 5 days, Sandra takes two shoes out and throws them out the window. If they are the same color, she treats herself to a practice problem from a past HMMT. What is the expected value (average number) of practice problems she gets to do?",
    "options": [],
    "answer": "5/9",
    "solution": "Solution:\n\nAnswer: $\\frac{5}{9}$\n\nOn any given day, there is a $\\frac{1}{9}$ chance that the second shoe that Sandra chooses makes a pair with the first shoe she chose. Thus the average number of problems she does in a day is $\\frac{1}{9}$, so, by the linearity of expectation, she does $\\frac{5}{9}$ problems total, on average.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor a positive integer $n$, let $\\theta(n)$ denote the number of integers $0 \\leq x < 2010$ such that $x^{2} - n$ is divisible by $2010$. Determine the remainder when $\\sum_{n=0}^{2009} n \\cdot \\theta(n)$ is divided by $2010$.",
    "options": [],
    "answer": "335",
    "solution": "Solution:\n\nAnswer: $335$\n\nLet us consider the sum $\\sum_{n=0}^{2009} n \\cdot \\theta(n) \\pmod{2010}$ in another way. Consider the sum $0^{2} + 1^{2} + 2^{2} + \\cdots + 2009^{2} \\pmod{2010}$. For each $0 \\leq n < 2010$, in the latter sum, the term $n$ appears $\\theta(n)$ times, so the sum is congruent to $\\sum_{n=0}^{2009} n \\cdot \\theta(n)$. In other words,\n$$\n\\sum_{n=0}^{2009} n \\cdot \\theta(n) = \\sum_{n=0}^{2009} n^{2} = \\frac{(2009)(2009+1)(2 \\cdot 2009+1)}{6} \\equiv (-1) \\cdot \\frac{2010}{6} \\cdot (-1) = 335 \\pmod{2010}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo mathematicians, Kelly and Jason, play a cooperative game. The computer selects some secret positive integer $n < 60$ (both Kelly and Jason know that $n < 60$, but they don't know what the value of $n$ is). The computer tells Kelly the unit digit of $n$, and it tells Jason the number of divisors of $n$. Then, Kelly and Jason have the following dialogue:\n\nKelly: I don't know what $n$ is, and I'm sure that you don't know either. However, I know that $n$ is divisible by at least two different primes.\n\nJason: Oh, then I know what the value of $n$ is.\n\nKelly: Now I also know what $n$ is.\n\nAssuming that both Kelly and Jason speak truthfully and to the best of their knowledge, what are all the possible values of $n$?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nThe only way in which Kelly can know that $n$ is divisible by at least two different primes is if she is given $0$ as the unit digit of $n$, since if she received anything else, then there is some number with that unit digit and not divisible by two primes (i.e., $1, 2, 3, 4, 5, 16, 7, 8, 9$). Then, after Kelly says the first line, Jason too knows that $n$ is divisible by $10$.\n\nThe number of divisors of $10, 20, 30, 40, 50$ are $4, 6, 8, 8, 6$, respectively. So unless Jason received $4$, he cannot otherwise be certain of what $n$ is. It follows that Jason received $4$, and thus $n = 10$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $x$ and $y$ are real numbers such that $\\frac{(x-4)^{2}}{4}+\\frac{y^{2}}{9}=1$, find the largest possible value of $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}$.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nThe first equation is an ellipse with major axis parallel to the $y$-axis. If the second expression is set equal to a certain value $c$, then it is also the equation of an ellipse with major axis parallel to the $y$-axis; further, it is similar to the first ellipse. So the largest value of $c$ occurs when both ellipses are tangent on the $x$-axis, at $x=6$, $y=0$, which gives $9$ as the largest value of $c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that $A B C$ is an isosceles triangle with $A B = A C$. Let $P$ be the point on side $A C$ so that $A P = 2 C P$. Given that $B P = 1$, determine the maximum possible area of $A B C$.",
    "options": [],
    "answer": "9/10",
    "solution": "Solution:\nAnswer: $\\sqrt{\\frac{9}{10}}$\n\nLet $Q$ be the point on $A B$ so that $A Q = 2 B Q$, and let $X$ be the intersection of $B P$ and $C Q$. The key observation is that, as we will show, $B X$ and $C X$ are fixed lengths, and the ratio of areas $[A B C] / [B C X]$ is constant. So, to maximize $[A B C]$, it is equivalent to maximize $[B C X]$.\n\nUsing Menelaus' theorem on $A B P$, we have\n$$\n\\frac{B X \\cdot P C \\cdot A Q}{X P \\cdot C A \\cdot Q B} = 1\n$$\nSince $P C / C A = 1 / 3$ and $A Q / Q B = 2$, we get $B X / X P = 3 / 2$. It follows that $B X = 3 / 5$. By symmetry, $C X = 3 / 5$.\n\nAlso, we have\n$$\n[A B C] = 3 [B P C] = 3 \\cdot \\frac{5}{3} [B X C] = 5 [B X C]\n$$\nNote that $[B X C]$ is maximized when $\\angle B X C = 90^\\circ$ (one can check that this configuration is indeed possible). Thus, the maximum value of $[B X C]$ is $\\frac{1}{2} B X \\cdot C X = \\frac{1}{2} \\left(\\frac{3}{5}\\right)^2 = \\frac{9}{50}$. It follows that the maximum value of $[A B C]$ is $\\frac{9}{10}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $f(x)=\\frac{1}{1-x}$. Let $f^{k+1}(x)=f\\left(f^{k}(x)\\right)$, with $f^{1}(x)=f(x)$. What is $f^{2008}(2008)$?",
    "options": [],
    "answer": "-1/2007",
    "solution": "Solution:\n\n$\\boxed{\\frac{-1}{2007}}$ Notice that, if $x \\neq 0,1$, then $f^{2}(x)=\\frac{1}{1-\\frac{1}{1-x}}=\\frac{x-1}{x}$, which means that $f^{3}(x)=\\frac{1}{1-\\frac{x-1}{x}}=x$. So $f^{n}$ is periodic with period $n=3$, which means that $f^{2007}(x)=x$ so $f^{2008}(2008)=f(2008)=\\frac{-1}{2007}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate the sum\n$$\n\\cos \\left(\\frac{2 \\pi}{18}\\right)+\\cos \\left(\\frac{4 \\pi}{18}\\right)+\\cdots+\\cos \\left(\\frac{34 \\pi}{18}\\right)\n$$",
    "options": [],
    "answer": "-1",
    "solution": "Solution:\nIf $k < 18$, then we can pair $\\cos \\left(\\frac{k \\pi}{18}\\right)$ with $\\cos \\left(\\frac{(18-k) \\pi}{18}\\right)$, and these two terms sum to $0$. If $k > 18$, then the pair $\\cos \\left(\\frac{k \\pi}{18}\\right)$ and $\\cos \\left(\\frac{(36-k) \\pi}{18}\\right)$ also sums to $0$. So, the only term in this series that is left over is $\\cos \\left(\\frac{18 \\pi}{18}\\right) = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny, which causes the cat to change the number of (magic) beans that Alice has from $n$ to $5n$, or (2) gives the cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears) as soon as the number of beans Alice has is greater than $2008$ and has last two digits $42$. What is the minimum number of cents Alice can spend to win the game, assuming she starts with $0$ beans?",
    "options": [],
    "answer": "35",
    "solution": "Solution:\n\nAnswer: $35$\n\nConsider the number of beans Alice has in base $5$. Note that $2008 = 31013_{5}$, $42 = 132_{5}$, and $100 = 400_{5}$. Now, suppose Alice has $d_{k} \\cdots d_{2} d_{1}$ beans when she wins; the conditions for winning mean that these digits must satisfy $d_{2} d_{1} = 32$, $d_{k} \\cdots d_{3} \\geq 310$, and $d_{k} \\cdots d_{3} = 4i + 1$ for some $i$.\n\nTo gain these $d_{k} \\cdots d_{2} d_{1}$ beans, Alice must spend at least $5\\left(d_{1} + d_{2} + \\cdots + d_{k}\\right) + k - 1$ cents (5 cents to get each bean in the \"units digit\" and $k-1$ cents to promote all the beans). We now must have $k \\geq 5$ because $d_{k} \\cdots d_{2} d_{1} > 2008$. If $k = 5$, then $d_{k} \\geq 3$ since $d_{k} \\cdots d_{3} \\geq 3100$; otherwise, we have $d_{k} \\geq 1$.\n\nTherefore, if $k = 5$, we have $5\\left(d_{1} + d_{2} + \\cdots + d_{k}\\right) + k - 1 \\geq 44 > 36$; if $k > 5$, we have $5\\left(d_{1} + d_{2} + \\cdots + d_{k}\\right) + k - 1 \\geq 30 + k - 1 \\geq 35$. But we can attain $36$ cents by taking $d_{k} \\cdots d_{3} = 1000$, so this is indeed the minimum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathcal{P}$ be a parabola, and let $V_{1}$ and $F_{1}$ be its vertex and focus, respectively. Let $A$ and $B$ be points on $\\mathcal{P}$ so that $\\angle A V_{1} B=90^{\\circ}$. Let $\\mathcal{Q}$ be the locus of the midpoint of $A B$. It turns out that $\\mathcal{Q}$ is also a parabola, and let $V_{2}$ and $F_{2}$ denote its vertex and focus, respectively. Determine the ratio $F_{1} F_{2} / V_{1} V_{2}$.",
    "options": [],
    "answer": "7/8",
    "solution": "Solution:\n$\\boxed{\\dfrac{7}{8}}$\n\nSince all parabolas are similar, we may assume that $\\mathcal{P}$ is the curve $y=x^{2}$. Then, if $A=\\left(a, a^{2}\\right)$ and $B=\\left(b, b^{2}\\right)$, the condition that $\\angle A V_{1} B=90^{\\circ}$ gives $a b+a^{2} b^{2}=0$, or $a b=-1$. Then, the midpoint of $A B$ is\n$$\n\\frac{A+B}{2}=\\left(\\frac{a+b}{2}, \\frac{a^{2}+b^{2}}{2}\\right)=\\left(\\frac{a+b}{2}, \\frac{(a+b)^{2}-2 a b}{2}\\right)=\\left(\\frac{a+b}{2}, \\frac{(a+b)^{2}}{2}+1\\right) .\n$$\n(Note that $a+b$ can range over all real numbers under the constraint $a b=-1$.) It follows that the locus of the midpoint of $A B$ is the curve $y=2 x^{2}+1$.\n\nRecall that the focus of $y=a x^{2}$ is $\\left(0, \\frac{1}{4 a}\\right)$. We find that $V_{1}=(0,0), V_{2}=(0,1), F_{1}=\\left(0, \\frac{1}{4}\\right)$, $F_{2}=\\left(0,1+\\frac{1}{8}\\right)$. Therefore, $F_{1} F_{2} / V_{1} V_{2}=\\frac{7}{8}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJohn M. is sitting at $(0, 0)$, looking across the aisle at his friends sitting at $(i, j)$ for each $1 \\leq i \\leq 10$ and $0 \\leq j \\leq 5$. Unfortunately, John can only see a friend if the line connecting them doesn't pass through any other friend. How many friends can John see?",
    "options": [],
    "answer": "36",
    "solution": "Solution:\n\n36\n\nThe simplest method is to draw a picture and count which friends he can see. John can see the friend on point $(i, j)$ if and only if $i$ and $j$ are relatively prime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCDE$ is a regular pentagon inscribed in a circle of radius $1$. What is the area of the set of points inside the circle that are farther from $A$ than they are from any other vertex?",
    "options": [],
    "answer": "pi/5",
    "solution": "Solution:\n\nAnswer: $\\frac{\\pi}{5}$\n\nDraw the perpendicular bisectors of all the sides and diagonals of the pentagon with one endpoint at $A$. These lines all intersect in the center of the circle, because they are the set of points equidistant from two points on the circle. Now, a given point is farther from $A$ than from point $X$ if it is on the $X$ side of the perpendicular bisector of segment $AX$. So, we want to find the area of the set of all points which are separated from $A$ by all of these perpendicular bisectors, which turns out to be a single $72^{\\circ}$ sector of the circle, which has area $\\frac{\\pi}{5}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCyclic pentagon $A B C D E$ has a right angle $\\angle A B C = 90^{\\circ}$ and side lengths $A B = 15$ and $B C = 20$. Supposing that $A B = D E = E A$, find $C D$.",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nBy Pythagoras, $A C = 25$. Since $\\overline{A C}$ is a diameter, angles $\\angle A D C$ and $\\angle A E C$ are also right, so that $C E = 20$ and $A D^{2} + C D^{2} = A C^{2}$ as well. Beginning with Ptolemy's theorem,\n\n$$\n\\begin{aligned}\n& (A E \\cdot C D + A C \\cdot D E)^2 = A D^2 \\cdot E C^2 = (A C^2 - C D^2) E C^2 \\\\\n& \\quad \\Longrightarrow C D^2 (A E^2 + E C^2) + 2 \\cdot C D \\cdot A E^2 \\cdot A C + A C^2 (D E^2 - E C^2) = 0 \\\\\n& \\quad \\Longrightarrow C D^2 + 2 C D \\left(\\frac{A E^2}{A C}\\right) + D E^2 - E C^2 = 0\n\\end{aligned}\n$$\n\nIt follows that $C D^2 + 18 C D - 175 = 0$, from which $C D = 7$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJohnny the grad student is typing all the integers from $1$ to $\\infty$, in order. The $2$ on his computer is broken however, so he just skips any number with a $2$. What's the $2008$th number he types?",
    "options": [],
    "answer": "3781",
    "solution": "Solution:\n\nAnswer: $3781$\n\nWrite $2008$ in base $9$ as $2671$, and interpret the result as a base $10$ number such that the base $9$ digits $2,3, \\ldots, 8$ correspond to the base $10$ digits $3,4, \\ldots, 9$. This gives an answer of $3781$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P$ be a polyhedron where every face is a regular polygon, and every edge has length $1$. Each vertex of $P$ is incident to two regular hexagons and one square. Choose a vertex $V$ of the polyhedron. Find the volume of the set of all points contained in $P$ that are closer to $V$ than to any other vertex.",
    "options": [],
    "answer": "sqrt(2)/3",
    "solution": "Solution:\nAnswer: $\\frac{\\sqrt{2}}{3}$\n\nObserve that $P$ is a truncated octahedron, formed by cutting off the corners from a regular octahedron with edge length $3$. So, to compute the value of $P$, we can find the volume of the octahedron, and then subtract off the volume of truncated corners.\n\nGiven a square pyramid where each triangular face is an equilateral triangle, and whose side length is $s$, the height of the pyramid is $\\frac{\\sqrt{2}}{2} s$, and thus the volume is $\\frac{1}{3} \\cdot s^{2} \\cdot \\frac{\\sqrt{2}}{2} s = \\frac{\\sqrt{2}}{6} s^{3}$.\n\nThe side length of the octahedron is $3$, and noting that the octahedron is made up of two square pyramids, its volume must be $2 \\cdot \\frac{\\sqrt{2}(3)^{3}}{6} = 9 \\sqrt{2}$.\n\nThe six \"corners\" that we remove are all square pyramids, each with volume $\\frac{\\sqrt{2}}{6}$, and so the resulting polyhedron $P$ has volume $9 \\sqrt{2} - 6 \\cdot \\frac{\\sqrt{2}}{6} = 8 \\sqrt{2}$.\n\nFinally, to find the volume of all points closer to one particular vertex than any other vertex, note that due to symmetry, every point in $P$ (except for a set with zero volume), is closest to one of the $24$ vertices. Due to symmetry, it doesn't matter which $V$ is picked, so we can just divide the volume of $P$ by $24$ to obtain the answer $\\frac{\\sqrt{2}}{3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 47471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p(x)$ be the polynomial of degree $4$ with roots $1, 2, 3, 4$ and leading coefficient $1$. Let $q(x)$ be the polynomial of degree $4$ with roots $1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}$ and leading coefficient $1$. Find $\\lim_{x \\rightarrow 1} \\frac{p(x)}{q(x)}$.",
    "options": [],
    "answer": "-24",
    "solution": "Solution:\n\nAnswer: $-24$\n\nConsider the polynomial $f(x) = x^4 q\\left(\\frac{1}{x}\\right)$—it has the same roots, $1, 2, 3, 4$, as $p(x)$. But this polynomial also has the same coefficients as $q(x)$, just in reverse order. Its leading coefficient is $q(0) = 1 \\cdot \\frac{1}{2} \\cdot \\frac{1}{3} \\cdot \\frac{1}{4} = \\frac{1}{24}$. So $f(x)$ is $p(x)$ scaled by $\\frac{1}{24}$, which means that $\\frac{p(x)}{f(x)}$ goes to $24$ as $x$ goes to $1$, and $\\frac{f(x)}{q(x)}$ goes to $-1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $(x, y)$ be a pair of real numbers satisfying\n$$\n56 x + 33 y = \\frac{-y}{x^{2} + y^{2}}, \\quad \\text{ and } \\quad 33 x - 56 y = \\frac{x}{x^{2} + y^{2}}\n$$\nDetermine the value of $|x| + |y|$.",
    "options": [],
    "answer": "11/65",
    "solution": "Solution:\nAnswer: $\\frac{11}{65}$ Observe that\n$$\n\\frac{1}{x + y i} = \\frac{x - y i}{x^{2} + y^{2}} = 33 x - 56 y + (56 x + 33 y) i = (33 + 56 i)(x + y i)\n$$\nSo\n$$\n(x + y i)^{2} = \\frac{1}{33 + 56 i} = \\frac{1}{(7 + 4 i)^{2}} = \\left(\\frac{7 - 4 i}{65}\\right)^{2}\n$$\nIt follows that $(x, y) = \\pm\\left(\\frac{7}{65}, -\\frac{4}{65}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA triangle (with non-zero area) is constructed with the lengths of the sides chosen from the set\n$$\n\\{2, 3, 5, 8, 13, 21, 34, 55, 89, 144\\}\n$$\nShow that this triangle must be isosceles (A triangle is isosceles if it has at least two sides the same length.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst note that the numbers in the list are an increasing sequence of Fibonacci numbers, i.e. starting with $5$, each number is the sum of the previous two numbers in the list. By the triangle inequality, we know that the longest side is always less than the sum of the other two sides. Suppose the triangle constructed is scalene with sides $a < b < c$. If $a$ and $b$ are consecutive numbers in the list then the longest side $c$ is the next number in the list or a larger one and $a + b \\leq c$, a contradiction of the triangle inequality. If $a$ and $b$ are not consecutive numbers in the list, let $d$ be the number just before $b$, so that $a < d < b$. We then have $a + b < d + b$. But by the previous argument $d + b \\leq c$, so by the transitivity of '$<$' we have $a + b < c$, again contradicting the triangle inequality. Therefore the supposition is false and the triangle cannot be scalene and must have at least two sides equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47474,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a pair $A = (x_1, y_1)$ and $B = (x_2, y_2)$ of points on the coordinate plane, let $d(A, B) = |x_1 - x_2| + |y_1 - y_2|$. We call a pair $(A, B)$ of (unordered) points *harmonic* if $1 < d(A, B) \\le 2$. Determine the maximum number of harmonic pairs among 100 points on the plane.\n\n(This problem was suggested by Zuming Feng and Oleg Golberg.)",
    "options": [],
    "answer": "3750",
    "solution": "We claim that there do not exist five points on the plane such that they are pairwise harmonic. Suppose such a set of five points exists. We say $(x_1, y_1)$ precedes $(x_2, y_2)$, denoted by $(x_1, y_1) \\prec (x_2, y_2)$, if and only if $x_1 \\le x_2$ and $y_1 \\le y_2$. This defines a partial order on the set. Because it has five elements, it has either a chain or an antichain of three elements. Let $A, B, C$ in the order of their $x$-coordinates form a chain or an antichain under the ordering we defined. Then $d(A, B) + d(B, C) = d(A, C)$ and all pairs of $A, B, C$ cannot be good. (This may also be proved by invoking the pigeon-hole principle to prove the more general result of Erdős-Szekeres: a sequence of $mn + 1$ real numbers either contains a non-increasing sequence of length $m+1$ or a non-decreasing sequence of length $n+1$.)\n\nConsider the graph with vertices at 100 points of the plane and edges connecting good pairs. Then due to the above statement the graph has no complete subgraph with 5 vertices. By Turan's theorem it has at most\n$$\n\\binom{100}{2} - 4 \\cdot \\binom{25}{2} = 3750 = 6 \\cdot 25^2\n$$\nedges. Therefore, every set of 100 points of the plane has at most 3750 harmonic pairs.\n\nTo complete our proof, we show that the bound 3750 can be attained. Consider the square with vertices at $A(0,1)$, $B(1,0)$, $C(0,-1)$, $D(-1,0)$. On each of the sides $AB, BC, CD, DA$ choose 25 points close to $A, B, C, D$, respectively. Among the chosen 100 points, any pair of two points lying on different sides of the square is good. Therefore, $6 \\cdot 25^2 = 3750$ is the maximum number of good pairs among 100 points of the plane.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Denote by $S_n$ the set of points $(x, y)$ with integer coordinates such that\n$$\n|x| + \\left\\lfloor y + \\frac{1}{2} \\right\\rfloor < n.\n$$\nA *path* is a sequence of distinct points $(x_1, y_1), (x_2, y_2), \\dots, (x_\\ell, y_\\ell)$ in $S_n$ such that, for $i = 2, \\dots, \\ell$, the distance between $(x_i, y_i)$ and $(x_{i-1}, y_{i-1})$ is 1 (in other words, the points $(x_i, y_i)$ and $(x_{i-1}, y_{i-1})$ are neighbors in the lattice of points with integer coordinates).\nProve that the points in $S_n$ cannot be partitioned into fewer than $n$ paths (a partition of $S_n$ into $m$ paths is a set $\\mathcal{P}$ of $m$ nonempty paths such that each point in $S_n$ appears in exactly one of the $m$ paths in $\\mathcal{P}$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Color the points in $S_n$ as follows (see Figure 1):\n- if $y \\ge 0$, color $(x, y)$ white if $x + y - n$ is even and black if $x + y - n$ is odd;\n- if $y < 0$, color $(x, y)$ white if $x + y - n$ is odd and black if $x + y - n$ is even.\n\n![](attached_image_1.png)\nFigure 1: Coloring of $S_3$\n\nConsider a path $(x_1, y_1), (x_2, y_2), \\dots, (x_\\ell, y_\\ell)$ in $S_n$. A pair of successive points $(x_{i-1}, y_{i-1})$ and $(x_i, y_i)$ in the path is called a pair of successive black points if both points in the pair are colored black.\n\nSuppose now that the points of $S_n$ are partitioned into $m$ paths and the total number of successive pairs of black points in all paths is $k$. By breaking the paths at each pair of successive black points,\n\nwe obtain $k+m$ paths in each of which the number of black points exceeds the number of white points by at most one. Therefore, the total number of black points in $S_n$ cannot exceed the number of white points by more than $k+m$. On the other hand, the total number of black points in $S_n$ exceeds the total number of white points by exactly $2n$ (there is exactly one more black point in each row of $S_n$). Therefore,\n$$\n2n \\leq k + m.\n$$\nThere are exactly $n$ adjacent black points in $S_n$ (call two points in $S_n$ *adjacent* if their distance is 1), namely the pairs\n$(x, 0)$ and $(x, -1)$,\nfor $x = -n+1, -n+3, \\dots, n-3, n-1$. Therefore $k \\leq n$ (the number of successive pairs of black points in the paths in the partition of $S_n$ cannot exceed the total number of adjacent pairs of black points in $S_n$) and we have $2n \\leq k+m \\leq n+m$, yielding\n$$\nn \\leq m.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all $y > 1$ satisfying\n$$\n\\int_{1}^{y} x \\ln x \\, dx = \\frac{1}{4}.\n$$",
    "options": [],
    "answer": "sqrt(e)",
    "solution": "Solution:\n\nApplying integration by parts with $u = \\ln x$ and $dv = x \\, dx$, we get\n$$\n\\int_{1}^{y} x \\ln x \\, dx = \\left.\\frac{1}{2} x^{2} \\ln x\\right|_{1}^{y} - \\frac{1}{2} \\int_{1}^{y} x \\, dx = \\frac{1}{2} y^{2} \\ln y - \\frac{1}{4} y^{2} + \\frac{1}{4}\n$$\nSo $y^{2} \\ln y = \\frac{1}{2} y^{2}$. Since $y > 1$, we obtain $\\ln y = \\frac{1}{2}$, and thus $y = \\sqrt{e}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways can you color the squares of a $2 \\times 2008$ grid in 3 colors such that no two squares of the same color share an edge?",
    "options": [],
    "answer": "2*3^2008",
    "solution": "Solution:\n\n$2 \\cdot 3^{2008}$\n\nDenote the colors $A$, $B$, $C$. The left-most column can be colored in $6$ ways. For each subsequent column, if the $k$th column is colored with $A B$, then the $(k+1)$th column can only be colored with one of $B A$, $B C$, $C A$. That is, if we have colored the first $k$ columns, then there are $3$ ways to color the $(k+1)$th column. It follows that the number of ways of coloring the board is $6 \\times 3^{2007}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 47478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHow many diagonals does a regular undecagon (11-sided polygon) have?",
    "options": [],
    "answer": "44",
    "solution": "Solution:\nThere are 8 diagonals coming from the first vertex, 8 more from the next, 7 from the next, 6 from the next, 5 from the next, etc., and 1 from the last, for $8+8+7+6+5+4+3+2+1=44$ total.\n\nThird method: Each vertex has 8 diagonals touching it. There are 11 vertices. Since each diagonal touches two vertices, this counts every diagonal twice, so there are $\\frac{8 \\cdot 11}{2}=44$ diagonals.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFarmer John has 5 cows, 4 pigs, and 7 horses. How many ways can he pair up the animals so that every pair consists of animals of different species? Assume that all animals are distinguishable from each other. (Please write your answer as an integer, without any incomplete computations.)",
    "options": [],
    "answer": "100800",
    "solution": "Solution:\n\nAnswer: $100800$\n\nSince there are $9$ cow and pigs combined and $7$ horses, there must be a pair with $1$ cow and $1$ pig, and all the other pairs must contain a horse. There are $4 \\times 5$ ways of selecting the cow-pig pair, and $7!$ ways to select the partners for the horses. It follows that the answer is $4 \\times 5 \\times 7! = 100800$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the largest $x$ such that $x^{2}$ divides $24 \\cdot 35 \\cdot 46 \\cdot 57$?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\nWe factor the product as $2^{4} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 23$. If $x^{2}$ divides this product, $x$ can have at most 2 factors of $2$, 1 factor of $3$, and no factors of any other prime. So $2^{2} \\cdot 3 = 12$ is the largest value of $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn acute triangle $ABC$, the three altitudes meet at $H$. Given that $AH = BC$, calculate at least one of the angles of $\\triangle ABC$.\n\nAnswer: $\\angle A = 45^{\\circ}$ (angles $B$ and $C$ cannot be determined).",
    "options": [],
    "answer": "Angle A = 45 degrees (angles B and C cannot be determined).",
    "solution": "Solution:\n\nIn the diagram, $\\angle AFH = \\angle CFB$ (both are right angles) and $\\angle FAH = \\angle FCB$ (both are complementary to $\\angle ABC$). We get $\\triangle AFH \\cong \\triangle CFB$ by AAS. It follows that $AF = FC$, so $\\triangle AFC$ is right isosceles. We find that $\\angle BAC = 45^{\\circ}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $DEF$ be a triangle and $H$ the foot of the altitude from $D$ to $EF$. If $DE = 60$, $DF = 35$, and $DH = 21$, what is the difference between the minimum and the maximum possible values for the area of $DEF$?",
    "options": [],
    "answer": "588",
    "solution": "Solution:\n\nObserve that the two possible configurations come from $DEF$ being obtuse and $DEF$ being acute. In either case, we have that $HF = \\sqrt{35^2 - 21^2} = 28$ and $EH = \\sqrt{60^2 - 21^2} = 9\\sqrt{39}$. This means that $EF$ can be either $FH + EH$ or $|EH - FH|$. The difference in area is hence $\\frac{1}{2} \\cdot 2 \\cdot FH \\cdot DH = 28 \\cdot 21 = 588$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n > 4$ be a positive integer. Determine the number of ways to walk from $(0,0)$ to $(n, 2)$ using only up and right unit steps such that the path does not meet the lines $y = x$ or $y = x - n + 2$ except at the start and at the end.",
    "options": [],
    "answer": "1/2*(n^2 - 5n + 2)",
    "solution": "Solution:\n\nIt is easy to see that the first two steps and the last two steps must all be right steps. So we need to compute the number of walks from $(2,0)$ to $(n-2,2)$ that do not pass through $(2,2)$ and $(n-2,0)$. There are $\\binom{n-2}{2}$ paths from $(2,0)$ to $(n-2,2)$, and exactly two of them are invalid. So the answer is\n$$\n\\binom{n-2}{2} - 2 = \\frac{1}{2}(n-2)(n-3) - 2 = \\frac{1}{2}\\left(n^2 - 5n + 2\\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p$ be a prime. Given a sequence of positive integers $b_{1}$ through $b_{n}$, exactly one of which is divisible by $p$, show that when\n$$\n\\frac{1}{b_{1}}+\\frac{1}{b_{2}}+\\ldots+\\frac{1}{b_{n}}\n$$\nis written as a fraction in lowest terms, then its denominator is divisible by $p$. Use this fact to explain why no prime $p$ is ever juicy.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe can assume that $b_{n}$ is the term divisible by $p$ (i.e. $b_{n}=k p$) since the order of addition doesn't matter. We can then write\n$$\n\\frac{1}{b_{1}}+\\frac{1}{b_{2}}+\\ldots+\\frac{1}{b_{n-1}}=\\frac{a}{b}\n$$\nwhere $b$ is not divisible by $p$ (since none of the $b_{i}$ are). But then\n$$\n\\frac{a}{b}+\\frac{1}{k p}=\\frac{k p a+b}{k p b}.\n$$\nSince $b$ is not divisible by $p$, $k p a + b$ is not divisible by $p$, so we cannot remove the factor of $p$ from the denominator. In particular, $p$ cannot be juicy as $1$ can be written as $\\frac{1}{1}$, which has a denominator not divisible by $p$, whereas being juicy means we have a sum $\\frac{1}{b_{1}}+\\ldots+\\frac{1}{b_{n}}=1$, where $b_{1}<b_{2}<\\ldots<b_{n}=p$, and so in particular none of the $b_{i}$ with $i<n$ are divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Write $1$ as a sum of $4$ distinct unit fractions.\n\nb. Write $1$ as a sum of $5$ distinct unit fractions.\n\nc. Show that, for any integer $k > 3$, $1$ can be decomposed into $k$ unit fractions.",
    "options": [],
    "answer": "a) 1 = 1/2 + 1/3 + 1/7 + 1/42\nb) 1 = 1/2 + 1/3 + 1/7 + 1/43 + 1/(43*42)\nc) For every integer k > 3, such a decomposition exists by repeatedly replacing a term 1/n with 1/(n+1) + 1/(n(n+1)) to increase the number of terms by one.",
    "solution": "Solution:\n\na.\n$1 = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{7} + \\frac{1}{42}$\n\nb.\n$1 = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{7} + \\frac{1}{43} + \\frac{1}{43 \\cdot 42}$\n\nc.\nIf we can do it for $k$ fractions, simply replace the last one (say $\\frac{1}{n}$) with $\\frac{1}{n+1} + \\frac{1}{n(n+1)}$. Then we can do it for $k+1$ fractions. So, since we can do it for $k=4$, we can do it for any $k > 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine all real numbers $a$ such that the inequality $|x^{2}+2 a x+3 a| \\leq 2$ has exactly one solution in $x$.",
    "options": [],
    "answer": "1, 2",
    "solution": "Solution:\n\nAnswer: $1, 2$\n\nLet $f(x) = x^{2} + 2 a x + 3 a$. Note that $f(-3/2) = 9/4$, so the graph of $f$ is a parabola that goes through $(-3/2, 9/4)$. Then, the condition that $|x^{2} + 2 a x + 3 a| \\leq 2$ has exactly one solution means that the parabola has exactly one point in the strip $-2 \\leq y \\leq 2$, which is possible if and only if the parabola is tangent to $y = 2$ or $y = -2$.\n\nThat is, $x^{2} + 2 a x + 3 a = 2$ has exactly one solution. Then, the discriminant $\\Delta = 4 a^{2} - 4(3 a - 2) = 4 a^{2} - 12 a + 8$ must be zero. Solving the equation yields $a = 1, 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 47487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$, $b$, and $c$ be positive real numbers satisfying $a^{b} > b^{a}$ and $b^{c} > c^{b}$. Does it follow that $a^{c} > c^{a}$?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nYes. We have\n$$\n\\left(a^{c}\\right)^{b} = \\left(a^{b}\\right)^{c} > \\left(b^{a}\\right)^{c} = \\left(b^{c}\\right)^{a} > \\left(c^{b}\\right)^{a} = \\left(c^{a}\\right)^{b} ;\n$$\nthe desired inequality follows by taking the $b$th root.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are $5$ dogs, $4$ cats, and $7$ bowls of milk at an animal gathering. Dogs and cats are distinguishable, but all bowls of milk are the same. In how many ways can every dog and cat be paired with either a member of the other species or a bowl of milk such that all the bowls of milk are taken?",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nAnswer: $20$\n\nSince there are $9$ dogs and cats combined and $7$ bowls of milk, there can only be one dog-cat pair, and all the other pairs must contain a bowl of milk. There are $4 \\times 5$ ways of selecting the dog-cat pair, and only one way of picking the other pairs, since the bowls of milk are indistinguishable, so the answer is $4 \\times 5 = 20$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $\\angle BAC = 90^\\circ$. A circle is tangent to the sides $AB$ and $AC$ at $X$ and $Y$ respectively, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on the side $BC$. Given that $AB = 6$, find the area of the portion of the circle that lies outside the triangle.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "π - 2",
    "solution": "Solution:\n\nLet $O$ be the center of the circle, and $r$ its radius, and let $X'$ and $Y'$ be the points diametrically opposite $X$ and $Y$, respectively. We have $OX' = OY' = r$, and $\\angle X' O Y' = 90^\\circ$. Since triangles $X' O Y'$ and $BAC$ are similar, we see that $AB = AC$. Let $X''$ be the projection of $Y'$ onto $AB$. Since $X'' B Y'$ is similar to $ABC$, and $X'' Y' = r$, we have $X'' B = r$. It follows that $AB = 3r$, so $r = 2$.\n\n![](attached_image_2.png)\n\nThen, the desired area is the area of the quarter circle minus that of the triangle $X' O Y'$. And the answer is $\\frac{1}{4} \\pi r^2 - \\frac{1}{2} r^2 = \\pi - 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriangle $ABC$ obeys $AB = 2AC$ and $\\angle BAC = 120^\\circ$. Points $P$ and $Q$ lie on segment $BC$ such that\n$$\n\\begin{aligned}\nAB^2 + BC \\cdot CP &= BC^2 \\\\\n3AC^2 + 2BC \\cdot CQ &= BC^2\n\\end{aligned}\n$$\n\nFind $\\angle PAQ$ in degrees.",
    "options": [],
    "answer": "40°",
    "solution": "Solution:\n\nFind $\\angle PAQ$ in degrees.\n\nAnswer: $40^\\circ$\n\nWe have $AB^2 = BC(BC - CP) = BC \\cdot BP$, so triangle $ABC$ is similar to triangle $PBA$.\n\nAlso, $AB^2 = BC(BC - 2CQ) + AC^2 = (BC - CQ)^2 - CQ^2 + AC^2$, which rewrites as $AB^2 + CQ^2 = BQ^2 + AC^2$.\n\nWe deduce that $Q$ is the foot of the altitude from $A$.\n\nThus, $\\angle PAQ = 90^\\circ - \\angle QPA = 90^\\circ - \\angle ABP - \\angle BAP$.\n\nUsing the similar triangles, $\\angle PAQ = 90^\\circ - \\angle ABC - \\angle BCA = \\angle BAC - 90^\\circ = 40^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice has an equilateral triangle $A B C$ of area $1$. Put $D$ on $B C$, $E$ on $C A$, and $F$ on $A B$, with $B D = D C$, $C E = 2 E A$, and $2 A F = F B$. Note that $A D$, $B E$, and $C F$ pass through a single point $M$. What is the area of triangle $E M C$?",
    "options": [],
    "answer": "1/6",
    "solution": "Solution:\n\nAnswer: $\\frac{1}{6}$\n\nTriangles $A C F$ and $B C F$ share a height, so the ratio of their areas is $A F / B F = 1 / 2$. By the same method, the ratio of the areas of $A M F$ and $B M F$ is $1 / 2$. So, the ratio of the areas of $A C M$ and $B C M$ is also $1 / 2$. Similarly, the ratio of the areas of $A B M$ and $B C M$ is $1 / 2$. But the sum of the areas of $A C M$, $B C M$, and $A B M$ is $1$, so the area of $A C M$ is $\\frac{1}{4}$. Then the area of $E M C$ is $2 / 3$ the area of $A C M$, because they share heights, so their areas are in the same ratio as their bases. The area of $E M C$ is then $\\frac{2 \\cdot 1}{3 \\cdot 4} = \\frac{1}{6}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of all primes $p$ for which there exists a prime $q$ such that $p^{2}+p q+q^{2}$ is a square.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n83 and 5 both work, because $3^{2}+3 \\cdot 5+5^{2}=49$. Now, say $p^{2}+p q+q^{2}=k^{2}$, for a positive integer $k$. Then $(p+q)^{2}-k^{2}=p q$, or:\n$$\n(p+q+k)(p+q-k)=p q\n$$\nSince $p+q+k$ is a divisor of $p q$, and it is greater than $p$ and $q$, $p+q+k=p q$. Then $p+q-k=1$. So:\n$$\n2 p+2 q=p q+1 \\Leftrightarrow p q-2 p-2 q+4=3 \\Leftrightarrow (p-2)(q-2)=3\n$$\nThis shows that one of $p$ and $q$ is 3 and the other is 5.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathcal{C}$ be the hyperbola $y^{2}-x^{2}=1$. Given a point $P_{0}$ on the $x$-axis, we construct a sequence of points $\\left(P_{n}\\right)$ on the $x$-axis in the following manner: let $\\ell_{n}$ be the line with slope 1 passing through $P_{n}$, then $P_{n+1}$ is the orthogonal projection of the point of intersection of $\\ell_{n}$ and $\\mathcal{C}$ onto the $x$-axis. (If $P_{n}=0$, then the sequence simply terminates.)\nLet $N$ be the number of starting positions $P_{0}$ on the $x$-axis such that $P_{0}=P_{2008}$. Determine the remainder of $N$ when divided by 2008.",
    "options": [],
    "answer": "254",
    "solution": "Solution:\nLet $P_{n}=(x_{n}, 0)$. Then the $\\ell_{n}$ meet $\\mathcal{C}$ at $(x_{n+1}, x_{n+1}-x_{n})$. Since this point lies on the hyperbola, we have $(x_{n+1}-x_{n})^{2}-x_{n+1}^{2}=1$. Rearranging this equation gives\n$$\nx_{n+1}=\\frac{x_{n}^{2}-1}{2 x_{n}}\n$$\nChoose a $\\theta_{0} \\in(0, \\pi)$ with $\\cot \\theta_{0}=x_{0}$, and define $\\theta_{n}=2^{n} \\theta_{0}$. Using the double-angle formula, we have\n$$\n\\cot \\theta_{n+1}=\\cot \\left(2 \\theta_{n}\\right)=\\frac{\\cot ^{2} \\theta_{n}-1}{2 \\cot \\theta_{n}}\n$$\nIt follows by induction that $x_{n}=\\cot \\theta_{n}$. Then, $P_{0}=P_{2008}$ corresponds to $\\cot \\theta_{0}=\\cot \\left(2^{2008} \\theta_{0}\\right)$ (assuming that $P_{0}$ is never at the origin, or equivalently, $2^{n} \\theta$ is never an integer multiple of $\\pi$ ). So, we need to find the number of $\\theta_{0} \\in(0, \\pi)$ with the property that $2^{2008} \\theta_{0}-\\theta_{0}=k \\pi$ for some integer $k$. We have $\\theta_{0}=\\frac{k \\pi}{2^{2008}-1}$, so $k$ can be any integer between 1 and $2^{2008}-2$ inclusive (and note that since the denominator is odd, the sequence never terminates). It follows that the number of starting positions is $N=2^{2008}-2$.\nFinally, we need to compute the remainder when $N$ is divided by 2008. We have $2008=2^{3} \\times 251$. Using Fermat's Little Theorem with 251, we get $2^{2008} \\equiv\\left(2^{250}\\right)^{4} \\cdot 256 \\equiv 1^{4} \\cdot 5=5\\ (\\bmod\\ 251)$. So we have $N \\equiv 3\\ (\\bmod\\ 251)$ and $N \\equiv-2\\ (\\bmod\\ 8)$. Using Chinese Remainder Theorem, we get $N \\equiv 254$ $(\\bmod\\ 2008)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCyclic pentagon $ABCD E$ has side lengths $AB = BC = 5$, $CD = DE = 12$, and $AE = 14$. Determine the radius of its circumcircle.",
    "options": [],
    "answer": "225 sqrt(11) / 88",
    "solution": "Solution:\n\nLet $C'$ be the point on minor arc $BCD$ such that $BC' = 12$ and $C'D = 5$, and write $AC' = BD = C'E = x$, $AD = y$, and $BD = z$. Ptolemy applied to quadrilaterals $ABC'D$, $BC'DE$, and $ABDE$ gives\n$$\n\\begin{aligned}\n& x^2 = 12y + 5^2 \\\\\n& x^2 = 5z + 12^2 \\\\\n& yz = 14x + 5 \\cdot 12\n\\end{aligned}\n$$\nThen\n$$\n\\left(x^2 - 5^2\\right)\\left(x^2 - 12^2\\right) = 5 \\cdot 12 yz = 5 \\cdot 12 \\cdot 14x + 5^2 \\cdot 12^2\n$$\nfrom which $x^3 - 169x - 5 \\cdot 12 \\cdot 14 = 0$. Noting that $x > 13$, the rational root theorem leads quickly to the root $x = 15$. Then triangle $BCD$ has area $\\sqrt{16 \\cdot 1 \\cdot 4 \\cdot 11} = 8\\sqrt{11}$ and circumradius $R = \\frac{5 \\cdot 12 \\cdot 15}{4 \\cdot 8 \\sqrt{11}} = \\frac{225 \\sqrt{11}}{88}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 47495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPirate ships Somy and Lia are having a tough time. At the end of the year, they are both one pillage short of the minimum required for maintaining membership in the Pirate Guild, so they decide to pillage each other to bring their counts up. Somy by tradition only pillages $28 \\cdot 3^{k}$ coins for integers $k$, and Lia by tradition only pillages $82 \\cdot 3^{j}$ coins for integers $j$. Note that each pillage can have a different $k$ or $j$. Somy and Lia work out a system where Somy pillages Lia $n$ times, Lia pillages Somy $n$ times, and after both sets of pillages Somy and Lia are financially even.\nWhat is the smallest $n$ can be?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nAnswer: 2\n\nClearly, $n=1$ cannot be achieved, because $28 \\cdot 3^{k}$ is never a multiple of $82$. However, two pillages is enough: Somy pillages $28$ and $28 \\cdot 81$ from Lia, and Lia pillages $81$ and $81 \\cdot 27$ from Somy. As is easily checked, both pillage $28 \\cdot 82$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 47496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe polynomial $a x^{2}-b x+c$ has two distinct roots $p$ and $q$, with $a$, $b$, and $c$ positive integers and with $0<p, q<1$. Find the minimum possible value of $a$.",
    "options": [],
    "answer": "5",
    "solution": "Solution:\nLet $x$ and $y$ be the roots. Then:\n\n$$\n\\begin{gathered}\n\\frac{b}{a}=x+y<2 \\Rightarrow b<2a \\\\\n\\frac{c}{a}=x y<1 \\Rightarrow c<a \\Rightarrow a>1 \\\\\nb^{2}>4 a c>4 c^{2} \\Rightarrow b>2c\n\\end{gathered}\n$$\n\nEvaluated at $1$, the polynomial must be greater than $0$, so $a+c>b$. Then:\n\n$$\n\\begin{gathered}\n2c < b < a+c \\\\\n2c+1 \\leq b \\leq a+c-1 \\\\\na \\geq c+2 \\geq 3\n\\end{gathered}\n$$\n\nIf $a=3$, then $c=1$ and $b=3$, by the above bounds, but this polynomial has complex roots. Similarly, if $a=4$, then $c=1$ and $b$ is forced to be either $3$ or $4$, again giving either $0$ or $1$ distinct real roots. So $a \\geq 5$. But the polynomial $5 x^{2}-5 x+1$ satisfies the condition, so $5$ is the answer.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 47497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, $c$ be nonzero real numbers such that $a+b+c=0$ and $a^{3}+b^{3}+c^{3}=a^{5}+b^{5}+c^{5}$. Find the value of $a^{2}+b^{2}+c^{2}$.",
    "options": [],
    "answer": "6/5",
    "solution": "Solution:\nLet $\\sigma_{1}=a+b+c$, $\\sigma_{2}=ab+bc+ca$ and $\\sigma_{3}=abc$ be the three elementary symmetric polynomials. Since $a^{3}+b^{3}+c^{3}$ is a symmetric polynomial, it can be written as a polynomial in $\\sigma_{1}$, $\\sigma_{2}$ and $\\sigma_{3}$. Now, observe that $\\sigma_{1}=0$, and so we only need to worry about the terms not containing $\\sigma_{1}$. By considering the degrees of the terms, we see that the only possibility is $\\sigma_{3}$. That is, $a^{3}+b^{3}+c^{3}=k \\sigma_{3}$ for some constant $k$. By setting $a=b=1$, $c=-2$, we see that $k=3$.\n\nBy similar reasoning, we find that $a^{5}+b^{5}+c^{5}=h \\sigma_{2} \\sigma_{3}$ for some constant $h$. By setting $a=b=1$ and $c=-2$, we get $h=-5$.\n\nSo, we now know that $a+b+c=0$ implies\n$$\na^{3}+b^{3}+c^{3}=3abc \\quad \\text{and} \\quad a^{5}+b^{5}+c^{5}=-5abc(ab+bc+ca)\n$$\nThen $a^{3}+b^{3}+c^{3}=a^{5}+b^{5}+c^{5}$ implies that $3abc=-5abc(ab+bc+ca)$. Given that $a$, $b$, $c$ are nonzero, we get $ab+bc+ca=-\\frac{3}{5}$.\n\nThen, $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)=0^{2}-2\\left(-\\frac{3}{5}\\right)=\\frac{6}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn 2000, the Clay Mathematics Institute named seven Millennium Prize Problems, with each carrying a prize of $1$ Million for its solution. Write down the name of ONE of the seven Clay Millennium Problems. If your submission is incorrect or misspelled, then your submission is disqualified. If another team wrote down the same Millennium Problem as you, then you get 0 points, otherwise you get 20 points.",
    "options": [],
    "answer": "Riemann Hypothesis",
    "solution": "Solution:\n\nThe seven Millennium Prize Problems are:\n(a) Birch and Swinnerton-Dyer Conjecture\n(b) Hodge Conjecture\n(c) Navier-Stokes Equations\n(d) P vs NP\n(e) Poincaré Conjecture\n(f) Riemann Hypothesis\n(g) Yang-Mills Theory",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 47499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe RSA Factoring Challenge, which ended in 2007, challenged computational mathematicians to factor extremely large numbers that were the product of two prime numbers. The largest number successfully factored in this challenge was RSA-640, which has 193 decimal digits and carried a prize of $\\$ 20,000$. The next challenge number carried prize of $\\$ 30,000$, and contains $N$ decimal digits. Your task is to submit a guess for $N$. Only the team(s) that have the closest guess(es) receives points. If $k$ teams all have the closest guesses, then each of them receives $\\left\\lceil\\frac{20}{k}\\right\\rceil$ points.",
    "options": [],
    "answer": "212",
    "solution": "Solution:\n\nAnswer: 212\n\nFor more information, see the Wikipedia entry at http://en.wikipedia.org/wiki/RSA_Factoring_Challenge.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 47500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind $\\max \\{\\operatorname{Perimeter}(T)\\}$ for $T$ a triangle contained in a regular septagon (7-sided figure) of unit edge length. Write your answer $N$ to 2 places after the decimal. If the correct answer rounded to 2 decimal places is $A$, you will receive 0 points if $N<A$ and $\\lfloor\\max \\{0,25-50 \\cdot(N-A)\\}\\rfloor$ points otherwise.",
    "options": [],
    "answer": "5.85",
    "solution": "Solution:\nAnswer: 5.85086\n\nLet the septagon be $A_{0} A_{1} \\ldots A_{6}$.\n\nIf $x$ is a point that can move along the x-axis, the distance from $x$ to a fixed point $p$ is a convex function in the x-coordinate. Therefore, the sum of the distances from $x$ to two other points is convex too, so if $x$ is constrained to lie on a closed line segment, its maximum value is attained at an endpoint. Therefore, the triangle of maximal perimeter has vertices at the vertices of the pentagon. The triangle with the largest such perimeter has almost evenly spaced vertices, so triangle $A_{0} A_{2} A_{4}$ has the maximal perimeter. It has area $5.85\\ldots$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  }
]