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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0c05 | Let $x$, $y$, $z$ be natural numbers so that $13x + 8y = 5z$. Prove that the number $(x + y)(y + z)(z + x)$ is divisible by $130$. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
073r | Let $ABC$ be a triangle and $D$, $E$, $F$ be points on the sides $BC$, $CA$, $AB$ respectively such that $AD = BE = CF$. Suppose the line segments $AD$, $BE$, $CF$ are not concurrent and enclose an equilateral triangle. Is $ABC$ necessarily equilateral? | [
"Yes. Draw $AK$ parallel to $BC$ and $AK = BD$. Join $K$ to $E$ and $F$. Choose $L$ on $BC$ with $L$ between $B$ and $C$, such that $BD = LC$. Note that $KB$ is parallel to $AD$ and $KB = AD$. Hence $\\angle KBE = \\angle APE = 60^\\circ$. On the other hand, $KB = AD = BE$. Thus $KBE$ is an equilateral triangle. Th... | India | Indija TS 2008 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k2e | Problem:
Find the largest positive integer $n$ for which there exist $n$ finite sets $X_{1}, X_{2}, \ldots, X_{n}$ with the property that for every $1 \leq a < b < c \leq n$, the equation
$$
\left|X_{a} \cup X_{b} \cup X_{c}\right| = \lceil \sqrt{a b c} \rceil
$$
holds. | [
"Solution:\nFirst, we construct an example for $N=4$. Let $X_{1}, X_{2}, X_{3}, X_{4}$ be pairwise disjoint sets such that $X_{1}=\\varnothing$, $|X_{2}|=1$, $|X_{3}|=2$, and $|X_{4}|=2$. It is straightforward to verify the condition.\n\nWe claim that there are no five sets $X_{1}, X_{2}, \\ldots, X_{5}$ for which ... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
00po | The incircle of a triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at the points $A_1$, $B_1$, $C_1$, respectively. Let the projections of the orthocenter $H_1$ of the triangle $A_1B_1C_1$ to the lines $AA_1$ and $BC$ be $P$ and $Q$, respectively. Show that the line $PQ$ bisects the line segment $B_1C_1$. | [
"Let $A_1S$, $B_1T$ and $C_1U$ be the altitudes of $A_1B_1C_1$. The circle $k$ with the diameter $A_1H_1$ contains the points $A_1$, $H_1$, $T$, $U$, $P$ and $Q$. Let $AA_1$ intersect the incircle of $ABC$ for the second time at $V$. Assume that $\\angle B \\ge \\angle C$.\n\nObserve that $\\angle C_1A_1V = \\angle... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
00qm | A positive integer $n$ is called *super special* if it can be represented in the form $n = \frac{x^3 + 2y^3}{u^3 + 2v^3}$ for some positive integers $x, y, u, v$. Prove that:
(a) There are infinitely many super special positive integers;
(b) 2014 is not super special. | [
"(a) Every perfect cube $k^3$ of a positive integer is super special because we can write\n$$\nk^3 = k^3 \\frac{x^3 + 2y^3}{x^3 + 2y^3} = \\frac{(kx)^3 + 2(ky)^3}{x^3 + 2y^3}\n$$\nfor some positive integers $x, y$.\n\n(b) Observe that $2014 = 2 \\cdot 19 \\cdot 53$. If $2014$ is super special, then we have,\n$$\nx^... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
0ikk | Problem:
Suppose we have a regular hexagon and draw all its sides and diagonals. Into how many regions do the segments divide the hexagon? (No proof is necessary.) | [
"Solution:\nAn accurate diagram and a careful count yields the answer.\n\n**Answer:** $24$"
] | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 24 | |
091h | Problem:
Let $A, B, C, D, E$ be points such that $A B C D$ is a cyclic quadrilateral and $A B D E$ is a parallelogram. The diagonals $A C$ and $B D$ intersect at $S$ and the rays $A B$ and $D C$ intersect at $F$. Prove that $\angle A F S=\angle E C D$. | [
"Solution:\n\nLet $M$ and $N$ be the feet of perpendicular from $S$ to $A B$ and $C D$, respectively. Then $S M F N$ is cyclic since it has two opposite right angles. Therefore $\\angle A F S=\\angle M F S=\\angle M N S$. We need to prove $\\angle M N S=\\angle E C D$. This will follow from similarity of triangles ... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ciu | Determine the finite commutative rings $A$ with at least four elements, with the property that for any choice of a proper subset $S$ of $A^*$ with at least two elements, the equality
$$
\sum_{x \in S} x = \prod_{x \in A^* \setminus S} x
$$
holds. | [] | Romania | 75th NMO | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Field Theory"
] | English | proof and answer | The finite field with four elements (F4). | |
0d5m | Find all integer solutions of the equation $x^{2} y^{5} - 2^{x} 5^{y} = 2015 + 4 x y$. | [
"Notice that if $x < 0$ or $y < 0$ then the term $2^{x} 5^{y}$ is not an integer while $x^{2} y^{5} - 2015 - 4 x y$ is an integer which is impossible. If $x = 0$ or $y = 0$ then the equation becomes $-2^{x} 5^{y} = 2015$ which has no solution. We deduce that both $x, y$ are positive integers. Moreover, because $2^{... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Intermediate Algebra > Exponential functions"
] | English, Arabic | proof and answer | (5, 3) | |
052z | Numbers $1, \dots, 200$ are written on a blackboard in one line. Juku has to write in front of each number plus or minus sign so that for any positive integer $n \le 100$ the number itself and one of its multiples have different signs. Which numbers must he assign a minus sign in order to get the maximal possible value... | [
"*Answer:* The numbers $51, \\dots, 100$.\n\nIf Juku writes a minus in front of the number $51, \\dots, 100$ and a plus in front of the others, then the conditions of the problem are satisfied: for $51 \\le n \\le 100$, the numbers $n$ and $2n$ have different signs; for $n \\le 50$ there is at least one multiple of... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | Minus signs on 51 through 100; plus signs on all other numbers. | |
05rh | Problem:
Au début, les 9 cases d'un échiquier $3 \times 3$ contiennent chacune un $0$. À chaque étape, Pedro choisit deux cases partageant un côté, et ajoute soit $1$ aux deux cases, soit $-1$ aux deux cases. Montrer qu'il est impossible d'atteindre en un nombre fini de coups la situation où toutes les cases sont remp... | [
"Solution:\n\nL'idée ici est de faire apparaître un invariant $I$.\n\nColorions l'échiquier naturellement en noir et blanc de telle sorte qu'il y a $4$ cases noires et $5$ blanches. Soit donc $I$ la somme des cases noires moins la somme des cases blanches. Au début, $I = 0$, et dans une hypothétique situation où to... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0kef | Problem:
Let $a_{0}, b_{0}, c_{0}, a, b, c$ be integers such that $\operatorname{gcd}\left(a_{0}, b_{0}, c_{0}\right)=\operatorname{gcd}(a, b, c)=1$. Prove that there exists a positive integer $n$ and integers $a_{1}, a_{2}, \ldots, a_{n}=a, b_{1}, b_{2}, \ldots, b_{n}=b, c_{1}, c_{2}, \ldots, c_{n}=c$ such that for al... | [
"Solution:\nThe problem statement is equivalent to showing that we can find a sequence of vectors, each with 3 integer components, such that the first vector is $\\left(a_{0}, b_{0}, c_{0}\\right)$, the last vector is $(a, b, c)$, and every pair of adjacent vectors has dot product equal to $1$.\n\nWe will show that... | United States | HMMT February 2020 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0isn | Problem:
What is the sum of all integers $x$ such that $|x+2| \leq 10$? | [
"Solution:\nThe inequality $|x+2| \\leq 10$ holds if and only if $x+2 \\leq 10$ and $x+2 \\geq -10$. So $x$ must be in the range $-12 \\leq x \\leq 8$.\n\nIf we add up the integers in this range, each positive integer cancels with its additive inverse, so the sum is equal to $-12 - 11 - 10 - 9 = -42$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | -42 | |
0a6s | Problem:
Show that there are infinitely many triples $(a, b, c)$ of positive integers such that
$$a^{2} + b^{2} + c^{2} + (a + b + c)^{2} = abc.$$ | [
"Solution:\n\nNote that $a = b = c = 12$ is a solution. Now, fix $c = 12$, the original equation becomes\n\n$$\n\\begin{array}{c}\n\\{a^{2} + b^{2} + 144 + (a + b + 12)^{2} = 12ab\\} \\\\\n\\{\\Rightarrow 2a^{2} + (24 - 10b)a + (2b^{2} + 24b + 288) = 0\\}\n\\end{array}\n\\quad (1)\n$$\n\nWe see that by Vieta's theo... | New Zealand | NZMO Round One | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0ldl | Let $B$, $C$ be two fixed points on the fixed circle $(O)$ ($BC$ is not the diameter of $(O)$). Point $A$ moves on $(O)$ such that $AB > BC$ and $M$ is the midpoint of $AC$. The circle of diameter $BM$ intersects $(O)$ at $R$. Suppose that $RM$ intersects $(O)$ at the second point $Q$ and cuts $BC$ at $P$. The circle o... | [
"1) It is easy to check that $BQ$ is the diameter of circle $(O)$. Denote $I$ as the intersection of $SR$ and $PK$. We have\n$$\n\\angle SPI = \\angle SBK = \\angle QCA\n$$\nand\n$$\n\\angle PSI = \\angle PBR = \\angle CQR.\n$$\nThis implies that two triangles $PSI$ and $CQM$ are similar. By the same way, we also h... | Vietnam | Vietnamese Team Selection Test for IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
0633 | Problem:
Man bestimme die kleinste positive Zahl $k$, für die folgendes der Fall ist: Wenn Kain in solcher Weise ganze Zahlen in die Zellen eines $2011 \times 2011$-Schachbrettes schreibt, dass die 4022 Summen, die man durch Addition aller Zahlen einer Zeile oder Spalte erhalten kann, paarweise übereinstimmen, so ist ... | [
"Solution:\n\n$k=2681$.\n\nBeweis von $k \\geq 2681$ : Abel muss mindestens 4021 der Summen ändern, o. B. d. A. alle Spaltensummen und alle bis auf höchstens eine Zeilensumme. Um die Spaltensummen zu ändern, muss Abel in jeder Spalte mindestens einen Eintrag abändern. Dies erledige Abel o. B. d. A. zuerst und pausi... | Germany | 1. Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2681 | |
0kgx | Problem:
Let $n>1$ be a positive integer. Each unit square in an $n \times n$ grid of squares is colored either black or white, such that the following conditions hold:
- Any two black squares can be connected by a sequence of black squares where every two consecutive squares in the sequence share an edge;
- Any two w... | [
"Solution:\n\nThe first two conditions also imply that there can be no $2 \\times 2$ checkerboards, so the boundary between black squares and white squares is either a lattice path or cycle (if one color encloses the other). Therefore, the set of squares of each color is the interior of a lattice polygon of genus $... | United States | HMMT Spring 2021 Team Round | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Maximum difference = 2n + 1 for odd n, and 2n − 2 for even n. | |
04ym | Let $a$, $b$ and $c$ be positive integers such that $ab$ is divisible by $2c$, $bc$ is divisible by $3a$ and $ca$ is divisible by $5b$. Find the least possible value of $abc$. | [
"Since $ab$ is divisible by $2c$ and $ca$ is divisible by $5b$, $ab \\cdot ca$ must be divisible by $2c \\cdot 5b$, hence $a^2$ is divisible by $2 \\cdot 5$. Therefore $a^2$ is divisible by $2$ and $5$, hence $a$ is divisible by $2$ and $5$. Similarly $b$ is divisible by $2$ and $3$, and $c$ is divisible by $3$ and... | Estonia | Estonija 2010 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 900 | |
0j9f | Problem:
Let $C$ denote the set of points $(x, y) \in \mathbb{R}^{2}$ such that $x^{2}+y^{2} \leq 1$. A sequence $A_{i}=\left(x_{i}, y_{i}\right) \mid i \geq 0$ of points in $\mathbb{R}^{2}$ is 'centric' if it satisfies the following properties:
- $A_{0}=\left(x_{0}, y_{0}\right)=(0,0),\ A_{1}=\left(x_{1}, y_{1}\righ... | [
"Solution:\n\nAnswer: $(-1006,1006 \\sqrt{3}),\\ (-1006,-1006 \\sqrt{3})$\n\nConsider any triple of points $\\triangle A_{n} A_{n+1} A_{n+2}$ with circumcenter $P_{n}$. By the Triangle Inequality we have $A_{n} P_{n} \\leq A_{n} A_{0}+A_{0} P_{n} \\leq A_{n} A_{0}+1$. Since $P_{n}$ is the circumcenter, we have $P_{... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry... | null | proof and answer | (-1006, 1006*sqrt(3)), (-1006, -1006*sqrt(3)) | |
0k0u | Problem:
You are out walking and see a group of rhinoceroses (which each have two horns and four legs) and triceratopses (which each have three horns and four legs). If you count 31 horns and 48 legs, how many triceratopses are there? | [
"Solution:\n\nSince each animal has 4 legs, there must be $48 / 4 = 12$ animals. Each triceratops has 3 horns and each rhinoceros has 2, so if there are $t$ triceratopses and $r$ rhinoceroses we get $t + r = 12$ and $3t + 2r = 31$. Subtracting twice the first equation from the second gives $t = 7$."
] | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 7 | |
0klv | The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$, where $m$ and $k$ are integers and $6$ is not a divisor of $m$. What is $m + k$?
(A) 47 (B) 58 (C) 59 (D) 88 (E) 90 | [
"The number of positive integer divisors of the positive integer whose prime factorization is $p_1^{e_1} p_2^{e_2} \\cdots p_n^{e_n}$ equals $(e_1 + 1)(e_2 + 1) \\cdots (e_n + 1)$. Because\n$$\n2021 = 2025 - 4 = 45^2 - 2^2 = (45 + 2)(45 - 2) = 47 \\cdot 43,\n$$\na number having $2021$ divisors must be of the form $... | United States | Fall 2021 AMC 10 B | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | MCQ | B | |
0fl2 | Problem:
Sea $I_{n}$ el conjunto de los $n$ primeros números naturales impares. Por ejemplo: $I_{3} = \{1, 3, 5\}$, $I_{6} = \{1, 3, 5, 7, 9, 11\}$, etc.
¿Para qué números $n$ el conjunto $I_{n}$ se puede descomponer en dos partes (disjuntas) de forma que coincidan las sumas de los números en cada una de ellas? | [
"Solution:\n\nLos primeros casos son:\n\n| $I_{1}=\\{1\\}$ | no descompone |\n| :--- | :--- |\n| $I_{2}=\\{1,3\\}$ | no descompone |\n| $I_{3}=\\{1,3,5\\}$ | no descompone |\n| $I_{4}=\\{1,3,5,7\\}$ | descompone $\\{1,7\\}$ y $\\{3,5\\}$ |\n| $I_{5}=$ | no descompone |\n| $I_{6}=\\{1,3,5,7,9,11\\}$ | descompone $\\... | Spain | Spain | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All even n with n ≥ 4 | |
05o4 | Problem:
Trouver tous les quadruplets $ (p, q, r, n) $ d'entiers strictement positifs vérifiant les trois conditions suivantes :
- $p$ et $q$ sont premiers,
- $p+q$ n'est pas divisible par $3$,
- $p+q = r (p-q)^n$. | [
"Solution:\n\nOn suppose d'abord que $p \\neq 3$ et $q \\neq 3$, et on considère $p$ et $q$ modulo $3$. Puisque $p+q \\not\\equiv 0 \\pmod{3}$, on n'a pas $p \\equiv 1 \\pmod{3}$ et $q \\equiv 2 \\pmod{3}$, ni $p \\equiv 2 \\pmod{3}$ et $q \\equiv 1 \\pmod{3}$. On est donc assuré que $p \\equiv q \\equiv 1 \\pmod{3... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2,3,5,2n) for all positive integers n; (3,2,5,n) for all positive integers n; (3,5,2,2); (5,3,1,3); (5,3,2,2); (5,3,4,1) | |
0g68 | 設 $S = \{1, 2, \dots, 2012\}$。求滿足下列條件之函數 $f : S \to S$ 的個數。
(1) $f$ 為一對一且映成函數。
(2) 對於所有數字 $1 \le a \le 2012$，皆會有 $f(a) + f^{-1}(a) = 2013$，其中 $f^{-1}$ 為 $f$ 的反函數。 | [
"考慮下列幾種情形：\n\n(1) 若存在 $a$ 使得 $f(a) = a$，則 $f(a) + f^{-1}(a) = 2a \\neq 2013$。所以不可能。\n\n(2) 若存在 $a \\neq b$ 使得 $f(a) = b$ 且 $f(b) = a$，則 $f(a) + f^{-1}(a) = 2b \\neq 2013$。所以不可能。\n\n(3) 存在三個相異數 $a, b, c$ 使得 $f(a) = b, f(b) = c$ 且 $f(c) = a$，則 $f(a) + f^{-1}(a) = b + c$ 且 $f(b) + f^{-1}(b) = c + a$。由於 $b + c = c + a ... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 1006!/503! | |
08dq | Problem:
Marcella, giocando, trova per puro caso due polinomi $p(x)$ e $q(x)$, non costanti e a coefficienti interi, verificanti la relazione:
$$
p(q(x+1)) = p\left(x^{3}\right) q(x+1)^{5}.
$$
Che cosa possiamo affermare con certezza dei due polinomi trovati da Marcella?
(A) Il coefficiente direttore di $p(x) q(x)$ è ... | [
"Solution:\n\nLa risposta è (E). Siano $m, n$ i gradi di $p(x)$ e $q(x)$, rispettivamente. Essi sono due interi positivi. La relazione tra i polinomi data nel problema implica la seguente equazione sui gradi, $m n = 3 m + 5 n$, che a sua volta equivale a $(m-5)(n-3) = 15$. Per ricavare $(m, n)$ dobbiamo dunque riso... | Italy | Progetto Olimpiadi della Matematica | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | E | |
0crz | Сфера $\omega$ проходит через вершину $S$ пирамиды $SABC$ и пересекает рёбра $SA$, $SB$ и $SC$ вторично в точках $A_1$, $B_1$ и $C_1$ соответственно. Сфера $\Omega$, описанная около пирамиды $SABC$, пересекается с $\omega$ по окружности, лежащей в плоскости, параллельной плоскости $(ABC)$. Точки $A_2$, $B_2$ и $C_2$ си... | [
"**Первое решение.** Утверждение задачи эквивалентно равенству $SA_2 \\cdot SA = SB_2 \\cdot SB = SC_2 \\cdot SC$. Значит, ввиду равенств $AA_1 = SA_2$ и двух аналогичных, достаточно доказать, что $AA_1 \\cdot AS = BB_1 \\cdot BS = CC_1 \\cdot CS$.\nПусть $\\ell$ — прямая, проходящая через центры сфер $\\Omega$ и $... | Russia | XL Russian mathematical olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0jsb | Problem:
How many subsets $S$ of the set $\{1,2, \ldots, 10\}$ satisfy the property that, for all $i \in [1,9]$, either $i$ or $i+1$ (or both) is in $S$? | [
"Solution:\nAnswer: 144\n\nWe do casework on the number of $i$'s not in $S$. Notice that these $i$'s that are not in $S$ cannot be consecutive, otherwise there exists an index $i$ such that both $i$ and $i+1$ are both not in $S$. Hence if there are $k$ $i$'s not in $S$, we want to arrange $k$ black balls and $10-k$... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | 144 | |
0drk | Let $A$ be a set of numbers chosen from $1, 2, \ldots, 2015$ with the property that any two distinct numbers, say $x$ and $y$, in $A$ determine a unique isosceles triangle (which is non-equilateral) whose sides are of length $x$ or $y$. What is the largest possible size of $A$? | [
"Let $x < y$ be two numbers in $A$. For them to determine a unique isosceles triangle, we must have $2x \\le y$. If $A = \\{2^0, 2^1, 2^2, \\ldots, 2^{10}\\}$, then any two of the numbers $x < y$ satisfy $2x \\le y$. So the maximum size is $\\ge 11$.\n\nNow suppose that there is a set $A$ with $|A| = 12$ that has t... | Singapore | Singapur 2015 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 11 | |
0dj8 | Let $ABC$ be an acute triangle. The line through $A$ perpendicular to $BC$ intersects $BC$ at $D$. Let $E$ be the midpoint of $AD$ and $\omega$ the circle of center $E$ and radius $AE$. The line $BE$ intersects $\omega$ at $X$ such that $X$ and $B$ are not on the same side of $AD$ and the line $CE$ intersects $\omega$ ... | [
"Note that the condition of the problem is that $AD$ is the radical axis of two circles ($BDX$) and ($CDY$) which implies that $E$ has the same power to these circles. This gives us\n$$\nEB \\cdot EX = EC \\cdot EY.\n$$\nHowever, $EX = EY$ because $E$ is the center of $\\omega$ and this means that $BE = CE$. From t... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
05nw | Problem:
Déterminer tous les entiers $a>0$ pour lesquels il existe des entiers strictement positifs $n, s, m_{1}, \cdots, m_{n}, k_{1}, \cdots, k_{s}$ tels que
$$
\left(a^{m_{1}}-1\right) \cdots\left(a^{m_{n}}-1\right)=\left(a^{k_{1}}+1\right) \cdots\left(a^{k_{s}}+1\right)
$$ | [
"Solution:\n\nOn va prouver que les entiers $a$ cherchés sont $a=2$ et $a=3$.\n\nTout d'abord, on constate que $2^{2}-1=2+1$ et que $(3-1)(3-1)=3+1$, ce qui assure que $a=2$ et $a=3$ sont effectivement des solutions du problème.\n\nRéciproquement, soit $a, n, s, m_{1}, \\cdots, m_{n}, k_{1}, \\cdots, k_{s}$ des ent... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2 and 3 | |
0jlt | Problem:
Given that $w$ and $z$ are complex numbers such that $|w+z|=1$ and $|w^{2}+z^{2}|=14$, find the smallest possible value of $|w^{3}+z^{3}|$. Here, $|\cdot|$ denotes the absolute value of a complex number, given by $|a+b i|=\sqrt{a^{2}+b^{2}}$ whenever $a$ and $b$ are real numbers. | [
"Solution:\n$|w^{3}+z^{3}| = |w+z|\\,|w^{2}-wz+z^{2}| = |w^{2}-wz+z^{2}| = \\left|\\frac{3}{2}(w^{2}+z^{2})-\\frac{1}{2}(w+z)^{2}\\right|$.\n\nBy the triangle inequality,\n$$\n\\left|\\frac{3}{2}(w^{2}+z^{2})-\\frac{1}{2}(w+z)^{2}+\\frac{1}{2}(w+z)^{2}\\right| \\leq \\left|\\frac{3}{2}(w^{2}+z^{2})-\\frac{1}{2}(w+z... | United States | HMMT 2014 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 41/2 | |
091l | Problem:
Let $n \geqslant 3$ be an integer. John and Mary play the following game: First John labels the sides of a regular $n$-gon with the numbers $1,2, \ldots, n$ in whatever order he wants, using each number exactly once. Then Mary divides this $n$-gon into triangles by drawing $n-3$ diagonals which do not interse... | [
"Solution:\n\nFor $n=3$ the answer is $6$. Suppose $n \\geqslant 4$. It is obvious that in each triangulation there are at least two triangles that share two sides with the polygon. We will prove that it is always best for Mary to choose a triangulation for which there is no more than two triangles of this kind.\n\... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | (n^2 + 3n - 6)/2 | |
048w | Let $ABCD$ be a unit square. The unit circle $k$ has center $C$. Determine the radius of the circle $k_1$ which touches the circle $k$ and line segments $\overline{AB}$ and $\overline{AD}$. | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3 - 2√2 | |
09bq | $a_1 = 2$, $a_{n+1} = \frac{n}{a_1 + a_2 + \cdots + a_n}$, $n \ge 1$ бол $a_{2010} > 0.9995$ болохыг батал. | [
"$S_n = a_1 + a_2 + \\dots + a_n > n$ гэж батлая.\n$n = 1$ үед $S_1 = a_1 = 2 > 1$\n$n = k$ үед $S_k > k$ гэе.\n$n = k + 1$ үед\n$$\n\\begin{align*}\nS_{k+1} - (k+1) &= S_k + a_{k+1} - (k+1) = S_k - (k+1) + \\frac{k}{S_k} = \\\\\n&= \\frac{S_k^2 - (k+1)S_k + k}{S_k} = \\frac{(S_k - k)(S_k - 1)}{S_k} > 0 \\\\\na_{n+... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Mongolian | proof only | null | |
0d6b | Call a positive integer $N \geq 2$ "special" if for every $k$ such that $2 \leq k \leq N$, $N$ can be expressed as a sum of $k$ positive integers that are relatively prime to $N$ (although not necessarily relatively prime to each other). Find all special positive integers. | [
"We claim that all odd numbers are special, and the only special even number is $2$. For any even $N > 2$, the number relatively to $N$ must be odd, and $N$ cannot be expressed as a sum of $3$ positive odd numbers.\n\nNow, suppose that $N$ is odd. We consider the binary decomposition of $N$\n$$\nN = 2^{a_{1}} + 2^{... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All odd positive integers and 2 | |
06n0 | a. Find the smallest number of lines drawn on the plane so that they produce exactly $2022$ points of intersection. (Note: For $1$ point of intersection, the minimum is $2$; for $2$ points, minimum is $3$; for $3$ points, minimum is $3$; for $4$ points, minimum is $4$; for $5$ points, minimum is $4$, etc.)
b. What hap... | [
"a.\nAnswer: $65$\nNote that\n$$\n\\binom{64}{2} = 2016 < 2022 < 2080 = \\binom{65}{2}.\n$$\nAs each pair of straight lines produces at most one intersection point, $64$ lines produce at most $2016$ intersection points, which is insufficient. Therefore, at least $65$ lines are needed.\nIt is possible to have $65$ l... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a) 65; b) 65 | |
0hou | Problem:
Let $a, b, c, d, e$, and $f$ be decimal digits such that the six-digit number $\overline{a b c d e f}$ is divisible by $7$. Prove that the six-digit number $\overline{b c d e f a}$ is divisible by $7$. | [
"Solution:\n\nWe have\n$$\n\\begin{aligned}\n\\overline{b c d e f a} & = \\overline{a b c d e f a} - \\overline{a 000000} \\\\\n& = \\overline{a b c d e f 0} - \\overline{a 000000} + a \\\\\n& = 10 \\cdot \\overline{a b c d e f} - 1000000 \\cdot a + a \\\\\n& = 10 \\cdot \\overline{a b c d e f} - 999999 \\cdot a \\... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
02bi | Problem:
Soma de razão $\frac{1}{2}$ - Se $S_{n}=\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n}}$, qual é o menor número inteiro positivo $n$ tal que $S_{n}>0,99$ ? | [
"Solution:\n\nComo\n$$\nS_{n}=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\cdots+\\frac{1}{2^{n}}\n$$\nsegue que\n$$\n\\frac{1}{2} S_{n}=\\frac{1}{2} \\times\\left(\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\cdots+\\frac{1}{2^{n}}\\right)=\\frac{1}{4}+\\frac{1}{8}+\\cdots+\\frac{1}{2^{n+1}}\n$$\nLogo,\n$$\n\\frac{1}{2} S_... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 7 | |
0gc7 | 有 $n$ 隻羊和一隻披著羊皮的狼。有些羊是好朋友(好友關係是互相的)。狼的目標是要吃掉所有的羊。首先牠從 $n$ 隻羊中挑一些建立好友關係。接下來的每一天, 牠從牠的好友羊中挑一隻吃掉。每當牠吃掉一隻羊 $A$ 時:
(i) 一隻 $A$ 的好友羊如果原本是狼的好友, 則會和狼絕交;
(ii) 一隻 $A$ 的好友羊如果原本不是狼的好友, 則會和狼建立好友關係。
重複以上動作, 直到狼再也沒有好友為止。
試求最大的正整數 $m$ (以 $n$ 表示), 滿足下列條件:
存在一種 $n$ 頭羊之間的好友關係, 使得狼總共有 $m$ 種不同的選擇起始好友羊的方式, 讓狼有方法可以吃完所有的羊。 | [
"答案：$2^{n-1}$。\n\n我們首先證明上界。令狼的好友數為 $a$，好友羊的對數為 $b$。注意到每次狼吃掉羊時，$a+b$ 會改變奇偶。因此若狼可以吃掉所有的羊，必須要有 $a+b+n-1 \\equiv 1 \\pmod{2}$ (狼在吃完 $n-1$ 隻羊後必可以吃掉最後一隻羊。) 因此，在狼的 $2^n$ 中選擇起始好友羊的方式中，至多只有 $2^{n-1}$ 種方法有機會吃掉所有的羊。\n\n以下構造達到上界的羊交友狀況。將羊編號 $1$ 到 $n$，並讓第 $i$ 跟 $i+1$ 號羊結為好友 $(i = 1, 2, \\dots, n-1)$，其他不交。我們將用數學歸納法證明，在此交友方式下，只要狼起... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 模擬競賽(一) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^{n-1} | |
07kv | A fixed point $O$ is joined to any point $P$ on a given line which does not contain $O$. On $OP$ a point $Q$ is taken such that $|OP| \cdot |OQ|$ is constant. Show $Q$ lies on a fixed circle if $P$ varies. Examine all possible cases. | [
"Let $R$ be the foot of the perpendicular from $O$ on the given line and let $S$ be chosen on $OR$ such that $|OR| \\cdot |OS| = |OP| \\cdot |OQ|$, i.e. $S$ is a point on the locus. The value of the constant determines the position of $S$, either between $O$ and $R$ or not. Then $P, Q, S$ and $R$ lie on a circle in... | Ireland | Irska | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0df6 | Find all quadruples $(p, a, b, x)$ that satisfy the equality $p^2 + 4a9^b = x^2$, where $p$ is a prime and $a, b, x$ are nonnegative integers. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All solutions arise by choosing nonnegative integers a, b and integers u, v ≥ 0 with uv = a·9^b and v − u being a prime p, then setting x = u + v and p = v − u. Equivalently, for any divisor d of a·9^b with p = a·9^b/d − d prime, set x = a·9^b/d + d. In particular, when a = 0, we have x = p for any prime p and any b ≥ ... | |
0j99 | Problem:
Square $ABCD$ has side length $2$, and $X$ is a point outside the square such that $AX = XB = \sqrt{2}$. What is the length of the longest diagonal of pentagon $AXB C D$? | [
"Solution:\n\nAnswer: $\\sqrt{10}$\n\nSince $AX = XB = \\sqrt{2}$ and $AB = 2$, we have $\\angle AXB = 90^{\\circ}$. Hence, the distance from $X$ to $AB$ is $1$ and the distance from $X$ to $CD$ is $3$. By inspection, the largest diagonals are thus $BX = CX = \\sqrt{3^{2} + 1^{2}} = \\sqrt{10}$."
] | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | sqrt(10) | |
00ov | In a soccer tournament each team plays exactly one game with all others. The winner gets 3 points, the loser zero and each team gets 1 point in case of a draw.
It is known that $n$ teams ($n \ge 3$) took part in a tournament and the final classification is given by an arithmetical progression of points, the last team h... | [
"a) The total number of matches is $n(n-1)/2$. Let $w$ be the number of games ended with a victory and $e$ the number of games ended in a draw ($e \\ge 1$ due to the last team). Thus, $w+e = n(n-1)/2$. If $r$ is the step of the arithmetical progression, we have that the total number of points in the final classific... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) Impossible for n = 12. b) The only possible value is n = 4, with scores 7, 5, 3, 1; up to relabeling, one configuration is: team 1 draws with team 2, defeats teams 3 and 4; team 2 defeats team 3 and draws with team 4; team 3 defeats team 4. | |
0dt1 | Let $ABCD$ be a square, $E$ be a point on the side $DC$, $F$ and $G$ be the feet of the altitudes from $B$ to $AE$ and from $A$ to $BE$, respectively. Suppose $DF$ and $CG$ intersect at $H$. Prove that $\angle AHB = 90^\circ$. | [
"Note that $A$, $F$, $G$, $B$ are concyclic, since they lie on the circle $\\Gamma$ with diameter $AB$. Let $DF$ intersect $\\Gamma$ again at $H'$; it suffices to show that $H = H'$.\n\n**Solution 1:**\nLet the diagonals $AC$ and $BD$ intersect at $O$; note that $O$ also lies on $\\Gamma$. Let $H'G \\cap AO = C'$. ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hja | Problem:
Six consecutive prime numbers have sum $p$. Given that $p$ is also prime, determine all possible values of $p$. | [
"Solution:\n\nWe consider two cases.\n- If $2$ is one of the six prime numbers, then the only possible sum is $p = 2 + 3 + 5 + 7 + 11 + 13 = 41$, which is indeed a prime.\n- If $2$ is not one of the six prime numbers, then all six primes are odd, so their sum must be even. Moreover, $p > 2$. Therefore $p$ is not pr... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 41 | |
03q7 | Arrange $1\,650$ students in $22$ rows by $75$ columns. It is known that for any two columns, the number of occasions that two students in the same row are of the same sex does not exceed $11$. Prove that the number of boy students does not exceed $928$. | [
"Let $a_i$ be the number of boy students in the $i$\\text{th}$ row, then the number of girl students in this row is $75 - a_i$. By the given condition, we have $\\sum_{i=1}^{22} \\left( C_{a_i}^2 + C_{75-a_i}^2 \\right) \\le 11 \\times C_{75}^2$. That is, $\\sum_{i=1}^{22} (a_i^2 - 75a_i) \\le -30,525$, implying $\... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0976 | Problem:
Fie că în tetraedru se intersectează două segmente, care vin din capetele unei laturi, în centrele cercurilor înscrise ale fețelor opuse. Demonstrați că două segmente, care vin din capetele laturi, care se încrucișează cu latura inițială, în centrele cercurilor înscrise a altor două fețe, prin urmare, la fel ... | [
"Solution:\n\nFie latura inițială este latura $AB$, $A_{1}$ este centrul cercului înscris în $\\triangle SBC$, $B_{1}$ este centrul cercului înscris în $\\triangle SAC$. Conform condiției problemei $[A A_{1}] \\cap [B B_{1}] \\neq \\varnothing$. Atunci punctele $A, A_{1}, B, B_{1}$ se află în același plan $(A A_{1}... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0dfa | A pile of $2^n - 1$ coins is placed on the player. Alice is allowed to make the following moves: If a pile has an even number of coins, she can throw half of those coins away. If a pile has an odd number of coins $k$ such that $k + 1$ is properly divisible by $2^l$ (i.e. $2^l|k + 1$ and $2^l < k + 1$) for some positive... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | n | |
057u | Every term of the sequence $a_1, a_2, a_3, \dots$ is either $0$ or $1$. It is known that both $0$ and $1$ occur at least $1010$ times among every $2021$ consecutive terms of the sequence. May one be sure that the sequence is periodic from some place on, i.e., there exist positive integers $n$ and $p$ such that $a_{n+i}... | [
"Consider the tuples $\\underbrace{11\\dots1}_{1011 \\text{ times}} \\underbrace{00\\dots0}_{1010 \\text{ times}}$ and $\\underbrace{11\\dots1}_{1010 \\text{ times}} \\underbrace{00\\dots0}_{1011 \\text{ times}}$. Concatenating infinitely many instances of these tuples in any order produces a sequence that satisfie... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | English | proof and answer | No | |
00l7 | Let *a*, *b*, *c* and *d* be real numbers with $a^2 + b^2 + c^2 + d^2 = 4$. Prove the inequality
$$(a+2)(b+2) \geq cd$$
and give four numbers *a*, *b*, *c* and *d* such that equality holds. | [
"The claimed inequality is equivalent to $2ab + 4a + 4b + 8 \\ge 2cd$, which can be written as\n$$\n2ab + 4a + 4b + a^2 + b^2 + c^2 + d^2 + 4 \\ge 2cd\n$$\non account of the condition $a^2 + b^2 + c^2 + d^2 = 4$. By the identity\n$$\na^2 + b^2 + 2ab + 4a + 4b + 4 = (a + b + 2)^2\n$$\nwe arrive at the equivalent and... | Austria | Regional Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | a = b = c = d = -1 | |
0d4f | We are given a lattice and two pebbles $A$ and $B$ that are placed at two lattice points. At each step we are allowed to relocate one of the pebbles to another lattice point with the condition that the distance between pebbles is preserved. Is it possible after finite number of steps to switch positions of the pebbles? | [
"Solution suggested by the student Mahdi Al-Shaikh Saleh. Consider the lattice $\\mathbb{Z}^{2}$ and $(x_{a}, y_{a}),(x_{b}, y_{b})$ the coordinates of positions of pebbles $A, B$ respectively. If we relocate one of the pebbles, say $B$, at a new position, $(x_{b}^{\\prime}, y_{b}^{\\prime})$, because $(x_{b}-x_{a}... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English, Arabic | proof and answer | No, it is impossible to switch the positions of the pebbles under the given moves. | |
043r | Given $2021$ distinct positive integers $a_1, a_2, \dots, a_{2021}$. Define the sequence $\{a_n\}$ inductively as follows: for each integer $n \ge 2022$, $a_n$ is the smallest positive integer different from $a_1, a_2, \dots, a_{n-1}$ and not dividing the product $a_{n-1}a_{n-2}\dots a_{n-2021}$. Prove: there exists a ... | [
"Let $k=2021$. In fact, we will prove the statement for any integer $k > 0$.\n\n**Lemma 1** There exists $C > 0$ independent of $k$, such that $\\tau(m) \\le C m^{\\frac{1}{k+1}}$ holds for all positive integer $m$, where $\\tau(m)$ is the number of positive factors of $m$.\n\n**Proof of lemma 1** Let $m = p_1^{\\a... | China | China National Team Selection Test | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0hl3 | Problem:
Find the number of real zeros of $x^{3}-x^{2}-x+2$. | [
"Solution:\n\nLet $f(x) = x^{3} - x^{2} - x + 2$, so $f'(x) = 3x^{2} - 2x - 1$.\n\nThe slope is zero when $3x^{2} - 2x - 1 = 0$, where $x = -\\frac{1}{3}$ and $x = 1$.\n\nNow $f\\left(\\frac{1}{3}\\right) > 0$ and $f(1) > 0$, so there are no zeros between $x = -\\frac{1}{3}$ and $x = 1$.\n\nSince $\\lim_{x \\righta... | United States | null | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 1 | |
0l1n | Let $n$ be the least prime number that can be written as the sum of 5 distinct prime numbers. What is the sum of the digits of $n$? | [
"The prime $2$ cannot be among the $5$ distinct primes chosen because, if it were, then the sum would be even. The first $5$ odd primes are $3$, $5$, $7$, $11$, and $13$, and their sum is $39$, which is not prime. The next smallest sum of $5$ distinct odd primes is $3 + 5 + 7 + 11 + 17 = 43$, which is prime. The re... | United States | AMC 10 A | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 7 | |
0c5y | Let $ABC$ be an acute triangle, let $D, E, F$ be the feet of the altitudes from $A, B, C$, respectively, and let $M, N, P$ be the midpoints of the sides $BC, CA, AB$, respectively. The circles $BDP$ and $CDN$ cross again at $X$, the circles $CEM$ and $AEP$ cross again at $Y$, and the circles $AFN$ and $BFM$ cross again... | [
"Clearly, it is sufficient to prove that $AX$ is the $A$-symmedian of the triangle $ABC$. To this end, we show that the triangles $XAB$ and $XCA$ are similar. It then follows that the line $AX$ bisects the angle $BXC$, and $XB/XC = AB^2/AC^2$, so $AX$ is indeed the $A$-symmedian of the triangle $ABC$.\nTo prove sim... | Romania | IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
06qc | For every $n \in \mathbb{N}$ let $d(n)$ denote the number of (positive) divisors of $n$. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ with the following properties:
(i) $d(f(x))=x$ for all $x \in \mathbb{N}$;
(ii) $f(x y)$ divides $(x-1) y^{x y-1} f(x)$ for all $x, y \in \mathbb{N}$. | [
"There is a unique solution: the function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ defined by $f(1)=1$ and\n$$\nf(n)=p_{1}^{p_{1}^{a_{1}}-1} p_{2}^{p_{2}^{a_{2}}-1} \\cdots p_{k}^{p_{k}^{a_{k}}-1} \\text{ where } n=p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{k}^{a_{k}} \\text{ is the prime factorization of } n>1.\n$$\n... | IMO | 49th International Mathematical Olympiad Spain | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | The unique function is given by f(1)=1 and, for n>1 with prime factorization n=∏ p_i^{a_i}, f(n)=∏ p_i^{p_i^{a_i}−1}. | |
0it2 | Problem:
Let $ABC$ be a triangle with $\angle A = 45^{\circ}$. Let $P$ be a point on side $BC$ with $PB = 3$ and $PC = 5$. Let $O$ be the circumcenter of $ABC$. Determine the length $OP$. | [
"Solution:\n\nUsing the extended Sine Law, we find the circumradius of $ABC$ to be $R = \\frac{BC}{2 \\sin A} = 4 \\sqrt{2}$.\n\nBy considering the power of point $P$, we find that $R^2 - OP^2 = PB \\cdot PC = 15$.\n\nSo $OP = \\sqrt{R^2 - 15} = \\sqrt{16 \\cdot 2 - 15} = \\sqrt{17}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | sqrt(17) | |
08pl | Problem:
Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{1}{a b(b+1)(c+1)}+\frac{1}{b c(c+1)(a+1)}+\frac{1}{c a(a+1)(b+1)} \geq \frac{3}{(1+a b c)^{2}}
$$ | [
"Solution:\nThe required inequality is equivalent to\n$$\n\\frac{c(a+1)+a(b+1)+b(c+1)}{a b c(a+1)(b+1)(c+1)} \\geq \\frac{3}{(1+a b c)^{2}}\n$$\nor equivalently to,\n$$\n(1+a b c)^{2}(a b+b c+c a+a+b+c) \\geq 3 a b c(a b+b c+c a+a+b+c+a b c+1)\n$$\nLet $m=a+b+c, n=a b+b c+c a$ and $x^{3}=a b c$, then the above can ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0j1k | Problem:
Let $a_{1}=3$, and for $n>1$, let $a_{n}$ be the largest real number such that
$$
4\left(a_{n-1}^{2}+a_{n}^{2}\right)=10 a_{n-1} a_{n}-9
$$
What is the largest positive integer less than $a_{8}$? | [
"Solution:\nAnswer: 335\nLet $t_{n}$ be the larger real such that $a_{n}=t_{n}+\\frac{1}{t_{n}}$. Then $t_{1}=\\frac{3+\\sqrt{5}}{2}$. We claim that $t_{n}=2 t_{n-1}$.\n\nWriting the recurrence as a quadratic polynomial in $a_{n}$, we have:\n$$\n4 a_{n}^{2}-10 a_{n-1} a_{n}+4 a_{n-1}^{2}+9=0\n$$\nUsing the quadrati... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 335 | |
02lo | The venusian prophet Zabruberson sent to his pupils a 10000-letter word, each letter being $A$ or $E$: the *Zabrubic word*. Their pupils decided that, for $1 \le k \le 10000$, each word comprised of $k$ consecutive letters of the *Zabrubic* word is a *prophetic word* of length $k$. It is known that there are at most 7 ... | [
"Let $f(n)$ be the maximum number of prophetic words of length $n$. Obviously, $f(1) = 2$, $f(2) = 4$ and $f(3) = 7$. Moreover, since there are exactly $2^3 = 8$ 3-letter words, each letter being $A$ or $E$, there is a forbidden substring $W$ of length three. So we can estimate $f(n)$ from above: in fact, each of t... | Brazil | XXX OBM | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 504 | |
03ny | Problem:
An acute triangle is a triangle that has all angles less than $90^{\circ}$ ($90^{\circ}$ is a Right Angle). Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$ meeting at $H$. The circle passing through points $D$, $E$, and $F$ meets $AD$, $BE$, and $CF$ again at $X$, $Y$, and $Z$ respectively.... | [
"Solution:\n\nLet the circumcircle of $ABC$ meet the altitudes $AD$, $BE$, and $CF$ again at $I$, $J$, and $K$ respectively.\n\n\n\nLemma (9-point circle). $I$, $J$, $K$ are the reflections of $H$ across $BC$, $CA$, $AB$. Moreover, $D$, $E$, $F$, $X$, $Y$, $Z$ are the midpoints of $HI$, $HJ... | Canada | CMO 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Equations and Inequalities > QM-... | null | proof only | null | |
026u | Problem:
A seguinte figura mostra um dodecágono regular.
Responda às seguintes perguntas:
a) Quantos triângulos equiláteros podem ser formados de modo que seus três vértices sejam vértices do dodecágono?
b) Quantos triângulos escalenos podem ser formados de modo que seus três vértices sejam vé... | [
"Solution:\nPodemos inscrever o dodecágono regular em um círculo como mostra a seguinte figura:\n\nEntão os arcos que correspondem a cada um dos lados do dodecágono devem medir $\\left(360 / 12\\right)^{\\circ}=30^{\\circ}$.\n\na) Os triângulos equiláteros que podemos formar estarão também ... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | a) 4; b) 168 | |
07xg | A Sudoku grid is a $9 \times 9$ table in which each row and each column contains each of the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ in some order. In addition, the nine $3 \times 3$ subgrids contain each of the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once. Suppose the product of all nine numbers on one diagonal is $M$... | [
"Because $2025 = 3^4 \\cdot 5^2$, we have $2025^3 = 3^{12} \\cdot 5^6$ and $2025^4 = 3^{16} \\cdot 5^8$. Each diagonal can have at most three $5$s. Therefore, $MN$ can have at most six factors $5$, and this only happens when the number in the centre of the grid is $5$. Hence $MN$ might be divisible by $2025^3$ but ... | Ireland | IRL_ABooklet_2025 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cap | Problem:
Fie $a, b, c, d$ numere naturale nenule, cu $a < b < c < d$ și $a d = b c$. Demonstrați că
$$
2 a + \sqrt{a} + \sqrt{d} < b + c + 1
$$ | [
"Solution:\n\nFie $b = a + x$, $c = a + y$, $d = a + z$ cu $0 < x < y < z$ numere naturale.\nAtunci relația din enunț se scrie $a(a + z) = (a + x)(a + y)$. Deducem că $z = x + y + \\frac{x y}{a} > x + y \\Rightarrow z \\geqslant x + y + 1$. Obținem astfel\n$$\na = \\frac{x y}{z - x - y} \\leqslant x y\n$$\nAvem că ... | Romania | Olimpiada Națională de Matematică | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0c7k | Find all positive integers $p$ for which there exists $n \in \mathbb{N}^*$ such that $p^n + 3^n$ divides $p^{n+1} + 3^{n+1}$. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 3 | |
0buf | Problem:
Să se calculeze $\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x^{2}}{x \sqrt{2}+\sqrt{1-x^{2}}+\sqrt{1+x^{2}}} d x$ | [] | Romania | Olimpiada Națională de Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | π/12 − √3/8 | |
0cek | Fix an integer $n \ge 2$. Determine the least possible value the sum
$$
\left\lfloor \frac{x_2 + x_3 + \cdots + x_n}{x_1} \right\rfloor + \left\lfloor \frac{x_1 + x_2 + \cdots + x_n}{x_2} \right\rfloor + \cdots + \left\lfloor \frac{x_1 + x_2 + \cdots + x_{n-1}}{x_n} \right\rfloor
$$
may achieve, as $x_1, x_2, \dots, x_... | [
"The minimum exists, as the summands are all non-negative integers; it is equal to $(n-1)^2$ and is achieved if, for instance, $x_1 = n$ and $x_2 = \\cdots = x_n = n+1$; the verification is routine.\n\nLet $s = x_1 + x_2 + \\cdots + x_n$ and let $S$ denote the sum in the statement. Note that\n$$\n\\left\\lfloor \\f... | Romania | Eighteenth STARS OF MATHEMATICS Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | (n-1)^2 | |
02xn | Problem:
Em uma sequência de inteiros positivos, uma inversão é um par de posições em que o elemento da posição mais à esquerda é maior que o elemento da posição mais à direita. Por exemplo, a sequência $2,5,3,1,3$ tem $5$ inversões: entre a primeira e a quarta posição, entre a segunda e todas as demais para a direita... | [
"Solution:\n\na) Primeiramente vamos mostrar que qualquer sequência maximizante do número de inversões precisa ser não-crescente. De fato, se existe um par de números consecutivos $a$ e $b$, com $a < b$, então a troca de posição desses elementos não altera a soma e aumenta o número de inversões em uma unidade. Para... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | a) 6; b) 509545 | |
034h | Problem:
Solve in integers the equation
$$
z^{2}+1=xy(xy+2y-2x-4).
$$ | [
"Solution:\nWe set $x = u - 1$, $y = v + 1$ and obtain the equation\n$$\nz^{2} + 1 = (u^{2} - 1)(v^{2} - 1)\n$$\nIt is easy to see that $u$, $v$ and $z$ must be even. Hence, if $|u| > 1$, then $u^{2} - 1$ has a prime divisor $p$ such that $p \\equiv 3 \\pmod{4}$. Therefore $z^{2} + 1 \\equiv 0 \\pmod{p}$, which is ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (-1, 1, 0) | |
0c1v | Let $n$ be a natural number with $n \ge 2$. Prove that, for any complex numbers $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$, the following statements are equivalent:
$$
a) \quad \sum_{k=1}^{n} |z - a_k|^2 \le \sum_{k=1}^{n} |z - b_k|^2, \text{ for any } z \in \mathbb{C};
$$
$$
b) \quad \sum_{k=1}^{n} a_k = \sum_{... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
099d | Let $a$, $b$, $c$ positive real numbers. Prove that
$$
\sqrt{\frac{3bc}{(a+b)(a+b+c)}} + \sqrt[4]{\frac{12a(a+b)}{(a+b+c)^2}} \le 2.
$$
Find the equality condition. | [
"It's easy to check, when $a = 1$, $b = 2$, $c = 3$ equality holds, hence by the Cauchy's inequality we have\n$$\n\\begin{aligned}\n& \\frac{1}{4} \\left( 2 \\sqrt{\\frac{3b}{a+b} \\cdot \\frac{4c}{a+b+c}} + 4 \\sqrt[4]{\\frac{3a}{a+b} \\left( \\frac{2(a+b)}{a+b+c} \\right)^2} \\cdot 1 \\right) \\le \\\\\n& \\le \\... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds exactly when a, b, c are in the ratio 1:2:3 (i.e., b = 2a and c = 3a with a > 0). | |
0fb1 | Problem:
Find all real $x$, $y$ such that $(1 + x)(1 + x^{2})(1 + x^{4}) = 1 + y^{7}$, $(1 + y)(1 + y^{2})(1 + y^{4}) = 1 + x^{7}$? | [
"Solution:\nIf $x = y$, then clearly $x \\neq 1$, so we have $(1 - x^{8}) = (1 - x)(1 + x^{7}) = 1 - x^{8} - x + x^{7}$, so $x = 0$ or $x^{6} = 1$, whose only real root (apart from the $x = 1$ we have discarded) is $x = -1$. That gives the two solutions above.\n\nSo assume $x \\neq y$. WLOG $x > y$.\nSo $(1 + x) > ... | Soviet Union | 1st CIS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | x = y = 0 or x = y = -1 | |
08ty | Determine the last 3 digits of the number obtained by multiplying all the odd numbers between $1$ and $100$. | [
"Let $N$ be the product of all the odd integers lying in between $1$ and $100$. It is clear that the last $3$ digits of $N$ equals the remainder obtained when $N$ is divided by $1000$. Note that $1000 = 10^3 = 2^3 \\times 5^3 = 8 \\times 125$.\n\nFrom the identity $(8x + a)(8y + b) = 8(axy + a + b) + ab$, valid for... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 875 | |
06ev | Given finitely many points in a plane, it is known that the area of the triangle formed by any three points of the set is less than $1$. Show that all points of the set lie inside or on the boundary of a triangle with area less than $4$. | [
"Let $A$, $B$, $C$ be points in $S$ such that $[ABC]$ is maximized. Let $\\ell_1$ be the line passing through $A$ which is parallel to $BC$. If there exists a point $X \\in S$ which lies on a different side of $\\ell_1$ as $B$, then the distance from $X$ to $BC$ is greater than that from $A$ to $BC$. This yields th... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and ... | null | proof only | null | |
0g4e | Problem:
In a garden, there are 2023 rose bushes planted in a row. Each bush contains either red or blue roses. Vicky is taking a walk and wants to pick some of the flowers. She starts at a bush of her choice, and picks a rose from it to add to her basket. She then continues walking down the row and picks a single flo... | [
"Solution:\n\n507\n\nWLOG, let there be more red than blue bushes, with $R > B$ total bushes of each colour, respectively. If $r = R$, Vicky is only forced to stop by at most $\\left\\lfloor \\frac{B}{2} \\right\\rfloor$, which gives a total of $R - \\left\\lfloor \\frac{B}{2} \\right\\rfloor \\geq 507$. This bound... | Switzerland | Switzerland Selection Solution | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 507 | |
03s0 | An integer $n$ is called good if $n \ge 3$ and there are $n$ lattice points $P_1, P_2, \dots, P_n$ in the coordinate plane satisfying the following conditions: If line segment $P_iP_j$ has a rational length, then there is $P_k$ such that both line segments $P_iP_k$ and $P_jP_k$ have irrational lengths; and if line segm... | [
"We claim that the minimum good number is $5$, and that $2005$ is good.\n\nIt is not difficult to see that $n=3$ is not a good number. Note that $n=4$ is also not a good number. Assume on the contrary that there are lattice points $P_1, P_2, P_3, P_4$ satisfying the conditions of the problem. Without loss of genera... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 5 | |
09kf | For all positive real numbers $u$ and $v$, prove
$$
\min \left\{ u, \frac{100}{v}, v + \frac{2023}{u} \right\} \le \sqrt{2123}.
$$
Here $\min X$ denotes the minimum element of $X$. | [
"Let us denote the left side of the given inequality by $S$.\n\nIf $u \\le \\sqrt{2123}$, then $S \\le u \\le \\sqrt{2123}$.\n\nIf $\\frac{100}{v} \\le \\sqrt{2123}$, then $S \\le \\frac{100}{v} \\le \\sqrt{2123}$.\n\nIf $u \\ge \\sqrt{2123}$ and $\\frac{100}{v} \\ge \\sqrt{2123}$, then we have\n$$\nS \\le v + \\fr... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0flt | Let $p$ be a prime number, and let $x$ and $n$ be integers such that $1 \le x < n$. We have $x + 1$ different boxes and $n - x$ identical balls. Let us call $f(n,x)$ the number of ways in which one can distribute the $n - x$ into the $x + 1$ boxes. Find all integers $n$ greater than $1$ such that $p$ divides $f(n,x)$ f... | [] | Spain | Spanija 2012 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | n is a power of p (n = p^k for some integer k ≥ 1) | |
05ac | The internal angle bisectors at vertices $B$ and $C$ of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at $E$ and $F$, respectively. Given that $BE = CF \neq 0$, may we be sure that triangle $ABC$ is isosceles? | [
"Let $\\angle ABC = 30^\\circ$ and $\\angle BCA = 90^\\circ$. Let $O$ be the circumcenter of triangle $ABC$; it is also the midpoint of the hypotenuse $AB$. We have $\\angle OAC = \\angle BAC = 60^\\circ$, and since $AO = CO$, it follows that $\\angle OCA = 60^\\circ$ (Fig. 30). Therefore, $\\angle FCO = 60^\\circ ... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | No | |
01sg | A straight line $l$ and a point $A$ are on a plane. $A$ does not belong to the $l$. For each point $M$ on the line $l$ a point $N$ is marked on the plane so that the triangle $AMN$ is equilateral (the vertices of the triangle $AMN$ are mentioned clockwise).
Find the locus of the vertices $N$ of the equilateral triangle... | [
"If $A$ lies under the line $\\ell$, then the required locus is the line $BC$, where $B \\in \\ell$, $C \\in m$, $m \\parallel \\ell$, $m \\nparallel A$, and the triangle $ABC$ is an equilateral triangle.\n\nWithout loss of generality suppose that $A$ lies under the line $\\ell$. Show that the required locus is the... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00nf | Let $a$, $b$, $c$ be pairwise distinct natural numbers.
Prove that
$$
\frac{a^3 + b^3 + c^3}{3} \ge abc + a + b + c.
$$
When does equality hold? | [
"It is well-known and easily verified that\n$$\na^3 + b^3 + c^3 - 3abc = \\frac{1}{2}(a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2). \\quad (1)\n$$\nAssume without loss of generality that $a > b > c \\ge 0$. Since the numbers are integers, we obtain $a - b \\ge 1$, $b - c \\ge 1$ and $a - c \\ge 2$.\nEquation (1) n... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Equality holds precisely when the three numbers are consecutive natural numbers, i.e., any permutation of t, t+1, t+2 for integer t ≥ 0. | |
0fli | Problem:
Se considera el polinomio de segundo grado $p(x) = a x^{2} + b x + c$, $(a \neq 0)$, cuyas raíces $x_{1}$ y $x_{2}$ se suponen distintas. Justifica que para que $p\left(x_{1}^{3}\right) = p\left(x_{2}^{3}\right)$ es suficiente que $a^{2} + 3 a c - b^{2} = 0$. ¿Es también necesaria esta condición? | [
"Solution:\n\n$$\n\\begin{gathered}\np\\left(x_{1}^{3}\\right) - p\\left(x_{2}^{3}\\right) = a\\left(x_{1}^{6} - x_{2}^{6}\\right) + b\\left(x_{1}^{3} - x_{2}^{3}\\right) = \\left(x_{1}^{3} - x_{2}^{3}\\right)\\left[a\\left(x_{1}^{3} + x_{2}^{3}\\right) + b\\right] \\\\\nx_{1}^{3} + x_{2}^{3} = \\left(x_{1} + x_{2}... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | No | |
098v | Problem:
Determinați toate funcțiile derivabile $F:(0 ;+\infty) \rightarrow \mathbb{R}$, pentru care: $F(1)=1$ și $F\left(\frac{1}{x}\right) F^{\prime}(x)=\frac{1}{x} \ln x$. | [
"Solution:\n\nFie $F^{\\prime}(x)=f(x)$. Atunci $\\left(F\\left(\\frac{1}{x}\\right)\\right)^{\\prime}=-\\frac{1}{x^{2}} f\\left(\\frac{1}{x}\\right)$.\nDeoarece $F\\left(\\frac{1}{x}\\right) f(x)=\\frac{1}{x} \\ln x$, obținem că $F(x) f\\left(\\frac{1}{x}\\right)=-x \\ln x$. Înmulțind fiecare parte a ultimei egali... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | F(x) = sqrt((ln x)^2 + 1) | |
02l0 | Problem:
Fatoriais - Se $n$ é um número natural, denotamos por $n!$ o produto de todos os inteiros de 1 a $n$. Por exemplo: $5! = 1 \times 2 \times 3 \times 4 \times 5$ e $13! = 1 \times 2 \times 3 \times 4 \times 5 \times \ldots \times 12 \times 13$. Por convenção, $0! = 1$. Encontre três números inteiros diferentes ... | [
"Solution:\n\nPrimeiramente observe que como o número tem 3 algarismos, então o maior dos algarismos tem que ser menor que ou igual a 6, já que $7! > 1000$. Como o número tem que ter 3 algarismos e $4! = 1 \\times 2 \\times 3 \\times 4 = 12$, então um dos algarismos tem que ser 5 ou 6, mas $6! = 720$ implicaria que... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 145 | |
0dmx | Problem:
Нека је $H$ ортоцентар, а $O$ центар описане кружнице оштроуглог троугла $ABC$. Тачке $D$ и $E$ су подножја висина из $A$ и $B$, редом. Обележимо са $K$ пресечну тачку правих $OD$ и $BE$, а са $L$ пресечну тачку правих $OE$ и $AD$. Нека је $X$ друга пресечна тачка кружница описаних око троуглова $HKD$ и $HLE$... | [
"Solution:\n\nАко је $X$ центар описаног круга $\\triangle ODE$, онда је $90^\\circ - \\angle KDE = 90^\\circ - \\angle ODE = \\angle XEO = \\angle XEL = \\angle XHD = \\angle XKD$ (сви углови су оријентисани), одакле следи да је $XK \\perp DE$; аналогно $XL \\perp DE$, тј. $K$ и $L$ леже на симетрали дужи $DE$, па... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
03uh | Given $\triangle ABC$ and $\triangle AEF$ such that $B$ is the midpoint of $EF$. Also, $AB = EF = 1$, $BC = 6$, $CA = \sqrt{33}$, and $\overrightarrow{AB} \cdot \overrightarrow{AE} + \overrightarrow{AC} \cdot \overrightarrow{AF} = 2$. The cosine of the angle between $\overrightarrow{EF}$ and $\overrightarrow{BC}$ is __... | [
"We have\n$$\n\\begin{aligned}\n2 &= \\overrightarrow{AB} \\cdot \\overrightarrow{AE} + \\overrightarrow{AC} \\cdot \\overrightarrow{AF} \\\\\n&= \\overrightarrow{AB} \\cdot (\\overrightarrow{AB} + \\overrightarrow{BE}) + \\overrightarrow{AC} \\cdot (\\overrightarrow{AB} + \\overrightarrow{BF}),\n\\end{aligned}\n$$... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | final answer only | 2/3 | |
0eel | Problem:
Naj bo $\mathcal{K}$ krožnica s središčem $O$ in $\mathcal{K}'$ krožnica, ki poteka skozi točko $O$ in ima polmer večji od dvakratnika polmera krožnice $\mathcal{K}$. Skupna tangenta krožnic $\mathcal{K}$ in $\mathcal{K}'$ se dotika krožnice $\mathcal{K}$ v točki $A$, krožnice $\mathcal{K}'$ pa v točki $B$. O... | [
"Solution:\n\n\nTangenta iz točke $B$ na krožnico $\\mathcal{K}$ različna od tangente $A B$ naj se krožnice $\\mathcal{K}$ dotika v točki $A'$. Dokazali bomo, da so točke $B, A'$ in $E$ kolinearne.\nKer je $|A B|=|A' B|$ in $|A O|=|A' O|$, se trikotnika $A B O$ in $A' B O$ ujemata v vseh tr... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | null | proof only | null | |
0007 | Sea $\lambda$ un número real tal que la desigualdad $0 < \sqrt{2002} - \frac{a}{b} < \frac{\lambda}{ab}$ se verifica para infinitos pares $(a, b)$ de números enteros positivos. Demostrar que $\lambda \ge 5$. | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | español | proof only | null | |
02lu | There are $2009$ pebbles in some points $(x, y)$ with both coordinates integer. An operation consists in choosing a point $(a, b)$ with four or more pebbles, removing four pebbles from $(a, b)$ and putting one pebble in each of the points
$$
(a, b - 1), (a, b + 1), (a - 1, b), (a + 1, b)
$$
Show that after a finite num... | [
"For each configuration that can be obtained, let $S$ be the sum of the squares of the coordinates of the pebbles. Each operation takes four pebbles at $(x, y)$ to one pebble at each of the points $(x-1, y)$, $(x+1, y)$, $(x, y-1)$, $(x, y+1)$, so each operation increases $S$ by\n$$(x-1)^2 + y^2 + (x+1)^2 + y^2 + x... | Brazil | XXXI Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
01c9 | Find integers $0 < a_1 < a_2 < a_3 < a_4$ such that for any $1 \le k < l \le 4$ number $a_k \cdot a_l + 1$ is a square of an integer. | [
"**Answer:** For example $2$, $4$, $12$, $420$.\n\nFor $a_1 = 2$ and $a_2 = 4$ we look for a number $a_3$ such that $2a_3 + 1$ and $4a_3 + 1$ are squares, say $b^2$ and $c^2$ respectively. Then we have $2b^2 - c^2 = 1$, which is Pell's equation.\n\nConsider two consecutive solutions of this Pell's equation: $(5, 7)... | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Pell's equations"
] | null | proof and answer | 2, 4, 12, 420 | |
0f9l | Problem:
A cube of side $100$ is divided into a million unit cubes with faces parallel to the large cube. The edges form a lattice. A prong is any three unit edges with a common vertex. Can we decompose the lattice into prongs with no common edges? | [] | Soviet Union | 24th ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Yes | |
08n8 | Problem:
Find all positive integers $n$ such that the equation $y^{2}+x y+3 x=n\left(x^{2}+x y+3 y\right)$ has at least a solution $(x, y)$ in positive integers. | [
"Solution:\nClearly for $n=1$, each pair $(x, y)$ with $x=y$ is a solution. Now, suppose that $n>1$ which implies $x \\neq y$. We have\n$$\n0 < n-1 = \\frac{y^{2}+x y+3 x}{x^{2}+x y+3 y}-1 = \\frac{(x+y-3)(y-x)}{x^{2}+x y+3 y}.\n$$\nSince $x+y \\geq 3$, we conclude that $x+y>3$ and $y>x$. Take $d=\\operatorname{gcd... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | {1, 3, 4, 9} | |
02sq | Problem:
Um hospital tem os seguintes funcionários:
Sara Dores da Costa: reumatologista
Iná Lemos: pneumologista
Ester Elisa: enfermeira
Ema Thomas: traumatologista
Ana Lisa: psicanalista
Inácio Filho: obstetra
a) De quantas maneiras os funcionários podem fazer uma fila?
b) De quantas maneiras os mesmos funcionários... | [
"Solution:\n\na) Para ser o primeiro da fila, podemos escolher qualquer um dos seis funcionários. Logo, há 6 possibilidades. Escolhido o primeiro da fila, restam cinco funcionários a serem escolhidos para ser o segundo da fila (porque um já foi escolhido). Para o terceiro lugar temos 4 possibilidades, e assim por d... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | a) 720; b) 120; c) 120 | |
044j | Find the smallest real number $\alpha$, such that for any convex polygon $P$ of area $1$, there exists a point $M$ in the plane, such that the area of the convex hull of $P \cup Q$ is at most $\alpha$, where $Q$ is the central-symmetric figure of $P$ about $M$.
(Contributed by Qu Zhenhua and Wu Yuchi) | [
"(i) If $M$ is outside $P$ or on the boundary. Draw a line $l$ through $M$ such that $P$, $Q$ lie on different sides of $l$ (they have no common interior points). Then $S(P \\cup Q) \\ge S(P) + S(Q) \\ge 2$.\n\n(ii) If $M$ is inside $P$. Let $A', B', C'$ be the respective symmetric points of $A, B, C$ about $M$. Th... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2 | |
0b7b | Given is a regular polygon $A_1A_2...A_{2010}$ centered at $O$. On each of the segments $OA_k$, with $k = 1, 2, ..., 2010$, lies the point $B_k$ such that
$$
\frac{OB_k}{OA_k} = \frac{1}{k}.
$$
Determine the ratio between the area of the polygon $B_1B_2...B_{2010}$ and that of $A_1A_2...A_{2010}$. | [
"In the sequel $[XYZ]$ will represent the area of a triangle $XYZ$. Denote by $S$ the area of the polygon $A_1A_2...A_{2010}$.\nNotice that $[OA_1A_2] = [OA_2A_3] = \\dots = [OA_{2010}A_1] = \\frac{1}{2010}S$, since the polygon $A_1A_2...A_{2010}$ is given as being regular. Then for each $k = 1, 2, ..., 2009$ we ha... | Romania | Local Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 1/2010 | |
05b2 | Prove that $2023^{10} > 10 \cdot (1^9 + 2^9 + 3^9 + \dots + 2022^9)$. | [
"Multiplying out $(k+1)^{10}$ yields among others the terms $k^{10}$ and $10k^9$. All of the other terms are positive for a positive $k$. Thus for all positive integers $k$ we have $(k+1)^{10} > k^{10} + 10k^9$. Combining this for $k = 2022, 2021, \\dots, 2, 1$ yields\n$$\n\\begin{aligned}\n2023^{10} &> 2022^{10} +... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
01pq | Find all pairs $(n; p)$ of natural numbers $n$ and prime numbers $p$ satisfying the equality $p^8 - p^4 = n^5 - n$. | [
"Answer: $(n; p) = (3; 2)$.\nIt is clear that $p \\ne n$. If $p=2$, then $n \\ge 3$ and we have $2^8 - 2^4 = 240 = 3^5 - 3$, i.e. $p=2$, $n=3$ is a solution. On the other hand, if $n > 3$, then $n^5 - n = n(n^4 - 1) > 3(3^4 - 1) = 240$, i.e. for $p=2$ there are no $n$ different from $3$ satisfying the initial equal... | Belarus | BelarusMO 2013_s | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 3, p = 2 | |
04ue | Let $ABCD$ be an isosceles trapezoid with longer base $AB$. Let $I$ be the incenter of triangle $ABC$ and $J$ the $C$-excenter of triangle $ACD$. Prove that $IJ$ and $AB$ are parallel. (Patrik Bak) | [
"Let $K$ be the incenter of triangle $ABD$. Since $IK \\parallel AB$, it suffices to show $JK \\parallel AB$. Let $\\angle ABD = \\angle ACD = \\varphi$. Then $\\angle AKD = 90^\\circ + \\frac{1}{2}\\varphi$ and $\\angle DJA = 90^\\circ - \\frac{1}{2}\\varphi$, implying that the quadrilateral $AKDJ$ is cyclic (Fig.... | Czech Republic | 67th Czech and Slovak 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 > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01l3 | Given cyclic quadrilateral $ABCD$ with $\frac{CD}{BD} > \frac{AB}{AC}$.
Prove that $CD \cdot BD > AB \cdot AC$. | [
"Since $ABCD$ is cyclic, $\\angle ABD = \\angle ACD$, $\\angle BAC = \\angle BDC$. For the area of the triangles $ABD$ and $ACD$ we have\n$$\nS(ABD) = AB \\cdot BD \\sin \\angle ABD = S_1 + S_4,\n$$\n$$\nS(ACD) = AC \\cdot CD \\sin \\angle ACD = S_3 + S_4.\n$$\nBy condition, $AC \\cdot CD > AB \\cdot BD$, so $S(ACD... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0a6k | Problem:
For which positive integers $n$, does there exist a sequence of real numbers $(x_{1}, x_{2}, \ldots, x_{n})$ such that
- $-2 < x_{i} < 2$ for all $i$,
- $x_{1} + x_{2} + x_{3} + \dots + x_{n} = 0$, and
- $x_{1}^{4} + x_{2}^{4} + x_{3}^{4} + \dots + x_{n}^{4} \geqslant 32$. | [
"Solution:\nNote that if $n = j$ works then $n > j$ also works for a positive integer $j$ as we can just set $x_{i} = 0$ for $n \\geq i > j$ and have $x_{1}, \\ldots, x_{n}$ be the sequence that worked for $n$.\n\nConsider $n = 4$. We take $x_{1}, x_{2} = \\sqrt[4]{8}$ and $x_{3}, x_{4} = -\\sqrt[4]{8}$ and all con... | New Zealand | NZMO Round Two | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | All integers n ≥ 4 |
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