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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kdb | Problem:
Let $ABC$ be a triangle inscribed in a circle $\omega$ and $\ell$ be the tangent to $\omega$ at $A$. The line through $B$ parallel to $AC$ meets $\ell$ at $P$, and the line through $C$ parallel to $AB$ meets $\ell$ at $Q$. The circumcircles of $ABP$ and $ACQ$ meet at $S \neq A$. Show that $AS$ bisects $BC$. | [
"Solution:\nIn directed angles, we have\n$$\n\\measuredangle CBP = \\measuredangle BCA = \\measuredangle BAP\n$$\nso $BC$ is tangent to the circumcircle of $ABP$. Likewise, $BC$ is tangent to the circumcircle of $ACQ$. Let $M$ be the midpoint of $BC$. Then $M$ has equal power $MB^{2} = MC^{2}$ with respect to the c... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03cf | Let $M$ be a set of $2017$ positive integers. For every non-empty $A \subset M$ we define
$$
f(A) = \{x \in M : x \text{ is divisible by odd number of elements of } A\}.
$$
Find the minimum number of colors such that it is possible to paint all nonempty subset of $M$ in such a way that whenever $A \neq f(A)$, the sets ... | [
"We first prove that the function $f$ is injective, i.e. $A \\neq B$ implies $f(A) \\neq f(B)$. Let $a$ be the smallest number which belongs to exactly one of the sets $A$ and $B$. We can assume, $a \\in A$, $a \\notin B$. Let $C = \\{b_1, b_2, \\dots, b_m\\}$ be the set (possibly empty) of the numbers from $B$ whi... | Bulgaria | BULGARIAN NATIONAL MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 2 | |
05gg | Problem:
Trouver tous les triplets d'entiers strictement positifs $(a, b, c)$ tels que $6^{a} = 1 + 2^{b} + 3^{c}$. | [
"Solution:\n\nRemarquons tout d'abord que $3$ divise $2^{b} + 1$ donc $b$ est impair (car $2^{b} + 1 \\equiv (-1)^{b} + 1 \\pmod{3}$).\n\n$\\triangleright$ Si $b = 1$, l'équation se réécrit $1 + 3^{c-1} = 2 \\cdot 6^{a-1}$ en divisant par $3$.\n\n- Si $a > 1$, alors $3$ divise $1 + 3^{c-1}$ ce qui est impossible.\n... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES, ENVOI No. 3 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(1,1,1), (2,3,3), (2,5,1)] | |
09o1 | Let triangle $ABC$ be acute-angled and inscribed in a circle $\omega$. Let $M$ be the midpoint of arc $BC$ that does not contain point $A$. Let point $D$ be chosen on segment $AM$ such that $D$ lies inside triangle $ABC$. Line $BD$ intersects side $AC$ at point $N$, and intersects the circle $\omega$ again at point $P$... | [] | Mongolia | MMO2025 Round 3 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Ge... | English | proof only | null | |
030k | Problem:
Em um tabuleiro $5 \times 5$, cada quadradinho possui uma peça em seu centro. O único movimento permitido para uma dessas peças é se deslocar para um quadradinho que compartilhe exatamente um vértice com o quadradinho em que ela está, como indicado na figura abaixo. Tanto é possível que várias peças ocupem um... | [
"Solution:\n\na) A figura a seguir mostra um conjunto de movimentos em que apenas 5 quadradinhos ficam vazios.\n\n\nb) Pinte as colunas do tabuleiro de forma alternada de preto e branco. Assim existem 15 quadradinhos pretos e 10 brancos. Veja que após um movimento de uma peça, ela fica em u... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 5 | |
0el9 | Problem:
Trapez $ABCD$ ima osnovnici dolgi $|AB| = 120\ \mathrm{mm}$ in $|CD| = 30\ \mathrm{mm}$ in prava kot pri ogliščih $B$ in $C$. Diagonali trapeza se sekata v točki $E$. Koliko milimetrov je točka $E$ oddaljena od stranice $BC$?
(A) 21
(B) 22
(C) 23
(D) 24
(E) 25 | [
"Solution:\n\nNaj bo $F$ pravokotna projekcija točke $E$ na stranico $BC$. Trikotnika $ABE$ in $CDE$ sta podobna. Razmerje med njunima stranicama nasproti oglišča $E$ je enako $|AB| : |CD| = 120 : 30 = 4 : 1$, zato je tudi razmerje med njunima višinama na ti dve stranici enako $|FB| : |FC| = 4 : 1$. Tudi trikotnika... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | D | |
0f4c | Problem:
Write down a row of arbitrary integers (repetitions allowed). Now construct a second row as follows. Suppose the integer $n$ is in column $k$ in the first row. In column $k$ in the second row write down the number of occurrences of $n$ in row 1 in columns 1 to $k$ inclusive. Similarly, construct a third row u... | [] | Soviet Union | 15th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
05k8 | Problem:
Soient $P$ et $Q$ deux polynômes à coefficients réels, de degrés $n \geqslant 0$. On suppose que le coefficient de $x^{n}$ de chacun de ces deux polynômes est égal à $1$ et que, pour tout réel $x$, on a $P(P(x)) = Q(Q(x))$.
Prouver que $P = Q$. | [
"Solution:\n\nLe résultat est évident si $n = 0$. Dans ce qui suit, on suppose donc que $n \\geqslant 1$.\n\nPar l'absurde : supposons que le polynôme $R = P - Q$ ne soit pas le polynôme nul.\n\nSoit $k$ le degré de $R$.\n\nPuisque $P$ et $Q$ sont tous deux de degré $n$, et de même coefficient dominant, on a $k \\i... | France | Olympiades Françaises de Mathématiques, Envoi Numéro 3 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
04ii | Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that
$$
\frac{a}{a+b^2} + \frac{b}{b+c^2} + \frac{c}{c+a^2} \le \frac{1}{4} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right). \quad (\text{Tonći Kokan})
$$ | [
"By using the condition $a+b+c=1$ and the inequality between arithmetic and geometric means, we have that\n$$\n\\frac{a}{a+b^2} = \\frac{a}{a(a+b+c)+b^2} = \\frac{a}{a^2+b^2+ab+ac} \\le \\frac{a}{2ab+ab+ac} = \\frac{1}{3b+c}.\n$$\nBy applying the inequality between harmonic and arithmetic means, it follows that\n$$... | Croatia | Croatia Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0j10 | The 2010 positive numbers $a_1, a_2, \dots, a_{2010}$ satisfy the inequality $a_i a_j \le i + j$ for all distinct indices $i, j$. Determine, with proof, the largest possible value of the product $a_1 a_2 \cdots a_{2010}$. | [
"(By Gabriel Carroll). Multiplying together the inequalities $a_{2i-1}a_{2i} \\le 4i - 1$ for $i = 1, 2, \\dots, 1005$, we get\n$$\na_1 a_2 \\cdots a_{2010} \\le 3 \\cdot 7 \\cdot 11 \\cdots 4019. \\qquad (1)\n$$\nIt remains to show that this bound can be attained.\nLet\n$$\na_{2008} = \\sqrt{\\frac{4017 \\cdot 401... | United States | USAMO 2010 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 3*7*11*...*4019 | |
0ijr | Problem:
Find
$$
\frac{\sqrt{31+\sqrt{31+\sqrt{31+\ldots}}}}{\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}
$$ | [
"Solution:\nLet the numerator be $x$ and the denominator $y$. Then $x^{2} = 31 + x$, so, as $x > 0$, we have\n$$\nx = \\frac{1 + \\sqrt{1 + 4 \\cdot 31}}{2} = \\frac{1 + 5 \\sqrt{5}}{2}.\n$$\nSimilarly we compute that\n$$\ny = \\frac{1 + \\sqrt{1 + 4 \\cdot 1}}{2} = \\frac{1 + \\sqrt{5}}{2}\n$$\nso that\n$$\n\\frac... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 6 - sqrt(5) | |
0kam | Problem:
You are traveling in a foreign country whose currency consists of five different-looking kinds of coins. You have several of each coin in your pocket. You remember that the coins are worth $1$, $2$, $5$, $10$, and $20$ florins, but you have no idea which coin is which and you don't speak the local language. Yo... | [
"Solution:\nThe answer is four.\n\nFirst we show that three candies are not always enough. If you only buy three candies, then it is possible the three coins you spend will be some combination of $1$-, $2$-, and $5$-florin coins, in which case you definitely won't receive any $10$- or $20$-florin coins in change. T... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 4 | |
0hzw | Problem:
Alex, Pei-Hsin, and Edward got together before the contest to send a mailing to all the invited schools. Pei-Hsin usually just stuffs the envelopes, but if Alex leaves the room she has to lick them as well and has a $25\%$ chance of dying from an allergic reaction before he gets back. Licking the glue makes E... | [
"Solution:\n\nThere are two possibilities: either Edward killed Pei-Hsin or the envelopes did. The envelope could only be responsible if Edward was not, so the chances of that would be $4/5 \\cdot 1/4 = 1/5$. This is the same as the probability that Edward killed her, so the events are equally likely and the answer... | United States | Harvard-MIT Math Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1/2 | |
07ho | For all positive integers $k$ find the smallest positive integer $n$ such that for every tournament with $n$ vertices, there exists a vertex $v$ such that both *in-degree* and *out-degree* of $v$ are at least $k$. (A tournament is a directed graph such that there is exactly one edge between each of its two vertices). | [
"We claim that the answer is $n = 4k - 1$. First and foremost, we provide an example of a graph with $4k - 2$ vertices with no desired vertex: divide the vertices into two parts with $2k - 1$ vertices. All the edges between the first and second part are directed toward the second part. Put the vertices at each part... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 4k - 1 | |
05am | Does there exist a function $f : \mathbb{R} \to \mathbb{R}$ that satisfies for all real numbers $x, y$ the equation
$$
f(x) + f(f(y)) + f(f(f(x))) = x + y?
$$ | [
"Taking $y = f(x)$ in the given equation, we get\n$$\nf(x) + f(f(f(x))) + f(f(f(x))) = x + f(x),\n$$\nfrom which, by expressing $f(f(f(x)))$, we find\n$$\nf(f(f(x))) = \\frac{x}{2}. \\qquad (7)\n$$\nTaking $y = 0$ in the given equation, we get $f(x) + f(f(0)) + f(f(f(x))) = x + 0$; using the equality (7) here, we g... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | No, there is no such function. | |
0dgs | Find an example of a sequence of natural numbers $1 \le a_1 < a_2 < \dots < a_n < a_{n+1} < \dots$ with the property that every positive integer $m$ can be uniquely written as $m = a_i - a_j$, with $i > j \ge 1$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Other",
"Discrete Mathematics > Other"
] | English | proof and answer | null | |
0et4 | Find the smallest positive multiple of $20$ with exactly $20$ positive divisors. | [
"Consider, for a positive integer $k$, the multiple $M_k = k \\times 20$. If $k = 2^{a_1} \\cdot 3^{a_2} \\cdot 5^{a_3} \\dots$ is the prime factorization of $k$, where $a_i \\ge 0$ for all $i \\ge 1$, then $M_k = 2^{a_1+2} \\cdot 3^{a_2} \\cdot 5^{a_3+1} \\cdot 7^{a_4} \\dots$ is the prime factorization of $M_k$.\... | South Africa | The South African Mathematical Olympiad Third Round | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 240 | |
05xv | Problem:
Soit $ABC$ un triangle équilatéral et $P$ un point sur le cercle circonscrit de ce triangle mais distinct de $A$, $B$ et $C$. Les droites passant par $P$ et parallèles à $(BC)$, $(CA)$ et $(AB)$ intersectent les droites $(CA)$, $(AB)$ et $(BC)$ en $M$, $N$ et $Q$ respectivement. Montrer que les points $M$, $N... | [
"Solution:\n\nOn se place dans le cas de la figure suivante :\n\n\nMontrons que $P$, $N$, $B$, $Q$ sont cocycliques. Comme $(PN)$ est parallèle à $(AC)$, et $(NB)$ est parallèle à $(AB)$, $\\widehat{ANP} = \\widehat{CAB} = 60^\\circ$. Comme $(PQ)$ est parallèle à $(AB)$, $\\widehat{PQB} = \... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k51 | Problem:
For how many pairs of sequences of nonnegative integers $\left(b_{1}, b_{2}, \ldots, b_{2018}\right)$ and $\left(c_{1}, c_{2}, \ldots, c_{2018}\right)$ does there exist a sequence of nonnegative integers $\left(a_{0}, \ldots, a_{2018}\right)$ with the following properties:
- For $0 \leq i \leq 2018, a_{i}<2^{... | [
"Solution:\n\nDefine the bitwise and of two nonnegative integers $x=\\cdots x_{3} x_{2} x_{1} x_{0}$ and $y=\\cdots y_{3} y_{2} y_{1} y_{0}$ expressed in binary to be $x \\& y=\\cdots z_{3} z_{2} z_{1} z_{0}$, where $z_{i}=1$ if both $x_{i}$ and $y_{i}$ are 1 , and 0 otherwise.\nNow, we can prove that from the defi... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (2^{2019}-1)^{2018} | |
09u6 | There are $n$ guests at a party. Any two guests are either friends or not friends. Every guest is friends with exactly four of the other guests. Whenever a guest is *not* friends with two other guests, those two other guests cannot be friends with each other either.
What are the possible values of $n$? | [
"We first consider the friends of one guest, say Marieke. We know that Marieke has exactly four friends at the party, say Aad, Bob, Carla, and Demi. The other guests (if there are any other guests) are not friends with Marieke. Hence, they cannot have any friendships among themselves and can therefore only be frien... | Netherlands | Final Round, September 2019 | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 5, 6, 8 | |
09jb | Let $E$ be the intersection of diagonals of a cyclic quadruple $ABCD$. A line through $E$ intersects with circumcircle of $BCE$ at $K$ ($K \neq E$), and intersects with the circumcircle of $CDE$ at $L$ ($L \neq E$). If the lines $KB, LD$ meet at $S$, then prove that the line $AS$ and $KL$ parallel. | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof only | null | |
0gun | Initially the equation
$$
\star \frac{1}{x-1} \star \frac{1}{x-2} \star \frac{1}{x-4} \dots \star \frac{1}{x-2^{2023}} = 0
$$
is written on the blackboard. Asli and Zehra alternatively make moves, Asli begins. A person making move replaces one of the stars in the equation with either $+$ or $-$. What is the maximal num... | [
"**Answer: 1011.**\n\nLet us describe the strategy of Aslı for guaranteeing at least 1011 solutions of the equation. Aslı makes her first move arbitrarily. After that, at each move she chooses any star with already replaced neighbour and replaces this star with the sign of this neighbour. She completes the game wit... | Turkey | Team Selection Test for IMO 2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 1011 | |
02q3 | Problem:
Um torneio de futebol com $57$ times será disputado com as seguintes regras:
a. Nenhum jogo pode terminar empatado.
b. O time que perder duas partidas será eliminado.
c. O torneio termina quando sobrar apenas um time, que será o campeão.
Se o time campeão perder uma vez, quantas partidas serão disputadas ... | [] | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 113 | |
0gsl | Let $ABC$ be a triangle such that the circle $\omega$ with diameter $BC$ is tangent to the exterior bisector of $\angle BAC$. The internal bisector of $\angle BAC$ intersects with the side $BC$ at point $K$ and with the circumcircle of $ABC$ at point $L \neq A$. Let $M$ be the midpoint of $BC$. Prove that the circumcir... | [
"Let $\\Gamma$ be the circumcircle of $ABC$. Let the intersection points of the external bisector of $\\angle BAC$ with $BC$ be $T$, with $\\Gamma$ be $R \\neq A$ and with $\\omega$ be $P$. Then, $AL$ and $PM$ are parallel since both are perpendicular to $AT$, which implies $TK/TM = TA/TP$. Looking at the power of ... | Turkey | 30th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gut | If $S$ is a set consisting of $12$ positive integers, then what is the maximum number of pairs $(a, b)$ such that $a, b \in S$ and $\frac{b}{a}$ is a prime number? | [
"Answer: $20$.\n\nExample: $S = \\{1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 42\\}$.\n\nConstruct a graph $G$ whose vertex set is $S$ and edge set consists of noble pairs. We will show that $|E(G)| = e \\le 20$.\n\nWe first prove that $G$ is bipartite. Assume that $G$ has an odd cycle. Start from one of the vertices an... | Turkey | Team Selection Test for IMO 2024 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 20 | |
0057 | Sea $n$ un entero positivo no divisible por $3$. Demuestre que $n$ admite una representación de la forma:
$$
n = \frac{3xy}{x+y}
$$
donde $x$, $y$ son enteros positivos,
si y sólo si $n$ tiene al menos un divisor de la forma $3k+2$ para algún $k = 0,1,2,...$. | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Spanish | proof only | null | |
01x1 | Given a positive integer $n$, determine the maximal constant $C_n$ satisfying the following condition: for any partition of the set $\{1, 2, \dots, 2n\}$ into two $n$-element subsets $A$ and $B$, there exist labellings $a_1, \dots, a_n$ and $b_1, \dots, b_n$ of $A$ and $B$, respectively, such that
$$
(a_1 - b_1)^2 + (a... | [
"See Romanian Masters of Mathematics - 2019 Shortlist."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | Cn = (13 n^3 - n) / 12 for odd n, and Cn = (13 n^3 - 4 n) / 12 for even n. | |
00p0 | Consider a cyclic quadrilateral such that the midpoints of its sides form another cyclic quadrilateral. Prove that the area of the smaller circle is less than or equal to half the area of the bigger circle. | [
"Let $ABCD$ be a cyclic quadrilateral with $AB = a$, $BC = b$, $CD = c$, $DA = d$, $AC = e$ and $BD = f$. Because the midpoints of the cyclic quadrilateral $ABCD$ form another cyclic quadrilateral, which is a parallelogram, we deduce that this parallelogram is a rectangle and $ABCD$ is orthogonal.\n\nIt follows tha... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | English | proof only | null | |
07mj | A direct common tangent touches two non-intersecting circles at $T$ and $T'$. The two transverse common tangents meet $TT'$ at $A$ and $B$. Prove
$$
|AT| = |BT'|
$$ | [
"Let contact points of transverse tangents be $S$, $S'$, $V$, $V'$ and let $O$, $O'$ be the centres of the circles (see figure below). Let $SS'$ meet $VV'$ at $Q$, which will also lie on $OO'$. Complete the rectangle $OVV'H$ and let $\\alpha = \\angle BO'V'$ and $\\beta = V'O'Q$. Note that $\\angle T'O'B = \\alpha$... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ggc | 設三角形 $ABC$ 的垂心為點 $H$。令直線 $BH$, $AC$ 交於點 $E$，且直線 $CH$, $AB$ 交於點 $F$。設點 $X$ 為直線 $BC$ 上任意一點。設三角形 $BEX$ 的外接圓與直線 $AB$ 再交於點 $Y$，且三角形 $CFX$ 的外接圓與直線 $AC$ 再交於點 $Z$。證明三角形 $AYZ$ 的外接圓與直線 $AH$ 相切。 | [
"By Miquel's theorem (applied to $X$, $F$, $E$ on sides of $HBC$), we know that the circumcircles of $HEF$, $BEX$, $CFX$ intersect at a point $M$. Now again, by Miquel's theorem (applied to $Y$, $E$, $A$ on sides of $HAB$ where $A$ is viewed as a point on $HA$), we know that the circumcircles of $HEA$, $BEY$ and th... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 國際競賽實作(二) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Tangents"
] | Chinese; English | proof only | null | |
05g6 | Problem:
Jules et Jim, deux colocataires, reçoivent 10 autres paires de colocataires chez eux. Jules fait un petit sondage au cours de la soirée et interroge les 21 autres personnes. Parmi elles, aucune n'a serré la main de son colocataire, et il n'y en a pas deux qui ont serré le même nombre de mains. Combien de main... | [
"Solution:\n\nAu cours de la soirée, parmi les 21, chaque personne serre entre 0 et 20 mains. Comme elles serrent toutes un nombre différent de mains, pour tout $n$ compris entre $0$ et $20$ il existe une et une seule personne $A_{n}$ parmi les 21 qui a serré exactement $n$ mains.\n\nComme $A_{20}$ a serré 20 mains... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 10 | |
0edr | Problem:
Naj bodo $a$, $b$ in $c$ dolžine stranic trikotnika in naj velja $c^{2}=a^{2}+b^{2}+a b$. Katera izmed navedenih trditev velja za tak trikotnik?
(A) Trikotnik je enakostraničen.
(B) Trikotnik je ostrokoten.
(C) Višinska točka leži izven trikotnika.
(D) Trikotnik je pravokoten.
(E) Središče trikotniku očrtane ... | [
"Solution:\n\nIz kosinusnega izreka $c^{2}=a^{2}+b^{2}-2 a b \\cos \\gamma$ in enakosti $c^{2}=a^{2}+b^{2}+a b$ dobimo $a b=-2 a b \\cos \\gamma$. Od tod sledi $\\cos \\gamma=-\\frac{1}{2}$ oziroma $\\gamma=120^{\\circ}$. Trikotnik je topokoten, torej leži višinska točka izven trikotnika."
] | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | MCQ | C | |
0ksc | Problem:
Let $f$ be a function from $\{1,2, \ldots, 22\}$ to the positive integers such that $m n \mid f(m)+f(n)$ for all $m, n \in \{1,2, \ldots, 22\}$. If $d$ is the number of positive divisors of $f(20)$, compute the minimum possible value of $d$.
Proposed by: Sheldon Kieren Tan | [
"Solution:\nLet $L=\\operatorname{lcm}(1,2, \\ldots, 22)$. We claim that the possible values of $f(20)$ are the multiples of $20 L$. If we can prove this, we will be done, since the minimum value of $d$ will be the number of divisors of $20 L=2^{6} \\cdot 3^{2} \\cdot 5^{2} \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\... | United States | HMMT February 2022 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number T... | null | proof and answer | 2016 | |
0c7o | Given a real number $a > 1$, determine all real numbers $b \ge 1$ such that
$$
\lim_{x \to \infty} \int_{0}^{x} (1 + t^a)^{-b} \, dt = 1.
$$ | [
"We show that $b_0 = 1 + 1/a$ is the only real number satisfying the required condition.\nLet $b \\ge 1$, and let $f_b, F_b: [0, \\infty) \\to \\mathbb{R}$, $f_b(x) = (1 + x^a)^{-b}$ and $F_b(x) = \\int_0^x f_b(t) \\, dt$. Since $f_b$ is positive, $F_b$ is increasing, so the limit $I(b) = \\lim_{x \\to \\infty} F_b... | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Integral Calculus > Applications"
] | English | proof and answer | b = 1 + 1/a | |
0chb | Let $n$ be a composite positive integer and let $1 = d_1 < d_2 < d_3 < \dots < d_k = n$ be the divisors of $n$, where $k \ge 3$. Assume that all the equations $d_{i+2}x^2 - 2d_{i+1}x + d_i = 0$, for $i \in \{1, 2, \dots, k-2\}$ have real solutions. Prove that $n = p^{k-1}$ for some prime number $p$. | [
"$$\n4d_{i+1}^2 - 4d_{i+2} \\cdot d_i \\ge 0 \\Leftrightarrow \\frac{d_{i+1}}{d_i} \\ge \\frac{d_{i+2}}{d_{i+1}} \\quad (*)\n$$\nfor any $i \\in \\{1, 2, \\dots, k-2\\}$.\nAs $d_2$ is the smallest proper divisor of $n$, it means that the number $\\frac{n}{d_2}$ is the greatest proper divisor of $n$, so $\\frac{n}{d... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
010z | Problem:
Determine all positive integers $n \geqslant 3$ such that the inequality
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n-1} a_{n}+a_{n} a_{1} \leqslant 0
$$
holds for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ which satisfy $a_{1}+\cdots+a_{n}=0$. | [
"Solution:\n$n=3$ and $n=4$.\n\nFor $n=3$ we have\n$$\n\\begin{aligned}\n& a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}=\\frac{\\left(a_{1}+a_{2}+a_{3}\\right)^{2}-\\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\\right)}{2} \\leqslant \\\\\n& \\quad \\leqslant \\frac{\\left(a_{1}+a_{2}+a_{3}\\right)^{2}}{2}=0 .\n\\end{aligned}\n$$\n\n... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | n = 3 and n = 4 | |
0559 | Let $A$ and $B$ be such points of the circle with centre $O$ that the triangle $AOB$ is right-angled. The perpendicular bisector of the segment $AO$ intersects the shorter arc $AB$ in point $K$. The lines $KO$ and $AB$ intersect in point $L$. Prove that the triangle $KBL$ is isosceles. | [
"\nFig. 4\n\nSince $K$ lies on the perpendicular bisector of the segment $AO$ (Fig. 4) we have $KA = KO$. On the other hand $KO = AO$ since $K$ and $A$ are points on the circle. Hence $AKO$ is an equilateral triangle from which we obtain that $\\angle AOK = 60^\\circ$. Hence $\\angle KOB = ... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0dua | Problem:
Poišči vsa racionalna števila $r$, za katera so vse rešitve enačbe
$$
r x^{2}+(r+1) x+r=1
$$
cela števila. | [
"Solution:\nPišimo $r=\\frac{a}{b}$, kjer sta $a$ in $b$ tuji števili, $a>0$, in predpostavimo, da ima enačba celoštevilske rešitve. Dobimo $a x^{2}+(a+b) x+(a-b)=0$, od koder sledi $a \\mid (a-b)$, zato je $a=1$.\n\nDa bi imela kvadratna enačba $x^{2}+(1+b) x+(1-b)=0$ kakšno celoštevilsko rešitev, mora biti njena ... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | r = 1 and r = -1/7 | |
0cur | A quadrilateral $ABCD$ is inscribed into a circle $\Gamma$ centered at $O$. Its diagonals $AC$ and $BD$ are perpendicular to each other; let $P$ be their meeting point (the point $O$ lies inside the triangle $BPC$). A point $H$ is chosen on the segment $BO$ so that $\angle BHP = 90^\circ$. The circumcircle $\omega$ of ... | [
"Let $BT$ be a diameter of $\\Omega$; then the points $P$, $H$, $T$, and $D$ are concyclic (see Fig. 9). Thus $\\angle PQD = 90^\\circ$, and the perpendicular bisectors to $PQ$, $TD$, and $AC$ coincide.\n\nConsider the common perpendicular bisector $l$ to the segments $DT$ and $PQ$. It passes through $O$, and there... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null | |
01tx | Let polynomials $P(x)$ and $Q(x)$ have the same degree. Let $P_Q(x)$ denote the polynomial such that its coefficients of even degree variables coincide with the corresponding coefficients of $P(x)$ and its coefficients of odd degree variables coincide with the corresponding coefficients of $Q(x)$. (For example, if $P(x... | [
"a)\nShow, for example, that the polynomials\n$$\nP(x) = 4x^4 + 4x^3 + 1 \\quad \\text{and} \\quad Q(x) = x^4 + 4x + 4\n$$\nsatisfy the condition.\nBy Cauchy's inequality, for any real $x$ the following inequalities hold:\n$$\nx^4 + x^4 + x^4 + 1 \\ge 4\\sqrt[4]{x^4 \\cdot x^4 \\cdot x^4 \\cdot 1} = 4|x|^3, \\quad ... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 4 | |
02go | Find the smallest positive prime that divides $n^2 + 5n + 23$ for some integer $n$. | [
"Let $f(n) = n^2 + 5n + 23$. For $n = -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4$ we get $f(n) = 47, 37, 29, 23, 19, 17, 17, 19, 23, 29, 37, 47, 59$. Observe that $f(-n-5) = f(n)$, so this only involves calculating 7 terms. None of these are divisible by any of $2, 3, 5, 7, 11, 13$. But all possible values mod $... | Brazil | XXV OBM | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 17 | |
0lbn | Let $ABC$ be a triangle inscribed in a circle $(O)$. Let $D$ be a point on the arc $BC$ not containing $A$. Suppose that the tangent of $(O)$ at $D$ meets $BC$ at $E$. The circumcircle of $ODE$ meets $AD$ at second point $F$ (different to $D$). The line $EF$ meets $AB$ at $M$, the line $DM$ meet $(O)$ at the second poi... | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0e8k | Problem:
a) Poenostavi izraz
$$
\left(\frac{\cos \alpha}{\cos \beta}\right)^2-(\cos \alpha \cdot \tan \beta)^2+\sin ^2 \alpha
$$
če je $\cos \beta \neq 0$.
b) Izračunaj vrednost $\alpha+\beta$, če je $\sin \alpha=\frac{\sqrt{5}}{5}$ in $\sin \beta=\frac{\sqrt{10}}{10}$ ter je $0^{\circ}<\alpha, \beta<45^{\circ}$. Nal... | [] | Slovenia | Državno tekmovanje | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | a) 1; b) 45° | |
0jcw | Problem:
For any positive integer $n$, let $N=\varphi(1)+\varphi(2)+\ldots+\varphi(n)$. Show that there exists a sequence
$$
a_{1}, a_{2}, \ldots, a_{N}
$$
containing exactly $\varphi(k)$ instances of $k$ for all positive integers $k \leq n$ such that
$$
\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{N... | [
"Solution:\n\nWe write all fractions of the form $b / a$, where $a$ and $b$ are relatively prime, and $0 \\leq b \\leq a \\leq n$, in ascending order. For instance, for $n=5$, this is the sequence\n$$\n\\frac{0}{1}, \\frac{1}{5}, \\frac{1}{4}, \\frac{1}{3}, \\frac{2}{5}, \\frac{1}{2}, \\frac{3}{5}, \\frac{2}{3}, \\... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
013z | Problem:
A sequence $\left(x_{n}\right), n \geq 0$, is defined as follows: $x_{0}=a, x_{1}=2$ and $x_{n}=2 x_{n-1} x_{n-2}-x_{n-1}-x_{n-2}+1$ for $n>1$. Find all integers $a$ such that $2 x_{3 n}-1$ is a perfect square for all $n \geq 1$. | [
"Solution:\nLet $y_{n}=2 x_{n}-1$. Then\n$$\n\\begin{aligned}\ny_{n} & =2\\left(2 x_{n-1} x_{n-2}-x_{n-1}-x_{n-2}+1\\right)-1 \\\\\n& =4 x_{n-1} x_{n-2}-2 x_{n-1}-2 x_{n-2}+1 \\\\\n& =\\left(2 x_{n-1}-1\\right)\\left(2 x_{n-2}-1\\right)=y_{n-1} y_{n-2}\n\\end{aligned}\n$$\nwhen $n>1$. Notice that $y_{n+3}=y_{n+2} y... | Baltic Way | Baltic Way 2005 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof and answer | a = ((2m - 1)^2 + 1)/2 for all positive integers m | |
07bd | Permutation $(a_1, a_2, \dots, a_n)$ of the set $\{1, 2, \dots, n\}$ is called **messy** if for any three indices $1 \le i < j < k \le n$, $a_i + a_k - 2a_j$ is not divisible by $n$. Find all natural numbers $n \ge 3$ for which there exists a messy permutation of the set $\{1, 2, \dots, n\}$. | [
"We claim that there exists a messy permutation of $\\{1, 2, \\dots, n\\}$ if and only if $n$ is a power of $2$.\n\nFirst suppose that there is some messy permutation $(a_1, a_2, \\dots, a_n)$ of $\\{1, 2, \\dots, n\\}$. There must be some $j$ such that $a_j \\equiv 2a_2 - a_1 \\pmod{n}$. If $j \\ge 3$, then $n \\m... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All n of the form 2^t with t ≥ 2 | |
0b51 | Problem:
Fie $ABC$ un triunghi ascuţitunghic, $AB < AC$, şi fie $\Gamma$ cercul său circumscris. Tangenta lui $\Gamma$ în $A$ intersectează dreapta $BC$ în $K$. Cercul $\Omega$, de rază $KA$ şi centru $K$, intersectează a doua oară $\Gamma$ în $L$. Cercul $\Omega$ intersectează a doua oară dreapta $BL$ în $M$ şi dreap... | [] | Romania | TESTUL 1 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Radical ax... | null | proof only | null | |
0aip | Prove that there exist pairwise disjoint sets $A_1, A_2, \dots, A_{2014}$ whose union is the set of natural numbers and for which the following condition holds:
For arbitrary natural numbers $a$ and $b$, at least two of the numbers $a, b, \text{gcd}(a, b)$ belong to one of the sets $A_1, A_2, \dots, A_{2014}$.
Докажи ... | [
"Let $v_2(n)$ be the greatest integer for which $2^{v_2(n)}$ is a divisor of $n$. Then $v_2(\\text{gcd}(a, b)) = \\min\\{v_2(a), v_2(b)\\}$. Therefore at least two of the numbers $v_2(a), v_2(b)$ and $v_2(\\text{gcd}(a, b))$ are equal.\n\nWe define sets $A_{i+1} = \\{n \\mid v_2(n) \\equiv i \\pmod{2014}\\}$ for $0... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0dj1 | Given is an $n \times n$ board, whose all cells are initially white. Khalid the Painter walks around the board and recolors the visited cells according to the following rules. Each *walk* of Khalid starts at the bottom-left corner of the board and continues as follows:
* if he is standing on a white cell, he paints it... | [
"(Base on the solution of Ali Alramdan, IMO 2023's team member)\n\nFor two times Khalid visits a square on two different directions, one of them he will go up and go right for the other. Let $f(i, j)$ be the number of time Khalid visited square $(i, j)$ so $f(0, 0) = s$ for some positive integer $s$ and\n$$\nf(i, j... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | s = 2^{2n - 1 - v_2\big(\binom{2n - 2}{n - 1}\big)} | |
019d | All ten-digit numbers composed of digits $1$ and $2$ are divided by $1024$ (with the remainder). How many different reminders are obtained by these calculations? | [
"Answer: $1024$.\nAll the reminders are pairwise distinct, because it is not difficult to see that the difference between any two numbers have an odd digit and several zeroes at the end of its decimal representation. Therefore it is divisible by $10^k$, $0 \\le k \\le 9$, and the quotient is odd. Therefore the diff... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 1024 | |
05ml | Problem:
Soit $ABCDEF$ un hexagone ayant tous ses angles égaux à $120^{\circ}$. Montrer que $AB + BC = DE + EF$. | [
"Solution:\n\nOn prolonge les côtés $[AB]$, $[CD]$ et $[EF]$, et on appelle $X$ l'intersection de $(AB)$ et $(CD)$, $Y$ celle de $(CD)$ et $(EF)$, et $Z$ celle de $(EF)$ et $(AB)$. Alors $\\widehat{XBC} = 180^{\\circ} - \\widehat{ABC} = 60^{\\circ}$ et de même pour $\\widehat{XCB}$, donc:\n\n$$\n\\widehat{ZXY} = \\... | France | French Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0g8k | 是否存在無窮多個整數 $a_1, a_2, a_3, \dots$ 及正整數 $N$, 其中 $0 < a_i < 10$, 使得對於所有正整數 $k > N$,
$$
\sum_{i=1}^{k} a_i 10^{i-1}
$$
都是完全平方數?
Determine whether there exists an infinite sequence of integers $a_1, a_2, a_3, \dots$ with $0 < a_i < 10$ and a positive integer $N$ such that, for every integer $k > N$, the number
$$
\sum_{i=... | [
"不可能。歸謬證法, 假設存在這樣的無窮序列和正整數 $N$. 令 $y_k = \\sum_{i=1}^{k} a_i 10^{i-1}$, 則對於所有 $k > N$, 存在正整數 $x_k$ 使得 $y_k = x_k^2$.\n\n1. 對所有 $n$, 令 $\\gamma_n$ 為滿足 $5^{\\gamma_n}|x_n$ 的最大正整數。以下證明: 對於所有 $n > N$, $2\\gamma_n \\ge n$.\n假設存在 $n < N$ 使得 $2\\gamma_n < n$, 則\n$$\ny_{n+1} = 10^n a_{n+1} + y_n = 5^{2\\gamma_n} \\left( 2^... | Taiwan | 2014 IMO Training Camp Stage 3 Mock Competition (1) | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0jy5 | Problem:
A polyhedron has $7n$ faces. Show that there exist $n+1$ of the polyhedron's faces that all have the same number of edges. | [
"Solution:\nLet $V$, $E$, and $F$ denote the number of vertices, edges, and faces respectively. Let $a_k$ denote the number of faces with $k$ sides, and let $M$ be the maximum number of sides any face has.\n\nSuppose that $a_k \\leq n$ for all $k$ and that $M > 8$. Note that each edge is part of exactly two faces, ... | United States | February 2017 | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
055q | Find all triples $(p, q, r)$ of primes such that $2018(p^2+q^2) = r^2+1$. | [
"Suppose that both $p$ and $q$ are odd. Then $p^2+q^2$ is even and the l.h.s. of the equation is divisible by 4. Squares of integers are congruent to 0 or 1 modulo 4 whence the r.h.s. is congruent to 1 or 2 modulo 4. The contradiction shows that one of $p$ and $q$ equals 2; let w.l.o.g. $p = 2$. Squares of integers... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | No such triples exist | |
07ub | Find all right-angled triangles with integer side lengths whose areas are numerically equal to their perimeters. | [
"Let $a, b, c$ be the side lengths of a right-angled triangle with hypotenuse $c$, i.e. $c^2 = a^2 + b^2$. The area of this triangle is equal to $ab/2$ and the perimeter is $a + b + c$. Area and perimeter agree exactly when $ab/2 = a + b + c$. Squaring both sides and using Pythagoras, this becomes\n$$\na^2 + b^2 = ... | Ireland | IRL_ABooklet | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | The only such right triangles are with side lengths 5, 12, 13 and 6, 8, 10. | |
0gqm | Find all pairs $(m, n)$ of positive integers satisfying $m^6 + 5n^2 = m + n^3$. | [
"Let us rewrite the equation: $n(n^2 - 5n) = m(m^5 - 1)$. If $n \\le m$ then $n^2 - 5n \\ge m^5 - 1 \\ge n^5 - 1$. Contradiction, Thus, $n > m$. Now $n^3 > n^3 - 5n^2 + n > n^3 - 5n^2 + m = m^6$. Therefore, $n^3 > m^6$, equivalently $n > m^2$. Let $n = m^2 + k$ for some $k > 0$. By putting $m^2 = n - k$ to the main... | Turkey | Team Selection Test | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1, 5) and (3, 11) | |
0iju | Problem:
Let $a_{1}, a_{2}, \ldots$ be a sequence defined by $a_{1}=a_{2}=1$ and $a_{n+2}=a_{n+1}+a_{n}$ for $n \geq 1$. Find
$$
\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}
$$ | [
"Solution:\nLet $X$ denote the desired sum. Note that\n$$\n\\begin{array}{rl}\nX & = \\frac{1}{4^{2}}+\\frac{1}{4^{3}}+\\frac{2}{4^{4}}+\\frac{3}{4^{5}}+\\frac{5}{4^{6}}+\\ldots \\\\\n4 X & =\\quad \\frac{1}{4^{1}}+\\frac{1}{4^{2}}+\\frac{2}{4^{3}}+\\frac{3}{4^{4}}+\\frac{5}{4^{5}}+\\frac{8}{4^{6}}+\\ldots \\\\\n1... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 1/11 | |
0569 | Do there exist five distinct prime numbers for which the sum of any three of them is a prime number as well? | [
"*Answer:* No.\n\nAssume that there exist five such prime numbers. If there are three among them pairwise incongruent modulo $3$, then by adding each of the three separately to the sum of the other two we get three distinct sums modulo $3$. One of those three is $0$ modulo $3$ and therefore is not a prime number. H... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | No | |
004k | Determinar el mayor número natural que tiene todas sus cifras distintas y es múltiplo de $5$, de $8$ y de $11$. | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Español | proof and answer | 9876513240 | |
07jg | Let $\overarc{ABCD}$ be a cyclic quadrilateral with incircle $\omega$. Let $M$ be the midpoint of arc $\overarc{ABC}$. Let $\Gamma$ be the circle with center $M$ and radius $MA$. $AD$ and $AB$ intersect $\Gamma$ at $X$ and $Y$, respectively. Let $Z$ be a point on line $XY$ (with $Z \neq Y$) such that $BY = BZ$. Prove t... | [
"Since $M$ is the center of the circumcircle of $AYC$ and $ABMC$ is also cyclic, we have:\n$$\n\\begin{align*}\n\\angle ABC &= \\angle AMC = 2 \\times \\angle AYC \\\\\n&= \\angle BYC + \\angle BCY \\\\\n&= \\angle AYC + \\angle BCY\n\\end{align*}\n$$\nThus $BY = BC$. Also, using the cyclicity of $AXYC$, we have:\n... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0k6w | Problem:
Let $n$ be a positive integer. Show that $2n+1$ and $4n^{2}+1$ are relatively prime, that is, their only common factor is $1$. | [
"Solution:\nAny common factor of the two numbers would also have to divide\n$$\n(4n^{2}+1)-(2n+1)(2n-1) = (4n^{2}+1)-(4n^{2}-1) = 2\n$$\nBut both numbers are odd, since they are $1$ more than an even number, so they are not divisible by $2$. Thus, their greatest common factor is $1$."
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0j4x | Problem:
given positive integers $a$ and $b$ such that $a > b$, define a sequence of ordered pairs $(a_{l}, b_{l})$ for nonnegative integers $l$ by $a_{0} = a$, $b_{0} = b$, and $(a_{l+1}, b_{l+1}) = (b_{l}, a_{l} \bmod b_{l})$, where, for all positive integers $x$ and $y$, $x \bmod y$ is defined to be the remainder l... | [
"Solution:\n\na.\nThe answer is $F_{m+1}$, where $F_{1} = 1$, $F_{2} = 2$, and $F_{i+1} = F_{i} + F_{i-1}$ for all $i \\geq 2$.\n\nWe consider a reverse sequence as follows: starting at $p_{0} = (k_{0}, 0)$ for some positive integer $k_{0}$, at each step we can take a pair $p_{i} = (r_{i}, s_{i})$ to any pair $p_{i... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a) n_m = F_{m+1}, where F_1 = 1, F_2 = 2, and F_{i+1} = F_i + F_{i-1}.
b) The second smallest is L_{m+1}, where L_1 = 1, L_2 = 3, and L_{i+1} = L_i + L_{i-1}; for m = 1 there is no second smallest. | |
07l4 | Let $S$ be a set of $n > 1$ distinct integers and suppose that
$$
T = \{x + y : x \neq y, x, y \in S\}
$$
has exactly $2n - 3$ distinct elements. Prove that, when arranged in increasing order, the elements of $S$ form an arithmetic progression. | [
"Write $S = \\{x_1, \\dots, x_n\\}$ where $x_1 < \\dots < x_n$. The $2n-3$ elements\n$$\nx_1 + x_2 < \\dots < x_1 + x_n < x_2 + x_n < x_3 + x_n < \\dots < x_{n-1} + x_n\n$$\naccount for all the elements of $T$. Now $x_2 + x_{n-1}$ lies between $x_1 + x_{n-1}$ and $x_2 + x_n$, so it must be $x_1 + x_n$. So $x_2 - x_... | Ireland | Irska | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
0i6l | Problem:
Find the sum of the even positive divisors of $1000$. | [
"Solution:\n$2184$. Notice that $2k$ is a divisor of $1000$ if and only if $k$ is a divisor of $500$, so we need only find the sum of the divisors of $500$ and multiply by $2$. This can be done by enumerating the divisors individually, or simply by using the formula: \n$$\n\\sigma\\left(2^{2} \\cdot 5^{3}\\right) =... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 2184 | |
0bq0 | Problem:
a) Să se arate că șirul $a_{n} = \int_{0}^{n} e^{-x^{2}} dx$, $n \in \mathbb{N}$, este convergent.
b) Dacă $f:[0, \infty) \rightarrow [0, \infty)$ este o funcție continuă cu $\lim_{x \rightarrow \infty} f(x) = 0$, rezultă în mod necesar că șirul $x_{n} = \int_{0}^{n} f(x) dx$, $n \in \mathbb{N}$, este converg... | [] | Romania | Olimpiada Natională de Matematică | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | null | proof and answer | a) Convergent (indeed bounded above by the improper integral ∫_0^∞ e^{-x^2} dx). b) Not necessarily; for example, f(x) = 1/(x+1) is continuous, nonnegative, tends to zero, but ∫_0^n f(x) dx = ln(n+1) diverges. | |
0eb2 | Find all real numbers $a \ge 0$ for which the equation $2|x-a| + 3|x+a| = 1$ has at least one real solution. | [
"We need to consider several possibilities.\n\nIf $x < -a$, then $-2(x - a) - 3(x + a) = 1$, which implies $x = -\\frac{a+1}{5}$. For this to be a valid solution, we must have\n\n$-\\frac{a+1}{5} < -a$, or $a < \\frac{1}{4}$.\n\nIf $-a \\le x \\le a$, then $-2(x-a) + 3(x+a) = 1$, so $x = 1-5a$. This solution is val... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities"
] | null | proof and answer | [0, 1/4] | |
0aqy | Problem:
Let $f(x)$ be a nonzero function whose domain and range is the set of complex numbers. Find all complex numbers $x$ such that $f\left(x^{2}\right)+x f\left(\frac{1}{x^{2}}\right)=\frac{1}{x}$. | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | no complex number x | |
0du1 | Find the smallest positive $m$ so that it is possible to choose 8 vertices from a regular polygon with $m$ sides such that no three of them are the vertices of an isosceles triangle. | [
"Let $P_1, P_2, \\dots, P_m$ be the vertices of a regular $m$-gon. Let $P_{t_1}, P_{t_2}, \\dots, P_{t_8}$ be 8 distinct vertices so that no 3 of them are the vertices of an isosceles triangle. Without loss of generality, we may assume that $1 = t_1 < t_2 < \\dots < t_8$. We'll show that $m \\ge 18$.\n\nLet $d_1 = ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 18 | |
0kry | Problem:
Let $ABCDEF$ be a hexagon and let $M, N, P, Q, R, S$ be the midpoints of $AB$, $BC$, $CD$, $DE$, $EF$, and $FA$, respectively. Show that $MQ \perp PS$ if and only if $RN^{2} = MQ^{2} + PS^{2}$. | [
"Solution:\n\nLet $A = (2a_1, 2a_2)$, $B = (2b_1, 2b_2)$, and so on. Then $M = (a_1 + b_1, a_2 + b_2)$ and $Q = (d_1 + e_1, d_2 + e_2)$, so the slope of $MQ$ is\n$$\n\\frac{(a_2 + b_2) - (d_2 + e_2)}{(a_1 + b_1) - (d_1 + e_1)}\n$$\nSimilarly, the slope of $PS$ is\n$$\n\\frac{(c_2 + d_2) - (f_2 + a_2)}{(c_1 + d_1) -... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0feg | Problem:
Averigua qué números de cuatro cifras significativas, $\overline{a b c d}$ (con $a \neq 0$), son iguales a $\overline{a b}^{2} + \overline{c d}^{2} - \overline{c d}$.
Nota: La notación $\overline{a b}$ representa, en este problema, el número que tiene $a$ decenas y $b$ unidades; en este caso se tiene que $a, ... | [
"Solution:\nTenemos $\\overline{a b c d} = \\overline{a b}^{2} + \\overline{c d}^{2} - \\overline{c d}$, por lo tanto\n$$\n\\overline{a b 00} = \\overline{a b}^{2} + \\overline{c d}^{2} - 2\\overline{c d} = \\overline{a b}^{2} + (\\overline{c d} - 1)^{2} - 1\n$$\ny de aquí se tiene:\n$$\n\\overline{a b 00} + 1 = \\... | Spain | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 5151, 4951, 9911, 6050, 4050 | |
09tw | Problem:
In een klas van minstens vier mensen geldt het volgende: als er vier van hen aan een ronde tafel gaan zitten, is er altijd iemand die allebei zijn buren kent of allebei zijn buren niet kent. Bewijs dat het mogelijk is om de mensen over twee groepen (waarvan er eentje leeg mag zijn) te verdelen, zodat in de en... | [
"Solution:\n\nVan alle mogelijke groepen mensen die we in deze klas kunnen maken, bekijken we de groepen waarin iedereen elkaar kent en daarvan nemen we er eentje met zoveel mogelijk mensen erin. (Deze grootste groep bestaat: er zijn eindig veel mensen en er is in elk geval een groep van mensen die elkaar allemaal ... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
03j8 | Problem:
Let $1 < x_{1} < 2$ and, for $n = 1, 2, \ldots$, define $x_{n+1} = 1 + x_{n} - \frac{1}{2} x_{n}^{2}$. Prove that, for $n \geq 3$, $\left| x_{n} - \sqrt{2} \right| < 2^{-n}$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
04t4 | There are different positive integers written on the board. Their (arithmetic) mean is a decimal number, with the decimal part exactly $0{,}2016$. What is the least possible value of the mean? (Patrik Bak) | [
"Let $s$ be the sum, $n$ the number and $p$ the integer part of the mean of the numbers on the board. Then we can write\n$$\n\\frac{s}{n} = p + \\frac{2{,}016}{10{,}000} = p + \\frac{126}{625},\n$$\nwhich gives\n$$\n625(s - pn) = 126n.\n$$\nNumbers $126$ and $625$ are coprime, thus $625 \\mid n$. Therefore $n \\ge ... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 313.2016 | |
09p1 | (1) Prove $t^3 \geq t^2 + 4$ for $t \geq 2$, and find the condition under which equality holds.
(2) For any numbers $t_1, t_2, \ldots, t_n \geq 2$, prove that
$$
\frac{t_1^3}{t_2^2 + 4} + \frac{t_2^3}{t_3^2 + 4} + \dots + \frac{t_{n-1}^3}{t_n^2 + 4} + \frac{t_n^3}{t_1^2 + 4} \geq n,
$$
and find the condition under whi... | [] | Mongolia | MMO2025 Round 3 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Part (1): Equality holds if and only if t equals 2. Part (2): Equality holds if and only if all the numbers equal 2. | |
0ca9 | Problem:
Fie $n \geq 2$ şi matricele $A, B \in M_{n}(\mathbb{R})$. Presupunem că există $x \in \mathbb{R} \backslash\left\{0, \frac{1}{2}, 1\right\}$ astfel încât $x A B + (1-x) B A = I_{n}$. Arătaţi că $(A B - B A)^{n} = O_{n}$. | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0bmc | Given two integers $h \ge 1$ and $p \ge 2$, determine the minimum number of pairs of opponents an *hp*-member parliament may have, if in every partition of the parliament into *h* houses of *p* member each some house contains at least one pair of opponents. | [
"Letting $N(h, p) = (h-1) \\cdot \\min(p, h/2+1)$, we now proceed to prove by induction on $h$ that if the number of edges of a graph on $hp$ vertices does not exceed $N(h, p)$, then the graph is $h$-partite on $p$-element classes. The base case $h=1$ is clear.\n\nNext, let $h \\ge 2$ and let $G = (V, E)$ be a grap... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | (h - 1) · min(p, h/2 + 1) + 1 | |
052f | Find the integral part of $A = \sqrt{2013 + \sqrt{2012 + \dots + \sqrt{2 + \sqrt{1}}}}$. | [
"On the one hand $A^2 > 2013 + \\sqrt{2012} > 2013 + 44 > 45^2$, therefore $A > 45$. On the other hand we can demonstrate with induction that $x_n = \\sqrt{n + \\sqrt{n-1} + \\dots + \\sqrt{1}} < \\sqrt{n} + 1$. This holds in case of $n = 1$. Suppose it holds for some $n$. Then $x_{n+1} = \\sqrt{n+1+x_n} < \\sqrt{n... | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 45 | |
0i9z | Problem:
In how many ways can 3 bottles of ketchup and 7 bottles of mustard be arranged in a row so that no bottle of ketchup is immediately between two bottles of mustard? (The bottles of ketchup are mutually indistinguishable, as are the bottles of mustard.) | [
"Solution:\n22\nConsider the blocks of consecutive bottles of ketchup in such an arrangement. A block of just one bottle must occur at the beginning or the end of the row, or else it would be between two bottles of mustard. However, a block of two or three bottles can occur anywhere. We cannot have three blocks of ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 22 | |
0d2f | The base-7 representation of number $n$ is $\overline{a b c}_{(7)}$, and the base-9 representation of number $n$ is $\overline{c b a}_{(9)}$. What is the decimal (base-10) representation of $n$? | [
"We have $n = c + 7b + 49a = a + 9b + 81c$. This implies that $20(2c - a) = (4a - b)$. Because $0 \\leq a, b, c \\leq 6$, either $4a - b = 2c - a = 0$ or $4a - b = 20$ and $2c - a = 1$.\n\n1. If $4a - b = 2c - a = 0$, then $b = 8c \\leq 6$ and therefore $a = b = c = 0$, that is $n = 0$.\n\n2. If $4a - b = 20$ and $... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 0 or 248 | |
011k | Problem:
Let $ABC$ be a triangle such that $\frac{|BC|}{|AB| - |BC|} = \frac{|AB| + |BC|}{|AC|}$. Determine the ratio $\angle A : \angle C$. | [
"Solution:\nAnswer: $1 : 2$.\n\nDenote $|BC| = a$, $|AC| = b$, $|AB| = c$. The condition $\\frac{a}{c - a} = \\frac{c + a}{b}$ implies $c^2 = a^2 + ab$ and\n$$\n\\frac{c}{a + b} = \\frac{a}{c}.\n$$\nLet $D$ be a point on $AB$ such that $|BD| = \\frac{a}{a + b} \\cdot c$ (see Figure 5). Then\n$$\n\\frac{|BD|}{|BC|} ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1:2 | |
0ajq | A $5 \times 5$ table is called *regular* if each of its cells contains one of four pairwise distinct real numbers, such that each of them occurs exactly once in every $2 \times 2$ subtable. The sum of all numbers of a *regular table* is called the *total sum* of the table. With any four numbers, one constructs all poss... | [
"We will prove that the maximum number of total sums is $60$.\nThe proof is based on the following claim.\n\n**Claim.** In a regular table either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.\n\n**Proof of the Claim.** Indeed, let $R$ be a row containing at least ... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 60 | |
0b2v | Problem:
Alice and Bob are playing a game with dice. They each roll a die six times, and take the sums of the outcomes of their own rolls. The player with the higher sum wins. If both players have the same sum, then nobody wins. Alice's first three rolls are $6$, $5$, and $6$, while Bob's first three rolls are $2$, $1... | [
"Solution:\n\nLet $a_{i}$ denote the value of Alice's $i$th roll, and $b_{i}$ denote the value of Bob's $i$th roll. For Bob to win, the following inequality must hold:\n$$\n6+5+6+a_{4}+a_{5}+a_{6}<2+1+3+b_{4}+b_{5}+b_{6}\n$$\nRearranging yields\n$$\n\\left(a_{4}-1\\right)+\\left(a_{5}-1\\right)+\\left(a_{6}-1\\righ... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 3895 | |
0kxl | Problem:
Betty has a $3 \times 4$ grid of dots. She colors each dot either red or maroon. Compute the number of ways Betty can color the grid such that there is no rectangle whose sides are parallel to the grid lines and whose vertices all have the same color. | [
"Solution:\n\nFirst suppose no $3 \\times 1$ row is all red or all blue. Then each row is either two red and one blue, or two blue and one red. There are $6$ possible configurations of such a row, and as long as no row is repeated, there's no monochromatic rectangle. This gives $6 \\cdot 5 \\cdot 4 \\cdot 3 = 360$ ... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 408 | |
0bmy | Let $A \in \mathcal{M}_5(\mathbb{C})$ be a matrix with $\text{Tr}(A) = 0$ and such that $I_5 - A$ is invertible. Prove that $A^5 \neq I_5$. | [
"Suppose, by way of contradiction, that $A^5 = I_5$. Let $\\lambda \\in \\mathbb{C}$ be an eigenvalue of $A$; then $\\lambda^5$ is an eigenvalue of $A^5$ and we obtain $\\lambda^5 = 1$.\nSince $I_5 - A$ is invertible, it follows that $\\det(I_5 - A) \\neq 0$. Consequently, the eigenvalues of $A$ belong to the set $... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0ihs | Problem:
A nonzero polynomial $f(x)$ with real coefficients has the property that $f(x) = f'(x) f''(x)$. What is the leading coefficient of $f(x)$? | [
"Solution:\nSuppose that the leading term of $f(x)$ is $c x^{n}$, where $c \\neq 0$. Then the leading terms of $f'(x)$ and of $f''(x)$ are $c n x^{n-1}$ and $c n(n-1) x^{n-2}$, respectively, so $c x^{n} = c n x^{n-1} \\cdot c n(n-1) x^{n-2}$, which implies that $n = (n-1) + (n-2)$, or $n = 3$, and $c = c n \\cdot c... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 1/18 | |
04sp | For a trapezoid $ABCD$ ($AB \parallel CD$) it holds $BC = AB + CD$. Prove that
(i) there is a point of a circle with diameter $BC$ on the leg $AD$,
(ii) there is a point of a circle with diameter $AD$ on the leg $BC$. | [
"(i) Let $M$, $N$ be the centers of the legs $BC$, $AD$. We show that the point $N$ lies on the circle with diameter $BC$.\nA well-known identity yields\n$$\nMN = \\frac{AB + CD}{2} = \\frac{1}{2} BC.\n$$\nIt means that the point $N$ has the same distance from the center $M$ of the circle with diameter $BC$ as radi... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
01w0 | The internal bisectors of angles $\angle DAB$ and $\angle BCD$ of a quadrilateral $ABCD$ intersect at the point $X_1$, and the external bisectors of these angles intersect at the point $X_2$. The internal bisectors of angles $\angle ABC$ and $\angle CDA$ intersect at the point $Y_1$, and the external bisectors of these... | [
"Denote by $A_1$ the intersection point of the lines $AX_1$ and $CX_2$, and by $C_1$ denote the intersection point of the lines $AX_2$ and $CX_1$. Let $B_1$ be the intersection point of the lines $BY_1$ and $DY_2$, and let $D_1$ be the intersection point of the lines $BY_2$ and $DY_1$. The lines $A_1A$ and $C_1C$ a... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
0lcy | Given a circle $(O)$ with a fixed chord $BC$ ($BC$ is not a diameter of the circle). Let $A$ move on the bigger arc $BC$ such that $ABC$ is an acute triangle and $AB < AC$. Let $I$ and $H$ respectively be the midpoint of $BC$ and the orthocenter of the triangle $ABC$. The ray $IH$ intersects the circle $(O)$ again at $... | [
"a) We rewrite the first part of the problem as following. Let the acute triangle $ABC$ inscribed in the circle $(O)$. $AD$ is the altitude and $H$ is orthocenter of the triangle $ABC$. $M$ is the midpoint of $BC$. The circle with the diameter $AH$ cuts $(O)$ again at $G$. $GD$ cuts $(O)$ again at $K$. The straight... | Vietnam | IMO 2015 Team Selection Tests | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0jw7 | Problem:
Let $Q(x) = a_{0} + a_{1} x + \cdots + a_{n} x^{n}$ be a polynomial with integer coefficients, and $0 \leq a_{i} < 3$ for all $0 \leq i \leq n$.
Given that $Q(\sqrt{3}) = 20 + 17 \sqrt{3}$, compute $Q(2)$. | [
"Solution:\n\nOne can evaluate\n$$\nQ(\\sqrt{3}) = \\left(a_{0} + 3 a_{2} + 3^{2} a_{4} + \\ldots\\right) + \\left(a_{1} + 3 a_{3} + 3^{2} a_{5} + \\ldots\\right) \\sqrt{3}\n$$\nTherefore, we have that\n$$\n\\left(a_{0} + 3 a_{2} + 3^{2} a_{4} + \\ldots\\right) = 20 \\text{ and } \\left(a_{1} + 3 a_{3} + 3^{2} a_{5... | United States | February 2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | null | proof and answer | 86 | |
01q4 | The graph of the parabola $y = x^2$ is drawn on the Cartesian plane $Oxy$. A triangle $ABC$ is inscribed in the parabola so that its side $AB$ is parallel to the $Ox$ axis and point $C$ lies between the line $AB$ and the $Ox$ axis. It is known that the length of the side $AB$ is 1 less than the length of the altitude $... | [
"Answer: $\\angle ACB = 45^\\circ$.\nLet $A(a; a^2)$ and $C(c; c^2)$ (see the Fig.). Since $AB \\parallel Ox$, we see that $A$ and $B$ are symmetric with respect to the axis $Oy$. Then their ordinates are equal and abscissae differ by sign.\nHence $B(-a; a^2)$. Since $H$ belongs to $AB$ and $CH \\parallel Oy$, we h... | Belarus | Final Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 45 degrees | |
04k3 | Determine the sum
$$
\frac{1}{2\sqrt{1} + 1\sqrt{2}} + \frac{1}{3\sqrt{2} + 2\sqrt{3}} + \dots + \frac{1}{100\sqrt{99} + 99\sqrt{100}}
$$ | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 9/10 | |
0asy | Problem:
Let $d$ and $n$ be integers such that $9 n + 2$ and $5 n + 4$ are both divisible by $d$. What is the largest possible value of $d$? | [
"Solution:\n\n26"
] | Philippines | Philippines Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 26 | |
071l | Problem:
If $a$, $b$, $c$, $x$ are real numbers such that $a b c \neq 0$ and
$$
\frac{x b + (1-x) c}{a} = \frac{x c + (1-x) a}{b} = \frac{x a + (1-x) b}{c}
$$
then prove that either $a + b + c = 0$ or $a = b = c$. | [
"Solution:\nSuppose $a + b + c \\neq 0$ and let the common value be $\\lambda$. Then\n$$\n\\lambda = \\frac{x b + (1-x) c + x c + (1-x) a + x a + (1-x) b}{a + b + c} = 1\n$$\nWe get two equations:\n$$\n-a + x b + (1-x) c = 0, \\quad (1-x) a - b + x c = 0\n$$\n(The other equation is a linear combination of these two... | India | INMO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0hqu | Problem:
Let $E$ be an ellipse that is not a circle. For which $n \geq 3$ is it possible to inscribe a regular $n$-gon in $E$? (For each $n$, either show how to construct such an $n$-gon or prove that none exists.) | [
"Solution:\nWe claim $n=3,4$ are the only solutions. To see that $3$ is possible, let $P$ be the endpoint of one of the axes, and draw two lines at angles of $\\pi / 6$ to that axis through $P$. By symmetry, these lines intersect the ellipse at points $Q, R$ equidistant from $P$, so $PQR$ is an equilateral triangle... | United States | Berkeley Math Circle Take-Home Contest #2 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | n = 3 and n = 4 | |
0cx8 | Find all pairs $(x, y)$ of positive integers such that
$$
x^{2}+y^{2}+33^{2}=2010 \sqrt{x-y}
$$ | [
"From $3 \\mid x^{2}+y^{2}$ it follows $3 \\mid x$ and $3 \\mid y$, hence $x=3u$ and $y=3v$, for some positive integers $u$ and $v$. Replace in the equation and get $3\\left(u^{2}+v^{2}+11^{2}\\right)=670 \\sqrt{3(u-v)}$, hence $u-v=3k^{2}$, for some positive integer $k$, i.e. $u^{2}+v^{2}+11^{2}=670k$. We have $u^... | Saudi Arabia | SAMC | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (54, 45) | |
0jwo | Problem:
Compute $\frac{x}{w}$ if $w \neq 0$ and $\frac{x+6y-3z}{-3x+4w}=\frac{-2y+z}{x-w}=\frac{2}{3}$. | [
"Solution:\nWe have $x+6y-3z=\\frac{2}{3}(-3x+4w)$ and $-2y+z=\\frac{2}{3}(x-w)$, so\n$$\n\\frac{x}{w}=\\frac{(x+6y-3z)+3(-2y+z)}{(-3x+4w)+3(x-w)}=\\frac{\\frac{2}{3}(-3x+4w)+3 \\cdot \\frac{2}{3}(x-w)}{(-3x+4w)+3(x-w)}=\\frac{\\frac{2}{3}[(-3x+4w)+3(x-w)]}{(-3x+4w)+3(x-w)}=\\frac{2}{3}\n$$"
] | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 2/3 | |
046v | (1) Prove: on the complex plane, for each line passing through the origin except for the real axis, there is at most one point $z$ such that $\frac{1+z^{23}}{z^{64}}$ is a real number.
(2) Prove: for any complex number $a \ne 0$ and any real number $\theta$, the equation
$$
1 + z^{23} + a z^{64} = 0
$$
has at least on... | [
"(1) For a non-zero complex number $z$, let $z = r(\\cos\\theta + i\\sin\\theta)$ where $r > 0$ and $\\theta \\in [0, 2\\pi)$, we have\n$$\nf(z) := \\frac{1+z^{23}}{z^{64}} = \\frac{1}{r^{64}} (\\cos(-64\\theta) + i\\sin(-64\\theta) + r^{23}(\\cos(-41\\theta) + i\\sin(-41\\theta))).\n$$\nTherefore,\n$$\n\\begin{ali... | China | 2023 Chinese IMO National Team Selection Test | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
060o | Problem:
Déterminer tous les entiers $n$ ayant la propriété suivante : si l'on pose $a_{k} = \operatorname{ppcm}(k, k+1, \ldots, k+n-1)$, alors la suite $(a_{k})$ est croissante. | [
"Solution:\n\nMontrons que les naturels ayant cette propriété sont exactement $1$ et $2$.\n\nDéjà, si $n=1$, $a_{k}=k$ est une suite croissante.\n\nEt si $n=2$, puisque deux entiers consécutifs sont premiers entre eux, $a_{k}=k(k+1)$ est une suite croissante également.\n\nSi $n=3$, on remarque que $a_{5}=210 > a_{6... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 1 and 2 | |
0139 | Problem:
It is known that $n$ is a positive integer, $n \leq 144$. Ten questions of type "Is $n$ smaller than $a$?" are allowed. Answers are given with a delay: The answer to the $i$'th question is given only after the $(i+1)$'st question is asked, $i=1,2, \ldots, 9$. The answer to the tenth question is given immediat... | [
"Solution:\n\nLet the Fibonacci numbers be denoted $F_{0}=1$, $F_{1}=2$, $F_{2}=3$ etc. Then $F_{10}=144$. We will prove by induction on $k$ that using $k$ questions subject to the conditions of the problem, it is possible to determine any positive integer $n \\leq F_{k}$.\n\nFirst, for $k=0$ it is trivial, since w... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0b1i | Problem:
Compute the sum of all possible distinct values of $m+n$ if $m$ and $n$ are positive integers such that
$$
\operatorname{lcm}(m, n)+\operatorname{gcd}(m, n)=2(m+n)+11
$$ | [
"Solution:\n\nLet $d=\\operatorname{gcd}(m, n)$. We consider the following cases:\n\na. Suppose $d=1$. Then the equation becomes $m n+1=2 m+2 n+11$ or $(m-2)(n-2)=14$. As $m$ and $n$ are coprime in this case, exactly one of them is even. This implies that exactly one of the factors $m-2$ and $n-2$ of $14$ is even. ... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 32 |
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