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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0f2y | Problem:
The set $\{S_0\}$ has the single member $(5, 19)$. We derive the set $\{S_{n + 1}\}$ from $\{S_n\}$ by adjoining a pair to $\{S_n\}$. If $\{S_n\}$ contains the pair $(2a, 2b)$, then we may adjoin the pair $(a, b)$. If $\{S\}$ contains the pair $(a, b)$ we may adjoin $(a+1, b+1)$. If $\{S\}$ contains $(a, b)$ a... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | Yes for (1,50); No for (1,100). In general, starting from (a,b) with a<b, let r be the greatest odd divisor of b−a. Then (1,n) is obtainable if and only if n≡1 (mod r), equivalently n−1 is divisible by r. | |
03r7 | The vertical cross-section of a circular cone with vertex $P$ is an isosceles right-angled triangle. Point $A$ is on the circumference of the base circle, point $B$ is interior to the base circle, $O$ is the center of the base circle, $AB \perp OB$ and intersecting at $B$, $OH \perp PB$ and intersecting at $H$, $PA = 4... | [
"Since $AB \\perp OB$, and $AB \\perp OP$, we have $AB \\perp PB$, and $PAB \\perp POB$. Moreover, from $OH \\perp PB$ we obtain that $OH \\perp HC$ and $OH \\perp PA$. Since $C$ is the midpoint of $PA$, $OC \\perp PA$. Thus, $PC$ is the altitude of the\n\n\ntetrahedron $O-HPC$ and $PC = 2$... | China | China Mathematical Competition (Hainan) | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | MCQ | D | |
08xr | Let $m$ be a positive integer of $1000$ digits, with the property that all of its digits are non-zero. For a positive integer $n$, consider $\left\lfloor \frac{m}{n} \right\rfloor$, where we define for any real number $r$, $\lfloor r \rfloor$ to be the largest integer less than or equal to $r$. Determine the largest po... | [
"Let $M$ be the maximum number we seek. First we show that $M \\le 939$ must hold.\nLet $m, n$ be positive integers satisfying the conditions of the problem, and let $k$ be the number of digits of $n$ (so, $1 \\le k \\le 1000$). Let us represent $\\frac{m}{n}$ as\n$$\n\\frac{m}{n} = a_1 10^{b_1} + a_2 10^{b_2} + \\... | Japan | Japan Mathematical Olympiad Initial Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Other"
] | English | proof and answer | 939 | |
0798 | Two circles $\omega_1$ and $\omega_2$ intersect at $P$ and $K$. $XY$ is the common tangent of them near to $P$ such that $X$ is on $\omega_1$ and $Y$ is on $\omega_2$. $XP$ intersects $\omega_2$ for the second time at $C$, and $YP$ intersects $\omega_1$ for the second time at $B$. $A$ is the intersection of $BX$ and $C... | [
"Since $Q$ is on both circumcircles of $ABC$ and $AXY$, so there is a spiral similarity about $Q$ carrying one of these circles to another such that it carries $X$ to $B$, and $Y$ to $C$. Suppose this similarity carries $K$ to a point $T$. It is enough that we prove that the points $P, K, T$ are collinear, because ... | Iran | 27th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
067w | In a School formed $112$ groups each contained $11$ students. Every pair of groups had exactly one common student. Prove that:
a. There exists a student belonging at least to $12$ groups.
b. There exists a student belonging to all groups. | [
"a. We consider an arbitrary group $O$. Each of the rest $111$ groups has exactly one student belonging to $O$. Since $111 = 11 \\cdot 10 + 1$, from the pigeonhole's rule we conclude that there exists one student $x \\in O$ who belongs at least to $11$ other groups. Hence $x$ belongs at least to $12$ groups, say $O... | Greece | SELECTION EXAMINATION | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0hqc | Problem:
Let $ABCD$ be a parallelogram. Suppose that the circumcenter of $\triangle ABC$ lies on diagonal $BD$. Prove that $ABCD$ is either a rectangle or a rhombus (or both). | [
"Solution:\n\nTo get a conclusion of the appropriate type (a rectangle OR a rhombus), we must divide up the problem into two cases. Here is one way of accomplishing this:\n\nCase 1. The circumcenter $O$ of $ABC$ is the center of $ABCD$, the common midpoint of diagonals $AC$ and $BD$. Then since radii $OA$ and $OB$ ... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plan... | null | proof only | null | |
0ixx | Problem:
Suppose $G$ is a graph with chromatic number $n$. Suppose there exist $k$ graphs $G_{1}, G_{2}, \ldots, G_{k}$ having the same vertex set as $G$ such that $G_{1} \cup G_{2} \cup \cdots \cup G_{k}=G$ and each $G_{i}$ has chromatic number at most $2$. Show that $k \geq \left\lceil \log_{2}(n) \right\rceil$, and... | [
"Solution:\n\nThe bound on $k$ follows from iterating part (a).\nLet $G$ be a graph with chromatic number $n$. Consider a coloring of $G$ using $n$ colors labeled $1,2, \\ldots, n$. For $i$ from $1$ to $\\left\\lceil \\log_{2}(n) \\right\\rceil$, define $G_{i}$ to be the graph on the vertices of $G$ for which two v... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0lc3 | Let $ABC$ be an acute triangle with circumcircle ($O$). $D$ is a point on arc $BC$ that does not contain $A$. A moving line $l$ through orthocenter $H$ of triangle $ABC$ cuts the circumcircles of triangle $ABH$ and triangle $ACH$ again at $M, N$ respectively ($M \neq H, N \neq H$).
a) Define the position of $l$ such t... | [
"a) Firstly, note that when $\\ell$ changes, the angles $\\angle AMN$ and $\\angle ANM$ both remain unchanged, so triangle $AMN$ is always self-congruent. Draw $AK$ perpendicular to $MN$ ($K \\in MN$), then $AK \\le AH$. Therefore, the area of triangle $AMN$ attains maximal value when $AH$ is the altitude or $MN \\... | Vietnam | VMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneo... | English | proof and answer | a) The area is maximal when the moving line is perpendicular to the line from the vertex to the orthocenter (equivalently, when MN is perpendicular to AH). b) The intersection point P lies on the fixed circumcircle of triangle AEF, where AE is parallel to DB and AF is parallel to DC. | |
048y | Let $a$, $b$ be real numbers such that all the zeros of the polynomial $P(x) = x^3 + a x^2 + b x - 8$ are real. Prove that $a^2 \ge 2b + 12$. (Kristina Ana Škreb) | [
"Polynomial $P(x)$ has three zeros, let us denote them by $x_1$, $x_2$ and $x_3$.\nAccording to Viète's formulas we have:\n$$\n\\begin{aligned}\nx_1 + x_2 + x_3 &= -a \\\\\nx_1 x_2 + x_2 x_3 + x_3 x_1 &= b \\\\\nx_1 x_2 x_3 &= 8.\n\\end{aligned}\n$$\nIt follows $x_1^2 + x_2^2 + x_3^2 = a^2 - 2b$, and from the A-G i... | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0gfx | 設三角形 $ABC$ 的外接圓為 $\omega$, 而其切於 $BC$ 邊的旁切圓為 $\Omega_A$。令 $\omega$ 與 $\Omega_A$ 的交點為 $X$ 和 $Y$。設 $P$ 為 $A$ 對 $\Omega_A$ 在點 $X$ 的切線的投影點, 而 $Q$ 為 $A$ 對 $\Omega_A$ 在點 $Y$ 的切線的投影點。設三角形 $APX$ 在點 $P$ 的切線, 及三角形 $AQY$ 在點 $Q$ 的切線在點 $R$ 相交。
證明直線 $AR$ 與 $BC$ 互相垂直。 | [
"設 $D$ 為 $BC$ 與 $\\Omega_A$ 的切點, 而 $D'$ 為 $D$ 在 $\\Omega_A$ 上的對徑點。令 $R'$ 為滿足 $AR' \\perp BC$ 及 $R'D' \\parallel BC$ 的 (唯一) 點。我們將證明 $R = R'$。\n\n設直線 $PX$ 分別與 $AB$, $D'R'$ 交於點 $S$, $T$。令 $U$ 為平行直線 $BC$, $D'R'$ 相交的無窮遠點。由於 (退化的) 六邊形 $ASXTUC$ 外切圓 $\\Omega_A$, 根據 Brianchon 定理知 $AT$, $SU$, $XC$ 三線共於一點, 設為 $V$。所以 $VS \\par... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 國際競賽實作(一) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese; English | proof only | null | |
09zr | The candy store sells chocolates in the flavours white, milk, and dark. You can buy them in three types of coloured boxes. The three boxes have the following contents:
* Gold: 2 white, 3 milk, 1 dark,
* Silver: 1 white, 2 milk, 4 dark,
* Bronze: 5 white, 1 milk, 2 dark.
Lavinia buys some boxes of chocolates (at least... | [] | Netherlands | Second Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 20 | |
0btd | Let $I \subset \mathbb{R}$ be an open interval and $f, g : I \to \mathbb{R}$ two functions that satisfy
$$
\frac{f(x) - g(y)}{x - y} + |x - y| \ge 0, \text{ for any } x, y \in I, x \ne y.
$$
i) Deduce that $f$ and $g$ are non-decreasing.
ii) Find $f, g : \mathbb{R} \to \mathbb{R}$, $f \ne g$, having the above property... | [
"i) The given relation can be rewritten as $f(x)+(x-z)^2 \\ge g(z) \\ge f(y)-(z-y)^2$, for all triple $x > z > y$, $x, y, z \\in I$; in particular, $f(x) + (x-y)^2 \\ge f(y)$, for all triplets $x > y$, $x, y \\in I$.\n\nFor $a, b \\in I$, $a > b$, taking $n \\in \\mathbb{N}^*$ and $x_k = a - \\frac{k}{n}(a-b)$, $k ... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | i) f and g are non-decreasing on I.
ii) One choice on R is f(x) = 0 for x < 0 and f(x) = 1 for x ≥ 0; g(x) = 0 for x ≤ 0 and g(x) = 1 for x > 0. | |
0b8w | Determine whether there exist a polynomial $f(x_1, x_2)$ in two variables, with integer coefficients, and two points $A = (a_1, a_2)$ and $B = (b_1, b_2)$ in the plane, satisfying all the following conditions
(i) $A$ is an integer point (i.e., $a_1$ and $a_2$ are integers);
(ii) $|a_1 - b_1| + |a_2 - b_2| = 2010$;
(iii... | [
"The triple $(f(x_1, x_2), A, B)$ does exist, so the answer is yes.\nLet $A = O = (0, 0)$, $B = (x_0, y_0) = (2009 + \\frac{2}{3}, \\frac{1}{3})$. The idea is to search for a polynomial $f$ such that $f(x, y) = 0$ is the equation of an ellipse centered at $B$, passing through $O$ and with tangent line $y = 0$ at $O... | Romania | Local Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
037j | Problem:
Find all positive integers $n$ for which the equality
$$
\frac{\sin (n \alpha)}{\sin \alpha}-\frac{\cos (n \alpha)}{\cos \alpha}=n-1
$$
holds true for all $\alpha \neq \frac{k \pi}{2}, \quad k \in \mathbb{Z}$. | [
"Solution:\nThe equality is equivalent to\n$$\n\\sin (n-1) \\alpha=\\frac{(n-1) \\sin 2 \\alpha}{2}\n$$\nWhen $n \\geq 4$ setting $\\alpha=\\frac{\\pi}{4}$ gives\n$$\n\\sin \\left((n-1) \\frac{\\pi}{4}\\right)=\\frac{n-1}{2} \\geq \\frac{3}{2}\n$$\na contradiction.\nWhen $n=1$ and $n=3$ the equality (1) is an ident... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | n = 1 and n = 3 | |
081n | Problem:
Gli interi da $1$ a $9$ sono scritti nelle nove caselle di una scacchiera $3 \times 3$, ogni intero in una casella diversa, in modo tale che ogni coppia di numeri consecutivi sia scritta in due caselle adiacenti (cioè aventi un lato in comune). Quanti sono i valori possibili del numero posto sulla casella cen... | [
"Solution:\n\nLa risposta è $5$. Perché la condizione data si possa realizzare, è necessario che si possa fare un percorso dalla casella con il numero $1$ alla casella con il numero $9$ muovendo successivamente da una casella ad una a lei adiacente. Coloriamo la scacchiera nel modo usuale, in modo tale che le casel... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 5 | |
0kn6 | When a certain unfair die is rolled, an even number is 3 times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?
(A) $\frac{3}{8}$ (B) $\frac{4}{9}$ (C) $\frac{5}{9}$ (D) $\frac{9}{16}$ (E) $\frac{5}{8}$ | [
"Suppose that the probability of rolling an odd number is $p$. The probability of rolling an even number is then $3p$. Because $p + 3p = 1$, it follows that $p = \\frac{1}{4}$, so the probability of rolling an odd number is $\\frac{1}{4}$ and the probability of rolling an even number is $1 - \\frac{1}{4} = \\frac{3... | United States | AMC 10 A | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | MCQ | E | |
03aj | Let $M$ be an infinite set of rational numbers such that the product of any 2009 of them (pairwise different) is an integer which is not divisible by 2009th powers of primes. Prove that all the numbers in $M$ are integers. | [
"Let $a_1, \\dots, a_{2008} \\in M$ and $A = a_1 \\dots a_{2008} = \\frac{p}{q}$, $(p, q) = 1$. Assume that $M$ contains infinitely many numbers $\\alpha_i = \\frac{p_i}{q_i}$ such that $(p_i, q_i) = 1$, $q_i > 1$ and $\\alpha_i \\ne a_1, \\dots, a_{2008}$. Since $\\alpha_i p$ is an integer, then $q_i$ divides $p$ ... | Bulgaria | Team selection test for 50. IMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
05an | In a triangle $ABC$, the internal angle bisectors at vertices $B$ and $C$ intersect at point $I$ and intersect the sides $CA$ and $AB$ at points $E$ and $F$, respectively. Let $M$ and $N$ be the midpoints of segments $BI$ and $CI$, respectively. The line $FM$ intersects the external angle bisector at vertex $B$ of tria... | [
"We have $\\angle KBE = 90^\\circ$ as $BE$ and $BK$ are the internal and external angle bisectors at the same vertex (Fig. 43). We will now show $\\angle BKC = 90^\\circ$. Let $X$ be a point on line $AB$ such that $CX \\parallel BE$; then the external angle bisector at vertex $B$ of triangle $ABC$ is also perpendic... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous ... | English | proof only | null | |
02p0 | Problem:
Um conjunto de números é chamado trilegal se pode ser dividido em subconjuntos com três elementos de tal modo que um dos elementos seja a soma dos outros dois. Por exemplo, o conjunto $\{1,2,3, \ldots, 11,12\}$ é trilegal pois pode ser dividido em $\{1,5,6\}$, $\{2,9,11\},\{3,7,10\}$ e $\{4,8,12\}$.
a. Mostr... | [
"Solution:\n\na. Para a primeira parte basta encontrar uma distribuição em subconjuntos com três elementos, por exemplo\n$$\n\\{1,6,7\\},\\{2,12,14\\},\\{3,8,11\\},\\{4,9,13\\},\\{5,10,15\\}\n$$\n\nb. Observemos que se um conjunto de três elementos cumpre a condição de ser trilegal, então ele tem de ser da forma\n$... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
01bq | Let $ABC$ be a triangle with $AB \neq AC$. The angle bisector of $\angle BAC$ intersects $BC$ in $D$. The circle with diameter $AD$ intersects $AC$ again in $P$, and $BC$ again in $Q$. The point $R \neq Q$ lies on the line parallel to $AD$ through $Q$. Suppose that $AQ = AR$. Prove that the points $B, P, Q$, and $R$ li... | [
"Let $g$ be the exterior angle bisector of $\\angle BAC$. Consider the reflection about $g$. As $QR$ is parallel to $AD$, and hence orthogonal to $g$, the condition $AQ = AR$ implies that this reflection maps $Q$ to $R$. It is clear that the reflection maps $B$ and $P$ to some points $B'$ on $AC$ and $P'$ on $AB$, ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
06zr | Problem:
Given a triangle $ABC$, take $A'$ on the ray $BA$ (on the opposite side of $A$ to $B$) so that $AA' = BC$, and take $A''$ on the ray $CA$ (on the opposite side of $A$ to $C$) so that $AA'' = BC$. Similarly take $B'$, $B''$ on the rays $CB$, $AB$ respectively with $BB' = BB'' = CA$, and $C'$, $C''$ on the rays... | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0f4v | Problem:
A tetrahedron $T'$ has all its vertices inside the tetrahedron $T$. Show that the sum of the edge lengths of $T'$ is less than $\frac{4}{3}$ times the corresponding sum for $T$. | [] | Soviet Union | 16th ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
06s5 | For a nonnegative integer $n$ define $\operatorname{rad}(n)=1$ if $n=0$ or $n=1$, and $\operatorname{rad}(n)=p_{1} p_{2} \cdots p_{k}$ where $p_{1}<p_{2}<\cdots<p_{k}$ are all prime factors of $n$. Find all polynomials $f(x)$ with nonnegative integer coefficients such that $\operatorname{rad}(f(n))$ divides $\operatorn... | [
"We are going to prove that $f(x)=a x^{m}$ for some nonnegative integers $a$ and $m$. If $f(x)$ is the zero polynomial we are done, so assume that $f(x)$ has at least one positive coefficient. In particular $f(1)>0$.\n\nLet $p$ be a prime number. The condition is that $f(n) \\equiv 0\\ (\\bmod\\ p)$ implies\n$$\nf\... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Ei... | null | proof and answer | f(x) = a x^m with nonnegative integers a and m (including the zero polynomial). | |
06jo | Let $\triangle ABC$ be an acute-angled triangle. Let $D$ be a point on the segment $BC$, $I$ the incentre of $\triangle ABC$. The circumcircle of $\triangle ABD$ meets $BI$ at $P$ and the circumcircle of $\triangle ACD$ meets $CI$ at $Q$. If the area of $\triangle PID$ and the area of $\triangle QID$ are equal, prove t... | [
"Since $\\angle AQI = \\angle AQC = \\angle ADC = 180^\\circ - \\angle ADB = 180^\\circ - \\angle APB = 180^\\circ - \\angle API$, the points $A, Q, I, P$ are concyclic. Let $DI$ meet $PQ$ at $M$, and let $AM$ meet $(AQIP)$ again at $X$. Since $[PID] = [QID]$, we know that $MP = MQ$. Also, note that $AP = PD$ and $... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane G... | null | proof only | null | |
0dju | At Hadi's birthday party, $2023$ friends came to give him $2023$ different gifts. Hadi wants to know exactly who gave which gifts, so he came up with a way: each time, choose a group of $11$ people and ask them what gifts they gave. When responding, the group only reported their set of gifts (not knowing exactly which ... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 337 | |
06cx | Let $a_1 = 1$, $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$ for $n = 1, 2, 3, \dots$. Find the greatest integer less than or equal to $a_{2000}$. Be sure to prove your claim. | [
"The answer is $44$.\nWe shall prove by induction that $\\sqrt{n} < a_n < b_n := \\frac{n}{\\sqrt{n-1}}$ for any integer $n \\ge 3$.\nFor the base case, we find $a_2 = 2$ and $a_3 = 2 \\in [\\sqrt{3}, \\frac{3}{\\sqrt{2}}]$.\nFor the inductive step, assume $\\sqrt{k} < a_k < b_k$ for some integer $k \\ge 3$. Note t... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 44 | |
0em6 | Solve $k(k + 5)^2 + 12 = n^3$ for positive integers $k$ and $n$. | [
"Since $n$ and $k$ are integers, their difference will also be an integer. Since $k(k + 5)^2 = k^3 + 10k^2 + 25k$,\n$$\n\\begin{aligned}\n(k + 2)^3 &= k^3 + 6k^2 + 12k + 8 < k(k + 5)^2 + 12 \\\\\n(k + 4)^3 &= k^3 + 12k^2 + 48k + 64 > k(k + 5)^2 + 12.\n\\end{aligned}\n$$\nThe only way to get equality is if $n = k + ... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | k=5, n=8 | |
0lgg | Problem:
A positive integer $n$ does not divide $2^{a} 3^{b}+1$ for any positive integers $a$ and $b$. Prove that $n$ does not divide $2^{c}+3^{d}$ for any positive integers $c$ and $d$. | [
"Solution:\n\nAssume the contrary: $n$ divides $2^{c}+3^{d}$. Clearly $n$ is not divisible by $3$; therefore $n$ divides $3^{k}-1$ for some $k$. Choosing $s$ so that $k s>d$ we see that $n$ divides $3^{k s-d}\\left(2^{c}+3^{d}\\right)=2^{c} 3^{k s-d}+3^{k s}$. Then $n$ also divides $2^{c} 3^{k s-d}+1=2^{c} 3^{k s-d... | Zhautykov Olympiad | XIV International Zhautykov Olympiad in Mathematics | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
03yh | Determine all possible values of integer $k$ for which there exist positive integers $a$ and $b$ such that $\frac{b+1}{a} + \frac{a+1}{b} = k$. | [
"choose any $(a, b)$ such that $b$ is the smallest. Then the quadratic equation\n$$\nx^2 + (1 - kb)x + b^2 + b = 0\n$$\nhas an integral root $x = a$. Let $x = a'$ be the second root, it follows from $a + a' = kb - 1$ that $a' \\in \\mathbb{Z}$, and from\n$$\na \\cdot a' = b(b + 1)\n$$\nthat $a' > 0$. Hence, we have... | China | China Western Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 3, 4 | |
0fjn | Problem:
Tenemos un conjunto de $221$ números reales cuya suma es $110721$. Los disponemos formando una tabla rectangular de modo que todas las filas y la primera y última columnas son progresiones aritméticas de más de un elemento. Probar que la suma de los elementos de las cuatro esquinas vale $2004$. | [
"Solution:\n\nDenotaremos por $a_{i}^{j}$ al elemento de la fila $i$-ésima y columna $j$-ésima del rectángulo.\nPongamos $n$ para el número de filas, $m$ para el de columnas y $S$ para la suma de los $n \\times m$ elementos.\nCon notación matricial queda\n$$\nM=\\left(\\begin{array}{cccc}\na_{1}^{1} & a_{1}^{2} & \... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2004 | |
0gst | In a school having $101$ pupils any pupil has at least one friend among remaining pupils of the school. Show that for each $1 < n < 101$ one can choose a group of $n$ school pupils such that each pupil of the group has at least one friend in the group. | [
"First of all let us show that for each even $1 < n < 101$ we can choose a group of $n$ pupils satisfying conditions. When $n = 2$ any pair of friends can be chosen. Assume that for $n = 2l$ we have already constructed a group $A_{2l}$ of $n = 2l$ pupils satisfying the conditions. Let $S$ be the set of all pupils o... | Turkey | 30th Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0e60 | Let a point $E$ lie on the side $CD$ of a square $ABCD$. Let a point $F$ lie on the line $AB$, but not on the line segment $AB$, and let $|BF| = |DE|$. Prove that the lines $AC$ and $EF$ are perpendicular. | [
"Let $T$ be the intersection point of the lines $AC$ and $EF$. Because $|BF| = |DE|$, $|BC| = |DA|$ and $\\angle CBF = \\angle ADE = \\frac{\\pi}{2}$, the triangles $ADE$ and $CBF$ are congruent. We thus have $|AE| = |CF|$, and the quadrilateral $AFCE$ is an isosceles trapezoid. Hence $\\angle EFA = \\angle CAF$, w... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0j7o | Let $c_n$ be a sequence which is defined recursively as follows: $c_0 = 1$, $c_{2n+1} = c_n$ for $n \ge 0$, and $c_{2n} = c_n + c_{n-2^e}$ for $n > 0$ where $e$ is the maximal nonnegative integer such that $2^e$ divides $n$. Prove that
$$
\sum_{i=0}^{2^n-1} c_i = \frac{1}{n+2} \binom{2n+2}{n+1}.
$$ | [
"**Solution** (By Josh Nichols-Barrer). Observe that the right-hand side of the given expression is the $(n+1)^{\\text{th}}$ Catalan number, which we recall is the number of well-formed strings of $(n+1)$ pairs of parentheses. Given such a string, let its signature $k$ be the integer represented in binary by the $n... | United States | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof only | null | |
0h71 | A triangle $ABC$ is given. The circle $w$ with the centre at the point $Q$ touches the side $BC$ and is internally tangent to the circumscribed circle of $\triangle ABC$ in the point $A$. Let $M$ be the midpoint of the side $BC$, and $N$ be the middle of the arc $BAC$ of the circumscribed circle of $\triangle ABC$. On ... | [
"Denote by $\\Gamma$ the circumscribed circle of $\\triangle ABC$. Let $W$ be the middle of the smaller arc $BC$ of the circle $\\Gamma$, and $O$ be the centre of $\\Gamma$ (Fig. 46). Obviously, points $N$, $M$, $W$, $O$ lie on the median perpendicular to the segment $BC$. By Lemma of Archimedes points of contact o... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nin... | null | proof only | null | |
0hgd | Let $AL$ be a bisector of triangle $ABC$. The circle centered at $B$ with radius $BL$ meets the ray $AL$ at point $E$, and the circle centered at $C$ with radius $CL$ meets the ray $AL$ at point $D$ (points $E$ and $D$ are different from point $L$). Prove that $AL^2 = AE \cdot AD$.
(Mykola Moroz) | [
"Clearly, triangles $BEL$ and $CDL$ are isosceles, and angles $\\angle CLD$ and $\\angle BLE$ are vertical (fig. 3). Then $\\angle BEL = \\angle BLE = \\angle CLD = \\angle CDL$. Then $\\angle AEB = \\angle ALC$ as adjacent to equal angles. Also $\\angle CAL = \\angle BAL$, as $AL$ is a bisector.\n\nNote that trian... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
00rj | There are $2016$ customers who entered a shop on a particular day. Every customer entered the shop exactly once (i.e. the customer entered the shop, stayed there for some time and then left the shop without returning back).
Find the maximal $k$ such that the following holds:
There are $k$ customers such that either all... | [
"First we show that no larger $k$ can be achieved: We break the day at $45$ disjoint time intervals and assume that at each time interval there were exactly $45$ customers who stayed in the shop only during that time interval (except in the last interval in which there were only $36$ customers). We observe that the... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 45 | |
0e5f | Find all integers $a$, $b$, $c$ and $d$ such that
$$
2a^2 + 3b^2 = c^2 + 6d^2.
$$ | [
"One solution is straightforward: $a = b = c = d = 0$. We will prove that it is the only one. Suppose there is another solution $(a, b, c, d)$. We may suppose that the integers $a$, $b$, $c$ and $d$ are coprime, otherwise their greatest common divisor could be deleted from the equation (because not all numbers are ... | Slovenia | National Math Olympiad 2012 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (0, 0, 0, 0) | |
0kab | Problem:
Let $p > 2$ be a prime number. $\mathbb{F}_p[x]$ is defined as the set of all polynomials in $x$ with coefficients in $\mathbb{F}_p$ (the integers modulo $p$ with usual addition and subtraction), so that two polynomials are equal if and only if the coefficients of $x^k$ are equal in $\mathbb{F}_p$ for each non... | [
"Solution:\nAnswer: $4p(p-1)$\n\nFirst, notice that $(\\operatorname{deg} f)(\\operatorname{deg} g) = p^{2}$ and both polynomials are clearly nonconstant. Therefore there are three possibilities for the ordered pair $(\\operatorname{deg} f, \\operatorname{deg} g)$, which are $(1, p^{2})$, $(p^{2}, 1)$, and $(p, p)$... | United States | HMMT February 2019 Team Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof and answer | 4p(p-1) | |
02r0 | Problem:
Um número inteiro positivo esconde outro número quando, apagando alguns de seus algarismos, aparece o outro. Por exemplo, o número $123$ esconde os números $1$, $2$, $3$, $12$, $13$ e $23$, mas não esconde $32$, $123$ e $213$.
a) Qual é o maior número de três algarismos escondido por $47239$?
b) Qual é o me... | [
"Solution:\n\na) Para obter o maior número possível de três algarismos escondido por $47239$, devemos primeiro fazer com que esse número tenha o maior algarismo possível na casa das centenas. Para isso, devemos apagar o $4$ e deixar o $7$ na casa das centenas. Após isso buscamos o maior algarismo possível na casa d... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 739; b) 290029; c) 200914063 | |
0e7t | Problem:
Za realno število $a$ naj $[a]$ označuje največje celo število, ki ni večje od $a$. Poišči vsa cela števila $y$, za katera obstaja realno število $x$, da velja $\left[\frac{x+23}{8}\right]=[\sqrt{x}]=y$. | [
"Solution:\n\nNaj bo $y$ tako število. Potem velja $\\sqrt{x} \\geq [\\sqrt{x}] = y$. Ker je $\\sqrt{x} \\geq 0$, je tudi $y = [\\sqrt{x}] \\geq 0$, torej lahko neenakost kvadriramo, da dobimo $x \\geq y^{2}$. Poleg tega je $\\frac{x+23}{8} < \\left[\\frac{x+23}{8}\\right] + 1 = y + 1$ oziroma $x < 8y - 15$. Od tod... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4 | |
0j9n | Problem:
Knot is on an epic quest to save the land of Hyruler from the evil Gammadorf. To do this, he must collect the two pieces of the Lineforce, then go to the Temple of Lime. As shown on the figure, Knot starts on point $K$, and must travel to point $T$, where $O K=2$ and $O T=4$. However, he must first reach both... | [
"Solution:\n\nAnswer: $2 \\sqrt{5}$\n\nLet $l_{1}$ and $l_{2}$ be the lines as labeled in the above diagram. First, suppose Knot visits $l_{1}$ first, at point $P_{1}$, then $l_{2}$, at point $P_{2}$. Let $K^{\\prime}$ be the reflection of $K$ over $l_{1}$, and let $T^{\\prime}$ be the reflection of $T$ over $l_{2}... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2√5 | |
025s | Problem:
Mostre que $M=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$ é um número inteiro. | [
"Solution:\n\nDenotemos $a=\\sqrt[3]{\\sqrt{5}+2}$ e $b=\\sqrt[3]{\\sqrt{5}-2}$. Então\n$$\na^{3}-b^{3}=(\\sqrt{5}+2)-(\\sqrt{5}-2)=4\n$$\ne\n$$\nab=\\sqrt[3]{(\\sqrt{5}+2)(\\sqrt{5}-2)}=\\sqrt[3]{5-4}=\\sqrt[3]{1}=1\n$$\nde modo que $M=a-b$ satisfaz $M^{3}=(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)=4-3M$.\nAssim, $M^{3}+3M-4=... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0175 | Let $ABCD$ be a square and let $S$ be the point of intersection of its diagonals $AC$ and $BD$. Two circles $k$, $k'$ go through $A$, $C$ and $B$, $D$; respectively. Furthermore, $k$ and $k'$ intersect in exactly two different points $P$ and $Q$. Prove that $S$ lies on $PQ$. | [
"It is clear that $PQ$ is the radical axis of $k$ and $k'$. The power of $S$ with respect to $k$ is $-|AS| \\cdot |CS|$ and the power of $S$ with respect to $k'$ is $-|BS| \\cdot |DS|$. Because $ABCD$ is a square, these two numbers are clearly the same. Thus, $S$ has the same power with respect to $k$ and $k'$ and ... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof only | null | |
0486 | Let $A_1A_2A_3A_4$ be a convex quadrilateral that is **not** cyclic and whose opposite sides are **not** parallel. For $1 \le i \le 4$, let $M_i$ be the midpoint of $A_{i-1}A_{i+1}$. Let $B_i$ be a point on the tangent to the circumcircle of triangle $A_{i-1}A_iA_{i+1}$ at $A_i$, such that the reflection of $M_i$ over ... | [
"For any point $X$ on line $A_{i+2}B_i$, we have:\n$$\n\\frac{d_{X-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}} = \\frac{d_{X-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}}.\n$$\nFor any point $X$ on line $A_iB_i$, we have:\n$$\n\\frac{d_{X-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{X-A_iA_{i+1}}}{A_iA_{i+1}} = 0.\n$$\nTherefore, for point $B_i$:... | China | China-TST-2025A | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geo... | English | proof only | null | |
05jx | Problem:
Trouver le plus grand entier $n \geqslant 3$, vérifiant:
"pour tout entier $k \in \{2,3, \cdots, n\}$ si $k$ et $n$ sont premiers entre eux alors $k$ est un nombre premier." | [
"Solution:\n\nOn remarque d'abord que $n=30$ vérifie la propriété. En effet, si $k>1$ est premier avec $n$, alors il est premier avec $2,3,5$. Si de plus $k$ n'est pas premier, alors il admet une factorisation non triviale $k=lm$ avec $\\ell, m>1$. Comme $\\ell$, $m$ sont premiers avec $n$, ils sont premiers avec $... | France | Olympiades Françaises de Mathématiques - Test du mercredi 9 janvier | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 30 | |
09fj | Let $p$ be a prime satisfying $p \equiv 1 \pmod 4$. Show that there exist infinitely many positive integers $n$ such that $2^n + n^2$ is divisible by $p$. | [
"Write $p = 2s + 1$ with $s$ even. Then by Wilson's theorem we have\n$$\n(s!)^2 \\equiv (-1)^s (p-1)! \\equiv -1 \\pmod{p}. \\qquad (1)\n$$\nFor any positive integer $k$, let $n_k := (p-1)^k s!$. Then $2^{n_k} \\equiv (2^{(p-1)})^{(p-1)^{k-1}s!} \\equiv 1 \\pmod{p}$ by Fermat's little theorem. On the other hand, we... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0hhh | Given $n \ge 4$ positive numbers. Consider all $\frac{n(n-1)}{2}$ pairwise sums of these numbers.
Show that there exist two sums that differ by no more than a factor of $\sqrt[4]{2}$. | [
"Let the numbers be arranged in non-increasing order: $x_1 \\ge x_2 \\ge \\dots \\ge x_n$. Consider the sums $2x_1 \\ge x_1 + x_2 \\ge x_1 + x_3 \\ge \\dots \\ge x_1 + x_n > x_1$. Therefore, some two of the sums $x_1 + x_2, x_1 + x_3, \\dots, x_1 + x_n$ differ by no more than a factor of $\\sqrt[4]{2}$."
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
05jp | Problem:
A l'extérieur du triangle $A B C$ on construit les deux points $X$ et $Y$ vérifiant :
- le triangle $A X B$ est isocèle de base $[A B]$;
- le triangle $B Y C$ est isocèle de base $[B C]$;
$-\widehat{A X B}+\widehat{B Y C}=180^{\circ}$.
Soit $Z$ le milieu de $[A C]$. Montrer que les droites $(X Z)$ et $(Y Z)$ ... | [
"Solution:\n\nNotons $D$ et $E$ les milieux de $[B C]$ et $[A B]$.\n\n\n\nOn a\n- $(X E) \\perp (D Z)$ car $(D Z) // (A B)$\n- $(E Z) \\perp (D Y)$ car $(E Z) // (B C)$\n- $\\frac{X E}{D Z}=\\frac{X E}{E A}=\\frac{C D}{D Y}=\\frac{E Z}{D Y}$ (la deuxième égalité provient du fait que les tri... | France | OFM | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0as9 | Problem:
Evaluate $\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}$. | [
"Solution:\n$\\frac{1}{3}+\\frac{1}{15}+\\frac{1}{35}+\\frac{1}{63}+\\frac{1}{99}+\\frac{1}{143}+\\frac{1}{195} =$\n$$\n\\frac{1}{2}\\left[\\left(1-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{5}\\right)+\\left(\\frac{1}{5}-\\frac{1}{7}\\right)+\\cdots+\\frac{1}{13}-\\frac{1}{15}\\right]=\\frac{1}{2}\\left(1-... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 7/15 | |
06yz | Problem:
$ABCD$ is a convex quadrilateral. $P$, $Q$ are points on the sides $AD$, $BC$ respectively such that $AP / PD = BQ / QC = AB / CD$. Show that the angle between the lines $PQ$ and $AB$ equals the angle between the lines $PQ$ and $CD$. | [
"Solution:\n\n\n\nIf $AB$ is parallel to $CD$, then it is obvious that $PQ$ is parallel to both. So assume $AB$ and $CD$ meet at $O$. Take $O$ as the origin for vectors. Let $\\mathbf{e}$ be a unit vector in the direction $OA$ and $\\mathbf{f}$ a unit vector in the direction $OC$. Take the ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08xf | Find all the factors $k$ of $10^{2013} - 1$, which satisfy $1 \le k \le 100$. | [
"1, 3, 9, 27, 37, 67\nLet us use the following notations in the subsequent argument to obtain the solution to this problem:\n* For a pair of positive integers $(a, b)$, denote by $\\gcd(a, b)$ the greatest common divisor of $a$ and $b$.\n* For integers $a, b$ and $p$, we write $a \\equiv b \\pmod p$ to mean that $a... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 1, 3, 9, 27, 37, 67 | |
0dd3 | Let 300 students participate to the Olympiad. Between each 3 participants there is a pair that are not friends. Hamza enumerates participants in some order and denotes by $x_{i}$ the number of friends of $i$-th participant. It occurs that
$$
\left\{x_{1}, x_{2}, \ldots, x_{299}, x_{300}\right\}=\{1,2, \ldots, N-1, N\} ... | [
"Firstly, we shall prove that if $A, B$ are friend then the sum of friends of each one does not exceed $300$. Indeed,\nSuppose that $A$ has $a \\leq N$ friends and $B$ has $b \\leq N$ friends. Note that $A$ and $B$ cannot have any common friend; otherwise, take $C$ as a friend of $A, B$ and then the triple ($A, B, ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 200 | |
0hkk | Problem:
Two thousand and eleven positive integers are chosen, all different and less than or equal to $4020$. Prove that two of them have no common factors except $1$. | [
"Solution:\n\nSplit the numbers from $1$ to $4020$ into the $2010$ pairs $\\{1,2\\}, \\{3,4\\}, \\ldots, \\{4019,4020\\}$. Since $2011$ numbers are chosen but there are only $2010$ pairs, two numbers have to lie in the same pair. These numbers cannot have any common factors except $1$, because any divisor of both n... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0fta | Problem:
In Genf sind 16 Geheimagenten am Werk. Jeder Agent überwacht mindestens einen anderen Agenten, aber keine zwei Agenten überwachen sich gegenseitig. Nehme an, dass je 10 Agenten so nummeriert werden können, dass der erste den zweiten überwacht, der zweite den dritten usw. und der zehnte den ersten. Zeige, dass... | [
"Solution:\n\nJeder Agent $A$ überwacht mindestens 7 andere Agenten. Nehme an, das sei nicht der Fall, dann gibt es mindestens 9 Agenten, die $A$ nicht überwacht. Nach Voraussetzung lassen sich diese 9 Agenten zusammen mit $A$ so nummerieren, dass jeder den nächsten überwacht. Insbesondere überwacht $A$ mindestens ... | Switzerland | IMO - Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0kgz | A regular hexagon of side length $1$ is inscribed in a circle. Each minor arc of the circle determined by a side of the hexagon is reflected over that side. What is the area of the region bounded by these $6$ reflected arcs?
(A) $\frac{5\sqrt{3}}{2} - \pi$
(B) $3\sqrt{3} - \pi$
(C) $4\sqrt{3} - \frac{3\pi}{2}$
(D) $\pi... | [] | United States | 2021 AMC 10 B Fall | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | B | |
0coc | Given $n \ge 3$ pairwise coprime positive integers. It is known that while dividing the product of any $n-1$ of them by the remaining number, a remainder equals to $r$ (the remainder $r$ is the same for all $(n-1)$-tuples). Prove that $r \le n-2$.
Даны $n \ge 3$ попарно взаимно простых чисел. Известно, что при делении... | [
"Если $r = 0$, то утверждение задачи, очевидно, истинно. Пусть $r > 0$. Пусть $a_1, \\dots, a_n$ — данные числа; положим $P = a_1a_2 \\dots a_n$, $P_i = P/a_i$ при $i = 1, 2, \\dots, n$. Заметим, что $a_i > r$, ибо число $P_i$ даёт остаток $r$ при делении на $a_i$.\n\nРассмотрим число $S = P_1 + P_2 + \\dots + P_n ... | Russia | Final round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English; Russian | proof only | null | |
0ffd | Problem:
Un triángulo equilátero inscrito en una circunferencia de centro $O$ y radio igual a $4~\mathrm{cm}$, se gira un ángulo recto en torno a $O$. Hallar el área de la parte común al triángulo dado y al obtenido en ese giro. | [
"Solution:\n\n\n\nEl área pedida es la diferencia entre el área del triángulo dado $ABC$ y los tres triángulos grises de la figura.\nTenemos\n$$\n[ABC] = 3 \\frac{1}{2} 16 \\frac{\\sqrt{3}}{2} = 12 \\sqrt{3}, \\quad BC = 4 \\sqrt{3}\n$$\nPara calcular el área de uno de los triángulos grises... | Spain | Olimpiadas Matemáticas Españolas | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 12(3 - sqrt(3)) | |
0ki7 | A farmer's rectangular field is partitioned into a $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want... | [] | United States | AMC 10 A | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Linear Algebra > Matrices"
] | null | MCQ | C | |
0200 | Problem:
If $k$ is an integer, let $\mathrm{c}(k)$ denote the largest cube that is less than or equal to $k$. Find all positive integers $p$ for which the following sequence is bounded:
$$
a_{0}=p \quad \text{ and } \quad a_{n+1}=3 a_{n}-2 c\left(a_{n}\right) \quad \text{ for } n \geqslant 0
$$ | [
"Solution:\nSince $\\mathrm{c}\\left(a_{n}\\right) \\leqslant a_{n}$ for all $n \\in \\mathbb{N}$, $a_{n+1} \\geqslant a_{n}$ with equality if and only if $\\mathrm{c}\\left(a_{n}\\right)=a_{n}$. Hence the sequence is bounded if and only if it is eventually constant, which is if and only if $a_{n}$ is a perfect cub... | Benelux Mathematical Olympiad | THIRD BENELUX MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | All positive perfect cubes | |
0hth | Problem:
A set $S$ of irrational real numbers has the property that among any subset of five numbers in $S$, one can find two with irrational sum. How large can $|S|$ be? | [
"Solution:\n\nThe answer is $|S| \\leq 8$. An example is $S=\\{n \\pm \\sqrt{2} \\mid n=1,2,3,4\\}$ (and any of its subsets).\n\nIn general, construct a graph with vertex set $S$ in which we join two numbers with rational sum. We claim this graph is bipartite; indeed if $a_{1}+a_{2}, a_{2}+a_{3}, \\ldots, a_{n}+a_{... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 8 | |
0ibh | Problem:
The Fibonacci numbers are defined by $F_{1} = F_{2} = 1$, and $F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 3$. If the number
$$
\frac{F_{2003}}{F_{2002}} - \frac{F_{2004}}{F_{2003}}
$$
is written as a fraction in lowest terms, what is the numerator? | [
"Solution:\nBefore reducing, the numerator is $F_{2003}^{2} - F_{2002} F_{2004}$. We claim $F_{n}^{2} - F_{n-1} F_{n+1} = (-1)^{n+1}$, which will immediately imply that the answer is $1$ (no reducing required). This claim is straightforward to prove by induction on $n$: it holds for $n=2$, and if it holds for some ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | 1 | |
0g5k | 考慮所有形如 $f(x) = (x - a_1)(x - a_2)(x - a_3)\cdots(x - a_{100})$ 的整係數多項式, 其中 $a_1, a_2, \cdots, a_{100}$ 是任意實數。試求 $\{a_1\} + \{a_2\} + \cdots + \{a_{100}\}$ 可能的最大值。
註：定義 $\{x\} = x - [x]$，其中 $[x]$ 為不大於 $x$ 的最大整數。 | [
"因為對任意 $x$ 都有 $\\{x\\} < 1$，因此 $\\{a_1\\} + \\{a_2\\} + \\cdots + \\{a_{100}\\} < 100$；又由根與係數關係知 $-\\sum_{i=1}^{100} a_i$ 為 $f(x)$ 的 $x^{99}$ 項係數，必然是個整數，因此\n$$\n\\{a_1\\} + \\{a_2\\} + \\cdots + \\{a_{100}\\} \\le 99.\n$$\n以下證明存在 $f(x)$ 使得 $\\{a_1\\} + \\{a_2\\} + \\cdots + \\{a_{100}\\} = 99$。考慮函數\n\n$f(x) = x(x-2... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 99 | |
01ym | Let $a_1, a_2, \dots, a_n$ be the sequence of positive integers. For each number $\ell$ from $1$ to $n-1$ the following collection was found:
$$(\gcd(a_1, a_{1+\ell}), \gcd(a_2, a_{2+\ell}), \dots, \gcd(a_n, a_{n+\ell})),$$
where all indices are taken modulo $n$, i.e. if $s > n$ then $a_s = a_{s-n}$. It turned out that... | [
"Let us prove the following\n**Statement:** $n = k^2 - k + 1$ for some positive integer $k$.\n**Proof:** Let $d$ be the largest of the greatest common divisors found. Then the equality $\\gcd(a_i, a_{i+\\ell}) = d$ is equivalent to the fact that each of the numbers $a_i$ and $a_{i+\\ell}$ is a multiple of $d$. Let ... | Belarus | Belarus2022 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a) yes; b) no | |
0fij | Problem:
Una caja contiene 900 tarjetas numeradas del 100 al 999. Se sacan al azar (sin reposición) tarjetas de la caja y se anota la suma de los dígitos de cada tarjeta extraída. ¿Cuál es la menor cantidad de tarjetas que se deben sacar, para garantizar que al menos tres de esas sumas sean iguales? | [
"Solution:\n\nHay 27 posibles resultados para la suma de dígitos (de 1 a 27). Las sumas 1 y 27 sólo se pueden obtener de un modo (100 y 999).\n\nEn el caso más desfavorable, al sacar 52 ($27 + 25$) tarjetas, todas repetirán suma dos veces y en la siguiente (extracción 53) una de ellas aparecerá por tercera vez.\n\n... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 53 | |
01xy | The bisectors of angles $\angle A$ and $\angle C$ of a convex quadrilateral $ABCD$ meet at the point $E$, and the bisectors of angles $\angle B$ and $\angle D$ meet at the point $F$ ($E$ and $F$ lie in the interior of $ABCD$). The point $M$ is the midpoint of the segment $EF$. The points $H_1, H_2, H_3$ and $H_4$ are t... | [
"Since $E$ lies on the bisector of $\\angle BAD$ ($\\angle BCD$), the point $E$ is equidistant from the sides $AB$ and $AD$ (respectively, $BC$ and $CD$). A similar statement holds for the point $F$. Denote the distances from the point $E$ to the lines $AB$ and $BC$ by $x_1$ and $y_1$, respectively, and the distanc... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0b1o | Problem:
An infinite geometric series has sum $2020$. If the first term, the third term, and the fourth term form an arithmetic sequence, find the first term. | [
"Solution:\n\nLet $a$ be the first term and $r$ be the common ratio. Thus, $\\frac{a}{1-r} = 2020$, or $a = 2020(1 - r)$.\n\nWe also have $a r^{2} - a = a r^{3} - a r^{2}$. Since the sum of the geometric series is nonzero, $a \\neq 0$, and so we have $r^{2} - 1 = r^{3} - r^{2}$, or $r^{3} - 2 r^{2} + 1 = 0$.\n\nSin... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1010(1 + sqrt(5)) | |
0g5v | 對於整數數對 $(a,b)$, 令 $P(x) = ax^3 + bx$。若對於某個正整數 $m$, 以下命題成立：
若整數 $x, y$ 滿足 $m$ 能整除 $P(x) - P(y)$，則 $m$ 能整除 $x - y$。
則稱數對 $(a,b)$ 為「$m$-充分」。若存在無限多個正整數 $k$ 使得 $(a,b)$ 為「$k$-充分」，則稱數對 $(a,b)$ 為「非常充分」。
試問：是否存在數對 $(a,b)$ 使得 $(a,b)$ 是「1110-充分」，但不是「非常充分」？ | [
"答案是否定的！我們以下證明若 $(a,b)$ 是「1110-充分」，則對於任意 $k = 37^r$，$(a,b)$ 均為 $k$-充分。\n\n(1) 若 $(a,b)$ 是 1110-充分，則 $(a,b)$ 是 37-充分。\n假設 $(a,b)$ 不是 37-充分，則存在 $x, y$ 使得 $37 \\mid P(x) - P(y)$ 但 $x - y$ 不是 37 的倍數。由中國剩餘定理，存在整數 $x', y'$ 使得 $x'$ 和 $y'$ 都是 30 的倍數，且 $37 \\mid x' - x$，$37 \\mid y' - y$。如此一來有 $30 \\mid P(x') - P(y')$ 且 $37... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No | |
0eeu | Problem:
a) Nariši graf funkcije $f$ s predpisom
$$
f(x)=\begin{cases}
2x+1 & \text{za } x>1 \\
-x+5 & \text{za } x \leq 1
\end{cases}
$$
b) Naj bo $g(x)=2x+1$. Zapiši vse vrednosti spremenljivke $x$, za katere velja enakost $g^{-1}(x)=g\left(x^{-1}\right)$?
 | [
"Solution:\n\na)\nNarisana premica z enačbo $y=2x+1$ oziroma del te premice.\nNarisana premica z enačbo $y=-x+5$ oziroma del te premice.\nPravilno označen graf funkcije $f$ v $x=1$.\n\nb)\nPostopek za izračun predpisa inverzne funkcije.\nZapis $g^{-1}(x)=\\frac{x-1}{2}$.\nZapis $g\\left(x^{-1}\\right)=2x^{-1}+1$.\n... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | x = 4 or x = -1 | |
09ox | Let us recall that the numbers $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{6}$ are irrational.
(1)
For all integers $s, t$, and $r$, if $s + \sqrt{2} t + \sqrt{3} r = 0$, then prove that $s = t = r = 0$.
(2)
For all integers $a, b, c$, and $d$, if the ratio $\frac{a + \sqrt{2} b + \sqrt{3} c}{b + \sqrt{2} c + \sqrt{3} d}$ is ... | [] | Mongolia | MMO2025 Round 3 | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Intermediate Algebra > Other"
] | English | proof only | null | |
0gq2 | Find all pairs $(m, n)$ of positive integers satisfying $2^n + n = m!$. | [
"$m = 1$ gives no solution, therefore $m!$ is even and $n$ should also be even. Let $n = 2^t \\cdot s$ ($t$ and $s$ are positive integers, $t \\ge 1$ and $s$ is odd). $t = 1$ readily leads to $n = 2$ and $m = 3$.\n\nNow let $t \\ge 2$. Then $m! = 2^n + n = 2^{2t \\cdot s} + 2^t \\cdot s \\ge 2^{2t} + 2^t$. By induc... | Turkey | 21st Turkish Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | (3, 2) | |
0fju | Problem:
¿Es posible colorear los puntos del plano cartesiano $Oxy$ de coordenadas enteras con tres colores, de tal modo que cada color aparezca infinitas veces en infinitas rectas paralelas al eje $Ox$ y tres puntos cualesquiera, cada uno de distinto color, no estén alineados? Justificar la contestación. | [
"Solution:\n\nProbemos que tal coloración es posible. Pintemos el punto $(x, y)$ de rojo si $x+y$ es par, de blanco si $x$ es impar e $y$ es par y de azul si $x$ es par e $y$ es impar.\n\nClaramente se satisface la condición de que cada color aparezca infinitas veces en infinitas rectas paralelas al eje $OX$.\n\nSu... | Spain | Spanish National Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Modular Arithmetic",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | Yes | |
0bxf | Let $n$ be a positive integer. For each of the numbers $1, 2, \dots, n$ we compute the difference between the number of its odd positive divisors and its even positive divisors. Prove that the sum of these differences is at least $0$ and at most $n$.
Kürschák Competition, 1999 | [
"We count how many times a number $d$ contributes to the sum of the differences. It appears $\\left\\lfloor \\frac{n}{d} \\right\\rfloor$ times; if $d$ is odd, then this term is to be added, while if $d$ is even this term is to be subtracted. Thus, the desired sum is\n$$\n\\sum_{k=1}^{n} (-1)^{k+1} \\left\\lfloor \... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0iit | Problem:
Find
$$
\frac{2^{2}}{2^{2}-1} \cdot \frac{3^{2}}{3^{2}-1} \cdot \frac{4^{2}}{4^{2}-1} \cdots \frac{2006^{2}}{2006^{2}-1}
$$ | [
"Solution:\n\n$$\n\\prod_{k=2}^{2006} \\frac{k^{2}}{k^{2}-1} = \\prod_{k=2}^{2006} \\frac{k^{2}}{(k-1)(k+1)} = \\prod_{k=2}^{2006} \\frac{k}{k-1} \\prod_{k=2}^{2006} \\frac{k}{k+1} = \\frac{2006}{1} \\cdot \\frac{2}{2007} = \\frac{4012}{2007}\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 4012/2007 | |
04oe | The incircle of the triangle $ABC$ is centred at $I$, and touches the sides $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ in points $D, E$ and $F$, respectively. Let $k$ be a circle centred at $A$ passing through the point $E$. Let $K$ be the second intersection of the line $DE$ with $k$. The line parallel to $D... | [
"We easily see that $\\angle AKE = \\angle KEA = \\angle DEC = \\angle CDE$. Thus $|AK| = |AE|$ and $AK \\parallel CD$.\n\n\n\nLet $L'$ be the intersection of the lines $CP$ and $AK$. Since the triangles $PBC$ and $PAL'$ are similar, by trigonometric calculus we get $|BP| : |AP| = |BC| : |A... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > A... | English | proof only | null | |
00jo | We are given an equilateral triangle $ABC$ with sides of length $2$. We consider all equilateral triangles $PQR$ with sides of length $1$ satisfying the following properties:
* $P$ lies on the side $AB$,
* $Q$ lies on the side $AC$ and
* $R$ lies in the interior or on the edge of the triangle $ABC$,
Describe the set of... | [
"Let $P$ and $Q$ be given fulfilling the conditions of the problem. Considering the circumcircle $k$ of $APQ$, we note that the centroid $S$ of $APQ$ must lie on $k$, since both $\\angle PAQ = 60^\\circ$ and $\\angle PSQ = 120^\\circ$ hold. Since $|SQ| = |SP|$, the arcs $SQ$ and $SP$ are of equal length, and we the... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | The middle third of the angle bisector from the chosen vertex of the large triangle (the altitude), i.e., the segment between its one-third and two-thirds points. | |
075l | Let $P(z) = a_n z^n + a_{n-1} z^{n-1} + \dots + a_m z^m$ be polynomial with complex coefficients such that $a_m \neq 0$, $a_n \neq 0$ and $n > m$. Prove that
$$
\max_{|z|=1}\{|P(z)|\} \ge \sqrt{\sum_{k=m}^{n} |a_k|^2 + 2|a_m a_n|}.
$$ | [
"Note that we may assume $m = 0$, since $P(z) = z^m Q(z)$ for some polynomial $Q(z)$ where $|P(z)| = |Q(z)|$ on the unit circle. Thus we may write\n$$\nP(z) = \\sum_{k=0}^{n} a_k z^k, \\quad a_n \\neq 0, a_0 \\neq 0,\n$$\nand we have to prove that\n$$\n\\max_{|z|=1}\\{|P(z)|\\} \\ge \\sqrt{\\sum_{k=0}^{n} |a_k|^2 +... | India | Indija TS 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
00dg | In Eventown all authentic coins weigh an even amount of grams and all fake coins weigh an odd amount of grams.
There are $2022$ coins and it is given that exactly $2$ of them are fake. We have an electronic scale which only shows if the total weight of the objects put on it is even or odd.
Find the least value of $k$ s... | [
"The answer is $k = 21$. First we show a strategy that allows us to identify the two fake coins using the scale $21$ times.\nWe label the coins with the numbers $1, 2, 3, \\dots, 2022$ written in binary. Since $2^{11} = 2048 > 2022$, every coin corresponds to an $11$-digit binary number.\nThe first $11$ weighings a... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 21 | |
04pn | Marko places coins on some of the unit squares of a $3 \times 3$ board, and then notes how many coins are in each row and each column. At least how many coins does Marko have to place on the board if he wants all six of those numbers to be mutually distinct? (United Kingdom) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 8 | |
0bre | A ring $(A, +, \cdot)$ has property (P) if $A$ is finite and the multiplicative group of units is isomorphic to a non-trivial subgroup of the additive group $(A, +)$. Show that:
(a) the number of elements of a ring having property (P) is even;
(b) there are $n$-element rings having property (P) for infinitely many posi... | [
"(a) Let $A$ be a ring having property (P), let $m = |U(A)|$, and notice that $(-1)^m = 1$. If $m$ is odd, then $-1 = 1$, so $1$ has order $2$ in the additive group $(A, +)$, and consequently $|A|$ is even. If $m$ is even, so must be $|A|$, since the former divides the latter.\n\n(b) Let $m$ be a positive integer, ... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
08un | It is known that there are exactly 2958 pairs $(m, n)$ of positive integers not exceeding 100 for which the inequalities $m < \sqrt{2n} < 2m$ are valid. How many pairs $(m, n)$ of positive integers not exceeding 100 are there for which $\sqrt{2n} < m$ is satisfied? | [
"Since $\\sqrt{2}$ is irrational, neither $m = \\sqrt{2} n$ nor $2m = \\sqrt{2} n$ can occur for a pair $(m, n)$ of positive integers. Therefore, the pairs $(m, n)$ of positive integers not exceeding 100 split into the following three types depending on the values of $\\sqrt{2} n$, $m$ and $2m$:\n\n* Type (1): sati... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3521 | |
0lfx | Problem:
Let $U=\{1,2,3, \ldots, 2014\}$. For all $a, b, c \in \mathbb{N}$ let $f(a, b, c)$ be the number of ordered sextuplets $(X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3})$ of subsets of $U$, satisfying the following conditions
(i) $Y_{1} \subseteq X_{1} \subseteq U$ and $|X_{1}|=a$
(ii) $Y_{2} \subseteq X_{2} \subs... | [
"Solution:\n\nIn order to avoid any confusion between the letters $a, b, c$ and their numerical values (as cardinalities of sets), the most convenient way will be to denote by $|\\ell|$ the cardinality symbolized by any such letter $\\ell$. We can now consider the true 3-element set $\\{a, b, c\\}$, and the canonic... | Zhautykov Olympiad | International Zhautykov Olympiad in Sciences | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0k89 | Problem:
A polynomial $P$ with integer coefficients is called tricky if it has $4$ as a root.
A polynomial is called $k$-tiny if it has degree at most $7$ and integer coefficients between $-k$ and $k$, inclusive.
A polynomial is called nearly tricky if it is the sum of a tricky polynomial and a $1$-tiny polynomial.
Le... | [
"Solution:\n\nA tricky $7$-tiny polynomial takes the form\n$$\n\\left(c_{6} x^{6}+\\ldots+c_{1} x+c_{0}\\right)(x-4)\n$$\nFor each fixed value of $k$, $c_{k}-4 c_{k+1}$ should lie in $[-7,7]$, so if we fix $c_{k}$, there are around $15 / 4$ ways of choosing $c_{k+1}$. Therefore if we pick $c_{0}, \\ldots, c_{6}$ in... | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Algorithms"
] | null | final answer only | 64912347 | |
07kx | Find all positive integers $n$ for which $n^8 + n + 1$ is a prime number. | [
"Let $f(x) = x^8 + x + 1$. Numerical values get large very quickly:\n$$\n\\begin{aligned}\nf(1) &= 3 \\\\\nf(2) &= 259 = 7 \\times 37 \\\\\nf(3) &= 6565 = 5 \\times 13 \\times 101 \\\\\nf(4) &= 65541 = 3 \\times 7 \\times 3121.\n\\end{aligned}\n$$\nThese numbers may suggest that $f(n)$ will be a prime number only i... | Ireland | Irska | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | n = 1 | |
02lz | Let $H$ be the hyperboloid $3x^2 + 3y^2 - z^2 - 1 = 0$.
(a) Prove that every point $(x, y, z) \in H$ belong to exactly two lines contained in $H$.
(b) Prove that all lines contained in $H$ form the same angle with the plane $z = 0$, and find that angle. | [
"The solution is based on the following\n**Theorem.** Let $P = (x_0, y_0, 0)$ be a point in the xy-plane and $(a, b, 1)$ a vector perpendicular to the vector $v = (x_0, y_0, 0)$, that is, such that $a x_0 + b y_0 = 0$.\nThen the set of points obtained by rotating the line $P + t \\cdot v$, $t \\in \\mathbb{R}$, aro... | Brazil | XXXI Brazilian Math Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Linear Algebra > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 60 degrees | |
00qt | For a polynomial $P \in \mathbb{R}[x]$, let $f(P) = n$ if $n$ is the smallest positive integer such that
$$
(\forall x \in \mathbb{R}) \underbrace{(P(P(\dots P(x))\dots))}_{n} > 0,
$$
and $f(P) = 0$ if such an integer $n$ does not exist. Does there exist a polynomial $P \in \mathbb{R}[x]$ of degree $2014^{2015}$ such t... | [
"The answer is that it does exist such a polynomial. Actually we shall prove a more general result: Let $s$ be an even integer and $t > 1$ be an arbitrary integer. Then for some constant $c > 0$ for the polynomial $P(x) = (x+1)^s + c-1$ (which is of degree $s$) we have $f(P) = t$. Indeed:\nThe polynomial $P$ is str... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | Yes | |
0av7 | Problem:
Find the smallest number $k$ such that for all real numbers $x$, $y$ and $z$
$$
\left(x^{2}+y^{2}+z^{2}\right)^{2} \leq k\left(x^{4}+y^{4}+z^{4}\right)
$$ | [
"Solution:\nNote that\n$$\n\\left(x^{2}+y^{2}+z^{2}\\right)^{2}=x^{4}+y^{4}+z^{4}+2 x^{2} y^{2}+2 x^{2} z^{2}+2 y^{2} z^{2}\n$$\nUsing the AM-GM Inequality, we find that\n$$\n2 x^{2} y^{2}+2 x^{2} z^{2}+2 y^{2} z^{2} \\leq 2\\left[\\left(x^{4}+y^{4}\\right) / 2\\right]+2\\left[\\left(x^{4}+z^{4}\\right) / 2\\right]... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3 | |
05jy | Problem:
Soit $ABCDEF$ un hexagone régulier et $M \in [A, C], N \in [C, E]$. On suppose que $\frac{AM}{AC}$ et $\frac{CN}{CE}$ sont égaux à un nombre $r > 0$, et que $B, M, N$ sont colinéaires. Déterminer la valeur de $r$.
 | [
"Solution:\n\nOn peut supposer que l'hexagone est inscrit dans un cercle de rayon $1$. On se place dans un repère tel que les coordonnées de $A, B, C$ sont respectivement $A = (0, 1)$, $B = \\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$ et $C = \\left(\\frac{\\sqrt{3}}{2}, -\\frac{1}{2}\\right)$.\n\nComme $CE =... | France | null | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 1/sqrt(3) | |
0ekn | Problem:
Katera tangenta na parabolo $z$ enačbo $y = x^{2} + x + 9$ je vzporedna premici $z$ enačbo $-4x + 2y - 5 = 0$?
(A) $4x - 8y + 37 = 0$
(B) $4x - 8y - 37 = 0$
(C) $8x - 4y - 35 = 0$
(D) $-4x - 8y + 37 = 0$
(E) $8x - 4y + 35 = 0$ | [
"Solution:\nZapišemo enačbo $x^{2} + x + 9 = 2x + n$ in jo preoblikujemo do oblike $x^{2} - x + 9 - n = 0$. Potem upoštevamo pogoj, da ima kvadratna enačba eno dvojno realno rešitev, če je vrednost diskriminante kvadratne enačbe enaka $0$. V ta pogoj vstavimo vrednosti parametrov in dobimo enačbo $1 - 4(9 - n) = 0$... | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | E | |
0f1d | Problem:
Let $p(x)$ be a polynomial with integer coefficients. Let $f(n)$ be the sum of the (decimal) digits in the value $p(n)$. Show that $f(n)$ takes some value $m$ infinitely many times. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Other"
] | null | proof only | null | |
0eys | Problem:
Given point $O$ inside the acute-angled triangle $ABC$, and point $O'$ inside the acute-angled triangle $A'B'C'$. $D$, $E$, $F$ are the feet of the perpendiculars from $O$ to $BC$, $CA$, $AB$ respectively, and $D'$, $E'$, $F'$ are the feet of the perpendiculars from $O'$ to $B'C'$, $C'A'$, $A'B'$ respectively.... | [
"Solution:\n\nLet $\\Gamma$ be the circumcircle of $DEF$. Let $OD$, $OE$, $OF$ meet it again at $A''$, $B''$, $C''$ respectively. Then the figure $O'A'B'C'$ must be similar to $OA''B''C''$. So to prove that $OD$ is parallel to $O'A'$, we have to prove that $AO$ is perpendicular to $B''C''$.... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0h4a | In a triangle $ABC$, $\angle C = 90^\circ$, $AC < BC$. Point $K$ inside $BC$ is such that $CK = CA$, and point $D$ inside $CK$ is such that $\angle DAK = \angle BAK$. $F$ and $P$ are, respectively, the feet of perpendiculars from $D$ to $AB$ and from $A$ to $FK$. Prove that $CP = \frac{1}{2}(AF + FD + DA)$. | [
"Нехай пряма $DF$ перетинає прямі $AC$ і $AP$ в точках $N$ і $M$ відповідно (нескладно довести, що такі точки перетину існують). Нехай $\\angle BAC = \\alpha$, $\\alpha > 45^\\circ$, $\\angle BAK = \\angle KAD = \\delta$. Тоді $\\angle FDB = \\alpha$, $\\angle NDC = \\alpha$. Далі, $\\alpha - \\delta = 45^\\circ$, ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0bmb | Let $ABC$ be an acute triangle with $AB \ne AC$ and $H$ its orthocenter. Consider a point $D$ on the side $BC$. The circumcircles of triangles $ABD$ and $ACD$ meet again $AC$ and $AB$ in $E$ and $F$, respectively. Lines $BE$ and $CF$ meet in point $P$. Prove that $HP$ is parallel to $BC$ if and only if the line $AD$ co... | [
"We only show the proof in the case when $E \\in (AC)$, $F \\in (AB)$, the other cases being similar (the diagram on the left shows such a case).\nFrom $\\angle PBC + \\angle PCB = \\angle EAD + \\angle FAD = \\angle BAC$ follows $\\angle BPC = 180^\\circ - \\angle BAC = \\angle BHC$, hence $B, C, H, P$ are concycl... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN 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 > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordin... | null | proof only | null | |
0jca | Problem:
Let $S$ be the set of the points $(x_{1}, x_{2}, \ldots, x_{2012})$ in 2012-dimensional space such that $|x_{1}| + |x_{2}| + \cdots + |x_{2012}| \leq 1$. Let $T$ be the set of points in 2012-dimensional space such that $\max_{i=1}^{2012} |x_{i}| = 2$. Let $p$ be a randomly chosen point on $T$. What is the pro... | [
"Solution:\n\nAnswer: $\\frac{1}{2^{2011}}$\n\nNote that $T$ is a hypercube in 2012-dimensional space, containing the rotated hyperoctahedron $S$. Let $v$ be a particular vertex of $S$, and we will consider the set of points $x$ on $T$ such that $v$ is the closest point to $x$ in $S$. Let $w$ be another point of $S... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 1/2^{2011} | |
0idz | Problem:
Urn $A$ contains $4$ white balls and $2$ red balls. Urn $B$ contains $3$ red balls and $3$ black balls. An urn is randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability t... | [
"Solution:\n\nThis is a case of conditional probability; the answer is the probability that the first ball is red and the second ball is black, divided by the probability that the second ball is black.\n\nFirst, we compute the numerator. If the first ball is drawn from Urn $A$, we have a probability of $2/6$ of get... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 7/15 | |
0duo | Problem:
Oče želi razdeliti trem sinovom 14560 SIT tako, da vsak naslednji sin dobi $20 \%$ večji znesek kot njegov mlajši brat. Koliko dobi vsak sin? Zapiši odgovor. | [
"Solution:\n\nUgotovitev: sinovi dobijo $x$, $1{,}2x$, $1{,}44x$.\n\n$\\begin{align*}\nx + 1{,}2x + 1{,}44x &= 14560 \\\\\n3{,}64x &= 14560 \\\\\nx &= \\frac{14560}{3{,}64} = 4000 \\text{ SIT}\n\\end{align*}$\n\nDrugi sin dobi $1{,}2 \\times 4000 = 4800$ SIT.\n\nTretji sin dobi $1{,}44 \\times 4000 = 5760$ SIT.\n\n... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 4000 SIT, 4800 SIT, 5760 SIT | |
0en3 | The acute-angled triangle $ABC$ has circumcentre $O$ and orthocentre $H$. The perpendicular bisector of $BH$ meets $AB$ at $Q$, the perpendicular bisector of $CH$ meets $AC$ at $P$ and these two bisectors meet at $R$.
a. Prove that the circumcircles of $AOP$, $POR$, $ROQ$ and $QOA$ have equal radii. When are they equa... | [
"a. Note that $R$ is the circumcentre of triangle $BHC$, hence $OR$ bisects $BC$. Also, $\\angle BRC = 2\\angle(180^\\circ - \\angle BHC) = 2\\angle QRP = 2\\angle A = \\angle BOC$, so it follows that $R$ is the reflection of $O$ across the line $BC$.\n\nNext, since $QR$ is perpendicular to $BH$, which is perpendic... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscella... | null | proof and answer | Part (a): The circumcircles of AOP, POR, ROQ, and QOA all have common radius R/(2 sin A), where R is the circumradius of triangle ABC and A is angle BAC. They equal R if and only if angle BAC = 30°. Part (b): If PQ passes through O, then PQ also passes through H; in this case angle BAC = 60°. | |
01rw | Find all pairs $(n; m)$ of positive integers $n$ and $m$ satisfying the equality
$$
n! + 505 = m^2.
$$ | [
"Answer: $(4; 23)$, $(5; 25)$, $(6; 35)$.\nIt is easy to see that the numbers $1! + 505 = 506$, $2! + 505 = 507$, $3! + 505 = 511$ are not perfect squares. Further, $4! + 505 = 529 = 23^2$, $5! + 505 = 625 = 25^2$, $6! + 505 = 1225 = 35^2$. So we have three pairs $(4; 23)$, $(5; 25)$, $(6; 35)$ satisfying the probl... | Belarus | FINAL ROUND | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (4, 23), (5, 25), (6, 35) | |
0ibk | Let $n$ be a positive integer. Consider sequences $a_0, a_1, \dots, a_n$ such that $a_i \in \{1, 2, \dots, n\}$ for each $i$ and $a_n = a_0$.
a. Call such a sequence *good* if for all $i = 1, 2, \dots, n$, $a_i - a_{i-1} \not\equiv i \pmod n$. Suppose that $n$ is odd. Find the number of good sequences.
b. Call such a... | [
"**First Solution:** The answer is $(n-1)^n - (n-1)$ for part (a) and $(n-1)((n-2)^{n-1} - 1)$ for part (b).\n\na. Observe that the number of good sequences is clearly the same for any choice of $a_0$. For fixed $a_0$, call the condition $a_i - a_{i-1} \\not\\equiv i \\pmod n$ *condition (i)*. Now let $S_i$ be the ... | United States | USA IMO | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers... | null | proof and answer | a: (n−1)^n − (n−1); b: (n−1)((n−2)^{n−1} − 1) | |
05ay | Juku and Miku are playing the following game. In the beginning, there is a positive integer on the board. Each turn, a player subtracts from the number on the board a non-zero digit that appears in his or his opponent's ID code, and replaces the number on the board with the result. Players take turns, Juku starts. The ... | [
"Let $a, a+1, \\dots, a+9$ be 10 arbitrary consecutive positive integers. If there exists a number $n$ among $a, a+1, \\dots, a+8$ for which Juku has a winning strategy, then we are done. We will now assume that if any of the numbers $a, a+1, \\dots, a+8$ is on the board, then the active player loses if his opponen... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
05q1 | Problem:
Soit $ABC$ un triangle d'orthocentre $H$. Soient $\left(d_{1}\right)$ et $\left(d_{2}\right)$ deux droites perpendiculaires se coupant en $H$. Soit $A_{1}$ (respectivement $B_{1}, C_{1}$) l'intersection de $\left(d_{1}\right)$ avec $(BC)$ (respectivement $(CA), (AB)$). Soit $A_{2}$ (respectivement $B_{2}, C_{... | [
"Solution:\n\nNous donnons une preuve avec des similitudes directes. Il existe d'autres preuves, un peu plus courtes, qui utilisent des nombres complexes.\n\nRappelons tout d'abord le résultat suivant :\n\n**Lemme 2.** Soit $ABC$ un triangle, $\\Gamma$ son cercle circonscrit et $H$ son orthocentre. Alors les symétr... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > An... | null | proof only | null |
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