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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0b0x | Problem:
In acute triangle $ABC$ with $\angle BAC > \angle BCA$, let $P$ be the point on side $BC$ such that $\angle PAB = \angle BCA$. The circumcircle of triangle $APB$ meets side $AC$ again at $Q$. Point $D$ lies on segment $AP$ such that $\angle QDC = \angle CAP$. Point $E$ lies on line $BD$ such that $CE = CD$. T... | [
"Solution:\n\nRefer to the figure shown below. Since $ABPQ$ is cyclic, we have $CP \\cdot CB = CQ \\cdot AC$. Also, we have $\\triangle CAD \\sim \\triangle CDQ$, so $CD^2 = CQ \\cdot AC$. This means that $CE^2 = CD^2 = CQ \\cdot AC = CP \\cdot CB$, so $\\triangle CDP \\sim \\triangle CBD$ and $\\triangle CEQ \\sim... | Philippines | 21st Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
026w | Problem:
Maria está planejando participar do Triatlon-Brasil que consta de $800~\mathrm{m}$ de nado, seguido de $20~\mathrm{km}$ de bicicleta e, finalmente, $4~\mathrm{km}$ de corrida. Maria corre a uma velocidade constante que é o triplo da velocidade com que nada e pedala 2,5 vezes mais rápido do que corre. Para ter... | [
"Solution:\n\nSeja $x$ a velocidade com que Maria nada, em metros por minuto. Então o tempo que Maria gasta nadando $800~\\mathrm{m}$ é dado por $800 / x$ minutos. Sabemos que sua velocidade na corrida é $3x$ e, na bicicleta, $2,5 \\times 3x = 7,5x$ metros por minuto. Assim, o tempo total que Maria gasta nas três e... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Swimming: 60 m/min; Running: 180 m/min; Cycling: 450 m/min | |
00yh | Problem:
Is there an integer $n$ such that $\sqrt{n-1} + \sqrt{n+1}$ is a rational number? | [
"Solution:\n\nInverting the relation gives\n$$\n\\frac{q}{p} = \\frac{1}{\\sqrt{n+1} + \\sqrt{n-1}} = \\frac{\\sqrt{n+1} - \\sqrt{n-1}}{(\\sqrt{n+1} + \\sqrt{n-1})(\\sqrt{n+1} - \\sqrt{n-1})} = \\frac{\\sqrt{n+1} - \\sqrt{n-1}}{2}.\n$$\nHence we get the system of equations\n$$\n\\left\\{\n\\begin{array}{l}\n\\sqrt{... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
05gj | Problem:
À l'intérieur du cercle $\Gamma$ se trouvent trois cercles $\gamma_{1}$, $\gamma_{2}$ et $\gamma_{3}$, et tangents à $\Gamma$ respectivement en $A$, $B$ et $C$, tous distincts. Les cercles $\gamma_{2}$ et $\gamma_{3}$ ont un point commun $K$ qui appartient à $[BC]$, les cercles $\gamma_{3}$ et $\gamma_{1}$ on... | [
"Solution:\n\n\n\nSoit $X$ un point de la tangente commune à $\\Gamma$ et $\\gamma_{1}$, et situé du côté opposé à $C$ par rapport à la droite $(AB)$. Alors $\\widehat{ALM}=\\widehat{XAM}=\\widehat{XAB}=\\widehat{ACB}$, ce qui assure que $(ML)$ et $(BC)$ sont parallèles. De même, on a $(KM)... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0cwa | A point $D$ is chosen on the leg $AC$ of an isosceles triangle $ABC$ with base $BC$. A point $K$ is chosen on the smaller arc $CD$ of the circumcircle of the triangle $BCD$. The ray $CK$ meets the line through $A$ parallel to $BC$ at $T$. Let $M$ be the midpoint of $DT$. Prove that $\angle AKT = \angle CAM$.
На сторон... | [
"Продлим отрезок $AM$ на его длину за точку $M$, получим точку $N$ такую, что $ADNT$ — параллелограмм. Поскольку $\\angle ANT = \\angle CAM$, для решения задачи достаточно показать, что $\\angle AKT = \\angle ANT$, или что точки $A, T, N, K$ лежат на одной окружности (см. рис. 16).\n\n\n\nП... | Russia | Final round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Russian | proof only | null | |
01h7 | Richard and Kaarel are taking turns to choose numbers from the set $\{1, \dots, p-1\}$ where $p > 3$ is a prime. Richard is the first one to choose. Any number which has been chosen by one of the players can be chosen again be neither of the players. Every number chosen by Richard is multiplied with the very next numbe... | [
"Answer: Yes, Kaarel.\n\nLet us split the numbers in the set to the following pairs: $(1, p-1)$, $(2, p-2)$, $\\dots$, $\\left(\\frac{p-1}{2}, \\frac{p+1}{2}\\right)$. If Richard chooses some number $a$, then let Kaarel choose the other number from the pair i.e. $p-a$. This forces Richard to choose a number from a ... | Baltic Way | Baltic Way 2020 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Yes, Kaarel. | |
0et6 | If $x$, $y$, $z$ are real numbers satisfying
$$
\begin{aligned}
(x + 1)(y + 1)(z + 1) &= 3 \\
(x + 2)(y + 2)(z + 2) &= -2 \\
(x + 3)(y + 3)(z + 3) &= -1,
\end{aligned}
$$
find the value of
$(x + 20)(y + 20)(z + 20)$. | [
"By putting $X = x+2$, $Y = y+2$ and $Z = z+2$, the system of equations translates to\n\n$$\n(X - 1)(Y - 1)(Z - 1) = 3 \\quad \\dots (1)\n$$\n\n$$\nXYZ = -2 \\quad \\dots (2)\n$$\n\n$$\n(X + 1)(Y + 1)(Z + 1) = -1. \\quad \\dots (3)\n$$\n\nBy expanding the left hand sides of (1) and (3), and substituting $XYZ = -2$,... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 6748 | |
09as | Prove that every even number not exceeding $2n(4n + 1)$ can be written as $\pm 1 \pm 2 \pm 3 \pm \cdots \pm 4n$, here we choose $+$ or $-$. | [
"Let us prove by induction. For $n = 1$, we have to construct even numbers not exceeding $2 \\cdot (4 + 1) = 10$.\n$$\n\\begin{aligned}\n+1 - 2 - 3 + 4 &= 0, & -1 + 2 - 3 + 4 &= 2, & +1 + 2 - 3 + 4 &= 4, \\\\\n+1 - 2 + 3 + 4 &= 6, & -1 + 2 + 3 + 4 &= 8, & +1 + 2 + 3 + 4 &= 10\n\\end{aligned}\n$$\n\nBy induction hyp... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
02ec | $A$ and $B$ wish to divide a cake into two pieces. Each wants the largest piece he can get. The cake is a triangular prism with the triangular faces horizontal. $A$ chooses a point $P$ on the top face. $B$ then chooses a vertical plane through the point $P$ to divide the cake. $B$ chooses which piece to take. Which poi... | [
"Given any triangle $ABC$, we have to find the point $P$ inside the triangle, such that any line through $P$ divides the triangle into two pieces which are as equal as possible. We take $P$ to be the centroid $G$. We show first that in that case the smallest piece is at least $\\frac{4}{9}$ of the total area. If we... | Brazil | IX OBM | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | Choose the centroid of the triangular face (this guarantees at least 4/9 of the cake). | |
0dm8 | Let $N$ be a positive integer. Ali and Hadi play a game in which they start by writing the numbers $1, 2, \dots, N$ on a board. They then take turns to make a move, starting with Ali. Each move consists of choosing a pair of integers $(k, n)$, where $k \ge 0$ and $n$ is one of the integers on the board, and then erasin... | [
"The answer is that Geoff wins when $N$ is of the form $2^n$ for $n$ odd or of the form $t2^n$ for $n$ even and $t > 1$ odd.\n\nCommon remarks. We will say that a set $S$ wins if the current player wins given $S$ as the current set of integers on the board. Otherwise, we will say that $S$ loses. We will let $J(S, \... | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Ali can force a win exactly when N = 2^n with n odd, or when N = t·2^n with n even and odd t > 1. | |
0kur | Problem:
Suppose $a$, $b$, and $c$ are real numbers such that
$$
\begin{aligned}
a^{2} - b c &= 14 \\
b^{2} - c a &= 14, \text{ and } \\
c^{2} - a b &= -3
\end{aligned}
$$
Compute $|a+b+c|$. | [
"Solution:\nSubtracting the first two equations gives $(a-b)(a+b+c)=0$, so either $a=b$ or $a+b+c=0$. However, subtracting first and last equations gives $(a-c)(a+b+c)=17$, so $a+b+c \\neq 0$. This means $a=b$.\n\nNow adding all three equations gives $(a-c)^{2}=25$, so $a-c= \\pm 5$. Then $a+b+c= \\pm \\frac{17}{5}... | United States | HMMT November 2023 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 17/5 | |
01uf | On the Cartesian plane $Oxy$ the parabola $y = x^2 - 2bx + a + 1$ meets the $x$-axis at points $A$ and $B$, and meets the $y$-axis at point $C$ (distinct from the origin). It appears that the point with the coordinates $(a, b)$ is a center of the circumscribed circle of the triangle $ABC$.
Find all possible values of t... | [
"Answer: $a = b = 2$.\nLet $A(\\alpha, 0)$, $B(\\beta, 0)$, $C(0, \\gamma)$, and $M(a, b)$. Since $M$ is the center of the circumscribed circle of the triangle $ABC$, $M$ lies on the perpendicular bisector of the segment $AB$. Therefore, the projection of $M$ on the $x$-axis is the midpoint of the segment $AB$, i.e... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermed... | English | proof and answer | a = b = 2 | |
018d | Compute the sum
$$
\sum_{n=1}^{\infty} \frac{F_n}{10^{n+1}}
$$
where $F_n$ is the $n$th Fibonacci number given by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for all $n \geq 2$. | [
"Let\n$$\nX = \\sum_{n=1}^{\\infty} \\frac{F_n}{10^{n+1}}\n$$\nThen\n$$\nX = \\frac{1}{10^2} + \\frac{1}{10^3} + \\frac{2}{10^4} + \\frac{3}{10^5} + \\frac{5}{10^6} + \\frac{8}{10^7} + \\frac{13}{10^8} + \\dots\n$$\nSo\n$$\n10X = \\frac{1}{10} + \\frac{1}{10^2} + \\frac{2}{10^3} + \\frac{3}{10^4} + \\frac{5}{10^5} ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/89 | |
0gjh | 給定一個整數的有限集 $S$，是否總是存在一個整係數多項式 $f(x)$，使得對於所有整數 $x$，$f(x)$ 為完全平方數若且唯若 $x \in S$？
Fix a finite set $S$ of integers. Is there a polynomial $f(x)$ of integral coefficients such that for all integers $x$, $f(x)$ is a perfect square if and only if $x \in S$? | [
"考慮多項式 $f(x) = (g(x))^2 + (x^2 + 1)^2$。若 $g(x) = 0$ 對所有 $x \\in S$ 皆成立，則 $f(x)$ 為平方數。若 $g(x) > (x^2 + 1)^2$ 對於所有 $x \\in \\mathbb{Z} \\setminus S$ 皆成立，則 $(g(x))^2 < f(x) < (g(x) + 1)^2$，也就是說 $f(x)$ 不為平方數。故我們可以取 $g(x) = M_S \\prod_{s \\in S} (x-s)^{10}$，其中 $M_S$ 是一個足夠大的常數使得 $g(x) > (x^2+1)^2$ 對於所有 $x \\in \\mathbb{Z... | Taiwan | IMO 1J, Independent Study 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Chinese; English | proof and answer | Yes | |
0a1e | We consider the number $666\ldots666$ which consists of $2023$ sixes. The square of this number has $4046$ digits.
How many of those digits are a $5$? | [
"5."
] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 2022 | |
03u4 | Let $f(x) = 3\sin x + 2\cos x + 1$. If real numbers $a, b, c$ are such that $af(x) + bf(x-c) = 1$ holds for any $x \in \mathbb{R}$, then $\frac{b\cos c}{a}$ equals ( ). | [
"Let $c = \\pi$. Then $f(x) + f(x-c) = 2$ for any $x \\in \\mathbb{R}$.\nNow let $a = b = \\frac{1}{2}$, and $c = \\pi$. We have\n$$\naf(x) + bf(x-c) = 1\n$$\nfor any $x \\in \\mathbb{R}$. Consequently, $\\frac{b\\cos c}{a} = -1$. So Answer is (C).\n\nMore generally, we have\n$$\nf(x) = \\sqrt{13}\\sin(x + \\varphi... | China | China Mathematical Competition | [
"Precalculus > Trigonometric functions"
] | English | MCQ | C | |
088f | Problem:
Un fabbro sta costruendo una cancellata orizzontale in ferro formata da tante barre verticali, parallele tra loro, ciascuna delle quali è posizionata a $18~\mathrm{cm}$ di distanza dalle 2 vicine. Il fabbro collega le estremità di ciascuna coppia di sbarre contigue con una barra incurvata ad arco di circonfer... | [
"Solution:\n\nLa risposta è (D). Con riferimento alla figura, siano $A$ e $B$ estremità di sbarre contigue, $V$ il vertice dell'arco $\\overparen{A B}$, $M$ il punto medio di $A B$ e $O$ il centro della circonferenza cui appartiene l'arco $\\widehat{A B}$. Deve valere $O B=O V$, posto dunque $O M=x$, per il Teorema... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D |
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