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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
015r | Find the maximum $n$ for which there exists a set of $n$ numbers such that these numbers are not divisible by $7$, $11$, and $13$ but the sum of any two of them is divisible by either $7$, or $11$, or $13$. | [
"Answer: $n = 8$.\n\nExample: take all the numbers $a$ such that $a \\equiv \\pm 1 \\pmod{7}$, $a \\equiv \\pm 1 \\pmod{11}$, $a \\equiv \\pm 1 \\pmod{13}$. Due to the Chinese remainder theorem we have exactly $8$ numbers in the interval from $1$ to $1001$ ($1$, $155$, $274$, $428$, $573$, $727$, $846$, $1000$). It... | Baltic Way | Baltic Way SHL | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
0ex4 | Problem:
Given $n$ odd and a set of integers $\{a_1\}$, $\{a_2\}$, ..., $\{a_n\}$, derive a new set $(\{a_1\} + \{a_2\}) / 2$, $(\{a_2\} + \{a_3\}) / 2$, ..., $(\{a_{n - 1}\} + \{a_n\}) / 2$, $(\{a_n\} + \{a_1\}) / 2$. However many times we repeat this process for a particular starting set we always get integers. Prove... | [
"Solution:\nLet the smallest value be $s$ and suppose it occurs $m$ times (with $m < n$). Then the values in the next stage are all at least $s$, and at most $m - 1$ equal $s$. So after at most $m$ iterations the smallest value is increased.\n\nWe can never reach a stage where all the values are equal, because if $... | Soviet Union | 4th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
047w | Let $p_1, p_2, \dots, p_{2025}$ be real numbers, and let $\{a_n^{(1)}\}_{n \ge 0}$, $\{a_n^{(2)}\}_{n \ge 0}$, $\dots$, $\{a_n^{(2025)}\}_{n \ge 0}$ be $2025$ real sequences satisfying:
(1) $a_0^{(i)}$ ($1 \le i \le 2025$) are all zero;
(2) $a_1^{(i)}$ ($1 \le i \le 2025$) are **not** all zero;
(3) For $i = 1, 2, \dots... | [
"**Proof:** We first prove a lemma.\n**Lemma:** Let $\\beta$ be a complex number. If a sequence $\\{a_n\\}$ satisfies $a_0 = 0$, $a_1 \\ne 0$, and for all $n \\ge 1$,\n$$\na_{n-1} + \\beta a_n + a_{n+1} = 0,\n$$\nthen $\\{a_n\\}$ does not converge to $0$.\n**Proof of Lemma:** Let $\\alpha_1, \\alpha_2$ be roots of ... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof only | null | |
0awk | Problem:
Find all pairs $(r, s)$ of real numbers such that the zeros of the polynomials
$$
f(x) = x^{2} - 2 r x + r
$$
and
$$
g(x) = 27 x^{3} - 27 r x^{2} + s x - r^{6}
$$
are all real and nonnegative. | [
"Solution:\nLet $x_{1}, x_{2}$ be the zeros of $f(x)$, and let $y_{1}, y_{2}, y_{3}$ be the zeros of $g(x)$.\nBy Viete's relation,\n$$\n\\begin{aligned}\nx_{1} + x_{2} & = 2 r \\\\\nx_{1} x_{2} & = r\n\\end{aligned}\n$$\nand\n$$\n\\begin{aligned}\ny_{1} + y_{2} + y_{3} & = r \\\\\ny_{1} y_{2} + y_{2} y_{3} + y_{3} ... | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | [(0, 0), (1, 9)] | |
0d1v | Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying $f(f(x)) = 4x + 1$ for all real number $x$. Prove that the equation $f(x) = x$ has a unique solution. | [
"Let $x_0$ be a solution of the equation $f(x) = x$. We have\n$$\nx_0 = f(x_0) = f(f(x_0)) = 4x_0 + 1.\n$$\nTherefore, $x_0 = -\\frac{1}{3}$. This proves the uniqueness of the solution.\n\nOn the other hand,\n$$\nf\\left(f\\left(-\\frac{1}{3}\\right)\\right) = 4\\left(-\\frac{1}{3}\\right) + 1 = -\\frac{1}{3}\n$$\n... | Saudi Arabia | Preselection tests for the full-time training | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
02pa | Problem:
Um professor e seus 30 alunos escreveram, cada um, os números de $1$ a $30$ em uma ordem qualquer. A seguir, o professor comparou as sequências. Um aluno ganha um ponto cada vez que um número aparece na mesma posição na sua sequência e na do professor. Ao final, observou-se que todos os alunos obtiveram quant... | [
"Solution:\n\nO número de acertos é um número entre $0$ e $30$ inclusive. Mas, observe que $29$ não pode ser obtido porque se $29$ números estão em posição certa, só há uma maneira de colocar o $30^{\\circ}$ número, que é em posição certa também.\n\nComo há $30$ alunos e $30$ possíveis resultados, $\\{0,1, \\ldots,... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
088u | Problem:
Siano $p(x)$ e $q(x)$ due polinomi distinti di grado minore o uguale a $3$, a coefficienti interi e tali che
$$
\begin{gathered}
p(1)=q(1), \quad p(2)=q(2), \quad p(3)=q(3), \\
p(-1)=-q(-1), \quad p(-2)=-q(-2), \quad p(-3)=-q(-3) .
\end{gathered}
$$
Qual è il minimo valore che può assumere $[p(0)]^{2}+[q(0)]^... | [
"Solution:\n\nLa risposta è $36$. Il polinomio $p(x)+q(x)$ si annulla per $x=-1$, $x=-2$ e $x=-3$; inoltre è di grado minore o uguale a $3$ ed è a coefficienti interi. Quindi, per il teorema di Ruffini, esso si scompone in questo modo: $p(x)+q(x)=k(x+1)(x+2)(x+3)$ con $k$ intero. In modo analogo osserviamo che $p(x... | Italy | Italian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 36 | |
0aze | Problem:
In how many ways can the letters of the word COMBINATORICS be arranged so that the letters $C$, $O$, $A$, $C$, $T$, $O$, $R$, $S$ appear in that order in the arrangement (although there may be letters in between)? | [
"Solution:\n\nThe word COMBINATORICS has 13 letters. The letters are: $C$, $O$, $M$, $B$, $I$, $N$, $A$, $T$, $O$, $R$, $I$, $C$, $S$.\n\nFirst, note that the letters $C$, $O$, $A$, $C$, $T$, $O$, $R$, $S$ appear in the word as follows:\n- $C$ appears 2 times\n- $O$ appears 2 times\n- $A$ appears 1 time\n- $T$ appe... | Philippines | 21st PMO Area Stage | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 77220 | |
0hbi | Suppose a country has a system of roads such that the roads intersect in towns only, and do not intersect in-between the towns. Furthermore, from any town one can reach any other town, if one can go in either of the two directions on each road. For no pair of towns, there is more than one direct road connecting them. T... | [
"Let us show such system of roads where such town does not exist. Denote by $A, B, C, D$ 3-tuples of towns, where the roads form a cycle, e.g. $A_1 \\to A_2 \\to A_3 \\to A_1$ (fig. 24). In this way, the condition that each town has one incoming and one outcoming road is satisfied. Now, we place additional roads wi... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Other"
] | English | proof and answer | No | |
04di | On a math test there are $40$ problems. For each correct answer one gets $15$ points, and for each incorrect answer $-4$ points. Dinko has solved all the problems, but he made some mistakes. How many incorrect answers he had, if he scored $353$ points in total? | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 13 | |
03xm | For any set $A = \{a_1, a_2, \dots, a_m\}$, denote $P(A) = a_1 a_2 \dots a_m$. Let $A_1, A_2, \dots$, and $A_n$ be all 99-element subsets of $\{1, 2, \dots, 2010\}$, $n = C_{2010}^{99}$. Prove that $2010 \mid \sum_{i=1}^{n} P(A_i)$. | [
"For each 99-element subset, $A_i = \\{a_1, a_2, \\dots, a_{99}\\}$ of $\\{1, 2, \\dots, 2010\\}$ uniquely corresponds to a 99-element subset $B_i = \\{b_1, b_2, \\dots, b_{99}\\}$ of $\\{1, 2, \\dots, 2010\\}$ by $b_k = 2011 - a_k$, $k = 1, 2, \\dots, 99$.\nSince $\\sum_{k=1}^{99} (a_k + b_k) = 99 \\times 2011$ is... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
0gb6 | 設三角形 $ABC$ 為銳角三角形，$A_1, B_1, C_1$ 分別位於 $BC, CA, AB$ 邊上，且 $AA_1, BB_1, CC_1$ 皆為三角形 $ABC$ 的內角平分線。令點 $I$ 為三角形 $ABC$ 的內心，點 $H$ 為三角形 $A_1B_1C_1$ 的垂心。證明：
$$
AH + BH + CH \ge AI + BI + CI.
$$ | [
"記 $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$, $\\angle ACB = \\gamma$。不失一般性，可設 $\\alpha \\le \\beta \\le \\gamma$。\n另將三角形 $ABC$ 的三邊長分別記為 $BC = a$, $CA = b$, $AB = c$。\n\n我們首先證明：$\\triangle A_1B_1C_1$ 也是銳角三角形。在 $BC$ 邊上取點 $D, E$ 滿足：$B_1D // AB$, 且 $B_1E$ 爲 $\\angle BB_1C$ 的內角平分線。由於 $\\angle B_1DB = 180^\\circ -... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Misc... | null | proof only | null | |
0bd1 | Prove that the equation
$$
\frac{1}{\sqrt{x} + \sqrt{1006}} + \frac{1}{\sqrt{2012 - x} + \sqrt{1006}} = \frac{2}{\sqrt{x} + \sqrt{2012 - x}}
$$
has 2013 integer solutions. | [
"One can easily check that the given relation holds for any admissible value of $x$. Since the number $x$ is subject to the conditions $0 \\le x \\le 2012$, the conclusion is easily reached."
] | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2013 | |
0l3d | Problem:
Compute the sum of all positive integers $n$ such that $50 \leq n \leq 100$ and $2 n+3$ does not divide $2^{n!}-1$. | [
"Solution:\nWe claim that if $n \\geq 10$, then $2 n+3 \\nmid 2^{n!}-1$ if and only if both $n+1$ and $2 n+3$ are prime.\n\nIf both $n+1$ and $2 n+3$ are prime, then assume $2 n+3 \\mid 2^{n!}-1$. By Fermat's Little Theorem, $2 n+3 \\mid 2^{2 n+2}+1$. However, since $n+1$ is prime, $\\gcd(2 n+2, n!)=2$, so $2 n+3 \... | United States | HMMT February 2024 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 222 | |
01w1 | The parabolas $y = x^2 - 2$ and $x = y^2 - 2$ intersect at the points $A$, $B$, $C$ and $D$, wherein $D$ lies in the third quadrant of the Cartesian plane.
Find the coordinates of the circumcenter of the triangle $ABC$. | [
"Answer: $(\\frac{1}{2}; \\frac{1}{2})$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | (1/2, 1/2) | |
07ue | Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that for all integers $a, b$
$$
f(a + f(b)) = b + f(a).
$$ | [
"The two solutions are $f(x) = x$ and $f(x) = -x$. We prove this in three stages. First we show that $f$ is self-inverse, that is, $f(f(x)) = x$ for all integers $x$. Secondly we show that $f$ is additive. Thirdly, we demonstrate that the stated solutions are the only self-inverse additive functions.\n\nTo show the... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x and f(x) = -x | |
0arr | Problem:
Find all nonnegative integer solutions of the system
$$
\begin{aligned}
& 5x + 7y + 5z = 37 \\
& 6x - y - 10z = 3
\end{aligned}
$$ | [
"Solution:\n(ans. $(x, y, z) = (4, 1, 2)$.\nEliminating $z$ by multiplying the first equation by $2$ and taking the sums, we obtain $16x + 13y = 77$. This is equivalent to $16(x - 4) + 13(y - 1) = 0$, hence $(x, y, z) = (4, 1, 2)$ is a solution. All other solutions are given by $x = 4 + 16t$, $y = 1 - 13t$, $t$ an ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (4, 1, 2) | |
0gx6 | Prove that
$$
\frac{a+b}{2b+c} + \frac{b+c}{2c+a} + \frac{c+a}{2a+b} \geq 2
$$
is true for any positive real numbers $a, b, c$. | [
"Applying the Cauchy-Bunyakowsky inequality to sets $\\frac{a_1}{\\sqrt{b_1}}, \\dots, \\frac{a_n}{\\sqrt{b_n}}$ and $\\sqrt{b_1}, \\dots, \\sqrt{b_n}$ in which numbers $a_1, \\dots, a_n, b_1, \\dots, b_n$ are positive we find that $\\frac{a_1^2}{b_1} + \\dots + \\frac{a_n^2}{b_n} \\ge \\frac{(a_1+\\dots+a_n)^2}{b_... | Ukraine | Ukrajina 2008 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
08m3 | Problem:
Determine all pairs $(m, n)$ for which it is possible to tile the table $m \times n$ with "corners" as in the figure below, with the condition that in the tiling there is no rectangle (except for the $m \times n$ one) regularly covered with corners.
 | [
"Solution:\nEvery \"corner\" covers exactly 3 squares, so a necessary condition for the tiling to exist is $3 \\mid m n$.\n\nFirst, we shall prove that for a tiling with our condition to exist, it is necessary that both $m, n$ for $m, n>3$ to be even. Suppose the contrary, i.e. suppose that $m>3$ is odd (without lo... | JBMO | 2009 Shortlist JBMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All and only boards of sizes 2×3, 3×2; 6×2k and 2k×6 for k ≥ 2; and 6k×4ℓ and 6k×(4ℓ+2) for k, ℓ ≥ 2. | |
029j | Problem:
O triângulo de latas - Um menino tentou alinhar 480 latas em forma de um triângulo com uma lata na $1^{a}$ linha, 2 latas na $2^{a}$ e assim por diante. No fim sobraram 15 latas. Quantas linhas tem esse triângulo? | [
"Solution:\n\nSuponhamos que o triângulo está composto por $n$ linhas, logo foram usadas $1+2+3+\\cdots+n$ latas, assim\n$$\n480-15=1+2+\\cdots+n=\\frac{n(n+1)}{2} \\Longrightarrow n^{2}+n-930=0\n$$\nResolvendo a equação $n^{2}+n-930=0$, obtemos:\n$$\nn=\\frac{-1 \\pm \\sqrt{1+4 \\times 930}}{2}=\\frac{-1 \\pm 61}{... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 30 | |
08fj | Problem:
Il robot "Mag-o-matic" manipola 101 bicchieri, disposti in una fila le cui posizioni sono numerate da 1 a 101. In ognuno dei bicchieri può trovarsi, oppure no, una pallina. Il robot Mag-o-matic accetta solo istruzioni elementari della forma $(a ; b, c)$, che interpreta come
"considera il bicchiere in posizion... | [
"Solution:\n\nRisolviamo direttamente il caso generale, dimostrando che un sottoinsieme $S$ è identificabile se e solo se $0 \\notin S$ e $101 \\in S$. Al termine descriveremo una scorciatoia che funziona nel caso dispari.\n\n**Condizione necessaria**\n\nDimostriamo intanto che le condizioni $0 \\notin S$ e $101 \\... | Italy | XXXVIII Olimpiade Italiana di Matematica, Cesenatico | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | A subset S is identifiable if and only if 0 is not in S and 101 is in S. In particular, the set of odd numbers is identifiable. | |
023e | Problem:
Um grupo de meninos e meninas se reúne para comer pizzas que são cortadas em 12 pedaços. Cada menino pode comer 6 ou 7 pedaços e cada menina pode comer 2 ou 3 pedaços. Sabemos que quatro pizzas nunca são suficientes para alimentar o grupo e que com cinco pizzas sempre há sobra. Quantos meninos e quantas menin... | [
"Solution:\n\nChamemos de $x$ o número de meninos e de $y$ o número de meninas. Pelas condições do problema sabemos que se eles comem o mínimo possível, ainda assim quatro pizzas não são suficientes, isto é,\n$$\n6x + 2y > 4 \\times 12 = 48\n$$\nPor outro lado, se eles comem o máximo possível, com cinco pizzas sobr... | Brazil | Nível 3 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 8 boys and 1 girl | |
0jrh | Problem:
Consider a $9 \times 9$ grid of squares. Haruki fills each square in this grid with an integer between $1$ and $9$, inclusive. The grid is called a super-sudoku if each of the following three conditions hold:
- Each column in the grid contains each of the numbers $1,2,3,4,5,6,7,8,9$ exactly once.
- Each row i... | [
"Solution:\n\nWithout loss of generality, suppose that the top left corner contains a $1$, and examine the top left $3 \\times 4$:\n\n| 1 | x | x | x |\n| :---: | :---: | :---: | :---: |\n| x | x | x | $\\{ \\}^{*}$ |\n| x | x | x | $*$ |\n\nThere cannot be another $1$ in any of the cells marked with an $x$, but th... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Logic"
] | null | proof and answer | 0 | |
01rp | A circle intersects a parabola at four distinct points. Let $M$ and $N$ be the midpoints of the arcs of the circle which are outside the parabola.
Prove that the line $MN$ is perpendicular to the axis of the parabola. | [
"We may assume that the parabola is defined by the equation $y = x^2$, while the circle is defined by the equation $(x-a)^2 + (y-b)^2 = R^2$. Let $A(a, b)$ be the center of the circle, $X_i(x_i, x_i^2)$, $i = 1, 2, 3, 4$, be common points of the circle and the parabola. Then $x_1, x_2, x_3, x_4$ are four roots of t... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Circles"
] | English | proof only | null | |
051f | Prove that for any positive integer $k$ there exist $k$ pairwise distinct integers for which the sum of their squares equals the sum of their cubes. | [
"For any integer $m > 1$ the numbers $2m^2 + 1$, $m(2m^2 + 1)$, $-m(2m^2 + 1)$ satisfy the conditions of the problem, because they are pairwise different and\n$$\n\\begin{aligned}\n& (2m^2 + 1)^2 + (m(2m^2 + 1))^2 + (-m(2m^2 + 1))^2 \\\\\n&= (1 + m^2 + m^2) \\cdot (2m^2 + 1)^2 = (2m^2 + 1)^3 = (1 + m^3 - m^3) \\cdo... | Estonia | Estonian Math Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0g13 | Problem:
Trouver toutes les fonctions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ telles que:
(i) $f(p)>0$ pour tout nombre premier $p$,
(ii) $p \mid (f(x)+f(p))^{f(p)}-x$ pour tout nombre premier $p$ et pour tout $x \in \mathbb{Z}$. | [
"Solution:\n\nPremière solution : Gardons les bons réflexes qui s'imposent avec les équations fonctionnelles. On va montrer (après avoir commencé par chercher les solutions potentielles !) que l'unique solution est l'identité. On comprend déjà que le petit Théorème de Fermat va jouer un rôle crucial.\n\nSoit donc $... | Switzerland | IMO-Selektion | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | f(x) = x for all integers x | |
0gcm | 有一個無限大的方格棋盤, 每個格子裡面放一個正整數, 任何一個長方形的內部總和都不是質數, 而且至少有一格放的是 $1$, 求所有格子最大的數至少是多少。 | [
"答案是 $9$。\n注意到 $1$ 旁邊可以放的最小數是 $8$, 但 $8 + 1 + 8 = 17$ 是質數, 所以格子裡一定要有 $9$。\n\n| 4 | 8 | 6 |\n|---|---|---|\n| 8 | 1 | 9 |\n| 6 | 9 | 9 |\n\n構造是在 $1$ 的周圍格利用模 $2$ 跟模 $3$, 然後除了這九格以外都放 $6$, 如此一來任何一個不只一個數的矩形內部的和都是 $2$ 或 $3$ 的倍數。"
] | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | 9 | |
00yl | Problem:
Show that for any integer $a \geq 5$ there exist integers $b$ and $c$, $c \geq b \geq a$, such that $a, b, c$ are the lengths of the sides of a right-angled triangle. | [
"Solution:\nWe first show this for odd numbers $a = 2i + 1 \\geq 3$. Put $c = 2k + 1$ and $b = 2k$. Then $c^{2} - b^{2} = (2k + 1)^{2} - (2k)^{2} = 4k + 1 = a^{2}$. Now $a = 2i + 1$ and thus $a^{2} = 4i^{2} + 4i + 1$ and $k = i^{2} + i$. Furthermore, $c > b = 2i^{2} + 2i > 2i + 1 = a$.\n\nSince any multiple of a Py... | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | proof only | null | |
062k | Problem:
Es sei $p$ eine Primzahl. Weiter seien $a, b, c$ ganze Zahlen, welche die Gleichungen $a^{2}+p b = b^{2}+p c = c^{2}+p a$ erfüllen.
Man beweise, dass dann $a = b = c$ gilt. | [
"Solution:\n\nWenn zwei der drei Zahlen $a, b, c$ gleich sind, können wir wegen der zyklischen Vertauschbarkeit oBdA $a = b$ annehmen. Damit wird die linke gegebene Gleichung zu $a^{2} + p b = a^{2} + p c$, woraus wegen $p \\neq 0$ direkt $b = c$ folgt, so dass die Behauptung erfüllt ist. Im Folgenden können wir da... | Germany | 2. IMO-Auswahlklausur | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0csc | Серёжа выбрал два различных натуральных числа $a$ и $b$. Он записал в тетрадь четыре числа: $a, a+2, b$ и $b+2$. Затем он выписал на доску все шесть попарных произведений чисел из тетради. Какое наибольшее количество точных квадратов может быть среди чисел на доске? | [
"**Ответ.** Два.\n\nЗаметим, что никакие два квадрата натуральных чисел не отличаются на 1, ибо $x^2 - y^2 = (x - y)(x + y)$, где вторая скобка больше единицы. Значит, числа $a(a+2) = (a+1)^2 - 1$ и $b(b+2) = (b+1)^2 - 1$ квадратами не являются. Более того, числа $ab$ и $a(b+2)$ не могут одновременно являться квадр... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2 | |
0bi8 | Let $a, b, c \in (0, \infty)$. Prove the inequality
$$
\frac{a - \sqrt{bc}}{a + 2(b + c)} + \frac{b - \sqrt{ca}}{b + 2(c + a)} + \frac{c - \sqrt{ab}}{c + 2(a + b)} \geq 0.
$$ | [
"The AM-GM inequality leads to\n$$\n\\frac{a - \\sqrt{bc}}{a + 2(b + c)} \\geq \\frac{a - \\frac{b+c}{2}}{a + 2(b + c)} = \\frac{2a - b - c}{2(a + 2b + 2c)}.\n$$\n\nSo, the left-hand part of the given inequality is at least\n$$\n\\frac{2a-b-c}{2(a+2b+2c)} + \\frac{2b-c-a}{2(2a+b+2c)} + \\frac{2c-a-b}{2(2a+2b+c)} = ... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
018y | Let $ABCD$ be a convex quadrilateral such that $\angle ADB = \angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
$$
AE \cdot ED + BE^2 = CD \cdot AE.
$$
Show that $\angle EBA = \angle DCB$. | [
"Let $F$ be the point symmetric to $E$ with respect to the line $DB$. Then the equality $\\angle ADB = \\angle BDC$ shows that $F$ lies on the line $DC$, on the same side of $D$ as $C$. Moreover, we have $AE \\cdot ED < CD \\cdot AE$, or $FD = ED < CD$, so in fact $F$ lies on the segment $DC$.\n\n![](attached_image... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
05ng | Problem:
Deux cercles $\Gamma_{1}$ et $\Gamma_{2}$ de centres $O_{1}$ et $O_{2}$ se coupent en $P$ et $Q$. Une droite passant par $O_{1}$ coupe $\Gamma_{2}$ en $A$ et $B$, et une droite passant par $O_{2}$ coupe $\Gamma_{1}$ en $C$ et $D$. Montrer que s'il existe un cercle passant par $A, B, C$ et $D$, alors le centre... | [
"Solution:\n\nOn note $\\Gamma$ le cercle passant par $A, B, C$ et $D$. D'après le théorème des axes radicaux, $(AB)$, $(CD)$ et $(PQ)$ sont concourantes en un point qu'on appelle $X$. De plus, $(AB)$ est perpendiculaire à $\\left(O_{1}O_{2}\\right)$, donc est la hauteur issue de $O_{1}$ dans $O_{1}O_{2}O$. De même... | France | French Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0129 | Problem:
Let $N$ be a positive integer. Two persons play the following game. The first player writes a list of positive integers not greater than $25$, not necessarily different, such that their sum is at least $200$. The second player wins if he can select some of these numbers so that their sum $S$ satisfies the con... | [
"Solution:\n\nIf $N=11$, then the second player can simply remove numbers from the list, starting with the smallest number, until the sum of the remaining numbers is less than $212$. If the last number removed was not $24$ or $25$, then the sum of the remaining numbers is at least $212-23=189$. If the last number r... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 11 | |
0c3w | Problem:
Fie $A B C$ un triunghi ascuţitunghic în care $A B \neq A C$. Fie $D$ mijlocul laturii $[B C]$, iar $E$ şi $F$ proiecţiile lui $D$ pe laturile $A B$, respectiv $A C$. Dacă $M$ este mijlocul segmentului $[E F]$, iar $O$ este centrul cercului circumscris triunghiului $A B C$, demonstraţi că dreptele $D M$ şi $A... | [
"Solution:\n\nFie $\\{S\\} = A O \\cap B C$ şi $T$ punctul în care dreapta $A D$ intersectează pentru a doua oară cercul circumscris triunghiului $A B C$. Patrulaterele $A B T C$ şi $A E D F$ sunt inscriptibile, deci $\\angle T B D \\equiv \\angle T A C \\equiv \\angle D E F$ şi $\\angle T C D \\equiv \\angle T A B... | Romania | Al treilea test de selecţie pentru OBMJ | [
"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 | |
0k0k | Problem:
Consider the $4 \times 4$ "multiplication table" below. The numbers in the first column multiplied by the numbers in the first row give the remaining numbers in the table. For example, the $3$ in the first column times the $4$ in the first row give the $12\ (=3 \cdot 4)$ in the cell that is in the $3$rd row a... | [
"Solution:\n\nThe minimum is $46$ and the maximum is $50$. To see this more easily, tilt the grid $45$ degrees:\n\n\n\nNow every path must include exactly one number from each row. The smallest and largest numbers in each row are respectively at the edge and in the middle, so the smallest a... | United States | 19th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Minimum sum = 46; Maximum sum = 50 | |
0ax1 | Problem:
Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. If $a_{n}\lfloor a_{n}\rfloor=49^{n}+2 n+1$, find the value of $2 S+1$, where $S=\left\lfloor\sum_{n=1}^{2017} \frac{a_{n}}{2}\right\rfloor$. | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (7^2018 - 7)/6 | |
09uq | Thomas and Nils are playing a game. They have a number of cards, numbered $1$, $2$, $3$, et cetera. At the start, all cards are lying face up on the table. They take alternate turns. The person whose turn it is, chooses a card that is still lying on the table and decides to either keep the card himself or to give it to... | [
"a.\nThomas and Nils both make $1009$ moves and Nils makes the last move. Nils can make sure that the last card on the table contains a number that is *not* divisible by $3$. Indeed, he could start taking cards with numbers that are divisible by $3$, until all these cards are gone. Because there are only $672$ such... | Netherlands | Final Round, September 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | a: Nils wins with certainty.
b: Nils wins with certainty. | |
0fyg | Problem:
Ein Schweizerkreuz besteht aus fünf Einheitsquadraten, einem zentralen und vier seitlich angrenzenden. Bestimme die kleinste natürliche Zahl $n$ mit folgender Eigenschaft: Unter je $n$ Punkten im Innern oder auf dem Rand eines Schweizerkreuzes gibt es stets zwei, deren Abstand kleiner als 1 ist. | [
"Solution:\n\nDie Menge aller Eckpunkte der fünf Einheitsquadrate ist ein Beispiel einer Menge von 12 Punkten, deren paarweise Abstände alle mindestens gleich 1 sind. Folglich ist $n \\geq 13$. Wir zeigen nun, dass unter 13 Punkten tatsächlich stets zwei einen Abstand $<1$ haben. Unterteile dazu das Schweizerkreuz ... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 13 | |
0kgi | Problem:
Leo the fox has a $5$ by $5$ checkerboard grid with alternating red and black squares. He fills in the grid with the numbers $1,2,3, \ldots, 25$ such that any two consecutive numbers are in adjacent squares (sharing a side) and each number is used exactly once. He then computes the sum of the numbers in the $... | [
"Solution:\n\nSince consecutive numbers are in adjacent squares and the grid squares alternate in color, consecutive numbers must be in squares of opposite colors. Then the odd numbers $1,3,5, \\ldots, 25$ all share the same color while the even numbers $2,4, \\ldots, 24$ all share the opposite color. Since we have... | United States | HMMT Spring 2021 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 169 | |
02cx | Problem:
Todos os vértices do pentágono $A B C D E$ estão sobre um mesmo círculo. Se $\angle D A C=50^{\circ}$, determine $\angle A B C+\angle A E D$.
 | [
"Solution:\n\nComo ângulos inscritos associados a um mesmo arco são iguais, temos $\\angle D A C=\\angle D B C$. Além disto, sabendo que a soma dos ângulos opostos de um quadrilátero inscritível é $180^{\\circ}$, segue que\n$$\n\\begin{aligned}\n\\angle A B C+\\angle A E D & = (\\angle A B D+\\angle A E D)+\\angle ... | Brazil | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 230° | |
0650 | In a circular ring with radii $R$ and $R-2r$, where $R = 11r$, we put non overlapping circles of radius $r$ tangent to the circles defining the circular ring. Determine the maximal number of these circles. (It is given that $9.94 < \sqrt{99} < 9.95$) | [
"Let we can put $N$ non overlapping circles $C_i(K_i, r)$ into the given circular ring tangent to its border. The circle $C(O, R-r)$ has length greater than the perimeter $\\ell_{K_1...K_N K_1}$ of the polygon with vertices the centers of the circles $C_i(K_i, r)$, and therefore:\n$$\nN \\cdot 2r < \\ell_{K_1...K_N... | Greece | 24th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 31 | |
0936 | Problem:
Determine all integers $n \geqslant 2$ such that there exists a permutation $x_{0}, x_{1}, \ldots, x_{n-1}$ of the numbers $0,1, \ldots, n-1$ with the property that the $n$ numbers
$$
x_{0}, \quad x_{0}+x_{1}, \quad \ldots, \quad x_{0}+x_{1}+\ldots+x_{n-1}
$$
are pairwise distinct modulo $n$. | [
"Solution:\nSuppose that $x_{0}, \\ldots, x_{n-1}$ is such a permutation.\nNote that $x_{0}=0$. Indeed, if $x_{i}=0$ for some $i>0$ then\n$$\nx_{0}+\\cdots+x_{i-1}=x_{0}+\\cdots+x_{i-1}+x_{i}\n$$\nwhich is a contradiction.\nOn the other hand\n$$\nx_{0}+x_{1}+\\cdots+x_{n-1}=0+1+2+\\cdots+n-1=n \\cdot \\frac{n-1}{2}... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | All even integers n ≥ 2 | |
065f | A triangle $AB\Gamma$ is given and let $O$ its circumcenter and $A_1, B_1, \Gamma_1$ the middles of its sides $B\Gamma, A\Gamma$ and $AB$, respectively. We consider the points $A_2, B_2, \Gamma_2$ such that $\overrightarrow{OA_2} = \lambda \cdot \overrightarrow{OA_1}$, $\overrightarrow{OB_2} = \lambda \cdot \overrighta... | [
"Let $H$ be the orthocenter of the triangle $AB\\Gamma$. Then $\\overrightarrow{AH} = 2 \\cdot \\overrightarrow{OA_1}$ and from $\\overrightarrow{OA_2} = \\lambda \\cdot \\overrightarrow{OA_1}$, we find: $\\overrightarrow{AH} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{OA_2}$.\nIf $AA_2$ meets $OH$ at $C$ (from t... | Greece | 26th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0058 | Se divide cada lado de un triángulo en 50 partes iguales, y cada punto de la división se une con el vértice opuesto mediante un segmento. Calcular el número de puntos de intersección determinados por estos segmentos.
Observación: Los vértices del triángulo original no se consideran puntos de intersección ni de divisió... | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | Spanish | proof and answer | 6913 | |
060n | Problem:
Soit $n$ un entier naturel. Joseph peut tirer $2n+1$ flèches. Chacun de ses tirs est un échec ou une réussite. Un tir est dit "équilibré" si le nombre d'échecs avant ce tir additionné au nombre de réussites après ce tir est égal à $n$. Déterminer si le nombre de tirs équilibrés est pair ou impair. | [
"Solution:\n\nPremière remarque : considérons une succession de deux tirs telle que le premier est réussi et le suivant raté. Alors il y a autant de tirs ratés avant pour les deux et autant de tirs réussis après donc soit les deux tirs sont équilibrés, soit aucun des deux ne l'est. De même, si le premier est raté e... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | odd | |
0go8 | Show that
$$
\sqrt[4]{\frac{(a^2 + b^2)(a^2 - ab + b^2)}{2}} + \sqrt[4]{\frac{(b^2 + c^2)(b^2 - bc + c^2)}{2}} + \sqrt[4]{\frac{(c^2 + a^2)(c^2 - ca + a^2)}{2}} \\
\le \frac{2}{3}(a^2 + b^2 + c^2) \left( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \right)
$$
for all positive real numbers $a$, $b$, $c$. | [
"We have $\\sqrt[4]{\\frac{(a^2 + b^2)(a^2 - ab + b^2)}{2}} \\le \\frac{a^2 + b^2}{a+b}$ for all nonnegative real numbers $a$, $b$ as this inequality is equivalent to $(a+b)^4(a^2-ab+b^2) \\le 2(a^2+b^2)^3$ which is in turn equivalent to $(a-b)^4(a^2+ab+b^2) \\ge 0$.\n\n$$\n\\frac{a^2+b^2}{a+b} + \\frac{b^2+c^2}{b+... | Turkey | Team Selection Test for IMO 2010 | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0aqw | Problem:
Two students, Lemuel and Christine, each wrote down an arithmetic sequence on a piece of paper. Lemuel wrote down the sequence $2, 9, 16, 23, \ldots$, while Christine wrote down the sequence $3, 7, 11, 15, \ldots$ After they have both written out 2010 terms of their respective sequences, how many numbers have... | [
"Solution:\n\nLet us first write the general term for each sequence.\n\nLemuel's sequence: $2, 9, 16, 23, \\ldots$\nThis is an arithmetic sequence with first term $a_1 = 2$ and common difference $d = 7$.\nSo the $n$th term is $a_n = 2 + 7(n-1) = 7n - 5$.\n\nChristine's sequence: $3, 7, 11, 15, \\ldots$\nThis is an ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 287 | |
0605 | Problem:
Soit $\mathbb{Z}[X]$ l'ensemble des polynômes à coefficients entiers. Trouver toutes les fonctions $f: \mathbb{Z}[X] \rightarrow \mathbb{Z}[X]$ telles que pour tous $P, Q \in \mathbb{Z}[X]$ et $r \in \mathbb{Z}$, on ait
$$
P(r)|Q(r) \Longleftrightarrow (f(P))(r)|(f(Q))(r)
$$ | [
"Solution:\n\nSoit $f$ une fonction solution de l'énoncé. Commençons par remarquer que si $P, Q, r$ sont tels que $|P(r)|=|Q(r)|$, alors $f(P)(r)$ et $f(Q)(r)$ se divisent mutuellement, et donc $|f(P)(r)|=|f(Q)(r)|$.\n\nRemarquons aussi que si $P$ et $Q$ sont deux polynômes tels que pour une infinité d'entiers $r$,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All such functions are exactly those of the form f(P)(X) = s(P) · A(X) · (P(X))^n, where n is a positive integer, A ∈ Z[X] satisfies A(r) ≠ 0 for every integer r, and s(P) ∈ {+1, −1} may depend on P (but not on X). | |
01ha | For a prime number $p$ and a positive integer $n$, denote by $f(p, n)$ the largest integer $k$ such that $p^k \mid n!$. Let $p$ be a given prime number and let $m$ and $c$ be given positive integers. Prove that there exists infinitely many positive integers $n$ such that $f(p, n) - c$ is divisible by $m$. | [
"We denote $v_p(n)$ for the largest power of $p$ dividing $n$.\nWe start with a lemma.\n**Lemma.** For any prime $q$ and modulus $m'$ not divisible by $q$, there exists infinitely many powers $q^n$ of $q$ such that $v_p(q^n!) \\equiv 1 \\pmod{m'}$.\n*Proof.* Define $a_k = v_q(q^k!)$. We then have $a_{k+1} = q a_k +... | Baltic Way | Baltic Way 2020 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0hhm | We will call a circle without boundary, i.e., a circle without the points on its circumference, a "hedgehog". The diameter of the hedgehog is the diameter of this circle. We will say that the hedgehog "sits" at a point where the center of the corresponding circle is located. Let us consider a triangle with sides $a, b,... | [
"Let us denote the width of the triangle as $ABC$, and the diameters of the hedgehogs as $d_a, d_b, d_c$, respectively. Then suppose $d_a + d_b > 2c$. This means that any point on the side $AB$ of the triangle is inside one of the hedgehogs, and therefore it is impossible to reach this side. Then we have $d_a + d_b... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof and answer | a + b + c | |
0dtf | Let $n$ be a positive integer. There are $n$ islands with $n - 1$ bridges connecting them such that one can travel from any island to another. One afternoon, a fire breaks out in one of the islands. Every morning, it spreads to all neighbouring islands. (Two islands are neighbours if they are connected by a bridge.) To... | [
"Solution:\n\nSuppose that there are integer solutions. By completing squares, the equation becomes\n$$\n(y+1)^2 = (x^2 + 10x + 2)^2 + 2000.\n$$\nLet $a = y + 1$, $b = x^2 + 10x + 2$, $u = a - b$, $v = a + b$. Then $uv = 2000$ and $b = \\frac{v-u}{2}$.\nThe equation $x^2 + 10x + 2 - b = x^2 + 10x + 2 + \\frac{u-v}{... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | floor(sqrt(n - 1)) | |
0fvv | Problem:
Seien $a$, $b$, $c$ nichtnegative reelle Zahlen mit arithmetischem Mittel $m = \frac{a+b+c}{3}$. Beweise, dass gilt
$$
\sqrt{a+\sqrt{b+\sqrt{c}}} + \sqrt{b+\sqrt{c+\sqrt{a}}} + \sqrt{c+\sqrt{a+\sqrt{b}}} \leq 3 \sqrt{m+\sqrt{m+\sqrt{m}}}
$$ | [
"Solution:\n\nSei $A$ die linke Seite der Ungleichung. Nach AM-QM gilt\n$$\nA \\leq 3 \\sqrt{\\frac{1}{3}(a+\\sqrt{b+\\sqrt{c}}+b+\\sqrt{c+\\sqrt{a}}+c+\\sqrt{a+\\sqrt{b}})} = 3 \\sqrt{m+\\frac{1}{3} B}\n$$\nWiederum nach AM-QM erhält man für $B$\n$$\nB \\leq 3 \\sqrt{\\frac{1}{3}(b+\\sqrt{c}+c+\\sqrt{a}+a+\\sqrt{b... | Switzerland | SMO Finalrunde | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0aht | Let $a$ and $n$ be integers. We define $a_n = 1 + a + a^2 + \dots + a^{n-1}$. Prove that if $a^p \equiv 1 \pmod p$ for every prime divisor $p$ of $n_2 - n_1$, then the number $\frac{a_{n_2} - a_{n_1}}{n_2 - n_1}$ is an integer. | [
"Lemma. Let $a$ and $n$ be integers such that $a \\equiv 1 \\pmod p$ for each prime $p \\nmid n$ and $a_n = 1 + a + a^2 + \\dots + a^{n-1}$. Then $n \\mid a_n$.\n\nProof of the lemma. Let $p^r$ be the largest power of the prime number $p$ such that $p^r \\mid n$. We will prove the equality\n$$\n1+a+a^2+\\dots+a^{n-... | North Macedonia | Team Selection Test for IMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0cmc | Problem:
Consider an integer $n \geq 2$ and write the numbers $1,2, \ldots, n$ down on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\{a+b,|a-b|\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is already on the board, do nothing. For all integers $n \geq ... | [
"Solution:\n\nThe answer is in the affirmative for all $n \\geq 2$. Induct on $n$. Leaving aside the trivial case $n=2$, deal first with particular cases $n=5$ and $n=6$.\n\nIf $n=5$, remove first the pair $(2,5)$, notice that $3=|2-5|$ is already on the board, so $7=2+5$ alone is written down. Removal of the pair ... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Yes, it is possible for all integers n at least two. | |
0e8u | Into a square with the side of length $2$ we draw two semicircles whose diameters are the sides of the square as shown in the figure. What is the area of the unshaded part of the square?
(A) $\frac{\pi}{2}$
(B) $2$
(C) $\frac{3}{2} + \frac{\pi}{4}$
(D) $\frac{3\pi}{4} - \frac{1}{2}$
(E) $\frac... | [
"Adding both diagonals onto the figure we notice that the parts $A$, $B$, $C$ and $D$ have equal area. The area of the unshaded part is therefore equal to one half of the area of the square, i.e. $\\frac{4}{2} = 2$. The correct answer is $B$.\n"
] | Slovenia | National Math Olympiad 2013 - First Round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | B | |
02g3 | $n$ tokens are to be placed on the squares of a $10 \times 10$ board such that no 4 tokens be the vertices of a rectangle with sides parallel to the sides of the board. Find the greatest value of $n$ for which this is possible. | [
"Let $A_i \\subset \\{1, 2, \\dots, 10\\}$ be the set of the positions of the tokens in the $i$$-$th line of the board, $1 \\le i \\le 10$. The problem is equivalent to finding $A_1, A_2, \\dots, A_{10}$ such that $|A_i \\cap A_j| \\le 1$ for $i \\ne j$ and $|A_1| + |A_2| + \\dots + |A_{10}|$ is maximum.\n\nLet $k_... | Brazil | XXI OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 34 | |
0elk | Problem:
Krog $K$ s polmerom $R$ razdelimo na tri krožne izseke, tako da je vsota ploščin manjših dveh izsekov enaka ploščini največjega izseka, razlika ploščin manjših dveh izsekov pa je enaka tretjini ploščine največjega izseka. V vsakega izmed izsekov včrtamo največji možen krog in te kroge označimo s $K_{1}, K_{2}... | [
"Solution:\n\nOznačimo središčne kote izsekov z $\\alpha_{1}, \\alpha_{2}$ in $\\alpha_{3}$, kjer je $\\alpha_{1}<\\alpha_{2}<\\alpha_{3}$. Ker je ploščina vsakega krožnega izseka premosorazmerna z njegovim s središčnim kotom, iz podatkov sledi $\\alpha_{1}+\\alpha_{2}+\\alpha_{3}=360^{\\circ}, \\alpha_{1}+\\alpha_... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | π R^2 (12√3 − 733/36) | |
0k3i | Problem:
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$
f(x+f(y+x y))=(y+1) f(x+1)-1
$$
for all $x, y \in \mathbb{R}^{+}$.
($\mathbb{R}^{+}$ denotes the set of positive real numbers.) | [
"Solution:\nLet $P(x, y)$ denote the assertion that\n$$\nf(x+f(y+x y))=(y+1) f(x+1)-1.\n$$\n\nClaim 1. $f$ is injective.\n\nProof: If $f(a)=f(b)$ then $P\\left(x, \\frac{a}{x+1}\\right), P\\left(x, \\frac{b}{x+1}\\right)$ yields $a=b$, since $f(x+1) \\in \\mathbb{R}^{+}$ so in particular is nonzero.\n\nNow $P\\left... | United States | HMIC 2018 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x for all x in the positive reals | |
0cs2 | Assume that $n$ is a positive integer, and a polynomial
$$
P(x) = a_{2n}x^{2n} + a_{2n-1}x^{2n-1} + \dots + a_1x + a_0,
$$
satisfies the conditions $100 \le a_i \le 101$ for all $0 \le i \le 2n$. Find the least possible $n$ such that this polynomial may have a real root. | [
"**Ответ.** $n = 100$.\n\nНазовём многочлен, удовлетворяющий условию задачи, **красивым**. Многочлен $P(x) = 100(x^{200} + x^{198} + \\dots + x^2 + 1) + 101(x^{199} + x^{197} + \\dots + x)$ красив и имеет корень $-1$. Значит, при $n = 100$ требуемое возможно.\n\nОсталось показать, что при $n < 100$ у красивого мног... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n = 100 | |
07v4 | The point $O$ is the circumcentre of triangle $ABC$. The point $E$ is on the extension of the side $AB$ such that $B$ is between $E$ and $A$, the point $F$ is on the extension of the side $AC$ such that $C$ is between $F$ and $A$, and the lines $BF$ and $CE$ intersect on the circumcircle of triangle $ABC$. The midpoint... | [
"Let $D$ be the intersection point of $BF$ and $CE$. The circumcircle $\\Omega$ of triangle $ABC$ has centre $O$ and radius $r = |ON|$. Let $G$ be the second intersection point of the circumcircle of triangle $ABF$ and the line $EF$. We then have $\\angle CAB = \\angle BGE$ and\n$$\n|EB| \\cdot |EA| = |EG| \\cdot |... | Ireland | IRL_ABooklet | [
"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",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Tr... | English | proof only | null | |
042h | Let $O$ be the circumcenter of triangle $ABC$. $H_A$ is the projection of $A$ onto $BC$. The extension of $AO$ intersects the circumcircle of $BOC$ at $A'$. The projections of $A'$ onto $AB$, $AC$ are $D$, $E$, and $O_A$ is the circumcenter of triangle $DH_AE$. Define $H_B, O_B, H_C, O_C$ similarly.
Prove: $H_AO_A$, $H... | [
"Let $T$ be the symmetry point of $A$ with regard to $BC$, $F$ be the projection of $A'$ onto $BC$, $M$ be the projection of $T$ onto $AC$.\nSince $AC = CT$, we have $\\angle TCM = 2\\angle TAM$. Since\n$$\n\\angle TAM = \\frac{\\pi}{2} - \\angle ACB = \\angle OAB, \\text{ we have}\n$$\n$$\n\\angle TCM = 2\\angle O... | China | China Team Selection Test | [
"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",
"Geometry > Plane Geometry > Analytic / C... | English | proof only | null | |
0369 | Problem:
Ivo writes consecutively the integers $1, 2, \ldots, 100$ on 100 cards and gives some of them to Yana. It is known that for every card of Ivo and every card of Yana, the card with the sum of the numbers on the two cards is not in Ivo and the card with the product of these numbers is not in Yana. How many card... | [
"Solution:\n\nYana has at least one card, say $k \\neq 1$. If Ivo has $1$, then the product $1 \\cdot k = k$ does not belong to Yana, a contradiction. Therefore Yana has $1$.\n\nIf $12$ is in Ivo, then the sum $13 = 1 + 12$ belongs to Yana, a contradiction. Therefore $12$ belongs to Yana. Since the sum $13 = 6 + 7$... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | 93 | |
0eyh | Problem:
$\{a_i\}$ and $\{b_i\}$ are permutations of $\{1/1, 1/2, \dots, 1/n\}$. $a_1 + b_1 \geq a_2 + b_2 \geq \ldots \geq a_n + b_n$. Prove that for every $m$ ($1 \leq m \leq n$), $a_m + b_m \geq \dfrac{4}{m}$. | [] | Soviet Union | 2nd ASU | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0gxt | Decimal representation of a number $a$ is written one or several times on the blackboard. As a result, the binary representation of the same number $a$ is obtained. Find all possible values of $a$. | [
"Let decimal representation of $a$ consist of exactly $k$ digits, and binary representation consists of exactly $l$ times more digits, i.e. of $kl$ digits. Since all digits of $a$ equal to either $0$ or $1$, then the number $a$ should belong to, from one side, the interval $[10^{k-1}, \\frac{1}{6}(10^k - 1)]$, and ... | Ukraine | The Problems of Ukrainian Authors | [
"Number Theory > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1 and 10 | |
0k40 | Problem:
Let $a$, $b$, $c$ be positive real numbers such that $a b c = 1$. Simplify
$$
\frac{1}{1+a+ab} + \frac{1}{1+b+bc} + \frac{1}{1+c+ca}.
$$ | [
"Solution:\nWe may let $a = y/x$, $b = z/y$, $c = x/z$ for some real numbers $x$, $y$, $z$. Then\n$$\n\\begin{aligned}\n\\frac{1}{1+a+ab} + \\frac{1}{1+b+bc} + \\frac{1}{1+c+ca} & = \\frac{1}{1 + y/x + z/x} + \\frac{1}{1 + z/y + x/y} + \\frac{1}{1 + x/z + y/z} \\\\\n& = \\frac{x}{x + y + z} + \\frac{y}{x + y + z} +... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 1 | |
03uu | In triangle $ABC$, we have $AB > AC$. The incircle $\omega$ touches $BC$ at $E$, and $AE$ intersects $\omega$ at $D$. Choose a point $F$ on $AE$ ($F$ is different from $E$), such that $CE = CF$. Let $G$ be the intersection point of $CF$ and $BD$. Prove that $CF = FG$. | [
"**Proof** Referring to the figure, draw a line from $D$, tangent to $\\omega$, and the line intersects $AB$, $AC$, $BC$ at points $M$, $N$, $K$ respectively.\n\nSince\n$$\n\\angle KDE = \\angle AEK = \\angle EFC,\n$$\nwe know $MK \\parallel CG$.\nBy Newton's theorem, the lines $BN$, $CM$, ... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
040r | Sequence $\{a_n\}$ is defined by:
$$
a_0 = \frac{1}{2},\ a_{n+1} = a_n + \frac{a_n^2}{2012},\ n = 0, 1, \dots
$$
Find the integer $k$ such that $a_k < 1 < a_{k+1}$. | [] | China | China Western Invitational Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 2012 | |
044s | Suppose $\alpha, \beta \ge 0$, $\alpha + \beta \le 2\pi$. Then the minimum of $\sin \alpha + 2 \cos \beta$ is ______. | [
"When $0 \\le \\alpha \\le \\pi$, $\\sin \\alpha + 2 \\cos \\beta \\ge 0 + 2 \\cdot (-1) = -2$.\n\nWhen $\\pi < \\alpha \\le 2\\pi$, there is $0 \\le \\beta \\le 2\\pi - \\alpha < \\pi$. At this point, as $\\beta$ gets bigger, $\\cos \\beta$ gets smaller. Therefore,\n$$\n\\begin{aligned}\n\\sin \\alpha + 2 \\cos \\... | China | China Mathematical Competition | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | -sqrt(5) | |
03lo | Problem:
Let $f(n, k)$ be the number of ways of distributing $k$ candies to $n$ children so that each child receives at most 2 candies. For example, if $n=3$, then $f(3,7)=0$, $f(3,6)=1$ and $f(3,4)=6$.
Determine the value of
$$
f(2006,1)+f(2006,4)+f(2006,7)+\cdots+f(2006,1000)+f(2006,1003) .
$$ | [
"Solution:\nThe number of ways of distributing $k$ candies to $2006$ children is equal to the number of ways of distributing $0$ to a particular child and $k$ to the rest, plus the number of ways of distributing $1$ to the particular child and $k-1$ to the rest, plus the number of ways of distributing $2$ to the pa... | Canada | 38th Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | ∑_{i=1}^{334} (-1)^i \binom{2005}{i} \binom{3008 - 3i}{2005} | |
0dtd | In triangle $ABC$, $\angle B = 90^\circ$, $AB > BC$, and $P$ is the point such that $BP = BC$ and $\angle APB = 90^\circ$, where $P$ and $C$ lie on the same side of $AB$. Let $Q$ be the point on $AB$ such that $AP = AQ$, and let $M$ be the midpoint of $QC$. Prove that the line through $M$ parallel to $AP$ passes throug... | [
"\nLet $\\angle BPC = \\angle BCP = \\alpha$, $\\angle CBP = \\beta$. As $\\angle QAP = 90^\\circ - \\angle PBA = \\angle CBP = \\beta$. Therefore $\\angle APQ = \\angle AQP = \\alpha$, since $AP = AQ$.\n\nAlso $\\angle CPQ = \\alpha + \\angle BPQ = \\angle APB = 90^\\circ$. Therefore the p... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06hg | Let $\lfloor x \rfloor$ denote the greatest integer not exceeding $x$. Find the last two digits of
$$
\left\lfloor \frac{1}{3} \right\rfloor + \left\lfloor \frac{2}{3} \right\rfloor + \left\lfloor \frac{2^2}{3} \right\rfloor + \cdots + \left\lfloor \frac{2^{2^{2^{14}}}}{3} \right\rfloor
$$ | [
"Note that the remainder when $2^n$ is divided by $3$ is $1$ when $n$ is even, and $2$ when $n$ is odd.\nHence $\\left[\\frac{2^n}{3}\\right] = \\frac{2^n-1}{3}$ when $n$ is even, and $\\left[\\frac{2^n}{3}\\right] = \\frac{2^n-2}{3}$ when $n$ is odd. It follows that\n$$\n\\begin{align*}\nS &= \\left[\\frac{1}{3}\\... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 15 | |
09es | Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that $f(m^3 + f(n)) = f(m)^3 + n$, $\forall n, m \in \mathbb{N}$. | [
"Let's see the given condition as a substitution $P(m, n)$. Adding to both sides $k^3$ and taking $f$ we get:\n\n$$\n\\left.\n\\begin{array}{l}\nf(k^3 + f(m^3 + f(n))) = f(k^3 + f(m)^3 + n) \\\\\nP(k, m^3 + f(n)) \\Rightarrow f(k^3 + f(m^3 + f(n))) = f(k)^3 + m^3 + f(n) \\\\\n\\qquad \\Rightarrow f(k^3 + f(m)^3 + n... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(n) = n for all natural n | |
094m | Problem:
Let $c \geq 4$ be an even integer. In some football league, each team has a home uniform and an away uniform. Every home uniform is coloured in two different colours, and every away uniform is coloured in one colour. A team's away uniform cannot be coloured in one of the colours from the home uniform. There a... | [
"Solution:\n\nWe first give an example of a league with $\\frac{n^{3}}{8}-\\frac{n^{2}}{4}$ teams.\n\nSplit the colours in two sets of size $n / 2$. Let $m = n / 2$ and let $c_{1}, \\ldots, c_{m}$ and $d_{1}, \\ldots, d_{m}$ be the colours in those sets.\n\nConsider all pairs of kits of the form $\\left(\\{c_{i}, c... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | c^3/8 - c^2/4 | |
0hm0 | Problem:
Prove that there exists a polynomial $f(x, y, z)$ with the following property: the numbers $|x|, |y|$, and $|z|$ are the sides of a triangle if and only if $f(x, y, z) > 0$. | [
"Solution:\n$$\nf(x, y, z) = (x + y + z)(-x + y + z)(x - y + z)(x + y - z) \\quad \\left[= x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} - x^{4} - y^{4} - z^{4}\\right]\n$$\nIt is easily seen that the transformation $x \\mapsto -x$, and symmetrically $y \\mapsto -y$ and $z \\mapsto -z$, do not change $f$, so it is enough... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0e0g | Problem:
Naj bo $p$ praštevilo, $a$, $b$ in $c$ pa taka cela števila, deljiva s $p$, da ima polinom
$$
q(x) = x^{3} + a x^{2} + b x + c
$$
vsaj dve različni celi ničli. Dokaži, da $p^{2}$ deli $b$ in $p^{3}$ deli $c$. | [
"Solution:\n\nNaj bosta $y$ in $z$ različni celi ničli polinoma $q$. Tedaj velja $y^{3} + a y^{2} + b y + c = 0$ in $z^{3} + a z^{2} + b z + c = 0$. Ker $p \\mid a$, $p \\mid b$ in $p \\mid c$ ter je $y^{3} = -c - b y - a y^{2}$, $p$ deli $y^{3}$. Podobno sledi, da $p \\mid z^{3}$. Ker je $p$ praštevilo, je zato de... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
09al | Let $M$ and $N$ be points on the side $BC$ of the triangle $ABC$. It is known that $\angle BAM = \angle CAN$ and $AL$ is bisector of the angle $A$. Prove that $\frac{BM}{MC} + \frac{NB}{NC} \ge 2 \cdot \frac{BL}{LC}$. | [
"Let $\\angle BAM = \\angle CAN$ and $\\angle MAN = x$. By the property of bisector $\\frac{BL}{LC} = \\frac{AB}{AC}$. If the area of $\\triangle ABM$ is $S_{ABM}$ and that of $\\triangle ANC$ is $S_{ANC}$ then $S_{ABM} = \\frac{1}{2}AB \\cdot AM \\sin \\varphi$, $S_{ANC} = \\frac{1}{2}AC \\cdot AN \\sin \\varphi$.... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0kc3 | Problem:
Determine all triplets of real numbers $(x, y, z)$ satisfying the system of equations
$$
\begin{aligned}
x^{2} y + y^{2} z & = 1040 \\
x^{2} z + z^{2} y & = 260 \\
(x - y)(y - z)(z - x) & = -540
\end{aligned}
$$ | [
"Solution:\nCall the three equations (1), (2), (3).\n\n(1)/(2) gives $y = 4z$.\n\n(3) $+$ (1) $-$ (2) gives\n$$\n\\left(y^{2} - z^{2}\\right)x = 15z^{2}x = 240\n$$\nso $z^{2} x = 16$.\n\nTherefore\n$$\n\\begin{aligned}\n& z(x + 2z)^{2} = x^{2} z + z^{2} y + 4z^{2} x = \\frac{81}{5} \\\\\n& z(x - 2z)^{2} = x^{2} z +... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (16, 4, 1) and (1, 16, 4) | |
0k47 | Problem:
Sarah stands at $(0,0)$ and Rachel stands at $(6,8)$ in the Euclidean plane. Sarah can only move 1 unit in the positive $x$ or $y$ direction, and Rachel can only move 1 unit in the negative $x$ or $y$ direction. Each second, Sarah and Rachel see each other, independently pick a direction to move at the same t... | [
"Solution:\n\nWe make the following claim: In a game with $n \\times m$ grid where $n \\leq m$ and $n \\equiv m \\pmod{2}$, the probability that Sarah wins is $\\frac{1}{2^{n}}$ under optimal play.\n\nProof: We induct on $n$. First consider the base case $n=0$. In this case Rachel is confined on a line, so Sarah is... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 63/64 | |
09sl | Problem:
Zij $n$ een natuurlijk getal. In een dorp wonen $n$ jongens en $n$ meisjes. Voor het jaarlijkse bal moeten $n$ danskoppels worden gevormd, die elk uit één jongen en één meisje bestaan. Elk meisje geeft een lijstje door, bestaande uit de naam van de jongen met wie ze het liefst zou willen dansen, plus nul of m... | [
"Solution:\n\nNoem bij elk meisje de jongen met wie ze het liefst zou willen dansen, haar lievelingsjongen. We bewijzen de opgave met inductie naar $n$.\n\nAls $n=1$, dan danst het meisje met de enige jongen, dus danst ze met haar lievelingsjongen.\n\nZij nu $k \\geq 1$ en neem aan dat we de opgave bewezen hebben v... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0di4 | Does there exist a function $f : \mathbb{Q} \to \mathbb{Q}$ satisfying
$$
f(x + y + 2f(y)) = \frac{2022}{2023} \cdot y + f(x)
$$
for all $x, y \in \mathbb{Q}$? | [
"The answer is No. Put $k = \\frac{2022}{2023}$ and let $x = 0$ then\n$$\nf(y + a f(y)) = k \\cdot \\frac{y}{a} + f(0)\n$$\nso $f$ is surjective over $\\mathbb{Q}$. Put $x = -a f(y)$, then $f(y) = k \\cdot \\frac{y}{a} + f(-a f(y))$ it is easy to see that $f$ is injective. So $f$ is bijective. Put $y = 0$, then $f(... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | No | |
0avc | Problem:
A railway passes through four towns $A$, $B$, $C$, and $D$. The railway forms a complete loop, as shown on the right, and trains go in both directions. Suppose that a trip between two adjacent towns costs one ticket. Using exactly eight tickets, how many distinct ways are there of traveling from town $A$ and ... | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 128 | |
01dz | Non-negative integers are written in some cells of $100 \times 100$ table. For each $k$, $1 \le k \le 100$, the $k$-th row of the table contains numbers from $1$ to $k$ written in increasing order (from left to right) but not necessarily in consecutive cells. The empty cells are filled with zeroes. Prove that there exi... | [
"Observe that the sum of numbers in the first column is at most $1 \\cdot 100 = 100$, the sum in the first and second columns is at most $1 \\cdot 100 + 2 \\cdot 99$, the sum in the first, second and third columns is at most $1 \\cdot 100 + 2 \\cdot 99 + 3 \\cdot 98$, etc. But the sum of all nonzero numbers equals ... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
00p1 | The incircle of a triangle $A_0B_0C_0$ touches the sides $B_0C_0$, $C_0A_0$, $A_0B_0$ at the points $A$, $B$, $C$, respectively, and the incircle of the triangle $ABC$ with incenter $I$ touches the sides $BC$, $CA$, $AB$ at the points $A_1$, $B_1$, $C_1$, respectively. Let $\sigma(ABC)$ and $\sigma(A_1B_1C)$ be the are... | [
"Let $BC = a$, $CA = b$, $AB = c$ and $2u = a+b+c$. Then $CA_1 = CB_1 = u-c$, $AC_1 = u-a$, $BC_1 = u-b$.\nWe have $\\sigma(ABC) = \\frac{1}{2}ab \\sin \\angle C$ and\n$$\n\\sigma(A_1B_1C) = \\frac{1}{2}(u-c)(u-c) \\sin \\angle C = \\frac{1}{8}(a+b-c)^2 \\sin \\angle C.\n$$\nTherefore $\\sigma(ABC) = 2\\sigma(A_1B_... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry ... | English | proof only | null | |
00ka | Determine all quadruples $(a, b, c, d)$ of real numbers satisfying the following system of equations.
$$
\begin{aligned}
ab + ac &= 3b + 3c \\
bc + bd &= 5c + 5d \\
ac + cd &= 7a + 7d \\
ad + bd &= 9a + 9b
\end{aligned}
$$ | [
"We first note that the first equation can be written in the form $a \\cdot (b+c) = 3 \\cdot (b+c)$ (and the others analogously)\n\n* Case I: $a+b \\ne 0$, $b+c \\ne 0$, $c+d \\ne 0$, $d+a \\ne 0$. In this case we have $(a, b, c, d) = (3, 5, 7, 9)$.\n\n* Case II: $a+b = b+c = c+d = d+a = 0$. In this case we obtain ... | Austria | Austria 2014 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (3, 5, 7, 9); (t, -t, t, -t) for any real t; (-9, 5, -5, 9); (3, -3, 7, -7) | |
099f | Let $M$ be the circumcenter of triangle $ABC$. $C_1, A_1, B_1$ the circumcenters of triangles $ABM, BCM, CAM$ respectively. Prove that, the lines $AA_1, BB_1, CC_1$ intersect in same point.
(proposed by M. Batbileg) | [
"Denote $BC \\cap AA_1 = A_2$, $AC \\cap BB_1 = B_2$, $AB \\cap CC_1 = C_2$. From well known property we have: $\\frac{AB_2}{B_2C} = \\frac{S_{ABB_1}}{S_{BCB_1}}$. In another way:\n$$\nS_{ABB_1} = \\frac{AB \\cdot AB_1 \\cdot \\frac{1}{2} \\sin(\\angle A + \\angle AMC - 90^\\circ)}{BC \\cdot B_1C \\cdot \\frac{1}{2... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Pla... | English | proof only | null | |
0avm | Problem:
Alice, Bob, Charlie and Eve are having a conversation. Each of them knows who are honest and who are liars. The conversation goes as follows:
Alice: Both Eve and Bob are liars.
Bob: Charlie is a liar.
Charlie: Alice is a liar.
Eve: Bob is a liar.
Who is/are honest? | [
"Solution:\n\nWe consider two cases:\n\nCase 1: Alice is honest.\nIf Alice is honest, both Eve and Bob must be liars. If Eve is a liar, then Bob must be honest. This cannot be the case.\n\nCase 2: Alice is a liar.\nIf Alice is liar, then either Eve is honest or Bob is honest. Suppose Eve is honest. Then, Bob is a l... | Philippines | 18th PMO National Stage Oral Phase | [
"Discrete Mathematics > Logic"
] | null | proof and answer | Charlie and Eve | |
0hxd | Problem:
Find all nonnegative integer solutions $(a, b, c, d)$ to the equation
$$
2^{a} 3^{b}-5^{c} 7^{d}=1.
$$ | [
"Solution:\nThe answer is $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$ and $(2,2,1,1)$. The solution involves several cases. It's clear that $a \\geq 1$, otherwise the left-hand side is even. The remainder of the solution involves several cases.\n\n- First, suppose $b=0$.\n- If $c \\geq 1$, then modulo 5 we discover $2^{a... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number... | null | proof and answer | (1,0,0,0), (3,0,0,1), (1,1,1,0), (2,2,1,1) | |
0d2n | The positive integer $a$ is relatively prime with $10$. Prove that for any positive integer $n$, there exists a power of $a$ whose last $n$ digits are $\underbrace{0 \cdots 0}_{n-1} 1$. | [
"This is equivalent to prove that there exists a positive integer $k$ such that $10^{n}$ divides $a^{k}-1$.\n\nFirst solution. Consider the remainders of the division of the $10^{n}+1$ powers $a^{1}, a^{2}, \\ldots, a^{10^{n}+1}$ of $a$ by $10^{n}$. Since there are at most $10^{n}$ possible remainders, by the pigeo... | Saudi Arabia | Preselection tests for the full-time training | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0inq | Problem:
Prove the identity
$$
\sum_{d \mid n} \tau(d)^{3}=\left(\sum_{d \mid n} \tau(d)\right)^{2}
$$ | [
"Solution:\nNote that $\\tau^{3}$ is multiplicative; in light of the convolution property just shown, it follows that both sides of the posed equality are multiplicative. Thus, it would suffice to prove the claim for $n$ a power of a prime. So, write $n=p^{k}$ where $p$ is a prime and $k$ is a nonnegative integer. ... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
02t8 | Problem:
Quantos dígitos possui o número $3^{100}$? Bom, podemos dar uma resposta aproximada para esta pergunta, sem usar a calculadora, simplesmente comparando-o com potências de $10$. Veja que $3^{2}<10$ nos permite concluir que $(3^{2})^{50}=3^{100}<10^{50}$. Então, $3^{100}$ tem no máximo $50$ dígitos pois, $10^{5... | [
"Solution:\n\nVeja que $2^{3}=8$ é menor que $10$, então\n$$\n2^{100}<2^{102}=(2^{3})^{34}<10^{34}\n$$\nComo $10^{34}$ é o primeiro número com $35$ dígitos, $2^{100}$ possui no máximo $34$ dígitos. Além disso, veja que $2^{7}=128$ é maior que $100$. Daí\n$$\n2^{100}>2^{98}=(2^{7})^{14}>(10^{2})^{14}=10^{28}\n$$\nCo... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 29 | |
0asj | Problem:
Find all real values of $x$ that satisfy the equation $x^{x^{2010}} = x^{2010}$. | [
"Solution:\n$\\sqrt[2010]{2010}$\n\n$-\\sqrt[2010]{2010}, 1$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | sqrt[2010]{2010}, -sqrt[2010]{2010}, 1 | |
0kw1 | Problem:
An HMMT party has $m$ MIT students and $h$ Harvard students for some positive integers $m$ and $h$. For every pair of people at the party, they are either friends or enemies. If every MIT student has 16 MIT friends and 8 Harvard friends, and every Harvard student has 7 MIT enemies and 10 Harvard enemies, comp... | [
"Solution:\n\nWe count the number of MIT-Harvard friendships. Each of the $m$ MIT students has 8 Harvard friends, for a total of $8m$ friendships. Each of the $h$ Harvard students has $m-7$ MIT friends, for a total of $h(m-7)$ friendships. So, $8m = h(m-7) \\Longrightarrow mh - 8m - 7h = 0 \\Longrightarrow (m-7)(h-... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 342 | |
01po | Find all pairs $(n; p)$ of natural numbers $n$ and prime numbers $p$ satisfying the equality $p^8 - p^4 = n^5 - n$. | [
"Answer: $(n; p) = (3; 2)$.\n\nIt is clear that $p \\ne n$. If $p = 2$, then $n \\ge 3$ and we have $2^8 - 2^4 = 240 = 3^5 - 3$, i.e. $p = 2$, $n = 3$ is a solution. On the other hand, if $n > 3$, then $n^5 - n = n(n^4 - 1) > 3(3^4 - 1) = 240$, i.e. for $p = 2$ there are no $n$ different from $3$ satisfying the ini... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (n, p) = (3, 2) | |
0jyy | Problem:
Alec wishes to construct a string of $6$ letters using the letters $A$, $C$, $G$, and $N$, such that:
- The first three letters are pairwise distinct, and so are the last three letters;
- The first, second, fourth, and fifth letters are pairwise distinct.
In how many ways can he construct the string? | [
"Solution:\nThere are $4! = 24$ ways to decide the first, second, fourth, and fifth letters because these letters can be selected sequentially without replacement from the four possible letters. Once these four letters are selected, there are $2$ ways to select the third letter because two distinct letters have alr... | United States | HMMT November | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 96 | |
0aau | An acute-angled scalene triangle $ABC$ is given, with $AC > BC$. Let $O$ be its circumcentre, $H$ its orthocentre, and $F$ the foot of the altitude from $C$. Let $P$ be the point (other than $A$) on the line $AB$ such that $AF = PF$, and $M$ be the midpoint of $AC$. We denote the intersection of $PH$ and $BC$ by $X$, t... | [
"It is enough to show that $OF \\perp FX$. Let $OE \\perp AB$, then it is trivial that $CH = 2OE$.\n\nSince from the hypothesis we have $PF = AF$ then we take $PB = PF - BF$ or $PB = AF - BF$. Also, $\\angle XPB = \\angle HAP$ and $\\angle HAP = \\angle HCX$ since $AFGC$ is inscribable (where $G$ is the foot of the... | North Macedonia | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h1m | $M$ and $N$ are chosen on the sides $AD$ and $BC$ of the square $ABCD$, such that $AM = BN$. Point $X$ is a feet of perpendicular from the point $D$ onto $AN$. Prove that angle $MXC$ is right. | [
"Consider the diagonals of $MNCD$, $O$ is the point of its intersection, which is a center of the circle with diameter $DN$ (fig. 17). We have the following equalities:\n$$\n\\angle NXC = \\angle NDC = \\angle MCD = \\angle MXD. \\text{ Hence,}\n$$\n$$\n\\angle MXC = \\angle MXD + \\angle DXC = \\angle CXN + \\angl... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0884 | Problem:
Sia $ABC$ un triangolo acutangolo, e siano $D, E$ i piedi delle altezze uscenti da $A, B$. Siano $A'$ il punto medio di $AD$, $B'$ il punto medio di $BE$. $CA'$ interseca $BE$ in $X$, $CB'$ interseca $AD$ in $Y$. Dimostrare che esiste una circonferenza passante per i punti $A', B', X, Y$. | [
"Solution:\n\nI triangoli $ADC$ e $BEC$ sono simili in quanto triangoli rettangoli con il medesimo angolo in $C$: segue che anche i triangoli $BB'C$ e $AA'C$ risultano simili, e che, in particolare, vale $\\widehat{BB'C} = \\widehat{CA'A}$. Vi sono ora due casi: o il quadrilatero $A'XB'Y$ è intrecciato, o non lo è.... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g85 | 令 $Z, N_0$ 分別表示整數、非負整數所成的集合。試求所有遞增函數 $f : N_0 \to Z$ 滿足
$$
f(2) = 7, f(mn) = f(m) + f(n) + f(m)f(n), \text{ 對所有的 } m, n \in N_0.
$$
Let $Z, N_0$ be the sets of all integers and non-negative integers respectively.
Find all increasing functions $f : N_0 \to Z$ such that
$$
f(2) = 7, \quad f(mn) = f(m) + f(n) + f(m)f(n),... | [
"答：$f(n) = n^3 - 1, n \\in N_0$.\n\n由題設觀察得：$f(0) = f(1) = 0$. 對 $n \\ge 2$, 定 $g(n) = f(n) + 1$. 則\n$g(2) = 8$ 且\n$$\n\\begin{aligned}\ng(mn) &= f(mn) + 1 = f(m) + f(n) + f(m)f(n) + 1 \\\\\n&= (f(m) + 1)(f(n) + 1) \\\\\n&= g(m)g(n), \\text{對所有的 } m, n \\ge 2.\n\\end{aligned}\n$$\n\n固定一整數 $n > 2$, 考慮一有理數列 $\\{p_k/q_... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(n) = n^3 - 1 for all n in N_0 | |
0evo | For an integer $n \ge 3$ and real numbers $a_1, \dots, a_n$ and $b_1, \dots, b_n$, show the following inequality.
$$
\sum_{i=1}^{n} a_i (b_i - b_{i+3}) \le \frac{3n}{8} \sum_{i=1}^{n} \left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \right] \\
(a_{n+1} = a_1 \text{ and } b_{n+1} = b_1 \text{ for } i = 1, 2, 3)
$$ | [
"It suffices to prove the following.\n$$\n\\sum_{i=1}^{n} a_i(b_i - b_{i+1}) \\le \\frac{n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\nBy replacing $b_i$ by $b_{i+j}$ in the above equation and adding up for $j = 0, 1, 2$, we can obtain our desired result. Let\n$$\n\\mathcal{R} = ... | South Korea | The 36th KOREAN MATHEMATICAL OLYMPIAD Final Round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Linear Algebra > Linear transformations",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Intermediate Algebra > Complex numbers"... | English | proof only | null |
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