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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
008d | Let a positive integer $n$ be called *apocalyptic* if among its positive divisors there are six of them which sum is equal to $3528$. For instance, $2012$ is apocalyptic since the sum of its six divisors, $1$, $2$, $4$, $503$, $1006$ and $2012$, is equal to $3528$. Determine the smallest apocalyptic positive integer. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | English | proof and answer | 2012 | |
078b | Let $n$ be any positive integer, and let $S(n)$ denote the number of permutations $\tau$ of $\{1, \dots, n\}$ such that $k^4 + (\tau(k))^4$ is prime for all $k = 1, \dots, n$. Show that $S(n)$ is always a square. | [] | India | IMOTC Practice Test 2 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Other"
] | null | proof only | null | |
0iek | Problem:
Two vertices of a cube are given in space. The locus of points that could be a third vertex of the cube is the union of $n$ circles. Find $n$. | [
"Solution:\nLet the distance between the two given vertices be $1$. If the two given vertices are adjacent, then the other vertices lie on four circles, two of radius $1$ and two of radius $\\sqrt{2}$. If the two vertices are separated by a diagonal of a face of the cube, then the locus of possible vertices adjacen... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 10 | |
0cic | Three lines in space are pairwise skew and are located at distance $1$ from each other. Prove that there exists a line situated at distance $1$ from each of the three lines. | [] | Romania | 75th NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
02gv | Determine all values of $n$ such that it is possible to divide a triangle in $n$ smaller triangles such that there are not three collinear vertices and such that each vertex belongs to the same number of segments. | [
"Consider the planar graph which vertices are the vertices of the triangles and edges are the sides of the triangles. Let $V$, $E$ and $F$ be the number of vertices, edges and faces of such graph and $d$ be the degree of each vertex. Note that $n = F - 1$. Then $E = \\frac{3F}{2} = \\frac{dV}{2} \\iff V = \\frac{2E... | Brazil | XXVI OBM | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 3, 7, 19 | |
01tb | Solve the equation $2^a - 5^b = 3$ in positive integers $a, b$. | [
"Answer: $(a; b) = (3; 1)$ or $(7; 3)$.\n\n(Solution by Y. Dubovik.) Note that if $a \\le 7$ or $b \\le 3$, then it is easy to verify that the only solutions are $(a; b) = (3; 1)$ and $(a; b) = (7; 3)$.\n\nNow it remains to prove that there are no solutions with $a > 7, b > 3$. In this case we can write $a = 7 + \\... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorizati... | English | proof and answer | (a, b) = (3, 1) or (7, 3) | |
0b3c | Problem:
Let $XZ$ be a diameter of circle $\omega$. Let $Y$ be a point on $XZ$ such that $XY = 7$ and $YZ = 1$. Let $W$ be a point on $\omega$ such that $WY$ is perpendicular to $XZ$. What is the square of the length of the line segment $WY$?
(a) 7
(b) 8
(c) 10
(d) 25 | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | a | |
05he | Problem:
$n$ joueurs participent à un tournoi d'échecs. Chaque joueur fait exactement une partie avec chacun des autres joueurs. Une victoire rapporte 1 point, un match nul un demipoint et une défaite aucun point. Une partie est dite anormale si le gagnant de cette partie obtient un score au tournoi strictement plus f... | [
"Solution:\n\na. Soit $m = [n / 2]$. On classe les joueurs suivant leur score (parmi les joueurs ayant le même score, on les classe arbitrairement). On dira qu'un joueur classé parmi les $m$ meilleurs est fort; sinon, on dira qu'il est faible.\n\nSoit $x$ le nombre de parties normales entre les joueurs forts et les... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | a) At most 75%.
b) Yes; it can be at least 70% (for example, a construction with a large enough number of players exceeds 70%). | |
0an8 | Problem:
If $4 + 12 \cdot 4^{x} = 16 \cdot 16^{x}$, what is the value of $2^{2x+4} - 2^{2x}$?
(a) 120
(b) 60
(c) 30
(d) 15 | [] | Philippines | Qualifying Round | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | d | |
03ub | Let $A$ and $B$ be two subsets of $\{1, 2, 3, \dots, 100\}$, satisfying $|A| = |B|$ and $A \cap B = \emptyset$. If $n \in A$ always implies $2n + 2 \in B$, then the maximum of $|A \cup B|$ is ( ). | [
"We will first prove that $|A \\cup B| \\le 66$, or equivalently $|A| \\le 33$. For this purpose, we only need to prove that, if $A$ is a subset of $\\{1, 2, \\dots, 49\\}$ with 34 elements, then there must exist $n \\in A$ such that $2n + 2 \\in A$. The proof is as follows.\n\nDivide $\\{1, 2, \\dots, 49\\}$ into ... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 66 | |
0eoh | The list of numbers
16; 19; 24; 25; 30; 31; 32; 46; $x$
has the same median and mean. If $x$ is greater than $46$, then the value of $x$ is
(A) 47 (B) 48 (C) 53 (D) 50 (E) 57 | [
"The median is the middle number, which is $30$, so the mean is also $30$. The sum of all nine numbers is therefore $9 \\times 30 = 270$, which means that $x = 270 - (16 + 19 + 24 + 25 + 30 + 31 + 32 + 46) = 270 - 223 = 47$."
] | South Africa | South African Mathematics Olympiad | [
"Statistics > Mathematical Statistics"
] | English | MCQ | A | |
02ik | Problem:
Se $n$ é um número natural e $\frac{n}{24}$ é um número entre $\frac{1}{6}$ e $\frac{1}{4}$, então $n$ é igual a:
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9 | [
"Solution:\n\nComo $\\frac{1}{6} = \\frac{4}{24}$ e $\\frac{1}{4} = \\frac{6}{24}$, então $n$ só pode ser igual a $5$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
0awj | Problem:
Given $n \in \mathbb{N}$, let $\sigma(n)$ denote the sum of the divisors of $n$ and $\varphi(n)$ denote the number of positive integers $m \leq n$ for which $\operatorname{gcd}(m, n)=1$. Show that for all $n \in \mathbb{N}$,
$$
\frac{1}{\varphi(n)}+\frac{1}{\sigma(n)} \geq \frac{2}{n}
$$
and determine when eq... | [
"Solution:\n\nWe note that equality holds for $n=1$. We prove the inequality when $n>1$ and show that it is strict in this case.\n\nBy the AM-GM inequality, $\\frac{1}{\\varphi(n)}+\\frac{1}{\\sigma(n)} \\geq \\frac{2}{\\sqrt{\\varphi(n) \\sigma(n)}}$. Hence, we need only show that $\\varphi(n) \\sigma(n)<n^{2}$, o... | Philippines | 19th Philippine Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Equality holds only for n = 1. | |
0do7 | Problem:
Конвексан четвороугао $A B C D$ је описан око кружнице $k$. Праве $A D$ и $B C$ се секу у тачки $P$, а кружнице описане око $\triangle P A B$ и $\triangle P C D$ се секу у тачки $X$. Доказати да тангенте из тачке $X$ на кружницу $k$ граде једнаке углове са правима $A X$ и $C X$. | [
"Solution:\n\nПошто је $\\varangle X A D=\\varangle X B C$ и $\\varangle X D P=\\varangle X C P$, важи $\\triangle X A D \\sim \\triangle X B C$.\n\nНека симетрала $s_{X}$ угла $A X C$ сече кругове $P A B$ и $P C D$ у тачкама $K$ и $R$, а симетрала $s_{P}$ угла $A P C$ сече кругове $P A B$ и $P C D$ у тачкама $L$ и... | Serbia | 13. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0hg4 | Represent the fraction $\frac{1}{2022}$ as a difference of two regular fractions with smaller denominators. | [
"One of the possible representations is\n$$\n\\frac{1}{2022} = \\frac{1}{2 \\cdot 3 \\cdot 337} = \\frac{3-2}{2 \\cdot 3 \\cdot 337} = \\frac{3}{2 \\cdot 3 \\cdot 337} - \\frac{2}{2 \\cdot 3 \\cdot 337} = \\frac{1}{2 \\cdot 337} - \\frac{1}{3 \\cdot 237} = \\frac{1}{674} - \\frac{1}{1011}.\n$$"
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | final answer only | 1/674 - 1/1011 | |
002x | Sea $O$ el circuncentro de un triángulo acutángulo $ABC$ y $A_1$ un punto en el arco menor $BC$ de la circunferencia circunscrita al triángulo $ABC$. Sean $A_2$ y $A_3$ puntos en los lados $AB$ y $AC$ respectivamente, tales que $\angle BAA_1 = \angle OAC$ y $\angle CA_1A_3 = \angle OAB$. Demuestre que la recta $A_2A_3$... | [] | Argentina | XX Olimpiada Iberoamericana de Matemáticas | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Simson l... | Español | proof only | null | |
0562 | Find all prime numbers $p$ such that $2p^3 + 4p^2 - 3p + 12$ is the fifth power of an integer. | [
"Denote $f(n) = 2n^3 + 4n^2 - 3n + 12$. The following table shows the remainders of $n^2, n^3, n^5$ and $f(n)$ upon division by 11:\n\n| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|-------------|---|---|---|---|---|---|---|---|---|---|----|\n| $n^2$ | 0 | 1 | 4 | 9 | 5 | 3 | 3 | 5 | 9 | 4 | 1... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | 11 | |
09fp | Do there exist four quadratic polynomials such that the sum of any three of them has a real root, but the sum of any two of them has no real root? | [
"Answer: no.\nAssume that there exist such four quadratic polynomials $f_1, f_2, f_3$ and $f_4$. If a quadratic polynomial $f(x)$ has no real root, then either $f(x) > 0$ for all $x \\in \\mathbb{R}$ (we write $f > 0$ and say $f$ is positive) or $f(x) < 0$ for all $x \\in \\mathbb{R}$ (we write $f < 0$ and say $f$ ... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | no | |
0ika | Problem:
Compute
$$
\int_{0}^{1} \frac{d x}{\sqrt{x}+\sqrt[3]{x}}
$$ | [
"Solution:\nWriting $x = u^{6}$ so that $d x = 6 u^{5} d u$, we have\n$$\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d x}{\\sqrt{x}+\\sqrt[3]{x}} & = \\int_{0}^{1} \\frac{6 u^{5} d u}{u^{3}+u^{2}} \\\\\n& = 6 \\int_{0}^{1} \\frac{u^{3} d u}{u+1} \\\\\n& = 6 \\int_{0}^{1}\\left(u^{2}-u+1-\\frac{1}{u+1}\\right) d u \\\\\... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | 5 - 6 ln(2) | |
07m9 | Suppose $a$, $b$, $c$ are the side lengths of a triangle $ABC$. Prove that
$$
2ab \sin \left(\frac{C}{2}\right) + 2bc \sin \left(\frac{A}{2}\right) + 2ca \sin \left(\frac{B}{2}\right) \le a^2 + b^2 + c^2;
$$
and that the inequality is strict unless the triangle $ABC$ is equilateral. | [
"Let $2s = a + b + c$. Using $\\cos(C) = \\frac{a^2 + b^2 - c^2}{2ab}$, we obtain\n$$\n\\begin{aligned}\n2 \\sin^2 \\left( \\frac{C}{2} \\right) &= 1 - \\cos(C) = \\frac{2ab - a^2 - b^2 + c^2}{2ab} \\\\\n&= \\frac{c^2 - (a-b)^2}{2ab} = \\frac{(c-a+b)(c+a-b)}{2ab} \\\\\n&= \\frac{2(s-a)(s-b)}{ab}\n\\end{aligned}\n$$... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0izp | Problem:
A triangle in the $xy$-plane is such that when projected onto the $x$-axis, $y$-axis, and the line $y = x$, the results are line segments whose endpoints are $(1,0)$ and $(5,0)$, $(0,8)$ and $(0,13)$, and $(5,5)$ and $(7.5,7.5)$, respectively. What is the triangle's area? | [
"Solution:\nAnswer: $\\frac{17}{2}$\n\nSketch the lines $x=1$, $x=5$, $y=8$, $y=13$, $y=10-x$, and $y=15-x$. The triangle has to be contained in the hexagonal region contained in all these lines. If all the projections are correct, every other vertex of the hexagon must be a vertex of the triangle, which gives us t... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 17/2 | |
08dt | Problem:
Agnese, Beatrice, Claudio e Dario giocano con 53 pile di monete. Comunque prese due pile, queste hanno un numero diverso di monete. Ad ogni turno, un giocatore sceglie una pila e toglie da questa una moneta. Perde chi togliendo una moneta a una pila rende questa pila di altezza uguale a un'altra presente sul ... | [
"Solution:\n\nLa risposta è (C). Chiamiamo $a_{1}, a_{2}, \\ldots, a_{53}$ le pile ordinate in ordine crescente di altezza.\n\nOsserviamo che il gioco finisce. Ad ogni mossa il numero totale di gettoni diminuisce di 1, e se ci sono 51 gettoni in totale allora il gioco è già finito perché ci sono almeno due pile con... | Italy | Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | C | |
0bba | Consider an increasing continuous function $f : [0, 1] \to \mathbb{R}$ and define the sequence $(a_n)_{n \ge 1}$ by $a_n = \frac{1}{2^n} \sum_{k=1}^{2^n} f(\frac{k}{2^n})$, for all integers $n \ge 1$.
a) Prove that the sequence $(a_n)_{n \ge 1}$ is increasing.
b) Given that there exists $p \in \mathbb{N}^*$ such that... | [
"a) Notice that\n$$\na_{n+1} = \\frac{1}{2^{n+1}} \\sum_{k=1}^{2^{n+1}} f\\left(\\frac{k}{2^{n+1}}\\right) = \\frac{1}{2^{n+1}} \\left( \\sum_{k=1}^{2^n} f\\left(\\frac{k}{2^n}\\right) + \\sum_{k=1}^{2^n} f\\left(\\frac{2k-1}{2^{n+1}}\\right) \\right)\n$$\nand $f(\\frac{2k-1}{2^{n+1}}) \\le f(\\frac{k}{2^n})$ to de... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kcy | Problem:
Let $S$ be the set of $3^{4}$ points in four-dimensional space where each coordinate is in $\{-1,0,1\}$. Let $N$ be the number of sequences of points $P_{1}, P_{2}, \ldots, P_{2020}$ in $S$ such that $P_{i} P_{i+1}=2$ for all $1 \leq i \leq 2020$ and $P_{1}=(0,0,0,0)$. (Here $P_{2021}=P_{1}$.) Find the largest... | [
"Solution:\nFrom $(0,0,0,0)$ we have to go to $( \\pm 1, \\pm 1, \\pm 1, \\pm 1)$, and from $(1,1,1,1)$ (or any of the other similar points), we have to go to $(0,0,0,0)$ or $(-1,1,1,1)$ and its cyclic shifts. If $a_{i}$ is the number of ways to go from $(1,1,1,1)$ to point of the form $( \\pm 1, \\pm 1, \\pm 1, \\... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof and answer | 4041 | |
0fd8 | Problem:
Probar que dados $4 n$ puntos en el espacio tridimensional, tales que no hay cuatro de ellos coplanarios, siempre se pueden formar $n$ pirámides de base triangular de modo que no hay intersecciones entre ellas. | [
"Solution:\n\nEvidentemente por hipótesis no hay 3 puntos alineados, pues en ese caso añadiendo un cuarto punto cualquiera a la terna violaríamos la hipótesis sobre el carácter no coplanario.\n\nConsideremos todas las posibles ternas de puntos y el plano que determina (unívocamente, al no haber 3 puntos alineados) ... | Spain | null | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0akx | At a chess tournament, every pair of contestants played each other at most once. If any two contestants, $A$ and $B$, failed to play each other, then exactly two other contestants, $C$ and $D$, played against both $A$ and $B$ during the tournament. Moreover, no four contestants played exactly five games between them. P... | [] | North Macedonia | Team Selection Test for BMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0krj | Problem:
How many solutions does $26 = $ twelve $+$ eleven $+$ two $+$ one have over the positive integers? (Each letter is a variable, and letters in the same word are multiplied.) | [
"Solution:\n\nFactoring, we get $26 = $ twelve $+$ eleven $+$ two $+$ one $= (elve + o)(tw + ne)$.\n\nBoth factors are at least $1 + 1 = 2$, so there are two cases: either $elve + o = 2$ and $tw + ne = 13$, or $elve + o = 13$ and $tw + ne = 2$.\n\nIn the first case, $elve = o = 1$, so $e = l = v = o = 1$. We are th... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 70 | |
0ggu | $\Delta P_1 P_2 P_3$ 為一正三角形。對於所有 $n \ge 4$, 小明可以選擇 $P_n$ 為 $\Delta P_{n-3} P_{n-2} P_{n-1}$ 的外心或垂心。試求所有的正整數 $n$, 使得小明可以經由適當地選取 $P_4, \dots, P_n$, 讓 $P_n$ 為 $\Delta P_1 P_2 P_3$ 的外心。 | [
"答案為所有 4 的倍數。\n\n令 $O$ 為 $\\Delta P_1 P_2 P_3$ 的外心，則小明只要一直選取垂心，便有 $P_{4k} = O, P_{4k+1} = P_1, P_{4k+2} = P_2$ 與 $P_{4k+3} = P_3$，對於所有 $k \\in \\mathbb{N}$ 皆成立。因此所以被 4 整除的 $n$ 都滿足題意。\n\n以下證明 $n$ 必須被 4 整除。這需要以下兩個 Lemma。\n\n**Lemma 1.** 對於所有 $n \\ge 3$, $\\Delta P_{n-2} P_{n-1} P_n$ 是頂角為 120° 的等腰三角形 (稱為型 A) 或正三角形 (稱為... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 獨立研究 (二) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese; English | proof and answer | all positive integers divisible by 4 | |
08r8 | Find the number of 20-tuple $(p_1, p_2, \dots, p_{10}, q_1, q_2, \dots, q_{10})$ of positive integers with $p_1 = q_{10} = 1$ and $p_i+1 < q_i - p_i$ for all $i = 1, 2, \dots, 9$. | [
"Generally, let $S_n$ be the number of $2n$-tuple $(p_1, p_2, \\dots, p_n; q_1, q_2, \\dots, q_n)$ of positive integers with $p_1 = q_n = 1$ and $p_{i+1}q_i - p_iq_{i+1} = 1$ for all $i = 1, 2, \\dots, n-1$. We use semicolons to make the boundary simple.\nLet $(p_1, p_2, \\dots, p_n; q_1, q_2, \\dots, q_n)$ a tuple... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | English | proof and answer | 16796 | |
0jqe | Problem:
For positive integers $x_{1}, x_{2}, \ldots, x_{n}$ satisfying $x_{1}+\cdots+x_{n}=101 n$, prove that
$$
\binom{x_{1}}{2}+\binom{x_{2}}{2}+\cdots+\binom{x_{n}}{2} \geq 5050 n
$$ | [
"Solution:\n\nNoting that $2\\binom{x}{2}=x^{2}-x$, we see that it suffices to prove that\n$$\nx_{1}^{2}+\\cdots+x_{n}^{2} \\geq 2(5050 n)+101 n=10201 n .\n$$\nThis follows immediately by Cauchy-Schwarz as\n$$\n(101 n)^{2}=\\left(x_{1}+\\cdots+x_{n}\\right)^{2} \\leq\\left(x_{1}^{2}+\\cdots+x_{n}^{2}\\right)(1+\\cd... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0hf9 | Find all triples of positive integers $a, b, c$, such that
$$
a + [a, b] = b + [b, c] = c + [c, a],
$$
where by $[x, y]$ we denote the smallest common multiple of $x, y$. | [
"Note that $b$, $[a, b]$ and $[b, c]$ are divisible by $b$, so $a$ is divisible by $b$. Similarly, $c$ is divisible by $a$, and $b$ is divisible by $c$, so $a = b = c$."
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | (n, n, n) for any positive integer n | |
06m0 | Let $f(n) = \prod_{k=1}^{n} \left(1 + 4 \cos^2\left(\frac{k\pi}{2n+1}\right)\right)$. Prove that $f(n)$ is an integer for all positive integers $n$. | [
"We shall prove that $f(n) = F_{2n+1}$, the $(2n+1)$th Fibonacci number. Let $\\zeta = e^{\\frac{2\\pi i}{2n+1}}$ be a primitive $(2n+1)$th root of unity. Then\n$$\n1 + 4 \\cos^2 \\left( \\frac{k\\pi}{2n+1} \\right) = 3 + 2 \\cos \\left( \\frac{2k\\pi}{2n+1} \\right) = 3 + 2 \\operatorname{Re}(\\zeta^k) = 3 + \\zet... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
09pr | Problem:
Vind alle functies $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ die voldoen aan
$$
f(f(f(n))) + f(f(n)) + f(n) = 3n
$$
voor alle $n \in \mathbb{Z}_{>0}$. | [] | Netherlands | TOETS TRAININGSKAMP | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(n) = n for all positive integers n | |
0avd | Problem:
The lengths of the two legs of a right triangle are in the ratio of $7:24$. The distance between its incenter and its circumcenter is $1$. Find its area. (Recall that the incenter of a triangle is the center of its inscribed circle and the circumcenter is the center of its circumscribing circle.) | [
"Solution:\n\nLet the legs of the right triangle be $7x$ and $24x$. The hypotenuse is then $c = \\sqrt{(7x)^2 + (24x)^2} = \\sqrt{49x^2 + 576x^2} = \\sqrt{625x^2} = 25x$.\n\nLet $A$, $B$, $C$ be the vertices of the triangle, with right angle at $A$. Let $AB = 7x$, $AC = 24x$, $BC = 25x$.\n\nThe incenter $I$ and cir... | Philippines | 19th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 336/325 | |
05id | Problem:
$ABCD$ est un trapèze dans lequel les côtés $AD$ et $BC$ sont parallèles, $K$ est un point du côté $AB$ et $L$ un point du côté $CD$. Montrer que si les angles $\widehat{BAL}$ et $\widehat{CDK}$ sont égaux alors les angles $\widehat{BLA}$ et $\overline{CKD}$ le sont aussi. | [
"Solution:\n\nOn suppose les angles $\\widehat{BAL}$ et $\\widehat{CDK}$ ont la même valeur. Nous allons montrer que les angles $KBL$ et $KCL$ sont égaux (ceci impliquera que les triangles $KDC$ et $BAL$ ont deux angles identiques, et donc que $\\widehat{BLA}=\\widehat{KCL}$).\n\n\n\nJe vai... | France | Olympiades Françaises de Mathématiques - Envoi Numéro 1 - Corrigé | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ei7 | Problem:
Kroglo prerežemo na dve polkrogli. Za koliko $\%$ je vsota površin obeh polkrogel večja od površine krogle?
(A) 25
(B) 50
(C) 75
(D) 22,5
(E) 40 | [
"Solution:\n\nPovršina krogle je enaka $P_{K} = 4 \\pi R^{2}$. Če kroglo prerežemo, dobimo še površini dveh glavnih krogov, ki sta enaki $2 \\pi R^{2}$. Nova površina je enaka $P_{PK} = 6 \\pi R^{2}$. Površina se torej poveča za $50\\%$."
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Solid Geometry > Surface Area"
] | null | MCQ | B | |
00r2 | Let $AB$ be a diameter of a circle ($\omega$) with centre $O$. From an arbitrary point $M$ on $AB$ such that $MA < MB$ we draw the circles ($\omega_1$) and ($\omega_2$) with diameters $AM$ and $BM$ respectively. Let $CD$ be an exterior common tangent of ($\omega_1$), ($\omega_2$) such that $C$ belongs to ($\omega_1$) a... | [
"Let $P$ be the second intersection of the circumcircles of the triangles $CED$ and $CFG$.\nFirst we will prove that $E, M, D$ are collinear. Indeed, if the common tangent of ($\\omega_1$), ($\\omega_2$) at $M$ intersects $CD$ at $S$ then $SC = SD = SM$, so $\\angle CMD = 90^\\circ$ and also $\\angle CME = 90^\\cir... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point... | null | proof only | null | |
0irq | Problem:
In 2000, the Clay Mathematics Institute named seven Millennium Prize Problems, with each carrying a prize of $1$ Million for its solution. Write down the name of ONE of the seven Clay Millennium Problems. If your submission is incorrect or misspelled, then your submission is disqualified. If another team wrot... | [
"Solution:\n\nThe seven Millennium Prize Problems are:\n(a) Birch and Swinnerton-Dyer Conjecture\n(b) Hodge Conjecture\n(c) Navier-Stokes Equations\n(d) P vs NP\n(e) Poincaré Conjecture\n(f) Riemann Hypothesis\n(g) Yang-Mills Theory"
] | United States | Harvard-MIT Mathematics Tournament | [
"Math Word Problems"
] | null | final answer only | Riemann Hypothesis | |
02uv | Problem:
Seja $ABCD$ um quadrilátero com $AD = BC$ e $\angle DAB + \angle ABC = 120^\circ$. Um triângulo equilátero $DEC$ é construído no exterior do quadrilátero. Prove que o triângulo $AEB$ também é equilátero. | [
"Solution:\n\n\n\nSejam $\\angle ADC = x$ e $\\angle DCB = y$. Como a soma dos ângulos internos de um quadrilátero é $360^\\circ$, temos\n$$\n\\begin{aligned}\n\\angle ADC + \\angle DCB + \\angle CBA + \\angle BAD & = 360^\\circ \\\\\nx + y & = 360^\\circ - 120^\\circ \\\\\n& = 240^\\circ\n... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0egw | Problem:
Naj bosta $a$ in $b$ naravni števili, za kateri velja $2^{a}-2^{b}=240$. Koliko je vrednost izraza $a+b$ ?
(A) 8
(B) 11
(C) 13
(D) 16
(E) Nič od naštetega. | [
"Solution:\n\nOčitno mora biti $a > b$. Enačbo preoblikujemo do $2^{b}\\left(2^{a-b}-1\\right) = 2^{4} \\cdot 15$. Od tod sledi $b = 4$ in $2^{a-b} - 1 = 15$. Drugo enačbo preuredimo do $2^{a-b} = 16$ in sklepamo, da je $a-b = 4$. Od tod izračunamo še $a = 8$. Torej je $a + b = 12$ in pravilen odgovor je $(\\mathbf... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | E | |
0by9 | Let $\triangle ABC$ be a triangle, let $M_A$ be the midpoint of the side $BC$, and let $P_A$ be the orthogonal projection of $A$ on the line $BC$; similarly, define $M_B, P_B$ and $M_C, P_C$. The lines $M_B M_C$ and $P_B P_C$ meet at $S_A$, and the tangent of the circle $ABC$ at $A$ meets the line $BC$ at $T_A$; simila... | [
"The three lines in question are concurrent at the center of the nine-point circle $\\omega$ of the triangle $ABC$. In what follows, polarity always refers to $\\omega$.\n\nTo prove that the center of $\\omega$ lies on the perpendicular through $A$ to the line $S_A T_A$, it is sufficient to show that the latter is ... | Romania | THE Tenth STARS OF MATHEMATICS COMPETITION | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | English | proof only | null | |
06km | The altitudes $AD$ and $BE$ of acute triangle $ABC$ intersect at $H$. Let $F$ be the intersection of $AB$ and a line that is parallel to the side $BC$ and goes through the circumcentre of $ABC$. Let $M$ be the midpoint of $AH$. Prove that $\angle CMF = 90^\circ$. | [
"Let $P$ and $N$ be the projection of $F$ and $O$ on $BC$ respectively. Recall that $ON = \\frac{1}{2} AH$. Therefore, $FP = AM = MH$.\nSince $FP$ and $AM$ are perpendicular to $BC$, they are parallel. Thus, $AFPM$ is a parallelogram. This implies $MP \\parallel AF$, and hence $MP \\perp CH$. Also, we have $MH \\pe... | Hong Kong | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0j2a | Problem:
Manya has a stack of $85 = 1 + 4 + 16 + 64$ blocks comprised of 4 layers (the $k$th layer from the top has $4^{k-1}$ blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensions half those of the larger block. Laura removes blocks one at a time from this stack, removing only bloc... | [
"Solution:\n\nEach time Laura removes a block, 4 additional blocks are exposed, increasing the total number of exposed blocks by 3. She removes 5 blocks, for a total of $1 \\cdot 4 \\cdot 7 \\cdot 10 \\cdot 13$ ways. However, the stack originally only has 4 layers, so we must subtract the cases where removing a blo... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Other"
] | null | proof and answer | 3384 | |
00zg | Problem:
Is it possible to partition all positive integers into disjoint sets $A$ and $B$ such that
(i) no three numbers of $A$ form arithmetic progression,
(ii) no infinite non-constant arithmetic progression can be formed by numbers of $B$? | [
"Solution:\n\nLet $\\mathbb{N}$ denote the set of positive integers. There is a bijective function $f: \\mathbb{N} \\rightarrow \\mathbb{N} \\times \\mathbb{N}$. Let $a_{0}=1$, and for $k \\geq 1$, let $a_{k}$ be the least integer of the form $m+t n$ for some integer $t \\geq 0$ where $f(k)=(m, n)$, such that $a_{k... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
082o | Problem:
Un'azienda dolciaria produce due tipi di torrone, usando la stessa pasta bianca e le stesse nocciole, ma in proporzioni diverse. Nel torrone di tipo $A$ le nocciole rappresentano il $30\%$ del peso ed il $40\%$ del volume; in quello di tipo $B$ le nocciole rappresentano il $60\%$ del peso. Quale percentuale d... | [
"Solution:\n\nLa risposta è $70$. Sia $P_{A}$ il peso del torrone $A$, $V_{A}$ il volume del torrone $A$, $P_{B}$ il peso del torrone $B$, $V_{B}$ il volume del torrone $B$. Si ha che $30\\%\\,P_{A}$ occupa $40\\%\\,V_{A}$, quindi il peso specifico delle nocciole è $\\frac{3}{4} \\frac{P_{A}}{V_{A}}$. Considerando ... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 70% | |
0gas | 一個 $2n \times 2n$ 的棋盤上的每一格都有一張椅子。現在有 $2n^2$ 對情侶要入座, 每個人坐一個座位。定義一對情侶之間的距離為他們座位相差的行數與相差的列數和 (舉例: 如果一對情侶分別坐在 (3,3) 和 (2,5), 則他們之間的距離為 $|3-2| + |3-5| = 3$)。定義所有情侶的總距離, 等於這 $2n^2$ 對情侶的距離總和。試求總距離的最大值。 | [
"最大值為 $4n^3$.\n\n1. 首先考慮水平方向的距離和的最大值: 將所有人投影到同一列上, 並將每一對情侶兩人之間連線。考慮兩種可能:\n- 存在兩對情侶的連線不重疊: 則兩對各取一人交換位置, 此時水平方向距離和更大。\n- 任兩對的連線都有重疊: 不失一般性, 假設每一對的男方都在女方左側。我們發現男方都必須在左側的 $n$ 行, 否則:\n * 若有一對情侶的男女方都在右側 $n$ 行中, 右側 $n$ 行共剩 $2n^2 - 2$ 個位子, 但還有另外 $2n^2 - 1$ 對情侶要坐。\n * 因此必然有一對情侶他們都在左側 $n$ 行中。然而, 都在左 $n$ 行的情侶, 其連線不可能和都在右 $n... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 4n^3 | |
00bv | Let $n$ be a positive integer. Find all $n$-tuples $(a_1, a_2, \dots, a_n)$ of distinct positive integers such that
$$
\frac{(a_1+d)(a_2+d)\dots(a_n+d)}{a_1 a_2 \dots a_n}
$$
is an integer for every integer $d \ge 0$. | [
"Let\n$$\nM(d) = \\frac{(a_1+d)(a_2+d)\\dots(a_n+d)}{a_1 a_2 \\dots a_n}.\n$$\nThe key to the proof is to notice that the assumption that $M(d)$ is an integer for every non-negative integer $d$ implies indeed that $M(d)$ is an integer for all integer $d$: if $d < 0$, consider $d' = |d| a_1 \\dots a_n + d$; then, $d... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | a_k = k for all k = 1, 2, ..., n (i.e., the tuple is (1, 2, ..., n)) | |
0igo | Problem:
Find the sum of the $x$-coordinates of the distinct points of intersection of the plane curves given by $x^{2}=x+y+4$ and $y^{2}=y-15 x+36$. | [
"Solution:\nSubstituting $y = x^{2} - x - 4$ into the second equation yields\n$$\n\\begin{aligned}\n0 &= \\left(x^{2} - x - 4\\right)^{2} - \\left(x^{2} - x - 4\\right) + 15x - 36 \\\\\n&= x^{4} - 2x^{3} - 7x^{2} + 8x + 16 - x^{2} + x + 4 + 15x - 36 \\\\\n&= x^{4} - 2x^{3} - 8x^{2} + 24x - 16 \\\\\n&= (x-2)\\left(x... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 0 | |
03ez | There are 11 points equally spaced on a circle. Some of the segments having endpoints among these vertices are drawn and colored in two colors, so that each segment meets at an internal point at most one other segment from the same color. What is the greatest number of segments that could be drawn?
(Mladen Vylkov) | [
"\n\nPut $n$ instead of 11 and let the points be $A_1, A_2, \\dots, A_n$. We may assume that the points are vertices of a regular $n$-gon. Let us first calculate the maximum number of diagonals that we can draw so that any diagonal meets no more than one other diagonal at an interior point.... | Bulgaria | Bulgarian Winter Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 35 | |
0eam | Problem:
Na tabli je napisano neko naravno število $n$. Na vsakem koraku lahko število na tabli nadomestimo z vsoto dveh naravnih števil, katerih zmnožek je enak številu na tabli. Določi najmanjše število, ki je lahko po končno korakih zapisano na tabli, in sicer v odvisnosti od začetnega števila $n$. | [
"Solution:\n\nPokažimo najprej, da če se število na tabli pri zamenjavi zmanjša, potem je novo število večje ali enako $5$. Števila $1$ po zamenjavi ne moremo dobiti, saj vsota dveh naravnih števil ni nikoli enaka $1$. Število $2=1+1$ lahko dobimo le iz števila $1$, število $3=1+2$ le iz števila $2$, število $4=1+3... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 5 if n >= 5; otherwise n | |
09m3 | One of the following three types of operations can be performed on a word. Let $x$, $y$, $z$, $a$, $b$, and $c$ be letters.
(1) Any subword of the form $xy$ can be changed to $xzzy$. For example, $abc \rightarrow azzbc$.
(2) Any subword of the form $xyz$ can be changed to $zyx$. For example, $cabc \rightarrow ccba$.
(3... | [
"Answer: No.\nSuppose the number of $a$'s in even positions is subtracted from the number of $a$'s in odd positions in the word. In that case, we obtain a quantity that remains invariant under the given operations. This invariant can be used to determine if one word can be transformed into another using the specifi... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
0hky | Problem:
January 3, 1911 was an odd date as its abbreviated representation, $1 / 3 / 1911$, can be written using only odd digits (note all four digits are written for the year). To the nearest month, how many months will have elapsed between the most recent odd date and the next odd date (today is $3 / 3 / 2001$, an e... | [
"Solution:\n\nThe most recent odd date was $11 / 19 / 1999$ (November has 30 days, but the assumption that it has 31 days does not change the answer), and the next odd date will be $1 / 1 / 3111$. From $11 / 19 / 1999$ to $1 / 1 / 2000$ is about 1 month. From 2000 to 3111 is 1111 years, or $12 \\cdot 1111 = 13332$ ... | United States | null | [
"Number Theory > Other"
] | null | proof and answer | 13333 | |
0eoa | The value of $\frac{2015 \times 2 + 4 \times 4030}{4030}$ is
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6 | [
"$$\n\\frac{2015 \\times 2 + 4 \\times 4030}{4030} = \\frac{4030 + 4 \\times 4030}{4030} = \\frac{5 \\times 4030}{4030} = 5.\n$$"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | D | |
0hxy | Problem:
Find the range of
$$
f(A) = \frac{(\sin A)\left(3 \cos^{2} A + \cos^{4} A + 3 \sin^{2} A + (\sin^{2} A)(\cos^{2} A)\right)}{(\tan A)(\sec A - (\sin A)(\tan A))}
$$
if $A \neq \frac{n \pi}{2}$. | [
"Solution:\nAnswer: $(3,4)$.\n\nWe factor the numerator and write the denominator in terms of fractions to get\n$$\n\\frac{(\\sin A)\\left(3+\\cos^{2} A\\right)\\left(\\sin^{2} A+\\cos^{2} A\\right)}{\\left(\\frac{\\sin A}{\\cos A}\\right)\\left(\\frac{1}{\\cos A}-\\frac{\\sin^{2} A}{\\cos A}\\right)} = \\frac{(\\s... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | (3,4) | |
0cjn | Let $(a_n)_{n \ge 1}$ be the sequence defined by $a_1 = 1$ and $a_{n+1} = \frac{a_n}{1+\sqrt{1+a_n}}$, for any $n \in \mathbb{N}^*$. Show that $\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \sum_{k=1}^{n} \log_2(1+a_k) = 2$. Traian Tămâian | [
"We have $a_n > 0$ and $a_{n+1} < a_n$, for any $n \\ge 1$. It turns out that the sequence $(a_n)_{n \\ge 1}$ is convergent, with the limit $\\ell \\in [0, 1)$. From the recurrence relation, we obtain $\\ell = \\frac{\\ell}{1+\\sqrt{1+\\ell}}$, so $\\ell = 0$. Then $\\lim_{n \\to \\infty} \\frac{a_n}{a_{n+1}} = \\l... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2 | |
0cp8 | A table consists of $n$ rows and 10 columns. Each cell of this table contains a digit (i.e. an integer from 0 to 9). It appears that for every row $A$ and every pair of columns $B$ and $C$ there exists a row that differs from $A$ exactly in columns $B$ and $C$. Prove that $n \ge 512$. (R. Karasev)
В каждой клетке табл... | [
"Пусть $R_0$ — первая строка таблицы. Рассмотрим любой набор из чётного количества столбцов и пронумеруем их слева направо: $C_1, \\dots, C_{2m}$. Тогда в таблице есть строка $R_1$, отличающаяся от $R_0$ ровно в столбцах $C_1$ и $C_2$; далее, есть строка $R_2$, отличающаяся от $R_1$ ровно в столбцах $C_3$ и $C_4$; ... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English, Russian | proof only | null | |
09xz | A parallelogram has two sides of length $4$ and two sides of length $7$. Also, one of the diagonals has length $7$. (Attention: the picture has not been drawn to scale.) What is the length of the other diagonal?
 | [
"$9$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | final answer only | 9 | |
0gu4 | Each edge of the complete graph $K_{2024}$ is coloured into one of the given 13 colours. Suppose that for any such colouring one can choose $k$ colours such that any two vertices of $K_{2024}$ are connected by some path such that its each edge is coloured to one of these $k$ colours. Find the minimal possible value of ... | [
"Answer: $k = 7$.\n\nLet us show that 7 colours are sufficient. Indeed, let us randomly divide 13 colours to two groups $A$ and $B$, consisting of 7 and 6 colours, respectively. Assume that by using edges coloured to colours of group $A$ some vertex $X$ is not connected to some other vertex $Y$. It means that the e... | Turkey | Team Selection Test for EGMO 2024 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 7 | |
0be2 | Let $A$ and $B$ be $2 \times 2$ real matrices such that $AB = A^2 B^2 - (AB)^2$ and $\det(B) = 2$. Evaluate $\det(A + 2B) - \det(B + 2A)$. | [
"Write $A(AB - BA - I_2)B = O_2$ to get $A(AB - BA - I_2) = O_2$, for $B$ is nonsingular. If $A$ is nonsingular, then $AB - BA = I_2$, false, for $\\det(AB - BA) = 0 \\ne 2 = \\det(I_2)$.\n\nSet $f(x) = \\det(A + xB)$, $x \\in \\mathbb{R}$. Since $\\det(A) = 0$, there exists $a \\in \\mathbb{R}$ such that $f(x) = a... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | 6 | |
0hld | Problem:
Find all prime numbers $p$ such that $p^{2}+8$ is prime number, as well. | [
"Solution:\n\nFor $p=3$ we have $p^{2}+8=17$, which is prime.\n\nIf $p \\neq 3$ then $p$ is not divisible by $3$. The remainder of $p$ when divided by $3$ is either $1$ or $2$. This means that $p=3k+1$ for some integer $k$, or $p=3l+2$ for some integer $l$.\n\nIn the first case we get $p^{2}=(3k+1)^{2}=9k^{2}+6k+1$... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 3 | |
0hg3 | What smallest value can be attained by the expression
$$
\frac{(x + y + |x - y|)^2}{xy}
$$
for positive $x$, $y$? | [
"We assume $x \\ge y$, then we can rewrite as:\n$$\n\\frac{(x + y + |x - y|)^2}{xy} = \\frac{(x + y + x - y)^2}{xy} = \\frac{(2x)^2}{xy} = \\frac{4x^2}{xy} = \\frac{4x}{y} \\ge 4\n$$\nWhen $x = y$, we get equality."
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 4 | |
0glu | Does there exist a function $f: \mathbb{N} \to \mathbb{Q}$ such that for any rational number $r$, there exists exactly one ordered pair $(m, n)$ of positive integers satisfying the equation $r = f(m) + \frac{1}{n}$? | [] | Thailand | T3MO 2017 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Functional equations"
] | English | proof only | null | |
0gzz | Prove that for all positive integers $n$ equation $a^n + 2010b^n = c^{n+1}$ has infinitely many natural solutions $a, b, c$. | [
"Consider $a = b = 2011k^{n+1}$, where $k$ is a positive integer, then\n$$\na^n + 2010b^n = 2011^n k^{n(n+1)} + 2010 \\cdot 2011^n k^{n(n+1)} = 2011^{n+1} k^{n(n+1)}.\n$$\nSince $2011 \\nmid 2011^n k^{n(n+1)}$ and $2011 \\nmid 2010$, we have\n$$\n2011^{n+1} k^{n(n+1)} = (2011k^n)^{n+1} = c^{n+1} \\Rightarrow c = 20... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Number Theory > Diophantine Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0cle | Let $ABC$ be a right triangle with right angle at $A$, and let $AD$ be its altitude from $A$ to $BC$. On the ray $[AD$, take points $E$ and $H$, such that $AE = AC$ and $AH = AB$. Construct squares $AEFG$ and $AHJI$, such that $C$ lies inside $AEFG$, and $B$ lies inside $AHJI$. Let $K = AC \cap EG$, $L = AB \cap IH$, $... | [
"a) We have $\\triangle AGK \\sim \\triangle AHL$ (1), because $\\angle AGK = \\angle AHL = 45^\\circ$, and $\\angle GAK = 90^\\circ - \\angle CAD = \\angle HAL$. Thus, $\\frac{AK}{AL} = \\frac{AG}{AH}$. Since $AG = AE = AC$ and $AH = AB$, it follows that $\\frac{AK}{AC} = \\frac{AL}{AB}$, therefore $LK \\parallel ... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0ccf | Let $XYZT$ be a parallelogram and $A, B, C, D$ variable points on the sides $XY, XT, TZ, ZY$, respectively, such that $ABCD$ is a cyclic quadrilateral with circumcenter $O$, $AC \parallel XT$, and $BD \parallel XY$. Let $P$ be the intersection of lines $AD$ and $BC$, and $Q$ be the intersection of lines $AB$ and $CD$. ... | [
"The key idea for this problem lies in the following lemma:\n\n**Lemma:** Let $ABC$ be a triangle and $X$ a point in the interior of angle $\\angle BAC$ such that $\\angle ABX = \\angle ACX$. Define $Y$ such that $BXC$ is a parallelogram. Then $AX$ and $AY$ are isogonal with respect to $\\angle BAC$.\n\n*Proof.* Co... | Romania | THE Eighteenth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"... | null | proof only | null | |
0kvn | Problem:
Let $A_{1} A_{2} \ldots A_{6}$ be a regular hexagon with side length $11 \sqrt{3}$, and let $B_{1} B_{2} \ldots B_{6}$ be another regular hexagon completely inside $A_{1} A_{2} \ldots A_{6}$ such that for all $i \in \{1,2, \ldots, 5\}$, $A_{i} A_{i+1}$ is parallel to $B_{i} B_{i+1}$. Suppose that the distance ... | [
"Solution:\n\n\nLet $X = A_{1} A_{2} \\cap A_{3} A_{4}$, and let $O$ be the center of $B_{1} B_{2} \\ldots B_{6}$. Let $p$ be the apothem of hexagon $B$. Since $O A_{2} X A_{3}$ is a convex quadrilateral, we have\n$$\n\\begin{aligned}\n\\left[A_{2} A_{3} X\\right] & = \\left[A_{2} X O\\righ... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3*sqrt(3) | |
0g9m | $\triangle ABC$ 中, $BC$ 邊的中點為 $M$, $AM$ 再交 $\triangle ABC$ 的外接圓 $\Gamma$ 於 $R$, 過 $R$ 且與 $BC$ 平行的直線再交 $\Gamma$ 於 $S$。自 $R$ 至 $BC$ 的垂線的垂足為 $U$, $T$ 為 $U$ 對 $R$ 的對稱點。$D$ 是 $BC$ 上的一點使得 $AD$ 為 $\triangle ABC$ 的高, $N$ 為 $AD$ 中點。最後令 $AS$, $MN$ 交於 $K$。證明: $AT$ 平分 $MK$。 | [
"(a) 令 $L$ 為 $MK$ 的中點, 過 $L$ 對 $BC$ 的垂線分別交 $BC$, $AM$ 於 $E$, $F$。\n因 $AD \\parallel EF$, $L$ 為 $EF$ 的中點。因此 $\\triangle LEM \\cong \\triangle LFK$, 並得\n$\\angle LKF = 90^\\circ$, $KF \\parallel BC$.\n\n\n\n(b) 令 $MN$ 交 $RU$ 於 $W$。因 $AD \\parallel RU$, $N$ 為 $AD$ 的中點, 故 $W$ 為 $RU$ 的中點。\n\n(c)... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gmj | Let $n$ be a positive integer and $\mathbb{R}^n$ be the set of ordered $n$-tuples of real numbers. Let $T$ denote the collection of $(x_1, x_2, \dots, x_n) \in \mathbb{R}^n$ for which there exists a permutation $\sigma$ of $1, 2, \dots, n$ such that $x_{\sigma(i)} - x_{\sigma(i+1)} \ge 1$ for each $1 \le i < n$. Prove ... | [] | Turkey | X. NATIONAL MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (n-1)/2 | |
0h30 | In the coordinate plane $xOy$, depict the set of all points such that $|y - x| = 2 - y - x$. | [
"Вихідне співвідношення рівносильне системі\n$$\n\\begin{cases}\n y = 1, \\\\\n x = 1; \\\\\n y \\le 2 - x.\n\\end{cases}\n$$\nСлід зобразити множину точок $\\{(x; 1) : x \\le 1\\} \\cup \\{(1; y) : y \\le 1\\}$."
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0fve | Problem:
Berechne die Quersumme der Zahl
$$
9 \times 99 \times 9999 \times \cdots \times \underbrace{99 \ldots 99}_{2^{n}}
$$
wobei sich die Anzahl Neunen in jedem Faktor verdoppelt. | [
"Solution:\nWir beweisen allgemeiner folgendes\nLemma 1. Sei $a$ eine Zahl mit höchstens $m$ Dezimalstellen, sodass die letzte Ziffer von $a$ nicht $0$ ist. Dann ist die Quersumme von $\\underbrace{99 \\ldots 99}_{m} \\times a$ gleich $9 m$.\nBeweis. Sei $a_{m-1} \\ldots a_{1} a_{0}$ die Dezimaldarstellung von $a$ ... | Switzerland | SMO Finalrunde | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 9 * 2^n | |
0ekz | Problem:
2023-mesto naravno število $n$ ima vse števke enake 1. Koliko je vsota števk naravnega števila $n \cdot 1111$?
(A) 8080
(B) 8083
(C) 8086
(D) 8092
(E) 8101 | [
"Solution:\n\nČe pisno množimo $n \\cdot 1111$, najprej v računu 4-krat podpišemo število $n$ in nato seštevamo števke po stolpcih. Pri izračunu vsot števk nikoli ne pride do prenosa števk. Torej je vsota števk produkta $n \\cdot 1111$ enaka 4-kratniku vsote števk števila $n$, to je $4 \\cdot 2023 = 8092$."
] | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0guq | Let $ABC$ be an acute triangle, $P$ be the midpoint of the side $BC$ and $K$ be the foot of the altitude from $A$. Let $D$ be a point on the segment $AP$ such that $\angle BDC = 90^\circ$. Let the second intersection point of the circumcircle of $ADK$ and line $BC$ be $E$. Let the second intersection point of the circu... | [
"\nSince $A$, $D$, $K$, $E$ are concyclic we have\n$$\n\\angle AKD = \\angle ADE = 90^\\circ\n$$\nand hence $\\triangle ADE$ is a right triangle. Since $D$ lies on the circle centered at $P$ and $ED \\perp PD$, we can see that $ED$ is tangent to the circumcircle of $BDC$. Using the power of... | Turkey | Team Selection Test for JBMO 2024 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k5p | Problem:
How many ways are there to place three rooks on an $8 \times 8$ chessboard such that the rooks are in different columns and different rows? | [
"Solution:\n\nThere are $\\binom{8}{3}$ ways to pick which three of the eight columns we wish to pick the rooks in. Once the set of three columns is fixed, then we have eight choices for where to pick the rook in the leftmost column, then seven choices for where to pick the rook in the middle column, and finally si... | United States | Berkeley Math Circle | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 18816 | |
084r | Problem:
Alberto e Barbara giocano con un dado. Dopo un po' si accorgono che il dado è truccato, e che il numero $1$ esce più frequentemente degli altri $5$ numeri (che invece restano equiprobabili). Decidono quindi che, quando esce $1$, quel tiro è annullato e si tira di nuovo. Se si continua a lanciare il dado fino ... | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Dato che se esce $1$ si tira di nuovo, è come se il dado avesse solo $5$ facce. Ora per fare $8$ qualunque uscita iniziale va bene, ma al secondo tiro valido solo il complemento a $8$ va bene. Quindi la probabilità richiesta è uguale a quella di ottenere un dato numero a... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | C | |
0i96 | Problem:
A quagga is an extinct chess piece whose move is like a knight's, but much longer: it can move 6 squares in any direction (up, down, left, or right) and then 5 squares in a perpendicular direction. Find the number of ways to place 51 quaggas on an $8 \times 8$ chessboard in such a way that no quagga attacks an... | [
"Solution:\nRepresent the 64 squares of the board as vertices of a graph, and connect two vertices by an edge if a quagga can move from one to the other. The resulting graph consists of 4 paths of length 5 and 4 paths of length 3 (given by the four rotations of the two paths shown, next page), and 32 isolated verti... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 68 | |
0jfg | Let the excircle of the triangle *ABC* opposite to the vertex *A* be tangent to side *BC* at $A_1$. Define the points $B_1$ and $C_1$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcenter of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle... | [
"Let $\\omega$ be the circumcircle of $ABC$, and let $O_1$ be the circumcenter of $A_1B_1C_1$. Because $A_1, B_1$, and $C_1$ are on the boundary of $ABC$ and $O_1$ is outside of $ABC$, $A_1B_1C_1$ is obtuse. Without loss of generality, assume that $\\angle B_1A_1C_1$ is obtuse so that $O_1$ and $A$ lie on the same ... | United States | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
012g | Problem:
Show that the sequence
$$
\left(\begin{array}{l}
2002 \\
2002
\end{array}\right),\left(\begin{array}{c}
2003 \\
2002
\end{array}\right),\left(\begin{array}{l}
2004 \\
2002
\end{array}\right), \ldots
$$
considered modulo $2002$, is periodic. | [
"Solution:\nDefine\n$$\nx_{n}^{k}=\\left(\\begin{array}{l}\nn \\\\\nk\n\\end{array}\\right)\n$$\nand note that\n$$\nx_{n+1}^{k}-x_{n}^{k}=\\left(\\begin{array}{c}\nn+1 \\\\\nk\n\\end{array}\\right)-\\left(\\begin{array}{l}\nn \\\\\nk\n\\end{array}\\right)=\\left(\\begin{array}{c}\nn \\\\\nk-1\n\\end{array}\\right)=... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0h64 | For every real numbers $a \ne b$ solve the system:
$$
\begin{cases}
3x+z = 2y+(a+b), \\
3x^2 + 3xz = y^2 + 2(a+b)y + ab, \\
x^3 + 3x^2z = y^2(a+b) + 2yab.
\end{cases}
$$
(Rublyov Bogdan) | [
"**Answer.** $x = y = a, z = b$ and $x = y = b, z = a$.\n\nLet $(x, y, z)$ be a solution of the system. Consider polynomials\n$$\n\\begin{aligned}\nP(t) &= (t-x)^3(t-z) = t^4 + p_1t^3 + q_1t^2 + r_1t + s_1 \\text{ and} \\\\\nQ(t) &= (t-y)^2(t-a)(t-b) = t^4 + p_2t^3 + q_2t^2 + r_2t + s_2,\n\\end{aligned}\n$$\nand le... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | x = y = a, z = b; and x = y = b, z = a | |
060g | Problem:
Déterminer tous les entiers $N$ tels que $2^{N} - 2N$ est un carré parfait. | [
"Solution:\n\nMontrons que les solutions sont $N = 0, 1$ ou $2$.\n\nOn remarque déjà que $2^{0} - 0 = 1^{2}$, $2^{1} - 2 = 0^{2}$ et $2^{2} - 4 = 0^{2}$. Donc les entiers $0, 1$ et $2$ sont solutions.\n\nSupposons par l'absurde qu'il existe $N$ un autre entier solution. On dispose alors de $x \\in \\mathbb{N}$ tel ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Exponential functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 0, 1, 2 | |
04h2 | Let $a$, $b$, $c$ and $d$ be positive real numbers such that $a^2 + b^2 = c^2 + d^2$.
Determine the largest possible area of a quadrilateral $ABCD$ with side-lengths $|AB| = a$, $|BC| = b$, $|CD| = c$, $|DA| = d$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | (ab + cd)/2 | |
0dv9 | Problem:
Premer prednjega kolesa je $1,1~\mathrm{m}$, zadnjega pa $0,8~\mathrm{m}$. Kolikšno razdaljo smo prevozili, če je prvo kolo naredilo 69 obratov manj kot zadnje? Rezultat zaokroži na centimeter natančno. Za $\pi$ uporabi približek $\frac{22}{7}$. Zapiši odgovor. | [
"Solution:\n\nPo enem obratu prevozimo $o = 2 \\pi r$.\n\nS prednjim kolesom prevozimo $x \\cdot 1{,}1 \\pi$.\n\nZ zadnjim kolesom prevozimo $(x + 69) \\cdot 0{,}8 \\pi$.\n\nZapisana enačba:\n$$\nx \\cdot 1{,}1 \\pi = (x + 69) \\cdot 0{,}8 \\pi\n$$\n\nIzračunan $x = 184$.\n\nRešitev in zapisan odgovor:\n$$\no = 636... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških and strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 636.11 m | |
0b42 | Problem:
Let $a$, $b$, and $c$ be positive real numbers such that $ab + bc + ca = 3$. Show that
$$
\frac{bc}{1 + a^{4}} + \frac{ca}{1 + b^{4}} + \frac{ab}{1 + c^{4}} \geq \frac{3}{2}
$$ | [
"Solution:\nIt can be shown that\n$$\n\\frac{1}{1 + a^{4}} \\geq \\frac{2 - a^{2}}{2}\n$$\nIndeed, simplifying yields\n$$\n\\begin{aligned}\n& 2 \\geq (1 + a^{4})(2 - a^{2}) \\\\\n& (a^{4} + 1)(a^{2} - 2) + 2 \\geq 0 \\\\\n& a^{6} - 2a^{4} + a^{2} - 2 + 2 \\geq 0 \\\\\n& a^{6} - 2a^{4} + a^{2} \\geq 0 \\\\\n& a^{2}... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0hfo | Petrik uses the computer program "Three", which converts the numbers written on the display. For one application of this program Petrik chooses 5 numbers from the written ones, and the program increases each of these 5 numbers in 3 times. At the beginning, the following 20 numbers are written on the display: 1, $3^1$, ... | [
"In one operation the product of all written numbers increases in $3^5$ times. At the beginning, this product equals $3^{0+1+2+...+19} = 3^{190}$. So, after using it $n$ times, the product will be equal to $3^{190+5n}$. By that time all numbers would have to become equal, and therefore at least $3^{19}$, so in the ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 38 | |
0eaz | Find all pairs of positive integers $(m, n)$, such that
$$
m + 3n - 5 = 2v - 11d.
$$
Here $v$ is the least common multiple of the numbers $m$ and $n$, and $d$ is the greatest common divisor of $m$ and $n$. | [
"Let us write $m = d a$ and $n = d b$. Here $a$ and $b$ are relatively prime integers and $v = d a b$. Plugging this into the given equation we get $d a + 3 d b - 5 = 2 d a b - 11 d$, so\n\n$$\nd a + 3 d b - 5 = 2 d a b - 11 d\n$$\n\nwhich rearranges to\n\n$$\nd a + 3 d b - 2 d a b + 11 d = 5\n$$\n\nor\n\n$$\nd(a +... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (m, n) = (9, 1) and (65, 5) | |
0dbx | Let $a, b, c \in \mathbb{R}_{+}$. Prove that
$$
\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
08zv | You are given that the following value is a rational number. Express it as an irreducible fraction.
$$
\sqrt{\frac{123! - 122!}{122! - 121!}}
$$ | [
"$$\n\\boxed{\\frac{122}{11}}\n$$\nWe have\n$$\n\\begin{align*}\n\\sqrt{\\frac{123! - 122!}{122! - 121!}} &= \\sqrt{\\frac{123 \\cdot 122! - 122!}{122 \\cdot 121! - 121!}} = \\sqrt{\\frac{(123 - 1) \\cdot 122!}{(122 - 1) \\cdot 121!}} \\\\\n&= \\sqrt{\\frac{122 \\cdot 122 \\cdot 121!}{121 \\cdot 121!}} = \\sqrt{\\f... | Japan | Japan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | final answer only | 122/11 | |
0kxa | Problem:
Zaineb makes a large necklace from beads labeled $290, 291, 292, \ldots, 2023$. She uses each bead exactly once, arranging the beads in the necklace any order she likes. Prove that no matter how the beads are arranged, there must be three beads in a row whose labels are the side lengths of a triangle. | [
"Solution:\n\nMore generally, we will prove that if there are $6n$ beads labeled $n+1, n+2, \\ldots, 7n$, there must be three beads in a row whose labels are the side lengths of a triangle. (When $n=289$, this coincides with the problem statement.)\n\nAiming for a contradiction, assume there are no three beads in a... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0knl | Carina has three pins, labeled **A**, **B**, and **C**, respectively, located at the origin of the coordinate plane. In a *move*, Carina may move a pin to an adjacent lattice point at distance $1$ away. What is the least number of moves that Carina can make in order for triangle $ABC$ to have area $2021$?
(A lattice p... | [] | United States | USA Junior MO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 133 | |
00c8 | In a math camp there are $2018$ children. The entertainer has $4036$ tokens. There are two tokens with each of the numbers from $1$ to $2018$; that is, there are two tokens with number $1$, two tokens with number $2$, and so on.
Two tokens with different numbers are given to every child. There cannot be two children r... | [
"Since every child holds hands with the two children sharing a number with him or her, when the children are arranged, they form several rounds (possibly more than one).\n\nIf the entertainer makes an exchange between two children in different rounds, the two rounds join in a single round. An exchange between two c... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 671 | |
09d7 | a, b, c эерэг бодит тоонууд ба
$$
\frac{1}{2} \le a, b, c \le 1 \text{ бол}
$$
$$
2 \le \frac{a+b}{1+c} + \frac{b+c}{1+a} + \frac{c+a}{1+b} \le 3
$$ | [
"Эхлээд зүүн гар талын тэнцэтгэл бишийг баталья. $a, b \\ge \\frac{1}{2} \\Rightarrow a+b \\ge 1 \\Rightarrow \\frac{a+b}{1+c} \\ge \\frac{a+b}{a+b+c}$\nАдилаар $\\frac{b+c}{1+a} \\ge \\frac{b+c}{a+b+c}$; $\\frac{c+a}{1+b} \\ge \\frac{c+a}{a+b+c}$ болох ба эдгээрийг нэмбэл $\\frac{a+b}{1+c} + \\frac{b+c}{1+a} + \\f... | Mongolia | ММО-48 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Mongolian | proof only | null | |
0fm3 | Problem:
Sea $ABC$ un triángulo acutángulo, $\omega$ su circunferencia inscrita de centro $I$, $\Omega$ su circunferencia circunscrita de centro $O$, y $M$ el punto medio de la altura $AH$, donde $H$ pertenece al lado $BC$. La circunferencia $\omega$ es tangente a este lado $BC$ en el punto $D$. La recta $MD$ corta a ... | [
"Solution:\n\nSupongamos que $b = c$. Entonces, el pie de la altura $H$ coincide con el punto de tangencia $D$, luego $DM$ es perpendicular a $BC$ y $N$ no está definido. Asumiremos entonces sin pérdida de generalidad que $b > c$. Sea $U$ el punto de la recta $BC$ cuya potencia es la misma respecto de $\\omega$ y $... | Spain | 48 aME | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonomet... | null | proof only | null | |
08pz | Problem:
Let $a$, $b$, $c$ be positive real numbers. Prove the inequality
$$
\left(a^{2}+a c+c^{2}\right)\left(\frac{1}{a+b+c}+\frac{1}{a+c}\right)+b^{2}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)>a+b+c
$$ | [
"Solution:\nBy the Cauchy-Schwarz Inequality, we have\n$$\n\\frac{1}{a+b+c}+\\frac{1}{a+c} \\geqslant \\frac{4}{2 a+b+2 c}\n$$\nand\n$$\n\\frac{1}{b+c}+\\frac{1}{a+b} \\geqslant \\frac{4}{a+2 b+c}\n$$\nSince\n$$\na^{2}+a c+c^{2}=\\frac{3}{4}(a+c)^{2}+\\frac{1}{4}(a-c)^{2} \\geqslant \\frac{3}{4}(a+c)^{2}\n$$\nthen,... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
060r | Problem:
Démontrer, pour tous réels $a$, $b$, $c$ strictement positifs, l'inégalité suivante :
$$
\frac{a}{9 b c+1}+\frac{b}{9 c a+1}+\frac{c}{9 a b+1} \geqslant \frac{a+b+c}{1+(a+b+c)^{2}} .
$$ | [
"Solution:\n\nVu la forme de l'inégalité, on est tenté d'appliquer l'inégalité des mauvais élèves. Seulement, si l'on applique l'inégalité en l'état, on trouve :\n$$\n\\frac{a}{9 b c+1}+\\frac{b}{9 c a+1}+\\frac{c}{9 a b+1} \\geqslant \\frac{(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})^{2}}{9(a b+b c+c a)+3}\n$$\nPour conclure,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03ql | Let $a$, $b$ and $c$ be positive real numbers. Determine the minimum value of
$$ \frac{a+3c}{a+2b+c} + \frac{4b}{a+b+2c} - \frac{8c}{a+b+3c} \text{ (posed by Li Shenghong)} $$ | [
"The answer is $12\\sqrt{2}-17$. Set\n$$\n\\begin{cases} x = a+2b+c, \\\\ y = a+b+2c, \\\\ z = a+b+3c. \\end{cases}\n$$\n\n$$\n\\begin{aligned}\n& \\frac{a+3c}{a+2b+c} + \\frac{4b}{a+b+2c} - \\frac{8c}{a+b+3c} \\\\\n&= \\frac{2y-x}{x} + \\frac{4(x+z-2y)}{y} - \\frac{8(z-y)}{z} \\\\\n&= -17+2\\frac{y}{x}+4\\frac{x}{... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 12*sqrt(2)-17 | |
08ey | Problem:
Sia $ABC$ un triangolo rettangolo in $B$. Sia $H$ il piede dell'altezza uscente da $B$ e sia $D$ l'intersezione fra la bisettrice dell'angolo in $A$ e il lato $BC$. Supponiamo che $HD$ sia perpendicolare a $BC$. Quanto vale $\left(\frac{AB}{BC}\right)^2$?
(A) $\frac{\sqrt{5}}{2}$
(B) $\frac{\sqrt{5}-1}{2}$
(... | [
"Solution:\n\nLa risposta è $\\mathbf{(B)}$.\n\n\n\nI triangoli $BAC$, $HBC$, $HAB$ sono simili, dunque ci forniscono le uguaglianze\n$$\n\\frac{AB}{BC} = \\frac{BH}{CH} = \\frac{AH}{BH}\n$$\nLe rette $DH$ e $AB$ sono parallele, perché entrambe perpendicolari a $BC$; per Talete e il teorema... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | B | |
0bhi | Denote $O$ the center of the square $ABCD$. The bisector of the angle $\angle OAB$ meets $OB$ in $N$ and $BC$ in $P$. Prove that $PC = 2ON$. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
01br | Let $ABCD$ be a convex quadrilateral such that the line $BD$ bisects the angle $ABC$. Suppose that the circumcircle of triangle $ABC$ intersects the sides $AD$ and $CD$ in the points $P$ and $Q$, respectively. The line through $D$ and parallel to $AC$ intersects the lines $BC$ and $BA$ at the points $R$ and $S$, respec... | [
"**Solution 1.** Since $\\angle SDP = \\angle CAP = \\angle RBP$, the quadrilateral $BRDP$ is cyclic (see Figure 1). Similarly, the quadrilateral $BSDQ$ is cyclic. Let $X$ be the second intersection point of the segment $BD$ with the circumcircle of the triangle $ABC$. Then\n$$\n\\angle AXB = \\angle ACB = \\angle ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geome... | null | proof only | null | |
04f4 | Determine all positive integers $n$ such that $\log_2(3^n + 7)$ is also a positive integer. | [
"Let $k = \\log_2(3^n + 7)$, where $k$ is a positive integer. Then $3^n + 7 = 2^k$.\n\nSo $2^k - 3^n = 7$.\n\nWe seek all positive integers $n$ such that this equation has a solution in positive integers $k$.\n\nLet us try small values of $n$:\n\nFor $n = 1$: $2^k - 3 = 7 \\implies 2^k = 10$ (not a power of $2$).\n... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | n = 2 | |
0i1q | Problem:
Define a monic irreducible polynomial with integral coefficients to be a polynomial with leading coefficient 1 that cannot be factored, and the prime factorization of a polynomial with leading coefficient 1 as the factorization into monic irreducible polynomials. How many not necessarily distinct monic irredu... | [
"Solution:\n\n$x^{8}+x^{4}+1=\\left(x^{8}+2 x^{4}+1\\right)-x^{4}=\\left(x^{4}+1\\right)^{2}-\\left(x^{2}\\right)^{2}=\\left(x^{4}-x^{2}+1\\right)\\left(x^{4}+x^{2}+1\\right)=\\left(x^{4}-x^{2}+1\\right)\\left(x^{2}+x+1\\right)\\left(x^{2}-x+1\\right)$, and $x^{8}+x+1=\\left(x^{2}+x+1\\right)\\left(x^{6}-x^{5}+x^{3... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 5 | |
09x0 | Problem:
Zij $ABC$ een scherphoekige driehoek en zij $P$ het snijpunt van de raaklijnen in $B$ en $C$ aan de omgeschreven cirkel van $\triangle ABC$. De lijn door $A$ loodrecht op $AB$ en de lijn door $C$ loodrecht op $AC$ snijden in $X$. De lijn door $A$ loodrecht op $AC$ en de lijn door $B$ loodrecht op $AB$ snijden... | [
"Solution:\n\nZij $M$ het midden van de omgeschreven cirkel van $\\triangle ABC$ en schrijf $\\alpha = \\angle BAC$. We bewijzen eerst dat $\\triangle BYA \\sim \\triangle BMP$ en vervolgens dat $\\triangle YBM \\sim \\triangle ABP$. Wegens de middelpunt-omtrekshoekstelling geldt $\\angle BMC = 2\\angle BAC = 2\\al... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null |
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