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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09m4 | Suppose $f(x)$ and $g(x)$ are quadratic trinomials with the property that
$$
\frac{f(-2)}{g(-2)} = \frac{f(3)}{g(3)} = 4.
$$
Given that $g(5) = 2$, $f(7) = 8$, and $g(7) = 6$, determine the value of $f(5)$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 16/9 | |
0c45 | Problem:
Fie $n \in \mathbb{N}^{*}, n \geq 2$. Demonstraţi că, pentru orice numere complexe $a_{1}, a_{2}, \ldots, a_{n}$ şi $b_{1}, b_{2}, \ldots, b_{n}$, următoarele afirmaţii sunt echivalente:
a) $\sum_{k=1}^{n}\left|z-a_{k}\right|^{2} \leq \sum_{k=1}^{n}\left|z-b_{k}\right|^{2}$, pentru orice $z \in \mathbb{C}$;
... | [
"Solution:\n\n$b) \\Rightarrow a)$ Avem\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n}\\left|z-a_{k}\\right|^{2} & =n|z|^{2}-z \\sum_{k=1}^{n} \\bar{a}_{k}-\\bar{z} \\sum_{k=1}^{n} a_{k}+\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2} \\\\\n& \\leq n|z|^{2}-z \\sum_{k=1}^{n} \\bar{b}_{k}-\\bar{z} \\sum_{k=1}^{n} b_{k}+\\sum_{k=1}^... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0gfo | Suppose that $p$ is an odd prime, $p \ge 7$ and $q = \frac{3p-7}{2}$.
Define the series
$$
S_q = \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7} + \cdots + \frac{1}{(q+1)(q+2)(q+3)}
$$
Express $1 + 2S_q - \frac{1}{p}$ as a rational number $\frac{m}{n}$ with $(m, n) = 1$.
Prove that $m$ is a multiple of $p$. | [
"We need the partial fraction decomposition\n$$\n\\frac{2}{(k+1)(k+2)(k+3)} = \\frac{1}{k+1} - \\frac{2}{k+2} + \\frac{1}{k+3}\n$$\nSum over $q = 1,4,7, \\dots$, we get\n$$\n\\begin{aligned}\n2S_q &= \\left(\\frac{1}{2} - \\frac{2}{3} + \\frac{1}{4}\\right) + \\left(\\frac{1}{5} - \\frac{2}{6} + \\frac{1}{7}\\right... | Taiwan | 國際奧林匹亞競賽第三次訓練營 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | Chinese; English | proof only | null | |
0ksr | Problem:
A certain rectangle can be tiled with a combination of vertical $b \times 1$ tiles and horizontal $1 \times a$ tiles. Show that the rectangle can be tiled with just one of the two types of tiles. | [
"Solution:\n\nLet $\\omega$ be a primitive $ab$-th root of unity. Fill the cells with complex numbers so that the $(i, j)$ cell (where the top left square is $(1,1)$) has entry $\\omega^{a(i-1)+b(j-1)}$.\n\nThen the sum of the entries in a horizontal $1 \\times a$ rectangle starting from $(i, j)$ is\n$$\n\\omega^{a... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0g9l | 固定一個銳角三角形 $ABC$。設 $E$, $F$ 點分別落在 $AC$, $AB$ 邊上, 並設 $M$ 點是 $EF$ 線段的中點。令 $EF$ 的中垂線與直線 $BC$ 交於 $K$ 點, 而 $MK$ 的中垂線分別交 $AC$, $AB$ 直線於 $S$, $T$ 點。若四邊形 $KSAT$ 共圓, 證明: $\angle KEF = \angle KFE = \angle A$.
Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respective... | [
"令四邊形 $KSAT$ 的外接圓為 $\\omega_1$。設直線 $AM$ 與直線 $ST$ 交於 $N$ 點，而設 $AM$ 與 $\\omega_1$ 的另一個交點為 $L$，如下圖所示。\n\n\n\n由於 $EF \\parallel TS$ 且 $M$ 為 $EF$ 的中點, $N$ 也會是 $ST$ 的中點。更由於 $K$ 與 $M$ 對直線 $ST$ 對稱, 可知 $\\angle KNS = \\angle MNS = \\angle LNT$。於是 $K, L$ 兩點對於 $ST$ 的中垂線對稱, 故 $KL \\parallel ST$。\n令 $G$... | Taiwan | 二〇一五數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bqv | Find all positive reals $x, y, z$ so that
$$
\begin{cases}
xyz = 6 \\
(x+1)(y+2)(z+3) = 48.
\end{cases}
$$ | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | x=1, y=2, z=3 | |
051n | Let $m$ be a positive integer. Prove that if Mari writes at least $m+3$ numbers on the board, then Jüri can choose 4 of those such that the sum of some two of those and the sum of the other two give the same remainder when divided by $m$. | [
"As Mari writes down $m+3$ numbers and there are only $m$ different remainders when dividing by $m$, there must be two that give equal remainders when divided by $m$; let those numbers be $a$ and $b$. The rest of the $m+1$ include two that also give equal remainders when divided by $m$; let those be $c$ and $d$. No... | Estonia | Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0k0f | Problem:
Let $m$ be a positive integer, and let $T$ denote the set of all subsets of $\{1,2, \ldots, m\}$. Call a subset $S$ of $T$ $\delta$-good if for all $s_{1}, s_{2} \in S$, $s_{1} \neq s_{2}$, $\left|\Delta\left(s_{1}, s_{2}\right)\right| \geq \delta m$, where $\Delta$ denotes symmetric difference (the symmetric ... | [
"Solution:\nAnswer: 2048\nLet $n=|S|$. Let the sets in $S$ be $s_{1}, s_{2}, \\ldots, s_{n}$. We bound the sum $\\sum_{1 \\leq i<j \\leq n}\\left|\\Delta\\left(s_{i}, s_{j}\\right)\\right|$ in two ways. On one hand, by the condition we have the obvious bound\n$$\n\\sum_{1 \\leq i<j \\leq n}\\left|\\Delta\\left(s_{i... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 2048 | |
0ilb | Problem:
Tim 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 ... | [
"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 $x$, $-12 \\leq x < 12$... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 33 | |
01q1 | Exactly one integer number from $1$ to $49$ is written in each cell of the $7 \times 7$ square table (see the figure). Per move it is allowed to choose any cell then simultaneously to increase (to decrease) the number in this cell by $1$ and to decrease (to increase respectively) the numbers in some two adjacent cells ... | [
"**a)** We show that using the allowed moves we can decrease any number in the table by $3$ so that all other numbers in the table keep their values. We will not consider the whole table but only the number $x$ which will be decreased by $3$ and three more numbers $a$, $b$, and $c$ which occupy (together with $x$) ... | Belarus | BelarusMO 2013_s | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) Yes. b) No. | |
0fmq | Determina todos los números enteros positivos $n$, para los cuales $S_n = x^n + y^n + z^n$ es constante, cualesquiera que sean $x, y, z$ reales tales que, $xyz = 1$ y $x + y + z = 0$. | [] | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Spanish | proof and answer | n = 1 and n = 3 | |
01sr | Students of two groups decided to organize a chess tournament where each student from the first group plays exactly one game with each student from the other group. But one student from the first group and one student from the other one, due to some reasons, failed to participate in the tournament, so the total number ... | [
"Answer: 17, 20, 29.\nLet $n$ and $m$ be the numbers of students of the first, and respectively the second group that were initially supposed to participate in the tournament. Then $S = (n-1)+(m-1)$ students have taken part in the tournament. By condition,\n$$\n(n-1)(m-1) = 0.8 \\text{ nm} \\iff (m-5)(n-5) = 20.\n$... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 17, 20, 29 | |
09er | Find all pair of natural numbers $(n, m)$ such that $2^{\varphi(n)} + 1 \mid m$ and $2^{\varphi(m)} + 1 \mid n$, where $\varphi(n)$ is Euler's function. | [
"Let $n, m > 1$. $\\varphi(m) = 2^{m_0} \\cdot m_1$, $\\varphi(n) = 2^{n_0} \\cdot n_1$ ($m_0, n_0 \\ge 0$, $m_1, n_1$-odd natural numbers.) Assume that $m_0 \\ge n_0$ and let $n$ be the least number such that $n \\mid 2^k - 1$.\nSet $k = 2^{k_0} \\cdot k_1$, where ($k_0$ is nonnegative whole number, $k_1$ is odd n... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (1,1), (1,3), (3,1) | |
05ru | Problem:
Soit $p$ un nombre premier.
Démontrer qu'il existe un nombre premier $q$ tel que $n^{p} \not\equiv p$ pour tout $n \in \mathbb{Z}$. | [
"Solution:\n\nTout d'abord, si $p$ ne divise pas $q-1$, alors $x \\mapsto x^{p}$ est une bijection de $\\mathbb{Z} / q \\mathbb{Z}$ dans lui-même, donc $q$ ne peut pas convenir. On en vient à chercher $q \\equiv 1\\ (\\bmod\\ p)$ tel que, pour tout $n \\not\\equiv 0\\ (\\bmod\\ q)$, $n$ soit d'ordre $\\omega_{q}(n)... | France | TEST DU 27 FÉVRIER 2019 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
00nh | Determine all triples $(a, b, c)$ of integers $a \ge 0$, $b \ge 0$ and $c \ge 0$ that satisfy the equation
$$
a^{b+20}(c-1) = c^{b+21} - 1.
$$ | [
"**Answer.** $\\{(1, b, 0) : b \\in \\mathbb{Z}_{>0}\\} \\cup \\{(a, b, 1) : a, b \\in \\mathbb{Z}_{>0}\\}$\n\nOne can first see that the right side factors:\n$$\na^{b+20}(c-1) = (c^{b+20} + c^{b+19} + \\dots + c + 1)(c-1).\n$$\nThe case $c=1$ will be handled separately (and is very simple). For $c \\ne 1$ the equa... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | {(1, b, 0) : b in Z_{>0}} ∪ {(a, b, 1) : a, b in Z_{>0}} | |
00nv | Determine all pairs of positive integers $(n, k)$ for which
$$
n! + n = n^k
$$
holds. | [
"*Answer.* The only solutions are $(2, 2)$, $(3, 2)$ and $(5, 3)$.\n\nBecause of $n! + n > n$, we immediately get $k \\ge 2$. We divide both sides of the equation by $n$ and get\n$$\n(n - 1)! + 1 = n^{k-1}.\n$$\nNow, we distinguish two cases:\n\n* $n$ is not a prime.\nSince $n$ is clearly not $1$, we can write $n$ ... | Austria | AUT_ABooklet_2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (2, 2), (3, 2), (5, 3) | |
08bf | Problem:
Sia $p(x)$ un polinomio a coefficienti interi tale che $p(0)=6$. Si sa che tra gli interi $m$ compresi fra 1 e 60 esattamente 40 sono tali che $p(m)$ sia multiplo di 3; inoltre, si sa che tra gli interi $m$ compresi fra 1 e 60 esattamente 30 sono tali che $p(m)$ sia multiplo di 4. Quanti sono gli interi $m$ c... | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. Per dimostrarlo, vogliamo mostrare che $p(m)$ è pari per ogni intero $m$, e che di conseguenza $p(m)$ è multiplo di 6 se e solo se è multiplo di 3. A questo punto possiamo sfruttare l'ipotesi che gli interi $m$ per cui $p(m)$ è multiplo di 3 sono precisamente 40 per otte... | Italy | Progetto Olimpiadi della Matematica | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
030e | Problem:
Com 5 algarismos não nulos, podemos formar 120 números, sem repetir algarismo em um mesmo número. Seja $S$ a soma de todos esses números. Determine a soma dos algarismos de $S$, sendo:
a) 1, 3, 5, 7 e 9 os 5 algarismos;
b) 0, 2, 4, 6 e 8 os 5 algarismos, lembrando que 02468 é um número com 4 algarismos e, p... | [
"Solution:\n\na) São 120 números ao todo, com todas as combinações possíveis. Assim, em cada uma das posições (unidade, dezena, centena, unidade do milhar, dezena do milhar), cada um dos algarismos aparece a mesma quantidade de vezes, ou seja, $\\frac{120}{5}=24$. Por exemplo, nas unidades, o algarismo 1 aparece 24... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) 30; b) 12 | |
02k1 | Problem:
Um padeiro quer gastar toda sua farinha para fazer pães. Trabalhando sozinho, ele conseguiria acabar com a farinha em 6 horas; com um ajudante, o mesmo poderia ser feito em 2 horas. O padeiro começou a trabalhar sozinho; depois de algum tempo, cansado, ele chamou seu ajudante e assim, após 150 minutos a farin... | [
"Solution:\n\nSeja $x$ a quantidade de farinha, em quilos, de que o padeiro dispõe. Trabalhando sozinho, ele usaria $\\frac{x}{6}$ quilos de farinha em 1 hora; trabalhando com seu ajudante, eles usariam $\\frac{x}{2}$ quilos de farinha em 1 hora. Seja $t$ o tempo, em horas, que o padeiro trabalhou sozinho. Como a f... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 45 minutes | |
01mp | Prove that if positive numbers $a$, $b$, $x$, $y$ satisfy the inequalities $ab \ge xa + yb$, then they satisfy the inequality $ab \ge 4xy$. | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0iqw | Problem:
Cyclic 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$. | [
"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... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 7 | |
096f | Problem:
Să se arate că dacă numerele reale $a, b, c$ satisfac relațiile
$$
\begin{aligned}
& a+b+c=4 \\
& a^{2}+b^{2}+c^{2}=6
\end{aligned}
$$
atunci
$$
\frac{86}{9} \leq a^{3}+b^{3}+c^{3} \leq 10
$$ | [
"Solution:\nNotăm\n$$\n\\begin{aligned}\n& a+b+c=p \\\\\n& ab+bc+ac=q \\\\\n& abc=r \\\\\n& s=a^{3}+b^{3}+c^{3}\n\\end{aligned}\n$$\nAtunci din condițile problemei obținem\n$$\n\\begin{gathered}\na+b+c=4 \\Rightarrow p=4 \\\\\n(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ac) \\Rightarrow p^{2}-2q=6 \\\\\n(a+b+c)^{3}=(-2) ... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 86/9 <= a^3 + b^3 + c^3 <= 10 | |
08nl | Problem:
Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = a^{2} + b^{2} + c^{2}$. Show that
$$
\frac{a^{2}}{a^{2} + ab} + \frac{b^{2}}{b^{2} + bc} + \frac{c^{2}}{c^{2} + ca} \geq \frac{a + b + c}{2}
$$ | [
"Solution:\nBy the Cauchy-Schwarz inequality it is\n$$\n\\begin{aligned}\n& \\left(\\frac{a^{2}}{a^{2} + ab} + \\frac{b^{2}}{b^{2} + bc} + \\frac{c^{2}}{c^{2} + ca}\\right)\\left((a^{2} + ab) + (b^{2} + bc) + (c^{2} + ca)\\right) \\geq (a + b + c)^{2} \\\\\n\\Rightarrow & \\frac{a^{2}}{a^{2} + ab} + \\frac{b^{2}}{b... | JBMO | JBMO Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05g0 | Problem:
Soit $x \geqslant 0$ un réel. Montrer que :
$$
1 + x^{2} + x^{6} + x^{8} \geqslant 4 x^{4}
$$
et trouver les cas d'égalité. | [
"Solution:\nOn utilise l'inégalité arithmético-géométrique, dans le cas $n = 4$. On obtient\n$$\n1 + x^{2} + x^{6} + x^{8} \\geqslant 4 x^{\\frac{2+6+8}{4}} = 4 x^{4}.\n$$\nSupposons qu'on a égalité. D'après le cas d'égalité, $x^{2} = 1$, donc $x = 1$. Réciproquement, si $x = 1$, on a $1 + x^{2} + x^{6} + x^{8} = 4... | France | ENVOI 2 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | x = 1 | |
0h0x | Real numbers $x, y$ satisfy the inequality:
$$
x^2 + 3xy + 4y^2 \leq \frac{7}{2}.
$$
Prove that $x + y \leq 2$. | [
"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 h... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
0h8k | What is the maximum length of a sequence of positive integers $a_1, a_2, \dots, a_n$, if following conditions hold:
* $a_1 > 1$ is a prime number;
* for any $i: 2 \le i \le n$, $a_i \vdash a_1 a_2 \dots a_{i-1}$ holds;
* $a_n = 2^2 \cdot 3^3 \cdot 5^5 \cdot 7^7 \cdot 11^{11} \cdot 13^{13} \cdot 17^{17}$. | [
"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 g... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 6 | |
08mj | Problem:
Consider a triangle $A B C$ and let $M$ be the midpoint of the side $B C$. Suppose $\angle M A C=\angle A B C$ and $\angle B A M=105^{\circ}$. Find the measure of $\angle A B C$. | [
"Solution:\nThe angle measure is $30^{\\circ}$.\n\n\nLet $O$ be the circumcenter of the triangle $A B M$. From $\\angle B A M=105^{\\circ}$ follows $\\angle M B O=15^{\\circ}$. Let $M', C'$ be the projections of points $M, C$ onto the line $B O$. Since $\\angle M B O=15^{\\circ}$, then $\\a... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"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 chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof and answer | 30° | |
0bl1 | Problem:
Fie $ABC$ un triunghi ascuţitunghic. Dreptele $\ell_{1}$ şi $\ell_{2}$ sunt perpendiculare pe dreapta $AB$ în punctele $A$, respectiv $B$. Perpendicularele duse din mijlocul $M$ al segmentului $[AB]$ pe dreptele $AC$ şi $BC$ intersectează $\ell_{1}$ şi $\ell_{2}$ în punctele $E$ şi respectiv $F$.
Dacă $D$ est... | [] | Romania | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Ge... | null | proof only | null | |
0kod | Problem:
A real number $x$ is chosen uniformly at random from the interval $[0,1000]$. Find the probability that
$$
\left\lfloor\frac{\left\lfloor\frac{x}{2.5}\right\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{x}{6.25}\right\rfloor .
$$ | [
"Solution:\nLet $y=\\frac{x}{2.5}$, so $y$ is chosen uniformly at random from $[0,400]$. Then we need\n$$\n\\left\\lfloor\\frac{\\lfloor y\\rfloor}{2.5}\\right\\rfloor=\\left\\lfloor\\frac{y}{2.5}\\right\\rfloor .\n$$\nLet $y=5 a+b$, where $0 \\leq b<5$ and $a$ is an integer. Then\n$$\n\\left\\lfloor\\frac{\\lfloor... | United States | HMMT November 2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 9/10 | |
0183 | Let $p \neq 3$ be a prime number. Show that there is a non-constant arithmetic sequence of positive integers $x_1, x_2, \dots, x_p$ such that the product of the terms of the sequence is a cube. | [
"Let $a_1, a_2, \\dots, a_p$ be any arithmetic sequence of positive integers and let $P$ be the product of the terms of this sequence. For any $n$, the sequence $P^n a_1, P^n a_2, \\dots, P^n a_p$ is also arithmetic, and the product of terms is $P^{np+1}$. Now either $p \\equiv 1 \\pmod 3$ or $p \\equiv -1 \\pmod 3... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0inr | 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$. Le... | [
"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}$,... | United States | USAMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle ch... | null | proof only | null | |
0l2e | Balls numbered $1, 2, 3, \ldots$ are deposited in $5$ bins, labeled $A$, $B$, $C$, $D$, and $E$, using the following procedure. Ball $1$ is deposited in bin $A$, and balls $2$ and $3$ are deposited in bin $B$. The next $3$ balls are deposited in bin $C$, the next $4$ in bin $D$, and so on, cycling back to bin $A$ after... | [
"**Answer (D):** After $n$ steps, a total of $1+2+3+\\dots+n = \\frac{1}{2}n(n+1)$ balls have been deposited. In particular, after step $n = 63$, a total of $\\frac{1}{2} \\cdot 63 \\cdot 64 = 2016$ balls have been deposited. The next batch of $64$ balls will include ball $2024$. Because $64$ has remainder $4$ when... | United States | AMC 10 B | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | MCQ | D | |
084n | Problem:
$AB$ e $CD$ sono due segmenti, entrambi lunghi $4$, aventi il punto medio $M$ in comune e tali che $B \widehat{M} D = 60^\circ$. Indichiamo con $X$ l'insieme di tutti e soli i punti che distano al più $1$ da almeno uno dei due segmenti. Quanto misura la superficie di $X$?
(A) $8 - \frac{4}{3} \sqrt{3}$
(B) $... | [
"Solution:\n\nLa risposta è (D). Sia $X_1$ (rispettivamente $X_2$) il luogo dei punti con distanza $\\leqslant 1$ dal solo segmento $AB$ (rispettivamente $CD$). Chiaramente $X_1$ e $X_2$ sono congruenti e $X$ è l'unione dei due. Allora $\\operatorname{Area}(X) = 2\\operatorname{Area}(X_1) - \\operatorname{Area}(X_1... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
04e1 | Solve the following equation in the set of real numbers
$$
\frac{1}{x^2} + \frac{1}{(1-x)^2} = 24.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = (1 + 1/√3)/2, x = (1 − 1/√3)/2, x = (1 + √2)/2, x = (1 − √2)/2 | |
08s0 | Find all positive integer pairs $(m, n)$ that satisfy the following conditions.
$$
(1)\ m, n \leq 20
$$
(2) $m$ and $n$ are relatively prime.
$$
(3) \quad \frac{5}{7} < \frac{m}{n} < \frac{3}{4}
$$ | [
"Now put $p = n - m$, $q = n$. Finding all positive integer pairs $(m, n)$ is equivalent to finding all integer pairs $(p, q)$. And then condition (1) and (3) is equivalent to the conditions that $1 \\leq q \\leq 20$, $\\frac{1}{4} < \\frac{p}{q} < \\frac{2}{7}$. And condition (2) is equivalent to the condition tha... | Japan | Japan 2007 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (8, 11), (11, 15), (13, 18), (14, 19) | |
01ec | There are $2019$ plates placed around a round table and on each of them there is one coin. Alice and Bob are playing a game that proceeds in rounds indefinitely as follows. In each round, Alice first chooses a plate on which there is at least one coin. Then Bob moves one coin from this plate to one of the two adjacent ... | [
"Answer: Yes, it is possible.\nWe provide a suitable strategy for Bob. Given a configuration of coins on the plates, let a block be any inclusion-wise maximal contiguous interval consisting of non-empty plates. The idea of Bob's strategy is to maintain the following invariant throughout the game: in every block, al... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Yes | |
08jv | Problem:
Prove that if $0 < \frac{a}{b} < b < 2a$ then
$$
\frac{2ab - a^2}{7ab - 3b^2 - 2a^2} + \frac{2ab - b^2}{7ab - 3a^2 - 2b^2} \geq 1 + \frac{1}{4}\left(\frac{a}{b} - \frac{b}{a}\right)^2
$$ | [
"Solution:\nIf we denote\n$$\nu=2-\\frac{a}{b}, \\quad v=2-\\frac{b}{a}$$\nthen the inequality rewrites as\n$$\n\\begin{aligned}\n& \\frac{u}{v+uv} + \\frac{v}{u+uv} \\geq 1 + \\frac{1}{4}(u-u)^2 \\\\\n& \\frac{(u-v)^2 + uv(1-uv)}{uv(uv+u+v+1)} \\geq \\frac{(u-u)^2}{4}\n\\end{aligned}\n$$\nOr\nSince $u > 0$, $v > 0... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0dp8 | a) In this country there are $1024$ cities, numbered with integers from $0$ to $1023$;
b) Two cities with numbers $m$ and $n$ are connected by a single road if and only if the binary notations of $m$ and $n$ differ in exactly one digit;
c) During the tourist's trip in that country, $8$ roads will be closed for repairin... | [
"Remind that $n$-dimensional *binary cube* (boolean) is a graph with vertices labeled by binary sequences of length $n$, and there is an edge between any two vertices if and only if their sequences differ only in one corresponding coordinate (digit).\n\nThen our problem is equivalent to the following: prove that if... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0j0w | Problem:
In the game of Galactic Dominion, players compete to amass cards, each of which is worth a certain number of points. Say you are playing a version of this game with only two kinds of cards, planet cards and hegemon cards. Each planet card is worth $2010$ points, and each hegemon card is worth four points per ... | [
"Solution:\n\nAnswer: $503$\n\nIf you have $P$ planets and $H$ hegemons, buying a planet gives you $2010+4H$ points while buying a hegemon gives you $4P$ points. Thus you buy a hegemon whenever $P-H \\geq 502.5$, and you buy a planet whenever $P-H \\leq 502.5$. Therefore $a_{i}=1$ for $1 \\leq i \\leq 503$. Startin... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 503 | |
03eb | An integer is written in each of the fields of a $9 \times 9$ square table. For every $k$ numbers in the same row (column), their sum is in the same row (column). Find the smallest possible number of zeros in the table if:
a) $k = 5$;
b) $k = 8$. | [
"a) Example: we number the rows and columns from $1$ to $9$. We write $1$ in the fields $(i, i)$ ($i = 1, \\dots, 9$); $-1$ in field $(1, 9)$ and in fields $(i, i-1)$ ($i = 2, \\dots, 9$); $0$ in other fields. Possible sums are $1$, $0$, and $-1$.\n\nEvaluation: Suppose there are at least $19$ non-zero numbers. Fro... | Bulgaria | 1 Autumn tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) 63; b) 0 | |
07t4 | Let $1 = d_1 < d_2 < d_3 < \dots < d_n = N$ be the list of all positive divisors of the integer $N$. Check that $N = 2020$ is a number for which
$$
N = d_3d_4d_7 \quad \text{and} \quad d_3d_4 < d_7.
$$
Find all numbers $N$ satisfying these conditions. | [
"The prime factorisation of $2020$ is $2020 = 4 \\cdot 5 \\cdot 101$ and so a complete list of divisors of $2020$ is:\n$$\n\\begin{aligned}\nd_1 &= 1,\\quad d_2 = 2,\\quad d_3 = 4,\\quad d_4 = 5,\\quad d_5 = 10,\\quad d_6 = 20,\\quad d_7 = 101, \\\\\nd_8 &= 202,\\quad d_9 = 404,\\quad d_{10} = 505,\\quad d_{11} = 1... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | All N of the following forms:
- N = p^11 (p prime)
- N = p^5 q with primes p, q and either q < p or q > p^5
- N = p^3 q^2 with primes p, q and q^2 < p
- N = p^2 q r with distinct primes p, q, r and p^2 q < r | |
0b40 | Problem:
Two tigers, Alice and Betty, run in the same direction around a circular track of circumference $400$ meters. Alice runs at a speed of $10~\mathrm{m}/\mathrm{s}$ and Betty runs at $15~\mathrm{m}/\mathrm{s}$. Betty gives Alice a $40$ meter headstart before they both start running. After $15$ minutes, how many ... | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | d | |
0iw0 | Problem:
Let $T$ be a right triangle with sides having lengths $3$, $4$, and $5$. A point $P$ is called awesome if $P$ is the center of a parallelogram whose vertices all lie on the boundary of $T$. What is the area of the set of awesome points? | [
"Solution:\nThe set of awesome points is the medial triangle, which has area $6 / 4 = 3 / 2$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 3/2 | |
08ab | Problem:
Una griglia con $m$ righe ed $n$ colonne ha ogni casella colorata in bianco o in nero in modo da rispettare le seguenti due condizioni:
a. ogni riga contiene tante caselle bianche quante nere;
b. se una riga incontra una colonna in una casella nera, allora quella riga e quella colonna hanno lo stesso numero... | [
"Solution:\n\nDalla prima condizione abbiamo che il numero di colonne è necessariamente pari. In generale le righe, di lunghezza $2a$, hanno $a$ caselle bianche ed $a$ caselle nere. Per ogni riga ci saranno almeno una colonna che la interseca in una casella bianca ed almeno una che la interseca in una casella nera.... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (m, n) are exactly the pairs (a, 2a) and (2a, 2a) for positive integers a | |
0cto | In a country, there are $n$ cities; some pairs of them are connected with two-way direct flights. There is a unique (perhaps, non-direct) route between every two cities. The mayor of each city $X$ found the number $f(X)$ of the enumerations of all cities by $1, 2, \ldots, n$ such that along each route starting at $X$, ... | [
"Choose an arbitrary capital $A$. Say that a city $C$ is even (resp., odd) if the route from $A$ to $C$ contains an even (resp., odd) number of flights. It suffices to prove that the sum of mayors' numbers in odd cities is equal to the sum of those in even cities. This claim can be proved by means of a bijection of... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English; Russian | proof only | null | |
0kqt | Problem:
For a nonnegative integer $n$, let $s(n)$ be the sum of digits of the binary representation of $n$. Prove that
$$
\sum_{n=0}^{2^{2022}-1} \frac{(-1)^{s(n)}}{2022+n}>0
$$ | [
"Solution:\n\nDefine\n$$\nf_k(x)=\\sum_{n=0}^{2^k-1} \\frac{(-1)^{s(n)}}{x+n}\n$$\nWe want to show that $f_{2022}(2022)>0$. We will in fact show something stronger.\n\nI claim that for all $x>0$, for all $k \\geq 0$, we have $f_k^{(i)}(x)>0$ for even $i$ and $f_k^{(i)}(x)<0$ for odd $i$, where $f^{(i)}$ denotes the... | United States | HMIC | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
07st | Let $f : \mathbb{Z} \to \mathbb{Z}$ be such that, for all $a, b \in \mathbb{Z}$
$$
f(a + b) = f(f(a)) + f(f(b)).
$$
Find all possible values of $f(2020)$. | [
"**Solution 1.** We first show that $f(f(x))$ must be affine. To see this, replace $(a, b)$ first by $(a - 1, a + 1)$ and then by $(a, a)$ to obtain\n$$\nf(f(a - 1)) + f(f(a + 1)) = f(2a) = 2f(f(a)).\n$$\nTherefore,\n$$\nf(f(a + 1)) - f(f(a)) = f(f(a)) - f(f(a - 1)).\n$$\nThis means that there is a constant $m$ so ... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 0 or 2020 | |
03tt | Given the set $P = \{1, 2, 3, 4, 5\}$, define $f(m, k) = \sum_{i=1}^{5} \lfloor m \sqrt{\frac{k+1}{i+1}} \rfloor$ for any $k \in P$ and positive integer $m$, where $\lfloor a \rfloor$ denotes the greatest integer less than or equal to $a$. Prove that for any positive integer $n$, there is $k \in P$ and positive integer... | [
"**Proof** Define set $A = \\{m\\sqrt{k+1} \\mid m \\in \\mathbb{N}^*, k \\in P\\}$, where $\\mathbb{N}^*$ denotes the set of all positive integers. It is easy to check that for any $k_1, k_2 \\in P, k_1 \\neq k_2$, $\\frac{\\sqrt{k_1+1}}{\\sqrt{k_2+1}}$ is an irrational number. Therefore, for any $k_1, k_2 \\in P$... | China | China Mathematical Competition (Extra Test) | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
04cx | Let $x$ and $y$ be real numbers such that $x + y \ge 0$. Prove that
$$
2^{n-1} (x^n + y^n) \ge (x + y)^n \quad \text{for all } n \in \mathbb{N}.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
01by | Do there exist pairwise distinct rational numbers $x$, $y$ and $z$ such that
$$
\frac{1}{(x - y)^2} + \frac{1}{(y - z)^2} + \frac{1}{(z - x)^2} = 2014?
$$ | [
"Let $a = x - y$ and $b = y - z$, then\n$$\n\\begin{aligned}\n\\frac{1}{(x-y)^2} + \\frac{1}{(y-z)^2} + \\frac{1}{(z-x)^2} &= \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{(a+b)^2} \\\\\n&= \\frac{b^2(a+b)^2 + a^2(a+b)^2 + a^2b^2}{a^2b^2(a+b)^2} \\\\\n&= \\left( \\frac{a^2 + b^2 + ab}{ab(a+b)} \\right)^2.\n\\end{alig... | Baltic Way | Baltic Way | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
067o | We consider $111$ mutually distinct points on the interior or on the circle of a unit disk. Prove that we can find at least $1998$ segments with ends from these points and length less than $\sqrt{3}$. | [
"We divide the circle into three equal sectors of $120^\\circ$ such that none of the points belong to their border, with the exception of the center of the circle. If the center of the circle is one of the points, then we consider that it belongs only to one of the sectors. This is possible because the number of th... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0eer | Problem:
Dan je ostrokoten trikotnik $ABC$ in taka točka $D$ v notranjosti tega trikotnika, da velja $\Varangle BAD = \Varangle DCB$ in $\Varangle CBD = \Varangle DAC$. Dokaži, da sta premici $AD$ in $BC$ pravokotni. | [
"Solution:\n\n\n\nOznačimo z $E$ presečišče premic $AD$ in $BC$, z $F$ presečišče premic $BD$ in $CA$ ter z $G$ presečišče premic $CD$ in $AB$. Ker je $\\Varangle BAD = \\Varangle DCB$, sta trikotnika $GAD$ in $ECD$ podobna, saj imata dva skladna kota. Torej je\n$$\n\\frac{|GD|}{|AD|} = \\f... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g5o | 對於圖 $G$ 與其中任意一點 $x$, 用符號 $G - \{x\}$ 表示去掉點 $x$ 與它相鄰的邊後所得到的新圖。給定兩個圖 $G$ 與 $H$, 他們的點都是編號為 $1, 2, \dots, n$ 而 $n \ge 4$。如果對於任意 $1 \le i < j \le n$, 圖 $G - \{i\} - \{j\}$ 與圖 $H - \{i\} - \{j\}$ 是同構的, 試證: 圖 $G$ 與圖 $H$ 是同構的。 | [
"三步驟：\n$$\n(1)\\ |E(G)| = |E(H)|: \\text{ 藉由考慮 } \\sum_{i \\neq j} |E(G - \\{i\\} - \\{j\\})| = C_2^{n-2} |E(G)|.\n$$\n\n$$\n(2)\\ \\deg_G(i) = \\deg_H(i) \\text{ for all } i: \\text{ 固定某個 } i_0, \\text{ 考慮 } \\sum_{j \\neq i_0} |E(G - \\{j\\} - \\{i_0\\})| = (n-3)(|E(G)| - \\deg_G(x_0)).\n$$\n\n$$\n(3)\\ \\text{定義... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0afu | Дропката $\frac{59}{143}$ да се претстави како збир на две прави нескратливи дропки. | [
"Бројот $143$ можеме да го запишеме како $143 = 11 \\cdot 13$. Според тоа дропката $\\frac{59}{143}$ може да се претстави во облик $\\frac{59}{143} = \\frac{59}{11 \\cdot 13} = \\frac{x}{11} + \\frac{y}{13}$. Ако десната страна на последното равенство го сведеме на најмал заеднички именител, добиваме $\\frac{13x+11... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Macedonian, English | final answer only | 2/11 + 3/13 | |
00bj | Which regular $n$-gons have a triangulation consisting of isosceles triangles? | [
"Call $n$ good if the regular $n$-gon can be triangulated with isosceles triangles. By *segments* we mean the sides and the diagonals of the $n$-gon; the sides are the shortest among all segments.\n\nLet $n$ be good and $T$ an isosceles triangulation of the regular $n$-gon $P$. Suppose that the base of a triangle $... | Argentina | Argentina_2018 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | English | proof and answer | All regular n-gons with n either a power of two at least four (n = 2^m, m ≥ 2) or a sum of two distinct powers of two (n = 2^u + 2^v with u > v ≥ 0). | |
05rw | Problem:
Soit $ABC$ un triangle dont les trois angles sont aigus, avec $AB > AC$, et soit $\Omega$ son cercle circonscrit. On note $M$ le milieu de $[BC]$. Les tangentes à $\Omega$ en $B$ et $C$ s'intersectent en $P$, et les droites $(AP)$ et $(BC)$ se coupent en $S$. On note $D$ le pied de la hauteur issue de $B$ dan... | [
"Solution:\n\n\n\nL'idée est de se rendre compte que la figure contient de nombreux points cocycliques. Pour commencer, on a $PB = PC$ donc $(MP)$ est la médiatrice de $[BC]$, et en particulier l'angle $\\widehat{BMP}$ est droit. Comme $\\widehat{BDP}$ l'est aussi, les points $B, D, M$ et $... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02t0 | Problem:
O personagem histórico mexicano Benito Juárez nasceu na primeira metade do século XIX (o século XIX vai do ano 1801 ao ano 1900). Sabendo que Benito Juárez completou $x$ anos no ano $x^{2}$, qual foi o ano do seu nascimento? | [
"Solution:\nOs quadrados perfeitos que estão mais próximos de 1801-1900 são:\n$$\n\\begin{aligned}\n& 42 \\times 42=1764 \\\\\n& 43 \\times 43=1849 \\\\\n& 44 \\times 44=1936\n\\end{aligned}\n$$\nSeja $x$ a idade de Benito Juárez no ano $x^{2}$. O número $x$ não pode ser 42, pois neste caso Benito não teria nascido... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 1806 | |
077y | Euclid has a tool called *cyclos* which allows him to do the following:
* Given three non-collinear marked points, draw the circle passing through them.
* Given two marked points, draw the circle with them as endpoints of a diameter.
* Mark any intersection points of two drawn circles or mark a new point on a drawn cir... | [] | India | INMO_2023 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09m9 | Find all positive integer solutions to the equation
$$
2^a + 2^b + 2^c + 2^d = 60 \cdot \min\{a, b, c, d\},
$$
where $\min\{a, b, c, d\}$ denotes the minimum of the numbers $a, b, c, d$. | [
"Answer: $\\{a, b, c, d\\} = \\{4, 5, 6, 7\\}$.\nIt is clear that the above is a solution, so we prove that there are no other solutions.\nWe may assume $a \\le b \\le c \\le d$. Since $S = 2^a + 2^b + 2^c + 2^d = 15 \\cdot 4a$, we have $2^a \\mid 4a$, thus $2^a \\le 4a$ and therefore $a \\le 4$.\n\nFirst, we prove... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | {4,5,6,7} | |
0a11 | The crab of a positive integer is the number you get when you write down its digits in reverse order. For example, the crab of $8267$ equals $7628$ and the crab of $15620$ equals $2651$ (because the leading zero is always dropped).
What is the smallest positive integer $n$ such that $n$ minus the crab of $n$ equals $12... | [
"$20406080$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 20406080 | |
08yz | Let $A$ be the number of cases such that each cell of a $2021 \times 2021$ table is filled with one of $1$, $2$, or $3$ in such a way that any $2 \times 2$ square in the table sums up to $8$. Answer the remainder after dividing $A$ by $100$. | [
"$\\boxed{3}$\n\nFor $1 \\le i \\le 2021$ and $1 \\le j \\le 2021$, let $(i, j)$ denote the cell in the $i$-th row and the $j$-th column, and let $f(i, j)$ denote the number filled in $(i, j)$. We also define $g(i, j)$ as\n$$\ng(i, j) = \\begin{cases} f(i, j) & (i + j \\text{ is even}), \\\\ 4 - f(i, j) & (i + j \\... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | 3 | |
0fp9 | Sean $C$ y $C'$ dos circunferencias tangentes exteriores con centros $O$ y $O'$ y radios $1$ y $2$, respectivamente. Desde $O$ se traza una tangente a $C'$ con punto de tangencia en $P'$ y desde $O'$ se traza la tangente a $C$ con punto de tangencia en $P$ en el mismo semiplano que $P'$ respecto de la recta que pasa po... | [
"Los triángulos $OPO'$ y $OP'O'$ son rectángulos en $P$ y $P'$, respectivamente y $\\angle PXO = \\angle P'XO'$, luego los triángulos $PXO$ y $P'XO'$ son semejantes con razón de semejanza $O'P'/OP = 2$. La razón entre sus áreas $S'$ y $S$ es entonces $S'/S = 4$. Por el Teorema de Pitágoras $OP' = \\sqrt{5}$ y $O'P ... | Spain | LII Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Spanish | proof and answer | (4*sqrt(2) - sqrt(5))/3 | |
09iz | Let $ABC$ isosceles triangle with $AB = AC$ and incenter $I$. Let $\omega$ be the circumcircle of $ABC$. The line $BI$ meets $\omega$ again at point $P$, and the line $CI$ meets $\omega$ again at point $Q$. Let $D$ be a point on the arc $BC$ of $\omega$ not containing $A$, different from $B$ and $C$. The line $BI$ meet... | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanc... | null | proof only | null | |
07o2 | Let $k \ge 2$ be an integer. Prove that for each positive integer $N < 40 \cdot 3^k$ the equation
$$
(x_1^2 - 1)(x_2^2 - 1) \cdots (x_k^2 - 1) = N
$$
has at most one integer solution $(x_1, x_2, \dots, x_k)$ such that $1 < x_1 \le x_2 \le \dots \le x_k$. | [
"Because $N > 0$, we cannot have $x_i = 1$ and so $2 \\le x_1$. Define $f(x) = (x^2 - 1)/3$ and for a given $N < 40 \\cdot 3^k$ we let $M = N/3^k$. The equation $(x_1^2 - 1)(x_2^2 - 1) \\cdots (x_k^2 - 1) = N$ is equivalent to $f(x_1)f(x_2) \\cdots f(x_k) = M$. If $x \\ge 2$ is an integer, $f(x) \\ge 1$ and so $f(x... | Ireland | Ireland | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
09ax | Is it possible to color all rational numbers with one of two colors, so that if $x, y \in \mathbb{Q}$, $x \neq y$, $xy = 1$ or $x + y \in \{0, 1\}$ then $x$ and $y$ must be colored with different colors. | [
"Let $x \\in \\mathbb{Q}^+$, $x = \\frac{a}{b}$, $(a, b) = 1$, $a > 0$, $b > 0$. Now let us apply Euclid's algorithm for $a$ and $b$. Here $r_0 = a$, $r_1 = b$. $r_{j-1} = q_j r_j + r_{j+1}$, $j = 1, 2, \\dots, n$. There exists $n = n(x)$, such that $r_n \\neq 0$ and $r_{n+1} = 0$.\n\nConsider the function $f: \\ma... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0k0e | Problem:
Let $n$ be a positive odd integer greater than $2$, and consider a regular $n$-gon $\mathcal{G}$ in the plane centered at the origin. Let a subpolygon $\mathcal{G}'$ be a polygon with at least $3$ vertices whose vertex set is a subset of that of $\mathcal{G}$. Say $\mathcal{G}'$ is well-centered if its centro... | [
"Solution:\n\n$\\Rightarrow$, i.e. $n$ has $\\geq 3$ prime divisors: Let $n=\\prod p_{i}^{e_{i}}$. Note it suffices to only consider regular $p_{i}$-gons. Label the vertices of the $n$-gon $0,1, \\ldots, n-1$. Let $S=\\left\\{\\frac{x n}{p_{1}}: 0 \\leq x \\leq p_{1}-1\\right\\}$, and let $S_{j}=S+\\frac{j n}{p_{3}... | United States | February 2017 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
082w | Problem:
Tre circonferenze passano per l'origine. Il centro della prima circonferenza sta nel primo quadrante, il centro della seconda sta nel secondo quadrante, il centro della terza sta nel terzo quadrante. Se $P$ è un punto interno alle tre circonferenze, allora
(A) $P$ sta nel secondo quadrante
(B) $P$ sta nel pri... | [
"Solution:\n\nLa risposta è (A). Osserviamo che se una circonferenza passa per l'origine ed ha il centro in un quadrante, allora nessun punto interno sta nel quadrante opposto. In particolare se $P$ è interno alle tre circonferenze, allora deve trovarsi nel secondo quadrante, poiché i tre centri si trovano nel prim... | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO BIENNIO | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | A | |
05sl | Problem:
Lors d'une fête, 2019 personnes s'assoient autour d'une table ronde, en se répartissant de façon régulière. Après s'être assises, elles constatent qu'un carton indiquant un nom est posé à chacune des places et que personne n'est assis à la place où figure son nom. Montrer qu'on peut tourner la table de telle ... | [
"Solution:\n\nOn commence par tester l'énoncé sur des valeurs plus petites, par exemple pour une fête de 5 personnes.\n\nEnsuite, on essaye de coder l'information. On appelle rotation une configuration obtenue après avoir tourné la table depuis sa position d'origine. À une rotation $r$, on associe $n(r)$ le nombre ... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
08bd | Problem:
Cinque amici, Aurelio, Ennio, Flaminia, Lucia e Regolo, hanno mangiato al ristorante. Il conto è di 180 euro, e viene pagato da Lucia, Ennio e Regolo: la prima paga 90 euro, il secondo 57 euro e il terzo 33 euro. Qual è il minimo numero di transazioni del tipo "Tizio dà $n$ euro a Caio" che devono essere effe... | [
"Solution:\n\nLa risposta è (C). Il conto del ristorante è 180 euro, i commensali sono 5, quindi ciascuno deve pagare 36 euro. Chi ha anticipato più di tale cifra deve ricevere soldi da chi ha anticipato meno (e in particolare dai due amici che non hanno pagato nulla). Tuttavia, siccome nessuno deve ricevere un mul... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Algorithms"
] | null | MCQ | C | |
00yi | Problem:
Let $p(x)$ be a polynomial with integer coefficients such that both equations $p(x)=1$ and $p(x)=3$ have integer solutions. Can the equation $p(x)=2$ have two different integer solutions? | [
"Solution:\n\nObserve first that if $a$ and $b$ are two different integers then $p(a)-p(b)$ is divisible by $a-b$. Suppose now that $p(a)=1$ and $p(b)=3$ for some integers $a$ and $b$. If we have $p(c)=2$ for some integer $c$, then $c-b= \\pm 1$ and $c-a= \\pm 1$, hence there can be at most one such integer $c$."
] | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No; at most one integer solution. | |
0106 | Problem:
The worlds in the Worlds' Sphere are numbered $1, 2, 3, \ldots$ and connected so that for any integer $n \geqslant 1$, Gandalf the Wizard can move in both directions between any worlds with numbers $n, 2n$ and $3n+1$. Starting his travel from an arbitrary world, can Gandalf reach every other world? | [
"Solution:\nAnswer: yes.\nFor any two given worlds, Gandalf can move between them either in both directions or none. Hence, it suffices to show that Gandalf can move to the world $1$ from any given world $n$. For that, it is sufficient for him to be able to move from any world $n>1$ to some world $m$ such that $m<n... | Baltic Way | Baltic Way 1997 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | yes | |
0few | Problem:
Calcular los números $p$ y $q$ tales que las raíces de la ecuación
$$
x^{2}+p x+q=0
$$
sean $D$ y $1 - D$, siendo $D$ el discriminante de esa ecuación de segundo grado. | [
"Solution:\nPor las fórmulas que relacionan las raíces y los coeficientes de la ecuación, se tiene\n$$\n\\begin{aligned}\n& D + 1 - D = -p \\\\\n& D \\cdot (1 - D) = q\n\\end{aligned}\n$$\nDe la primera se obtiene inmediatamente $p = -1$; de la segunda, $q = D - D^{2}$.\nPero $D = p^{2} - 4q = 1 - 4q = 1 - 4\\left(... | Spain | TANDA II | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | p = -1 with q = 0 or q = 3/16 | |
06gr | In $\triangle ABC$, $\angle A = 30^\circ$ and $AB = 4$. $D$ is the midpoint of $AB$. Determine the maximum possible value of $\tan \angle DCB$. | [] | Hong Kong | Year 2012 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | (4√2 + 3√3)/5 | |
009h | On the table there are $2013$ cards with $1, 2, \ldots, 2013$ written on them; the cards are face down (the numbers on them cannot be seen). It is allowed to select any set of cards, to ask if the arithmetic mean of the numbers on them is an integer, and to receive a truthful answer.
a) Find all numbers that can be de... | [
"Replace $2013$ by a general odd number $2k-1$, $k \\ge 2$. The sum $S = 1+2+\\dots+(2k-1)$ equals $k(2k-1)$. For part a), the only number that can be determined with certainty is $k$, the one in the middle. To find $k$, ignore a card with an unknown number $x$ on it, thus forming a set of $2k-2$ cards, and ask abo... | Argentina | NATIONAL XXX OMA | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Logic"
] | null | proof and answer | a) Only the middle number 1007 can be determined with certainty. b) The maximum number of groups is 1007. | |
08qn | Problem:
The positive integer $k$ and the set $A$ of different integers from $1$ to $3k$ inclusive are such that there are no distinct $a, b, c$ in $A$ satisfying $2b = a + c$. The numbers from $A$ in the interval $[1, k]$ will be called small; those in $[k+1, 2k]$ - medium and those in $[2k+1, 3k]$ - large. Is it alw... | [
"Solution:\n\nA counterexample for a) is $k=3$, $A=\\{1,2,9\\}$, $x=2$ and $d=8$.\nA counterexample for c) is $k=3$, $A=\\{1,8,9\\}$, $x=8$ and $d=1$.\n\nWe will prove that b) is true.\n\nSuppose the contrary and let $x, d$ have the above properties. We can assume $0 < d < 3k$, $0 < x \\leq 3k$ (since for $d=3k$ th... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) No (counterexample: k=3, A={1,2,9}, x=2, d=8); b) Yes; c) No (counterexample: k=3, A={1,8,9}, x=8, d=1) | |
0j4e | Problem:
Let $ABCD$ be a square of side length $13$. Let $E$ and $F$ be points on rays $AB$ and $AD$, respectively, so that the area of square $ABCD$ equals the area of triangle $AEF$. If $EF$ intersects $BC$ at $X$ and $BX=6$, determine $DF$. | [
"\nLet $Y$ be the point of intersection of lines $EF$ and $CD$. Note that $[ABCD]=[AEF]$ implies that $[BEX]+[DYF]=[CYX]$. Since $\\triangle BEX \\sim \\triangle CYX \\sim \\triangle DYF$, there exists some constant $r$ such that $[BEX]=r \\cdot BX^{2}$, $[YDF]=r \\cdot CX^{2}$, and $[CYX]=... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | √13 | |
05w2 | Problem:
Trouver tous les entiers $n \geqslant 1$ ayant la propriété suivante : il existe une permutation $d_{1}, d_{2}, \ldots, d_{k}$ des diviseurs positifs de $n$ telle que, pour tout $i \leqslant k$, la somme $d_{1}+d_{2}+\ldots+d_{i}$ soit un carré parfait. | [
"Solution:\n\nSoit $n$ un des entiers recherchés, et $d_{1}, d_{2}, \\ldots, d_{k}$ une permutation adéquate des diviseurs positifs de $n$. Pour tout entier $i \\leqslant k$, on pose $s_{i}=\\sqrt{d_{1}+d_{2}+\\ldots+d_{i}}$. On dit qu'un entier $\\ell$ est bon si $s_{i}=i$ et $d_{i}=2 i-1$ pour tout $i \\leqslant ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n = 1 and n = 3 | |
0003 | Los números enteros del $1$ al $2002$, ambos inclusive, se escriben en una pizarra en orden creciente $1$, $2$, $\ldots$, $2001$, $2002$. Luego, se borran los que ocupan el primer lugar, cuarto lugar, séptimo lugar, etc., es decir, los que ocupan los lugares de la forma $3k+1$.
En la nueva lista se borran los números q... | [] | Argentina | XVII Olimpíada Iberoamericana de Matemática | [
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | español | proof and answer | 1598 | |
01fq | Let $n \ge 4$, and consider a non-intersecting $n$-gon $P_1P_2...P_n$ in the plane. Suppose that, to each $P_k$, there is a unique other vertex $Q_k$ among $P_1, ..., P_n$ that lies closest to it. The polygon is said to be *hostile* if $Q_k \ne P_{k \pm 1}$ for all $k$ (counting cyclically).
a. Prove that there exist ... | [
"(a) As an auxiliary result, we prove the following. There is no convex quadrilateral $ABCD$ in which $C$ is the closest neighbour of $A$, and $D$ is the closest neighbour of $B$. See Figure below:\n\nConvex quadrilateral.\nIndeed, the diagonals $AC$ and $BD$ cross at a point $P$ by convexi... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof and answer | a) No convex hostile polygons exist. b) Hostile concave polygons exist for all integers n ≥ 4. | |
0ge9 | Find all triples of positive integers $(x, y, z)$ satisfying
$$
x^2 + 4^y = 5^z.
$$
試求出所有正整數組 $(x, y, z)$ 滿足
$$
x^2 + 4^y = 5^z.
$$ | [
"所有的解為 $(1, 1, 1)$, $(11, 1, 3)$, $(3, 2, 2)$。\n\n我們先處理 $y \\ge 2$ 的情形。我們有\n$$\nx^2 \\equiv 5^z \\pmod{8},\n$$\n而 $x^2 \\equiv 0, 1, 4$, $5^z \\equiv 1, 5$, 因此 $z$ 為偶數。令 $z = 2z'$, 則\n$$\nx^2 + (2^y)^2 = (5^{z'})^2.\n$$\n因為 $5^{z'}$ 與 $2^y$ 互質且 $2 \\mid 2^y$, 我們由畢氏三元數公式有\n$$\nx = m^2 - n^2, \\quad 2^y = 2mn, \\quad... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence re... | null | proof and answer | (1, 1, 1), (11, 1, 3), (3, 2, 2) | |
04rk | Find all real $a$, $b$, $c$, such that
$$
a^2 + b^2 + c^2 = 26, \quad a+b=5 \quad \text{and} \quad b+c \ge 7.
$$ | [
"We show that the only solution is $a=1$, $b=4$ and $c=3$.\nLet $s = b + c \\ge 7$. Substituting $a = 5-b$ and $c = s-b$ the first condition gives\n$$\n(5-b)^2 + b^2 + (s-b)^2 = 26,\n$$\nthus\n$$\n3b^2 - 2(s+5)b + s^2 - 1 = 0.\n$$\nThe equation has a real solution iff the discriminant $4(s+5)^2 - 12(s^2-1) \\ge 0$.... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a=1, b=4, c=3 | |
08on | Problem:
If $x^{3}-3 \sqrt{3} x^{2}+9 x-3 \sqrt{3}-64=0$, find the value of $x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015$. | [
"Solution:\n$x^{3}-3 \\sqrt{3} x^{2}+9 x-3 \\sqrt{3}-64=0 \\Leftrightarrow (x-\\sqrt{3})^{3}=64 \\Leftrightarrow (x-\\sqrt{3})=4 \\Leftrightarrow x-4=\\sqrt{3} \\Leftrightarrow x^{2}-8 x+16=3 \\Leftrightarrow x^{2}-8 x+13=0$\n\n$x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015=\\left(x^{2}-8 x+13\\right)\\left(x^... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1898 | |
04e6 | From $n^3$ unit cubes Ivica assembled a large cube with edge length $n$ and then he coloured some of the six sides of the large cube. When he disassembled the large cube, he found that exactly $1000$ unit cubes don't have any coloured side. Show that this is indeed possible and determine the number of sides of the larg... | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 3 | |
0k0g | Problem:
Let $ABC$ be a triangle, and let $BCDE$, $CAFG$, $ABHI$ be squares that do not overlap the triangle with centers $X$, $Y$, $Z$ respectively. Given that $AX = 6$, $BY = 7$, and $CZ = 8$, find the area of triangle $XYZ$. | [
"Solution:\n\nBy the degenerate case of Von Aubel's Theorem we have that $YZ = AX = 6$ and $ZX = BY = 7$ and $XY = CZ = 8$ so it suffices to find the area of a $6$-$7$-$8$ triangle which is given by $\\frac{21 \\sqrt{15}}{4}$.\n\nTo prove that $AX = YZ$, note that by LoC we get\n$$\nYX^2 = \\frac{b^2}{2} + \\frac{c... | United States | February 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 21 sqrt 15 / 4 | |
0862 | Problem:
Vi sono $10000$ lampadine numerate da $1$ in poi, ciascuna delle quali viene accesa e spenta con un normale interruttore. All'inizio tutte le lampadine sono spente; poi si premono una volta tutti gli interruttori delle lampadine contrassegnate dai multipli di $1$ (di conseguenza tutte le lampadine vengono acc... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. L'interruttore di posto $n$ viene toccato una volta per ogni divisore positivo di $n$. Quindi la $n$-esima lampadina rimane accesa alla fine se e solo se $n$ ha un numero dispari di divisori. Questo succede solo per i quadrati perfetti: se infatti $n$ non è un quadrato... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | MCQ | E | |
0k4g | Problem:
Let $a$, $b$, $x$, $y$, $z$ be positive real numbers. Prove the inequality
$$
\frac{x}{a y+b z}+\frac{y}{a z+b x}+\frac{z}{a x+b y} \geq \frac{3}{a+b} .
$$ | [
"Solution:\nApplying Cauchy-Schwarz inequality to the triples\n$$\n\\sqrt{\\frac{x}{a y+b z}}, \\sqrt{\\frac{y}{a z+b x}}, \\sqrt{\\frac{z}{a x+b y}} \\text{ and } \\sqrt{x(a y+b z)}, \\sqrt{y(a z+b x)}, \\sqrt{z(a x+b y)} \\text{, }\n$$\nwe get that\n$$\n\\frac{x}{a y+b z}+\\frac{y}{a z+b x}+\\frac{z}{a x+b y} \\g... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
03rz | The curve represented by the equation $$\frac{x^2}{\sin\sqrt{2} - \sin\sqrt{3}} + \frac{y^2}{\cos\sqrt{2} - \cos\sqrt{3}} = 1$$ is ( ).
(A) An ellipse with the foci on the x-axes
(B) A hyperbola with the foci on the x-axes
(C) An ellipse with the foci on the y-axes
(D) A hyperbola with the foci on the y-axes | [
"Since $\\sqrt{2} + \\sqrt{3} > \\pi$, so $0 < \\frac{\\pi}{2} - \\sqrt{2} < \\sqrt{3} - \\frac{\\pi}{2} < \\frac{\\pi}{2}$ and\n$$\n\\cos(\\frac{\\pi}{2} - \\sqrt{2}) > \\cos(\\sqrt{3} - \\frac{\\pi}{2}), \\text{ i.e. } \\sin\\sqrt{2} > \\sin\\sqrt{3}.\n$$\n\nSince\n$$\n(\\sin\\sqrt{2} - \\sin\\sqrt{3}) - (\\cos\\... | China | China Mathematical Competition (Jiangxi) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | MCQ | C | |
0bxd | Alina and Bogdan play a game on a $2 \times n$ rectangular grid ($n \ge 2$) whose sides of length $2$ are glued together to form a cylinder. Alternating moves, each player cuts out a unit square of the grid. A player loses if his/her move causes the grid to lose circular connection (two unit squares that only touch at ... | [
"If $n = 2j + 1$ is odd, Alina's strategy is the following: she cuts out a unit square and labels the columns from $-j$ to $j$, the column from which the first unit square has been removed receiving the label $0$. Starting at this point of the game, whenever Bogdan removes a square from column number $k \\in \\{-j,... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Alina wins when the number of columns is odd; Bogdan wins when the number of columns is even. | |
0d17 | Consider $S = \{(x, y, z) \mid x, y, z \in \{1, 2, \dots, 2012\}\}$ as a set of $2012^3$ points in three-dimensional space. For any segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the space, we define its *distance triplet* to be the ordered triple
$$
(|x_1 - x_2|, |y_1 - y_2|, |z_1 - z_2|).
$$
Al... | [
"We claim that Alice can draw up to $K = \\frac{2012^3}{2}$ segments. Since there are $2012^3$ points, conditions (a) and (b) guarantee that Alice can draw at most $K$ segments. We will prove that she can do so.\n\nLet $T = \\{1, 2, \\dots, 2012\\}$. We will define a bijection $f: T \\to T$ such that for all distin... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | 2012^3/2 | |
06hi | In $\triangle ABC$, $AB = 13$ and $BC = 7$. $D$ and $E$ are points on $AB$ and $AC$ respectively such that $BD = BC$ and $\angle DEB = \angle CEB$. Find the product of all possible values of the length of $AE$.
在 $\triangle ABC$ 中，$AB = 13$ 及 $BC = 7$。設 $D$ 和 $E$ 分別為 $AB$ 和 $AC$ 上的點，使得 $BD = BC$ 及 $\angle DEB = \angle... | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English; Chinese | proof and answer | 507/10 | |
0gue | 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... | [
"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; ... | Turkey | Team Selection Test for IMO 2023 | [
"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 chasing"
] | English | proof only | null | |
06ax | If $a, b, c$ are real numbers such that two of them have difference greater than $\frac{1}{2\sqrt{2}}$, prove that there exists an integer $x$ such that
$$
x^2 - 4(a + b + c)x + 12(ab + bc + ca) < 0.
$$ | [
"The discriminant equals to\n$$\n\\begin{aligned}\n\\Delta &= 16(a + b + c)^2 - 48(ab + bc + ca) \\\\\n&= 8((a - b)^2 + (b - c)^2 + (c - a)^2).\n\\end{aligned}\n$$\nSince two of the numbers are at least $\\frac{1}{2\\sqrt{2}}$ apart, the square of their difference will be greater than $1/8$, so $\\Delta > 1$. Since... | Greece | Hellenic Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0jaz | Problem:
Let $N$ be the number of distinct roots of $\prod_{k=1}^{2012}\left(x^{k}-1\right)$. Give lower and upper bounds $L$ and $U$ on $N$. If $0<L \leq N \leq U$, then your score will be $\left\lfloor\frac{23}{(U / L)^{1.7}}\right\rfloor$. Otherwise, your score will be 0. | [
"Solution:\nFor $x$ to be such a number is equivalent to $x$ being a $k^{\\text{th}}$ root of unity for some $k$ up to $2012$. For each $k$, there are $\\varphi(k)$ primitive $k^{\\text{th}}$ roots of unity, so the total number of roots is $\\sum_{k=1}^{2012} \\varphi(k)$.\n\nWe will give a good approximation of th... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > Möbius inversion",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | null | |
0d1e | For any positive integer $n$ denote by $a_n$ the number of quadratic functions $f(x) = ax^2 + bx + c$, $a, b, c \in \{1, 2, \dots, n\}$, having only integer roots. Prove that for every $n \ge 4$,
$$
n < a_n < n^2.
$$ | [
"The equations $x^2 + kx + k - 1 = 0$, $k = 2, \\dots, n$, and $2x^2 + 4x + 2 = 0$, $x^2 + 4x + 4 = 0$, satisfy the property, hence $n+1 \\le a_n$.\nIf $f$ is a quadratic function satisfying our property, then $f(x) = a(x + x_1)(x + x_2)$, where $x_1, x_2 \\in \\mathbb{Z}_+$, and $a, a(x_1 + x_2)$, $a x_1 x_2 \\in ... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number The... | English | proof only | null | |
02q2 | Problem:
Um poliedro convexo $\mathcal{P}$ tem 26 vértices, 60 arestas e 36 faces. 24 faces são triangulares e 12 são quadriláteros. Uma diagonal espacial é um segmento de reta unindo dois vértices não pertencentes a uma mesma face. $\mathcal{P}$ possui quantas diagonais espaciais? | [
"Solution:\n\nOs 26 vértices determinam exatamente $\\binom{26}{2} = 26 \\times 25 / 2 = 325$ segmentos. Destes segmentos, 60 são arestas e como cada quadrilátero tem duas diagonais, então temos $12 \\times 2 = 24$ diagonais que não são espaciais.\n\nPortanto, o número de diagonais espaciais é $325 - 60 - 24 = 241$... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 241 | |
0ch9 | Determine all positive integers $a, b, c, d, e, f$ satisfying the following condition: for any two of them, $x$ and $y$, two of the remaining four numbers, $z$ and $t$, exist such that $\frac{x}{y} = \frac{z}{t}$. | [
"If $x = a$ and $y = f$, for all $z, t \\in \\{b, c, d, e\\}$ we have $\\frac{x}{y} = \\frac{a}{f} \\le \\frac{z}{t}$, where the equality holds for $z = a$ and $t = f$. Since $z \\in \\{b, c, d, e\\}$, it follows that $z \\ge b$, hence $a \\ge b$. Therefore, $a = b$. Similarly, we infer that $e = f$.\n\nWe now choo... | Romania | 74th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All solutions are exactly the 6-tuples that are permutations of either (a, a, b, b, c, c) with a ≤ b ≤ c or (a, a, a, b, b, b) with a ≤ b. | |
0l7q | Find the sum of all integer bases $b > 9$ for which $17_b$ is a divisor of $97_b$. | [
"If $17_b$ is a divisor of $97_b$, then $\\frac{9b+7}{b+7}$ is a positive integer. Note that $\\frac{9b+7}{b+7} = 9 - \\frac{56}{b+7}$. Hence $17_b$ is a divisor of $97_b$ if and only if $b > 9$ and $b + 7$ is a divisor of $56$. Because $56 = 2^3 \\cdot 7$, the two possibilities for $b$ are $b = 21$ and $b = 49$. T... | United States | 2025 AIME I | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 70 | |
03an | Let $a_1, \dots, a_n, b_1, \dots, b_n$ be real numbers and $c_1, \dots, c_n$ be positive real numbers. Prove that
$$
\left( \sum_{i,j=1}^{n} \frac{a_i a_j}{c_i + c_j} \right) \left( \sum_{i,j=1}^{n} \frac{b_i b_j}{c_i + c_j} \right) \ge \left( \sum_{i,j=1}^{n} \frac{a_i b_j}{c_i + c_j} \right)^2 .
$$ | [
"*First solution.* Let $d_1, \\dots, d_n$ be real numbers and\n$$\nf(x) = \\sum_{i,j=1}^{n} \\frac{d_i d_j}{c_i + c_j} x^{c_i+c_j}, \\quad x > 0.\n$$\nSince $x f'(x) = (\\sum_{i=1}^n d_i x^{c_i})^2 \\ge 0$, it follows that $f(x) \\ge f(0+) = 0$. In particular,\n$$\nf(1) = \\sum_{i,j=1}^{n} \\frac{d_i d_j}{c_i + c_j... | Bulgaria | 58. National mathematical olympiad Final round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0jce | Problem:
The game of rock-scissors is played just like rock-paper-scissors, except that neither player is allowed to play paper. You play against a poorly-designed computer program that plays rock with $50\%$ probability and scissors with $50\%$ probability. If you play optimally against the computer, find the probabi... | [
"Solution:\n\nAnswer: $\\frac{163}{256}$\n\nSince rock will always win against scissors, the optimum strategy is for you to always play rock; then, you win a game if and only if the computer plays scissors. Let $p_{n}$ be the probability that the computer plays scissors $n$ times; we want $p_{0}+p_{1}+p_{2}+p_{3}+p... | United States | HMMT November 2012 | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 163/256 | |
09zx | Problem:
Vind alle viertallen $(a, b, c, d)$ van niet-negatieve gehele getallen zodat $a b = 2(1 + c d)$ en er een niet-ontaarde driehoek bestaat met zijden van lengte $a-c$, $b-d$ en $c+d$. | [
"Solution:\n\nEr geldt $a > c$ en $b > d$ omdat $a-c$ en $b-d$ zijden van een driehoek moeten zijn. Dus $a \\geq c+1$ en $b \\geq d+1$, aangezien het om gehele getallen gaat. We onderscheiden nu twee gevallen: $a > 2c$ en $a \\leq 2c$.\n\nStel dat $a > 2c$ geldt. Dan is $a b > 2 b c \\geq 2c \\cdot (d+1) = 2 c d + ... | Netherlands | IMO-selectietoets III | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | (1, 2, 0, 1) and (2, 1, 1, 0) |
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