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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
05fd | Problem:
Soit $p$ un nombre premier impair, $h < p$ un entier, $e \in \{1,2\}$.
On pose $n = h \cdot p^{e} + 1$ et on suppose que :
$$
\left\{\begin{array}{r}
n \mid 2^{n-1} - 1 \\
n \nmid 2^{h} - 1
\end{array}\right.
$$
Montrer que $n$ est premier. | [
"Solution:\n\nSoit $\\omega$ l'ordre de $2$ modulo $n$. Par hypothèse, $\\omega$ divise $n-1$ mais $\\omega$ ne divise pas $h = \\frac{n-1}{p^{e}}$, donc $p$ divise $\\omega$.\n\n$\\omega$ divise $\\varphi(n)$ par le théorème d'Euler donc $p$ divise $\\varphi(n)$. Évidemment, $p$ ne divise pas $n$ donc il existe $q... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0e1i | Problem:
Tabela velikosti $8 \times 8$ je razdeljena na 64 kvadratkov. Določi najmanjše število barv, s katerimi moramo pobarvati kvadratke tabele, da bodo vsaki štirje kvadratki, ki tvorijo lik oblike
(lik lahko zrcalimo ali zasukamo),
pobarvani z različnimi barvami. | [
"Solution:\n\nZadostuje 8 barv. Označimo barve s številkami od 1 do 8 in tabelo pobarvajmo kot prikazuje prva slika. Vsak lik predpisane oblike je pobarvan z različnimi barvami, saj najbližja kvadratka enake barve ležita diagonalno z enim vmesnim kvadratkom in obeh hkrati ne moremo pokriti z enim samim likom. \n\nD... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
02ch | Problem:
Uma linha de trem está dividida em 10 trechos pelas estações $A, B, C, D, E, F, G, H, I, J$ e $K$. A distância de $A$ até $K$ é igual a $56~\mathrm{km}$. O trajeto de dois trechos consecutivos é sempre menor ou igual a $12~\mathrm{km}$ e o trajeto de três trechos consecutivos sempre é maior ou igual a $17~\ma... | [
"Solution:\n\na) Podemos escrever $AK = 56$ e $AK = AD + DG + GJ + JK$. Como $AD, DG$ e $GJ \\geq 17$, então $JK \\leq 5$. Daí, para $HK \\geq 17$, devemos ter $HJ \\geq 12$. Mas, sabemos que $HJ \\leq 12$, assim $HJ = 12$. A partir de $HK \\geq 17$ e $HJ = 12$, concluímos $JK \\geq 5$ e a única possibilidade é $JK... | Brazil | null | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) 5 km; b) 22 km; c) 29 km | |
07hv | A $m \times N$ table is filled with $2 \times 2$ and $1 \times 3$ tiles. Different tiles don't have common cells. Prove that the number of ways for choosing a $1 \times 2$ rectangle (vertical or horizontal) so that one of its cells is filled with a $2 \times 2$ tile and one of them is filled with a $1 \times 3$ tile is... | [
"Assume that in this table we use $s \\times 2$ tiles. Each tile has eight segments at its border. Assume that the number of $2 \\times 2$ tiles adjacent to the borders of the table is $b$ and the number of shared segments between $2 \\times 2$ and $1 \\times 3$ tiles is $f$. Assume that there are $g$ segments in t... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0hg2 | Let $F$, $F_A$, $F_B$ and $F_C$ be the Feuerbach points of non-equilateral triangle $ABC$, and point $I$ be its incenter. Prove that lines $AF_A$, $BF_B$, $CF_C$ and $IF$ pass through one point. Feuerbach points are the tangent points of the nine-point circle with incircle (touches internally) and three excircles (touc... | [
"Denote by $O_B$ the center of the excircle that touches $AC$, and by $E$ the center of the nine-point circle (fig. 23). Points $E$, $I$ and $F$, as well as the points $E$, $F_B$ and $O_B$ lie on the same line since $F$ and $F_B$ are the tangent points of the circles with the corresponding centers. Besides, $E \\ne... | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0bdg | Determine the largest integer $r$ satisfying the following condition: Amongst every five 500-element subsets of the set $\{1, 2, \dots, 1000\}$, there exist two sharing at least $r$ elements. | [
"The required integer is $r = 200$. To prove it, we first show that, amongst every five 500-element subsets of $\\{1, 2, \\dots, 1000\\}$, there are two sharing at least 200 elements, and then provide an example of five such every two of which share exactly 200 elements.\n\nThe first part is a special case of the l... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 200 | |
0dnw | Problem:
Нека је $a$ природан број такав да за сваки природан број $n$ број $n^{2} a-1$ има бар један делилац већи од 1 који даје остатак 1 при дељењу са $n$. Доказати да је $a$ потпун квадрат.
(Душан Ђукић) | [
"Solution:\n\nНека је $n^{2} a-1 = (n x_{n} + 1) d_{n}$, где $x_{n}, d_{n} \\in \\mathbb{N}$. Тада је $d_{n} \\equiv -1 \\pmod{n}$, па је\n$$\nn^{2} a-1 = (n x_{n} + 1)(n y_{n} - 1) \\quad \\text{за неке } x_{n}, y_{n} \\in \\mathbb{N}\n$$\nшто се своди на $n a - n x_{n} y_{n} = y_{n} - x_{n} > -x_{n} y_{n}$. Одавд... | Serbia | 11. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
098t | Problem:
Fie $(3+\sqrt{8})^{2023}=a+b \sqrt{8}$, $a, b \in \mathbb{N}$. Arătați că $a$ este produsul a două numere naturale, a căror diferență este egală cu $2$. | [
"Solution:\n\nAvem că\n$$\n(3 \\pm \\sqrt{8})^{2023}=3^{2023} \\pm C_{2023}^{1} \\cdot 3^{2022} \\cdot \\sqrt{8}+C_{2023}^{2} \\cdot 3^{2021} \\cdot \\sqrt{8}^{2} \\pm \\cdots \\pm \\sqrt{8}^{2023}\n$$\nAtunci $(3+\\sqrt{8})^{2023}=a+b \\sqrt{8}$, $a, b \\in \\mathbb{N}$, implică $(3-\\sqrt{8})^{2023}=a-b \\sqrt{8}... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0jr8 | Problem:
Let $S$ be the set of all 3-digit numbers with all digits in the set $\{1,2,3,4,5,6,7\}$ (so in particular, all three digits are nonzero). For how many elements $\overline{a b c}$ of $S$ is it true that at least one of the (not necessarily distinct) "digit cycles"
$$
\overline{a b c}, \overline{b c a}, \overl... | [
"Solution:\n\nAnswer: 127\n\nSince the value of each digit is restricted to $\\{1,2, \\ldots, 7\\}$, there is exactly one digit representative of each residue class modulo $7$.\n\nNote that $7 \\mid \\overline{a b c}$ if and only if $100 a + 10 b + c \\equiv 0 \\pmod{7}$ or equivalently $2a + 3b + c \\equiv 0$. So ... | United States | HMMT February 2015 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 127 | |
09p5 | A triangular fortress has guard towers at each vertex and at the midpoint of each side. A guard stationed at a vertex can defend both adjacent sides, while a guard stationed at a midpoint can defend only that side. Determine the number of ways to assign guards to the six towers so that each side is defended by exactly ... | [] | Mongolia | MMO2025 Round 3 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | If n is even: (2n^3 + 9n^2 + 14n + 8) / 8; if n is odd: (2n^3 + 9n^2 + 14n + 7) / 8. Equivalently, floor((2n^3 + 9n^2 + 14n + 8) / 8). | |
014l | Problem:
Suppose that the real numbers $a_{i} \in [-2,17]$, $i=1,2, \ldots, 59$, satisfy $a_{1}+a_{2}+\cdots+a_{59}=0$. Prove that
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{59}^{2} \leq 2006 .
$$ | [
"Solution:\nFor convenience denote $m=-2$ and $M=17$. Then\n$$\n\\left(a_{i}-\\frac{m+M}{2}\\right)^{2} \\leq \\left(\\frac{M-m}{2}\\right)^{2},\n$$\nbecause $m \\leq a_{i} \\leq M$. So we have\n$$\n\\begin{aligned}\n\\sum_{i=1}^{59}\\left(a_{i}-\\frac{m+M}{2}\\right)^{2} & =\\sum_{i} a_{i}^{2}+59 \\cdot\\left(\\fr... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
054g | Do there exist distinct positive integers $x$ and $y$ such that the number $x+y$ is divisible by $2016$, the number $x-y$ is divisible by $2017$ and the number $xy$ is divisible by $2018$? | [
"For example, numbers $x = 2016 \\cdot 2015 - 2018$ and $y = 2018$ meet the conditions. As $2016 \\cdot 2015 > 4 \\cdot 1009 = 2 \\cdot 2018$ implies $x > y$, they are distinct. The sum $2016 \\cdot 2015$ is divisible by $2016$ and the product is obviously divisible by $2018$. Furthermore, $x - y = 2016 \\cdot 2015... | Estonia | National Olympiad Final Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Yes; for example, x = 2016*2015 − 2018 and y = 2018. | |
04fm | Let $a \le b \le c$ be real numbers. Prove that $c^2 - b^2 + a^2 \ge (c - b + a)^2$. (Santos J. Prob. Seminar) | [
"We need to prove that\n$$\nc^2 - b^2 + a^2 \\ge (c - b + a)^2.\n$$\n\nExpand the right-hand side:\n$$(c - b + a)^2 = (c + a - b)^2 = (c + a)^2 - 2b(c + a) + b^2 = c^2 + 2ac + a^2 - 2bc - 2ab + b^2.$$\n\nSo the inequality becomes:\n$$\nc^2 - b^2 + a^2 \\ge c^2 + 2ac + a^2 - 2bc - 2ab + b^2.\n$$\n\nBring all terms t... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
082d | Problem:
La pianta di una casa ha la forma di una L ottenuta affiancando in modo opportuno quattro quadrati il cui lato misura 10 metri. Le pareti laterali sono tutte alte 10 metri e il tetto della casa ha sei facce partenti dai sei muri laterali e inclinate di $30^{\circ}$ rispetto ad un piano orizzontale.
Determinar... | [
"Solution:\n\nIl sottotetto visto dall'alto ha la forma della figura 1, poichè i punti di incontro delle varie facce del tetto sono equidistanti dagli spigoli della base poichè tutte le facce hanno la stessa inclinazione rispetto all'orizzontale.\nOsserviamo che possiamo trasformare il sottotetto come in figura 2 s... | Italy | Cesenatico Gara Individuale | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2750/(3√3) m^3 | |
0bkp | Prove that if the sides of a triangle are directly proportional to $4$, $5$ and $6$, then its largest angle is twice its smallest angle. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
02xx | Problem:
Sobre uma reta $r$, marcam-se os pontos $A$ e $B$, e sobre uma reta $s$, paralela à $r$, marcam-se os pontos $C$ e $D$, de maneira que $A B C D$ seja um quadrado. Marca-se também o ponto $E$ no segmento $C D$.
a) Qual a razão entre as áreas dos triângulos $A B E$ e $B C D$, se $E$ f... | [
"Solution:\n\na) Seja $2k$ a área do quadrado $A B C D$, então a área do triângulo $A B E$ é igual a $k$ e a área do triângulo $B C D$ também é igual a $k$, portanto, a razão entre as áreas é $1$.\n\nb) Se $\\frac{A_{B F E}}{A_{D F E}}=2$, então $\\frac{B F}{F D}=2$. Se $r \\parallel s$, então $\\triangle A B F \\s... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | a) 1; b) 1 | |
028o | Problem:
Amélia, Bruno, Constância e Denise são 4 amigos que moram em Estados diferentes e se encontram sentados numa mesa quadrada, cada um ocupa um lado da mesa.
- À direita de Amélia está quem mora no Amazonas;
- Em frente à Constância está a pessoa que mora em São Paulo;
- Bruno e Denise estão um ao lado do outro;... | [
"Solution:\n\nBruno ou Amélia (O desafio tem duas soluções)."
] | Brazil | null | [
"Discrete Mathematics > Logic"
] | null | final answer only | Bruno ou Amélia | |
0fjg | Problem:
Se consideran $2002$ segmentos en el plano, tales que la suma de sus longitudes es la unidad. Probar que existe una recta $r$ tal que la suma de las longitudes de las proyecciones de los $2002$ segmentos dados sobre $r$ es menor que $2/3$. | [
"Solution:\n\nCada segmento determina dos vectores de igual módulo y sentido opuesto.\nConsideramos los $2 \\cdot 2002 = 4004$ vectores así obtenidos y los ordenamos por sus direcciones entre $0$ y $2\\pi$ respecto de un sistema de referencia ortonormal arbitrario.\n\nConstruimos ahora un polígono convexo de $4004$... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0585 | Prove that
$$
sin 10^\circ \cdot \cos 20^\circ \cdot \sin 30^\circ \cdot \cos 40^\circ \cdot \sin 50^\circ \cdot \cos 60^\circ \cdot \sin 70^\circ \cdot \cos 80^\circ = \frac{1}{256}.
$$ | [
"As $\\sin 10^\\circ = \\cos 80^\\circ$, $\\sin 30^\\circ = \\cos 60^\\circ$, $\\sin 50^\\circ = \\cos 40^\\circ$, and $\\sin 70^\\circ = \\cos 20^\\circ$, the desired equality is equivalent to\n$$\n(\\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 60^\\circ \\cdot \\cos 80^\\circ)^2 = \\frac{1}{256}.\n$$\nIt is... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0ka8 | Problem:
For dinner, Priya is eating grilled pineapple spears. Each spear is in the shape of the quadrilateral $PINE$, with $PI = 6~\mathrm{cm}$, $IN = 15~\mathrm{cm}$, $NE = 6~\mathrm{cm}$, $EP = 25~\mathrm{cm}$, and $\angle NEP + \angle EPI = 60^\circ$. What is the area of each spear, in $\mathrm{cm}^2$? | [
"Solution:\n\nWe consider a configuration composed of 2 more quadrilaterals congruent to $PINE$. Let them be $P' I' N' E'$, with $E' = P$ and $N' = I$, and $P'' I'' N'' E''$ with $P'' = E$, $E'' = P'$, $N'' = I'$, and $I'' = N$. Notice that this forms an equilateral triangle of side length $25$ since $\\angle P P' ... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 100√3/3 | |
06fs | In a school there are $2008$ students. Students are members of certain committees. A committee has at most $1004$ members and every two students join a common committee.
a. Determine the smallest possible number of committees in the school.
b. If it is further required that the union of any two committees consists of... | [
"a. The smallest number of committees in the school is $6$.\nIf a student joins at most $2$ committees, that student shares a common committee with at most $2(1004 - 1) = 2008 < 2007$ students, which contradicts the assumption. Therefore, each student joins at least $3$ committees. Thus, there are at least\n$$\n\\f... | Hong Kong | null | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 6; b) Yes | |
09f9 | Let $m$ be a positive integer greater than one. Prove that the number $m^4 + 1$ does not have a divisor between $m^2$ and $(m+1)^2$. | [
"Assume that there exists a divisor $n$ of $m^4 + 1$ such that $m^2 < n < (m+1)^2$. Let $n = m^2 + x$ where $1 \\le x \\le 2m$. Then $m^4 + 1 = (m^2 + x)(m^2 - y)$ for some $y \\in \\mathbb{Z}_{\\ge 1}$. Hence $1 + xy = m^2(x - y)$ and it is obvious that $x > y \\ge 1$.\n\n(i) If $x - y = 1$, then $1 + (y+1)y = m^2... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0f6o | Problem:
Call a side or diagonal of a regular $n$-gon a segment. How many colors are required to paint all the segments of a regular $n$-gon, so that each segment has a single color and every two segments with a vertex in common have different colors. | [] | Soviet Union | 19th ASU | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | n−1 if n is even; n if n is odd | |
0cah | Problem:
Fie $N \geq 4$ un număr natural fixat.
Doi jucători, $A$ şi $B$, formează o mulţime ordonată $x_{1}, x_{2}, x_{3}, \ldots$, adăugând alternativ elemente: $A$ alege $x_{1}$ egal cu $1$ sau cu $-1$, apoi $B$ îl adaugă pe $x_{2}$ egal cu $2$ sau cu $-2$, apoi $A$ îl adaugă pe $x_{3}$ egal cu $3$ sau cu $-3$, ş.a... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ Barajul 3 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Player A has a winning strategy if N is odd or divisible by 4; player B has a winning strategy if N is congruent to 2 modulo 4. | |
0agu | Find all the prime numbers $p$ for which there exist positive integers $x$ and $y$ that satisfy the equation
$$
x(y^2 - p) + y(x^2 - p) = 5p
$$ | [
"Given equation is equivalent to\n$$\n(x + y)(xy - p) = 5p.\n$$\nWe consider the following cases:\n1. Let $x + y = 1$ and $xy = 6p$. For prime $p \\geq 2$ the equation $x^2 - x + 6p = 0$ has no solutions.\n\n2. Let $x + y = 5$ and $xy = 2p$. For prime $p \\geq 2$ the equation $x - 5x + 2p = 0$ has the determinant $... | North Macedonia | XV-th Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | {2, 3, 7} | |
03wn | Suppose that $k$, $l$ are two positive integers.
Prove that:
There are infinitely many positive integers $m \ge k$ such that $\binom{m}{k}$ and $l$ are relatively prime. | [
"Let $m = k + t \\times l \\times (k!)$, where $t$ is any positive integer. To prove that $\\binom{m}{k}$ and $l$ are relatively prime, we only need to prove that for any prime factor $p$ of $l$, $p \\nmid \\binom{m}{k}$.\n\nIf $p \\nmid k!$, we have\n$$\n\\begin{aligned}\nk! \\binom{m}{k} &= \\prod_{i=1}^{k} (m-k+... | China | China Mathematical Competition (Complementary Test) | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
03xz | The code setting of a cipher lock is established on an $n$-regular-polygon with vertices $A_1, A_2, \dots, A_n$: each vertex is assigned a number (0 or 1) and a color (red or blue), such that either the numbers or the colors on each pair of adjacent vertices are the same. We ask: How many code-sets can be realized for ... | [
"Given an arbitrary code-set for the lock, if two adjacent vertices have different numbers, we label the sides linking them by letter $a$; if they have different colors, we label it by $b$; if both the numbers and colors are the same, we label it by $c$. Once the number and color on vertex $A_1$ are set (there are ... | China | China Mathematical Competition (Complementary Test) | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | Odd n: 3^n + 1; Even n: 3^n + 3 | |
033s | Problem:
The bisectors of $\angle A$, $\angle B$ and $\angle C$ of $\triangle ABC$ meet its circumcircle at points $A_1$, $B_1$ and $C_1$, respectively. Set $AA_1 \cap CC_1 = I$, $AA_1 \cap BC = N$ and $BB_1 \cap A_1C_1 = P$. Denote by $O$ the circumcenter of $\triangle IPC_1$ and let $OP \cap BC = M$. If $BM = MN$ an... | [
"Solution:\n\nSet $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$ and $\\angle BCA = \\gamma$. Then\n$$\n\\begin{aligned}\n\\angle IPC_1 & = \\frac{1}{2}\\left(\\overparen{BA_1} + \\overparen{C_1B_1}\\right) \\\\\n& = \\frac{1}{2}\\left(\\overparen{BA_1} + \\overparen{AC_1} + \\overparen{AB_1}\\right) \\\\\n& = \\f... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | ∠A = 72°, ∠B = 36°, ∠C = 72° | |
0cra | Let $\Omega$ and $O$ be the circumcircle and the circumcenter of a triangle $ABC$. A circle with diameter $AO$ intersects the circumcircle of the triangle $OBC$ at point $S \neq O$. The tangents to $\Omega$ at points $B$ and $C$ meet at point $P$. Prove that the points $A$, $S$, and $P$ are collinear. (R. Sadykov) | [
"Поскольку $CP$ и $BP$ — касательные к $\\Omega$, имеем $\\angle OBP = \\angle OCP = 90^\\circ$; значит, точка $P$ лежит на описанной окружности треугольника $OBC$, и $PO$ — диаметр этой окружности (см. рис. 4). Поэтому $\\angle OSP = 90^\\circ$.\n\nДалее, поскольку $AO$ — диаметр окружности, проходящей через $A$, ... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dp4 | In triangle $ABC$ let $A_0, B_0, C_0$ are the midpoints of the sides $BC, CA, AB$, respectively, and $A_1, B_1, C_1$ are the midpoints (by length) of the broken lines $BAC, CBA, BCA$, respectively. Prove that the lines $A_0A_1, B_0B_1, C_0C_1$ are concurrent. | [
"\n\nW.l.o.g. assume that $AC \\le AB$. Then $A_1$ belongs to the segment $AB$, moreover, it lies between $A$ and $C_0$. Then notice that $AC + AA_1 = BA_1 \\iff AC + AC_0 - A_1C_0 = C_0B + A_1C_0 \\iff A_1C_0 = \\frac{1}{2}AC \\Rightarrow A_1C_0 = A_0C_0$ i.e. $\\angle C_0A_1A_0 = \\angle ... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0is8 | Problem:
Positive real numbers $x$, $y$ satisfy the equations $x^{2} + y^{2} = 1$ and $x^{4} + y^{4} = \frac{17}{18}$. Find $x y$. | [
"Solution:\n\nLet $s = x^2 + y^2 = 1$ and $p = x^2 y^2$.\n\nWe have:\n\n$$\nx^4 + y^4 = (x^2 + y^2)^2 - 2x^2 y^2 = s^2 - 2p\n$$\n\nGiven $x^4 + y^4 = \\frac{17}{18}$, so:\n\n$$\ns^2 - 2p = \\frac{17}{18}\n$$\n\nBut $s = 1$, so:\n\n$$\n1 - 2p = \\frac{17}{18}\n$$\n$$\n2p = 1 - \\frac{17}{18} = \\frac{1}{18}\n$$\n$$\... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 1/6 | |
0glr | Let $H$ be the orthocenter of an acute triangle $ABC$. The circumcircle of $\triangle BCH$ intersects $AB$ and $AC$ again at points $A_1$ and $A_2$ respectively. Define points $B_1, B_2, C_1$ and $C_2$ analogously. Prove that the circumcenter of the triangle formed by lines $A_1A_2, B_1B_2$, and $C_1C_2$ is on the Eule... | [
"Let $D, E$, and $F$ be the feet of the altitudes from $A, B$, and $C$, respectively with respect to $\\triangle ABC$. Let $B_1B_2$ intersect $C_1C_2$ at $K$, $C_1C_2$ intersects $A_1A_2$ at $L$, and $A_1A_2$ intersects $B_1B_2$ at $M$. Let $O$ be the circumcenter of $\\triangle ABC$.\n---\n\nWe have\n$$\n\\begin{a... | Thailand | T3MO 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00a9 | Find all primes $p$ such that $p^3 - 4p + 9$ is a perfect square. | [
"We check directly that $p=2$ is a solution $(2^3 - 4 \\cdot 2 + 9 = 3^2)$, and $p=3$ is not. Henceforth let $p > 3$. If $p^3 - 4p + 9 = n^2$ for some $n \\in \\mathbb{N}$ then $p^3 - 4p = n^2 - 9$,\n$$(p-2)p(p+2) = (n-3)(n+3).$$\nOne of the numbers $p-2$, $p$, $p+2$ is divisible by $3$, hence $n$ is also a multipl... | Argentina | Argentine National Olympiad 2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | p = 2, 7, 11 | |
0ds1 | Determine the minimum number of lines that can be drawn on the plane so that they intersect in exactly $200$ distinct points.
(Note that for $3$ distinct points, the minimum number of lines is $3$ and for $4$ distinct points, the minimum is $4$.) | [
"Let $m$ be the integer so that $\\left(\\frac{m}{2}\\right) < n \\le \\left(\\frac{m+1}{2}\\right)$. Then since $m$ lines intersect in at most $\\left(\\frac{m}{2}\\right)$ points, we have $n > m$. We shall show that there exist $m + 1$ lines that intersect in exactly $n$ points. Let $p = n - \\left(\\frac{m}{2}\\... | Singapore | Singapur | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 21 | |
09me | Let $a, b, c \ge 0$ be non-negative numbers satisfying $a^3 + b^3 + c^3 = abc + 2$. Prove that
$$
a^4 + b^4 + c^4 \ge a + b + c.
$$ | [
"It suffices to prove\n$$\nF(a, b, c) = 2(a^4 + b^4 + c^4) - (a + b + c)(a^3 + b^3 + c^3 - abc) \\ge 0\n$$\nfor all $a, b, c \\ge 0$. We may assume $b \\ge a$ and $c \\ge a$ and then we have\n$$\nF(a, b, c) = (b-c)^2(b(b-a) + bc + c(c-a)) + a^2(c-a)(b-a) \\ge 0.\n$$\nEquality holds for $(a, b, c) = (0, 1, 1), (1, 0... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
09us | A square with area $4$ is divided into two grey and two transparent squares, each having an area of size $1$; see the left figure. Another such square is put on top of this square. The side of the second square is lying exactly on the middle of the diagonal of the first square; see the right figure.
![](attached_image_... | [
"$3\\frac{1}{2}$"
] | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | final answer only | 7/2 | |
06q9 | Let $a_{1}, a_{2}, \ldots, a_{n}$ be distinct positive integers, $n \geq 3$. Prove that there exist distinct indices $i$ and $j$ such that $a_{i}+a_{j}$ does not divide any of the numbers $3 a_{1}, 3 a_{2}, \ldots, 3 a_{n}$. | [
"Without loss of generality, let $0<a_{1}<a_{2}<\\cdots<a_{n}$. One can also assume that $a_{1}, a_{2}, \\ldots, a_{n}$ are coprime. Otherwise division by their greatest common divisor reduces the question to the new sequence whose terms are coprime integers.\n\nSuppose that the claim is false. Then for each $i<n$ ... | IMO | 49th International Mathematical Olympiad Spain | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof only | null | |
0kp8 | Problem:
Compute the number of positive real numbers $x$ that satisfy
$$
\left(3 \cdot 2^{\left\lfloor\log _{2} x\right\rfloor}-x\right)^{16}=2022 x^{13}
$$ | [
"Solution:\nLet $f(x)=3 \\cdot 2^{\\left\\lfloor\\log _{2} x\\right\\rfloor}-x$. Note that for each integer $i$, if $x \\in\\left[2^{i}, 2^{i+1}\\right)$, then $f(x)=3 \\cdot 2^{i}-x$. This is a line segment from $\\left(2^{i}, 2^{i+1}\\right)$ to $\\left(2^{i+1}, 2^{i}\\right)$, including the first endpoint but no... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 9 | |
0h9s | There are several gentlemen in the club.
Every two are either friends, or enemies. It is known that each of the gentlemen has exactly $4$ enemies. In addition, for each of them, the enemy of his friend is his enemy. How many gentlemen can be present at the club? | [
"Note that the condition implies that each of the gentlemen has equal number of friends. Let $n$ denote the number of gentlemen in the club. Each of them has exactly $4$ enemies, therefore exactly $n-5$ friends.\n\nConsider a single gentleman $A_1$. Let $B_1, B_2, B_3, B_4$ denote his enemies. Since each friend of ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 5, 6, 8 | |
0gh8 | 設 $Q$ 為一個由若干個質數所成的集合 (不必然為有限集)。對於一個正整數 $n$，考慮其質因數分解，並定義 $p(n)$ 為這個分解中的指數和，而 $q(n)$ 為這個分解中在 $Q$ 中質數的指數和。若 $n$ 為一整數，且 $p(n)+p(n+1)$ 和 $q(n)+q(n+1)$ 都是偶數，則稱 $n$ 是特別的。證明存在一個與 $Q$ 無關的常數 $c > 0$，使得對於任意正整數 $N > 100$，在 $[1, N]$ 的特別整數至少有 $cN$ 個.
(舉例來說，若 $Q = \{3, 7\}$，則 $p(42) = 3$，$q(42) = 2$，$p(315) = 4$，$q(315) = 3$)
Let $Q... | [
"**Fact 1:** For any 5 integers, there are at least 2 of them that have same parity for both $p$ and $q$ by the pigeonhole principle.\n\n**Fact 2:** $p(m) + p(n) \\equiv p(m/d) + q(m/d) \\pmod{2}$ if $d \\mid \\gcd(m, n)$.\n\nConsider the set\n$$\nA_k = \\{72k, 72k + 6, 72k + 8, 72k + 9, 72k + 12\\}\n$$\nBy Fact 1,... | Taiwan | 2023 數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Chinese (Traditional) | proof only | null | |
092n | Problem:
Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $f(a)+f(b)$ divides $2(a+b-1)$ for all $a, b \in \mathbb{N}$. | [
"Solution:\nWe will first prove that $f$ is either injective or bounded.\nAssume that we have $f(m)=f(n)=t$ for some $m$ and $n$. If we plug in $m$ and $n$ as $b$, we get respectively:\n$$\n\\begin{aligned}\n& f(a)+t \\mid 2(a+m-1) \\\\\n& f(a)+t \\mid 2(a+n-1)\n\\end{aligned}\n$$\nSince the divisor of two numbers ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | f(n) = 1 for all n, and f(n) = 2n - 1 for all n | |
08cf | Problem:
Luca scrive su una lavagna tutte le possibili sequenze costituite da 2017 interi positivi distinti la cui somma è $2016 \cdot 2017 \cdot 2018$. Fatto ciò, sostituisce ognuna di tali sequenze con il massimo comun divisore dei suoi elementi. Quando questa lunga operazione è terminata, quanto vale il massimo dei... | [
"Solution:\n\nLa risposta è (B). Consideriamo una sequenza $b_{1}, b_{2}, \\ldots, b_{2017}$ di numeri interi positivi tutti distinti la cui somma sia $2016 \\cdot 2017 \\cdot 2018$. Sia $d$ il massimo comun divisore degli elementi di questa sequenza. Per definizione, i numeri $a_{1}:=b_{1} / d, a_{2}:=b_{2} / d, \... | Italy | Gara di Febbraio | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | B | |
038k | The sequence $\{a_i\}_{i=1}^\infty$ is such that $a_1 > 0$ and $a_{n+1} = \frac{a_n}{1+a_n^2}$ for $n \ge 1$. Prove that:
a) $a_n \le \frac{1}{\sqrt{2n}}$ for $n \ge 2$;
b) there exists $n$ such that $a_n > \frac{7}{10\sqrt{n}}$. | [] | Bulgaria | First selection test for IMO 2007, Vietnam | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0dhf | Find all pairs of integers $(m, n)$ such that $\binom{n}{m} = 1984$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | (m, n) = (1, 1984) and (1983, 1984) | |
0gop | Find the number of partitions of $\{1, 2, \dots, 2012\}$ into two sets such that none of the sets contains two distinct elements whose sum is a power of $2$. | [
"**The answer is $1024$.**\nLet us call a partition of $\\{1, 2, \\dots, n\\}$ into two sets *nice partition for $n$*, if none of the sets contains two distinct elements whose sum is a power of $2$. Let $p_n$ be the number of nice partitions for $n$. We observe that removing $n$ from a nice partition for $n$ gives ... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 1024 | |
0dwd | Problem:
Če zmnožek treh zaporednih naravnih števil $n-1$, $n$ in $n+1$ povečamo za srednje število, dobimo število med 3000 in 4000. Določi ta števila. | [
"Solution:\n\nNajprej ugotovimo, da je $(n-1) n (n+1) + n = n^{3}$.\n\nZapišemo neenačbo $3000 < n^{3} < 4000$.\n\nSklepamo, da je $\\sqrt[3]{3000} < n$, oziroma $14,42 < n$, in da je $n^{3} < 4000$, oziroma $n < 15,87$.\n\nTako je $n = 15$.\n\nIskana zaporedna naravna števila so $14$, $15$ in $16$."
] | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 14, 15, 16 | |
0kee | Problem:
Compute the sum of all positive integers $a \leq 26$ for which there exist integers $b$ and $c$ such that $a+23 b+15 c-2$ and $2 a+5 b+14 c-8$ are both multiples of $26$. | [
"Solution:\n\nAssume $b$ and $c$ exist. Considering the two values modulo $13$, we find\n$$\n\\begin{cases}a+10 b+2 c \\equiv 2 & (\\bmod\\ 13) \\\\ 2 a+5 b+c \\equiv 8 & (\\bmod\\ 13)\\end{cases}\n$$\nSubtracting twice the second equation from the first, we get $-3 a \\equiv -14\\ (\\bmod\\ 13)$. So, we have $a \\... | United States | HMMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 31 | |
08lj | Problem:
Let $ABC$ be a triangle, $(BC < AB)$. The line $\ell$ passing through the vertex $C$ and orthogonal to the angle bisector $BE$ of $\angle B$, meets $BE$ and the median $BD$ of the side $AC$ at points $F$ and $G$, respectively. Prove that segment $DF$ bisects the segment $EG$.
 | [
"Solution:\nLet $CF \\cap AB = \\{K\\}$ and $DF \\cap BC = \\{M\\}$. Since $BF \\perp KC$ and $BF$ is angle bisector of $\\varangle KBC$, we have that $\\triangle KBC$ is isosceles, i.e. $BK = BC$, also $F$ is midpoint of $KC$. Hence $DF$ is midline for $\\triangle ACK$, i.e. $DF \\parallel AK$, from where it is cl... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
076t | Let $a$, $b$, $c$ be distinct positive real numbers such that $abc = 1$. Prove that
$$
\sum_{\text{cyclic}} \frac{a^6}{(a-b)(a-c)} > 15.
$$ | [
"Let us consider a cubic polynomial whose roots are $a$, $b$, $c$. We get $P(x) = x^3 - px^2 + qx - r$, where $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$. We observe that\n$$\n\\frac{a^n}{(a-b)(a-c)} + \\frac{b^n}{(b-c)(b-a)} + \\frac{c^n}{(c-a)(c-b)} = \\sum_{\\text{cyclic}} \\frac{-a^n(b-c)}{(a-b)(b-c)(c-a)... | India | India_2017 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ger | Let $S$ be a set of positive integers such that for every $a, b \in S$, there always exists $c \in S$ such that $c^2$ divides $a(a+b)$. Show that there exists an $a \in S$ such that $a$ divides every element of $S$.
設 $S$ 為一個正整數的非空子集, 其中對於任意 $a, b \in S$, 都可以找到一個 $c \in S$ 使得 $c^2$ 整除 $a(a+b)$. 證明存在 $a \in S$, 使得 $a$ ... | [
"Let $a$ be the minimum element in $S$. We will show that this $a$ satisfies the condition. Towards contradiction, suppose $b \\in S$ be the minimum element in $S$ such that $a \\nmid b$. Define $b_0 = b$ and let $b_{i+1}$ be an element in $S$ so that $b_{i+1}^2 \\mid a(a+b_i)$ for $i \\ge 0$.\nNote that if $b_{i+1... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 獨立研究 (一) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theor... | null | proof only | null | |
0lcf | Let be given an alphabet consisting of $29$ characters. A sequence of characters is called word. For each positive integer $n$, let $X_n$ be the set all words with $n$ characters. Consider function $f: X_n \to X_2$ defined as follow: for each word from $X_n$, we erase any $n-2$ characters in this word to get a word fro... | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 421; b) 281 | |
0ggi | 令 $n$ 與 $k$ 為滿足 $n > k \ge 1$ 的正整數。有 $2n+1$ 位學生站成一圈。對於每位學生, 他的 $2k$ 位鄰居, 指的是他左手邊離他最近的 $k$ 位同學與右手邊離他最近的 $k$ 位同學。
已知學生中恰有 $n+1$ 位女生。證明：存在一個女生, 她的 $2k$ 位鄰居中有至少 $k$ 位女生。 | [
"**解. 考慮**\n$$\na_m = \\begin{cases} 1, & \\text{第 } t \\text{ 位學生是女生, 其中 } t \\equiv m \\pmod{2n+1}, \\\\ 0, & \\text{otherwise.} \\end{cases}\n$$\n考慮 $b_i = a_i + a_{i-k-1} - 1 \\in \\{-1, 0, 1\\}$。易知對於所有 $m$,\n$$\nb_{m+1} + \\cdots + b_{m+2n+1} = 2(a_1 + \\cdots + a_{2n+1}) - (2n+1) = 2(n+1) - (2n+1) = 1. \\quad... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 獨立研究(一) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Chinese; English | proof only | null | |
0iwm | Problem:
Von Neumann's Poker: The first step in Von Neumann's game is selecting a random number on $[0,1]$. To generate this number, Chebby uses the factorial base: the number $0 . A_{1} A_{2} A_{3} A_{4} \ldots$ stands for $\sum_{n=0}^{\infty} \frac{A_{n}}{(n+1)!}$, where each $A_{n}$ is an integer between $0$ and $n... | [] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2√e - e | |
0g30 | Problem:
Sei $k$ ein Kreis mit Mittelpunkt $O$ und seien $A$, $B$ und $C$ drei Punkte auf $k$ mit $\angle ABC > 90^{\circ}$. Die Winkelhalbierende von $\angle AOB$ schneide den Umkreis des Dreiecks $BOC$ ein zweites Mal in $D$. Zeige, dass $D$ auf der Geraden $AC$ liegt. | [
"Solution:\n\n\n\nSei $\\angle ACB = \\alpha$. Wir wollen zeigen, dass auch $\\angle DCB = \\alpha$ gilt, dann liegen $A$, $D$ und $C$ auf einer Geraden.\n\nAufgrund des Zentriwinkelsatzes über der Sehne $AB$ gilt $\\angle AOB = 2 \\angle ACB = 2\\alpha$. Weil $OD$ die Winkelhalbierende des... | Switzerland | Vorrunde 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0780 | Let $f, g$ be functions $\mathbb{R} \to \mathbb{R}$ such that
$$
f(g(x) + y) = g(x + y) \quad \forall x, y \in \mathbb{R}
$$
then either $f$ is the identity¹ function or $g$ is periodic². | [] | India | EGMO TST Day 2 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
0ajf | Let $H$ be the orthocentre and $G$ be the centroid of acute-angled triangle $\triangle ABC$ with $AB \neq AC$. The line $AG$ intersects the circumcircle of $\triangle ABC$ at $A$ and $P$. Let $P'$ be the reflection of $P$ in the line $BC$. Prove that $\angle CAB = 60^\circ$ if and only if $HG = GP'$. | [
"Let $\\omega$ be the circumcircle of $\\triangle ABC$. Reflecting $\\omega$ in line $BC$, we obtain circle $\\omega'$ which, obviously, contains points $H$ and $P'$. Let $M$ be the midpoint of $BC$. As triangle $\\triangle ABC$ is acute-angled, then $H$ and $O$ lie inside this triangle.\n\nLet us assume that $\\an... | North Macedonia | Girls European 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 > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geom... | English | proof only | null | |
05sf | Problem:
Clara et Isabelle jouent au jeu suivant. Au début du jeu, elles choisissent un entier $n \geqslant 1$, puis mettent $n$ bonbons dans un saladier. Puis elles jouent à tour de rôle, en commençant par Clara. À chaque tour, si le saladier contient $k$ bonbons, la joueuse dont c'est le tour choisit un entier $\ell... | [
"Solution:\n\nTout d'abord, si $n$ est impair, Clara peut commencer par choisir $\\ell = n-2$. En effet, puisque $n$ est impair, on sait que $\\operatorname{PGCD}(n, n-2) = \\operatorname{PGCD}(n, 2) = 1$. Mais alors Isabelle se retrouve avec deux bonbons, elle est obligée d'en manger un seul, et laisse le dernier ... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | all odd positive integers n | |
0gb7 | 三角形 $ABC$ 中, $AB = AC \ne BC$, 而點 $I$ 為其內心。直線 $BI$ 交 $AC$ 於點 $D$。已知過 $D$ 且與 $AC$ 垂直的直線交 $AI$ 於點 $E$。證明: $I$ 對直線 $AC$ 的反射點落在三角形 $BDE$ 的外接圓上。 | [
"令 $\\Gamma$ 為以 $E$ 為圓心、過 $B$, $C$ 兩點的圓。因為 $DE \\perp AC$, $C$ 對 $D$ 的反射點 $F$ 會落在 $\\Gamma$ 上。由 $\\angle DCI = \\angle ICB = \\angle CBI$, 知直線 $DC$ 為三角形 $IBC$ 外接圓的切線。設 $J$ 為 $I$ 對 $D$ 的反射點。使用有向線段, 知\n$$\nDC \\cdot DF = -DC^2 = -DI \\cdot DB = DJ \\cdot DB,\n$$\n得 $J$ 亦落在 $\\Gamma$ 上。\n\n令 $... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j84 | Problem:
Square $ABCD$ is inscribed in circle $\omega$ with radius $10$. Four additional squares are drawn inside $\omega$ but outside $ABCD$ such that the lengths of their diagonals are as large as possible. A sixth square is drawn by connecting the centers of the four aforementioned small squares. Find the area of t... | [
"Solution:\n\nLet $DEGF$ denote the small square that shares a side with $AB$, where $D$ and $E$ lie on $AB$. Let $O$ denote the center of $\\omega$, $K$ denote the midpoint of $FG$, and $H$ denote the center of $DEGF$. The area of the sixth square is $2 \\cdot OH^{2}$.\n\nLet $KF = x$. Since $KF^{2} + OK^{2} = OF^... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 144 | |
0ff2 | Problem:
Determinar todas las ternas de números reales $(a, b, c)$, con $a \neq b$, $a \neq 0$, $b \neq 0$, tales que las parábolas
$$
y = a x^{2} + b x + c, \quad y = b x^{2} + c x + a
$$
tienen el mismo vértice. | [
"Solution:\nSe observa en primer lugar que las parábolas pasan por el punto común $N(1, a + b + c)$.\n\nPrimera solución\nSea $V\\left(x_{0}, y_{0}\\right)$ el vértice de las dos parábolas. El cambio de variable\n$$\nX = x - x_{0}, \\quad Y = y - y_{0}\n$$\ntransforma $V$ en el punto $O'$ que es el origen de coorde... | Spain | TANDA III | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (a, -2a, 4a) with a ≠ 0 | |
06a2 | Find all values of the positive integer $v$ for which there exist triads $(\alpha, \beta, \gamma)$ of positive integers satisfying the equation
$$
\alpha + \beta + \gamma = v\alpha\beta\gamma. \qquad (E)
$$
For these values find all solutions of the equation (E). | [
"Since the equation is symmetric with respect to $\\alpha, \\beta, \\gamma$ we suppose that $\\alpha \\ge \\beta \\ge \\gamma$. Then we have:\n$$\n\\alpha \\le \\alpha + \\beta + \\gamma \\le 3\\alpha \\Leftrightarrow \\alpha \\le v\\alpha\\beta\\gamma \\le 3\\alpha \\Rightarrow 1 \\le v\\beta\\gamma \\le 3.\n$$\n\... | Greece | 37th Hellenic Mathematical Olympiad 2020 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | Valid v are 1, 2, and 3. For v = 3: the only solution is (1, 1, 1). For v = 2: all permutations of (2, 1, 1). For v = 1: all permutations of (3, 2, 1). | |
0dg0 | At a gala banquet, $12n + 6$ chairs, where $n \in \mathbb{N}$, are equally arranged around a large round table. A seating will be called a proper seating of rank $n$ if a gathering of $6n + 3$ married couples sit around this table such that each seated person also has exactly one sibling (brother/sister) of the opposit... | [
"We will call a woman unusual if she sits closer to her husband than her brother. Our goal is to find the smallest possible number of unusual women. Let us call this number $k$. We note that going from each man to his sister and from each woman to her husband we obtain an oriented graph which breaks up into oriente... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 6n | |
09fq | The diagonals $AC$, $BD$ of a convex quadrilateral $ABCD$ intersect at the point $O$. Let $M$, $N$ be the midpoints of the sides $AB$, $CD$ respectively. The point $P$ is chosen so that the quadrilateral $MONP$ is a parallelogram. Prove that the areas of the triangles $APD$, $BPC$ are equal. | [
"Let $Q$ be the intersection of the lines $OP$ and $MN$.\n\n\nThen $S_{BOC} + S_{BPC} = 2S_{BQC} = S_{BMC} + S_{CNB}$. Hence\n$$\n\\begin{align*}\nS_{BPC} &= \\frac{1}{2}S_{ABC} + \\frac{1}{2}S_{BCD} - S_{BOC} \\\\\n&= \\frac{1}{2}(S_{AOB} + S_{COD} + S_{BOC} + S_{COD} - 2S_{BOC}) \\\\\n&= ... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0e0p | a. Find the maximum real number $C$, such that the inequality
$x^2 + y^2 + 1 \ge C(x + y)$
holds for all real $x$ and $y$.
b. Find the maximum real number $C$, such that the inequality
$x^2 + y^2 + xy + 1 \ge C(x + y)$
holds for all real $x$ and $y$. | [
"a.\nFirst, rewrite the inequality as $x^2 - Cx + y^2 - Cy + 1 \\ge 0$ and then form perfect squares:\n$$\n\\left(x - \\frac{C}{2}\\right)^2 - \\frac{C^2}{4} + \\left(y - \\frac{C}{2}\\right)^2 - \\frac{C^2}{4} + 1 \\ge 0.\n$$\nThis implies\n$$\n\\left(x - \\frac{C}{2}\\right)^2 + \\left(y - \\frac{C}{2}\\right)^2 ... | Slovenia | Selection Examinations for the IMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | a) sqrt(2); b) sqrt(3) | |
0flu | With several identical spheres, we build up a tetrahedron using $n$ spheres for each side. Determine, in terms of $n$, the number of tangency points between the spheres in the construction. | [] | Spain | Spanija 2012 | [
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | n^3 - n | |
0jrn | Problem:
Members of a parliament participate in various committees. Each committee consists of at least 2 people, and it is known that every two committees have at least one member in common. Prove that it is possible to give each member a colored hat (hats are available in black, white or red) so that every committee... | [
"Solution:\n\nPick a committee $C$ of smallest size and give one of its members a black hat and the rest of its members a white hat. Give everyone else who is not in this committee a red hat. Committee $C$ contains two colors of hats (black and white) by choice. Any other committee with the exact same members as $C... | United States | BAMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hmo | Problem:
The sum of the digits of all counting numbers less than $13$ is
$$
1+2+3+4+5+6+7+8+9+1+0+1+1+1+2=51
$$
Find the sum of the digits of all counting numbers less than $1000$. | [
"Solution:\nThe counting numbers less than $1000$ are simply the three-digit numbers, as long as we allow numbers to start with $0$ (except the number $000$, but including this does not affect the digit sum).\nEach of the ten digits appears as a hundreds digit $100$ times, because it can be paired with any of the $... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 13500 | |
0iyg | Problem:
Compute
$$
\int_{1}^{\sqrt{3}} x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x
$$ | [
"Solution:\n\nUsing the fact that $x=e^{\\ln (x)}$, we evaluate the integral as follows:\n$$\n\\begin{aligned}\n\\int x^{2 x^{2}+1}+\\ln \\left(x^{2 x^{2 x^{2}+1}}\\right) d x & =\\int x^{2 x^{2}+1}+x^{2 x^{2}+1} \\ln \\left(x^{2}\\right) d x \\\\\n& =\\int e^{\\ln (x)\\left(2 x^{2}+1\\right)}\\left(1+\\ln \\left(x... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | 13 | |
0id9 | Problem:
A rectangle has perimeter $10$ and diagonal $\sqrt{15}$. What is its area? | [
"Solution:\nIf the sides are $x$ and $y$, we have $2x + 2y = 10$, so $x + y = 5$, and $\\sqrt{x^{2} + y^{2}} = \\sqrt{15}$, so $x^{2} + y^{2} = 15$.\n\nSquaring the first equation gives $x^{2} + 2xy + y^{2} = 25$, and subtracting the second equation gives $2xy = 10$, so the area is $xy = 5$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 5 | |
095l | Problem:
Găsiți domeniul de valori $E_{f}$ al funcției $f: \mathbb{R} \setminus \{-1,0,1\} \rightarrow \mathbb{R}$,
$$
f(x)=\operatorname{arctg}\left(\frac{x}{x+1}\right)+\operatorname{arctg}\left(\frac{x}{x-1}\right)-\operatorname{arctg}\left(\frac{1}{2 x^{2}}\right)
$$ | [
"Solution:\nFie $x \\in \\mathbb{R} \\setminus\\{-1,0,1\\}=D_{f}$. Găsim derivata $f^{\\prime}(x)$ pe $D_{f}$. Cum $(\\operatorname{arctg} u)^{\\prime}=\\frac{u^{\\prime}}{1+u^{2}}$, atunci $\\forall x \\in D_{f}$ :\n$$\n\\begin{aligned}\nf^{\\prime}(x) & =\\left(\\operatorname{arctg}\\left(\\frac{x}{x+1}\\right)\\... | Moldova | A 62 - A OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Limits",
"Precalculus > Trigonometric functions"
] | null | proof and answer | {-π/2, π/2} | |
0bgd | Problem:
Fie $m$ şi $n$ numere naturale, $m, n \geq 2$. Considerăm matricele $A_{1}, A_{2}, \ldots, A_{m} \in \mathcal{M}_{n}(\mathbb{R})$, nu toate nilpotente. Demonstraţi că există un număr întreg $k>0$ astfel încât $A_{1}^{k}+A_{2}^{k}+\cdots+A_{m}^{k} \neq O_{n}$.
Notă: Numim nilpotentă o matrice pătratică având ... | [
"Solution:\n\nNotăm cu $\\lambda_{i 1}, \\lambda_{i 2}, \\ldots, \\lambda_{i n}$ valorile proprii ale matricei $A_{i}$, $i=1,2, \\ldots, m$. Presupunem prin absurd că $A_{1}^{k}+A_{2}^{k}+\\cdots+A_{m}^{k}=O_{n}$, oricare ar fi $k \\geq 1$. Atunci $\\operatorname{tr}\\left(A_{1}^{k}\\right)+\\operatorname{tr}\\left... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
05vb | Problem:
Soient $a, b, c$ trois entiers tels que $7$ divise $a^{2} + b^{2} + c^{2}$. Montrer que $7$ divise $a^{4} + b^{4} + c^{4}$. | [
"Solution:\nLes carrés modulo $7$ valent $0, 1, 2, 4$. On cherche donc toutes les sommes de trois de ces nombres étant divisible par $7$. Après avoir testé toutes les possibilités, on s'aperçoit que les seuls ensembles de trois carrés modulo $7$ dont la somme est divisible par $7$ sont $\\{0, 0, 0\\}$ et $\\{1, 2, ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0e80 | Let $ABCD$ be a parallelogram. Let $E$ and $F$ denote such points on the sides $AB$ and $BC$ that satisfy $|AE| = |CF|$. Let $G$ be obtained by reflection of the point $E$ through the point $A$. Denote by $M$ the point of intersection of the lines $AF$ and $EC$, and denote by $N$ the point of intersection of the lines ... | [
"We prove that the lines $DM$ and $DN$ are the interior and the exterior bisector of the angle $\\angle ADC$. Let $X$ be the point of intersection of the lines $AF$ and $CD$.\n\nMenelaus' theorem for the triangle $EBC$ and the collinear points $A, M$ and $F$ says\n$$\n\\frac{|EA| \\ |BF| \\ |CM|}{|AB| \\ |FC| \\ |M... | Slovenia | Selection Examinations for the IMO 2013 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09s4 | Problem:
Laat $m$ en $n$ positieve gehele getallen zijn zodat $5 m+n$ een deler is van $5 n+m$. Bewijs dat $m$ een deler is van $n$. | [
"Solution:\nEr is een positieve gehele $k$ met $(5 m+n) k=5 n+m$. Dus $5 k m-m=5 n-k n$, oftewel $(5 k-1) m=(5-k) n$. De linkerkant is positief, dus de rechterkant ook, waaruit volgt dat $k<5$. Als $k=1$, dan is $4 m=4 n$, dus $m=n$, dus $m \\mid n$. Als $k=2$, dan is $9 m=3 n$, dus $3 m=n$, dus $m \\mid n$. Als $k... | Netherlands | Selectietoets | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0b82 | Let $G$ be a group in which $a^2b = ba^2$ implies $ab = ba$.
a) Show that if $G$ has $2^n$ elements, then $G$ is an abelian group.
b) Exhibit an example of a nonabelian group with the given property. | [
"a) For $a \\in G$, let $C(a) = \\{b \\mid b \\in G$ and $ab = ba\\}$. The hypothesis shows that $C(a^2) \\subset C(a)$. Since $C(a) \\subset C(a^2)$, we see that $C(a) = C(a^2)$, for every $a \\in G$. Consequently, $C(a) = C(a^2) = \\dots = C(a^{2^n}) = C(e) = G$, for every $a \\in G$, that is $G$ is abelian.\n\nb... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
0k04 | Problem:
Kelvin and 15 other frogs are in a meeting, for a total of 16 frogs. During the meeting, each pair of distinct frogs becomes friends with probability $\frac{1}{2}$. Kelvin thinks the situation after the meeting is cool if for each of the 16 frogs, the number of friends they made during the meeting is a multip... | [
"Solution:\n\nConsider the multivariate polynomial\n$$\n\\prod_{1 \\leq i<j \\leq 16}\\left(1+x_{i} x_{j}\\right)\n$$\nWe're going to filter this by summing over all $4^{16}$ 16-tuples $\\left(x_{1}, x_{2}, \\ldots, x_{16}\\right)$ such that $x_{j}= \\pm 1, \\pm i$. Most of these evaluate to 0 because $i^{2}=(-i)^{... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | final answer only | 1167 | |
0kf0 | Problem:
A simple graph $G$ on $2020$ vertices has its edges colored red and green. It turns out that any monochromatic cycle has even length. Given this information, what is the maximum number of edges $G$ could have? | [
"Solution:\n\nNote that $G$ has no $K_{5}$; indeed, it's well-known that the only triangle-free coloring of the edges of $K_{5}$ consists of two monochromatic $5$-cycles. Therefore, the number of edges of $G$ is at most $\\binom{4}{2} \\cdot 505^{2} = 1530150$ by Turán's theorem.\n\nTo show this occurs, we split th... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1530150 | |
0drw | Let $b$ be a number with $-2 < b < 0$. Prove that there exists a positive integer $n$ such that all the coefficients of the polynomial $(x+1)^n(x^2+bx+1)$ are positive. | [
"Expanding $(x+1)^n(x^2+bx+1)$, we obtain\n$$\nx^{n+2} + (n+b)x^{n+1} + \\sum_{k=0}^{n-2} \\left[ \\binom{n}{k+2} + b\\binom{n}{k+1} + \\binom{n}{k} \\right] x^{n-k} + (n+b)x + 1.\n$$\nWe require $n > 2$ so that $n+b > 0$. We also require $\\binom{n}{k+2} + b\\binom{n}{k+1} + \\binom{n}{k} > 0$ for all $k=0, \\dots... | Singapore | Singapur | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0cnl | Eight teams participated in a soccer tournament, and each pair of teams have played exactly once. It appeared that if two teams $A$ and $B$ played a draw then the resulting numbers of points of $A$ and $B$ are different. Find the greatest possible number of draws in this tournament. (Each win is worth 3 points, each dr... | [
"Докажем, что ровно по 6 ничьих может быть не более, чем у двух команд. Действительно, любая такая команда имеет либо 6, либо $6+3$ очка (в зависимости от того, выиграла или проиграла она свой результативный матч). Если таких команд три, то у двух из них поровну очков, значит, между собой они сыграли не вничью; это... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 22 | |
0gj3 | 一、有 2024 位妹妹各有一隻洋娃娃。考慮將這 2024 隻洋娃娃分給這些妹妹，使得每位妹妹各揹一隻洋娃娃的所有配對方法。對於所有 $k \ge 0$，令 $p(k)$ 為這些方法中，恰有 $k$ 位妹妹揹著自己的洋娃娃的方法數量。證明：
$$
\sum_{k=0}^{2024} k \times p(k) = 2024!
$$
I. 2024 girls each has her own doll. Consider all the ways to distribute a doll to each of the 2024 girls. For any $k \ge 0$, let $p(k)$ be the number o... | [
"一、原命題等價於對 $\\{1,2,...,2024\\}$ 的重排, $p(k)$ 等價於恰有 $k$ 個不動點的重排數量。注意到等號左式為所有重排的不動點總數。又注意到讓 $1$ 為不動點的重排數為 $(2024-1)!$, 其餘點雷同, 因此所有重排的不動點總數必須為 $2024 \\times (2024-1)! = 2024!$, 故等式成立。"
] | Taiwan | APMO Taiwan Preliminary Round 1 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | Chinese; English | proof only | null | |
0jbh | Problem:
Steph and Jeff each start with the number $4$, and Travis is flipping a coin. Every time he flips a heads, Steph replaces her number $x$ with $2x-1$, and Jeff replaces his number $y$ with $y+8$. Every time he flips a tails, Steph replaces her number $x$ with $\frac{x+1}{2}$, and Jeff replaces his number $y$ w... | [
"Solution:\n\nAnswer: $137$\n\nSuppose that $a$ heads and $b$ tails are flipped. Jeff's number at the end is $4 + 8a - 3b$. Note that the operations which Steph applies are inverses of each other, and as a result it is not difficult to check by induction that her final number is simply $1 + 3 \\cdot 2^{a-b}$.\n\nWe... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 137 | |
0bu6 | Problem:
Fie $p$ un număr prim impar şi fie $G$ un grup care are exact $p+1$ elemente. Arătaţi că, dacă $p$ divide numărul automorfismelor lui $G$, atunci $p \equiv 3(\bmod 4)$. | [
"Solution:\n\nÎntrucât $p$ este un divizor prim al $|\\operatorname{Aut} G|$, există un $f$ în $\\operatorname{Aut} G$ de ordin $p$. Deoarece $f$ este o permutare a mulţimii $G \\setminus \\{e\\}$, rezultă că $f$ este un ciclu de lungime $p$, deci $G \\setminus \\{e\\} = \\{x, f(x), \\ldots, f^{p-1}(x)\\}$, oricare... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Abstract Algebra > Group Theory",
"Number Theory > Other"
] | null | proof only | null | |
0bz5 | Let $A$ and $B$ be two finite sets. Find the number of functions $f : A \to A$ satisfying the property that there exist two functions $g : A \to B$ and $h : B \to A$ such that $g(h(x)) = x$, $\forall x \in B$, and $h(g(x)) = f(x)$, $\forall x \in A$. | [
"Let $f$, $g$, $h$ be three functions satisfying the properties in the problem. Because $g \\circ h = 1_B$, it is clear that $g$ is surjective and $h$ is injective. As $f = h \\circ g$, we also get that $|B| = |\\text{Im } f| \\le |A|$.\n\nAs $f(f(x)) = h(g(h(g))) = h(g(x)) = f(x)$ we get $y \\in \\text{Im } f$, $f... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | binom(|A|,|B|) * |B|^{|A|-|B|} | |
0jfu | Problem:
Suppose that $x$ and $y$ are chosen randomly and uniformly from $(0,1)$. What is the probability that $\left\lfloor\sqrt{\frac{x}{y}}\right\rfloor$ is even? Hint: $\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$. | [
"Solution:\nAnswer: $1-\\frac{\\pi^{2}}{24}$ OR $\\frac{24-\\pi^{2}}{24}$\n\nNote that for every positive integer $n$, the probability that $\\left\\lfloor\\sqrt{\\frac{x}{y}}\\right\\rfloor=n$ is just the area of the triangle formed between $(0,0),\\left(1, \\frac{1}{n^{2}}\\right),\\left(1, \\frac{1}{(n+1)^{2}}\\... | United States | HMMT November 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1 - π^2/24 | |
0f0s | Problem:
$ABC$ is an acute-angled triangle. $D$ is the reflection of $A$ in $BC$, $E$ is the reflection of $B$ in $AC$, and $F$ is the reflection of $C$ in $AB$. Show that the circumcircles of $DBC$, $ECA$, $FAB$ meet at a point and that the lines $AD$, $BE$, $CF$ meet at a point. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformati... | null | proof only | null | |
09av | Prove that there are infinitely many prime numbers $p$ such that $x^p \equiv 1 \pmod p$ equation has at least 2010 solutions in $\mathbb{Z}_p = \{0, 1, \dots, p-1\}$.
(proposed by U. Batzorig) | [
"Using the fact that $2^{2^{k+1}} - 1 \\equiv 0 \\pmod{2^{2^k} + 1}$, we conclude the following. If $l \\cdot 2^l \\equiv 0 \\pmod{2^{k+1}}$ then $(2^l)^{2^l} - 1$ is divisible by $2^{2^k} + 1$.\n$$\n\\begin{cases} l \\cdot 2^l \\equiv 0 \\pmod{2^{k+1}} \\\\ l < k+1 \\end{cases} \\quad (1)\n$$\nDenoting $l = k+1-c$... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0fjd | Problem:
En un triángulo $ABC$, $A'$ es el pie de la altura relativa al vértice $A$, y $H$ el ortocentro.
a) Dado un número real positivo $k = \frac{AA'}{HA'}$, hallar la relación entre los ángulos $B$ y $C$ en función de $k$.
b) Si $B$ y $C$ son fijos, hallar el lugar geométrico del vértice $A$ para cada valor de $... | [
"Solution:\n\nPara resolver la primera cuestión tenemos,\n$$\nBA' = c \\cos B, \\quad \\tan HBA' = \\cot C = \\frac{HA'}{BA'}, \\quad AA' = c \\sen B\n$$\nDeducimos\n$$\nk = \\frac{AA'}{HA'} = \\frac{c \\sen B}{c \\cos B \\cot C}, \\text{ de donde } \\quad \\tan B \\cdot \\tan C = k\n$$\n\nResolvamos la segunda cue... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geom... | null | proof and answer | a) tan(B) · tan(C) = k.
b) With BC = a and coordinates centered at the midpoint of BC, the locus of A(x, y) satisfies y^2 = k(a^2/4 − x^2), i.e., x^2/(a^2/4) + y^2/(k a^2/4) = 1, an ellipse centered on the midpoint of BC; for k < 1 the major axis lies along BC, and for k > 1 it lies perpendicular to BC. | |
06sr | Let $\nu$ be an irrational positive number, and let $m$ be a positive integer. A pair $(a, b)$ of positive integers is called good if
$$
a\lceil b \nu\rceil-b\lfloor a \nu\rfloor=m
$$
A good pair $(a, b)$ is called excellent if neither of the pairs $(a-b, b)$ and $(a, b-a)$ is good. (As usual, by $\lfloor x\rfloor$ and... | [
"For positive integers $a$ and $b$, let us denote\n$$\nf(a, b)=a\\lceil b \\nu\\rceil-b\\lfloor a \\nu\\rfloor\n$$\nWe will deal with various values of $m$; thus it is convenient to say that a pair $(a, b)$ is $m$-good or $m$-excellent if the corresponding conditions are satisfied.\nTo start, let us investigate how... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Number-Theoretic Functions > σ (sum of divi... | English | proof only | null | |
0946 | Problem:
Let $n$ be a positive integer. We are given a $2n \times 2n$ table. Each cell is coloured with one of $2n^{2}$ colours such that each colour is used exactly twice. Jana stands in one of the cells. There is a chocolate bar lying in one of the other cells. Jana wishes to reach the cell with the chocolate bar. A... | [
"Solution:\n\nFix the colouring of the cells and the starting position. We prove that Jana can reach any cell. Call a series of moves legal, if she starts from the starting cell with a teleport move, and uses the two types of moves alternately. Divide the cells into four categories.\n\n- Call a cell teleport reacha... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Yes, Jana can always reach the chocolate for any colouring and starting positions when starting with teleportation and alternating moves. | |
077g | Problem:
Find all pairs of integers $(a, b)$ so that each of the two cubic polynomials
$$
x^{3}+a x+b \text{ and } x^{3}+b x+a
$$
has all the roots to be integers. | [
"Solution:\nThe only such pair is $(0,0)$, which clearly works. To prove this is the only one, let us prove an auxiliary result first.\n\nLemma If $\\alpha, \\beta, \\gamma$ are reals so that $\\alpha+\\beta+\\gamma=0$ and $|\\alpha|,|\\beta|,|\\gamma| \\geq 2$, then\n$$\n|\\alpha \\beta+\\beta \\gamma+\\gamma \\al... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | (0, 0) | |
01lm | Three distinct real numbers satisfy the following condition: the sum of cubes of any two of them is $\frac{3}{4}$ smaller than the square of the remaining number.
Find all possible values of the product of these numbers. | [] | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 1/8 | |
0l3a | Problem:
Let $F E L D S P A R$ be a regular octagon, and let $I$ be a point in its interior such that $\angle F I L = \angle L I D = \angle D I S = \angle S I A$. Compute $\angle I A R$ in degrees. | [
"Solution:\n\n\n\nObserve that $I$ lies on line $D R$ due to symmetry, so $I D \\parallel F L$. Thus $\\angle F L I = \\angle L I D = \\angle F I L$, implying that triangle $F I L$ is isosceles with $F I = F L$. Similarly, $A I = A S$. Since $F L S A$ is a square, $F I = F L = A S = A I = F... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof and answer | 82.5° | |
0ifb | Problem:
In the base 10 arithmetic problem $H M M T + G U T S = R O U N D$, each distinct letter represents a different digit, and leading zeroes are not allowed. What is the maximum possible value of $R O U N D$? | [
"Solution:\n\nClearly $R = 1$, and from the hundreds column, $M = 0$ or $9$. Since $H + G = 9 + O$ or $10 + O$, it is easy to see that $O$ can be at most $7$, in which case $H$ and $G$ must be $8$ and $9$, so $M = 0$. But because of the tens column, we must have $S + T \\geq 10$, and in fact since $D$ cannot be $0$... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 16352 | |
009d | Given a set of several non-negative integers, a legal move consists in selecting a positive integer $a$ out of the set and perform one of these operations:
* If $a$ is odd, it is replaced by $a-1$.
* If $a$ is even, it is replaced by either $a-1$ or $a-2$.
Two players, A and B, make legal moves in turns, starting from ... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Player B wins if and only if n is congruent to 6 or 7 modulo 8; otherwise Player A wins. | |
0l19 | Problem:
Let $f(x) = x^{2} + 6x + 6$. Compute the greatest real number $x$ such that $f(f(f(f(f(x))))) = 0$. | [
"Solution:\nObserve that $f(x) = (x+3)^2 - 3$. Now, we claim that\n\nClaim 1. $f^{k}(x) = (x+3)^{2^{k}} - 3$ for all positive integers $k$.\n\nProof. We use induction. The base case $k=1$ is clear. To show the inductive step, note that $f^{k}(x) = (x+3)^{2^{k}} - 3$ implies\n\n$$\nf^{k+1}(x) = f(f^{k}(x)) = f((x+3)... | United States | HMMT November 2024 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 3^(1/32) - 3 | |
0jjm | A finite sequence of integers $a_1, a_2, \dots, a_n$ is called *regular* if there exists a real number $x$ satisfying
$$
\lfloor kx \rfloor = a_k \quad \text{for } 1 \le k \le n.
$$
Given a regular sequence $a_1, a_2, \dots, a_n$, for $1 \le k \le n$ we say the term $a_k$ is *forced* if the following condition is satis... | [
"The maximum is 985. To prove this, we start with two lemmas.\n\n**Lemma 2.** Suppose we begin with the pair of integers $(1, 1)$, and each step we are allowed to replace one of the integers with the sum of both. Then after $k$ steps, the maximum possible sum of the two numbers is $F_{k+3}$, where $F_n$ is the $n$t... | United States | IMO Team Selection Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Number Theory > Other"
] | null | proof and answer | 985 | |
063p | Problem:
Man bestimme alle Funktionen $f: \mathbb{Z} \rightarrow \mathbb{Z}$, die für alle $x, y \in \mathbb{Z}$ die Gleichung
$$
f(x-f(y)) = f(f(x)) - f(y) - 1
$$
erfüllen. | [
"Solution:\n\nDie Gleichung (1) wird genau von den Funktionen $f_1: x \\rightarrow -1$ und $f_2: x \\rightarrow x+1$ erfüllt. Durch Einsetzen wird leicht bestätigt, dass beide Funktionen Lösungen sind.\n\nNun sei $f$ eine Funktion, die (1) für alle $x, y \\in \\mathbb{Z}$ erfüllt. Durch Einsetzen von $x=0$ und $y=f... | Germany | 2. Auswahlklausur | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(x) = -1 and f(x) = x + 1 | |
06o5 | Find all nonnegative real numbers $a$, $b$, $c$ such that
$$
\frac{4a + 9b + 25c}{2a + 3b + 5c} + \frac{4b + 9c + 25a}{2b + 3c + 5a} + \frac{4c + 9a + 25b}{2c + 3a + 5b} = 10
$$ | [
"By the Cauchy-Schwarz inequality, we have\n$$\n(a + b + c)(4a + 9b + 25c) \\geq (2a + 3b + 5c)^2.\n$$\nThis implies $\\frac{4a + 9b + 25c}{2a + 3b + 5c} \\geq \\frac{2a + 3b + 5c}{a + b + c}$. Adding similar inequalities, we obtain\n$$\n\\begin{aligned}\n& \\frac{4a + 9b + 25c}{2a + 3b + 5c} + \\frac{4b + 9c + 25a... | Hong Kong | Hong Kong Team Selection Test 1 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Exactly those triples where two of a, b, c are zero and the remaining one is positive; i.e., (a>0, b=0, c=0) or (a=0, b>0, c=0) or (a=0, b=0, c>0). | |
0ayj | Problem:
Let $x$ be a real number satisfying $x^{2}-\sqrt{6} x+1=0$. Find the numerical value of $\left|x^{4}-\frac{1}{x^{4}}\right|$. | [
"Solution:\nNote that $x+\\frac{1}{x}=\\sqrt{6}$. Then\n$$\n\\begin{aligned}\n\\left|x^{4}-\\frac{1}{x^{4}}\\right| &=\\left(x^{2}+\\frac{1}{x^{2}}\\right)\\left|x^{2}-\\frac{1}{x^{2}}\\right| \\\\\n&=\\left(x^{2}+\\frac{1}{x^{2}}\\right)\\left(x+\\frac{1}{x}\\right)\\left|x-\\frac{1}{x}\\right| \\\\\n&=\\left(\\le... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 8√3 | |
0kn7 | Triangle $ABC$ is equilateral with side length $6$. Suppose that $O$ is the center of the inscribed circle of this triangle. What is the area of the circle passing through $A$, $O$, and $C$?
(A) $9\pi$
(B) $12\pi$
(C) $18\pi$
(D) $24\pi$
(E) $27\pi$ | [] | United States | AMC 12 B | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Dis... | null | MCQ | B |
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