id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0iwr | Let $ABP$, $BCQ$, $CAR$ be three non-overlapping triangles erected outside of acute triangle $ABC$. Let $M$ be the midpoint of segment $AP$. Given that $\angle PAB = \angle CQB = 45^\circ$, $\angle ABP = \angle QBC = 75^\circ$, $\angle RAC = 105^\circ$, and $RQ^2 = 6CM^2$, compute $AC^2/AR^2$. | [
"Because $\\angle BAP = \\angle BQC = 45^\\circ$ and $\\angle PBA = \\angle CQB = 75^\\circ$, triangles $BQC$ and $BAP$ are similar to each other, from which it follows that triangles $BCP$ and $BQA$ are similar to each other. Hence, the law of sines gives\n$$\n\\frac{CP}{AQ} = \\frac{BP}{BA} = \\frac{\\sin \\angle... | United States | Team Selection Test 2009 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2/3 | |
0iz5 | Problem:
David, Delong, and Justin each showed up to a problem writing session at a random time during the session. If David arrived before Delong, what is the probability that he also arrived before Justin? | [
"Solution:\n\nLet $t_{1}$ be the time that David arrives, let $t_{2}$ be the time that Delong arrives, and let $t_{3}$ be the time that Justin arrives. We can assume that all times are pairwise distinct because the probability of any two being equal is zero. Because the times were originally random and independent ... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 2/3 | |
05sc | Problem:
Soit $ABCD$ un trapèze avec $(AB)$ parallèle à $(CD)$. On suppose qu'il y a deux cercles $\omega_{1}$ et $\omega_{2}$ à l'intérieur du trapèze tels que $\omega_{1}$ est tangent aux côtés $[DA]$, $[AB]$ et $[BC]$ et $\omega_{2}$ est tangent aux côtés $[BC]$, $[CD]$ et $[DA]$. Soit $(d_{1})$ la seconde tangente... | [
"Solution:\n\n\n\nOn note $X$ le point d'intersection de $(AD)$ et $(BC)$. On note respectivement $E$ et $F$ les centres de $\\omega_{1}$ et $\\omega_{2}$. Notons que $\\omega_{1}$ est le cercle inscrit à $XAB$ et que $\\omega_{2}$ est le cercle $X$-exinscrit à $XCD$. Enfin, on note $G$ le ... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarit... | null | proof only | null | |
0eck | First $n$ positive integers are written on the whiteboard. Ana and Bine are playing a game. In each move one of the players must erase 3 numbers whose average value is an integer. Ana starts and then they take turns after each move. The player that can not make a move loses. Determine who has the winning strategy and j... | [
"In each step average value of chosen three numbers will be integer if and only if their sum will be divisible by 3. Therefore it is enough to know only residues of numbers modulo 3. Let us denote with $x_0, x_1$ and $x_2$ how many numbers on board give residue 0, 1 or 2 modulo 3, respectively.\nWe can present curr... | Slovenia | Selection Examinations for the IMO 2015 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Ana wins if n ≡ 3, 4, 5 (mod 6) and also when n = 7; otherwise Bine wins. | |
04kf | Karlo and Lovro play the following game. Karlo cuts the paper of dimensions $9 \times 9$ into rectangles of integer dimensions having at least one side of length $1$. Lovro then chooses a positive integer $k \in \{1, \dots, 9\}$, after which Karlo gives him as many coins as the total area of $1 \times k$ and $k \times ... | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 12 | |
0g7z | 若整數 $a$ 使方程式 $(m^2 + n)(n^2 + m) = a(m - n^3)$ 有正整數解 $m, n$, 則稱整數 $a$ 為友善的 (friendly).
a. 證明集合 $\{1, 2, \dots, 2013\}$ 中至少有 500 個友善的整數 (friendly integers).
b. 決定 $a = 2$ 是否為友善的。 | [
"(a) 我們取 $a = 4k - 3$, $k \\ge 2$，再取 $m = 2k - 1$, $n = k - 1$，我們得到\n$$ (m^2+n)(n^2+m) = ((2k-1)^2+(k-1))((k-1)^2+(2k-1)) = (4k-3)k^3 = a(m-n)^3. $$\n因此 $5, 9, \\dots, 2009, 2013$ 是友善的且 $\\{1, 2, \\dots, 2013\\}$ 包含至少 503 個友善的整數。\n\n(b) 我們證明 $a = 2$ 不是友善的。我們考慮當 $a = 2$ 時的方程式 $(m^2+n)(n^2+m) = 2(m-n^3)$，並把左式寫成平方差的形式... | Taiwan | 二〇一三數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | At least 503 integers in the set are friendly (for example, all numbers congruent to one modulo four from five to two thousand thirteen), and 2 is not friendly. | |
0fzw | Problem:
Seien $a, b, c \in \mathbb{R}$ mit $a, b, c \geq 1$. Zeige, dass gilt:
$$
\min \left(\frac{10 a^{2}-5 a+1}{b^{2}-5 b+10}, \frac{10 b^{2}-5 b+1}{c^{2}-5 c+10}, \frac{10 c^{2}-5 c+1}{a^{2}-5 a+10}\right) \leq a b c
$$ | [
"Solution:\n\nNous étudions d'abord le cas $a=b=c$. L'inéquation à prouver dans ce cas est\n$$\n\\begin{aligned}\n& \\frac{10 a^{2}-5 a+1}{a^{2}-5 a+10} \\leq a^{3} \\Leftrightarrow 10 a^{2}-5 a+1 \\leq a^{5}-5 a^{4}+10 a^{3} \\\\\n& \\Leftrightarrow a^{5}-5 a^{4}+10 a^{3}-10 a^{2}+5 a-1 \\geq 0 \\Leftrightarrow (a... | Switzerland | IMO-Selektion | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
07wm | Find all possible values of $2^n + n^3$, when $n$ is an integer satisfying $2^n - n^3 = 4$! | [
"Integers $n < 2$ obviously do not satisfy $2^n - n^3 = 4! = 24$. For $2 \\le n \\le 9$ we easily check that they satisfy $2^n < n^3$, hence $2^n - n^3 \\ne 24$.\n$$\n2^2 < 3^2 \\qquad 2^4 < 2^6 = 4^3 \\qquad 2^6 < 2^3 \\cdot 3^3 = 6^3 \\qquad 2^9 = 8^3 < 9^3\n$$\n$$\n2^3 < 3^3 \\qquad 2^5 < 2^6 = 4^3 < 5^3 \\qquad... | Ireland | IRL_ABooklet_2024 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 2024 | |
0e1g | Find all rational numbers $r$ and all integers $k$, such that the equation $r(5k - 7r) = 3$ is satisfied. | [
"Obviously, $r \\ne 0$. Let us write $r$ as a reduced fraction $r = \\frac{m}{n}$ and let us assume that $n$ is a positive integer. Then $\\frac{m}{n}(5k - 7m) = 3$ or, equivalently, $m(5kn - 7m) = 3n^2$. Hence, $m$ divides $3n^2$. Since $m$ and $n$ are coprime, we conclude that $m$ divides $3$. Let us consider fou... | Slovenia | National Math Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (k, r) in {(2, 1), (-2, -1), (2, 3/7), (-2, -3/7)} | |
0102 | Problem:
In a sequence $u_{0}, u_{1}, \ldots$ of positive integers, $u_{0}$ is arbitrary, and for any non-negative integer $n$,
$$
u_{n+1}= \begin{cases}\frac{1}{2} u_{n} & \text{ for even } u_{n} \\ a+u_{n} & \text{ for odd } u_{n}\end{cases}$$
where $a$ is a fixed odd positive integer. Prove that the sequence is per... | [
"Solution:\n\nSuppose $u_{n}>a$. Then, if $u_{n}$ is even we have $u_{n+1}=\\frac{1}{2} u_{n}<u_{n}$, and if $u_{n}$ is odd we have $u_{n+1}=a+u_{n}<2 u_{n}$ and $u_{n+2}=\\frac{1}{2} u_{n+1}<u_{n}$. Hence the iteration results in $u_{n} \\leqslant a$ in a finite number of steps. Thus for any non-negative integer $... | Baltic Way | Baltic Way 1997 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
09oj | Let $ABC$ be a triangle with $AB \ne AC$, and let $H$ be its orthocenter. Let $T$ be a point on the arc $BC$ of the circumcircle of triangle $ABC$ that does not contain point $A$. Let $l$ be the line through $H$ parallel to $BC$, and let $l$ intersect lines $TB$ and $TC$ at points $P$ and $Q$, respectively. Let the cir... | [
"(1) By the cyclic quadrilateral property,\n$$ \\angle PSA = \\angle PBA = 180^\\circ - \\angle ABT = \\angle ACT = 180^\\circ - \\angle ACQ = 180^\\circ - \\angle ASQ, $$\nso point $S$ lies on line $PQ$.\n\nFrom the given condition, we have $\\angle PAB + \\angle CAQ = \\angle BAC$. Hence, $\\angle PSB + \\angle C... | Mongolia | MMO2025 Round 4 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous ... | English | proof only | null | |
03yz | Let $n \ge 2$ be a given integer.
(1) Prove that one can arrange all the subsets of the set $\{1, 2, \dots, n\}$ as a sequence of subsets $A_1, A_2, \dots, A_{2^n}$, such that $|A_{i+1}| = |A_i| + 1$ or $|A_i| - 1$, where $i = 1, 2, \dots, 2^n$ and $A_{2^n+1} = A_1$.
(2) Determine, with proof, all possible values of th... | [
"(1) We prove by mathematical induction that there exists a sequence $A_1, A_2, \\dots, A_{2^n}$ such that $A_1 = \\{1\\}, A_{2^n} = \\emptyset$ and satisfies the condition in (1).\nWhen $n = 2$, the sequence $\\{1\\}, \\{1, 2\\}, \\{2\\}, \\emptyset$ of $\\{1, 2\\}$ works.\nAssume that when $n = k$, there exists s... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | 0 | |
07pg | The quadrilateral $ABCD$ is inscribed in a circle. It is known that the lines $DA$ and $BC$ intersect at an angle of $60^\circ$ and that $|DA| = |BC| = 2$, $|AB| = 4$. Find the radius of the circle. | [
"Let the lines $DA$ and $CB$ intersect at $E$. We have to distinguish two possibilities: either $|CD| < |AB|$ or $|CD| > |AB|$, see diagram below.\n\n\n\nSince $AD = BC$, by symmetry we get $DE = CE$ and $AB \\parallel DC$. Since $\\angle AEB = 60^\\circ$ this implies that $\\triangle ABE$ ... | Ireland | Ireland | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | R = 2 or R = 2*sqrt(7/3) | |
0kk4 | Problem:
$O$ is the center of square $ABCD$, and $M$ and $N$ are the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Points $A'$, $B'$, $C'$, $D'$ are chosen on $\overline{AO}$, $\overline{BO}$, $\overline{CO}$, $\overline{DO}$, respectively, so that $A'B'MC'D'N$ is an equiangular hexagon. The ratio $\... | [
"Solution:\n\nAssume without loss of generality that the side length of $ABCD$ is $1$ so that the area of the square is also $1$. This also means that $OM=ON=\\frac{1}{2}$. As $A'B'MC'D'N$ is equiangular, it can be seen that $\\angle A'NO=60^{\\circ}$, and also by symmetry, that $A'B' \\parallel AB$, so $\\angle OA... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 8634 | |
0ay9 | Problem:
In triangle $A B C$ with $\angle A B C=60^{\circ}$ and $5 A B=4 B C$, points $D$ and $E$ are the feet of the altitudes from $B$ and $C$, respectively. $M$ is the midpoint of $B D$ and the circumcircle of triangle $B M C$ meets line $A C$ again at $N$. Lines $B N$ and $C M$ meet at $P$. Prove that $\angle E D ... | [
"Solution:\n\nFrom the given, $A B=4 l$ and $B C=5 l$ for some constant $l>0$. Since $\\angle A B C=60^{\\circ}$, $B E=\\frac{5 l}{2}$ and $C E=\\frac{5 \\sqrt{3} l}{2}$. Also, by the cosine law, $A C=\\sqrt{21}$. Since $B E D C$ is cyclic, $\\angle E D A=\\angle A B C=60^{\\circ}$. Consequently, $\\angle E D B=30^... | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler li... | null | proof only | null | |
0449 | In the plane rectangular coordinate system, given hyperbola $\Gamma: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ ($a, b > 0$), a line with inclination angle $\frac{\pi}{4}$ passes through a vertex of $\Gamma$ and another point $(2, 3)$ on it. Then the eccentricity of $\Gamma$ is ______. | [
"The slope of the line described in the question is $1$ and it passes through point $(2, 3)$, so its equation is $y = x + 1$. This line intersects with the $x$-axis at point $(-1, 0)$, and thus $(-1, 0)$ is a vertex of $\\Gamma$. Hence, $a = 1$.\n\nAnd since point $(2, 3)$ is on $\\Gamma$, we know that $\\frac{2^2}... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 2 | |
0fgr | Problem:
Para cada número natural $n$ se considera el polinomio
$$
P_{n}(x)=x^{n+2}-2x+1
$$
a) Demostrar que la ecuación $P_{n}(x)=0$ tiene una raíz $c_{n}$ y sólo una en el intervalo $(0,1)$.
b) Calcular
$$
\lim_{n \rightarrow \infty} c_{n}
$$ | [
"Solution:\n\na) Tenemos $P_{n}(0)=1$ y\n$$\nP_{n}\\left(\\frac{3}{4}\\right)=\\left(\\frac{3}{4}\\right)^{n+2}-\\frac{3}{2}+1=\\left(\\frac{3}{4}\\right)^{n+2}-\\frac{1}{2}\n$$\npero $\\left(\\frac{3}{4}\\right)^{n+2}$ es decreciente y para $n=1$ es $\\left(\\frac{3}{4}\\right)^{1+2}=\\frac{27}{64}<\\frac{1}{2}$, ... | Spain | OME 23 | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 1/2 | |
010s | Problem:
Let $ABC$ be a triangle with $\angle C = 60^{\circ}$ and $|AC| < |BC|$. The point $D$ lies on the side $BC$ and satisfies $|BD| = |AC|$. The side $AC$ is extended to the point $E$ where $|AC| = |CE|$. Prove that $|AB| = |DE|$. | [
"Solution:\n\nConsider a point $F$ on $BC$ such that $|CF| = |BD|$ (see Figure 14).\nSince $\\angle ACF = 60^{\\circ}$, triangle $ACF$ is equilateral. Therefore $|AF| = |AC| = |CE|$ and $\\angle AFB = \\angle ECD = 120^{\\circ}$. Moreover, $|BF| = |CD|$. This implies that triangles $AFB$ and $ECD$ are congruent, an... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0jhf | Problem:
Find the number of positive integer divisors of $12!$ that leave a remainder of $1$ when divided by $3$. | [
"Solution:\n\nFirst we factor $12! = 2^{10} 3^{5} 5^{2} 7^{1} 11^{1}$, and note that $2, 5, 11 \\equiv -1 \\pmod{3}$ while $7 \\equiv 1 \\pmod{3}$. The desired divisors are precisely $2^{a} 5^{b} 7^{c} 11^{d}$ with $0 \\leq a \\leq 10$, $0 \\leq b \\leq 2$, $0 \\leq c \\leq 1$, $0 \\leq d \\leq 1$, and $a + b + d$ ... | United States | HMMT November 2013 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | final answer only | 66 | |
0118 | Problem:
A sequence of positive integers $a_{1}, a_{2}, \ldots$ is such that for each $m$ and $n$ the following holds: if $m$ is a divisor of $n$ and $m<n$, then $a_{m}$ is a divisor of $a_{n}$ and $a_{m}<a_{n}$. Find the least possible value of $a_{2000}$. | [
"Solution:\nAnswer: $128$.\nLet $d$ denote the least possible value of $a_{2000}$. We shall prove that $d=128$.\n\nClearly the sequence defined by $a_{1}=1$ and $a_{n}=2^{\\alpha_{1}+\\alpha_{2}+\\cdots+\\alpha_{k}}$ for $n=p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\cdots p_{k}^{\\alpha_{k}}$ has the required proper... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 128 | |
0h7r | 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 ... | [
"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$."
] | Ukraine | 56th Ukrainian National Mathematical Olympiad, Third Round | [
"Geometry > Solid Geometry > 3D Shapes"
] | English | proof and answer | 6 | |
0i3t | Problem:
Parallelogram $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$. | [
"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 peri... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 7/400 | |
05km | Problem:
Soit $n$ un entier strictement positif. Trouver le plus petit entier $k$ ayant la propriété suivante : pour tous réels $a_{1}, \ldots, a_{d}$ vérifiant $a_{1}+a_{2}+\cdots+a_{d}=n$ et $0 \leqslant a_{i} \leqslant 1$ pour $i=1,2, \ldots, d$, il est possible de regrouper les nombres en $k$ paquets (éventuelleme... | [
"Solution:\n\n$\\triangleright$ Les réels $a_{1}=\\ldots=a_{2 n-1}=\\frac{n}{2 n-1}$ appartiennent à l'intervalle $\\left.] \\frac{1}{2}, 1\\right]$ et leur somme vaut $n$. Comme chaque réel est strictement plus grand que $\\frac{1}{2}$, chaque paquet contient au plus un de ces $2 n-1$ réels (car la somme de deux d... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2n - 1 | |
0fh3 | Problem:
Sea $\mathcal{D}$ el conjunto de los números complejos que se pueden escribir en la forma $a+b \sqrt{-13}$, con $a, b$ enteros. El número $14=14+0 \sqrt{-13}$ puede escribirse como producto de dos elementos de $\mathcal{D}$: $14=2 \cdot 7$. Expresar 14 como producto de dos elementos de $\mathcal{D}$ de todas ... | [
"Solution:\n\nSi $a, b, c, d \\in \\mathbb{Z}$, entonces\n$$\n14=(a+b \\sqrt{-13})(c+d \\sqrt{-13})=(a c-13 b d)+(b c+a d) \\sqrt{-13}\n$$\nhay que resolver el sistema\n$$\n\\begin{aligned}\na c-13 b d & =14 \\\\\nb c+a d & =0\n\\end{aligned}\n$$\nSi $b=0$ puede ser $a=14,7,2,1$ y entonces $c=1,7,14,2$, y $d=0$ y l... | Spain | OME 25 | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Algebraic Number Theory > Unique factorization",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 14 = (14 + 0·√-13)(1 + 0·√-13) = (7 + 0·√-13)(2 + 0·√-13) = (1 + √-13)(1 − √-13) = (−1 + √-13)(−1 − √-13) | |
06vy | Let $\mathcal{A}$ denote the set of all polynomials in three variables $x, y, z$ with integer coefficients. Let $\mathcal{B}$ denote the subset of $\mathcal{A}$ formed by all polynomials which can be expressed as
$$
(x+y+z) P(x, y, z)+(x y+y z+z x) Q(x, y, z)+x y z R(x, y, z)
$$
with $P, Q, R \in \mathcal{A}$. Find the... | [
"We start by showing that $n \\leqslant 4$, i.e., any monomial $f=x^{i} y^{j} z^{k}$ with $i+j+k \\geqslant 4$ belongs to $\\mathcal{B}$. Assume that $i \\geqslant j \\geqslant k$, the other cases are analogous.\nLet $x+y+z=p$, $x y+y z+z x=q$ and $x y z=r$. Then\n$$\n0=(x-x)(x-y)(x-z)=x^{3}-p x^{2}+q x-r,\n$$\nthe... | IMO | IMO 2020 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof and answer | 4 | |
02xj | Problem:
Os denominadores de duas frações irredutíveis são $600$ e $700$. Qual é o menor valor possível do denominador de sua soma quando escrita como uma fração irredutível?
Observação: Dizemos que a fração $p / q$ é irredutível se os inteiros $p$ e $q$ não possuem fatores primos em comum em suas fatorações. Por exem... | [
"Solution:\nSejam $a / 600$ e $b / 700$ as duas frações irredutíveis. Assim, $\\mathrm{mdc}(a, 600) = \\mathrm{mdc}(b, 700) = 1$. A soma das duas frações pode ser escrita como\n$$\n\\begin{aligned}\n\\frac{a}{600} + \\frac{b}{700} &= \\frac{7a + 6b}{6 \\cdot 7 \\cdot 100} \\\\\n&= \\frac{7a + 6b}{3 \\cdot 7 \\cdot ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 168 | |
07ev | $\{a_n\}_{n \ge 0}$ and $\{b_n\}_{n \ge 0}$ are two sequences of positive integers that $a_i, b_i \in \{0, 1, 2, \dots, 9\}$. There is an integer number $M$ such that $a_n, b_n \ne 0$ for all $n \ge M$ and for each $n \ge 0$
$$
(\overline{a_n \cdots a_1 a_0})^2 + 999 \mid (\overline{b_n \cdots b_1 b_0})^2 + 999
$$
(No... | [
"Define\n$$\nA_n := \\overline{a_n \\dots a_1 a_0}, \\quad B_n := \\overline{b_n \\dots b_1 b_0}, \\quad K_n := \\frac{B_n^2 + 999}{A_n^2 + 999}.\n$$\n**Claim 1.** $\\{K_n\\}$ is eventually constant or $10^n \\mid A_{n-1}^2 + 999$ for each positive integer $n$.\n\n---\n\n*Proof.* For each $n \\ge 1$\n$$\nB_n^2 + 99... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
04m8 | Let $f: \mathbb{N} \to \mathbb{N}$ be a function such that for all positive integers $a$ and $b$,
$$
f(a) + f(b) - ab \mid af(a) + bf(b).
$$
Find all such functions $f$. | [
"It is given that\n$$\nf(a) + f(b) - ab \\mid af(a) + bf(b). \\qquad (3)\n$$\nTaking $a = b = 1$ in (3), we have $2f(1) - 1 \\mid 2f(1)$. Then $2f(1) - 1 \\mid 2f(1) - (2f(1) - 1) = 1$ and hence $f(1) = 1$.\n\nLet $p \\ge 7$ be a prime. Taking $a = p$ and $b = 1$ in (3), we have $f(p) - p + 1 \\mid pf(p) + 1$ and h... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(n) = n^2 for all positive integers n | |
02hl | Problem:
1) Determine o valor de $(666666666)^2-(333333333)^2$. | [
"Solution:\n\n1. Usando a fatoração $x^2-y^2=(x-y)(x+y)$, obtemos:\n$$\n\\begin{aligned}\n666\\,666\\,666^2 - 333\\,333\\,333^2 &= (666\\,666\\,666 - 333\\,333\\,333)(666\\,666\\,666 + 333\\,333\\,333) \\\\\n&= 333\\,333\\,333 \\times 999\\,999\\,999 \\\\\n&= 333\\,333\\,333 \\times (1\\,000\\,000\\,000 - 1) \\\\\n... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 333333332666666667 | |
0fzb | Problem:
Seien $f$, $g$ zwei Polynome mit ganzen Koeffizienten und seien $a$, $b$ ganzzahlige Fixpunkte von $f \circ g$. Beweise, dass ganzzahlige Fixpunkte $c$, $d$ von $g \circ f$ existieren mit $a+c=b+d$. | [
"Solution:\n\nSetze $c = g(a)$ und $d = g(b)$, dann sind $c$, $d$ ganze Fixpunkte von $g \\circ f$. Falls nun $a = b$ gilt, sehen wir sofort, dass die Gleichung erfüllt ist. Für $a \\neq b$ gilt nun $a-b \\mid g(a)-g(b) = c-d \\mid f(c)-f(d) = a-b$, und da es sich um ganze Zahlen handelt, muss $|a-b| = |c-d|$ gelte... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0jpc | Problem:
Find the sum of squares of all distinct complex numbers $x$ satisfying the equation
$$
0 = 4 x^{10} - 7 x^{9} + 5 x^{8} - 8 x^{7} + 12 x^{6} - 12 x^{5} + 12 x^{4} - 8 x^{3} + 5 x^{2} - 7 x + 4
$$ | [
"Solution:\nAnswer: $-\\frac{7}{16}$\n\nFor convenience denote the polynomial by $P(x)$. Notice $4+8=7+5=12$ and that the consecutive terms $12 x^{6}-12 x^{5}+12 x^{4}$ are the leading terms of $12 \\Phi_{14}(x)$, which is suggestive. Indeed, consider $\\omega$ a primitive $14$-th root of unity; since $\\omega^{7}=... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | -7/16 | |
0a6n | Problem:
The function $r_{n}(x)$ is the remainder when $x$ is divided by $n$, where $0 \leq r_{n}(x) < n$. For which $n$ does there exists some ordering $\{a_{1}, \ldots , a_{n - 1}\}$ of $\{1, 2, \ldots , n - 1\}$ such that $\{r_{n}(a_{1}), r_{n}(2 \times a_{2}), \ldots , r_{n}((n - 1) \times a_{n - 1})\}$ is an order... | [
"Solution:\nNotice that $r_{n}(x)$ is just $x$ modulo $n$. Therefore\n$$\n\\prod_{i} r_{n}(i a_{i}) \\equiv \\prod_{i} i a_{i} \\pmod{n}\n$$\nFor primes $p$, apply Wilson's theorem to see that we must have\n$$\n\\prod_{i} i a_{i} \\equiv -1 \\pmod{p}\n$$\nHowever,\n$$\n\\prod_{i} i a_{i} = \\prod_{i} i \\prod_{i} a... | New Zealand | NZMO Round Two | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | n = 2 | |
0lei | Let $ABC$ be an acute, scalene triangle with orthocenter $H$ and $D$, $E$, $F$ be the feet of altitudes from vertices $A$, $B$, $C$ respectively. Let $(I)$ be the circumcircle of triangle $HEF$ with centre $I$ and $K$, $J$ be the midpoints of $BC$, $EF$ respectively. $HJ$ meets $(I)$ again at $G$, $GK$ meets $(I)$ agai... | [
"a) It is well known that $KE$, $KF$ are both tangent to $(I)$. Thus, $GK$ is the symmedian of $\\angle GEF$, it follows that $\\overarc{LE} = \\overarc{HF}$. Hence, $AH$, $AL$ are isogonal with respect to angle $BAC$. It is clear that $AH$ is the diameter of $(I)$. Therefore $AL$ is the altitude of $\\angle AEF$.\... | Vietnam | VMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar... | English | proof only | null | |
0brb | Let $ABCD$ be a square and $E$ be a point on its diagonal $BD$, different from its midpoint. Denote $H$ and $K$ the orthocenters of the triangles $ABE$, respectively $ADE$. Prove that $\overline{BH} + \overline{DK} = 0$.
Mihaela Berindeanu | [
"Notice that the points $H$ and $K$ are on the diagonal $AC$, because $AC$ is perpendicular on $BE$ and $DE$. Also, $H$ and $K$ are on the altitudes from $E$ in the two triangles, which are perpendicular on the sides of the initial square.\n\nIt follows that the triangle $EHK$ is right and isosceles, so $H$ and $K$... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Quadrilaterals... | English | proof only | null | |
0hrm | Problem:
We have a row of boxes that is infinite in one direction, as shown.
Determine 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. | [
"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$$... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0hvs | Problem:
Vandal 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 V... | [
"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$... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0gr0 | There are some number of green crawlers in the lowest leftmost unit square and some number of brown crawlers in the highest leftmost unit square of the grid $2014 \times 2014$. Each green crawler at each move can pass to the neighboring square located at its up or at its right. Each brown crawler at each move can pass ... | [
"The answer is $1343$. We denote the lowest leftmost, the highest leftmost, the highest rightmost and the lowest rightmost unit squares by $A$, $B$, $C$, $D$, respectively. Example: Let us $672$ green crawlers to $A$ and $671$ brown crawlers to $B$. One green crawler from $A$ makes $2013$ up moves and after that ma... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1343 | |
00n7 | Let $ABCDE$ be a convex pentagon having a circumcircle and satisfying $AB = BD$. The point $P$ is the intersection of the diagonals $AC$ and $BE$. The lines $BC$ and $DE$ intersect in point $Q$.
Show that the line $PQ$ is parallel to the diagonal $AD$. | [
"\nFigure 2: Problem 6\n\nSolution:\n\nWe denote the circumcircle of the pentagon $ABCDE$ by $k$, see Figure 2. By assumption, the triangle $ABD$ is isosceles, which implies that the tangent $t_B$ to $k$ in $B$ is parallel to $AD$.\n\nWe apply Pascal's theorem to the inscribed hexagon $BEDA... | Austria | AUT_ABooklet_2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English | proof only | null | |
0ahp | Let $x$ be a real number such that the numbers $x^3$ and $x^2 + x$ are rational. Prove that $x$ is rational. | [
"Let $a = x^3$, $b = x^2 + x$. Then $a = x^3 = x(x^2 + x) - (x^2 + x) = x b - b = b(x - 1)$. It is clear that $b \\neq -1$, since, if this wasn't the case, $x^2 + x = -1$ or $(x + \\frac{1}{2})^2 - \\frac{1}{4} = -1$. Then $(x + \\frac{1}{2})^2 = -\\frac{3}{4}$ which is impossible. We get that $x = \\frac{a + b}{b ... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
08xm | On the circumference of a circle, 6 points $A$, $B$, $C$, $D$, $E$, $F$ are placed in this order counter-clockwise, and three lines $AD$, $BE$ and $CF$ intersect at a single point. If
$$
AB = 1,\ BC = 2,\ CD = 3,\ DE = 4,\ EF = 5,
$$
find the value of $FA$. Here we denote the length of a line segment $XY$ also by $XY$. | [
"$$\n\\boxed{\\frac{15}{8}}\n$$\nLet $P$ be the point of intersection of lines $AD$, $BE$ and $CF$. Then, we have $\\angle PBA = \\angle PDE$, since they are subtended by the same arc $\\overarc{EA}$ of the circle at the points $B$ and $D$ on the circumference. We also have $\\angle BPA = \\angle DPE$ so that the t... | Japan | Japan Mathematical Olympiad Initial Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof and answer | 15/8 | |
0hk3 | Problem:
Find all positive integers $n$ such that $n(n+1)$ is a perfect square. | [
"Solution:\nSince $n$ and $n+1$ are coprime numbers, if $n(n+1)$ is a perfect square, then each of $n$ and $n+1$ has to be a perfect square itself. However, that is impossible since if $n = x^{2}$ and $n+1 = y^{2}$ we would have $1 = y^{2} - x^{2} = (y - x)(y + x)$ and $1$ can't be expressed as a product of two dif... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | no positive integers | |
0jib | Problem:
I have 8 unit cubes of different colors, which I want to glue together into a $2 \times 2 \times 2$ cube. How many distinct $2 \times 2 \times 2$ cubes can I make? Rotations of the same cube are not considered distinct, but reflections are. | [
"Solution:\n\nOur goal is to first pin down the cube, so it can't rotate. Without loss of generality, suppose one of the unit cubes is purple, and let the purple cube be in the top left front position. Now, look at the three positions that share a face with the purple cube. There are $\\binom{7}{3}$ ways to pick th... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 1680 | |
0jro | Problem:
Which number is larger, $A$ or $B$, where
$$
A=\frac{1}{2015}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2015}\right) \quad \text{and} \quad B=\frac{1}{2016}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2016}\right) ?
$$
Prove that your answer is correct. | [
"Solution:\nWe claim that:\n$$\nA=\\frac{1}{2015}\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{2015}\\right)>B=\\frac{1}{2016}\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{2016}\\right) .\n$$\nTo prove this, let $S=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{2015}$. Then $A=\\frac{1}{2015} S$ and $B... | United States | BAMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | A is larger than B | |
09bm | a, b сөрөг биш бодит тоонуудын хувьд
$$
\left(\frac{a+b}{2}\right)^{9} \geq a^{3}b^{3}\left(\frac{a^{3}+b^{3}}{2}\right)
$$
тэнцэтгэл биш биелэхийг батал. | [
"$$\n\\left(\\frac{a+b}{2}\\right)^{12} = \\left(\\frac{a^3 + b^3 + ab(a+b) + ab(a+b) + ab(a+b)}{8}\\right)^4\n$$\n\n$$\n\\geq \\frac{a^3 + b^3}{2} \\cdot \\left( \\frac{ab(a+b)}{2} \\right)^3\n$$\n\n$$\n\\Rightarrow \\left( \\frac{a+b}{2} \\right)^9 \\geq \\frac{a^3 + b^3}{2} \\cdot a^3 b^3 \\blacktriangle\n$$\n\n... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Mongolian | proof only | null | |
00r6 | Let $a$ be a positive integer. For all positive integers $n$, we define
$$
a_n = 1 + a + a^2 + \dots + a^{n-1}.
$$
Let also $s, t$ be two different positive integers satisfying the following property: If $p$ is a prime divisor of $s-t$ then $p$ also divides $a-1$. Prove that the number
$$
\frac{a_s - a_t}{s - t}
$$
is ... | [
"Without loss of generality, we assume that $s > t$. Then we have that\n$$\na_s - a_t = a^t + a^{t+1} + \\dots + a^{s-1} = a^t a_{s-t}.\n$$\nSo, in order to prove that $s-t|a_s - a_t$, it is enough to prove that $s-t|a_{s-t}$, or equivalently that for every prime power $p^k|s-t$, then $p^k|a_{s-t}$. We write $s-t =... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
04hg | Let $N \ge 3$ be an odd positive integer. At the beginning in each square of an $N \times N$ board there is number $0$. In one move one can choose two squares with a common side and increase or decrease by $1$ the numbers in those two squares. If after $K$ moves, the sums of numbers in every row and every column are al... | [
"In each move the sum of all numbers increases or decreases by two, so the sum of all numbers remains even. Let us assume that after $K$ moves the sums of numbers in each row and each column is equal and let us denote it by $S$. The sum of all numbers is then equal to $N \\cdot S$. Since this must be even and $N$ i... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0715 | Problem:
Two players play a game on a $2000 \times 2001$ board. Each has one piece and the players move their pieces alternately. A short move is one square in any direction (including diagonally) or no move at all. On his first turn each player makes a short move. On subsequent turns a player must make the same move ... | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes, the first player can always win. | |
03cz | Solve $(x+1)\log_3 x + 4x \log_3 x - 16 = 0$. | [
"We need $x > 0$. Substitute $y = \\log_3 x$ to obtain $(x+1)y^2 + 4x y - 16 = 0$, which is equivalent to $(y+4)(x y + y - 4) = 0$.\n\nThus either $\\log_3 x = -4$ and $x = \\frac{1}{81}$ or $\\log_3 x = \\frac{4}{x+1}$, with the obvious solution $x = 3$.\n\nDue to monotonicity, it is easy to check that the latter ... | Bulgaria | Bulgaria 2022 | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | x = 3 and x = 1/81 | |
0jg0 | Problem:
Let $P$ be the number of ways to partition $2013$ into an ordered tuple of prime numbers. What is $\log_{2}(P)$? If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor\frac{125}{2}\left(\min \left(\frac{C}{A}, \frac{A}{C}\right)-\frac{3}{5}\right)\right\rflo... | [
"Solution:\n\nAnswer: $614.519\\ldots$\n\nWe use the following facts and heuristics.\n\n(1) The ordered partitions of $n$ into any positive integers (not just primes) is $2^{n-1}$. This can be guessed by checking small cases and finding a pattern, and is not difficult to prove.\n\n(2) The partitions of $\\frac{2013... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 614.519... | |
0fyl | Problem:
Seien $m$, $n$ natürliche Zahlen, sodass $m+n+1$ prim ist und ein Teiler von $2\left(m^{2}+n^{2}\right)-1$. Zeige, dass $m=n$ gilt. | [
"Solution:\n\nNach Voraussetzung ist $p = m+n+1$ auch ein Teiler von\n$$\n2(m+n) \\cdot (m+n+1) - \\left(2\\left(m^{2}+n^{2}\\right)-1\\right) = 4mn + 2m + 2n + 1 = (2m+1)(2n+1)\n$$\nDa $p$ prim ist, muss es daher einen der beiden Faktoren rechts teilen. Ausserdem ist $p$ kein echter Teiler dieser Faktoren wegen $p... | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0bt1 | Let $A$ be a ring and let $D$ be the set of all non-invertible elements of $A$. Assuming $a^2 = 0$ for all $a$ in $D$, prove that:
a) $axa = 0$ for all $a$ in $D$ and all $x$ in $A$; and
b) If $D$ is finite and $|D| \ge 2$, there exists $a$ in $D \setminus \{0\}$ such that $ab = ba = 0$ for all $b$ in $D$. | [
"a) Let $a$ be a member of $D$ and let $x$ be a member of $A$. If $x$ is invertible, then $ax$ is a member of $D$, so $axax = 0$, and $axa = 0$. If $x$ is a member of $D$, then $1+x$ is invertible, so $a + ax = a(1+x)$ is also a member of $D$. Hence $0 = (a+ax)^2 = a^2 + a^2x + axa + axax = axa(1+x)$, and consequen... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Ring Theory"
] | English | proof only | null | |
0a99 | Problem:
The difference between the cubes of two consecutive positive integers is a square $n^{2}$, where $n$ is a positive integer. Show that $n$ is the sum of two squares. | [
"Solution:\n\nAssume that $(m+1)^{3}-m^{3}=n^{2}$. Rearranging, we get $3(2m+1)^{2} = (2n+1)(2n-1)$. Since $2n+1$ and $2n-1$ are relatively prime (if they had a common divisor, it would have divided the difference, which is $2$, but they are both odd), one of them is a square (of an odd integer, since it is odd) an... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 22 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
091s | Problem:
Let $A B C D E$ be a convex pentagon with all five sides equal in length. The diagonals $A D$ and $E C$ meet in $S$ with $\varangle A S E=60^{\circ}$. Prove that $A B C D E$ has a pair of parallel sides. | [
"Solution:\n\nLet $F$ be such that $D E F$ is an equilateral triangle and the points $B$ and $F$ lay in the opposite half-planes determined by $D E$. Denote $\\varangle D A E=\\alpha$. Then $\\varangle A D E=\\alpha$.\n\nSince $\\varangle E S D=120^{\\circ}$, we have $\\varangle D E C=60^{\... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geom... | null | proof only | null | |
076h | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
f(x^2 + x f(y)) = x f(x + y)
$$
for all reals $x, y$. | [
"It is easy to see that $f(0) = 0$ and $f(x^2) = x f(x)$ for all $x \\in \\mathbb{R}$. If $f(\\alpha) = 0$ for some $\\alpha$, then\n$$\nf(x^2 + x f(\\alpha)) = x f(x + \\alpha),\n$$\nfor all $x \\in \\mathbb{R}$. Therefore\n$$\nx f(x + \\alpha) = f(x^2) = x f(x),\n$$\nfor all $x \\in \\mathbb{R}$. For $x \\neq 0$,... | India | IND_TSExams | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = 0 for all real x; and f(x) = x for all real x | |
0lap | Given a triangle $ABC$, not right-angled in $A$. Let $M$ be the midpoint of $BC$. Consider a point $D$ moving on the line $AM$, in such a way that $D$ is not coincident with $M$. Denote by $(O_1)$ the circle passing through $D$ and touching $BC$ in $B$; by $(O_2)$ the circle passing through $D$ and touching $BC$ in $C$... | [] | Vietnam | Vietnamese Team Selection for IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0i7j | Problem:
$AD$ and $BC$ are both perpendicular to $AB$, and $CD$ is perpendicular to $AC$. If $AB = 4$ and $BC = 3$, find $CD$.
 | [
"Solution:\n\nBy Pythagoras in $\\triangle ABC$, $AC = 5$. But $\\angle CAD = 90^{\\circ} - \\angle BAC = \\angle ACB$, so right triangles $CAD$ and $BCA$ are similar, and $\\frac{CD}{AC} = \\frac{BA}{CB} = \\frac{4}{3} \\Rightarrow CD = \\frac{20}{3}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 20/3 | |
037h | Problem:
Find all real numbers $a$, such that the inequality
$$
x^{4}+2 a x^{3}+a^{2} x^{2}-4 x+3>0
$$
holds true for all real numbers $x$. | [
"Solution:\nFirst Solution. Write the equation in the form\n$$\nx^{2}(x+a)^{2}>4 x-3\n$$\nThen for $x=1$ we get $(a+1)^{2}>1$, i.e. $a>0$ or $a<-2$. If $a<-2$, then $x=-a$ gives a contradiction $0>-4 a-3$. Thus, $a>0$.\n\nConversely, if $a>0$, then (1) is satisfied for all $x$. Indeed, when $x \\leq 0$ this is obvi... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a > 0 | |
07c5 | a) Prove that there exist two regular tetrahedra such that each edge of one is perpendicular to an edge of the other.
b) For an arbitrary regular tetrahedron, prove that there exists a unique tetrahedron with the above property. | [
"a) The following figure of a cube $ABCDEFGH$ shows that the two regular tetrahedra $ACHF$ and $BDEG$ are perpendicular.\n\n\n\nb) Consider two perpendicular tetrahedra $ABCD$ and $XYZT$. One of the following cases involves those three edges of $ABCD$ that are perpendicular to edges $XY$, $... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
05r7 | Problem:
Patricia a placé 2018 points dans le plan, de sorte que les distances entre 2 points quelconques soient deux à deux distinctes. Elle colorie alors chacun de ses 2018 points, en faisant attention à ce que ; pour chaque point $P$, les points $Q$ et $R$ placés le plus près et le plus loin de $P$ sont de la même ... | [
"Solution:\n\nOn va montrer que Patricia a pu utiliser un maximum de 504 couleurs. Tout d'abord, dès qu'une couleur sert à colorier un sommet $v$, elle sert aussi à colorier les sommets le plus proche et le plus éloigné de $v$, c'est-à-dire 3 sommets au moins.\n\nSupposons maintenant que deux couleurs, au moins, n'... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics >... | null | proof and answer | 504 | |
0kt9 | Problem:
Let $\mathcal{E}$ be an ellipse with foci $A$ and $B$. Suppose there exists a parabola $\mathcal{P}$ such that
- $\mathcal{P}$ passes through $A$ and $B$,
- the focus $F$ of $\mathcal{P}$ lies on $\mathcal{E}$,
- the orthocenter $H$ of $\triangle F A B$ lies on the directrix of $\mathcal{P}$.
If the major and ... | [
"Solution:\nLet $D$ and $E$ be the projections of $A$ and $B$ onto the directrix of $\\mathcal{P}$, respectively. Also, let $\\omega_{A}$ be the circle centered at $A$ with radius $A D = A F$, and define $\\omega_{B}$ similarly.\nIf $M$ is the midpoint of $\\overline{D E}$, then $M$ lies on the radical axis of $\\o... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2402 | |
07wx | There is a $n \times n$ square forest grid in which you wish to grow a forest. You have $n^2$ saplings (young trees) so that the $i$-th sapling, once planted, will grow to a tree of height exactly $i$ metres tall. Each sapling has to be planted in a different square of the forest grid. A balanced forest is one in which... | [
"The number of ways to grow a forest that may or may not be balanced is equal to $n^2!$. To answer the question we will first count the unbalanced forests, then subtract.\n\nThe main observation is that an unbalanced forest contains exactly one tree that violates both conditions. Indeed, if there were two such tree... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n^2! - n^2 * binom(n^2, 2n - 1) * ((n - 1)!)^2 * (n - 1)^2! | |
0i9d | Problem:
The 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$. | [
"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 cente... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | c(a + b)/(a + b + c) | |
01of | $N$ boys ($N \ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *tall* if he is taller than both of his neighbors.
Find all possible numbers of tall boys in the arrangement. | [
"Answer: any integer number from $1$ to $[N/2]$.\n\nConsider arbitrary arrangement of the boys along the circle. We put the signs \"+\" or \"-\" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign \"+\" before the boy if he is taller than the previous boy and we... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All integers from 1 to floor(N/2) | |
04wg | Find all sets $S$ of 3 or more primes with the following property: The elements of $S$ can be written around a circle such that if you compute the largest prime factor of the sum of each consecutive pair, you again obtain all elements of $S$ (in some order). (Michal Janík) | [] | Czech Republic | National Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | {2, 3, 5, 7} | |
07ju | Let $n$, $m$ be positive integers such that $n + m$ is odd. Assume that the edges of a complete bipartite graph $K_{n,m}$ are labeled by $1$ and $-1$ such that the sum of the numbers written on all edges is zero. Show that this graph has a spanning tree such that the sum of the numbers on its edges is $0$.
*A spanning... | [
"We define an operation on spanning trees: we add an edge to it. This will certainly create a cycle. Now, we remove one of the edges of the cycle to obtain another spanning tree. The sum of the numbers written on the edges of the spanning tree changes by at most two. With this operation, it is possible to reach any... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0cpt | An acute-angled triangle $ABC$ is inscribed into circle $\omega$. The tangents to $\omega$ passing through $B$ and $C$ intersect the tangent to $\omega$ passing through $A$ at points $K$ and $L$ respectively. The line $k$ passing through $K$ is parallel to $AB$; the line $\ell$ passing through $L$ is parallel to $AC$. ... | [
"Первое решение. Докажем, что точка $P$ лежит на серединном перпендикуляре к отрезку $BC$. Пусть $O$ — центр окружности $\\omega$, а $X$ — точка пересечения прямых $BC$ и $PL$ (см. рис. 7).\n\n\n\nТак как $O$ лежит на серединном перпендикуляре к отрезку $BC$, достаточно доказать, что $OP \\... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English, Russian | proof only | null | |
096y | Problem:
Determinați cea mai mare valoare posibilă a raportului dintre suma cifrelor unui număr de patru cifre și însuși numărul. | [
"Solution:\n\nFie $\\overline{abcd}$ un număr de patru cifre. Atunci\n$$\n\\begin{gathered}\n\\frac{\\overline{abcd}}{a+b+c+d}=\\frac{1000a+100b+10c+d}{a+b+c+d}=1+\\frac{999a+99b+9c}{a+b+c+d} \\\\\n\\text{Deoarece } d \\leq 9, \\text{ obținem } \\frac{\\overline{abcd}}{a+b+c+d} \\geq 1+\\frac{999a+99b+9c}{a+b+c+9}=... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 19/1099 | |
0ea8 | In the parallelogram $ABCD$ we have $|AB| = |BD|$. Let $K$ be a point on the line $AB$ different from $A$, such that $|KD| = |AD|$. Denote the reflection of the point $C$ over $K$ by $M$, and the reflection of $B$ over $A$ by $N$. Prove that $MDN$ is an isosceles triangle with the apex at $D$. | [
"Triangle $ADK$ is isosceles with the apex at $D$. So, $\\angle BKD = 180^\\circ - \\angle DKA = 180^\\circ - \\angle KAD = \\angle CBK$ and $|DK| = |DA| = |BC|$. The triangles $DKB$ and $CBK$ are congruent because they have a common side $KB$, congruent angles $\\angle BKD = \\angle CBK$ and $|DK| = |BC|$. Thus, $... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01qe | We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets: if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a+b$ on both sheets.
Prove that after $m2^{m-1}$ steps, the su... | [
"1. See IMO-2014 Shortlist, Problem C2."
] | Belarus | SELECTION and TRAINING SESSION | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0kia | Set $u_0 = \frac{1}{4}$, and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence $u_{k+1} = 2u_k - 2u_k^2$. This sequence tends to a limit; call it $L$. What is the least value of $k$ such that
$$
|u_k - L| \le \frac{1}{2^{1000}}?
$$
(A) 10 (B) 97 (C) 123 (D) 329 (E) 401 | [] | United States | AMC 12 B | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | A | |
0guc | Is it possible that a set consisting of $23$ real numbers has exactly $2422$ non-empty subsets such that the product of all elements of each subset is a rational number? | [
"Answer: Yes.\nLet $n$ be a positive integer whose value will be determined later on. Let $S$ be the set consisting of $2^n$'s for $i = 0, 1, \\dots, 22$. Let $Q$ be the set of subsets of $S$ whose product of own elements is a rational number together with the empty set. Since every positive integer has a unique re... | Turkey | 31st Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | Yes | |
04sa | Find all four-digit numbers $n$ satisfying the following conditions:
i) number $n$ is product of three different primes;
ii) sum of the two smallest of these prime numbers is equal to the difference of largest two of them;
iii) sum of three primes is equal to the square of another prime. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2015 | |
085v | Problem:
Dall'insieme $\{1,2, \ldots, 100\}$ scegliamo 50 numeri distinti, la cui somma è 3000. Come minimo, quanti numeri pari abbiamo scelto?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6. | [
"Solution:\n\nLa risposta è (E). La somma di tutti i numeri positivi dispari minori di 100 è 2500. Quindi in ogni sottoinsieme $X$ di $\\{1, \\ldots, 100\\}$ la somma dei cui elementi faccia 3000 il contributo dei numeri pari di $X$ alla somma deve essere uguale ad almeno 500. Cinque numeri pari non sono sufficient... | Italy | Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | E | |
05vx | Problem:
Soient $x, y, z$ des réels strictement positifs tels que
$$
x+\frac{y}{z}=y+\frac{z}{x}=z+\frac{x}{y}=2
$$
Déterminer toutes les valeurs possibles que peut prendre le nombre $x+y+z$. | [
"Solution:\n\nDans un tel problème, il faut chercher à examiner chaque équation séparément mais aussi à les mettre en relation. En pratique, cela consiste à regarder l'équation obtenue lorsque l'on effectue la somme ou le produit de deux ou plusieurs équations. Une autre idée est d'appliquer des inégalités connues ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 3 | |
04i1 | Let $\overline{AC}$ be the diameter of the circle $k_1$ with the centre $B$. Circle $k_2$ touches the line $AC$ at the point $B$ and the circle $k_1$ at the point $D$. Tangent from $A$ (different from $AC$) to circle $k_2$ touches that circle at the point $E$ and intersects the line $BD$ in the point $F$. Determine the... | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 5:3 | |
0hng | Problem:
Evaluate $\sum_{i=1}^{\infty} \frac{(i+1)(i+2)(i+3)}{(-2)^{i}}$. | [
"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 $96$."
] | United States | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | final answer only | 96 | |
026z | Problem:
Considere o seguinte hexágono regular $ABCDEF$, cujo lado mede $4$, e onde os pontos $P$ e $Q$ são os pontos médios dos lados $\overline{BC}$ e $\overline{DE}$, respectivamente.
Calcule o perímetro do hexágono $ABPQEF$. | [
"Solution:\nProlongamos os lados $\\overline{BC}$ e $\\overline{ED}$ do modo indicado na figura seguinte:\n\nSendo $ABCDEF$ um hexágono regular, cada ângulo interno mede $120^\\circ$. Portanto, $\\measuredangle RCD = 60^\\circ$, $\\measuredangle RDC = 60^\\circ$, e concluímos que o triângul... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 22 | |
0i23 | Problem:
Find $(x+1)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right) \cdots$, where $|x|<1$. | [
"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 $S = \\frac{1}{1-x}$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/(1 - x) | |
09wy | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ die voldoen aan
$$
f\left(x^{2} y\right)+2 f\left(y^{2}\right)=\left(x^{2}+f(y)\right) \cdot f(y)
$$
voor alle $x, y \in \mathbb{R}$. | [
"Solution:\n\nOplossing I. Invullen van $x=1$ geeft $f(y)+2 f\\left(y^{2}\\right)=(1+f(y)) f(y)$, dus\n$$\n2 f\\left(y^{2}\\right)=f(y)^{2}\n$$\nWe kunnen hiermee in de oorspronkelijke functievergelijking de term $2 f\\left(y^{2}\\right)$ links wegstrepen tegen $f(y)^{2}$ rechts:\n$$\nf\\left(x^{2} y\\right)=x^{2} ... | Netherlands | Selectietoets | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all x; f(x) = 2x for all x; f(x) = 2|x| for all x | |
0jna | Problem:
The sequences of real numbers $\{a_{i}\}_{i=1}^{\infty}$ and $\{b_{i}\}_{i=1}^{\infty}$ satisfy $a_{n+1} = (a_{n-1} - 1)(b_{n} + 1)$ and $b_{n+1} = a_{n} b_{n-1} - 1$ for $n \geq 2$, with $a_{1} = a_{2} = 2015$ and $b_{1} = b_{2} = 2013$. Evaluate, with proof, the infinite sum
$$
\sum_{n=1}^{\infty} b_{n}\lef... | [
"Solution:\n\nAnswer: $\\quad 1+\\frac{1}{2014 \\cdot 2015}$ OR $\\frac{4058211}{4058210}$\n\nFirst note that $a_{n}$ and $b_{n}$ are weakly increasing and tend to infinity. In particular, $a_{n}, b_{n} \\notin\\{0,-1,1\\}$ for all $n$.\n\nFor $n \\geq 1$, we have $a_{n+3} = (a_{n+1} - 1)(b_{n+2} + 1) = (a_{n+1} - ... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 1 + 1/(2014*2015) | |
07wh | A trapezium $ABCD$ with $AB \parallel DC$ and $|AB| < |DC|$ is inscribed in a circle, centre $O$. The diagonals $AC$ and $BD$ are mutually perpendicular at $P$. If $E$ is the midpoint of $AB$ and $F$ is the midpoint of $DC$, prove $|OF| = |PE|$. | [
"First note that $\\angle BAC = \\angle BDC$ (inscribed angles) and $\\angle BAC = \\angle ACD$ (alternate angles at $AB \\parallel CD$). Hence, $\\angle BDC = \\angle ACD$ and triangle $CDP$ is isosceles with $|CP| = |DP|$. This implies that $P$ is on the perpendicular bisector of $CD$ and that $\\angle PBA = \\an... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
03ot | Assume that $n$ is a positive integer, and $A_1, A_2, \dots, A_{n+1}$ are $n+1$ nonempty subsets of the set $\{1, 2, \dots, n\}$. Prove that there are two disjoint and nonempty subsets $\{i_1, i_2, \dots, i_k\}$ and $\{j_1, j_2, \dots, j_m\}$ such that
$$
A_{i_1} \cup A_{i_2} \cup \dots \cup A_{i_k} = A_{j_1} \cup A_{j... | [
"**Proof I** We prove by induction for $n$.\n\nWhen $n = 1$, $A_1 = A_2 = \\{1\\}$, the proposition holds.\n\nSuppose it holds for $n$. We consider the case of $n+1$.\nSuppose $A_1, A_2, \\dots, A_{n+2}$ are nonempty subsets of $\\{1, 2, \\dots, n+1\\}$. Let $B_i = A_i \\setminus \\{n+1\\}, i = 1, 2, \\dots, n+2$. ... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
0efe | Problem:
S katerim izmed navedenih števil je deljivo število $5^{2017} + 5^{2016} + 5^{2015}$?
(A) 15
(B) 31
(C) 2015
(D) 39
(E) 2017 | [
"Solution:\n\n$5^{2017} + 5^{2016} + 5^{2015} = 5^{2015}(5^2 + 5 + 1) = 31 \\cdot 5^{2015}$. Število je deljivo z $31$. Pravilen odgovor je (B)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | B | |
0i8x | Problem:
a.
Let $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 consisten... | [
"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... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a) S1(101) = 101, S2(101) = 101, S3(101) = 0; b) 6; c) 89 | |
0gpk | In an isosceles triangle $ABC$ with $AB = AC$, let $D$ be the foot of the perpendicular of $A$ and $P$ be an interior point of the triangle $ADC$ such that $\angle APB > 90^\circ$ and $\angle PBD + \angle PAD = \angle PCB$. Let $Q$ be the intersection of the lines $CP$ and $AD$, and $R$ be the intersection of the lines... | [
"First observe that $\\angle PAR = \\angle QBR$. Let $\\angle PAR = \\alpha$, $\\angle PBC = \\beta$ and $\\angle BAD = \\theta$. Applying trigonometric form of Ceva Theorem for the point $P$ inside the triangle $ABC$ gives\n$$\n\\frac{\\sin(\\theta + \\alpha)}{\\sin(\\theta - \\alpha)} \\cdot \\frac{\\cos(\\alpha ... | Turkey | 20th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00qd | Triangle $ABC$ is said to be perpendicular to triangle $DEF$ if the perpendiculars from $A$ to $EF$, from $B$ to $FD$ and from $C$ to $DE$ are concurrent. Prove that if $ABC$ is perpendicular to $DEF$ then $DEF$ is perpendicular to $ABC$. | [
"Let $U, V, W$ be the feet of the perpendiculars from $A, B, C$ to $EF, FD, DE$ respectively, and let $X, Y, Z$ be the feet of the perpendiculars from $D, E, F$ to $BC, CA, AB$ respectively. Since $U$ and $Z$ both subtend a right angle from $AF$, $AFUZ$ is concyclic and so (with an appropriate sign convention) $\\a... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09s8 | Problem:
Bepaal alle positieve gehele getallen $n$ waarvoor er positieve gehele getallen $a_{1}, a_{2}, \ldots, a_{n}$ bestaan met
$$
a_{1}+2 a_{2}+3 a_{3}+\ldots+n a_{n}=6 n
$$
en
$$
\frac{1}{a_{1}}+\frac{2}{a_{2}}+\frac{3}{a_{3}}+\ldots+\frac{n}{a_{n}}=2+\frac{1}{n}
$$ | [
"Solution:\n\nAls we de ongelijkheid van het rekenkundig en harmonisch gemiddelde toepassen op $a_{1}$, twee keer $a_{2}$, drie keer $a_{3}, \\ldots, n$ keer $a_{n}$, dan vinden we\n$$\n\\frac{6 n}{\\frac{1}{2} n(n+1)}=\\frac{a_{1}+2 a_{2}+3 a_{3}+\\ldots+n a_{n}}{\\frac{1}{2} n(n+1)} \\geq \\frac{\\frac{1}{2} n(n+... | Netherlands | IMO-selectietoets II | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 3 | |
06vi | Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that... | [
"Notice that $\\angle TFB = \\angle FDA$ because $FT$ is tangent to circle $BDF$, and moreover $\\angle FDA = \\angle CGA$ because quadrilateral $ADFG$ is cyclic. Similarly, $\\angle TGB = \\angle GEC$ because $GT$ is tangent to circle $CEG$, and $\\angle GEC = \\angle CFA$. Hence,\n$$\n\\begin{equation*}\n\\angle ... | IMO | IMO 2019 Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dsh | Given $\triangle ABC$, let $I$, $O$, $\Gamma$ denote its incenter, circumcenter and circumcircle respectively. Let $AI$ intersect $\Gamma$ at $M\ (\neq A)$. Circle $\omega$ is tangent to $AB$, $AC$ and $\Gamma$ internally at $T$ (i.e. the mixtilinear incircle opposite $A$). Let the tangents at $A$ and $T$ to $\Gamma$ m... | [] | Singapore | Singapore International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > ... | null | proof only | null | |
02bx | Problem:
Os pontos $M$, $N$ e $P$ são escolhidos sobre os lados $BC$, $CA$ e $AB$ do triângulo $ABC$ de modo que $BM = BP$ e $CM = CN$. A perpendicular baixada de $B$ à $MP$ e a perpendicular baixada de $C$ à $MN$ se intersectam em $I$. Prove que os ângulos $\angle IPA$ e $\angle INC$ são congruentes.
![](attached_im... | [
"Solution:\n\nComo o triângulo $MNC$ é isósceles, $CI$ além de altura é também a bissetriz relativa ao vértice $C$. Consequentemente, $\\angle ICN = \\angle ICM$. Pelo caso de congruência $LAL$, segue que $\\triangle ICM \\equiv \\triangle ICN$. Daí, $\\angle IMC = \\angle INC$. De forma semelhante, temos $\\angle ... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0d1d | Find all pairs of prime numbers $p, q$ such that $p^2 - p - 1 = q^3$. | [
"The only such pair is $(p, q) = (37, 11)$.\nWe have $p(p-1) = (q+1)(q^2-q+1)$. Since $p$ is prime and $p > p-1$ and $q^2-q+1 > q+1$, there exists an integer $m > 0$ such that\n$$\nq^2 - q + 1 = mp, \\quad (1)\n$$\n$$\np-1 = m(q+1). \\quad (2)\n$$\nSince $q^3 = p^2 - p + 1 > (p-1)^2 = m^2(q+1)^2 > m^2q^2$ we deduce... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | (37, 11) | |
0h85 | 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.
b. 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 connec... | [
"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 ... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
034f | Problem:
Let $ABC$ be a triangle with altitude $CH$, where $H$ is an interior point of the side $AB$. Denote by $P$ and $Q$ the incenters of $\triangle AHC$ and $\triangle BHC$, respectively. Prove that the quadrilateral $ABQP$ is cyclic if and only if either $AC = BC$ or $\angle ACB = 90^\circ$. | [
"Solution:\n($\\Leftarrow$) If $AC = BC$, then the quadrilateral $ABQP$ is cyclic since it is an isosceles trapezoid. If $\\angle ACB = 90^\\circ$, then we have $\\angle ACI = \\angle BCI = 45^\\circ$, where $I$ is the incenter of $\\triangle ABC$. We have also $\\angle APC = \\angle BQC = 135^\\circ$, i.e. $\\angl... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0lb4 | A triangle $ABC$ is not isosceles at $A$, and its angles $\angle ABC$, $\angle ACB$ are acute. Consider a point $D$ moving on edge $BC$, so that $D$ does not coincide with $B$, $C$ and with the perpendicular projection of $A$ on $BC$. The line $d$, perpendicular with $BC$ at $D$, intersects the lines $AB$ and $AC$ at $... | [
"Since $\\angle ABC$ and $\\angle ACB$ are acute, $E$ lies on the opposite ray to ray $AB$ on the edge $AB$, at the same time, $F$ lies on the edge $AC$ or on the opposite ray to ray $AC$. Hence, it follows from the definition of the points $M, N, P$ that $E, M, N$ are collinear and $M, F, P$ are collinear.\nHence ... | Vietnam | Vijetnam 2011 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Pla... | English | proof only | null | |
05ek | Problem:
Un polyèdre a 6 sommets et 12 arêtes. Montrer que chaque face est un triangle. | [
"Solution:\n\nNotons $F$ le nombre de faces, $A$ le nombre d'arêtes et $S$ le nombre de sommets. La formule d'Euler donne $F - A + S = 2$, donc $F = A - S + 2 = 8$.\n\nNotons $x_{i}$ ($i = 1, 2, \\ldots, 8$) le nombre d'arêtes de la face $i$. On a $x_{1} + \\cdots + x_{8} = 2A = 24$ car chaque arête appartient à ex... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0cyt | Let $a_{1} + a_{2} + \ldots + a_{n} = 0$ and $|a_{1}| + |a_{2}| + \ldots + |a_{n}| = 1$.
Prove that
$$
\left|a_{1} + 2 a_{2} + \ldots + n a_{n}\right| \leq \frac{n-1}{2}
$$ | [
"Since $\\sum_{k=1}^{n} a_{k} = 0$, it follows that\n$$\n\\sum_{k=1}^{n} k a_{k} = \\sum_{k=1}^{n} (k - x) a_{k}\n$$\nfor every $x \\in \\mathbb{R}$. We obtain\n$$\n\\begin{aligned}\n\\left|\\sum_{k=1}^{n} k a_{k}\\right| & = \\left|\\sum_{k=1}^{n} (k - x) a_{k}\\right| \\leq \\sum_{k=1}^{n} |k - x| \\left|a_{k}\\r... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0jn2 | Problem:
The complex numbers $x, y, z$ satisfy
$$
\begin{aligned}
x y z & = -4 \\
(x+1)(y+1)(z+1) & = 7 \\
(x+2)(y+2)(z+2) & = -3
\end{aligned}
$$
Find, with proof, the value of $(x+3)(y+3)(z+3)$. | [
"Solution:\n\nConsider the cubic polynomial $f(t) = (x + t)(y + t)(z + t)$. By the theory of finite differences, $f(3) - 3 f(2) + 3 f(1) - f(0) = 3! = 6$, since $f$ is monic. Thus\n$$\nf(3) = 6 + 3 f(2) - 3 f(1) + f(0) = 6 + 3(-3) - 3(7) + (-4) = -28.\n$$\n\n\nSolution 2:\n\nAlternatively, note that the system of e... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | -28 | |
0aa0 | Problem:
An encyclopedia consists of $2000$ numbered volumes. The volumes are stacked in order with number $1$ on top and $2000$ on the bottom. One may perform two operations with the stack:
(i) For $n$ even, one may take the top $n$ volumes and put them in the bottom of the stack without changing the order.
(ii) Fo... | [
"Solution:\n\nLet the positions of the books in the stack be $1,2,3, \\ldots, 2000$ from the top (and consider them modulo $2000$). Notice that both operations fix the parity of the number of the book at any given position. Operation (i) subtracts an even integer from the number of the book at each position. If $A$... | Nordic Mathematical Olympiad | The 29th Nordic Mathematical Contest | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | (1000!)^2 | |
069e | Let $\triangle ABC$ be a triangle inscribed in a circle $\Gamma$ with center $O$. Let $I$ be the incenter of $\triangle ABC$ and $D, E, F$ the contact points of the incircle of $\triangle ABC$ with $BC, AC, AB$, respectively. If $S$ is the foot of the perpendicular from $D$ to the line $EF$, prove that the line $SI$ pa... | [
"Let $X$ be the second intersection point of $\\Gamma$ with $IA'$, where $A'$ is the antipodal of $A$ with respect to $\\Gamma$. We will prove that the points $X, S, I$ are collinear.\n\nWe have $\\widehat{IXA} = 90^\\circ$, so $X$ belongs to the circle of diameter $AI$. The same holds for the points $E$ and $F$, t... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry >... | English | proof only | null | |
0dl4 | Find all non-constant functions $f : \mathbb{R} \to \mathbb{R}$ such that:
(i) $f(x)f(x+y) = f(x)^2 + x \cdot f_{2025}(y)$ for all $x, y \in \mathbb{R}$,
(ii) $f_{2025}(\cos x) + f_{2025}(\cos(\pi - x)) = 0$ for all $x \in \mathbb{R}$.
with $f_n(x) = f(f(\ldots(f(x))\ldots))$ is the composition $n$ times of the functio... | [] | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x |
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