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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07q8 | Let $[x]$ be the greatest integer not exceeding $x$. Consider the statements
a. $[x+y] = [x] + [y]$.
b. $[-x - y] = [-x] + [-y]$.
Characterise the pairs $x, y$ of real numbers such that at least one of (a) or (b) is true and characterise the pairs $x, y$ such that both (a) and (b) are true. | [
"At least one of (a) or (b) is true unless $x + y$ is an integer, but $x$ is not an integer. Exactly one of (a) and (b) is true if none of $x$, $y$, $x + y$ are integers. Both (a) and (b) are true if at least one of $x$, $y$ is an integer.\n\nMethod: for $x$ a real number, write $x = [x] + \\{x\\}$, where $0 \\le \... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | Both (a) and (b) are true if and only if at least one of x or y is an integer. At least one of (a) or (b) is true for all pairs except those with x and y both non-integers but x + y an integer (these are exactly the pairs for which neither holds). Equivalently: (a) holds iff {x} + {y} < 1; (b) holds iff x or y is an in... | |
0c5e | Find all reciprocal polynomials with real coefficients, of degree at least $2$, whose derivative has all the roots of modulus $1$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | All such polynomials are real constant multiples of (x + 1)^n for any n ≥ 2, and of (x − 1)^{2m} for any m ≥ 1. | |
0jop | Problem:
Let $r_{1}, \ldots, r_{n}$ be the distinct real zeroes of the equation
$$
x^{8}-14 x^{4}-8 x^{3}-x^{2}+1=0
$$
Evaluate $r_{1}^{2}+\cdots+r_{n}^{2}$. | [
"Solution:\nAnswer: 8\nObserve that\n$$\n\\begin{aligned}\nx^{8}-14 x^{4}-8 x^{3}-x^{2}+1 & =\\left(x^{8}+2 x^{4}+1\\right)-\\left(16 x^{4}+8 x^{3}+x^{2}\\right) \\\\\n& =\\left(x^{4}+4 x^{2}+x+1\\right)\\left(x^{4}-4 x^{2}-x+1\\right) .\n\\end{aligned}\n$$\nThe polynomial $x^{4}+4 x^{2}+x+1$ has no real roots. On ... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 8 | |
0eqv | Water pours into a tank at a constant rate. After 2 hours the tank was 36% full and after 10 hours it is 100% full. How full was the tank to begin with?
(A) 10%
(B) 12%
(C) 15%
(D) 18%
(E) 20% | [
"In 8 hours (from 2 hours to 10 hours), the tank filled up a further $64\\%$ (from $36\\%$ to $100\\%$). That is $8\\%$ per hour.\n\nSo in the first two hours it filled up $16\\%$, meaning it started with $36 - 16 = 20\\%$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | E | |
0e1d | Problem:
Dan je ostrokotni trikotnik $ABC$, v katerem je $|AB| > |BC| > |AC|$. Naj bo $D$ od $C$ različna točka na daljici $BC$, da je $|AC| = |AD|$. Označimo s $H$ višinsko točko trikotnika $ABC$, z $A_{1}$ in $B_{1}$ pa nožišči višin iz točk $A$ in $B$. Premica $DH$ seka premico $AC$ v točki $E$, premico $A_{1}B_{1}... | [
"Solution:\n\nTrikotnik $CAD$ je zaradi $|AC| = |AD|$ enakokrak. Premica $AA_{1}$ je višina tudi v tem enakokrakem trikotniku, zato je $\\angle HDA = \\angle ACH$. V štirikotniku $HA_{1}CB_{1}$ velja $\\angle CA_{1}H = \\frac{\\pi}{2} = \\angle CB_{1}H$, zato je tetiven in je $\\angle B_{1}A_{1}H = \\angle B_{1}CH$... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g5x | 給三個正整數 $K$, $B$, $W$, 其中 $B > W > 1$。有兩堆球: 一堆有黑球 $B$ 個; 另一堆有白球 $W$ 個。依照以下方法來分割:
操作：所有堆依照 **非遞增** 的方式排列，若黑球堆與白球堆數量一樣則白球堆置於前。現在選出前面 $K$ 堆，若全部少於 $K$ 堆則選擇全部。然後將所選擇的每一堆分成兩堆，而且這兩堆的球數最多只差一個。
(例如：令 $K = 4$。現在有四堆黑球分別是 $5, 4, 4, 2$ 個, 三堆白球分別是 $8, 4, 2$ 個。所以依序排列是 $(w8, b5, w4, b4, w2, b2)$, 其中 $w8$ 指白球一堆 $8$ 個, $b5$ 指黑球一堆 $5$ 個... | [
"起始狀態是 $A_0 = (bB, wW)$，而 $A_{i-1} \\to A_i$ 代表進行第 $i$ 次的「排列-分割」。\n依照操作模式，直到所有的堆都是一個球。我們標示以下幾個重要的階段：\n$A_s$ : 第一次出現有一堆一個球時 (無論是黑球或白球)。\n$A_t$ : 總堆數大於 $K$ 時; 然而有可能: 當所有的堆都是一個球時還是不超過 $K$ 堆, 則令 $t = \\infty$.\n$A_f$ : 當所有的黑球都是一球一堆時。\n我們只需要證明：題意所要求的停止時刻是在 $A_{f-1}$ 或更早之前，也就是 $A_{f-1}$ 中一定有至少一堆一個白球。\n明顯地, $s \\le f$. (*... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0ay1 | Problem:
Let $A$ be the sum of the decimal digits of the largest 2017-digit multiple of $7$ and let $B$ be the sum of the decimal digits of the smallest 2017-digit multiple of $7$. Find $A - B$. | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 18144 | |
00qr | Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that
$$
(x + y)f(2yf(x) + f(y)) = x^3 f(yf(x)), \quad \forall x, y \in \mathbb{R}^+.
$$ | [
"First we show that such a function should be injective. Indeed, if $f(a) = f(b)$ then $\\forall y > 0$\n$$\n\\frac{f(yf(a))}{f(2yf(a) + f(y))} = \\frac{f(yf(b))}{f(2yf(b) + f(y))}\n$$\nand by the given relation\n$$\n\\begin{align*}\n\\frac{a+y}{a^3} &= \\frac{b+y}{b^3} \\\\\n& \\Leftrightarrow b^3(a+y) = a^3(b+y) ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | no such function exists | |
0edw | Problem:
Smetana predstavlja $7 \%$ mase mleka, iz $61 \%$ mase smetane pa nastane maslo. Iz koliko mleka nastane $3 \mathrm{~kg}$ masla?
(A) manj kot $5 \mathrm{~kg}$
(B) $17 \mathrm{~kg}$
(C) $54 \mathrm{~kg}$
(D) $69 \mathrm{~kg}$
(E) več kot $70 \mathrm{~kg}$ | [
"Solution:\nOznačimo z $x$ maso mleka v kg. V $x$ kg mleka je $0{,}07 x \\mathrm{~kg}$ smetane. Iz te smetane nastane $0{,}61 \\cdot 0{,}07 x \\mathrm{~kg}$ masla. Velja $0{,}61 \\cdot 0{,}07 x=3$. Sledi, da je $x \\doteq 70{,}258 \\mathrm{~kg}$. $3 \\mathrm{~kg}$ masla nastane iz več kot $70 \\mathrm{~kg}$ mleka."... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0kwi | Problem:
Suppose that $a$ and $b$ are real numbers such that the line $y=a x+b$ intersects the graph of $y=x^{2}$ at two distinct points $A$ and $B$. If the coordinates of the midpoint of $A B$ are $(5,101)$, compute $a+b$. | [
"Solution:\n\nLet $A=\\left(r, r^{2}\\right)$ and $B=\\left(s, s^{2}\\right)$. Since $r$ and $s$ are roots of $x^{2}-a x-b$ with midpoint $5$, $r+s=10=a$ (where the last equality follows by Vieta's formula).\n\nNow, as $-r s=b$ (Vieta's formula), observe that\n\n$$\n202=r^{2}+s^{2}=(r+s)^{2}-2 r s=100+2 b.\n$$\n\nT... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 61 | |
0eej | Problem:
Za trikotnik $ABC$ velja $|AB| = 2|AC|$, $D$ pa je taka točka na poltraku $CA$, da velja $|CD| = 3|AC|$. Dokaži, da pravokotnica iz točke $C$ na premico $BD$ razpolavlja daljico $AB$. | [
"Solution:\n\n\n\nOznačimo presečišče pravokotnice iz točke $C$ na premico $BD$ in premice $BD$ z $E$, presečišče premic $CE$ in $AB$ pa z $M$. Velja $|AD| = 2|AC| = |AB|$, zato je trikotnik $DBA$ enakokrak z vrhom pri $A$, torej velja $\\angle BDA = \\angle ABD$. Sledi $\\angle ACM = 90^{\... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
05yx | Problem:
Lucile a écrit au tableau $s$ 2023-uplets d'entiers. Elle s'autorise alors des opérations de la forme suivante : elle choisit deux 2023-uplets $\mathbf{u}=\left(u_{1}, u_{2}, \ldots, u_{2023}\right)$ et $\mathbf{v}=\left(v_{1}, v_{2}, \ldots, v_{2023}\right)$, non nécessairement distincts, parmi ceux qu'elle ... | [
"Solution:\n\nTout d'abord, si $s=1$, Lucile a écrit un seul vecteur $\\mathbf{u}$. Tout vecteur $\\mathbf{x}$ qu'elle pourra ensuite écrire aura des coordonnées de même signe que $\\mathbf{u}$, et elle ne pourra donc pas écrire tous les vecteurs possibles.\n\nEnsuite, si $s=2$, soit $\\mathbf{u}$ et $\\mathbf{v}$ ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | all integers s ≥ 3 | |
0fow | Problem:
Encuentra todos los enteros positivos $n$, que verifican
$$
n=2^{2x-1}-5x-3=\left(2^{x-1}-1\right)\left(2^{x}+1\right)
$$
para algún entero positivo $x$. | [
"Solution:\nRealizando el producto del miembro de la derecha y reorganizando términos, la igualdad entre los miembros segundo y tercero se puede escribir como\n$$\n2^{x-1}=5x+2\n$$\n\nSe comprueba fácilmente que ni $x=1$ ni $x=2$ son soluciones, mientras que si $x \\geq 3$, entonces $5x+2$ ha de ser múltiplo de 4, ... | Spain | null | [
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 2015 | |
07sy | Let $n$ be a positive integer. An $n$-level honeycomb is a plane region covered with regular hexagons of side-length $1$ connected along edges, such that the centres of the boundary hexagons are lined up along a regular hexagon of side-length $n\sqrt{3}$. The diagram shows a $2$-level honeycomb from which the central h... | [
"We will first prove that coverings can be found if $n = 3k$ and if $n = 3k - 1$. When $k = 1$ these are the cases $n = 2$ and $n = 3$ for which the diagrams below show a covering by trexes. In these diagrams, the unit hexagons are represented by their centres.\n\n\n\nTo extend this for all... | Ireland | IRL_ABooklet_2020 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | All positive integers n with n ≡ 0 or 2 (mod 3). | |
08fo | Problem:
Poniamo $a_{1}=1$ e, per ogni $n \geq 2$,
$$
a_{n}=n\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) .
$$
Qual è il più piccolo valore di $n$ per cui $a_{n}$ è divisibile per 2022?
(A) 47
(B) 289
(C) 337
(D) 2022
(E) 2023 | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Consideriamo le equazioni\n$$\n\\left\\{\n\\begin{array}{l}\na_{n}=n\\left(a_{1}+a_{2}+\\cdots+a_{n-1}\\right) \\\\\na_{n+1}=(n+1)\\left(a_{1}+a_{2}+\\cdots+a_{n-1}+a_{n}\\right)\n\\end{array}\n\\right.\n$$\nvalide per ogni $n \\geq 2$. Riscriviamo ora la seconda nella f... | Italy | Italian Mathematical Olympiad - February Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | C | |
01c7 | Let $\omega(n)$ denote the number of distinct prime factors of a positive integer $n$. Show that there are infinitely many positive integers $n$ such that $\omega(n) < \omega(n+1) < \omega(n+2)$. | [
"Choose $n = 2^m$. Then we want to have $\\omega(2^m) = 1 < \\omega(2^m+1) < \\omega(2^m+2) = 1 + \\omega(2^{m-1} + 1)$ for infinitely many positive integers $m$.\n\nSuppose there are only finitely many such $m$. Then there exists $N$ such that for all $n \\ge N$ either $2^m + 1$ is prime or $\\omega(2^m + 1) \\ge ... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other"
] | null | proof only | null | |
07iy | In triangle *ABC* the points *N* and *M* are on the side $BC$ such that $\angle BAM = \angle MAN = \angle NAC$. The points *P* and *Q* are on the angle bisector such that *A*, *P*, and *Q* are all on the same side with respect to $BC$ and $\frac{1}{3}\angle BAC = \frac{1}{2}\angle BPC = \angle BQC$. Let $AM$ and $CQ$ m... | [
"We shall prove the following lemma.\n\n\n\n**Lemma 1.** *The points $E$, $F$, $B$, and $C$ are on a circle centered at $T'$.*\n\n*Proof.* An easy angle chasing shows that $BAEQ$ and $CAFQ$ are indeed cyclic.\nThus,\n$$\n\\angle FBE = \\angle QBE = \\angle QAE = \\angle QAF = \\angle QCF = ... | Iran | 41th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar triang... | null | proof only | null | |
02mv | Problem:
Quantos são os pares de números inteiros positivos $(x, y)$ tais que
$$
\frac{x y}{x+y}=144 ?
$$ | [
"Solution:\nA equação dada é equivalente a $x y = 144(x + y) = 144 x + 144 y$, portanto, isolando $x$, obtemos $x = \\frac{144 y}{y - 144}$. Como $x$ e $y$ devem ser inteiros positivos, o denominador $y - 144$ deve ser um número inteiro positivo, digamos, $y - 144 = n$. Substituindo essa expressão no valor de $x$, ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 45 | |
09mu | A country consists of $n$ islands. Some of the islands are connected by bridges, and the bridges intersect only on the islands. Each bridge has a positive integer toll. A path is defined as a sequence of bridges such that each pair of consecutive bridges share an endpoint on the same island.
Prove that $n$ or $n-2$ is ... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0ihc | Problem:
Compute
$$
\sum_{k=0}^{\infty} \frac{4}{(4 k)!}
$$ | [
"Solution: $e + 1 / e + 2 \\cos 1$\nThis is the power series\n$$\n4 + \\frac{4 x^{4}}{4!} + \\frac{4 x^{8}}{8!} + \\cdots\n$$\nevaluated at $x = 1$. But this power series can be written as the sum\n$$\n\\begin{aligned}\n& \\left(1 + \\frac{x}{1!} + \\frac{x^{2}}{2!} + \\frac{x^{3}}{3!} + \\frac{x^{4}}{4!} + \\frac{... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | e + 1/e + 2 cos 1 | |
0env | Let $PQ$ be the diameter of semicircle $H$. Circle $\omega$ is internally tangent to $H$ and tangent to $PQ$ at $C$. Let $A$ be a point on $H$ and $B$ a point on $PQ$ such that $AB$ is perpendicular to $PQ$ and tangent to $\omega$. Prove that $AC$ bisects $\angle PAB$. | [
"In version centred at $C$ with any radius gives $\\triangle CAP \\sim \\triangle CP'A'$ and $\\triangle CAB \\sim \\triangle CB'A'$. Clearly the line $PQ$ inverts to itself. $\\omega$ inverts to $\\omega'$, a line. $H$ inverts to semicircle $H'$ tangent to $\\omega'$ with diameter $P'Q'$. $AB$ inverts to arc $A'B'... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08xk | Let $n$ be a positive integer. For every pair of students enrolled in a certain school having $n$ students, either the pair are mutual friends or not mutual friends. Let $N$ be the smallest possible sum $a+b$ of positive integers $a$ and $b$ satisfying the following 2 conditions concerning students in this school:
(1) ... | [
"Call a team a good team if any pair of students in the team are mutual friends, a bad team if any pair of students in the team are not mutual friends. By agreement, we consider a team consisting of only one student to be both good and bad. Let us show that the minimum value of $N$ we seek is $n+1$.\n\nIf, in a cer... | Japan | Japan Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | n+1 | |
06aj | Each student of a class has finite number of cards. Each card has a number on it from the interval $[0,1]$. Find the smallest possible constant $c > 0$, such that the following holds, independently from the distribution of the cards to the students:
Each student that has a total sum of numbers less than 1000, shares th... | [
"Amongst all possible arrangements into boxes, pick one, where the maximum value inside a box is as small as possible. If there are several arrangements, achieving this smallest maximum value, pick one where the number of boxes achieving this value is as small as possible.\nSay that the boxes have total values equa... | Greece | Selection examinations | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 1000/91 | |
0c29 | In a box there are $50$ cards on which the first $100$ positive integers are written as follows: on the first card there are $1$ (on one side) and $2$ (on the other side), on the second card there are $3$ (on the one side) and $4$ (on the other side) and so on, up to the $50$th card, on which the numbers $99$ (on one s... | [] | Romania | 69th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 185 | |
0iet | Problem:
Let $S=\{s_{0}, \ldots, s_{n}\}$ be a finite set of integers, and define $S+k=\{s_{0}+k, \ldots, s_{n}+k\}$. We say that $S$ and $T$ are equivalent, written $S \sim T$, if $T=S+k$ for some $k$. Given a (possibly infinite) set of integers $A$, we say that $S$ tiles $A$ if $A$ can be partitioned into subsets eq... | [
"Solution:\n\nConsider the set $S_{0}$ equivalent to $S$ that contains $3$. If it contains $5$ but not $7$, then the set $S_{1}$ equivalent to $S$ containing $7$ must contain $9$, which is not prime. Likewise, $S_{0}$ cannot contain $7$ but not $5$, because then the set $S_{1}$ containing $5$ must contain $9$. Supp... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0716 | Problem:
Show that a square with side $1$ cannot be covered by five squares with side less than $1/2$. | [
"Suppose, for contradiction, that five squares, each with side less than $1/2$, can cover a unit square.\n\nThe diagonal of each small square is less than $1/2 \\sqrt{2} < 0.71$.\n\nConsider the four corners of the unit square. Each small square can cover at most one corner, since the distance between any two corne... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0l1s | Problem:
Let $\Omega$ and $\omega$ be circles with radii $123$ and $61$, respectively, such that the center of $\Omega$ lies on $\omega$. A chord of $\Omega$ is cut by $\omega$ into three segments, whose lengths are in the ratio $1:2:3$ in that order. Given that this chord is not a diameter of $\Omega$, compute the len... | [
"Solution:\nDenote the center of $\\Omega$ as $O$. Let the chord intersect the circles at $W, X, Y, Z$ so that $WX = t$, $XY = 2t$, and $YZ = 3t$. Notice that $Y$ is the midpoint of $WZ$; hence $\\overline{OY} \\perp \\overline{WXYZ}$.\n\nThe fact that $\\angle OYX = 90^\\circ$ means $X$ is the antipode of $O$ on $... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 42 | |
0162 | Six circular mint cookies, each of radius greater than $1$, are given. Show that it is impossible to place them all upon a circular plate of radius $3$ without overlaps. | [
"Place the cookies upon the plate, denote the centre of the plate by $O$, and the centres of the cookies, in anticlockwise order, by $P_j$, $1 \\leq j \\leq 6$. Since the radii are greater than $1$, the points $P_j$ are at a distance at least $1$ from the edge of the plate, which means $|OP_j| \\leq 2$. Moreover, b... | Baltic Way | Baltic Way SHL | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
07qf | Show that the number
$$
\left( \frac{251}{\frac{1}{\sqrt{252-5\sqrt{3}}}} - \frac{10\sqrt{3}}{63} \right) + \frac{1}{\frac{251}{\sqrt{252+5\sqrt{3}}}} + \frac{10\sqrt{3}}{63}
$$
is an integer and find its value. | [
"Let $a^3 = 252$ and $b^3 = 250$, then $ab = \\sqrt[3]{252^3/250} = 10\\sqrt[3]{63}$. Using that $251 = \\frac{a^3+b^3}{2}$ and $1 = \\frac{a^3-b^3}{2}$, the given expression inside the bracket becomes\n\n$$\n\\frac{\\frac{a^3+b^3}{2}}{\\frac{a^3-b^3}{2} - ab} + \\frac{\\frac{a^3-b^3}{2}}{\\frac{a^3+b^3}{2} + ab}\n... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2016 | |
0fw0 | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck mit $AB \neq AC$ und Höhenschnittpunkt $H$. Der Mittelpunkt der Seite $BC$ sei $M$. Die Punkte $D$ auf $AB$ und $E$ auf $AC$ seien so, dass $AE = AD$ ist und $D, H, E$ auf einer Geraden liegen. Zeige, dass $HM$ und die gemeinsame Sehne der Umkreise der beiden Dreiecke $ABC... | [
"Solution:\n\nSei $k$ der Umkreis von $\\triangle ABC$ und sei $O$ Mittelpunkt von $k$. Der zweite Schnittpunkt von $AO$ mit $k$ sei $P$. Wir zeigen als erstes, dass die Punkte $H, M, P$ auf einer Geraden liegen. Die Höhenfusspunkte der Höhen von $\\triangle ABC$ durch $B$ bzw. $C$ seien $B_1$ bzw. $C_1$. Da $AP$ e... | Switzerland | IMO Selektion | [
"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 | |
02y2 | Problem:
As cidades $A$, $B$ e $C$ estão posicionadas nos vértices de um triângulo equilátero com $60\ \mathrm{km}$ de lado. Entre elas, ligando duas a duas, existem três rodovias, $AB$, $AC$ e $BC$, todas em linha reta. Uma ferrovia será construída, também em linha reta, devendo interceptar $AB$ a $30\ \mathrm{km}$ d... | [
"Solution:\n\n\n\na) Na figura anterior, os triângulos retângulos $\\triangle ADI$ e $\\triangle BGD$ possuem os mesmos ângulos, pois $\\angle ADI = \\angle GDB$ e $\\angle AID = \\angle DGB = 90^\\circ$. Portanto, eles são semelhantes e daí\n$$\n\\frac{AD}{BD} = \\frac{h_1}{h_2}\n$$\nAs de... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 60 km | |
00mv | Find all pairs $(a, b)$ of real numbers such that
$$
a \cdot \lfloor b \cdot n \rfloor = b \cdot \lfloor a \cdot n \rfloor
$$
for all positive integers $n$. | [
"**Answer.** The solutions are all pairs $(a, b)$ with $a = 0$ or $b = 0$ or $a = b$ or both $a$ and $b$ integers.\n\nLet $a_0 = \\lfloor a \\rfloor$ and $a_i$ be the binary digits of the fractional part of $a$ such that $a = a_0 + \\sum_{i=1}^{\\infty} \\frac{a_i}{2^i}$ with $a_0 \\in \\mathbb{Z}$ and $a_i \\in \\... | Austria | Austria2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All real pairs with a = 0 or b = 0 or a = b or both a and b integers. | |
0ctq | The medians $AM_A$, $BM_B$, and $CM_C$ of a triangle $ABC$ meet at $M$. Construct a circle $\Omega_A$ passing through the midpoint of $AM$ and tangent to $BC$ at $M_A$. Construct the circles $\Omega_B$ and $\Omega_C$ similarly. Prove that the circles $\Omega_A$, $\Omega_B$, and $\Omega_C$ share a common point.
В треуг... | [
"Let $K_A$, $K_B$, and $K_C$ be the midpoints of the segments $AM$, $BM$, and $CM$, respectively (see Fig. 24). By angle chasing, one sees that the circles $MK_AK_BM_B$, $M_BK_CM_A$, and $MAK_BM_C$ are concurrent at some point $X$. This point belongs to the circles $\\Omega_A$, $\\Omega_B$, and $\\Omega_C$: for exa... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English; Russian | proof only | null | |
0k3w | Problem:
Let $n$ be a positive integer. Is it possible to arrange the numbers $1,2, \ldots, n$ in a row so that the arithmetic mean of any two of these numbers is not equal to some number between them? | [
"Solution:\nThis is possible for every $n$. Note that if it is possible for some number greater than $n$, then it is also possible for $n$. Thus, it suffices to prove the statement for powers of $2$. We will proceed by induction.\n\nFirst, the statement is true for $2$, since $1,2$ works.\n\nNow suppose it is possi... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Yes; it is possible for all positive integers. | |
0cnp | In an acute triangle $ABC$, the altitudes $AA_1$, $BB_1$, $CC_1$ are drawn. A perpendicular to $AC$ passing through $A_1$ intersects line $B_1C_1$ at $D$. Prove that $\angle ADC = 90^\circ$. | [
"Let $K$ be the intersection point of segments $A_1D$ and $AC$ (see Figure 5).\n\nFrom the equality $\\angle AB_1B = \\angle AA_1B = 90^\\circ$, it follows that the quadrilateral $AB_1A_1B$ is inscribed in a circle (with diameter $AB$), so $\\angle A_1B_1K = 180^\\circ - \\angle A_1B_1A = \\angle ABC$.\n\nSimilarly... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English; Russian | proof only | null | |
0ig8 | Problem:
There are 42 stepping stones in a pond, arranged along a circle. You are standing on one of the stones. You would like to jump among the stones so that you move counterclockwise by either 1 stone or 7 stones at each jump. Moreover, you would like to do this in such a way that you visit each stone (except for t... | [
"Solution:\nNumber the stones $0, 1, \\ldots, 41$, treating the numbers as values modulo $42$, and let $r_n$ be the length of your jump from stone $n$. If you jump from stone $n$ to $n+7$, then you cannot jump from stone $n+6$ to $n+7$ and so must jump from $n+6$ to $n+13$. That is, if $r_n = 7$, then $r_{n+6} = 7$... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 63 | |
0aa8 | Problem:
A set of different positive integers is called meaningful if for any finite nonempty subset the corresponding arithmetic and geometric means are both integers.
a) Does there exist a meaningful set which consists of 2019 numbers?
b) Does there exist an infinite meaningful set?
Note: The geometric mean of the... | [
"Solution:\n\na) Notice that $\\{2019!\\cdot 1^{2019!}, 2019!\\cdot 2^{2019!}, \\ldots, 2019!\\cdot 2019^{2019!}\\}$ is such a set. Observe that if all the elements are divisible by $2019!$ then the arithmetic means will be integer for all the subsets. Also, if $A$ is a set such that the geometric means are integer... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) Yes. b) No. | |
0gph | Let $p$ be a prime, $n$ be a positive integer, and let $Z_{p^n}$ denote the set of congruence classes modulo $p^n$. Determine the number of functions $f : Z_{p^n} \to Z_{p^n}$ satisfying the condition
$$
f(a) + f(b) \equiv f(a + b + pab) \pmod{p^n}
$$
for all $a, b \in Z_{p^n}$. | [
"The answer is $p^n$.\nLet $a_i = \\frac{(1+p)^i - 1}{p}$ for $i \\ge 0$. By induction we obtain $f(a_i) \\equiv i f(1) \\pmod{p^n}$ for all $i \\ge 0$.\nIf $p > 2$ or $p = 2$ and $n = 1$, then\n$$\na_i \\equiv a_j \\pmod{p^n} \\Rightarrow (1+p)^{j-i} \\equiv 1 \\pmod{p^{n+1}} \\Rightarrow i \\equiv j \\pmod{p^n}.\... | Turkey | Team Selection Test for IMO 2011 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | p^n | |
0ci4 | a) Prove that the numbers of the positive divisors of $137$, $138$ and $139$ are integer powers of $2$.
b) What is the largest number of consecutive positive integers such that each of them has the number of its positive divisors a power of $2$? | [
"a) The numbers $137$ and $139$ are prime ($137 < 139 < 13^2$ and they have no positive divisors less than $13$), so each of them have $2^1$ divisors.\nSince $138 = 2^1 \\cdot 3^1 \\cdot 23^1$, it results that $138$ has $2 \\cdot 2 \\cdot 2 = 2^3$ positive divisors.\n\nb) Extend the sequence $137$, $138$, $139$ wit... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 7 | |
02n1 | Problem:
Bolinha de gude - Três amigos jogam uma partida de bolinha de gude, convencionando que o perdedor de cada rodada dobra as bolinhas dos outros jogadores, ou seja, ele dá aos outros dois um número tal de bolinhas que eles fiquem com o dobro do que tinham no início da rodada. O primeiro jogador perdeu a primeira... | [
"Solution:\n\nDenotemos por $x$, $y$ e $z$ o número de bolinhas que cada um tinha no início da partida. Temos\n\n| | $1^{o}$ | $2^{o}$ | $3^{o}$ |\n| :---: | :---: | :---: | :---: |\n| Início | $x$ | $y$ | $z$ |\n| $1^{a}$ rodada | $x-y-z$ | $2y$ | $2z$ |\n| $2^{a}$ rodada | $2(x-y-z)$ | $2y-2z-(x-y-z)$ | $4z$ |\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | First: 104, Second: 56, Third: 32 | |
0i07 | Problem:
A fair coin is flipped every second and the results are recorded with $1$ meaning heads and $0$ meaning tails. What is the probability that the sequence $10101$ occurs before the first occurrence of the sequence $010101$? | [
"Solution:\nCall it a win if we reach $10101$, a loss if we reach $010101$. Let $x$ be the probability of winning if the first flip is a $1$, let $y$ be the probability of winning if the first flip is a $0$. Then the probability of winning is $(x + y) / 2$ since the first flip is $1$ or $0$, each with probability $... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 21/32 | |
0de5 | Let $P(x)$ be a nonconstant integer polynomial and positive integer $n$. The sequence $a_0, a_1, \ldots$ is defined by $a_0 = n$ and $a_k = P(a_{k-1})$ for $k \ge 1$. Given that for each positive integer $b$, the sequence contains a $b$-th power of some positive integer greater than 1. Prove that $\deg P = 1$. | [
"Suppose $\\deg P \\ge 2$.\n\nLet $d = \\deg P \\ge 2$. Then for large $k$, $a_k$ grows very rapidly, since $a_k = P(a_{k-1})$ and $P$ is a degree $d$ polynomial with integer coefficients.\n\nLet $b$ be any positive integer. By assumption, there exists $k$ such that $a_k = m^b$ for some integer $m > 1$.\n\nLet us f... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ds8 | There are $2017$ distinct points in the plane. For each pair of these points, construct the midpoint of the segment joining the pair of points. What is the minimum number of distinct midpoints among all possible ways of placing the points? | [
"Suppose the points are placed on the $x$-axis with coordinates $(i, 0)$, $i = 0, \\dots, 2016$. Then midpoints are $(i/2, 0)$, $i = 1, 2, \\dots, 4031$. Thus there are $4031$ distinct midpoints.\n\nNext we shall prove that there are at least $4031$ distinct midpoints. Let $A_1, \\dots, A_{2017}$ be the points and ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4031 | |
0hci | Determine all positive $x, y, z$, that satisfy the following system of inequalities:
$$
\begin{cases} (x+1)(y+1) \le (z+1)^2, \\ \left(\frac{1}{x}+1\right)\left(\frac{1}{y}+1\right) \le \left(\frac{1}{z}+1\right)^2. \end{cases}
$$ | [
"Consider the following substitutions: $a = x+1 > 1$, $b = y+1 > 1$ and $c = z+1 > 1$. Then the first inequality is $ab \\le c^2$. The second inequality can be rewritten the following way (using the first one):\n$$\n\\frac{ab}{(a-1)(b-1)} \\le \\left(\\frac{c}{c-1}\\right)^2 = \\left(1+\\frac{1}{c-1}\\right)^2 \\le... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | All positive triples with x = y = z | |
01o1 | a) Is it possible for some $n$ to arrange 100 pawns in the cells of an $n \times n$ table so that at most one pawn stands in each cell and exactly 2 pawns stand in each $2 \times 2$ square?
b) Is it possible to arrange 110 pawns in the same way? | [
"a) It is impossible.\n\nFor $n = 14$ we can divide the table into 49 of $2 \\times 2$ squares. If there are exactly 2 pawns in each of them, then the total number of the pawns in the table is equal to $2 \\cdot 49 = 98$. Since we have more than 98 pawns, we have $n > 14$. On the other hand, if $n \\geq 15$, then t... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) No; b) Yes | |
0aer | За бројот $a$, е исполнето равенството $a + \frac{1}{a} = 1$. Пресметaj ја вредноста на $a^5 + \frac{1}{a^5}$. | [
"Ќе воведеме ознака $b = \\frac{1}{a}$. Тогаш $a+b=1$ и $ab=1$, па според тоа\n$$\na^2 + b^2 = a^2 + 2ab + b^2 - 2ab = (a+b)^2 - 2ab = 1^2 - 2 \\cdot 1 = -1\n$$\n$$\na^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)(a^2 + b^2 - ab) = 1 \\cdot (-1 - 1) = -2\n$$\n$$\na^5 + b^5 = (a^2 + b^2)(a^3 + b^3) - a^2 b^2 (a+b) = (-1)(... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Macedonian, English | proof and answer | 1 | |
06nh | $ABCD$ is a square. $P$ is a point inside $ABCD$ such that $\angle APD + \angle BPC = 180^\circ$ and $\angle BPC$ is acute. If $PB = 3$ and $PC = 4$, find $BC$. | [
"Answer: $1 + 2\\sqrt{2}$\n\nWe translate $\\triangle APD$ to $\\triangle BQC$ so that the condition becomes $\\angle BQC + \\angle BPC = 180^\\circ$. This means $BQCP$ is a cyclic quadrilateral. Draw a line through $P$ parallel to $AB$. This line passes through $Q$, and let $E$ and $F$ be the points where this lin... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1 + 2√2 | |
0fvx | Problem:
Längs eines Kreises stehen die Zahlen $1, 2, \ldots, 2006$ in beliebiger Reihenfolge. Es können nun wiederholt zwei auf dem Kreis benachbarte Zahlen miteinander vertauscht werden. Nach einer Folge solcher Vertauschungen steht jede der Zahlen diametral gegenüber ihrer Anfangsposition. Beweise, dass mindestens ... | [
"Solution:\n\nFür $1 \\leq k \\leq 2006$ sei $a_{k}$ wie folgt definiert: Am Anfang ist $a_{k}=0$ und jedes Mal, wenn die Zahl $k$ eine Position im (gegen den) Uhrzeigersinn wandert, erhöht (erniedrigt) sich $a_{k}$ um $1$. Die Größe $a_{1} + \\ldots + a_{2006}$ ist am Anfang gleich $0$ und ändert sich bei einer Ve... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
00bh | The rows and the columns of a $16 \times 16$ table are labeled from $1, 2, \dots, 16$, and the product $i \cdot j$ is written in the square in row $i$, column $j$. Several rows are chosen (at least $2$) and also several columns (at least $2$). Then the numbers at their intersections are deleted.
a) Can the sum of all ... | [
"a) The sum of the deleted numbers is always composite. Let the chosen rows be $i_1, \\dots, i_p$, $p \\ge 2$ and the chosen columns $j_1, \\dots, j_q$, $q \\ge 2$. The deleted numbers in row $i_1$ are $i_1j_1, i_1j_2, \\dots, i_1j_q$.\nLikewise the deleted numbers in row $i_2$ are $i_2j_1, i_2j_2, \\dots, i_2j_q$ ... | Argentina | Argentina_2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) No. b) Yes, for example 271. | |
0kq6 | Problem:
Diana is playing a card game against a computer. She starts with a deck consisting of a single card labeled $0.9$. Each turn, Diana draws a random card from her deck, while the computer generates a card with a random real number drawn uniformly from the interval $[0,1]$. If the number on Diana's card is large... | [
"Solution:\n\nBy linearity of expectation, we can treat the number of turns each card contributes to the total independently. Let $f(x)$ be the expected number of turns a card of value $x$ contributes (we want $f(0.9)$). If we have a card of value $x$, we lose it after $1$ turn with probability $1-x$. If we don't l... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 100 | |
0eqz | At a height of *h* metres above sea level, the distance $d$, in kilometres, to the visible horizon is given approximately by the formula $d = 8\sqrt{\frac{h}{5}}$. A mountaineer on the side of Table Mountain can see a distance of $80$ kilometres to the horizon. How high, in metres, is the mountaineer above sea level?
... | [
"Since $d = 8\\sqrt{\\frac{h}{5}}$, it follows that $d^2 = \\frac{64h}{5}$, so $h = \\frac{5}{64}d^2$. With $d = 80$, this gives $d^2 = 6400$, so $h = \\frac{5}{64} \\times 6400 = 500$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | B | |
09mi | We define a triangle as *large* if its area is greater than the area of a given convex pentagon, with the vertices of the triangle chosen from those of the pentagon. Determine the maximum number of large triangles that can be formed within a convex pentagon
(Gonchigdorj Radnaasumberel) | [
"Answer: 4.\n\nFirst, we construct a convex pentagon with 4 large triangle. Let $AB_1B_2C_1C_2$ be the convex pentagon such that $AB_1 = B_2C_1 = C_2A = 1$ and $B_1B_2 = C_1C_2 = \\epsilon > 0$. When $\\epsilon$ is sufficiently small each triangle $AB_iC_j$ is big for each $i$ and $j$.\n\nNow we show that the numbe... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | English | proof and answer | 4 | |
0k5d | In the nation of Onewaynia, certain pairs of cities are connected by one-way roads. Every road connects exactly two cities (roads are allowed to cross each other, e.g., via bridges), and each pair of cities has at most one road between them. Moreover, every city has exactly two roads leaving it and exactly two roads en... | [
"In the language of graph theory, we have a simple digraph $G$ which is 2-regular and we seek the number of sub-digraphs which are 1-regular. We now present two solution paths.\n\n**First solution, combinatorial** We construct a simple undirected bipartite graph $\\Gamma$ as follows:\n* the vertex set consists of t... | United States | USA TSTST | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
02ih | Problem:
Há $1002$ balas de banana e $1002$ balas de maçã numa caixa. Lara tira, sem olhar o sabor, duas balas da caixa. Se $q$ é a probabilidade das duas balas serem de sabores diferentes e $p$ é a probabilidade das duas balas serem do mesmo sabor, qual o valor de $q-p$?
A) $0$
B) $1/2004$
C) $1/2003$
D) $2/2003$
E)... | [
"Solution:\n\nA primeira bala pode ser de qualquer sabor; para fixar idéias suponhamos que seja de banana. Depois que esta bala é retirada, sobram $1002 + 1001$ balas na caixa — no nosso caso, $1002$ de maçã e $1001$ de banana.\n\nA probabilidade $q$ de que a segunda bala seja diferente (no nosso exemplo, de maçã) ... | Brazil | Brazilian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | C | |
09i1 | Let $ABC$ be a triangle and let $M$ be a point on the side $AB$ and let $N$ be a point on the side $AC$. Suppose that $CM$ and $BN$ intersect at point $D$ and suppose that the circumcircles of $BDM$ and $BNC$ intersect again at point $E$. Show that $M, N, E$ are collinear if $\angle BDM = \angle ACB$.
(Proposed by Khu... | [] | Mongolia | Round 2 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0cj3 | Let $f : [0, \infty) \to \mathbb{R}$ be a differentiable function, with continuous derivative, such that $f(0) > 0$, $f'(x) + (f(x))^2 > 0$, for any $x > 0$ and
$$
\lim_{x \to \infty} [f'(x) + (f(x))^2] = 0.
$$
Show that $\lim_{x \to \infty} f(x) = 0$. | [] | Romania | 75th NMO | [
"Calculus > Differential Equations > ODEs",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Limits"
] | English | proof only | null | |
0cg7 | Find all the integers $n$ such that there is a permutation $a_1, a_2, \dots, a_{1013}$ of the numbers $1012, 1013, \dots, 2024$, so that $a_1n^{1012} + a_2n^{1011} + \dots + a_{1012}n + a_{1013} = 0$. | [
"If $n \\in \\mathbb{N}$, then $a_1n^{1012} + a_2n^{1011} + \\dots + a_{1012}n + a_{1013} > 0$, hence $n \\in \\mathbb{N}$ is not suitable.\n\nIf $n = -k$, with $k \\ge 2$, then $a_1n^{1012} + a_2n^{1011} + \\dots + a_{1012}n + a_{1013} = k^{1011}(a_1k - a_2) + k^{1009}(a_3k - a_4) + \\dots + k(a_{1011}k - a_{1012}... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | n = -1 | |
0ar3 | Problem:
Suppose $x(x-b-3) = -2(b+1)$. Find $x$. | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | x = 2 or x = b + 1 | |
01ew | An infinite set $B$ consisting of non-negative integers has the following property. For each $a, b \in B$ ($a > b$) the number $\frac{a-b}{(a,b)} \in B$. Prove that $B$ contains all non-negative integers. Here $(a, b)$ is the greatest common divisor of numbers $a$ and $b$. | [
"If $d$ is the greatest common divisor of all the numbers in set $B$, let $A = \\{b/d : b \\in B\\}$. Then for each $a, b \\in A$ ($a > b$) we have\n$$\n\\frac{a-b}{d(a,b)} \\in A. \\quad (*)\n$$\nObserve that the greatest common divisor of the set $A$ equals $1$, therefore we can find a finite subset $A_1 \\subset... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
02e0 | A convex quadrilateral is inscribed in a circle of radius $1$. Show that its perimeter minus the sum of its two diagonals lies between $0$ and $2$. | [
"Let the quadrilateral be $ABCD$. We have $AB + BC > AC$ and $CD + DA > AC$, so the perimeter is bigger than $2 \\cdot AC$ and, similarly, bigger than $2 \\cdot BD$. Hence it's bigger than $AC + BD$, and the perimeter minus the sum of the diagonals is bigger than zero. Notice that this inequality is sharp: just con... | Brazil | VII OBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
00oo | On a table, we have ten thousand matches, two of which are inside a bowl.
Anna and Bernd play the following game: They alternate taking turns and Anna begins. A turn consists of counting the matches in the bowl, choosing a proper divisor $d$ of this number and adding $d$ matches to the bowl. The game ends when more th... | [
"Anna's strategy consists of always adding a single match to the bowl while there are less than 1350 matches inside.\nWith this strategy, she will always change an even number of matches to an odd number of matches, so that Bernd is forced to choose an odd divisor and give her an even number of matches again.\nBern... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
00b0 | In a convex quadrilateral $ABCD$ we have that:
* $R$ and $S$ are points in the interior of the segments $CD$ and $AB$ respectively, with $AD = CR$ and $BC = AS$.
* $P$ and $Q$ are the midpoints of $DR$ and $SB$ respectively.
* $M$ is the midpoint of $AC$.
If it is known that $MPC + MQA = 90°$, prove that $ABCD$ is a cy... | [
"Let $E$ and $F$ be points in the prolongations of $AB$ and $CD$ such that $DF = DA = CR$ and $BE = BC = AS$, as in the picture.\n\nNote that $P$ is the midpoint of $FC$. Then, $MP$ is a midsegment of the triangle $AFC$, which implies that $A\\hat{F}C = M\\hat{P}C$. Since the triangle $ADF$... | Argentina | 29° Olimpiada Matemática del Cono Sur | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08j6 | Problem:
$n$ points are given in a plane, not three of them colinear. One observes that no matter how we label the points from $1$ to $n$, the broken line joining the points $1,2,3, \ldots, n$ (in this order) does not intersect itself.
Find the maximal value of $n$.
Problem:
Fie $n$ puncte în plan, oricare trei neco... | [
"Solution:\n\nNotice that $n=4$ satisfies the condition. Indeed, for a concave quadrilateral, this can be checked immediately.\n\nThen, observe that for $n \\geq 5$ one can choose four points $A, B, C, D$ such that $ABCD$ is a convex quadrilateral. The diagonals $AC$ and $BD$ intersect at a point, hence labeling $A... | JBMO | 7th JBMO | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof and answer | 4 | |
05mq | Problem:
Soit $ABC$ un triangle non isocèle en $A$ et $\Gamma$ son cercle circonscrit. Soit $N$ le milieu de l'arc de $\Gamma$ entre $B$ et $C$ qui contient $A$. Soient $I$ le centre du cercle inscrit à $ABC$, $J$ le centre du cercle $A$-exinscrit à $ABC$ et $K$ le point d'intersection de $BC$ avec la bissectrice exté... | [
"Solution:\n\n\n\nNotons $J_b$ et $J_c$ les centres des cercles exinscrits dans les angles $\\widehat{B}$ et $\\widehat{C}$. Alors $I$ est l'orthocentre de $J_b J_c$. En utilisant le fait que $B$ et $C$ sont sur le cercle de diamètre $J_b J_c$, et que $A, B, C, N$ sont cocycliques, on a $\\... | France | TEST DU GROUPE A ET DES CANDIDATES À L'ÉPREUVE EGMO | [
"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 > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
055e | Let $n$ be a natural number, $n \ge 3$. Find the maximal number of diagonals of a regular $n$-gon one can select in such a way that every two selected diagonals that intersect each other inside the polygon are perpendicular. | [
"If $n$ is odd, one can select all $n - 3$ diagonals connecting one fixed vertex to others. In order to prove that the conditions of the problem do not allow more, it suffices to show that no two diagonals are perpendicular. Fix one diagonal arbitrarily; it partitions the boundary of the polygon into two halves, ou... | Estonia | IMO Team Selection Contest II | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | English | proof and answer | If n is odd: n − 3. If n is even: n − 2. | |
0ayo | Problem:
In the figure below, the incircle of the isosceles triangle has radius $3$. The smaller circle is tangent to the incircle and the two congruent sides of the triangle. If the smaller circle has radius $2$, find the length of the base of the triangle.
 | [] | Philippines | 21st PMO Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Tr... | null | proof and answer | 3√6 | |
00yb | Problem:
In a certain kingdom, the king has decided to build 25 new towns on 13 uninhabited islands so that on each island there will be at least one town. Direct ferry connections will be established between any pair of new towns which are on different islands. Determine the least possible number of these connections... | [
"Solution:\n\nLet $a_{1}, \\ldots, a_{13}$ be the numbers of towns on each island. Suppose there exist numbers $i$ and $j$ such that $a_{i} \\geq a_{j} > 1$ and consider an arbitrary town $A$ on the $j$-th island. The number of ferry connections from town $A$ is equal to $25 - a_{j}$. On the other hand, if we \"mov... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 222 | |
087o | Problem:
Antonio, Beppe, Carlo e Duccio si distribuiscono casualmente le 40 carte di un mazzo, 10 a testa. Antonio ha l'asso, il due e il tre di denari. Beppe ha l'asso di spade e l'asso di bastoni. Carlo ha l'asso di coppe. Chi è più probabile che abbia il 7 di denari?
(A) Antonio
(B) Beppe
(C) Carlo
(D) Duccio
(E) ... | [
"Solution:\n\nLa risposta è (D). Intuitivamente, le posizioni delle carte rimanenti sono equiprobabili, quindi è più probabile che il 7 di denari finisca a Duccio \"perché ha più spazio libero\". Più formalmente: contiamo i modi di distribuire 34 carte tra 4 persone, in modo che uno ne riceva 10, uno 9, uno 8 e uno... | Italy | Olimpiadi di Matematica | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | D | |
0jgs | Problem:
a. Prove that there are $2012$ points on the unit circle such that the distance between any two of them is rational.
b. Does there exist an infinite set of points on the unit circle such that the distance between any two of them is rational? | [
"Solution:\n\nThe answer to part (b) is yes.\nFor brevity, we use the notation of complex numbers. For any integers $x$ and $y$, not both zero, let $z = x + y i$ and\n$$\nP(z) = \\frac{z^{2}}{\\bar{z}^{2}}\n$$\nClearly, $|P(z)| = 1$ so $P(z)$ is a point on the unit circle. We claim that for any $z_{1} = x_{1} + y_{... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | Yes | |
0cue | Assume that $n$ is a composite positive integer. For each its proper divisor $d$, we write down the number $d+1$ on the board. Find all values of $n$ for which the written numbers appear to be exactly all the proper divisors of some positive integer $m$. (A proper divisor of a positive integer $a > 1$ is any its positi... | [
"$n = 4$ and $n = 8$.\n\nThe number $2$ does not appear on the board, so $m$ is odd. Therefore, all numbers on the board are odd, so $n$ is a power of $2$. Assume that $a \\ge b \\ge c$. Set $G(x) = P(x)/x^n$ and notice that $G(a) \\le G(b) \\le G(c)$.",
"$n = 4$ или $n = 8$.\n\nПервое решение. Заметим, что число... | Russia | XLIII Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English; Russian | proof and answer | n = 4 or n = 8 | |
01gj | Alice and Bob are playing hide and seek. Initially, Bob chooses a secret fixed point $B$ in the unit square. Then Alice chooses a sequence of points $P_0, P_1, \ldots, P_N$ in the plane. After choosing $P_k$ (but before choosing $P_{k+1}$) for $k \ge 1$, Bob tells "warmer" if $P_k$ is closer to $B$ than $P_{k-1}$, othe... | [
"Let $S_0$ be the set of all points in the square, and for each $1 \\le k \\le N$, let $S_k$ be the set of possible points $B$ consistent with everything Bob has said. For each $k$, we then have that $S_k$ is the disjoint union of the two possible values $S_{k+1}$ can take for each of Bob's possible answers. Hence ... | Baltic Way | Baltic Way 2020 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0a1m | Ikram has a large bowl with little balls in it. On each ball a positive integer is written. If he randomly picks three balls from the bowl and takes the difference between the largest and smallest number on these three balls, it turns out that the outcome is always also on one of the balls in the bowl (or on one of the... | [
"$8$."
] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 8 | |
0ftd | Problem:
Gegeben sei ein Parallelogramm $ABCD$ und ein Punkt $O$ in dessen Innern, sodass $\Varangle AOB + \Varangle DOC = \pi$. Zeige dass gilt
$$
\Varangle CBO = \Varangle CDO
$$ | [
"Solution:\n\nKonstruiere den Punkt $Q$ auf der anderen Seite von $CD$ wie $O$, sodass $CQ \\parallel BO$ und $DQ \\parallel AO$. Nun gilt $\\Varangle DQC = \\Varangle AOB = \\pi - \\Varangle DOC$ und damit ist $CODQ$ ein Sehnenviereck. Da $BCQO$ ein Parallelogramm ist, gilt $\\Varangle CBO = \\Varangle CQO$. Da $C... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0jw2 | Problem:
The distinct prime factors of an integer are its prime factors listed without repetition. For example, the distinct prime factors of $40$ are $2$ and $5$.
Let $A = 2^{k} - 2$ and $B = 2^{k} \cdot A$, where $k$ is an integer $(k > 1)$.
Show that, for every choice of $k$,
a. $A$ and $B$ have the same set of ... | [
"Solution:\n\na. Since $B$ is given as a multiple of $A$, every prime that divides $A$ also divides $B$.\n\nConversely, suppose $p$ is a prime that divides $B$. Since $B = 2^{k} \\cdot A$, either $p$ divides $2^{k}$ or $p$ divides $A$. If $p$ divides $2^{k}$, then $p = 2$. But then $p$ divides $A$ anyway, because $... | United States | BAMO-8 and BAMO-12 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
07r1 | The images under reflection of the circumcentre of triangle $ABC$ in the sides of the triangle are $X$, $Y$, and $Z$. Prove $\triangle XYZ$ is congruent to $\triangle ABC$ and corresponding sides are parallel. | [
"Let $X$, $Y$ and $Z$ be the reflections in $BC$, $CA$ and $AB$ respectively and let $E$ and $F$ be the midpoints of $CA$ and $AB$ respectively.\n\n\n\nSince $Z$ is the reflection of $O$ in $AB$, $AZ = AO = OB$. Similarly $AY = OC$.\nAlso $ZZAB = LOAB$ and $ZYAC = LOAC$, hence $LBAC = \\fra... | Ireland | Ireland_2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
08ml | Problem:
Find all positive integers $n$ such that $36^{n}-6$ is a product of two or more consecutive positive integers. | [
"Solution:\nAnswer: $n=1$.\n\nAmong each four consecutive integers there is a multiple of $4$. As $36^{n}-6$ is not a multiple of $4$, it must be the product of two or three consecutive positive integers.\n\nCase I. If $36^{n}-6 = x(x+1)$ (all letters here and below denote positive integers), then $4 \\cdot 36^{n} ... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebr... | null | proof and answer | n = 1 | |
061m | Problem:
Beim Schachspiel erhält der Sieger 1 Punkt und der Besiegte 0 Punkte. Bei Unentschieden (Remis) erhält jeder der Spieler $\frac{1}{2}$ Punkt.
Vierzehn Schachspieler, von denen keine zwei gleich alt waren, trugen einen Wettbewerb aus, in dem jeder gegen jeden spielte. Nach Abschluss des Wettbewerbs wurde eine ... | [
"Solution:\n\nDie Gesamtzahl der Punkte der letzten neun Spieler beträgt mindestens $(9 \\cdot 8) : 2 = 36$ Punkte (denn wenn nur jeder der neun gegen einen anderen der neun spielen würde, dann wären es, da in jeder Partie ein Punkt vergeben wird, insgesamt schon 36 Punkte). Die Gesamtzahl der drei Erstplatzierten ... | Germany | Auswahlwettbewerb zur IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 40 | |
0kpo | Problem:
Find, with proof, the maximum positive integer $k$ for which it is possible to color $6k$ cells of a $6 \times 6$ grid such that, for any choice of three distinct rows $R_{1}, R_{2}, R_{3}$ and three distinct columns $C_{1}, C_{2}, C_{3}$, there exists an uncolored cell $c$ and integers $1 \leq i, j \leq 3$ so... | [
"Solution:\nThe answer is $k=4$. This can be obtained with the following construction:\n\n\n\nIt now suffices to show that $k=5$ and $k=6$ are not attainable. The case $k=6$ is clear. Assume for sake of contradiction that $k=5$ is attainable. Let $r_{1}, r_{2}, r_{3}$ be the rows of three d... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 4 | |
0f5u | Problem:
A number is prime however we order its digits. Show that it cannot contain more than three different digits. For example, $337$ satisfies the conditions because $337$, $373$ and $733$ are all prime. | [] | Soviet Union | 18th ASU | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0isw | Problem:
Suppose that $j(0) j(1) \cdots j(n-1)$ is a valid juggling sequence. For $i=0,1, \ldots, n-1$, let $a_{i}$ denote the remainder of $j(i)+i$ when divided by $n$. Prove that $(a_{0}, a_{1}, \ldots, a_{n-1})$ is a permutation of $(0,1, \ldots, n-1)$. | [
"Solution:\n\nSuppose that $a_{i}=j(i)+i-b_{i} n$, where $b_{i}$ is an integer. Note that $f\\left(i-b_{i} n\\right)=i-b_{i} n+j(i)=a_{i}$. Since $\\{i-b_{i} n \\mid i=0,1, \\ldots, n-1\\}$ contains $n$ distinct integers (as their residue $\\bmod n$ are all distinct), and $f$ is a permutation, we see that after app... | United States | 11th Annual Harvard-MIT Mathematics Tournament - Team Round: B Division | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
03ik | Problem:
$P(x)$ and $Q(x)$ are two polynomials that satisfy the identity $P(Q(x)) \equiv Q(P(x))$ for real numbers $x$. If the equation $P(x)=Q(x)$ has no real solution, show that the equation $P(P(x))=Q(Q(x))$ also has no real solution. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
05wb | Problem:
Trouver les entiers $n$ pour lesquels l'égalité
$$
\sum_{i=1}^{n} \sum_{j=1}^{n}\left\lfloor\frac{i \times j}{n+1}\right\rfloor=\frac{n^{2}(n-1)}{4}
$$
est vérifiée.
On rappelle que, pour tout nombre réel $x$, la notation $\lfloor x\rfloor$ désigne la partie entière de $x$, c'est-à-dire le plus grand entier i... | [
"Solution:\n\nPour tout entier $k$, on note $r_{n+1}(k)$ le reste de la division euclidienne de $k$ par $n+1$, c'est-à-dire le plus petit entier naturel tel que $n+1$ divise $k-r_{n+1}(k)$. La double somme de gauche se réécrit alors comme\n$$\n\\begin{aligned}\n\\mathcal{S} & =\\sum_{i=1}^{n} \\sum_{j=0}^{n} \\frac... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof and answer | All integers such that n+1 is prime | |
0h76 | There is number in every cell of the table $5 \times 5$ written under such conditions: not all number should be different, however, there is no row or column where all five numbers are equal; the middle number (the third one) in every row and column equals the mean of the numbers in its row or column. What is the minim... | [
"Let us prove that there cannot be less than three such numbers. Let the middle number of the table be $a$. Then there exists at least one number that is less than $a$ in the third row and column. If there is no such number then all the numbers in the row or column are the same, which contradicts the conditions. Wi... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3 | |
0kfl | Problem:
Let $P(x)$ be the monic polynomial with rational coefficients of minimal degree such that $\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{4}}, \ldots, \frac{1}{\sqrt{1000}}$ are roots of $P$. What is the sum of the coefficients of $P$? | [
"Solution:\nFor irrational $\\frac{1}{\\sqrt{r}}, -\\frac{1}{\\sqrt{r}}$ must also be a root of $P$. Therefore\n$$\nP(x) = \\frac{\\left(x^{2} - \\frac{1}{2}\\right)\\left(x^{2} - \\frac{1}{3}\\right) \\cdots \\left(x^{2} - \\frac{1}{1000}\\right)}{\\left(x + \\frac{1}{2}\\right)\\left(x + \\frac{1}{3}\\right) \\cd... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 1/16000 | |
0ctc | Let $ABCD$ be a trapezoid with $BC \parallel AD$ and $AB = CD$. A circle $\omega$ centered at $I$ is tangent to the segments $AB$, $CD$, and $DA$. The circle $BIC$ meets the side $AB$ at points $B$ and $E$. Prove that $CE$ is tangent to $\omega$.
Внутри равнобокой трапеции $ABCD$ с основаниями $BC$ и $AD$ расположена ... | [
"Notice that $\\angle ICD = \\angle IBA = \\angle IBE = \\angle ICE$ (see Fig. 7). So the lines $CD$ and $CE$ are symmetric in $IC$.\n\n",
"Заметим, что $I$ лежит на оси симметрии трапеции, поэтому $\\angle ICD = \\angle IBA$. Пользуясь вписанностью четырехугольника $CBEI$, получаем $\\an... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0ah8 | A total of $2^n$ coins are distributed among several children. If one of the children has at least half of the coins, the coins are redistributed: coins are transferred from such a child to each of the other children in such a way that each of them gets as many coins as it had. In the case when one child possesses all ... | [
"At most $n$ consecutive redistributions. We will start with an example showing that $n$ consecutive redistributions are possible. Let $2^n$ coins be distributed among $3$ children initially as follows: $1$, $2^{n-1} + 2^{n-2} + \\dots + 2$, $1$. The successive redistributions (a total of $n$) will be:\n$$\n2^1, 2^... | North Macedonia | Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | n | |
0c7w | a) Determine the four digit numbers in base 10 which are written in base two only with the digit $1$.
b) How many four digit numbers in base 10 are, such that in base two they contain exactly one zero digit. | [] | Romania | 70th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (a) 1023, 2047, 4095, 8191; (b) 38 | |
00gc | For a real number $x$, let $\lfloor x\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that
$$
\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor
$$
is even for every positive integer $n$. | [
"Consider four cases:\n- $n \\leq 5$. Then $\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor=0$ is an even number.\n\n- $n$ and $n+1$ are both composite (in particular, $n \\geq 8$). Then $n=ab$ and $n+1=cd$ for $a, b, c, d \\geq 2$. Moreover, since $n$ and $n+1$ are coprime, $a, b, c, d$ are all distinct and sm... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0jyp | Problem:
a. Does $\sum_{i=1}^{p-1} \frac{1}{i} \equiv 0 \pmod{p^2}$ for all odd prime numbers $p$? (Note that $\frac{1}{i}$ denotes the number such that $i \cdot \frac{1}{i} \equiv 1 \pmod{p^2}$)
b. Do there exist 2017 positive perfect cubes that sum to a perfect cube?
c. Does there exist a right triangle with ratio... | [] | United States | February 2017 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Pythagorean triples",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Other"
] | null | proof and answer | a: No. (It holds for primes at least five, but fails for three.)
b: Yes. For example, by scaling 3^3 + 4^3 + 5^3 = 6^3 and splitting into equal cubes using 2017 = 11^3 + 7^3 + 7^3.
c: Yes. Example sides 3/2, 20/3, 41/6 give area 5.
d: No. A prime-only 3×3 magic square cannot exist.
e: No. The product equals 90/π^4, whi... | |
0bi6 | Let $n$ be a strictly positive integer. For any natural number $k$, we denote by $a(k)$ the number of natural divisors $d$ of $k$ such that $k \le d^2 \le n^2$. Compute the sum $\sum_{k=1}^{n^2} a(k)$. | [
"$$\nS(n) = \\{(k, d) \\mid d \\text{ divides } k, k \\le d^2 \\le n^2, 1 \\le k \\le n^2\\}.\n$$\nFor $1 \\le k \\le n^2$, let $A(k)$ be the set of the natural divisors $d$ of $k$ such that $k \\le d^2 \\le n^2$.\nA natural number $d \\in \\{1, 2, \\dots, n\\}$ belongs to the sets $A(d)$, $A(2d)$, $\\dots$, $A(d^2... | Romania | 65th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | n(n+1)/2 | |
0fst | Problem:
Es sei $q(n)$ die Quersumme der natürlichen Zahl $n$. Bestimme den Wert von
$$
q\left(q\left(q\left(2000^{2000}\right)\right)\right)
$$ | [
"Solution:\n\nEs gilt $q\\left(2000^{2000}\\right)=q\\left(2^{2000}\\right)$, denn die beiden Zahlen unterscheiden sich nur um angehängte Nullen. Wir schätzen nun ab. Es gilt $2^{2000}=4 \\cdot 8^{666}<10^{667}$ und somit\n$$\nq\\left(2^{2000}\\right) \\leq 9 \\cdot 667=6003\n$$\nWeiter ist damit\n$$\nq\\left(q\\le... | Switzerland | IMO - Selektion | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4 | |
01xo | Find all functions $f(x): \mathbb{R} \to \mathbb{R}$ satisfying for all $x \in \mathbb{R}$ the inequality
$$
x = \frac{3}{4}f(|x|) + |f(x)|.
$$
(I. Voronovich) | [
"Answer: $f(x) = -4x$ for all $x \\ge 0$ and $f(x) = \\pm 2x$ for all $x < 0$ (signs are chosen independently)."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All functions f: R -> R such that f(x) = -4x for x >= 0, and for each x < 0 one may choose independently f(x) = 2x or f(x) = -2x. | |
0bc8 | The function $f : [0, 1] \to \mathbb{R}$ is differentiable and
$$
\int_{0}^{1} f(x) \, dx = \int_{0}^{1} x f(x) \, dx = 0.
$$
Prove that there exists $c \in (0, 1)$ such that $f'(c) = 0$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Calculus > Differential Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof only | null | |
0ffa | Problem:
Representar gráficamente la función
$$
y = |||x-1|-2|-3|
$$
en el intervalo $-8 \leq x \leq 8$. | [
"Solution:\n\nPrimera solución\n\nTomemos en primer lugar la variable independiente auxiliar $\\xi = x-1$. Se trata de representar la función\n$$\ny(\\xi) = |||\\xi|-2|-3|\n$$\nen el intervalo $-9 \\leq \\xi \\leq 7$.\nPero la función $y(\\xi)$ es par, así que basta trazar su gráfica en el intervalo $[0,9]$, para t... | Spain | Olimpiadas Matemáticas Españolas | [
"Precalculus > Functions"
] | null | final answer only | null | |
01h3 | For $n = \prod_i p_i^{e_i}$ ($p_i$ primes, $e_i$ positive integers), define $\psi(n) = \prod_i (p_i + 1)^{e_i - 1}$. Show that by repeated application of $\psi$, we may transform any starting number into a number of the form $2^m$ for $m \ge 0$. | [
"As $p + 1$ is even when $p$ is odd, we have that the greatest prime divisor of $p + 1$ is strictly less than $p$ for all primes $p \\neq 2$. Therefore, the greatest prime divisor of the number on the blackboard decreases at every step, unless it is $2$, in which case the number is evidently of the desired form."
] | Baltic Way | Baltic Way 2020 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
04er | Determine all positive integers $n$ such that
$$
\lfloor \sqrt[4]{1} \rfloor + \lfloor \sqrt[4]{2} \rfloor + \dots + \lfloor \sqrt[4]{n} \rfloor = \frac{3}{2}n + 1.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 32 | |
016r | Find all positive integers $n$ for which there exists a positive real number $C$ such that
$$
\sum_{1 \le i < j \le n} x_i x_j \le C (x_1 x_2 + \dots + x_{n-1} x_n + x_n x_1)
$$
for all positive real numbers $x_1, \dots, x_n$. | [
"Answer: $1$, $2$, $3$.\nFor $n = 1$, $n = 2$ or $n = 3$, the inequality holds trivially for all positive $C$, for all $C \\ge \\frac{1}{2}$, and for all $C \\ge 1$, respectively.\n\nSuppose $n \\ge 4$. Let two positive real numbers $s$ and $t$ such that $s < t$ be fixed, and choose numbers $x_1, x_2, \\dots, x_n$ ... | Baltic Way | BALTIC WAY | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 1, 2, 3 | |
0c92 | Let $\mathbb{N}$ be the set of all positive integers. A subset $A$ of $\mathbb{N}$ is *sum-free* if, whenever $x$ and $y$ are (not necessarily distinct) members of $A$, their sum $x+y$ does not belong to $A$. Determine all *surjective* functions $f: \mathbb{N} \to \mathbb{N}$ such that, for each sum-free subset $A$ of ... | [
"The identity is the only surjection of the positive integers onto themselves sending every sum-free set onto a sum-free set (no verification is needed, of course).\nTo prove this, fix a function $f$ satisfying the conditions in the statement, and proceed in several steps.\n\n*Step 1.* Notice that a 2-element set $... | Romania | Romanian Master of Mathematics | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Other",
"Number Theory > Other"
] | English | proof and answer | f(n) = n for all n in N | |
0lcv | Let $x$, $y$ and $z$ be positive real numbers. Find the maximum value of the following expression
$$
\frac{x^3 y^4 z^3}{(x^4 + y^4)(xy + z^2)^3} + \frac{y^3 z^4 x^3}{(y^4 + z^4)(yz + x^2)^3} + \frac{z^3 x^4 y^3}{(z^4 + x^4)(zx + y^2)^3}
$$ | [
"We will prove that the maximal value of the given expression $P$ is $\\frac{3}{16}$. Indeed, applying these inequalities\n$$\nx^4 + y^4 \\geq xy(x^2 + y^2) \\text{ and } (xy + z^2)^2 \\geq 4xyz^2,\n$$\nwe have\n$$\n(x^4 + y^4)(xy + z^2)^3 \\geq 4x^2y^2z^2(x^2 + y^2)(xy + z^2) \\\\ \\geq 4x^2y^2z^2(z^2x^2 + z^2y^2 ... | Vietnam | VMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 3/16 | |
0eyy | Problem:
A wolf is in the center of a square field and there is a dog at each corner. The wolf can run anywhere in the field, but the dogs can only run along the sides. The dogs' speed is $\frac{3}{2}$ times the wolf's speed. The wolf can kill a single dog, but two dogs together can kill the wolf. Prove that the dogs ... | [] | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null |
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