id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
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values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0i0z | Problem:
There are 30 members of the math faculty at UC Berkeley. The evil Gastropod places a red or green stamp on each member's forehead, then places them all in separate rooms. Each day, the Gastropod convenes all the faculty in one room and asks any member who has conclusively determined that his/her stamp is red ... | [
"Solution:\n\nWe prove the following generalization: if exactly $n$ faculty members were given red stamps and the rest have green stamps (and the procedure is as before), no hands will be raised for the first $n-1$ days, and all $n$ faculty members will raise their hands on the $n$th day. This will be shown by indu... | United States | Berkeley Math Circle Monthly Contest #1 | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
090a | Let $N_0$ be the set of nonnegative integers and $\mathbb{Z}$ be the set of integers. Find the number of possible tuples $(f(0), f(1), \dots, f(2024))$ for $f: N_0 \to \mathbb{Z}$ satisfying
$$
f(m+n)^2 = f(m|f(n)|) + f(n^2)
$$
for any $m, n \in N_0$. | [
"Denote the equation $f(m + n)^2 = f(m|f(n)|) + f(n^2)$ by $P(m, n)$. $P(0, 0)$ implies $f(0)^2 = 2f(0)$, thus $f(0) \\in \\{0, 2\\}$.\n\nWhen $f(0) = 0$, $P(m, 0)$ implies $f(m) = 0$ for any $m \\in \\mathbb{N}_0$. Since this $f$ satisfies the condition, there is just one possible tuple $(f(0), f(1), \\dots, f(202... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | 2^990 + 1 | |
00ie | For a positive integer $n$, let $f_n(x)$ be defined by
$$
f_n(x) = \sum_{k=1}^{n} |x - k|.
$$
Determine the solution to the inequality $f_n(x) < 41$ for every two-digit integer $n$ (in decimal notation).
G. Baron, Vienna | [
"Note first that the function $f_n(x)$ satisfies $f_n(x) = f_n(n+1-x)$:\n$$\nf_n(n+1-x) = \\sum_{k=1}^{n} |n+1-x-k| = \\sum_{k=1}^{n} |x - (n+1-k)| = \\sum_{k=1}^{n} |x-k| = f_n(x)\n$$\nby reversing the order of summation. Let us first consider the case $x < 1$: then, $x-k < 0$ for every $k \\ge 1$ and thus\n$$\nf_... | Austria | Austria 2010 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | For n = 10: (3/2, 19/2); for n = 11: (19/7, 65/7); for n = 12: (17/4, 35/4); for n ≥ 13: no solutions. | |
03uv | Let $S_n$ denote the sum of the first $n$ terms in a number sequence $\{a_n\}$, satisfying
$$
S_n + a_n = \frac{n-1}{n(n+1)}, \quad n = 1, 2, \dots
$$
Then $a_n = \underline{\hspace{2cm}}$. | [
"As\n$$\n\\begin{aligned}\na_{n+1} &= S_{n+1} - S_n \\\\\n&= \\frac{n}{(n+1)(n+2)} - a_{n+1} - \\frac{n-1}{n(n+1)} + a_n,\n\\end{aligned}\n$$\nwe have\n$$\n\\begin{aligned}\n2a_{n+1} &= \\frac{n+2-2}{(n+1)(n+2)} - \\frac{1}{n+1} + \\frac{1}{n(n+1)} + a_n \\\\\n&= \\frac{-2}{(n+1)(n+2)} + a_n + \\frac{1}{n(n+1)}.\n\... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a_n = 1/2^n - 1/(n(n+1)) | |
0boc | Let $a, b, c, d \ge 0$ be real numbers such that $a + b + c + d = 1$. Prove that
$$
\sqrt{a + \frac{(b-c)^2}{6} + \frac{(c-d)^2}{6} + \frac{(d-b)^2}{6}} + \sqrt{b} + \sqrt{c} + \sqrt{d} \le 2.
$$ | [
"Observe that $(b-c)^2 \\le 2(\\sqrt{b} - \\sqrt{c})^2$, since $(\\sqrt{b} + \\sqrt{c})^2 = b + c + 2\\sqrt{bc} \\le 2b + 2c \\le 2$. Thus, $a + \\sum \\frac{(b-c)^2}{6} \\le a + \\frac{1}{3} \\sum (\\sqrt{b} - \\sqrt{c})^2 = 1 - \\frac{1}{3}(\\sum \\sqrt{b})^2$.\n\nTo end the proof, it suffices to show that $S + \... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
060u | Problem:
Soient $a, b, c, d$ quatre réels tels que $|a|>1, |b|>1, |c|>1$ et $|d|>1$. Supposons que $(a+1)(b+1)(c+1)(d+1) = (a-1)(b-1)(c-1)(d-1)$. Montrer que
$$
\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} + \frac{1}{d-1} > 0
$$ | [
"Solution:\nOn sait d'après l'énoncé qu'aucun des $a, b, c, d$ ne vaut $1$. La condition de l'énoncé s'écrit alors\n$$\n\\frac{a+1}{a-1} \\frac{b+1}{b-1} \\frac{c+1}{c-1} \\frac{d+1}{d-1} = 1\n$$\nRemarquons que comme $|a|>1$, alors on a soit $a>1$ et donc $a+1>0$ et $a-1>0$, ou alors $a<-1$ et donc $a+1<0$ et $a-1... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
09kv | Let $a$ and $b$ be positive reals satisfying
$$
\begin{cases} a^2 = 2 + b \\ b^3 = 3 - a \end{cases}.
$$
Find all positive integer $n$ such that the equation $x^n + a x + b = 0$ has no real solutions. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | n = 2 and n = 4 | |
021s | Problem:
Dois divisores - O número $2^{48} - 1$ é divisível por dois números compreendidos entre 60 e 70. Quais são esses números?
(a) 61 e 63
(b) 61 e 65
(c) 63 e 65
(d) 63 e 67
(e) 67 e 69 | [
"Solution:\n\nLembre que\n$$\na^{4}-1=(a-1)\\left(a^{3}+a^{2}+a+1\\right)\n$$\nLogo, se $a=2^{12}$, temos:\n$$\n2^{48}-1=\\left(2^{12}\\right)^{4}-1=\\left(2^{12}-1\\right)\\left(2^{36}+2^{24}+2^{12}+1\\right)\n$$\ne $2^{12}-1=\\left(2^{6}+1\\right)\\left(2^{6}-1\\right)=65 \\times 63$. A opção correta é (c)."
] | Brazil | Lista 6 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | (c) | |
0cd0 | Let $\mathcal{F}$ be the set of all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + f(y)) = f(x) + f(y)$ for all real numbers $x$ and $y$. Determine all real numbers $r$ satisfying the following condition: For every function $f$ in $\mathcal{F}$, there exists some real number $z$ such that $f(z) = rz$. | [
"We first prove that $S$ is a subset of $R$. Let $f$ be a member of $\\mathcal{F}$ and write $P(x, y)$ for $f(x + f(y)) = f(x) + f(y)$.\nBegin by showing that $1 + 1/n$ belongs to $R$ for all positive integers $n$. To this end, induct on $n$ to prove that $f(nf(0)) = (n + 1)f(0)$; letting $z = nf(0)$, then $f(z) = ... | Romania | THE 73rd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS - FOURTH SELECTION TEST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | r = 1 + 1/n or r = 1 − 1/n for any positive integer n | |
02ou | Problem:
Determine o maior múltiplo de $36$ que possui todos os algarismos pares e diferentes. | [
"Solution:\nPara um número ser divisível por $36 = 4 \\times 9$, deve ser divisível por $4$ e por $9$. Assim, a soma dos algarismos do número $n$ procurado deve ser divisível por $9$.\n\nPor outro lado, como todos os algarismos são pares, a soma dos algarismos também é par. Assim, a soma dos algarismos é no mínimo ... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 8640 | |
006i | Julián elige $2007$ puntos del plano entre los que no haya $3$ alineados, y traza con rojo todos los segmentos que unen dos de esos puntos. A continuación, Roberto traza varias rectas. Su objetivo es que cada segmento rojo sea cortado en un punto interior por (al menos) una de las rectas. Determinar el menor $l$ tal qu... | [] | Argentina | Argentina 2008 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Spanish | proof and answer | 63 | |
0gi9 | 設圓內接四邊形 $ABCD$ 的外接圓為 $\omega$, 半徑為 $r$ 且對角線 $AC$ 和 $BD$ 相交於 $P$。
假設 $AD = DP$, $S$ 為從 $P$ 到 $AB$ 的垂足而點 $Q$ 位於直線 $SP$ 上, 使得 $PQ = r$ 且 $S, P, Q$ 依序位於直線上。令通過 $A$ 且垂直 $CQ$ 的直線與通過 $B$ 垂直於 $DQ$ 的直線相交於 $E$, 證明 $E$ 位於 $\omega$ 上。
Let $ABCD$ be a cyclic quadrilateral with circumcircle $\omega$ and circumradius $r$. The diagon... | [
"First observe that\n$$\n\\angle DPA = \\angle BPC = \\angle CBP = \\angle CBD = \\angle CAD = \\angle PAD\n$$\nso $DP = DA$. Thus there is a symmetry in the problem statement swapping $(A, D) \\leftrightarrow (B, C)$. Let $O$ be the centre of $\\omega$ and let $E$ be the reflection of $P$ in $CD$ which, by\n$$\n\\... | Taiwan | IMO 1J, Mock Exam 1 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | Chinese; English | proof only | null | |
0h49 | Let $H$ be the orthocenter of a triangle $ABC$, $M$ be the midpoint of $AB$, $N$ be the midpoint of $AC$. The rays $MH$ and $NH$ intersect the circumcircle of $ABC$ at points $P$ and $Q$ respectively. Prove that the lines $BQ$, $CP$ and $AH$ are either concurrent or parallel. | [
"Let $AA_1$ be the altitude of triangle $ABC$. Consider the case when the lines $BQ$ and $CP$ intersect at some point $T$. We need to prove that the point $T$ lies on the line $AA_1$. Suppose first that points $A$ and $P$ lie on the same side of line $BC$, and points $A$ and $Q$ lie on different sides of line $BC$ ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
05uv | Problem:
Soit $n$ un entier naturel. On dit qu'un ensemble fini d'entiers est $n$-équilibré s'il y a exactement $n$ manières de le partitionner en deux sous-ensembles de même somme. Par exemple, l'ensemble $\{1,3,4,5,6,7\}$ est 2-équilibré, car les seules manières de le partitionner en deux sous-ensembles de même somm... | [
"Solution:\n\nTout d'abord, l'ensemble $\\{1\\}$ est 0-équilibré et l'ensemble $\\{1,2,3\\}$ est 1-équilibré.\n\nÉtant donné un entier $n \\geqslant 2$, nous allons démontrer que l'ensemble\n$$\nE=\\{k: 3 n \\leqslant k \\leqslant 4 n-1\\} \\cup\\{k: 4 n+1 \\leqslant k \\leqslant 5 n\\} \\cup\\{8 n(n-2)\\}\n$$\nest... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | All integers greater than or equal to zero | |
0dpr | Polynomial $Q(x) = k_n x^n + k_{n-1} x^{n-1} + \dots + k_1 x + k_0$ with real coefficients is called *mighty* if $|k_0| = |k_1| + |k_2| + \dots + |k_{n-1}| + |k_n|$, and *non-increasing* if $k_0 \ge k_1 \ge \dots \ge k_{n-1} \ge k_n$.
Let $P(x) = a_d x^d + a_{d-1} x^{d-1} + \dots + a_1 x + a_0$ be a polynomial with rea... | [
"Note that if for some real numbers $x_1, x_2, \\dots, x_m$ the following equality holds:\n$$\n|x_1| + |x_2| + \\dots + |x_m| = |x_1 + x_2 + \\dots + x_m|,\n$$\nthen they are of the same sign.\nLet\n$$\nQ(x) = P(x)(x-1)^t (x+1)^s = b_n x^n + b_{n-1} x^{n-1} + \\dots + b_0,\n$$\nwhere $b_n = a_d > 0$. From the probl... | Silk Road Mathematics Competition | XIX Silk Road Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
005i | Se tiene una hoja cuadrada de $9 \times 9$ cuadriculada en cuadritos de $1 \times 1$. Se corta la hoja con el objetivo de dividirla en cuadritos de $1 \times 1$.
Cada corte debe ser recto y seguir una línea de la cuadrícula.
Después de efectuar cada corte, está permitido reacomodar convenientemente los pedazos en una p... | [] | Argentina | Argentina 2008 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof and answer | 8 | |
075p | Suppose $2n$ real numbers are placed in the cells of a $2 \times n$ grid such that the sum of the numbers in each of the $n$ columns is $1$. Prove that one can erase one of the two numbers in each column such that the sum of the remaining numbers in each of the rows does not exceed $\frac{n+1}{4}$. | [
"Assume that the numbers in the first row are $a_1 \\le a_2 \\le \\dots \\le a_n$ in some order. The numbers in the second row are $b_j = 1 - a_j$, $1 \\le j \\le n$. Hence $b_1 \\ge b_2 \\ge \\dots \\ge b_n$.\n\nIf $a_1 + a_2 + \\dots + a_n \\le \\frac{n+1}{4}$, we are done (we can erase all the numbers in the sec... | India | Indija TS 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0cde | Find all real numbers $a > 0$ such that there exist real numbers $x, y > 0$ with the properties
$$
x = a - \sqrt{y}, \quad y = a + \sqrt{x}, \quad \frac{x}{y} + \frac{y}{x} = 2023.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 44/43 | |
0im7 | Problem:
A parallelogram has 3 of its vertices at $(1,2)$, $(3,8)$, and $(4,1)$. Compute the sum of all possible $x$ coordinates of the 4th vertex. | [
"Solution:\nAnswer: $8$. There are three possibilities: the 4th vertex must be opposite one of the three given vertices. These three possibilities have as a medial triangle the three given vertices, so the sum of their $x$ coordinates is the same as the sum of the $x$ coordinates of the given triangle."
] | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | final answer only | 8 | |
0d43 | Let $ABC$ be a triangle, $I$ its incenter, and $\omega$ a circle of center $I$. Points $A'$, $B'$, $C'$ are on $\omega$ such that rays $IA'$, $IB'$, $IC'$ starting from $I$ intersect perpendicularly sides $BC$, $CA$, $AB$, respectively. Prove that lines $AA'$, $BB'$, $CC'$ are concurrent. | [
"Define $A''$, $B''$, $C''$ to be the intersection points of the lines $AA'$, $BB'$, $CC'$ with the sides $BC$, $CA$, $AB$, respectively, $B_a$, $C_a$ the intersection points of the tangent line to $\\omega$ at $A'$ with the lines $AB$, $AC$, respectively, $C_b$, $A_b$ the intersection points of the tangent line to... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coord... | English, Arabic | proof only | null | |
07pb | A certain country is taking part in the IMO this year for the 24th time. The team would like to mark the occasion by finding positive integers $n_1, \dots, n_{24}$ such that
$$
n_1^{24} + n_2^{24} + \dots + n_{24}^{24} = 2015^{2015}.
$$
Determine whether or not such a set of numbers exists. | [
"**0.**\nWe first claim that if $n$ is odd, then $n^{2k} \\equiv 1 \\pmod{2^{k+2}}$. This can be verified inductively: first prove it for $k=1$, and the inductive step reduces to checking that $(1+2^{k+2})^2-1$ is divisible by $2^{k+3}$.\nIt follows that $n^{24} = (n^3)^8$ is congruent to either 0 or 1 (mod 32), de... | Ireland | Ireland | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No such set exists. | |
08rq | Let $ABCD$ be a convex quadrilateral with $AB = 3$, $BC = 4$, $CD = 5$, $DA = 6$ and $\angle ABC = 90^\circ$. Find the area of $ABCD$. | [
"Since $AB = 3$, $BC = 4$ and $\\angle ABC = 90^\\circ$, we get $AC = 5$. Let $M$ be the midpoint of $AD$. Because $AM = DM$, $MC = MC$, $AC = 5 = DC$ and $\\angle ABC = \\angle AMC = 90^\\circ$, $AB = AM$, $AC = AC$, the triangles $ABC$, $AMC$ and $DMC$ are all congruent.\n\nTherefore, the area of $ABCD$ is $\\fra... | Japan | Japan 2007 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 18 | |
0c5h | Let $(a_n)_{n \ge 1}$ be a sequence of real positive numbers, such that the sequence $(a_{n+1} - a_n)_{n \ge 1}$ converges to a non-zero limit. Compute the limit
$$
\lim_{n \to \infty} \left( \frac{a_{n+1}}{a_n} \right)^n .
$$ | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | e | |
0dxv | Problem:
V ravnini leži 16 črnih točk, kot prikazuje slika. Najmanj koliko izmed teh točk moramo pobarvati rdeče, da ne bo obstajal kvadrat z oglišči v preostalih črnih točkah in s stranicami, katerih dolžine so naravna števila ali $\sqrt{2}$? Odgovor utemelji.
 | [
"Solution:\n\nDokazali bomo, da moramo pobarvati pet točk.\n\n\n\nKer mora biti v vsakem izmed 4 kvadratkov $A, B, C, D$, označenih na prvi sliki, vsaj eno oglišče rdeče, potrebujemo vsaj 4 rdeče točke. Denimo, da je to že dovolj. Kvadrat velikosti $3 \\times 3$ ima vsaj eno rdeče oglišče. ... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 5 | |
0eu8 | For the vertex $A$ of $\triangle ABC$, define $\ell_A$ by the distance between the feet of perpendiculars drawn to the sides $AB$ and $AC$ from the intersection of the angle bisector of $\angle A$ and the side $BC$. Similarly, define $\ell_B$ and $\ell_C$ for the vertices $B$ and $C$, respectively. Prove the inequality... | [
"Let $a$, $b$ and $c$ be the lengths of the opposite sides of vertices $A$, $B$ and $C$, respectively. Also, let $D$ be the intersection of $BC$ and the angle bisector of $\\angle A$. Put $p = BD$ and $q = CD$. Since $AD$ is the angle bisector of $\\angle A$, we have $bp = cq$ and thus a little manipulation yields\... | South Korea | 20th Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0bzv | Determine the increasing functions $f : \mathbb{N} \to \mathbb{N}$ having the property that the number $\frac{f(x) + f(y)}{1 + f(x + y)}$ is a non-negative integer, for all $x, y \in \mathbb{N}$. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All functions of the form f(n) = a n + 1 with a a positive integer. | |
07c7 | In quadrilateral $ABCD$, $AC$ is the bisector of angle $A$ and $\angle ADC = \angle ACB$. $X$ and $Y$ are the feet of the altitudes from $A$ to $BC$ and $CD$, respectively. Prove that the orthocenter of triangle $AXY$ is located on line $BD$ (A triangle's orthocenter is the intersection point of its altitudes). | [
"Let $E$ be the foot of the perpendicular line from $Y$ to $AX$ and $P$ be the intersection point of $YE$ and $BD$. It is sufficient to prove that $XP \\perp AY$, or equivalently, $XP \\parallel CD$. Note that\n$$\nYE \\parallel CB \\Rightarrow \\frac{DY}{YC} = \\frac{DP}{PB}. \\quad (1)\n$$\nOn the other hand, tri... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08t6 | Determine the number which can be represented in the form $n^2 + 4n$ for some positive integer $n$ and for which the absolute value of its difference from the number $10000$ is the smallest. | [
"[9996]\n\nNote that $n^2 + 4n = (n + 2)^2 - 4$ and that if $m < n$, then $m^2 + 4m < n^2 + 4n$.\n\nSince $100^2 = 10000$, we see that the desired solution is either $100^2 - 4 = 9996$ or $101^2 - 4 = 10197$, whichever has the smaller difference (in absolute value) from $10000$. So, the answer is $9996$."
] | Japan | Japan Mathematical Olympiad First Round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 9996 | |
0jbi | Problem:
Let $ABC$ be a triangle with incenter $I$. The circumcircle of $\triangle AIB$ meets the lines $CA$ and $CB$ again at $P$ (different from $A$) and $Q$ (different from $B$) respectively. Prove that $A$, $B$, $P$, and $Q$ are (in some order) the vertices of a trapezoid. | [
"Solution:\n\nLet $\\omega$ and $k$ be the circumcircles of $\\triangle ABC$ and $\\triangle AIB$ respectively, and let $CI$ meet $k$ again at $M$. Computing the angles of $\\triangle AIM$ shows that it is isosceles with $MA = MI$. Similarly $MB = MI$. Therefore $M$ is the center of $k$.\n\nAssume for the moment th... | United States | Berkeley Math Circle Monthly Contest | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry >... | null | proof only | null | |
05b8 | Find all functions $f: \mathbb{R} \setminus \{1\} \to \mathbb{R} \setminus \{0\}$ which satisfy the equations $f(0) = 1$ and $f(f(xy)) = 1 - \frac{1}{y f(f(x))}$ for any real numbers $x$ and $y$ such that $xy \neq 0,\ xy \neq 1$ and $x \neq 1$. | [
"Note that if $x \\neq 0$ and $x \\neq 1$ then $f(x) \\neq 0$ and $f(x) \\neq 1$. Indeed, difference from 0 is given in the problem and, if we supposed $f(x) = 1$, taking $y = 1$ would yield $f(f(xy)) = f(f(x)) = f(1)$, meaning that $f(f(xy))$ were undefined and could not satisfy the equation given in the problem. ... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 1/(1 - x) | |
0cgd | There are 2024 cards of the same size, face-down on a table, on which the integers $1$, $2$, $3$, $\dots$, $2024$ are written. We say that a card is a *winner* if it has a number divisible by $13$ or by $100$. What is the minimum number of cards we need to turn face up to make sure that we obtain at least one winner? | [
"There are $155$ multiples of $13$ not larger than $2024$: $13 \\cdot 1$, $13 \\cdot 2$, $\\dots$, $13 \\cdot 155$.\n\nThere are $20$ multiples of $100$ not larger than $2024$: $100 \\cdot 1$, $100 \\cdot 2$, $\\dots$, $100 \\cdot 20$.\n\nThere is only one common multiple of $13$ and $100$ not larger than $2024$, n... | Romania | 74th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | 1851 | |
091j | Problem:
We are given a positive integer $n$ which is not a power of $2$. Show that there exists a positive integer $m$ with the following two properties:
(i) $m$ is the product of two consecutive positive integers;
(ii) the decimal representation of $m$ consists of two identical blocks of $n$ digits. | [
"Solution:\n\nLemma. Let $x$ and $k$ be integers greater than $2$. If $k$ is odd then the number $x^{k}+1$ is the product of two coprime numbers.\n\nProof. Let $m=\\operatorname{gcd}(x+1, k)$. There is a polynomial $Q(x) \\in \\mathbb{Z}[x]$ such that\n$$\nx^{k}+1=(x+1)\\left(x^{k-1}-x^{k-2}+\\cdots+x^{2}-x+1\\righ... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ioi | Problem:
Determine the largest positive integer $n$ such that there exist positive integers $x, y, z$ so that
$$
n^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x+3 x+3 y+3 z-6
$$ | [
"Solution:\n\nThe given equation rewrites as $n^{2}=(x+y+z+1)^{2}+(x+y+z+1)-8$. Writing $r=x+y+z+1$, we have $n^{2}=r^{2}+r-8$. Clearly, one possibility is $n=r=8$, which is realized by $x=y=1, z=6$. On the other hand, for $r>8$, we have $r^{2}<r^{2}+r-8<(r+1)^{2}$."
] | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 8 | |
09qc | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ waarvoor geldt dat
$$
f(x)=\max_{y \in \mathbb{R}}(2 x y-f(y))
$$
voor alle $x \in \mathbb{R}$. | [
"Solution:\nVoor alle $x \\in \\mathbb{R}$ geldt\n$$\nf(x)=\\max_{y \\in \\mathbb{R}}(2 x y-f(y)) \\geq 2 x^{2}-f(x)\n$$\ndus $2 f(x) \\geq 2 x^{2}$, oftewel $f(x) \\geq x^{2}$.\nOmdat $(x-y)^{2} \\geq 0$, geldt $x^{2} \\geq 2 x y-y^{2}$ voor alle $x, y \\in \\mathbb{R}$. Omdat we al hebben laten zien dat $f(y) \\g... | Netherlands | toets | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | f(x) = x^2 | |
0fe8 | Problem:
Se colorean los números $1,2, \ldots, n$ de dos colores, azul y rojo. Probar que si $n=2017$ existe una coloración tal que la ecuación
$$
8(x+y)=z
$$
no tiene soluciones monocromáticas. Determinar el menor $n$ para el que nunca es posible colorear los números de forma tal que no haya soluciones monocromáticas... | [
"Solution:\n\nPara la segunda pregunta, observamos que si el $1$ es azul, entonces $8(1+1)=16$ es rojo, y por tanto $8(16+16)=256$ azul y $8(256+1)=2056$ es rojo. En general, si un número $i$ es azul, $8(i+1)$ es rojo, y por consiguiente $i-15$ no puede ser rojo ya que en ese caso tendríamos $8((i-15)+16)=8(i+1)$. ... | Spain | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2056 | |
04um | Find all triples $(x, y, z)$ of real numbers such that
$$
\begin{aligned}
yz + 2 &= x^2, \\
zx + 3 &= y^2.
\end{aligned}
$$ | [] | Czech Republic | First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | All solutions are given as follows. Either x = 0 with y = ±√3 and z = ∓2/√3, or y = 0 with x = ±√2 and z = ∓3/√2. Otherwise, let t = y − x ≠ 0 and let S = x + y. Then S and t satisfy 3 t S^2 − 2 S + (t^3 − 10 t) = 0, so
S = [1 ± sqrt(1 + 30 t^2 − 3 t^4)] / (3 t),
and the corresponding triple is
x = (S − t)/2,
y = (S + ... | |
0h9q | $n$ students arrived to the summer camp. Every child (boy or girl) in the camp knows exactly one boy and one girl. Is it possible if
a) $n = 2000$;
b) $n = 2018$. | [
"a) Suppose $1000$ boys and $1000$ girls arrived to the camp. We can split them into groups with $4$ students in each: $2$ boys and $2$ girls. Moreover, they know each other in the following way: $B_1 \\leftrightarrow B_2 \\leftrightarrow G_2 \\leftrightarrow G_1 \\leftrightarrow B_1$. One can check that all condit... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | a) Yes. b) No. | |
0ea5 | Problem:
Simetrala notranjega kota $\angle A C B$ ostrokotnega trikotnika $A B C$ seka stranico $A B$ v točki $D$. Očrtana krožnica trikotnika $A D C$ seka stranico $B C$ v različnih točkah $C$ in $E$. Premica skozi točko $B$, vzporedna premici $A E$, seka premico $C D$ v točki $F$. Dokaži, da je trikotnik $A F B$ ena... | [
"Solution:\n\n\n\nOznačimo $\\angle A C B=\\gamma$. Potem je $\\angle A C F=\\angle F C B=\\frac{\\gamma}{2}$, saj je $C F$ simetrala kota $\\angle A C B$. Zaradi koncikličnosti točk $A, D, E$ in $C$ sledi $\\angle D A E=\\angle D C E=\\frac{\\gamma}{2}$. Ker je premica $B F$ vzporedna prem... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ghq | 證明對於任何正整數 $n$，$5^n - 3^n$ 都無法被 $2^n + 65$ 整除。
Prove that $5^n - 3^n$ is not divisible by $2^n + 65$ for any positive integer $n$. | [
"解. 我們先證明 D', H, X 共線：由於 $\\angle GXB = \\angle ACB = \\angle GFB$, 得 G, F, B, X 共圓。\n結合 $AG \\cdot AX = AF \\cdot AB = AH \\cdot AD$ 我們得到 G, X, D, H 共圓。注意到 $\\triangle DFH \\sim \\triangle DAE$ 且 $\\triangle DFG \\sim \\triangle DD'E$, 因此 $\\triangle DGH \\sim \\triangle DAD'$, 故 $\\angle DXA = \\angle DHG = \\ang... | Taiwan | 2023 數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | Chinese (Traditional) | proof only | null | |
0ege | Problem:
Dana je funkcija s predpisom $f(x)=\frac{x^{2}-2x+1}{x^{2}+1}$.
a) Izračunaj ničle, pole, asimptoto in nariši graf funkcije $f$.
b) Izračunaj presečišča grafa funkcije $f$ s premico $y=2$.
c) Določi zalogo vrednosti funkcije $f$. | [
"Solution:\n\na.\nFunkcija ima dvojno ničlo v $x=1$. Pola nima. Vodoravna asimptota je $y=1$. Funkcija seka ordinatno os v točki $N(0,1)$. Narišemo graf funkcije $f$.\n\n\n\nb.\nZapišemo enačbo: $\\frac{x^{2}-2x+1}{x^{2}+1}=2$. Po ureditvi dobimo enačbo: $x^{2}+2x+1=0$. Dvojna rešitev te en... | Slovenia | 18. tekmovanje v znanju matematike za dijake srednjih tehniških i strokovnih šol, Odbirno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) Zero at x = 1 (double). No real poles. Horizontal asymptote y = 1.
b) Intersection with y = 2 at the point (−1, 2) (double contact).
c) Range: [0, 2]. | |
0ehf | Problem:
Naj bo $T$ težišče trikotnika $ABC$ in $D$ razpolovišče stranice $BC$. Premice $AT$, $BT$ in $CT$ naj drugič sekajo trikotniku $ABC$ očrtano krožnico zaporedoma v točkah $P$, $Q$ in $R$. Denimo, da je $|AD|=\frac{\sqrt{3}}{2}|AC|$. Dokaži, da je trikotnik $PQR$ enakokrak. | [
"Solution:\n\nOznačimo dolžine stranic trikotnika $ABC$ in njegove kote kot običajno. Naj bo $E$ razpolovišče stranice $AC$. Potem je $|AE|=\\frac{b}{2}$, $|AC|=b$ in po predpostavki $|AD|=\\frac{\\sqrt{3}}{2}b$. Ker težišče deli težiščnico v razmerju $2:1$, lahko izračunamo še $|AT|=\\frac{2}{3}|AD|=\\frac{1}{\\sq... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > ... | null | proof only | null | |
02ep | $ABCD$ is a convex quadrilateral. $E, F, G, H$ are the midpoints of sides $AB, BC, CD, DA$ respectively. Find the point $P$ such that area $PHAE = \text{area } PEBF = \text{area } PFCG = \text{area } PGDH$. | [
"Let $M$ be the midpoint of $AC$, $N$ the midpoint of $BD$. Take $P$ so that $PM$ is parallel to $BD$ and $PN$ is parallel to $AC$. Now $EM$ is a midline of $\\triangle ABC$, so $\\text{area } AEM = \\frac{\\text{area } ABC}{4}$. Similarly, $\\text{area } AHM = \\frac{\\text{area } ADC}{4}$. So $\\text{area } MHAE ... | Brazil | XII OBM | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Let M and N be the midpoints of AC and BD. Take P as the intersection of the line through M parallel to BD and the line through N parallel to AC. For this P, the four areas are equal (each equals one quarter of the area of ABCD). | |
0583 | Let $D$ and $E$ be the midpoints of sides $AB$ and $AC$, respectively, of a triangle $ABC$. Prove that the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if the line $AC$ is tangent to the circumcircle of the triangle $BED$. | [
"The conditions of the problem imply that $DE$ is the midsegment parallel to the side $BC$ of the triangle $ABC$ (Fig. 30).\n\n\n\nBy properties of inscribed angle, the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if $\\angle ECB = \\angle DBE$, and the line $A... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0euj | Two players play a game on a $(m+1)$ by $m$ grid, which has $m+1$ horizontal lines, $m$ vertical lines, and hence $m(m+1)$ intersection points. One pebble is put on an intersection point. Two players alternatively move the pebble to an adjacent point along an edge, but they are not allowed to use edges used previously ... | [
"Let $\\{(x, y) : x = 0, 1, \\dots, m, y = 1, 2, \\dots, m\\}$ be the set of all points on the grid. Let $(a, 0)$ be the initial position of the pebble. Let\n$$\nP = (a, 0), \\quad Q = (0, a), \\quad R = (m+1-a, m+1) \\text{ and } S = (m+1, m+1-a).\n$$\n\nLet $X$ be the set of all points of the grid on the lines $P... | South Korea | 22nd Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0gpb | Let $K$ be a point in the interior of an acute triangle $ABC$ and $ARBPCQ$ be a convex hexagon whose vertices lie on the circumcircle $\Gamma$ of the triangle $ABC$. Let $A_1$ be the second point where the circle passing through $K$ and tangent to $\Gamma$ at $A$ intersects the line $AP$. The points $B_1$ and $C_1$ are... | [
"Let $O$ be the center of $\\Gamma$. Since $ABC$ is an acute triangle $O$ lies inside $ABC$. Assume that $K$ lies on the same side of the lines $AO$ and $BO$ as $C$, and on the same side of the bisector of the line segment $AB$ as $B$. Then $KA \\ge OA$.\n\nLet $\\omega$ be the circle passing through $K$ and tangen... | Turkey | Team Selection Test for IMO 2011 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
012o | Problem:
The following solitaire game is played on an $m \times n$ rectangular board, $m, n \geqslant 2$, divided into unit squares. First, a rook is placed on some square. At each move, the rook can be moved an arbitrary number of squares horizontally or vertically, with the extra condition that each move has to be m... | [
"Solution:\n\nFirst, consider any row that is not the row where the rook starts from. The rook has to visit all the squares of that row exactly once, and on its tour around the board, every time it visits this row, exactly two squares get visited. Hence, $m$ must be even; a similar argument for the columns shows th... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | All m × n boards with m even and n even | |
09qx | Problem:
Voor positieve gehele getallen $a$ en $b$ definiëren we $a \ominus b=\frac{a-b}{\operatorname{ggd}(a, b)}$. Bewijs dat voor elk geheel getal $n>1$ geldt: $n$ is een priemmacht (d.w.z. dat $n$ te schrijven is als $n=p^{k}$ met $p$ een priemgetal en $k$ een positief geheel getal) dan en slechts dan als voor all... | [
"Solution:\n\nStel eerst dat $n=p^{k}$ met $p$ priem en $k>0$. We moeten bewijzen dat $\\operatorname{ggd}(n, n \\ominus m)=1$ voor alle $m<n$. Dus bekijk een willekeurig positief geheel getal $m<n$. We schrijven $m=p^{l} s$ met $p \\nmid s$ en $0 \\leq l<k$. Nu is $\\operatorname{ggd}(n, m)=\\operatorname{ggd}\\le... | Netherlands | Dutch TST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0392 | Find all values of the real parameter $a$ such that the inequality
$$
\log_a(a^x + 1) + \frac{1}{\log_{a^2-1} a} \le x - 1 + \log_a(a^2 - 1)
$$
holds true for every $x \in (0, 1]$. | [
"We have that $a > 1$, $x > 0$ and $x \\neq \\log_a 2$. Then the inequality is equivalent to\n$$\n(a^{x-1} - 1)(a^{x+1} + 1) \\le 0,\n$$\ngiving $x \\le 1$. Since $\\log_a 2 > 0$ and $\\log_a 2 \\le 1$ when $a \\ge 2$ we obtain:\n\n1. For $1 < a < 2$ the inequality holds true iff $0 < x \\le 1$.\n\n2. For $a \\ge 2... | Bulgaria | Spring Mathematical Tournament | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | (1, 2) | |
04wz | For any real number $p \ge 1$ let us consider the set of all real numbers $x$ with
$$
p < x < \left(2 + \sqrt{p + \frac{1}{4}}\right)^2
$$
Prove that from such a given set one can select four mutually different natural numbers $a, b, c, d$ with $ab = cd$. | [
"The numbers $a = (k-1)k$, $b = (k+1)k$, $c = (k-1)(k+1)$, $d = k^2$ clearly satisfy the equality $ab = cd$ and the inequalities $a < c < d < b$ for any $k > 1$. Let thus $k$ be the least natural number for which $p < a$, i.e. $p < (k-1)k$ (for a given $p$). We will show that for this $k$ necessarily $b = (k+1)k \\... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
002b | Carlos y Yue juegan al siguiente juego: Primero Carlos escribe un signo $+$ o un signo $-$ delante de cada uno de los $50$ números $1$, $2$, $\ldots$, $50$. Luego, por turnos, cada uno elige un número de la sucesión obtenida; comienza eligiendo Yue. Si el valor absoluto de la suma de los $25$ números que eligió Carlos ... | [] | Argentina | XX Olimpiada Matemática Argentina | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | español | proof only | null | |
035a | Problem:
Let $a, b_{1}, c_{1}, \ldots, b_{n}, c_{n}$ be real numbers such that
$$
x^{2 n}+a x^{2 n-1}+a x^{2 n-2}+\cdots+a x+1=\left(x^{2}+b_{1} x+c_{1}\right) \ldots\left(x^{2}+b_{n} x+c_{n}\right)
$$
for every real number $x$. Prove that $c_{1}=\cdots=c_{n}=1$. | [
"Solution:\nWe first show that the assertion follows from the following lemma.\n\nLEMMA. The polynomial $P(z)=z^{2 n}+a z^{2 n-1}+a z^{2 n-2}+\\cdots+a z+1$ has at least $2 n-2$ complex zeros lying on the unit circle and different from $\\pm 1$.\n\nSince the coefficients of $P(z)$ are real numbers, its non-real zer... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0ccp | Let $f : [0, 1] \to \mathbb{R}$ be a continuous function.
a) Show that
$$
\lim_{n \to \infty} \int_{0}^{1} f(x^n) dx = f(0).
$$
b) If $f(0) = 0$ and $f$ is differentiable at the right in $0$, show that the limits exist, are finite and equal.
$$
\lim_{\varepsilon \to 0} \int_{\varepsilon}^{1} \frac{f(x)}{x} dx \quad \... | [
"a) Since $f$ is continuous, $f$ is bounded, with $\\text{Im}(f) \\subseteq [-M, M]$, where $M > 1$, and, for any $\\varepsilon > 0$, there is a $\\delta > 0$, such that $|f(x) - f(0)| < \\frac{\\varepsilon}{2}$, $\\forall x \\in [0, \\delta]$, and, for any $x \\in [0, 1 - \\frac{\\varepsilon}{4M}]$, there is a pos... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Limits",
"Precalculus > Functions"
] | null | proof only | null | |
0iyh | Problem:
The incircle $\omega$ of equilateral triangle $ABC$ has radius $1$. Three smaller circles are inscribed tangent to $\omega$ and the sides of $ABC$, as shown. Three smaller circles are then inscribed tangent to the previous circles and to each of two sides of $ABC$. This process is repeated an infinite number ... | [
"Solution:\n\nOne can find using the Pythagorean Theorem that, in each iteration, the new circles have radius $1/3$ of that of the previously drawn circles. Thus the total circumference is $2\\pi + 3 \\cdot 2\\pi \\left(\\frac{1}{1-1/3} - 1\\right) = 5\\pi$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 5π | |
0581 | Let $a$, $b$, $c$, $d$ be positive real numbers satisfying the system of equations
$$
\begin{cases} a^2 + \frac{1}{b^2} = \frac{1}{2}, \\ b^2 + \frac{4}{c^2} = 8, \\ c^2 + \frac{16}{d^2} = 2, \\ d^2 + \frac{4}{a^2} = 32. \end{cases}
$$
Determine the product $abcd$. | [
"Multiplying all equations gives\n$$\n\\left(a^2 + \\frac{1}{b^2}\\right) \\left(b^2 + \\frac{4}{c^2}\\right) \\left(c^2 + \\frac{16}{d^2}\\right) \\left(d^2 + \\frac{4}{a^2}\\right) = 2^8.\n$$\nBy AM-GM, $a^2 + \\frac{1}{b^2} \\ge 2 \\cdot \\frac{a}{b}$, where the equality holds if and only if $a = \\frac{1}{b}$. ... | Estonia | Estonian Math Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 4 | |
0e3j | Let $\triangle ABC$ be an equilateral triangle with the side of $20$ units. Vid divides this triangle into $400$ smaller equilateral triangles with the sides of $1$ unit. Eva then picks $4$ of the vertices of these smaller triangles. The vertices lie inside the triangle $ABC$ and form a parallelogram with sides paralle... | [
"Denote the sides of the parallelogram by $a$ and $b$. We may assume that $a \\ge b$. The small triangles which share at least one point with the sides of the parallelogram are marked in the figure. We can find their number by subtracting $2$ triangles in the corners and the number of the triangles inside the white... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | 3√3, 4√3, 9√3/2 | |
07cs | In the country of Sugarland, there are $13$ students in the IMO team selection camp. $6$ team selection tests were taken and the results have come out. Assume that no students have the same score on the same test. To select the IMO team, the national committee of math Olympiad have decided to choose a permutation of th... | [
"The answer of the problem is yes.\nAlthough the statement is discussed on $13$ students, here is an example for $14$ students, all having a chance of being a team member. (students are labelled by $1, 2, \\ldots, 14$.)\n\n| Test Rank | #1 | #2 | #3 | #4 | #5 | #6 |\n|-----------|----|----|----|----|----|----|\n| #... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Yes | |
01ur | A point $X$ lies inside an isosceles right triangle $ABC$ with $\angle B = 90^\circ$.
Prove the inequality $AX + BX + \sqrt{2}CX \ge \sqrt{5}AB$ and find all points $X$ for which the equality holds. | [
"First solution. Consider the rotation around point $C$ by $90^\\circ$ such that the images of point $A$ and point $B$ lie in the same half-plane with respect to the line $CB$ (see the Fig.). Denote the images of points $A$ and $X$ by $D$ and $Y$ respectively. By construction, the triangles $ACD$ and $CXY$ are isos... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plan... | English | proof and answer | Inequality: AX + BX + sqrt(2)·CX ≥ sqrt(5)·AB. Equality holds for exactly one interior point X. In a coordinate model with A at (0,1), B at (0,0), and C at (1,0), this point is X = (2/5, 1/5). Geometrically, it is the image of C under a ninety-degree rotation about H, where H is the foot of the perpendicular from C to ... | |
05gl | Problem:
Trouver toutes les fonctions $f$ de $\mathbb{R}$ dans $\mathbb{R}$ telles que pour tous réels $x, y$,
$$
f(x) f(y)=f(x y)+x+y
$$ | [
"Solution:\n\nEn évaluant en $y=0$ on obtient $f(x) f(0)=f(0)+x$ pour tout réel $x$.\n\nPour $x=1$, on obtient que $f(0) \\neq 0$, sinon on aurait $1=0$.\n\nEn évaluant en $x=0$ on obtient $f(0)^2=f(0)$ donc $f(0)=1$.\n\nDonc $f(x)=1+x$ pour tout $x$ réel.\n\nRéciproquement, soit $f$ la fonction telle que $f(x)=1+x... | France | ENVOI 2 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1 + x for all real x | |
02h4 | Let $n$ be an integer, $n \ge 3$. Let $f(n)$ be the largest number of isosceles triangles whose vertices belong to some set of $n$ points in the plane without three colinear points. Prove that there exists positive real constants $a$ and $b$ such that $an^2 < f(n) < bn^2$ for every integer $n$, $n \ge 3$. | [
"First consider $n-1$ points in the circumference of a circle and its center. Since any two points in the circumference and the center determine an isosceles triangle, the total number of isosceles triangles in this set of points is at least $(n-1)(n-2)/2 > en^2$ for some small $e$.\n\nIn another hand, let's consid... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0il4 | Problem:
Take a unit sphere $S$, i.e., a sphere with radius $1$. Circumscribe a cube $C$ about $S$, and inscribe a cube $D$ in $S$, so that every edge of cube $C$ is parallel to some edge of cube $D$. What is the shortest possible distance from a point on a face of $C$ to a point on a face of $D$? | [
"Solution:\n\nUsing the Pythagorean theorem, we know that the length of a diagonal of a cube of edge length $s$ is $s \\sqrt{3}$. Since $D$ is inscribed in a sphere that has diameter $2$, this means that its side length is $2 / \\sqrt{3}$.\n\nThe distance from a face of $D$ to a face of $C$ will be the distance bet... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 1 - 1/√3 | |
0bfr | Consider the regular hexagon $ABCDEF$ and a point $S$ outside the plane $(ABC)$, so that $SB = SO = SF$, where $O$ is the center of the hexagon. Let $M, N, P, Q$, respectively $R$ be the orthogonal projections of $A$ onto the straight lines $SB, SC, SD, SE$ respectively $SF$.
Prove that the points $A, M, N, P, Q$ and $... | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 90° | |
0jr3 | Problem:
Let $n$ be a fixed positive integer. Initially, $n$ $1$'s are written on a blackboard. Every minute, David picks two numbers $x$ and $y$ written on the blackboard, erases them, and writes the number $(x+y)^4$ on the blackboard. Show that after $n-1$ minutes, the number written on the blackboard is at least $2... | [
"Solution:\n\nWe proceed by strong induction $n$, with the base case $n=1$ being vacuous. For the inductive step, consider the situation in which we have two numbers $x$ and $y$ remaining on the blackboard. Suppose the first one was written after $a-1$ operations, and the second one was written after $b-1$ operatio... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0b7j | Let $X$ and $Y$ be two finite subsets of the half-open interval $[0, 1)$ such that $0 \in X \cap Y$ and $x + y \neq 1$ for all $x \in X$ and $y \in Y$. Prove that the set $\{x + y - \lfloor x + y \rfloor \mid x \in X \text{ and } y \in Y\}$ has at least $|X| + |Y| - 1$ elements. | [
"Let $S(X, Y) = \\{x + y - \\lfloor x + y \\rfloor : x \\in X \\text{ and } y \\in Y\\}$ and proceed by induction on $|Y|$.\n\nIf $|Y| = 1$, the statement is clear.\n\nAssume henceforth $|Y| > 1$ and let $y_0 \\in Y$, $y_0 \\neq 0$. The conditions $y_0 \\neq 0$ and $x + y_0 = 1$ for no $x \\in X$ imply that $0 \\no... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0e34 | Find all non-zero real numbers $x$ such that
$$
\min \left\{ 4, x + \frac{4}{x} \right\} \ge 8 \min \left\{ x, \frac{1}{x} \right\}.
$$ | [
"Let us check when $4 \\le x + \\frac{4}{x}$. If $x > 0$, then this inequality is equivalent to $4x \\le x^2 + 4$ or $0 \\le (x-2)^2$. This is always true. If $x < 0$, then we get $0 \\ge (x-2)^2$. This is never the case, so $4 \\ge x + \\frac{4}{x}$. Hence,\n$$\n\\min \\left\\{ 4, x + \\frac{4}{x} \\right\\} = \\b... | Slovenia | National Math Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (-∞, 0) ∪ (0, 1/2] ∪ [2, ∞) | |
0jhl | Prove that the ratio
$$
\frac{1^1 + 3^3 + 5^5 + \dots + (2^{2012} - 1)^{(2^{2012} - 1)}}{2^{2012}}
$$
is an odd integer. | [
"We will prove by induction that, for all $n \\ge 2$,\n$$\n1^1 + 3^3 + 5^5 + \\dots + (2^n - 1)^{(2^n - 1)} \\equiv 2^n \\pmod{2^{n+1}}. \\quad (40)\n$$\nThe base case $n = 2$ is straightforward. Assume that (40) holds for a given $n \\ge 2$; we will prove it for $n + 1$; that is, to prove that\n$$\n1^1 + 3^3 + 5^5... | United States | TSTST | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0jfp | Problem:
Let $ABC$ be a triangle with $AB = 5$, $BC = 4$, and $CA = 3$. Initially, there is an ant at each vertex. The ants start walking at a rate of $1$ unit per second, in the direction $A \rightarrow B \rightarrow C \rightarrow A$ (so the ant starting at $A$ moves along ray $\overrightarrow{AB}$, etc.). For a posi... | [
"Solution:\n\nAnswer: $\\frac{47}{24}$\n\nWe instead maximize the area of the remaining triangles. This area (using $\\frac{1}{2} x y \\sin \\theta$) is\n\n$$\n\\frac{1}{2}(t)(5-t) \\frac{3}{5} + \\frac{1}{2}(t)(3-t) \\frac{4}{5} + \\frac{1}{2}(t)(4-t) \\cdot 1 = \\frac{1}{10}(-12 t^2 + 47 t)\n$$\n\nwhich has a max... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 47/24 | |
0bhn | Let $I \subset \mathbb{R}$ be an interval and $f : I \to \mathbb{R}$ be a continuous function. Prove that the following properties are equivalent:
i) $f(x) - f(y) < |x - y|$, for every $x, y \in I$;
ii) $\int_x^y f(t)dt - f(x)(y-x) \le \frac{1}{2}(x-y)^2$, for every $x, y \in I$. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Calculus > Integral Calculus > Applications",
"Precalculus > Functions"
] | null | proof only | null | |
0ik1 | Let $P(x)$ be a polynomial of degree $n > 1$ with integer coefficients and let $k$ be a positive integer. Consider the polynomial
$$
Q(x) = \underbrace{P(P(\dots(P(x)\dots)))}_{k \text{ P's}}
$$
Prove that there are at most $n$ integers $t$ such that $Q(t) = t$. | [
"Let $\\mathbb{Z}$ denote the set of integers. We define\n$$\nS_P = \\{t \\mid t \\in \\mathbb{Z} \\text{ and } P(t) = t\\} \\quad \\text{and} \\quad S_Q = \\{t \\mid t \\in \\mathbb{Z} \\text{ and } Q(t) = t\\}.\n$$\nClearly, $S_P$ is a subset of $S_Q$. Also note that there are at most $n$ elements in $S_P$. This ... | United States | IMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0db9 | Find all pairs of positive integers $(m, n)$ such that
$$
125 \cdot 2^{n} - 3^{m} = 271.
$$ | [
"Considering the equation $\\bmod\\ 5$ we get\n$$\n3^{m} \\equiv -1 \\pmod{5},\n$$\nso $m = 4k + 2$ for some positive integer $k$. Then, considering the equation $\\bmod\\ 7$ we get\n$$\n-2^{n} - 9^{2k+1} \\equiv 5 \\pmod{7},\n$$\nwhich means\n$$\n2^{n} + 2^{2k+1} \\equiv 2 \\pmod{7}.\n$$\nSince $2^{s} \\equiv 1, 2... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (6, 3) | |
01ae | $A$ and $B$ are points on a given circle. Points $C$ and $D$ move along the circle such that $C$ and $D$ are on the same side of the line $AB$ and the length of the segment $CD$ does not change. $I_1$ and $I_2$ are incenters of the triangles $ABC$ and $ABD$. Prove that there exists a circle such that in every moment th... | [
"\nLet $M$ be a midpoint of the second arc $AB$. Then it is well-known, that $MI_1 = MA = MB$ (lemma of P. Mansion) and similarly $MI_2 = MA = MB$. Observe that $\\angle I_1MI_2 = \\angle CMD = \\text{const.}$ Therefore the points $I_1$ and $I_2$ move along the circle with the center $M$ an... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions an... | null | proof only | null | |
0bo1 | Find all functions $f : \mathbb{N}^* \to \mathbb{N}^*$ fulfilling the relation
$$
d(x, f(y)) \cdot m(f(x), y) = d(x, y) \cdot m(f(x), f(y)), \text{ for every } x, y \in \mathbb{N}^*,
$$
where $d(a, b)$ and $m(a, b)$ signify the greatest common divisor, respectively the lowest common multiple of the positive integers $a... | [
"Plugging $x = 1$ yields $m(a, y) = m(a, f(y))$, $\\forall y \\in \\mathbb{N}^*$, where $a = f(1)$. Also, taking $y = 1$ leads to $d(x, a)f(x) = m(a, f(x))$, $\\forall x \\in \\mathbb{N}^*$. This implies $d(x, a)f(x) = m(x, a)$, $\\forall x \\in \\mathbb{N}^*$. (*)\n\nRelation (*) implies $af(a) = a$ and $af(a^2) =... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(n) = n for all positive integers n | |
0ghw | 令 $m, n \ge 2$ 為整數, $X = \{1, 2, \dots, n\}$, 且 $X_1, X_2, \dots, X_m$ 為 $m$ 個 $X$ 的相異非空子集 (其兩兩交集不必然非空)。一個函數 $f : X \to \{1, 2, \dots, n+1\}$ 是好棒棒的, 若且唯若存在指標 $k$ 使得
$$
\sum_{x \in X_k} f(x) > \sum_{x \in X_i} f(x) \text{對於所有} i \neq k \text{皆成立。}
$$
證明好棒棒函數的總數不少於 $n^n$.
Let $m, n \ge 2$ be integers, $X = \{1, 2, \cdo... | [
"對於 $Y \\subset X$ 與任何定義域在 $X$ 上的函數 $f$，讓我們定義 $f(Y) = \\sum_{x \\in Y} f(x)$，則 $f$ 是好棒棒的等價於 $f(X_i)$ 在唯一的指標 $i \\in \\{1, 2, \\dots, m\\}$ 時取到最大值。\n令 $\\mathcal{G}$ 為所有函數 $g : X \\to \\{1, 2, \\dots, n\\}$ 所成集合；注意到 $|\\mathcal{G}| = n^n$。現在，對於每一個 $g \\in \\mathcal{G}$，取任一極大化 $g(X_\\ell)$ 的指標 $\\ell$，並定義\n$$\nf_g(x)... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Chinese (Traditional) | proof only | null | |
07p0 | Let $n > 2$ be an integer and
$$
f(x) = x^n + x^{n-1} - x^{n-2} - 3.
$$
Prove that $f(x)$ cannot be factored as the product of two polynomials with integer coefficients and degree less than $n$. | [
"Suppose, for the sake of contradiction, that $f(x) = g(x)h(x)$ where $g(x)$ and $h(x)$ both have integer coefficients and degree at least one. Since\n$$\ng(0)h(0) = f(0) = -3,\n$$\none of $g(0)$, $h(0)$ must be $\\pm 1$, say $g(0) = \\pm 1$. The absolute value of the product of the roots (in the complex numbers) o... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
09ik | 66 students are participating in an olympiad. Each student has at least one friend and for each student, the sum of the number of their friends and the arithmetic mean of the number of friends of their friends is 11. Prove that the students can be divided into two classrooms where no two friends are in the same room. H... | [
"Let $d_X$ denote the number of friends of a student $X$ and let $m_X$ denote the arithmetic mean of the number of friends of $X$'s friends.\nLet $A$ be a student with a minimal number of friends and let $B$ be a student with a maximal number of friends. Clearly, we have $d_A \\le m_B$ and $m_A \\le d_B$, and since... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
029c | Problem:
Seja $Q$ um ponto na circunferência de diâmetro $AB$, sendo $Q$ diferente de $A$ e $B$. Seja $QH$ a reta perpendicular a $AB$ que passa por $Q$, sendo $H$ pertencente a $AB$. Os pontos de interseção da circunferência de diâmetro $AB$ e a circunferência de centro $Q$ e raio $QH$ são $C$ e $D$. Prove que $CD$ p... | [
"Solution:\n\nConsideremos as duas figuras a seguir. Sejam $\\angle CQH = \\alpha$, $\\angle QCD = \\beta$ e $E$ a interseção de $QH$ com a circunferência de diâmetro $AB$. Delas, podemos concluir que:\n\ni) Temos $\\angle CDH = \\frac{\\alpha}{2}$, pois na circunferência de centro em $Q$ o seu ângulo central $\\an... | Brazil | null | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f19 | Problem:
(1) $O$ is the circumcenter of the triangle $ABC$. The triangle is rotated about $O$ to give a new triangle $A'B'C'$. The lines $AB$ and $A'B'$ intersect at $C''$, $BC$ and $B'C'$ intersect at $A''$, and $CA$ and $C'A'$ intersect at $B''$. Show that $A''B''C''$ is similar to $ABC$.
(2) $O$ is the center of t... | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
048h | Today it's Marko's birthday. His father and grandfather have this conversation:
– Now Marko's, your and my age are all prime numbers!
– Yes, and five years from now all the three ages will be perfect squares.
How old was his grandfather when Marko was born? | [] | Croatia | CroatianCompetitions2011 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 48 | |
02ie | Problem:
Na figura temos $\hat{B}=50^{\circ}$, $AD$ e $CD$ são as bissetrizes dos ângulos $\hat{A}$ e $\hat{C}$ respectivamente.
Qual a medida do ângulo $A \hat{D} C$?
A) $90^{\circ}$
B) $100^{\circ}$
C) $115^{\circ}$
D) $122.5^{\circ}$
E) $125^{\circ}$
 | [
"Solution:\n\nNesta questão, usaremos o seguinte importante teorema da Geometria Plana:\nTeorema: A soma dos ângulos internos de um triângulo é $180^{\\circ}$.\n\nDo teorema acima temos $\\hat{A}+\\hat{B}+\\hat{C}=180^{\\circ}$, e como $\\hat{B}=50^{\\circ}$, segue que\n\n\n\n$$\n\\hat{A}+5... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
0caw | Determine all primes $p$ and all positive integers $x$ and $y$ satisfying $x^3 + y^3 = p(xy + p)$. | [
"Up to a swap of the first two entries, the only solutions are $(x, y, p) = (1, 8, 19)$, $(x, y, p) = (2, 7, 13)$ and $(x, y, p) = (4, 5, 7)$. The verification is routine.\n\nSet $s = x + y$. Rewrite the equation in the form $s(s^2 - 3xy) = p(p + xy)$, and express $xy$:\n$$\nxy = \\frac{s^3 - p^2}{3s + p}. \\qquad ... | Romania | THE Fifteenth ROMANIAN MASTER OF MATHEMATICS | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | All solutions, up to swapping the first two entries, are (x, y, p) = (1, 8, 19), (2, 7, 13), (4, 5, 7). | |
0chr | Let $H$ be the orthocenter of the acute triangle $ABC$ and let $X$ be the midpoint of the side $BC$. The perpendicular at $H$ to $HX$ intersects the sides $AB$ and $AC$ at the points $Y$ and $Z$ respectively. Let $O$ be the circumcenter of the triangle $ABC$ and $O'$ be the circumcenter of the triangle $BHC$. Prove tha... | [
"a) Let $A'$ be the point diametrically opposite to $A$ on the circumcircle of the triangle $ABC$. It is known that $H$ and $A'$ are symmetric with respect to $X$, so $H$, $X$ and $A'$ are collinear. Then $\\angle ABA' = \\angle ACA' = 90^\\circ$. Thus, the quadrilaterals $A'BYH$ and $A'CZH$ are cyclic, therefore $... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Mi... | English | proof only | null | |
05yy | Problem:
Soit $n \geqslant 1$ un entier. Morgane écrit au tableau, en base 10, les nombres $2023, 2023 \times 2, \ldots, 2023 \times n$. Pour tout chiffre $c$ compris entre 1 et 9, elle note alors $\mathrm{d}_{c}(n)$ le nombre d'apparitions du chiffre $c$ sur le tableau. Par exemple, si $n=3$, elle écrit les nombres $... | [
"Solution:\n\nSoit $k$ un multiple de $\\varphi(2023)$, soit $\\ell = (10^{k} - 1) / 2023$. Nous allons démontrer que $\\ell$ est une solution. Pour ce faire, on introduit l'ensemble\n$$\n\\Omega = \\{2023 n : 1 \\leqslant n \\leqslant \\ell\\}\n$$\nLes éléments de $\\Omega$ décrivent précisément l'ensemble des rés... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
0h0y | An equilateral triangle with the side length $7$ is divided into $49$ small equilateral triangles with the side length $1$, as it is shown on fig. 28. Parallelograms with side lengths $1$ and $2$ are cut from the triangle along the grid lines. What is the greatest number of parallelograms that can be cut in this way?
... | [
"Paint the small triangles in black and white as it is shown on fig. 29. Since every parallelogram with the side lengths $1$ and $2$ contains two white triangles, and the total number of white triangles is $21$, the number of parallelograms does not exceed $10$.\n\nOn the fig. 30 it is shown how one can cut $10$ pa... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 10 | |
01a2 | Does there exist a figure on the squared plane that can be split onto dominoes by exactly 17 ways? | [
"Answer: yes, this figure exists. It is shown on the picture.\n\n\n\nIf cell $A$ belongs to the horizontal domino, then we have the following forced picture.\n\n\n\nEach of the $3 \\times 2$ boxes can be split on dominoes by $3$ ways, so we have here $9$ ways.\n\nIf... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | Yes | |
024g | Problem:
Retângulos - As medidas dos lados de um retângulo são números pares. Quantos retângulos desses existem com área igual a 96? | [
"Solution:\n\nSe $a$ e $b$ denotam o comprimento e a largura do retângulo, temos $a \\times b = 96$. Logo, $a$ e $b$ são divisores pares de $96$ e, portanto, temos quatro retângulos satisfazendo as condições dadas, a saber, os retângulos de lados medindo $2$ e $48$; $4$ e $24$; $6$ e $16$; e $8$ e $12$."
] | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 4 | |
0bul | Problem:
Se consideră unghiurile adiacente $\Varangle A O B$ și $\Varangle B O C$, astfel încât bisectoarele lor $[OM$ și $[ON$ să formeze un unghi de $75^{\circ}$.
a) Să se determine $m(\Varangle A O B)$ și $m(\Varangle B O C)$ știind că $3 \cdot m(\Varangle A O B)=2 \cdot m(\Varangle B O C)$.
b) Dacă semidreapta $[O... | [
"Solution:\n\na) Notăm $m(\\Varangle A O B)=\\alpha$ și $m(\\Varangle B O C)=\\beta$. Avem:\n$[OM$ bisectoarea $\\Varangle A O B \\Rightarrow m(\\Varangle A O M)=m(\\Varangle M O B)=\\frac{\\alpha}{2}$;\n$[ON$ bisectoarea $\\Varangle B O C \\Rightarrow m(\\Varangle B O N)=m(\\Varangle N O C)=\\frac{\\beta}{2}$.\n\n... | Romania | Olimpiada de Matematică - Etapa Locală | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a) m(∠AOB) = 60°, m(∠BOC) = 90°. b) m(∠TON) = 15°, m(∠BON) = 45°, m(∠BOT) = 60°, m(∠COT) = 30°. | |
02km | Problem:
Você já viu um truque numérico? Aqui vão os passos de um truque numérico:
(I) Escolha um número qualquer.
(II) Multiplique-o por 6.
(III) Do resultado subtraia 21.
(IV) Divida agora este novo resultado por 3.
(V) Deste último resultado subtraia o dobro do número que você escolheu.
a. Experimente fazer esses... | [
"Solution:\n\na.\nVamos fazer o experimento com os números $0$, $5$ e $-4$.\n\nPara $0$:\n\n- Escolha $0$.\n- Multiplique por $6$: $0 \\times 6 = 0$.\n- Subtraia $21$: $0 - 21 = -21$.\n- Divida por $3$: $-21 \\div 3 = -7$.\n- Subtraia o dobro do número escolhido: $-7 - 2 \\times 0 = -7$.\n\nPara $5$:\n\n- Escolha $... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -7 | |
0f9h | Problem:
Two unequal circles intersect at $X$ and $Y$. Their common tangents intersect at $Z$. One of the tangents touches the circles at $P$ and $Q$. Show that $ZX$ is tangent to the circumcircle of $PXQ$. | [] | Soviet Union | 24th ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians"
] | null | proof only | null | |
0kmp | Problem:
Let $f(x) = x^{3} + 3x - 1$ have roots $a, b, c$. Given that
$$
\frac{1}{a^{3} + b^{3}} + \frac{1}{b^{3} + c^{3}} + \frac{1}{c^{3} + a^{3}}
$$
can be written as $\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n) = 1$, find $100m + n$. | [
"Solution:\nWe know that $a^{3} = -3a + 1$ and similarly for $b, c$, so\n$$\n\\frac{1}{a^{3} + b^{3}} = \\frac{1}{2 - 3a - 3b} = \\frac{1}{2 + 3c} = \\frac{1}{3(2/3 + c)}.\n$$\nNow,\n$$\nf(x - 2/3) = x^{3} - 2x^{2} + \\frac{13}{3}x - \\frac{89}{27}\n$$\nhas roots $a + 2/3$, $b + 2/3$, and $c + 2/3$. Thus the answer... | United States | HMMT November 2021 Team Round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 3989 | |
07c1 | Suppose $a$, $b$, $c$ and $d$ are positive real numbers such that $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1} = 2$. Prove that
$$
\sqrt{\frac{a^2+1}{2}} + \sqrt{\frac{b^2+1}{2}} + \sqrt{\frac{c^2+1}{2}} + \sqrt{\frac{d^2+1}{2}} \geq 3(\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}) - 8.
$$ | [
"$$\n\\sum \\left( \\sqrt{\\frac{a^2+1}{2}} - \\sqrt{a} \\right) = \\sum \\frac{\\frac{a^2+1}{2} - a}{\\sqrt{\\frac{a^2+1}{2}} + \\sqrt{a}} = \\frac{1}{2} \\sum \\frac{(a-1)^2}{\\sqrt{\\frac{a^2+1}{2}} + \\sqrt{a}}\n$$\nAccording to Cauchy-Schwarz inequality, $\\sqrt{\\frac{a^2+1}{2}} + \\sqrt{a} \\le \\sqrt{2(\\fr... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00br | A natural number is written on each face of a cube. To each vertex of the cube assign the product of the numbers of the three faces that have this vertex in common. Let the sum of these 8 products be 315. Determine the sum of the numbers on the faces (find all possibilities). | [
"Let the numbers on the pair of opposite faces be $a_1, a_2, b_1, b_2, c_1, c_2$. Note that $a_1$ participates in 4 products: $a_1b_1c_1, a_1b_1c_2, a_1b_2c_1, a_1b_2c_2$. Likewise $a_2$ participates in the remaining 4 products and in a completely analogous fashion the products are $a_2b_1c_1, a_2b_1c_2, a_2b_2c_1,... | Argentina | Argentina_2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 21, 25, 29, 41 | |
0873 | Problem:
Il perimetro di un rombo è $32~\mathrm{cm}$ e ciascuno dei due angoli acuti misura $30^\circ$. Quanto vale il volume del solido ottenuto facendo ruotare il rombo intorno a un suo lato?
(A) $128 \sqrt{3} \pi$
(B) $128 \pi$
(C) $64(\sqrt{3}-1) \pi$
(D) $64 \pi$
(E) $32 \sqrt{3} \pi$. | [
"Solution:\n\nLa risposta è (B). Si osserva che il solido di rotazione che si ottiene ruotando un rombo intorno a un suo lato può essere visto come un cono sovrapposto a un cilindro \"scavato\". La regione di spazio che viene scavata nel cilindro è uguale al cono sopra di esso, di conseguenza il volume $V$ richiest... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | MCQ | B | |
0ivs | Problem:
Six men and their wives are sitting at a round table with 12 seats. These men and women are very jealous - no man will allow his wife to sit next to any man except for himself, and no woman will allow her husband to sit next to any woman except for herself. In how many distinct ways can these 12 people be sea... | [
"Solution:\n\nAnswer: 288000\n\nThink of this problem in terms of \"blocks\" of men and women, that is, groups of men and women sitting together. Each block must contain at least two people; otherwise you have a man sitting next to two women (or vice-versa).\n\nWe will define the notation $[a_{1}, b_{1}, a_{2}, b_{... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 288000 | |
06fx | Let $n$ be a positive integer. Determine the number of sequences $x_1, x_2, \dots, x_{2n-1}, x_{2n}$, with $x_i = 1$ or $-1$, and such that $\left|\sum_{i=2k-1}^{2m} x_i\right| \le 2$ for all integers $k, m$, with $1 \le k \le m \le n$. | [
"The answer is $2 \\cdot 3^n - 2^n$.\nLet $y_j = x_{2j-1} + x_{2j}$ for $j = 1, 2, \\dots, n$. Every desired sequence $\\{x_j\\}$ corresponds to a sequence $\\{y_j\\}$ such that $y_j = -2, 0, 2$ for each $j$, and $|y_k + y_{k+1} + \\dots + y_m| \\le 2$ for any $1 \\le k \\le m \\le n$. Suppose there are $r$ terms o... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 2*3^n - 2^n | |
041y | For any integer $n$ with $n > 1$, let $n = p_1^{a_1} \cdots p_t^{a_t}$ be its standard factorization, write
$$
\omega(n) = t, \quad \Omega(n) = \alpha_1 + \cdots + \alpha_t.
$$
Prove or disprove the following statement: Given any positive integer $k$ and any positive real numbers $\alpha$ and $\beta$, there exists a po... | [
"The answer is YES.\nFrom the definition of $\\omega$ and $\\Omega$, we have\n$$\n\\omega(ab) \\le \\omega(a) + \\omega(b), \\qquad \\textcircled{1}\n$$\n$$\n\\Omega(ab) = \\Omega(a) + \\Omega(b), \\qquad \\textcircled{2}\n$$\nfor any positive integers $a, b$. Given a fixed positive integer $k$ and positive real nu... | China | China Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | YES | |
0e2i | Problem:
Ničla linearne funkcije je $2$, začetna vrednost pa $\frac{5}{2}$. Zapiši enačbo premice, ki je vzporedna grafu dane funkcije in seka os $x$ pri $\frac{4}{3}$, v implicitni obliki. | [
"Solution:\n\nUpoštevamo, da je $n=\\frac{5}{2}$ in $f(2)=0$, kar vstavimo v predpis za linearno funkcijo $f(x)=k \\cdot x+n$. Dobimo $k \\cdot 2+\\frac{5}{2}=0$ in izračunamo $k=-\\frac{5}{4}$. Upoštevamo lastnost vzporednih premic, torej da imata enak smerni koeficient. Enačba premice je $y=-\\frac{5}{4} x+n_{1}$... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 15x + 12y - 20 = 0 | |
0ec3 | Problem:
Naj bosta $D$ in $E$ zaporedoma razpolovišči stranic $B C$ in $C A$ trikotnika $A B C$. Premici $A D$ in $B E$ sekata trikotniku $A B C$ očrtano krožnico zaporedoma še v točkah $P$ in $Q$. Denimo, da je $|D P|=|E Q|$. Dokaži, da je trikotnik $A B C$ enakokrak z vrhom $C$.
 | [
"Solution:\n\n1. način. Ker sta $D$ in $E$ razpolovišči daljic $B C$ in $A C$, sta premici $D E$ in $A B$ vzporedni. Od tod sledi $\\angle E D A = \\angle B A D$, po izreku o obodnih kotih pa velja $\\angle B A D = \\angle B A P = \\angle B Q P$. Torej je $\\angle E Q P = \\angle E D A = \\pi - \\angle P D E$, kar ... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
01dg | Let $ABCD$ be a convex quadrilateral. Let $P$ be a point such that $\angle APC = \angle BPD = 30^\circ$. Prove that
$$
2(AB + AC + AD + BC + BD + CD) \geq PA + PB + PC + PD.
$$ | [
"**Lemma.** If $\\angle XPZ = 30^\\circ$, then $PY$ does not exceed the perimeter of $XYZ$.\n\n**Proof:** We reflect $Y$ about $XP$ and $PZ$ getting $Y'$ and $Y''$. We observe that $\\angle Y''PY' = 60^\\circ$ and $PY = PY' = PY''$, so $PY'Y''$ is equilateral, so\n$$\nPY = PY' = Y'Y'' \\leq Y'X + XZ + ZY'' = XY + Y... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chas... | null | proof only | null | |
02id | Problem:
Toda vez que Joãozinho vai ao cinema, ele toma 2 refrigerantes. Ele gastou toda a sua mesada de $R\$ 50,00$ indo ao cinema 6 vezes e tomando um total de 20 refrigerantes, incluindo os que ele tomou quando foi ao cinema. Se Joãozinho tivesse tomado só um refrigerante cada vez que foi ao cinema, com essa econom... | [
"Solution:\n\n(C) A economia teria sido equivalente a 6 refrigerantes, permitindo a Joãozinho mais um cinema e mais um refrigerante. Logo o ingresso do cinema é 5 vezes o valor do refrigerante."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
06mr | Given a $2023 \times 2023$ square grid, there are beetles on some of the unit squares, with at most one beetle on each unit square. In the first minute, every beetle will move one step to its right or left adjacent square, or to its top or bottom adjacent square. In the second minute, every beetle will move again, only... | [
"The answer is $2022^2 + 1 = 4088485$.\nFirst we show that $2022^2$ is insufficient. Suppose there are $2022^2$ beetles in the top left $2022 \\times 2022$ square grid. This square grid can be partitioned into many $2 \\times 2$ square grids in the natural way. If the beetles in each $2 \\times 2$ square grid move ... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 4088485 | |
024o | Problem:
Num tabuleiro $5 \times 5$, um cavaleiro do jogo de xadrez está na casa marcada com $A$. Depois ele se move marcando as casas por onde passa:
$A \rightarrow B \rightarrow C \rightarrow D \rightarrow E \rightarrow F \rightarrow G \rightarrow H$.
| $A$ | | | | $G$ |
| :--- | :--- | :--- | :--- | :--- |
... | [
"Solution:\n\n| $A$ | $X$ | $M$ | $R$ | $G$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $N$ | $S$ | $H$ | $Y$ | $L$ |\n| $I$ | $B$ | $W$ | $F$ | $Q$ |\n| $T$ | $O$ | $D$ | $K$ | $V$ |\n| $C$ | $J$ | $U$ | $P$ | $E$ |"
] | Brazil | Desafios | [
"Discrete Mathematics > Other"
] | null | final answer only | A → B → C → D → E → F → G → H → I → J → K → L → M → N → O → P → Q → R → S → T → U → V → W → X → Y |
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