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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0dtw | For any 4-digit positive integer $n$, define $f(n) = (a+b)^2$, where $a, b$ are the numbers formed by the first two and last two digits of $n$, respectively (leading zeroes are allowed). Find all 4-digit positive integers $n$ such that $f(n) = n$. | [
"Since $(a+b)^2 = f(n) = n = 100a + b$, we have\n$$\n(a+b)(a+b-1) = 99a.\n$$\nBy Euclid's lemma, one of the factors in the LHS is divisible by 11. Also, note that the two factors are coprime, so one of them is not divisible by 3, and thus the other must be divisible by 9.\nHence $a+b \\equiv 0 \\text{ or } 1 \\pmod... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2025, 3025, 9801 | |
0jpy | Problem:
For which prime numbers $p$ is $p^{2}+2$ also prime? Prove your answer. | [
"Solution:\n\nThe answer is $p=3$. This indeed works, since $3^{2}+2=11$.\n\nConsider any other prime number $p \\neq 3$. Then it follows that $p^{2} \\equiv 1 \\pmod{3}$; i.e. that $p$ leaves remainder $1$ or $2$ when divided by $3$. Consequently, $p^{2}+2$ is divisible by $3$. Since $p \\geq 2$, we have $p^{2}+2 ... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | p = 3 | |
04yw | Three players $A$, $B$ and $C$ play the following game. At the beginning of the game, each player has a sheet of paper with the name of the player written on it. Player $A$ chooses one of the other players and replaces the name on this player's sheet with the name on his own sheet. Then player $B$ makes a similar move,... | [
"Player $B$ does not have a winning strategy, since on the first move player $A$ can write the name $A$ on his sheet, after that the name $B$ is not on any of the sheets. Similarly player $C$ does not have a winning strategy.\n\nTo prove that even player $A$ does not have a winning strategy, we show that players $B... | Estonia | Estonija 2010 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No. None of the players has a winning strategy. | |
0i17 | Problem:
Alice plays the following game of solitaire on a $20 \times 20$ chessboard. She begins by placing 100 pennies, 100 nickels, 100 dimes, and 100 quarters on the board so that each of the 400 squares contains exactly one coin. She then chooses 59 of these coins and removes them from the board. After that, she re... | [
"Solution:\n\nAssume that the squares have unit length. Consider, at any time during the game, the perimeter of the region of empty squares (this region may or may not be connected). For example, suppose that at some time 64 squares are empty. If the empty squares are packed together forming an $8 \\times 8$ square... | United States | 2nd Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
070x | Problem:
Find all solutions to $ (m+1)^a = m^b + 1 $ in integers greater than $1$. | [
"Solution:\n\nTaking the equation modulo $m+1$ we get $(-1)^b = -1$, so $b$ is odd. Hence we can divide the right-hand side by $m+1$ to get $m^{b-1} - m^{b-2} + \\ldots - m + 1$. This has an odd number of terms. If $m$ is odd, then each term is odd and so the total is odd, but $(m+1)^{a-1}$ is even (note that $a > ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials ... | null | proof and answer | (m,a,b) = (2, 2, 3) | |
0l88 | For a positive real number $c$, the sequence $a_1, a_2, \dots$ of real numbers is defined as follows. Let $a_1 = c$, and for $n \ge 2$, let
$$
a_n = \sum_{i=1}^{n-1} (a_i)^{n-i+1}.
$$
Find all positive real numbers $c$ such that $a_i > a_{i+1}$ for all positive integers $i$. | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 0 < c < (\sqrt{5} - 1)/2 | |
04e8 | Let $0 < a < b < c < d$ and let each of the quadratic functions $p(x) = x^2 + dx + a$ and $q(x) = x^2 + cx + b$ have two different real roots. Prove that all four roots are mutually different. (Russia 1996) | [
"Let the roots of $p(x) = x^2 + dx + a$ be $x_1$ and $x_2$, and the roots of $q(x) = x^2 + cx + b$ be $y_1$ and $y_2$.\n\nSince $p(x)$ has two different real roots, its discriminant is positive:\n$$\nD_p = d^2 - 4a > 0\n$$\nSimilarly, $q(x)$ has two different real roots, so\n$$\nD_q = c^2 - 4b > 0\n$$\n\nSuppose, f... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0bsn | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that $f^2$ is differentiable and $(f^2)' = f$. | [
"$$\na = \\inf \\{ t \\in (-\\infty, x_0) \\mid f^2(x) > 0, \\forall x \\in [t, x_0] \\},\n$$\n\n$$\nb = \\sup \\{t \\in (x_0, \\infty) \\mid f^2(x) > 0, \\forall x \\in [x_0, t]\\}.\n$$\nWe have $-\\infty \\le a < x_0 < b \\le \\infty$ and $f(x) \\ne 0, \\forall x \\in (a, b)$. Since $f = (f^2)'$ has the intermedi... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All solutions are:
- f(x) ≡ 0;
- f(x) = { 0, for x < a; (x − a)/2, for x ≥ a }, for any real a;
- f(x) = { (x − b)/2, for x ≤ b; 0, for x > b }, for any real b;
- f(x) = { (x − b)/2, for x ≤ b; 0, for x ∈ (b, a); (x − a)/2, for x ≥ a }, for any reals b < a;
- f(x) = x/2 + c for all real x, for any real c. | |
06qm | For any integer $n \geq 2$, we compute the integer $h(n)$ by applying the following procedure to its decimal representation. Let $r$ be the rightmost digit of $n$.
(1) If $r=0$, then the decimal representation of $h(n)$ results from the decimal representation of $n$ by removing this rightmost digit $0$.
(2) If $1 \leq ... | [
"We identify integers $n \\geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from $0$ to $9$. We recursively define ten functions $f_{0}, \\ldots, f_{9}$ that map some strings into integers for $k=9,8, \\ldots, 1,0$.... | IMO | IMO Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
00j5 | Determine all pairs $(a, b)$ of non-negative integers, such that $a^b + b$ divides $a^{2b} + 2b$. (Note that $0^0 = 1$ holds.) | [
"For brevity, we name $n = a^b + b$ and $m = a^{2b} + 2b$.\n\nFor $a = b = 0$, we obtain $n = m = 1$, and therefore $n|m$. We see that $(0, 0)$ is a solution.\n\nFor $a = 0$ and $b > 0$, we obtain $n = b$ and $m = 2b$, and again $n|m$. We see that $(0, b)$ is, in fact a solution for all values of $b$.\n\nFor $a = 1... | Austria | AustriaMO2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | {(a,0) for all nonnegative integers a} ∪ {(0,b) for all nonnegative integers b} ∪ {(2,1)} | |
0ccn | Let $a, b \in \mathbb{R}$, with $a < b$, be two arbitrary real numbers. We say that a function $f : [a, b] \to \mathbb{R}$ has property ($\mathcal{P}$) if it is an integrable function on $[a, b]$, such that
$$
f(x) - f\left(\frac{x+a}{2}\right) = f\left(\frac{x+b}{2}\right) - f(x) \quad \text{for any } x \in [a, b].
$$... | [
"We will show that the functions with property ($\\mathcal{P}$) are precisely the constant functions on the interval $[a, b]$.\n\nThe equality in the statement can be rewritten equivalently as\n$$\nf(x) = \\frac{1}{2} \\left( f\\left(\\frac{x+a}{2}\\right) + f\\left(\\frac{x+b}{2}\\right) \\right) \\quad (1)\n$$\nW... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = t/(b - a) for all x in [a, b] | |
0gyw | On a plane are placed two triangles $ABC$ and $BKL$ such that segment $AK$ is divided into three equal parts by the intersection point of the medians of triangle $ABC$ and the intersection point of the bisectors of triangle $BKL$ ($AK$ is a median of $ABC$, $KA$ is a bisector of $ABKL$), and quadrilateral $KALC$ is a t... | [
"Let $I$ be the center of $ABKL$, contextual from the characteristic of the centroid of the triangle and from the condition of the task: $AI = x$, $KI = 2x$. Let $BA = x$, then from the characteristic of the bisector $BK = 2x$ and $KC = 2x$. Then $KALC$ is a trapezium. That's why $KA \\parallel LC$ (see Fig. 20). F... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | All angles of triangle BKL are 60° | |
0avs | Problem:
Find all real numbers $x$ that satisfies
$$
\frac{x^{4}+x+1}{x^{4}+x^{2}-x-4}=\frac{x^{4}+1}{x^{4}+x^{2}-4}
$$ | [
"Solution:\nLet $f(x)=x^{4}+x+1$ and $g(x)=x^{4}+x^{2}-x-4$. The equation is then equivalent to\n$$\n\\frac{f(x)}{g(x)}=\\frac{f(x)-x}{g(x)+x} \\Longleftrightarrow x(f(x)+g(x))=0\n$$\nHence, $x=0$ or $f(x)+g(x)=2x^{4}+x^{2}-3=(2x^{2}+3)(x^{2}-1)=0$ which gives $-1,0,1$ as acceptable values of $x$."
] | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | [-1, 0, 1] | |
08hu | Problem:
2003 denars were divided in several bags and the bags were placed in several pockets. The number of bags is greater than the number of denars in each pocket. Is it true that the number of pockets is greater than the number of denars in one of the bags? | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Yes | |
05ax | A point $X$ is chosen on the median $AD$ of triangle $ABC$. The circumcircle of triangle $ABX$ intersects the median $BE$ of triangle $ABC$ at point $Y \neq B$. The circumcircle of triangle $EXY$ intersects the line $DE$ at point $K \neq E$. Prove that the location of $K$ does not depend on $X$. | [
"We will use directed angles (Fig. 3 and 4 depict both possible configurations). As $DE$ is a midline in $ABC$, we have $DE \\parallel AB$. Thus\n$$\n\\angle DEY = \\angle DEB = \\angle ABE = \\angle ABY = \\angle AXY = \\angle DXY.\n$$\nTherefore points $D$, $E$, $X$, $Y$ are concyclic. This means that the circumc... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dfg | Prove that for any 2 positive integers $m$ and $n$ with $(m, n) \neq (1, 1)$ the value of expression
$$
\frac{1}{m} + \frac{1}{m+1} + \dots + \frac{1}{m+n-1}
$$
is not an integer. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0gby | 設 $S$ 為一個有限集合, 且令 $\mathcal{A}$ 為所有由 $S$ 映至 $S$ 之函數所成的集合。令 $f$ 是 $\mathcal{A}$ 的一個元素且令 $T = f(S)$ 表示 $S$ 在 $f$ 之下所對應之值域。假設對每一個 $\mathcal{A}$ 中的元素 $g$ 且 $g \neq f$, 皆滿足 $f \circ g \circ f \neq g \circ f \circ g$.
試證: $f(T) = T$.
Let $S$ be a finite set, and let $\mathcal{A}$ be the set of all functions from $S$ to $S$.... | [
"For $n \\ge 1$, denote the $n$-th composition of $f$ with itself by\n$$\nf^n \\stackrel{\\text{def}}{=} f \\circ f \\circ \\dots \\circ f.\n$$\nBy hypothesis, if $g \\in \\mathcal{A}$ satisfies $f \\circ g \\circ f = g \\circ f \\circ g$, then $g = f$. A natural idea is to try to plug in $g = f^n$ for some $n$ in ... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Other"
] | null | proof only | null | |
093z | Problem:
Let $ABC$ be a triangle and let $M$ be the midpoint of the segment $BC$. Let $X$ be a point on the ray $AB$ such that $2 \angle CXA = \angle CMA$. Let $Y$ be a point on the ray $AC$ such that $2 \angle AYB = \angle AMB$. The line $BC$ intersects the circumcircle of the triangle $AXY$ at $P$ and $Q$, such that... | [
"Solution:\n\nLet $P'$ and $Q'$ be points on line $BC$ such that $P'M = AM = Q'M$ and $B$ and $P'$ are on the same side of the line $AM$.\nWe have $\\angle P'AM = \\angle MP'A = \\frac{1}{2} \\angle CMA = \\angle CXA$. Therefore points $P', A, C, X$ are concyclic.\n\n\n\nThis gives us that ... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bpz | Problem:
Determinaţi valorile lui $x \in \mathbb{Z}$ pentru care $\sqrt[3]{x^{3}-6 x^{2}+12 x+29} \in \mathbb{Q}$. | [
"Solution:\n\nPentru ca radicalul de ordin 3 dintr-un număr întreg să fie raţional, trebuie ca numărul de sub radical să fie cub perfect.\n\nDeci există $y \\in \\mathbb{Z}$ astfel încât $x^{3}-6 x^{2}+12 x+29 = y^{3} \\Leftrightarrow (x-2)^{3} + 37 = y^{3} \\Leftrightarrow y^{3} - (x-2)^{3} = 37 \\Leftrightarrow (... | Romania | Olimpiada de Matematică | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | {-2, 5} | |
01mj | Find all positive integer $m$ so that the number $7 \cdot 2^m$ may be presented as the sum of the squares of three positive integers. | [] | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | All odd positive integers m | |
0dz0 | Let $\triangle ABC$ be an isosceles triangle with the apex at $B$ and choose a point $D$ on the altitude from $B$ such that $AC$ is tangent to the circumcircle $\mathcal{K}$ of the triangle $ABD$. Let $E$ be a point on $\mathcal{K}$ such that the chord $DE$ is perpendicular to the chord $AB$. Prove that triangles $ABE$... | [
"Let $T$ be the intersection of the chords $DE$ and $AB$. We know that $DTB$ is a right triangle. Let $\\angle CAB = \\angle ACB = \\alpha$. The line $AC$ is tangent to the circumcircle, so the angle $\\angle CAB$ is equal to the corresponding angle $\\angle AEB$ over the chord $AB$. Thus, $\\angle AEB = \\alpha$.\... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ais | Let $AH_A, BH_B, CH_C$ be the heights in $\triangle ABC$. We draw perpendiculars $p_A, p_B, p_C$ through the vertices $A, B, C$ to $H_B H_C, H_C H_A, H_A H_B$, respectively. Prove that $p_A, p_B, p_C$ pass through the same point. | [
"Let $O$ be the center of the circumscribed circle around $\\triangle ABC$. We will show that each of the lines $p_A, p_B, p_C$ passes through $O$.\n\nBecause of symmetry, it is enough to show that $OC \\perp H_A H_B$.\n\nLet $D$ be the point of intersection of these two lines. We restrict ourselves to the case whe... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous... | English | proof only | null | |
0fjs | Problem:
Colocamos, formando una circunferencia, 2004 fichas bicolores: blancas por una cara y negras por la otra. Un movimiento consiste en elegir una ficha con la cara negra hacia arriba y dar la vuelta a tres fichas: la elegida, la de su derecha, y la de su izquierda. Supongamos que inicialmente hay una sola ficha ... | [
"Solution:\n\nNumeremos las fichas desde $1$ hasta $2004$: la $1$ es negra y las restantes son blancas. Cada ficha inicialmente blanca debe ser tocada un número par de veces, para que al final del proceso siga teniendo la cara blanca hacia arriba. Cada movimiento posible cambia el número de fichas negras en un núme... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | For 2004 tokens: impossible. For 2003 tokens: possible. | |
01b4 | Find all functions $f$ defined on all real numbers and taking real values such that
$$ f(f(y)) + f(x - y) = f(xf(y) - x) $$
for all real numbers $x, y$. | [
"Answer: $f(x) = 0$.\n\nSubstituting $x = y = 0$ to the original equality gives $f(f(0)) + f(0) = f(0)$, implying\n$$\nf(f(0)) = 0. \\tag{1}\n$$\nTaking $x = \\frac{f(0)}{2}$ and $y = f(0)$ in the original equality gives\n$$\nf(f(f(0))) + f\\left(-\\frac{f(0)}{2}\\right) = f\\left(\\frac{f(0)}{2} \\cdot f(f(y)) - \... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 | |
0iv7 | Problem:
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log_{2}(S)$? | [
"Solution:\nThe sum of all the coefficients is $(1+i)^{2009}$, and the sum of the real coefficients is the real part of this, which is $\\frac{1}{2}\\left((1+i)^{2009}+(1-i)^{2009}\\right)=2^{1004}$. Thus $\\log_{2}(S)=1004$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1004 | |
0gbk | 有 $n$ 座島, 其中 $n \ge 3$. 一間渡輪公司在其中若干對島嶼間架設航線, 使得當我們將這些島嶼分成非空的兩群時, 總是可以從兩組中各找出一座島, 使得這兩座島之間有航線。每年年底, 公司將會挑選有航線相連的島 $X$ 和 $Y$, 並關閉其間的航線。同時, 對於所有其他島嶼 $A$, 若 $A$ 和 $X$ 有航線但和 $Y$ 沒有航線, 則公司將增設 $A$ 和 $Y$ 之間的航線; 反之, 若 $A$ 和 $Y$ 有航線但和 $X$ 沒有航線, 則公司將增設 $A$ 和 $X$ 之間的航線。
對於兩群島 $A$ 和 $B$, 若 $A \in A$ 且 $B \in B$, 則我們稱連結 $A$ 和 $B$ 的航... | [
"我們稱兩群島 $\\mathcal{A}$ 和 $\\mathcal{B}$ 構成一個網路, 若且唯若對於任何 $A \\in \\mathcal{A}$ 和 $B \\in \\mathcal{B}$, $A$ 和 $B$ 之間都有航班。\n\n1. 先任選一條航線, 令兩端點為島 $A$ 和 $B$. 由於 $n \\ge 3$, 必存在第三座島 $C$; 且由題目假設, 它必然與 $A$ 或 $B$ 相連, 不失一般性假設為與 $A$ 相連。定義 $\\mathcal{A}_0 = \\{A\\}$, $\\mathcal{B}_0 = \\{B, C\\}$; 易知 $(\\mathcal{A}_0, \\math... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
04yu | Juku drew a regular hexagon and chose three triangles with different areas whose vertices were among the vertices of the hexagon. Prove that the sum of the areas of the triangles is equal to the area of the hexagon. | [
"Any triangle whose vertices are among the vertices of a regular hexagon is one of the following:\n* a triangle $\\Delta_1$ whose vertices are three consecutive vertices of the hexagon;\n* a triangle $\\Delta_2$ whose two vertices are adjacent vertices of the hexagon and the third one is adjacent to none of the fir... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
09kq | Let $ABC$ be a triangle with points $D$ and $E$ taken on side $BC$ such that $D$ is between $B$ and $E$. A point $F$ lies on segment $AD$ and satisfies the condition $\frac{AF}{FD} = \frac{BE}{EC}$. The circumcircle of triangle $ABD$ intersects segment $AE$ at a point $G$ (distinct from $A$). If lines $EF$ and $GC$ are... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
05ir | Problem:
Trois sauterelles se trouvent aux points $(0,0)$, $(0,1)$ et $(1,0)$ d'une feuille quadrillée. Chaque minute une sauterelle saute sur un autre point de la grille d'une telle façon que son saut soit parallèle à la droite passant par les deux autres sauterelles. Est-il possible qu'au bout d'un certain temps les... | [
"Solution:\n\nSi $AB$ et $CD$ sont deux droites parallèles, alors l'aire du triangle $ABC$ est égale à celle du triangle $ABD$, car ces deux triangles ont la même base $[AB]$ et la même hauteur. Il découle de cette remarque que l'aire du triangle formé par les trois sauterelles ne change pas lors de leurs sauts. Or... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | No | |
03li | Problem:
Let $T$ be the set of all positive integer divisors of $2004^{100}$. What is the largest possible number of elements that a subset $S$ of $T$ can have if no element of $S$ is an integer multiple of any other element of $S$ ? | [
"Solution:\nAssume throughout that $a, b, c$ are nonnegative integers. Since the prime factorization of $2004$ is $2004 = 2^{2} \\cdot 3 \\cdot 167$,\n$$\nT = \\left\\{ 2^{a} 3^{b} 167^{c} \\mid 0 \\leq a \\leq 200, 0 \\leq b, c \\leq 100 \\right\\}\n$$\nLet\n$$\nS = \\left\\{ 2^{200-b-c} 3^{b} 167^{c} \\mid 0 \\le... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 10201 | |
0adu | Во координатната рамнина $xOy$ е дадена права $4x+3y=n$, $n>0$ која од координатниот почеток е на растојание 12. Определи ја плоштината на триаголникот што ја гради правата со координатните оски.
 | [
"Правата со координатните оски гради правоаголен триаголник со катети $\\overline{OA} = a = \\frac{n}{4}$ и $\\overline{OB} = b = \\frac{n}{3}$ (види цртеж). Според тоа должината на хипотенузата е еднаква на\n\n\n\n$\\overline{AB} = c = \\sqrt{a^2 + b^2} = \\sqrt{\\frac{n^2}{16} + \\frac{n^... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | Macedonian, English | proof and answer | 150 | |
0hjk | Problem:
Find all positive prime numbers $p$ such that $p+2$ and $p+4$ are prime as well. | [
"Solution:\n\nFor $p=3$, $p+2=5$, $p+4=7$ and these are obviously prime.\n\nFor $p>3$, we know that $p$ is not divisible by $3$. The remainder of $p$ when divided by $3$ can be either $1$ or $2$. If it is $1$, then $p+2$ is divisible by $3$; if it is $2$, then $p+4$ is divisible by $3$. Hence $p=3$ is the only solu... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 3 | |
04jq | Let $n$ be a positive integer. Each of the numbers $n$, $n+1$, $n+2$, $\dots$, $2n-1$ has a largest odd divisor. Determine the sum of these largest odd divisors. | [] | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n^2 | |
02nw | Let $n$ be an integer and $n_1$ be one of its divisors. Let $A$ be a $n \times n$ symmetric matrix defined by $a_{i,i} = 4$, $a_{i,i+1} = a_{i+1,i} = -1$ for all $i$ such that $1 \le i \le n-1$ and $i+1$ is not a multiple of $n_1$, $a_{i,i+n_1} = a_{i+n_1,i} = -1$ and $a_{i,j} = 0$ otherwise.
Prove that $A$ has an inve... | [
"Let's find the inverse of $A/4 = I - X$, where all entries in $X$ are either $0$ or $1/4$. We will use the series\n$$\n(I - X)^{-1} = I + X + X^2 + X^3 + \\dots\n$$\nFirst let's prove that this series converges. It suffices to show that the maximum $M$ such that $\\|Xw\\| \\le M\\|w\\|$ for all column vectors $w$ ... | Brazil | Brazilian Math Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
00vq | Let $\triangle ABC$ be a triangle with $AB < AC$, and let $D$ be a point on the side $AC$ such that $AD = \frac{AC-AB}{2}$. Points $X$ and $Y$ are chosen on the line through $A$ parallel to $BC$ such that $BX = CY$ and line $AC$ is tangent to the circumcircle of $\triangle XDY$. Prove that the tangents to the circumcir... | [
"\nDenote by $l$ the line through $A$ parallel to $BC$.\nAssume without loss of generality that $X$ is closer to $A$ than $Y$. First we show that points $X$ and $Y$ are uniquely defined. Suppose that $X'$ and $Y'$ are also points on $l$ such that $BX' = CY'$ and that $AC$ is tangent to $(X'... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point c... | English | proof only | null | |
03e7 | Given is an acute triangle $ABC$ with incenter $I$ and the incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$. The circle with center $C$ and radius $CE$ meets $EF$ for the second time at $K$. If $X$ is the $C$-excircle touchpoint with $AB$, show that $CX$, $KD$, $IF$ concur.
(Kristyan Vasilev) | [
"We claim the concurrency point is the $F$-antipode $F'$. It is well-known that this is $\\overline{CX} \\cap \\overline{IF}$.\nLet $P = \\overline{EF} \\cap \\overline{DF'}$ and $Q = \\overline{DF} \\cap \\overline{EF'}$. Then since $\\angle PEQ = \\angle PDQ = 90^\\circ$, $DEPQ$ is cyclic.\nNow, we have\n$$\n\\be... | Bulgaria | Autumn tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0ik7 | Problem:
If $a$, $b$ are nonzero real numbers such that $a^{2} + b^{2} = 8 a b$, find the value of $\left|\frac{a+b}{a-b}\right|$. | [
"Solution:\nNote that\n$$\n\\left|\\frac{a+b}{a-b}\\right| = \\sqrt{\\frac{(a+b)^2}{(a-b)^2}} = \\sqrt{\\frac{a^2 + b^2 + 2 a b}{a^2 + b^2 - 2 a b}} = \\sqrt{\\frac{10 a b}{6 a b}} = \\frac{\\sqrt{15}}{3}\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | √15/3 | |
0jj9 | Problem:
Given a regular pentagon of area $1$, a pivot line is a line not passing through any of the pentagon's vertices such that there are $3$ vertices of the pentagon on one side of the line and $2$ on the other. A pivot point is a point inside the pentagon with only finitely many non-pivot lines passing through it... | [
"Solution:\n\nLet the pentagon be labeled $A B C D E$. First, no pivot point can be on the same side of $A C$ as vertex $B$. Any such point $P$ has the infinite set of non-pivot lines within the hourglass shape formed by the acute angles between lines $P A$ and $P C$. Similar logic can be applied to points on the s... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 1/2 (7 - 3√5) | |
09vj | Problem:
Vind alle functies $f: \mathbb{Z} \rightarrow \mathbb{Z}$ die voldoen aan de volgende twee voorwaarden:
(i) voor alle gehele getallen $x$ geldt $f(f(x))=x$;
(ii) voor alle gehele getallen $x$ en $y$ zodat $x+y$ oneven is geldt er dat $f(x)+f(y) \geq x+y$. | [
"Solution:\n\nDe functie $f(x)=x$ voor alle $x$ voldoet. Neem nu verder aan dat niet voor alle $x$ geldt $f(x)=x$. Vanwege (i) is er dan zowel een waarde van $x$ met $f(x)>x$ als een waarde van $x$ met $f(x)<x$. We nemen nu een $a \\in \\mathbb{Z}$ met $f(a)<a$ en bekijken een willekeurige $x$ met $x \\not\\equiv a... | Netherlands | IMO-selectietoets III | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All solutions are either the identity function f(x) = x for all integers x, or the family f(x) = x + d for even x and f(x) = x − d for odd x, where d is any odd integer. | |
0iwh | Problem:
A circle $\omega_{1}$ of radius $15$ intersects a circle $\omega_{2}$ of radius $13$ at points $P$ and $Q$. Point $A$ is on line $PQ$ such that $P$ is between $A$ and $Q$. $R$ and $S$ are the points of tangency from $A$ to $\omega_{1}$ and $\omega_{2}$, respectively, such that the line $AS$ does not intersect... | [
"Solution:\n\nLet $O_{1}$ be the center of $\\omega_{1}$ and $O_{2}$ be the center of $\\omega_{2}$. Then $O_{1}O_{2}$ and $PQ$ are perpendicular. Let their point of intersection be $X$. Using the Pythagorean theorem, the fact that $PQ = 24$, and our knowledge of the radii of the circles, we can compute that $O_{1}... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | 14 + sqrt(97) | |
0dwp | Problem:
Poišči vsa realna števila $x$ in $y$, za katera je
$$
x^{3}-y^{3}=7(x-y) \quad \text{ in } \quad x^{3}+y^{3}=5(x+y)
$$ | [
"Solution:\n\nEnačbi preoblikujemo v $(x-y)\\left(x^{2}+x y+y^{2}\\right)=7(x-y)$ in $(x+y)\\left(x^{2}-x y+y^{2}\\right)=5(x+y)$. Če je $x=y$, iz druge enačbe dobimo $2 x^{3}=10 x$, torej $x=0$ ali $x= \\pm \\sqrt{5}$, v nasprotnem primeru pa lahko prvo enačbo delimo z $x-y$ in dobimo\n$$\nx^{2}+x y+y^{2}=7\n$$\nČ... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (0,0), (sqrt(5), sqrt(5)), (-sqrt(5), -sqrt(5)), (sqrt(7), -sqrt(7)), (-sqrt(7), sqrt(7)), (sqrt(2)+1, sqrt(2)-1), (sqrt(2)-1, sqrt(2)+1), (-sqrt(2)-1, -sqrt(2)+1), (-sqrt(2)+1, -sqrt(2)-1) | |
0jo8 | Problem:
Determine, with proof, the value of $\log_{2} 3 \log_{3} 4 \log_{4} 5 \ldots \log_{255} 256$. | [
"Solution:\nWe use the fact that $\\log_{b} a = \\frac{\\log a}{\\log b}$. Thus, the product equals\n$$\n\\frac{\\log 3}{\\log 2} \\frac{\\log 4}{\\log 3} \\ldots \\frac{\\log 256}{\\log 255} = \\frac{\\log 256}{\\log 2} = \\log_{2}(256) = 8\n$$"
] | United States | Berkeley Math Circle | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 8 | |
0ejo | Problem:
Dan je paralelogram $ABCD$ z dolžinama stranic $|AB|=7~\mathrm{cm}$ in $|BC|=5~\mathrm{cm}$. Višina na stranico $AB$ meri $4~\mathrm{cm}$ in kot $\alpha=\angle BAD$ je oster.
a) Z ravnilom in šestilom konstruiraj dani paralelogram in trikotniku $BCD$ očrtaj krožnico. Središče trikotniku očrtane krožnice ozna... | [
"Solution:\n\na)\nS šestilom in ravnilom konstruiramo paralelogram. Trikotniku $BCD$ očrtamo krog in središče kroga označimo s $S$.\n\nb)\nUdrti kot $\\angle BSD$ meri $254^\\circ$, vbočeni kot $\\angle BSD$ meri $106^\\circ$, saj skupaj tvorita polni kot. Vbočeni kot $\\angle BSD$ je središčni kot, kot $\\angle BC... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Odbirno tekmovanje | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Interior angles of the parallelogram: 53° and 127°. Obtuse angle between the perpendicular bisector of AD and the altitude to AB: 127°. | |
04ik | Let $O$ be the circumcentre and $H$ be the orthocentre of the triangle $ABC$. Ray $AO$ intersects the circumcircle of the triangle $ABC$ in point $D$. Prove that the line $HD$ passes through the midpoint of the segment $\overline{BC}$. | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
06vp | Let $\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\ell \in \mathcal{L}$ a point $f(\ell)$ on $\ell$. Suppose that for any point $X$, and for any three lines $\ell_{1}, \ell_{2}, \ell_{3}$ passing through $X$, the points $f\left(\ell_{1}\right), f\left(\ell_{2}\r... | [
"Write $\\angle\\left(\\ell_{1}, \\ell_{2}\\right)$ for the directed angle modulo $180^{\\circ}$ between the lines $\\ell_{1}$ and $\\ell_{2}$. Given a point $P$ and an angle $\\alpha \\in\\left(0,180^{\\circ}\\right)$, for each line $\\ell$, let $\\ell^{\\prime}$ be the line through $P$ satisfying $\\angle\\left(\... | IMO | IMO 2019 Shortlisted Problems | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
039v | Find all pairs of positive integers $(m, n)$, $m \le n$, such that there exists an $m \times n$ table of zeros and ones which satisfy the following condition:
If there is a zero (resp. one) in a cell, then the number of the zeros (resp. ones) in the row of that cell is equal to the number of the zeros (resp. ones) in t... | [
"Denote by $a_{pq}$ the number in the $p$-th row and $q$-th column, by $A_0(p)$ (resp. $A_1(p)$) the number of the zeros (resp. ones) in the $p$-th row, and by $B_0(q)$ (resp. $B_1(q)$) the number of the zeros (resp. ones) in the $q$-th column.\n\n**Lemma.** If exactly three of the numbers $a_{ik}$, $a_{il}$, $a_{j... | Bulgaria | Fall Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | English | proof and answer | All pairs of positive integers with n = m or n = 2m. | |
0adw | Правоаголна метална плочка има маса 10g. Таа треба да се раздели (расече) на три дела кои имаат целобројна маса. Со добиените три дела може да се измери во грамови секоја маса од 1 до 10 грама која е природен број. Колкава треба да биде масата на секој од делбените делови делови ? | [
"Деловите на кои треба да се пресече плочката се $2$g, $3$g, $5$g ($2+3+5=10$). Масите од $2$g, $3$g, $5$g, $7$g, $8$g, $10$g можат да се измерат директно. Масата од $1$g ќе биде најмала од сите, па кога ќе ја ставиме на вага со $2$g, страната со $2$g ќе натежне. Масата од $4$g кога ќе ја ставиме на вага со маса од... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Math Word Problems"
] | Macedonian, English | proof and answer | 2 g, 3 g, 5 g | |
07xp | $$
a_1 = 24, \quad a_2 = 2025, \quad \text{and} \quad a_{n+2} = 3 \cdot a_{n+1} \cdot a_n^4 \quad \text{for all } n \ge 1.
$$
Determine all $n \ge 1$ for which there exists an integer $x$ such that $a_n = x^3$. | [
"After factorising $a_1$ and $a_2$ into prime numbers, $a_1 = 2^3 \\cdot 3$ and $a_2 = 3^4 \\cdot 5^2$, we see that we can write $a_n = 2^{\\alpha_n} \\cdot 3^{\\beta_n} \\cdot 5^{\\gamma_n}$, where for all $n \\ge 1$\n$$\n\\begin{align*}\n\\alpha_1 &= 3 & \\alpha_2 &= 0 & \\alpha_{n+2} &= \\alpha_{n+1} + 4\\alpha_... | Ireland | IRL_ABooklet_2025 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All n of the form n = 8k + 5 for k ≥ 0 | |
0cas | Let $a$ and $b$ be two distinct positive integers having the same parity.
Prove that $\frac{a!+b!}{2^a}$ is not an integer (note: $n! = 1 \cdot 2 \cdot \dots \cdot n$, for any positive integer $n$). | [
"First, we prove that for any positive integer $n$, $2^n$ is not a divisor of $n!$.\nAssume the opposite; then, in the prime factorization of $n!$ there should be at least $n$ factors equal to $2$. The exponent of $2$ in $n!$ is equal to $\\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{2^2} \\rfloor + \\dots + ... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FOURTH SELECTION TEST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
07at | Let $I$ be the incenter of triangle $ABC$. $X$ is a point on arc $BC$ of the circumcircle of triangle $ABC$ such that if $E$ and $F$ are the feet of the perpendiculars from $X$ to $BI$ and $CI$, respectively, and $M$ is the midpoint of $EF$, then $MB = MC$. If $D$ is the foot of the perpendicular from $I$ to $BC$, show... | [
"First, we prove two lemmas.\n\n**Lemma 1.** Let $l_1$ and $l_2$ be two lines and $A$, $B$ and $C$ three points in the plane. Denote by $A_1$ and $A_2$ the perpendicular projections of $A$ on $l_1$ and $l_2$, respectively. Furthermore, let $A'$ be the midpoint of $A_1A_2$. $B'$ and $C'$ are defined similarly. $A$, ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tang... | English | proof only | null | |
0edf | Find all positive integers $a, b, c$ and $d$ such that
$$
4^a \cdot 5^b - 3^c \cdot 11^d = 1.
$$ | [
"Consider the equation modulo $3$. We get $(-1)^b \\equiv 1 \\pmod{3}$, so $b$ must be even and we can write $b = 2b_1$. The equation can now be rearranged into\n$$\n2^{2a} \\cdot 5^{2b_1} - 1 = 3^c \\cdot 11^d \\Rightarrow (2^a \\cdot 5^{b_1} - 1)(2^a \\cdot 5^{b_1} + 1) = 3^c \\cdot 11^d.\n$$\nWe have $\\gcd(2^a ... | Slovenia | Slovenija 2016 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (1, 2, 2, 1) | |
0fdj | Problem:
Se considera la función $f: \mathbb{N} \rightarrow \mathbb{Z}$ definida como sigue:
$$
f(n)= \begin{cases}-f\left(\frac{n}{2}\right) & \text{ si } n \text{ es par } \\ f(n-1)+1 & \text{ si } n \text{ es impar }\end{cases}
$$
para $n \geq 0$. Demostrar que $f(n)$ es múltiplo de 3 si, y sólo si, $n$ es múltiplo... | [
"Solution:\n\nDe la definición de $f$ se sigue que $f\\left(2^{a}\\right)=(-1)^{a}$, para todo $a \\geq 0$. Siendo $a>b \\geq 0$, entonces\n$$\nf\\left(2^{a}+2^{b}\\right)=f\\left(2^{b}\\left(2^{a-b}+1\\right)\\right)=(-1)^{b}\\left[(-1)^{a-b}+1\\right]=(-1)^{a}+(-1)^{b}\n$$\ny, en general, si $a_{1}>\\cdots>a_{k} ... | Spain | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof and answer | (4^{2017}-1)/3 | |
04dl | Let $a$ and $b$ be the legs, and $c$ the hypotenuse of a right triangle.
Prove that the following inequality holds
$$
\left(1 + \frac{c}{a}\right) \left(1 + \frac{c}{b}\right) \geqslant 3 + 2\sqrt{2}.
$$ | [
"By rearranging the left hand side of the inequality we get\n$$\n\\begin{aligned}\n\\left(1 + \\frac{c}{a}\\right) \\left(1 + \\frac{c}{b}\\right) &= \\frac{(a+c)(b+c)}{ab} = \\frac{c^2 + ab + c(a+b)}{ab} \\\\\n&= \\frac{c^2}{ab} + 1 + \\frac{c(a+b)}{ab}\n\\end{aligned}\n$$\nSince the triangle is right it follows t... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0gez | (a) 證明：對於任意兩個互質的正整數 $a, b$, 總是存在正整數 $x, y$ 使得 $ax^n + by^n$ 是 $n$-好數。
(b) 證明：對於任意滿足 $\gcd(a_1, \dots, a_k) = 1$ 的 $k$ 個正整數 $a_1, \dots, a_k$, 總是存在正整數 $x_1, \dots, x_k$ 使得 $a_1x_1^n + a_2x_2^n + \dots + a_kx_k^n$ 是 $n$-好數。(註：$a_1, \dots, a_k$ 不必兩兩相異。)
Problem:
II-N. Let $n$ be a given positive integer. We say that a po... | [
"**解.** We first prove (a). Let $N, C_1, C_2$ be some positive integers to be determined. Let $P$ be the product of all primes at most $\\max\\{a, b, 2n+4\\}$. Let $S_1 = \\{Pn + C_1 : n \\in [N]\\}$ and $S_2 = \\{Pn + C_2 : n \\in [N]\\}$. For each prime $p \\in (\\max\\{a, b, 2n+4\\}, \\sqrt{N})$, we first count ... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"... | null | proof only | null | |
09fv | Let $p$ be a given prime number. We call a set $A$ of positive integers a *p-set* if the cardinality of $A_p$, the set of residues of elements of $A$ modulo $p$, is $p-1$. Determine the smallest value of $n$ such that, for any *p-set* $A$ of $n$ elements, there exists a *p*-element subset of $A$ with sum divisible by $... | [
"*Answer:* The smallest value is $p+2$ for $p$ odd and $2$ for $p=2$.\n\nThe $p=2$ case is easy so we assume $p$ is odd.\nTake $A$ to be the set $\\{1, 2, \\dots, p-1, p+1, 2p-1\\}$. Then $A$ is a $p$-set and the sum of all elements of $A$ is divisible by $p$. Hence the sum of any $p$ elements in $A$ is not divisib... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | p + 2 for odd p, and 2 for p = 2 | |
04mr | Determine all positive integers $n$ such that there exist positive integers $a$ and $b$ which satisfy
$$
S(a) = S(b) = S(a + b) = n.
$$
Here $S(a)$ denotes the sum of digits of $a$. (Romania 1999) | [
"Notice that, for each positive integer $a$, the numbers $a$ and $S(a)$ give the same remainder when divided by $9$.\nApplying this observation to $S(a) = S(b) = S(a+b)$, we conclude that $a$, $b$ and $a+b$ give the same remainder when divided by $9$. This implies that $(a+b) - b = a$ and $(a+b) - a = b$ are both d... | Croatia | Croatia_2018 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | All positive integers divisible by 9 | |
0j68 | Problem:
Alice and Bob play a game in which two thousand and eleven $2011 \times 2011$ grids are distributed between the two of them, 1 to Bob, and the other 2010 to Alice. They go behind closed doors and fill their grid(s) with the numbers $1,2, \ldots, 2011^{2}$ so that the numbers across rows (left-to-right) and do... | [
"Solution:\n\nAnswer: 1\nConsider the grid whose entries in the $j$th row are, in order, $2011j-2010, 2011j-2009, \\ldots, 2011j$. Call this grid $A_{0}$. For $k=1,2, \\ldots, 2010$, let grid $A_{k}$ be the grid obtained from $A_{0}$ by swapping the rightmost entry of the $k$th row with the leftmost entry of the $k... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1 | |
0krb | Problem:
Alice and Bob stand atop two different towers in the Arctic. Both towers are a positive integer number of meters tall and are a positive (not necessarily integer) distance away from each other. One night, the sea between them has frozen completely into reflective ice. Alice shines her flashlight directly at t... | [
"Solution:\n\nLet Alice's tower be of a height $a$, and Bob's tower a height $b$. Reflect the diagram over the ice to obtain an isosceles trapezoid. Then we get that by Ptolemy's Theorem, $4 a b = 26^{2} - 16^{2} = 4 \\cdot 105$, thus $a b = 105$. Hence $a \\in \\{1, 3, 5, 7, 15, 21, 35, 105\\}$. But $\\max(a, b) \... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | 7 or 15 | |
004n | Sea $ABC$ un triángulo e $I$ su incentro. La circunferencia de centro $I$ que pasa por $B$ intersecta a $AC$ en los puntos $E$ y $F$, con $E$ y $F$ entre $A$ y $C$ y distintos de ellos. La circunferencia circunscrita al triángulo $IEF$ intersecta a los segmentos $EB$ y $FB$ en $Q$ y $R$, respectivamente. La recta $QR$ ... | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof and answer | BP = BS = (a - b + c)/2 | |
0hys | Problem:
$\triangle ABC$ is inscribed in a circle $k$ with center $O$ so that $\angle ACB = 120^\circ$.
a. If $H$ is the orthocenter of $\triangle ABC$, prove that $A, B, O, H$ lie on a circle with center the midpoint of the arc $ACB$. (The orthocenter of $\triangle ABC$ is the intersection point of its three altitud... | [
"Solution:\n\nLet $O_1$ be the midpoint of the arc $ACB$ and let $R$ be the radius of $k$.\n\n$\\triangle AOO_1$ and $\\triangle BOO_1$ are equilateral ($\\angle ACB = 120^\\circ \\Rightarrow \\angle AOB = 120^\\circ$). The segments $AB$ and $OO_1$ intersect each other in their midpoint, $D$. If line $AO$ intersect... | United States | BAMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous... | null | proof only | null | |
0ckt | The isosceles triangle $ABC$ has $\angle BAC = 30^\circ$ and $AB = AC$. Take the point $D$ on the side $AC$ and the distinct points $E$, $F$, $G$ on the side $AB$ so that $BC = BD = DE = EF$ and $DG = DF$.
a) Prove that $BF = GE$.
b) Find the measure of $\angle BCG$.
 | [
"a) From $\\triangle DFG$ isosceles follows $\\angle DFG = \\angle DGF$, hence $\\angle DFB = \\angle DGE$. From $\\triangle BDE$ isosceles follows $\\angle DBF = \\angle DEG$.\n\nThe above and $BD = DE$ yield $\\triangle BDF \\equiv \\triangle EDG$ (S.A.A.), whence $BF = GE$.\n\nb) Since $\\triangle ABC$ is isosce... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 52°30' | |
0ea9 | Problem:
Dana je funkcija $f$ s predpisom $f(x)=\frac{12}{x^{2}+x-6}-2$. Zapiši ničle, pole, enačbo asimptote in začetno vrednost funkcije $f$ ter nariši graf funkcije $f$. Nariši še graf funkcije $g$ s predpisom $g(x)=|f(x)|$.
 | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | Zeros: x = 3 and x = −4; Poles: x = 2 and x = −3; Asymptote: y = −2; Initial value: f(0) = −4. | |
0big | Let $n$ be a positive integer, and let $A_n$, respectively $B_n$, be the set of non-negative integers $k < n$ such that the number of distinct prime factors of $\gcd(k, n)$ is even, respectively odd. Show that $|A_n| = |B_n|$ if $n$ is even, and $|A_n| > |B_n|$ if $n$ is odd. | [
"Since $\\gcd(k, n)$ depends only upon the residue class of $k$ modulo $n$, $|A_n| - |B_n| = \\sum_k (-1)^{s(k,n)}$, where $s(k, n)$ is the number of distinct prime factors of $\\gcd(k, n)$, and $k$ ranges over any complete residue system modulo $n$.\nWe shall prove that the above sum equals $n \\prod_{p|n} (1 - 2/... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof only | null | |
0kda | Problem:
Jody has 6 distinguishable balls and 6 distinguishable sticks, all of the same length. How many ways are there to use the sticks to connect the balls so that two disjoint non-interlocking triangles are formed? Consider rotations and reflections of the same arrangement to be indistinguishable. | [
"Solution:\n\nFor two disjoint triangles to be formed, three of the balls must be connected into a triangle by three of the sticks, and the three remaining balls must be connected by the three remaining sticks.\n\nThere are $\\binom{6}{3}$ ways to pick the 3 balls for the first triangle. Note that once we choose th... | United States | HMMO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 7200 | |
0dgh | Find all pairs of positive integers $(x, y)$ such that
$$
\frac{1}{x} + \frac{1}{y} + \frac{1}{\text{lcm}(x, y)} + \frac{1}{\text{gcd}(x, y)} = \frac{1}{2}.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (x, y) ∈ {(8,8), (5,20), (20,5), (6,12), (12,6), (8,12), (12,8), (9,24), (24,9), (12,15), (15,12)} | |
07ym | Problem:
Dato un triangolo isoscele $ABC$ con $AB = AC$ e $\widehat{BAC} < 60^\circ$, sia $D$ il punto su $AC$ tale che $\widehat{DBC} = \widehat{BAC}$, sia $E$ l'intersezione dell'asse di $BD$ con la retta parallela a $BC$ passante per $A$, e sia $F$ il punto sulla retta $AC$, dalla parte di $A$ rispetto a $C$, tale ... | [
"Solution:\n\nSi ponga, per comodità di notazione, $\\overline{BAC} = \\alpha$ e $\\widehat{ABC} = \\widehat{ACB} = \\beta$.\n\nSi tracci la circonferenza circoscritta al triangolo $BDA$ e sia $E'$ la sua intersezione con la parallela a $BC$ passante per $A$; mostreremo che $E'$ si trova sull'asse di $BD$, e deve d... | Italy | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ajv | Let $p$ is a prime number and let $3p+10$ is the sum of the squares of six consecutive positive integers. Prove that $36|p-7$. | [
"From the conditions of the problem, we have that\n$$3p+10=(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2+(n+3)^2=6n^2+6n+19,$$\nso, we have that\n$$3p=6n^2+6n+9,$$\nand\n$$p=2n^2+2n+3=2n(n+1)+3.$$\nIf one of the numbers $n$ or $n+1$ is divisible with $3$, then we have a contradiction with the condition that $p$ is a prime nu... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0jt9 | Problem:
Let the function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ take only integer inputs and have integer outputs. For any integers $x$ and $y$, $f$ satisfies
$$
f(x)+f(y)=f(x+1)+f(y-1)
$$
If $f(2016)=6102$ and $f(6102)=2016$, what is $f(1)$? | [
"Solution:\nWe have\n$$\nf(x+1)=f(x)+f(y)-f(y-1)\n$$\nIf $y$ is fixed, we have\n$$\nf(x+1)=f(x)+\\text{ constant }\n$$\nimplying $f$ is linear. Using our two points, then, we get $f(x)=8118-x$, so $f(1)=8117$"
] | United States | HMMT November | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 8117 | |
0j8l | Problem:
For any positive real numbers $a$ and $b$, define $a \circ b = a + b + 2 \sqrt{a b}$. Find all positive real numbers $x$ such that $x^{2} \circ 9x = 121$. | [
"Solution:\n\nSince $a \\circ b = (\\sqrt{a} + \\sqrt{b})^{2}$, we have\n$$\nx^{2} \\circ 9x = (x + 3\\sqrt{x})^{2}.\n$$\nMoreover, since $x$ is positive, we have $x + 3\\sqrt{x} = 11$, and the only possible solution is that $\\sqrt{x} = \\frac{-3 + \\sqrt{53}}{2}$, so\n$$\nx = \\frac{31 - 3\\sqrt{53}}{2}.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | (31 - 3√53)/2 | |
0gyk | A convex $2009$-gon is given.
a) Find the maximal number of vertices of this polygon which can be marked in such a way that there are no two marked vertices connected by the side of the polygon.
b) Find the maximal number of vertices of this polygon which can be marked in such a way that among any three of marked verti... | [
"a) Let us enumerate all the vertices from $1$ to $2009$. Now mark those which have even numbers: $2, 4, 6, \\ldots, 2008$. This marking obviously satisfies the above condition. We'll now show that more than $1004$ vertices cannot be marked. Indeed, if we, for example, mark the first vertex, then we cannot mark ver... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | a) 1004; b) 1339 | |
0cl7 | Determine the complex numbers $z$ și $w$ with the property that
$$
|z^{2n} + z^n w^n + w^{2n}| = 2^{2n} + 2^n + 1,
$$
for any positive integer $n$. | [
"Applying the modulus to both members of the identity $(z^2 + zw + w^2) \\cdot (z^2 - zw + w^2) = z^4 + z^2w^2 + w^4$ and, using the problem hypothesis, we obtain that $|z^2 - zw + w^2| = 3$.\n\nApplying the modulus to both members of the identity $(z^4 + z^2w^2 + w^4) \\cdot (z^4 - z^2w^2 + w^4) = z^8 + z^4w^4 + w... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities"
] | English | proof and answer | All solutions are the pairs (z, w) with w = 2z and |z| = 1, or z = 2w and |w| = 1; equivalently, (z, w) = (t, 2t) or (2t, t) with |t| = 1. | |
0la7 | Let $m = 2007^{2008}$. How many natural numbers $n$ are there such that $n < m$ and $n(2n + 1)(5n + 2)$ is a multiple of $m$? | [] | Vietnam | Vijetnam 2008 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 9 | |
0kct | Problem:
Let $A$ be a set of integers such that for each integer $m$, there exists an integer $a \in A$ and positive integer $n$ such that $a^{n} \equiv m \pmod{100}$. What is the smallest possible value of $|A|$? | [
"Solution:\nWork in $R = \\mathbb{Z} / 100 \\mathbb{Z} \\cong \\mathbb{Z} / 4 \\mathbb{Z} \\times \\mathbb{Z} / 25 \\mathbb{Z}$.\nCall an element $r \\in R$ type $(s, t)$ if $s = \\nu_{2}(r) \\leq 2$ and $t = \\nu_{5}(r) \\leq 2$. Also, define an element $r \\in R$ to be coprime if it is of type $(0,0)$, powerful i... | United States | HMMT February 2020 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Algebra > Abstract Algebra > Group Theory",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 41 | |
0e42 | Problem:
Določi vsa praštevila $p$ in $q$, za katera je tudi število $2^{2}+p^{2}+q^{2}$ praštevilo. | [
"Solution:\n\nČe je $(p, q)$ rešitev naloge, je tudi $(q, p)$ rešitev. Zato je dovolj obravnavati primer $p \\leq q$. Očitno $p=q=2$ ni rešitev. Če sta $p$ in $q$ lihi praštevili, je število $2^{2}+p^{2}+q^{2}$ sodo in večje od $2$, zato ni praštevilo. Torej je $p=2$.\n\nUgotovimo, kdaj je število $8+q^{2}$ praštev... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | (p, q) = (2, 3) or (3, 2) | |
0e8p | Function $f$ satisfies $3f(x) + f(-x) = 4 \sin x \cos x$ for any real number $x$. Find the correct functional form of $f$.
(A) $\sin x$
(B) $\cos x$
(C) $\cos x \sin x$
(D) $\sin 2x$
(E) $\cos 2x$ | [
"Rearrange the equality to obtain $f(x) = \\frac{4}{3} \\sin x \\cos x - \\frac{1}{3} f(-x)$. Inserting $-x$ instead of $x$ and noting that the sine function is odd and the cosine function is even, we get $f(-x) = \\frac{4}{3} \\sin(-x) \\cos(-x) - \\frac{1}{3} f(x) = -\\frac{4}{3} \\sin x \\cos x - \\frac{1}{3} f(... | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | MCQ | D | |
04uo | Consider an infinite sequence $a_0, a_1, a_2, \dots$ of integers that satisfies $a_0 \ge 2$ and $a_{n+1} \in \{2a_n - 1, 2a_n + 1\}$ for all indices $n \ge 0$. Prove that any such infinite sequence contains infinitely many composite numbers. (Martin Melicher, Josef Tkadlec) | [] | Czech Republic | First Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
06l4 | Find the maximal value of
$$
S = \sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}},
$$
where $a$, $b$, $c$, $d$ are nonnegative real numbers which satisfy $a + b + c + d = 100$. | [
"(IMO Shortlist 2018 A7) See the official solution."
] | Hong Kong | HKG TST | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 49^(1/3) + 2 / 56^(1/3) | |
03g3 | Find all functions $f : \mathbb{R} \to \mathbb{R}$, that satisfy the following condition
$$
f(f(x) + x f(y)) = x f(y + 1), \forall x, y \in \mathbb{R}.
$$
(Alexander Ivanov) | [
"Let $f(0) = a, a \\in \\mathbb{R}$. We set $x = 0$ and obtain $f(a) = 0$. Next, $y = a$ yields\n$$\nf(f(x)) = x f(a + 1) \\quad (1).\n$$\nAssume first that $f(a+1) \\neq 0$. Then $f$ is an injection. Indeed, assuming $f(x_1) = f(x_2)$ for some $x_1 \\neq x_2$, after putting these values in (1) we get contradiction... | Bulgaria | 6 TST for BMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all real x, and f(x) = 0 for all real x | |
07dn | In triangle $ABC$ points $P, Q$ lie on the exterior angle bisector of vertex $A$ such that $P, B$ lie on the same side of line $AC$ and $Q$ is on the other side. Perpendicular line from $P$ to $AB$ cuts the perpendicular line from $Q$ to $AC$ at $X$. Points $P', Q'$ are respectively chosen on $PB, QC$ such that $QX = Q... | [
"First we start with two lemmas.\n\n**Lemma.** Let $ABCD$ be a quadrilateral and $P$ be a point on segment $AB$. Let $X, Y, Z$ be points on $BC, CD, AD$ respectively such that $PX \\parallel AC, XY \\parallel BD, YZ \\parallel AC$. Then we have $PZ \\parallel BD$.\n\n*Proof.* From $PX \\parallel AC, XY \\parallel B... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01bw | Determine whether $712! + 1$ is a prime number. | [
"Answer: It is composite.\n\nWe will show that $719$ is a prime factor of given number. All congruences are considered modulo $719$. By Wilson's theorem $718! \\equiv -1$. What is more $713 \\cdot 714 \\cdot 715 \\cdot 716 \\cdot 717 \\cdot 718 \\equiv (-6)(-5)(-4)(-3)(-2)(-1) \\equiv 720 \\equiv 1$. Hence $712! \\... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | composite | |
01wn | The rhombus $ABCD$ is given. Let $E$ be one of the points of intersection of the circles $\Gamma_B$ and $\Gamma_C$, where $\Gamma_B$ is the circle centered at $B$ and passing through $C$, and $\Gamma_C$ is the circle centered at $C$ and passing through $B$. The line $ED$ intersects $\Gamma_B$ at point $F$.
Find the val... | [
"**Answer: 60°.**\nWe will count the angle $AFB$ as the sum of angles $AFE$ and $BFE$. Note that $\\angle AFE = \\angle ACE$ since they share the arc $AE$ in $\\Gamma_B$. And $\\angle BFE = \\angle BEF$ in the isosceles triangle $BEF$. The angle $BEF$ equals to the half of the arc $BD$, which equals to the angle $B... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 60° | |
01ye | Do there exist six pairwise distinct positive integers $a$, $b$, $c$, $d$, $e$ and $m$, such that
$$
\begin{cases}
a+b+c = d+e+m, \\
ab+bc+ac = de+em+dm, \\
abc = dem + 3^{2021} \cdot 2^{2022}?
\end{cases}
$$
(Igor Voronovich) | [
"**Answer:** yes, such numbers exist.\nConsider the triplets $(a, b, c) = (13, 4, 3)$ and $(d, e, m) = (12, 7, 1)$. These numbers satisfy the system\n$$\n\\begin{cases}\na+b+c = 20 = d+e+m, \\\\\nab+bc+ac = 103 = de+em+dm, \\\\\nabc - dem = 3^2 \\cdot 2^3.\n\\end{cases}\n$$\nClearly, for $k = 2^{673} \\cdot 3^{673}... | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | yes | |
0l67 | Problem:
Let $a$, $b$, and $c$ be real numbers satisfying the system of equations
$$a\sqrt{1 + b^{2}} + b\sqrt{1 + a^{2}} = \frac{3}{4},$$
$$b\sqrt{1 + c^{2}} + c\sqrt{1 + b^{2}} = \frac{5}{12},$$
and
$$c\sqrt{1 + a^{2}} + a\sqrt{1 + c^{2}} = \frac{21}{20}.$$
Compute $a$. | [
"Solution:\nRecall that the functions $\\sinh(x) = \\frac{e^{x} - e^{-x}}{2}$ and $\\cosh(x) = \\frac{e^{x} + e^{-x}}{2}$ satisfy the relation\n$$\\sinh(x + y) = \\sinh(x)\\cosh(y) + \\cosh(x)\\sinh(y) = \\sinh(x)\\sqrt{1 + \\sinh(y)^{2}} + \\sinh(y)\\sqrt{1 + \\sinh(x)^{2}}.$$ \nSince $\\sinh$ is surjective, we c... | United States | HMMT February | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 7/(2*sqrt(30)) | |
00i9 | Let
$$
f(n) = \sum_{k=0}^{2010} n^k = 1 + n + n^2 + \dots + n^{2010}.
$$
Prove that for every integer $m$ with $2 \le m \le 2010$, there is no non-negative integer $n$ such that $f(n)$ is divisible by $m$. | [
"Assume that $m$ divides $f(n)$ for some integer $n$ and some $2 \\le m \\le 2010$. As $f(1) = 2011$ and $2011$ is a prime number, $m$ cannot be a divisor of $f(1)$, so we may restrict ourselves to the case $n \\ne 1$.\nIn this case, we can write $f(n)$ as\n$$\nf(n) = \\frac{n^{2011} - 1}{n - 1}.\n$$\nLet $p$ be a ... | Austria | Austria 2010 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
00or | Find all integer solutions of the equation
$$
3^{x} - 5^{y} = z^{2}
$$ | [
"We start by observing that $z$ must be even, so $z^{2} = 3^{x} - 5^{y} \\equiv (-1)^{x} - 1 \\pmod{4}$ is divisible by $4$, which implies that $x$ is even, say $x = 2t$. Then our equation can be rewritten as $(3^{t} - z)(3^{t} + z) = 5^{y}$, which means that both $3^{t} - z = 5^{k}$ and $3^{t} + z = 5^{y - k}$ for... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (2, 1, 2) | |
0cpq | 2011 storehouses are connected by some roads so that it is possible to reach each storehouse starting from any other one (possibly, using many roads). The *i*-th storehouse contains $x_i$ kg of cement. By one turn, it is allowed to transport any amount of cement from one storehouse to another one if they are connected ... | [
"**Answer.** $2010$ turns.\n\nFirst, let us show that it is not always possible to complete the plan in $2009$ turns. Suppose (for any road system) that all the cement is initially located at one storehouse $S$, and it needs to be distributed equally among all storehouses. Then, for each storehouse except $S$, ceme... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English, Russian | proof and answer | 2010 | |
06s7 | Prove that for every prime $p > 100$ and every integer $r$ there exist two integers $a$ and $b$ such that $p$ divides $a^{2} + b^{5} - r$. | [
"Fix $p$, and let $\\mathcal{P} = \\{0, 1, \\ldots, p-1\\}$ be the set of residue classes modulo $p$. For every $r \\in \\mathcal{P}$, let $S_{r} = \\{(a, b) \\in \\mathcal{P} \\times \\mathcal{P} : a^{2} + b^{5} \\equiv r\\}$, and let $s_{r} = |S_{r}|$. Our aim is to prove $s_{r} > 0$ for all $r \\in \\mathcal{P}$... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM... | null | proof only | null | |
077s | Suppose $a_0, \dots, a_{100}$ are positive reals. Consider the following polynomial for each $k$ in $\{0, 1, \dots, 100\}$:
$$
a_{100+k}x^{100} + 100a_{99+k}x^{99} + a_{98+k}x^{98} + a_{97+k}x^{97} + \dots + a_{2+k}x^2 + a_{1+k}x + a_k,
$$
where indices are taken modulo 101, i.e., $a_{100+i} = a_{i-1}$ for any $i$ in $... | [
"Suppose for contradiction that all 101 polynomials have all real roots. Let $P_k(x)$ denote the $k$-th polynomial:\n$$\nP_k(x) = a_{100+k}x^{100} + 100a_{99+k}x^{99} + a_{98+k}x^{98} + a_{97+k}x^{97} + \\dots + a_{2+k}x^2 + a_{1+k}x + a_k,\n$$\nwhere all indices are modulo 101.\n\nNote that all coefficients $a_i$ ... | India | INMO_2023 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
053h | A class consists of $7$ boys and $13$ girls. During the first three months of the school year, each boy has communicated with each girl at least once. Prove that there exist two boys and two girls such that both boys communicated with both girls first time in the same month. | [
"*Solution 1:* Call the first communication between a boy and a girl their *acquaintance*. During the $3$ months, there are $7 \\cdot 13 = 91$ acquaintances in total. Thus there exists a month when there was at least $31$ acquaintances. Let the boys be denoted by $p_1$ through $p_7$ and let $T_i$, $i = 1, \\dots, 7... | Estonia | IMO Team Selection Contest | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0e1s | Let $a$, $b$ and $c$ be positive integers. Prove that $a^2 + b^2 + c^2$ is divisible by $4$ if and only if $a$, $b$ and $c$ are even. | [
"First, let $a$, $b$ and $c$ be even: $a = 2a_1$, $b = 2b_1$, $c = 2c_1$. Then the number\n$$\na^2 + b^2 + c^2 = 4(a_1^2 + b_1^2 + c_1^2)\n$$\nis divisible by $4$.\n\nNow, let us prove the converse. Assume that $a^2 + b^2 + c^2$ is divisible by $4$. If exactly one of the numbers $a$, $b$ and $c$ were odd or if all ... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0014 | Sea $ABC$ un triángulo acutángulo tal que el ángulo $B$ mide $60^\circ$. La circunferencia de diámetro $AC$ corta a las bisectrices interiores de los ángulos $A$ y $C$ en los puntos $M$ y $N$, respectivamente ($M \neq A$, $N \neq C$). La bisectriz interior del ángulo $B$ corta a $MN$ y $AC$ en los puntos $R$ y $S$, res... | [] | Argentina | XIV Olimpiada Matemática de Países del Cono Sur | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | español | proof only | null | |
074c | Problem:
Let $ABC$ be an acute-angled triangle and let $H$ be its orthocentre. Let $h_{\max}$ denote the largest altitude of the triangle $ABC$. Prove that
$$
AH + BH + CH \leq 2 h_{\max}
$$ | [
"\n\nLet $\\angle C$ be the smallest angle, so that $CA \\geq AB$ and $CB \\geq AB$. In this case the altitude through $C$ is the longest one. Let the altitude through $C$ meet $AB$ in $D$ and let $H$ be the orthocentre of $ABC$. Let $CD$ extended meet the circumcircle of $ABC$ in $K$. We h... | India | Indian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometri... | null | proof only | null | |
06f6 | The incircle (with centre $O$) of an isosceles triangle $ABC$ with $AB = AC$ meets $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. Let $N$ be the intersection of the lines $OL$ and $KM$, and let $Q$ be the intersection of the lines $BN$ and $CA$. Let $P$ be the foot of the perpendicular from $A$ to $BQ$. Suppose $BP = ... | [
"$\\frac{AB}{BC}$ can be $\\frac{\\sqrt{2}}{2}$ or $\\frac{\\sqrt{10}}{2}$.\nIt is well-known that $Q$ is the midpoint of $AC$. We first consider the case when $P$ lies inside $\\triangle ABC$. Let $X$ be the projection of $C$ on $BQ$, and let $Y$ be the point on the line $BQ$ such that $XY = XC$, and $X$ lies betw... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof and answer | sqrt(2)/2 or sqrt(10)/2 | |
06sb | Let $\mathbb{Z}_{\geqslant 0}$ be the set of all nonnegative integers. Find all the functions $f: \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ satisfying the relation
$$
f(f(f(n)))=f(n+1)+1
$$
for all $n \in \mathbb{Z}_{\geqslant 0}$. | [
"Answer. There are two such functions: $f(n)=n+1$ for all $n \\in \\mathbb{Z}_{\\geqslant 0}$, and\n$$\nf(n)=\\left\\{\\begin{array}{ll}\nn+1, & n \\equiv 0(\\bmod 4) \\text{ or } n \\equiv 2(\\bmod 4), \\\\\nn+5, & n \\equiv 1(\\bmod 4), \\\\\nn-3, & n \\equiv 3(\\bmod 4)\n\\end{array} \\quad \\text{ for all } n ... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Functional equations",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Two functions: (1) f(n) = n + 1 for all nonnegative integers n; (2) the piecewise function by residue mod four: f(n) = n + 1 if n is congruent to 0 or 2 modulo 4, f(n) = n + 5 if n is congruent to 1 modulo 4, and f(n) = n − 3 if n is congruent to 3 modulo 4. | |
0kaq | Problem:
Every positive integer is either nice or naughty, and the Oracle of Numbers knows which are which. However, the Oracle will not directly tell you whether a number is nice or naughty. The only questions the Oracle will answer are questions of the form "What is the sum of all nice divisors of $n$?", where $n$ is... | [
"Solution:\nLet $f(u)$ denote the sum of all nice divisors of $u$. Note that $f(u) \\leq \\sigma(u)$, where $\\sigma(u)$ denotes the sum of all the divisors of $u$, including $1$ and $u$.\n\nWe also note the following facts about the $\\sigma$ function:\n\n1. $\\sigma$ is multiplicative: if $a$ and $b$ are relative... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0bnl | The set of positive integers is divided into subsets in the following way:
$$
\{1, 2\}, \{3, 4, 5\}, \{6, 7, 8, 9\}, \{10, 11, 12, 13, 14\}, \dots
$$
a) Find the smallest element of the 100th subset.
b) Is 2015 the largest element of such a subset? | [
"a) The first 99 subsets contain $2 + 3 + \\dots + 100 = 5049$ elements. In the first 99 subsets are written the numbers $1, 2, 3, \\dots, 5049$, so the smallest element of the 100th subset is $5050$.\n\nb) If $2015$ is the largest element of the $n$-th subset, then $2 + 3 + \\dots + (n + 1) = 2015$. Adding $1$ in ... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 5050; b) Yes — 2015 is the largest element of the 62nd subset. | |
0crt | A plane $\alpha$ intersects the edges $AB$, $BC$, $CD$, and $DA$ of a triangular pyramid $ABCD$ at points $K$, $L$, $M$, and $N$, respectively. It appears that the dihedral angles $\angle(KLA, KLM)$, $\angle(LMB, LMN)$, $\angle(MNC, MNK)$, and $\angle(NKD, NKL)$ are equal. Prove that the projections of the vertices $A$... | [
"Обозначим через $A', B', C', D'$ проекции вершин $A, B, C, D$ соответственно на плоскость $\\alpha$. Пусть $X$ — произвольная точка на продолжении отрезка $KL$ за точку $K$. Тогда имеем $\\angle(KXA, KXN) = \\angle(KLA, KLM)$ и $\\angle(KNA, KNX) = \\angle(NKD, NKL)$. По условию, эти углы равны; значит, в трехгран... | Russia | XL Russian mathematical olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05ud | Problem:
On note $\mathbb{Z}[x, y, z]$ l'ensemble des polynômes à coefficients entiers en les trois variables $x, y$ et $z$. On dit ensuite qu'un polynôme $P \in \mathbb{Z}[x, y, z]$ est olympique si $\mathbb{Z}[x, y, z]$ contient des polynômes $A, B$ et $C$ tels que
$$
P(x, y, z)=(x+y+z) A(x, y, z)+(x y+y z+z x) B(x,... | [
"Solution:\n\nPosons $s_{1}=x+y+z$, $s_{2}=x y+y z+z x$ et $s_{3}=x y z$. On constate que\n$$\n\\begin{cases}\nx^{3} \\equiv -x^{2}(y+z) \\equiv -x(x y+x z) \\equiv x y z \\equiv 0 & (\\bmod\\ s_{1}, s_{2}, s_{3}) \\\\\nx^{2} y^{2} \\equiv x y(-x z-y z) \\equiv -x y z(x+y) \\equiv 0 & (\\bmod\\ s_{1}, s_{2}, s_{3})... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 3 | |
0cov | Given three quadratic polynomials $f_1(x) = x^2 + 2a_1x + b_1$, $f_2(x) = x^2 + 2a_2x + b_2$, $f_3(x) = x^2 + 2a_3x + b_3$ such that $a_1a_2a_3 = b_1b_2b_3 > 1$. Prove that one of these polynomials has two distinct real roots.
Даны квадратные трёхчлены $f_1(x) = x^2 + 2a_1x + b_1$, $f_2(x) = x^2 + 2a_2x + b_2$, $f_3(x... | [
"Предположим противное; тогда дискриминанты всех трёхчленов неположительны, то есть $a_k^2 \\le b_k$ $(k = 1, 2, 3)$. Левые (а значит, и правые) части этих неравенств неотрицательны, поэтому их можно перемножить, получая $(a_1a_2a_3)^2 \\le b_1b_2b_3$, то есть $N^2 \\le N$, где $N = a_1a_2a_3$. Но это противоречит ... | Russia | Regional round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Russian | proof only | null | |
0j3y | Problem:
Let $S$ be the set of ordered pairs of integers $(x, y)$ with $1 \leq x \leq 5$ and $1 \leq y \leq 3$. How many subsets $R$ of $S$ have the property that all the points of $R$ lie on the graph of a single cubic? A cubic is a polynomial of the form $y=a x^{3}+b x^{2}+c x+d$, where $a, b, c$, and $d$ are real n... | [
"Solution:\n\nAnswer: 796\n\nWe observe that $R$ must contain at most 1 point from each column of $S$, because no function can contain more than 1 point with the same $x$-coordinate. Therefore, $|R| \\leq 5$ ($|R|$ denotes the number of elements of $R$). Note that 4 points determine a cubic, so if $R$ is any subset... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 796 |
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