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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
00qj | Let $n$ be a positive integer, $g(n)$ be the number of positive divisors of $n$ of the form $6k + 1$ and $h(n)$ be the number of positive divisors of $n$ of the form $6k - 1$, where $k$ is a nonnegative integer. Find all positive integers $n$ such that $g(n)$ and $h(n)$ have different parity. | [
"Let $n = 2^a \\cdot 3^b p_1^{\\alpha_1} \\dots p_s^{\\alpha_s}$ where $p_i \\neq 2, 3$ for $i = 1, 2, \\dots, s$ are distinct prime numbers. If $t$ is a divisor of $n$ of the form $6k \\pm 1$, then $t$ is a divisor of $p_1^{\\alpha_1} \\dots p_s^{\\alpha_s}$ (in other words, $t$ is not divisible by $2$ or by $3$).... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All n of the form n = 2^a · 3^b · m^2, with a, b ≥ 0 and m a positive integer. | |
07y2 | How many crosswords can we make from the word “CROSSWORD”, assuming the following rules?
1. A crossword has two words, one horizontal and one vertical, intersecting at a single letter.
2. A 'word' is any ordered set of at least two letters. It does not need to be a real word in any language.
3. Letters should be used a... | [
"We consider first the sequence of the letters, and then the configuration in which they are arranged. There are 9 letters on CROSSWORD, consisting of 3 unique letters and three duplicate pairs. If all letters were distinct there would be $9! = 362,880$ permutations, but because of the three duplicates, this counts... | Ireland | IRL_ABooklet_2025 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 6667920 | |
025o | Problem:
Uma noite, Wanderson sonhou com dois números de três algarismos:
$$
a b c \text{ e } d e f,
$$
de modo que a soma
$$
a b c + d e f + a b c d e f
$$
coincidia com a soma de todos os números de três algarismos. Note que $a b c$ não é o produto dos algarismos $a, b$ e $c$, e sim o número de três algarismos $a, b... | [
"Solution:\n\nPodemos calcular a soma de todos os números de 3 algarismos\n$$\n100 + 101 + 102 + \\cdots + 999 = \\frac{(100 + 999) \\times 900}{2} = 1099 \\times 450\n$$\n\nPor outro lado,\n$$\na b c + d e f + a b c d e f = a b c + d e f + 1000 \\times a b c + d e f\n$$\nO sonho de Wanderson pode então ser escrito... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 494550; b) Impossible | |
0gpm | Show that for all positive real numbers $a, b, c$ satisfying $a^3 + b^3 + c^3 = a^4 + b^4 + c^4$ the following inequality is held:
$$
\frac{a}{a^2 + b^3 + c^3} + \frac{b}{a^3 + b^2 + c^3} + \frac{c}{a^3 + b^3 + c^2} \ge 1
$$ | [
"Let $f(a, b, c) = \\frac{a}{a^2 + b^3 + c^3} + \\frac{b}{a^3 + b^2 + c^3} + \\frac{c}{a^3 + b^3 + c^2}$ and $g(a, b, c) = a(a^2 + b^3 + c^3) + b(a^3 + b^2 + c^3) + c(a^3 + b^3 + c^2)$. Observe that the Cauchy-Schwarz Inequality gives $f(a, b, c) \\cdot g(a, b, c) \\ge (a+b+c)^2$.\n\nObserve that $g(a, b, c) = (a+b... | Turkey | 17th Junior Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
04iv | Let $\lfloor x \rfloor$ denote the largest integer not greater than real number $x$. Determine all real solutions of the equation
$$
11 \lfloor x \rfloor + \lfloor x + \frac{1}{2} \rfloor = 9x.
$$ | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 0, 4/3, 25/9 | |
0i9c | Problem:
Let $z = 1 - 2i$. Find $\frac{1}{z} + \frac{2}{z^{2}} + \frac{3}{z^{3}} + \cdots$. | [
"Solution:\nLet $x = \\frac{1}{z} + \\frac{2}{z^{2}} + \\frac{3}{z^{3}} + \\cdots$, so $z \\cdot x = \\left(1 + \\frac{2}{z} + \\frac{3}{z^{2}} + \\frac{4}{z^{3}} + \\cdots\\right)$.\n\nThen $z \\cdot x - x = 1 + \\frac{1}{z} + \\frac{1}{z^{2}} + \\frac{1}{z^{3}} + \\cdots = \\frac{1}{1 - 1/z} = \\frac{z}{z - 1}$.\... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | (2i - 1)/4 | |
0esj | What is the value of $(2+4+6+...+198+200) - (1+3+5+...+197+199)$? | [
"Rearranging, the value is $(2-1) + (4-3) + (6-5) + \\dots + (200-199)$ which has $100$ brackets and therefore totals $100 \\times 1 = 100$."
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 100 | |
04en | 27 points are arranged in 9 columns and 3 rows, and each is coloured blue or red. Prove that there exists a rectangle whose vertices are all of the same colour. | [
"Consider the $3 \\times 9$ grid, with each point coloured blue or red.\n\nFor each row, consider the sequence of colours in the 9 columns. For a fixed row, there are $2^9 = 512$ possible colourings. However, we only have 3 rows.\n\nNow, focus on the columns. For each column, consider the triple of colours in the 3... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0e5e | Let $[x]$ be the greatest integer that is not greater than the real number $x$, and let $[[x]] = x - [x]$. Solve the equation
$$
3[[x]] = x[x] + 1.
$$ | [
"Let $x = n + a$ where $n \\in \\mathbb{Z}$ and $a \\in [0, 1)$. Then $3a = n^2 + an + 1$.\n\nIf $n \\ge 0$, then $3 > 3a = n^2 + an + 1 \\ge n^2 + 1$, hence $n \\le 1$.\n\nIf $n = 0$, we get the equation $3a = 1$, hence $a = x = \\frac{1}{3}$.\n\nIf $n = 1$, we get the equation $2a = 2$, which has no solutions str... | Slovenia | Selection Examinations for the IMO 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = -1/2 and x = 1/3 | |
0dsp | Find all functions $f : \mathbb{N} \setminus \{1\} \rightarrow \mathbb{N}$ such that for all distinct $x, y \in \mathbb{N}$ with $y \ge 2018$,
$$
\gcd(f(x), y) \cdot \operatorname{lcm}(x, f(y)) = f(x)f(y)
$$ | [] | Singapore | Singapore International Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | f(n) = n for all n ≥ 2 | |
08v7 | Let $I$ and $O$ be the incenter and circumcenter of a triangle $ABC$, respectively. If $AB = 2$, $AC = 3$ and $\angle AIO = 90^\circ$, what is the area of the triangle $ABC$? Here for a line segment $XY$ its length also is denoted by $XY$. | [
"Let $M$, $N$ be the midpoints of the sides $AB$, $AC$, respectively. Since $\\angle AMO = \\angle ANO = 90^\\circ = \\angle AIO$, we see that the points $A$, $M$, $N$, $I$ lie on the circle having $AO$ as a diameter. In particular, the quadrilateral $AMIN$ is inscribed in this circle, and hence we have $\\angle AN... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > D... | null | proof and answer | 15√7/16 | |
0fpi | Sea $ABC$ un triángulo rectángulo en $C$ no isósceles con catetos $b > a$.
1. Hallar el lado del cuadrado $AXYZ$ que circunscribe al triángulo $ABC$ (los vértices $B$ y $C$ tienen que estar en lados distintos del cuadrado).
2. Explicar paso a paso cómo construir el cuadrado $AXYZ$ con regla y compás. | [
"(i) Sea $l$ la longitud del cuadrado y $x$ la longitud del segmento $XC$. Los triángulos rectángulos $AXC$ y $BYC$ son semejantes (puesto que $\\angle BCY = \\pi/2 - \\angle ACX = \\angle CAX$), de donde $l/b = (l - x)/a$, es decir, $x/l = (b - a)/b$. Entonces, aplicando el Teorema de Pitágoras,\n$$\nb^2 = l^2 + x... | Spain | LII Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Spanish | proof and answer | l = b^2 / sqrt((b - a)^2 + b^2) | |
01h8 | Let $N$ be a positive integer. Determine the number of pairs, $(a, b)$, of positive integers $a$ and $b$ such that the number
$$
\frac{ab}{a+b}
$$
is a divisor of $N$. | [
"In what follows $d(n)$ denotes the number of positive divisors of a positive integer $n$. We show that the number of pairs is $d(N)^2$. Let $m$ be a fixed positive divisor of $N$. We claim that the number of pairs, $(a, b)$, such that $ab/(a+b) = m$ is $d(m)^2$. This becomes clear if we rewrite the equation as\n$$... | Baltic Way | Baltic Way 2020 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | d(N)^2 | |
07ab | Suppose $S$ is a convex figure in plane with area $10$. Consider a chord of length $3$ in $S$ and let $A$ and $B$ be two points on this chord which divide it into three equal parts. For a variable point $X$ in $S - \{A, B\}$, let $A'$ and $B'$ be the intersection points of rays $AX$ and $BX$ with the boundary of $S$. L... | [
"The idea is to remove some neighborhoods of $A$ and $B$, because near these points we cannot bound $\\frac{AA'}{BB'}$. Let $Z$ be the intersection of the ray $AB$ with the boundary of $S$ and let $ZB'$ intersect $AA'$ in $A''$. $A''$ is between $A$ and $A'$ since $S$ is convex. So, if we let $S''$ be the set of th... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ghu | 給定正整數 $k > 1$ 以及 $n$ 個 $k$ 維座標中的相異點 $a^{(1)} = (a_1^{(1)}, \dots, a_k^{(1)})$, $\dots, a^{(n)} = (a_1^{(n)}, \dots, a_k^{(n)})$, 我們定義 $a^{(i)}$ 的『分數』如下：
$$
\prod_{j=1}^{k} \#\{i' \mid 1 \le i' \le n \text{ 使得 } \pi_j(a^{(i')}) = \pi_j(a^{(i)})\},
$$
其中 $\#S$ 表示集合 $S$ 的元素個數, 而 $\pi_j : \mathbb{R}^k \to \mathbb{R}^{k-1}$... | [
"The answer is $\\frac{1}{k-1}$. To see that $t \\le \\frac{1}{k-1}$, consider the sequences where all but one of the components are equal to 1, and the remaining one component is $1, 2, \\dots, m$. Then $n = km + 1$ and the score of the zero sequence is $m^k$. In order for the statement to hold, we must have\n$$\n... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Chinese (Traditional) | proof and answer | 1/(k-1) | |
0h54 | It is known that $\frac{a}{b+c+d} + \frac{b}{c+d+a} + \frac{c}{d+a+b} + \frac{d}{a+b+c} = 1$. Find the value of the expression
$$
\frac{a^2}{b+c+d} + \frac{b^2}{c+d+a} + \frac{c^2}{d+a+b} + \frac{d^2}{a+b+c}.
$$ | [] | Ukraine | Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | 0 | |
0dc9 | Suppose that $a, b, c, d$ are pairwise distinct positive integers such that $a+b=c+d=p$ for some odd prime $p>3$. Prove that $a b c d$ is not a perfect square. | [
"Suppose that $a b c d = n^{2}$ for some $n \\in \\mathbb{Z}^{+}$. We can suppose that $a < c < d < b$. From this, we have $b d + a c - a d - b c = (b - a)(d - c) > 0$, thus\n$$\na d + b c < \\frac{1}{2}(a d + b c + b d + a c) = \\frac{(a + b)(c + d)}{2} = \\frac{p^{2}}{2} .\n$$\nDenote $\\gcd(a d, b c) = k \\in \\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0d3e | Prove that $2014$ divides $53 n^{55} - 57 n^{53} + 4 n$ for all integer $n$. | [
"Notice first that $2014 = 2 \\times 19 \\times 53$. Applying Fermat's little theorem for the prime numbers $2, 19, 53$ we obtain:\n$$\n\\begin{gathered}\n53 n^{55} - 57 n^{53} + 4 n \\equiv n - n + 0 \\equiv 0 \\quad \\bmod 2 \\\\\n53 n^{55} - 57 n^{53} + 4 n \\equiv 15 n^{17} (n^{19})^{2} - 0 + 4 n \\equiv 15 n^{... | Saudi Arabia | SAMC | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English, Arabic | proof only | null | |
0ead | Problem:
Dana je pokončna prizma, katere osnovna ploskev je enakostranični trikotnik s stranico, dolgo $12~\mathrm{cm}$. Izračunaj višino pokončne piramide, ki ima enako višino, enako osnovno ploskev in enako površino plašča kot dana pokončna prizma. | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 2 cm | |
013f | Problem:
A circle is divided into 13 segments, numbered consecutively from 1 to 13. Five fleas called $A, B, C, D$ and $E$ are sitting in the segments 1, 2, 3, 4 and 5. A flea is allowed to jump to an empty segment five positions away in either direction around the circle. Only one flea jumps at the same time, and two ... | [
"Solution:\nWrite the numbers from 1 to 13 in the order $\\mathbf{1}, 6, 11, \\mathbf{3}, 8, 13, 5, 10, 2, 7, 12, 4, 9$. Then each time a flea jumps it moves between two adjacent numbers or between the first and the last number in this row. Since a flea can never move past another flea, the possible permutations ar... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Exactly the cyclic permutations of A, B, C, D, E: ABCDE, BCDEA, CDEAB, DEABC, EABCD. | |
0831 | Problem:
Una stella a 6 punte viene disegnata costruendo sui lati di un esagono regolare sei triangoli isosceli con angolo al vertice di $30$ gradi. Sapendo che la circonferenza che passa per le punte della stella ha raggio $1$, calcolare l'area della stella stessa.
(A) $\sqrt{3}$
(B) $\frac{3(\sqrt{3}-1)}{2}$
(C) $\... | [
"Solution:\n\nLa risposta è (B). I sei triangoli isosceli hanno due angoli di $75^{\\circ}$. L'angolo compreso fra due lati di due triangoli isosceli aventi un vertice in comune è quindi $360^{\\circ}-\\left(75^{\\circ}+75^{\\circ}+120^{\\circ}\\right)=90^{\\circ}$. Con riferimento alla figura a fianco, si osserva ... | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO BIENNIO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | B | |
0fom | Un conjunto de rectas en el plano está en posición general si no hay dos que sean paralelas ni tres que pasen por el mismo punto. Un conjunto de rectas en posición general separa el plano en regiones, algunas de las cuales tienen área finita; a estas las llamamos sus regiones finitas. Demostrar que para cada $n$ sufici... | [
"**Solución parcial por Daniel Lasaosa Medarde, Pamplona, España.** Llamemos vértice a cada punto donde se cortan dos rectas del conjunto, vértice azul a cada vértice donde las dos rectas que se corten están coloreadas de azul, y polígono a cada región finita del plano delimitada por las rectas trazadas. Decimos qu... | Spain | LV Olimpiada Internacional de Matemáticas | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Spanish | proof only | null | |
0b3j | Problem:
Suppose that the polynomial $P(x) = x^{3} + 4x^{2} + bx + c$ has a single root $r$ and a double root $s$ for some distinct real numbers $r$ and $s$. Given that $P(-2s) = 324$, what is the sum of all possible values of $|c|$? | [
"Solution:\nBy Vieta's formula, we have $r + 2s = -4$ and writing $P(x) = (x - r)(x - s)^{2}$, we have $324 = P(-2s) = (-2s - r)(-3s)^{2} = 36s^{2}$. Thus, $s^{2} = 9$ and $s \\in \\{-3, 3\\}$.\n\nWe next observe that $|c| = |P(0)| = |r| s^{2} = |-4 - 2s| s^{2}$. Hence, the sum of all possible values of $|c|$ is $9... | Philippines | 24th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 108 | |
0huj | Problem:
Let $S > 0$. If $a, b, c, x, y, z$ are positive real numbers such that $a + x = b + y = c + z = S$, prove that
$$
a y + b z + c x < S^{2}
$$ | [
"Solution:\nDenote $T = S / 2$. One of the triples $(a, b, c)$ and $(x, y, z)$ has the property that at least two of its members are greater than or equal to $T$. Assume that $(a, b, c)$ is the one, and choose $\\alpha = a - T$, $\\beta = b - T$, and $\\gamma = c - T$. We then have $x = T - \\alpha$, $y = T - \\bet... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0b2z | Problem:
In how many ways can the letters of the word PANACEA be arranged so that the three As are not all together?
(a) 540
(b) 576
(c) 600
(d) 720 | [
"Solution:\n\nThe word PANACEA has 7 letters, with the letter A appearing 3 times, and the other letters P, N, C, E each appearing once.\n\nFirst, find the total number of arrangements of the letters:\n\nNumber of arrangements $= \\dfrac{7!}{3!} = \\dfrac{5040}{6} = 840$\n\nNow, count the number of arrangements whe... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | d | |
00ha | We say that a triangle $ABC$ is great if the following holds: for any point $D$ on the side $BC$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $AB$ and $AC$, respectively, then the reflection of $D$ in the line $PQ$ lies on the circumcircle of the triangle $ABC$.
Prove that triangle $ABC$ is ... | [
"For every point $D$ on the side $BC$, let $D'$ be the reflection of $D$ in the line $PQ$. We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at $A$.\n\nChoose $D$ to be the point where the angle bisector from $A$ meets $BC$. Note that $P$ and $Q$ lie on the rays ... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2016 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0khn | Problem:
In triangle $ABC$, $\angle A = 2 \angle C$. Suppose that $AC = 6$, $BC = 8$, and $AB = \sqrt{a} - b$, where $a$ and $b$ are positive integers. Compute $100a + b$. | [
"Solution:\n\nLet $x = AB$, and $\\angle C = \\theta$, then $\\angle A = 2\\theta$ and $\\angle B = 180^\\circ - 3\\theta$.\n\nExtend ray $BA$ to $D$ so that $AD = AC$. We know that $\\angle CAD = 180^\\circ - 2\\theta$, and since $\\triangle ADC$ is isosceles, it follows that $\\angle ADC = \\angle ACD = \\theta$,... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 7303 | |
0af2 | Во триаголникот $ABC$, $\angle ACB = 40^{\circ}$. Симетралите на внатрешниот и надворешниот агол во темето $C$ ја сечат правата $AB$ во точките $D$ и $E$, така што $\triangle CDE$ е рамнокрак. Определи ги аглите на триаголникот $ABC$.
 | [
"Нека $CD$ е симетрала на надворешниот агол, а $CE$ на внатрешниот во темето $C$. Симетралите $CD$ и $CE$ се нормални, тоа се симетрали на два напоредни агли, $\\angle DCE = 90^{\\circ}$. Од условот $\\triangle CDE$ е рамнокрак, па следува дека тој е рамнокрак правоаголен со хипотенуза $DE$. Оттука следува дека $\\... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | Macedonian, English | proof and answer | ∠A = 115°, ∠B = 25°, ∠C = 40° | |
0d8k | There are $4950$ ants. Assume that, for any three ants $A$, $B$ and $C$, if the ant $A$ is the boss of the ant $B$, and the ant $B$ is the boss of the ant $C$ then the ant $A$ is also the boss of the ant $C$. We want to divide the ants into $n$ groups so that in any group, either any two ants have the boss relationship... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 141 | |
0keg | Problem:
Alice, Bob, and Charlie roll a 4-, 5-, and 6-sided die, respectively. What is the probability that a number comes up exactly twice out of the three rolls? | [
"Solution:\n\nThere are $4 \\cdot 5 \\cdot 6 = 120$ different ways that the dice can come up. The common number can be any of $1, 2, 3, 4$, or $5$: there are $3 + 4 + 5 = 12$ ways for it to be each of $1, 2, 3$, or $4$, because we pick one of the three people's rolls to disagree, and there are $3, 4$, and $5$ ways ... | United States | HMMT February 2020 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 13/30 | |
057s | For every positive integer $x$, denote by $\kappa(x)$ the number of composite numbers not greater than $x$. Find all positive integers $n$ such that
$$
(\kappa(n))! \cdot \text{lcm}(1,2,\dots,n) > (n-1)!
$$ | [
"The inequality holds for $n = 2,3,4,5,7,9$ and does not hold for $n = 1,6,8,10,11,12$. Assume in the rest that $n \\ge 13$. By definition of $\\kappa(n)$, there exists exactly $n-1-\\kappa(n)$ prime numbers not greater than $n$; these are the primes dividing $\\text{lcm}(1,2,\\dots,n)$. We have $n-1-\\kappa(n) \\g... | Estonia | IMO Team Selection Contest | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2, 3, 4, 5, 7, 9 | |
08xe | Let $ABCD$ be a convex quadrilateral for which the line segments $AC$ and $BD$ intersect perpendicularly at a point $X$. Suppose $AX = 5$, $BX = 6$ and $CX = 20$ are satisfied. Let us denote by $O_1, O_2, O_3$ and $O_4$, respectively, the circle with center at $A$ and radius $AX$, the circle with center at $B$ and radi... | [
"Let $DX = d$. Choose xy-coordinate axis in such a way that $X = (0,0)$, $A = (5,0)$, $B = (0,6)$, $C = (-20,0)$, $D = (0,-d)$ are satisfied. Let $\\Gamma$ be the circle tangent to each of the circles $O_1, O_2, O_3, O_4$, and let $P = (x, y)$ be the center of $\\Gamma$ and $r$ be its radius. Since the circle $O_1$... | Japan | Japan 2013 Initial Round | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 12 | |
001o | Se definen las sucesiones $(a_n)_{n \ge 0}$ y $(b_n)_{n \ge 0}$ por:
$a_0 = 1$, $b_0 = 4$ y
$a_{n+1} = a_n^{2001} + b_n$, $b_{n+1} = b_n^{2001} + a_n$ para $n \ge 0$.
Demuestre que 2003 no divide a ningún uno de los terminos de estas sucesiones. | [] | Argentina | XVIII Olimpiada Iberoamericana de Matemática | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | español | proof only | null | |
0f7l | Problem:
What is the smallest number of subsets of $S = \{1, 2, \ldots, 33\}$, such that each subset has size $9$ or $10$ and each member of $S$ belongs to the same number of subsets? | [] | Soviet Union | 21st ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 10 | |
0b5r | Each of 2009 (distinct) points in the plane is coloured blue or red, so that on every blue-centered unit circle lie exactly two red points. Determine the largest possible number of blue points. | [
"Suppose there are $r$ red points among some $n$ points. Since any pair of them can lie on at most two blue-centered unit circles, it means that the number $b$ of blue points can be at most $2\\binom{r}{2} = r(r-1)$. Since $b + r = n$, this leads to condition $r + r(r-1) = r^2 \\ge n$, i.e. $r \\ge \\lceil \\sqrt{n... | Romania | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 1964 | |
0i77 | Problem:
Evaluate $\int_{0}^{(\sqrt{2}-1) / 2} \frac{\mathrm{~d} x}{(2 x+1) \sqrt{x^{2}+x}}$. | [
"Solution:\n\nLet $u=\\sqrt{x^{2}+x}$. Then $d u=\\frac{2 x+1}{2 \\sqrt{x^{2}+x}} d x$. So the integral becomes $2 \\int \\frac{d u}{4 x^{2}+4 x+1}$, or $2 \\int \\frac{d u}{4 u^{2}+1}$. This is $\\tan^{-1}(2 u)$, yielding a final answer of $\\tan^{-1}\\left(2 \\sqrt{x^{2}+x}\\right)+C$ for the indefinite integral.... | United States | Harvard-MIT Math Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | π/4 | |
0266 | Problem:
Diferença de quadrados - Se a diferença dos quadrados de dois números inteiros consecutivos é $2000$, então os dois números são:
(A) menores que $100$.
(B) menores que $1000$, porém maiores que $99$.
(C) menores que $10000$, porém maiores que $999$.
(D) menores que $100000$, porém maiores que $9999$.
(E) não... | [
"Solution:\n\nA resposta correta é (E).\n\nInicialmente, observe que o quadrado de um número par é par, e o quadrado de um número ímpar é ímpar. Se os dois números são consecutivos, então um número é par e o outro é ímpar. Portanto, elevando ao quadrado, um deles é par e o outro é ímpar. Mas, a diferença entre um n... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | E | |
08wv | Each of two people, $A$ and $B$, chose 3 positive one-digit integers (which need not be distinct). The sum of the 3 numbers $A$ chose is equal to the product of the 3 numbers $B$ chose, and the product of the 3 numbers $A$ chose is equal to the sum of the 3 numbers $B$ chose. How many possibilities are there for triple... | [
"Let us denote by $S_A$ and $P_A$ the sum and product of the 3 numbers chosen by $A$, respectively, and by $S_B$ and $P_B$ those for $B$. Then the condition of the problem states that $S_A = P_B$ and $S_B = P_A$ hold.\nWe then have $P_B = S_A \\le 9+9+9 \\le 27$. In view of the fact that for any pair of real number... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | 7 | |
034k | Problem:
Find all values of the real parameter $a$ such that the number of the solutions of the equation
$$
3\left(5 x^{2}-a^{4}\right)-2 x=2 a^{2}(6 x-1)
$$
does not exceed the number of the solutions of the equation
$$
2 x^{3}+6 x=\left(3^{6 a}-9\right) \sqrt{2^{8 a}-\frac{1}{6}}-(3 a-1)^{2} 12^{x}
$$ | [
"Solution:\nThe first equation is quadratic with discriminant $D=(9 a^{2}-1)^{2}$. Therefore it has two different solutions for $a \\neq \\pm \\frac{1}{3}$ and exactly one solution for $a= \\pm \\frac{1}{3}$.\n\nSince the function\n$$\n2 x^{3}+6 x+(3 a-1)^{2} 12^{x}\n$$\nis strictly increasing, the second equation ... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 1/3 | |
03d7 | Let $ABCDV$ be a regular-based right pyramid with apex at $V$. The plane $\lambda$ intersects the edges $VA$, $VB$, $VC$, and $VD$ at points $M$, $N$, $P$, and $Q$, respectively. Compute the ratio $VQ:QD = p:q$, if $VM:MA = 2:1$, $VN:NB = 1:1$, and $VP:PC = 1:2$. | [
"Let $l = VA = VB = VC = VD$, $\\angle AVC = 2\\varphi$, and denote by $E$ the intersection point of the altitude $VO$ with $\\lambda$. Since $\\lambda \\cap (ACV) = MP$, $\\lambda \\cap (BDV) = NQ$, $(ACV) \\cap (BDV) = VO$, $E$ will be the intersection point of the diagonals of $MNPQ$ ($E = MP \\cap NQ$). Then $V... | Bulgaria | Bulgaria 2022 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2:3 | |
0i4j | Problem:
How many sequences of $0$s and $1$s are there of length $10$ such that there are no three $0$s or $1$s consecutively anywhere in the sequence? | [
"Solution:\nWe can have blocks of either $1$ or $2$ $0$s and $1$s, and these blocks must be alternating between $0$s and $1$s. The number of ways of arranging blocks to form a sequence of length $n$ is the same as the number of omino tilings of a $1$-by-$n$ rectangle, and we may start each sequence with a $0$ or a ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 178 | |
0g53 | Problem:
Trouver tous les triplets $(a, b, p)$ d'entiers strictement positifs où $p$ est premier et l'équation
$$
(a+b)^{p}=p^{a}+p^{b}
$$
est vérifiée. | [
"Solution:\n$(a, b, p)=(1,1,2)$ est le seul triplet qui vérifie l'équation. Séparons le problème en deux cas.\n\n- Cas $1: a=b$\n\nL'équation se simplifie en $2^{p} \\cdot a^{p}=2 \\cdot p^{a}$, et comme $4\\mid 2^{p}, 2\\mid p^{a}$ ce qui implique que $p=2$. Substituant cela dans l'équation de base donne l'égalité... | Switzerland | Deuxième tour 2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (1, 1, 2) | |
0deo | Let $ABC$ be a triangle inscribed in circle $(O)$ with diameter $KL$ passes through the midpoint $M$ of $AB$ such that $L$, $C$ lie on the different sides with respect to $AB$. A circle passes through $M$, $K$ cuts $LC$ at $P$, $Q$ (point $P$ lies between $Q$, $C$). The line $KQ$ cuts $(LMQ)$ at $R$. Prove that $ARBP$ ... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry ... | null | proof only | null | |
0fdx | Problem:
Un grupo de chicos y chicas han comido en un restaurante en el que sólo se sirven pizzas cortadas en 12 raciones. Cada chico comió 6 o 7 raciones y cada chica 2 o 3 raciones.
Se sabe que 4 pizzas no fueron suficientes y que con 5 pizzas hubo de sobra.
Calcular el número de chicos y de chicas del grupo. | [
"Solution:\nSean $x$ e $y$ el número de chicos $y$ de chicas, respectivamente.\nTenemos\n$$\n7x + 3y \\leq 59\n$$\ny\n$$\n6x + 2y \\geq 49\n$$\nRestando miembro a miembro obtenemos\n$$\nx + y \\leq 10\n$$\ny por (2),\n$$\n6x + 2(10 - x) \\geq 49\n$$\nde donde $x \\geq 8$.\nPero (1) implica que $x \\leq 8$.\nLuego $... | Spain | null | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 8 boys and 1 girl | |
04rs | Let $ABCD$ be a parallelogram such that the projections $K$, $L$ of $D$ onto the sides $AB$, $BC$, respectively, are their interior points. Prove that $KL \parallel AC$ if and only if
$$
\angle BCA + \angle ABD = \angle BDA + \angle ACD.
$$ | [
"Alternate angles $ABD$ and $CDB$ are equal (Fig. 2), hence $\\angle BCA + \\angle ABD + \\angle BDA + \\angle ACD = 180^\\circ$. The equality $\\angle BCA + \\angle ABD = \\angle BDA + \\angle ACD$ thus holds if and only if\n$$\n\\angle BCA + \\angle ABD = 90^\\circ.\n$$\n(1)\n\nFig. 2\nPo... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0l05 | Problem:
Let $A_{1} A_{2} \ldots A_{19}$ be a regular nonadecagon. Lines $A_{1} A_{5}$ and $A_{3} A_{4}$ meet at $X$. Compute $\angle A_{7} X A_{5}$.
Proposed by: Nithid Anchaleenukoon | [
"Solution:\n\nInscribing the nonadecagon in a circle, note that\n$$\n\\angle A_{3} X A_{5}=\\frac{1}{2}\\left(\\widehat{A_{1} A_{3}}-\\widehat{A_{4} A_{5}}\\right)=\\frac{1}{2} \\widehat{A_{5} A_{3} A_{4}}=\\angle A_{5} A_{3} X\n$$\nThus $A_{5} X=A_{5} A_{3}=A_{5} A_{7}$, so\n$$\n\\begin{al... | United States | HMMT February 2024 Guts Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | 1170°/19 | |
026y | Problem:
Bruno tem uma lista com todos os números naturais de 10 dígitos que se podem formar utilizando apenas os dígitos 1, 2, 3 e 4 e além disto que possuem igual quantidade de algarismos 1 e 2, por exemplo, 3333333333, 1111342222 etc. Bernardo tem a lista de todos os números naturais de 20 dígitos formados por 10 d... | [
"Solution:\n\nAbordemos inicialmente uma versão simplificada do problema. Suponha que Bruno quisesse listar todos os números de dois dígitos utilizando os mesmos quatro dígitos e também possuindo iguais quantidades de algarismos $1$ e $2$. Ele obteria a seguinte lista:\n$12, 21, 33, 44, 34, 43$.\n\nSuponha agora qu... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
01yy | There are $n \ge 2$ cities in a country. Some pairs of cities are connected with an air communication, and for any such pair the connection is mutual. It is possible to travel from one city to another (possibly with a couple of flights). It is also known that the minimal number of flights that are needed to travel from... | [
"Answer: $1, 2, \\ldots, [n/2]$.\n\nWe reformulate the problem in the language of Graph Theory. Consider a graph, the vertices of which correspond to cities and the edges correspond to air connections. It is given that the graph is connected and the eccentricity of all of its vertices (the greatest of the minimal d... | Belarus | SELECTION and TRAINING SESSION | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | all integers m with 1 ≤ m ≤ floor(n/2) | |
0gq6 | In a triangle $ABC$ with incenter $I$, the incircle of $ABC$ touches the side $[BC]$ at the point $D$ and let $T$ be the midpoint of the line segment $[ID]$. The line passing through $I$ and perpendicular to $AD$ intersects the lines $AB$ and $AC$ at the points $K$ and $L$, respectively. The line passing through $T$ an... | [
"Let the incircle of $ABC$ touch the sides $[AB]$ and $[AC]$ at the points $F$ and $E$, respectively. Let $AE = AF = x$, $BD = BF = y$ and $CD = CE = z$. Let the line passing through $D$ and parallel to $AC$ intersect $AB$ at $P$ and the line passing through $D$ and parallel to $AB$ intersect $AC$ at $Q$. Then $BPD... | Turkey | Team Selection Test for IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English | proof only | null | |
0hnp | Problem:
Let $n$ be an integer greater than $12$. Points $P_{1}, P_{2}, \ldots, P_{n}, Q$ in the plane are distinct. Prove that for some $i$, at least $n / 6 - 1$ of the distances
$$
P_{1} P_{i}, P_{2} P_{i}, \ldots, P_{i-1} P_{i}, P_{i+1} P_{i}, \ldots, P_{n} P_{i}
$$
are less than $P_{i} Q$. | [
"Solution:\n\nCut the plane into six $60^{\\circ}$ \"pizza slices\" with vertex $Q$. Rotating if necessary, we may assume that none of the $P_{j}$ lie on the cuts. By the pigeonhole principle, one slice contains at least $n / 6$ of the $P_{j}$. Let $P_{i}$ be a point in this slice farthest from $Q$. It remains to s... | United States | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01la | Points $K$ and $L$ are marked on the side $AB$ of the triangle $ABC$ so that $\angle ACK = \angle KCL = \angle LCB$. Point $M$ is marked on the side $BC$ so that $\angle MKC = \angle BKM$.
Find the value of $\angle MLC$, if $ML$ is a bisector of $\angle KMB$.
(S. Mazanik) | [
"Answer: $30^\\circ$.\n\n\n\nSince $L$ lies on the bisector of $\\angle KCB$, $L$ is equidistant from the lines $CP$ and $CB$. Similarly, since $L$ lies on the bisector of $\\angle KMB$, $L$ is equidistant from the rays $MK$ and $MB$. Therefore, $L$ is an equidistant point for the rays $KM$... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 30° | |
0fla | Problem:
Sea $P$ un punto cualquiera de la bisectriz del ángulo $A$ en el triángulo $ABC$, y sean $A'$, $B'$, $C'$ puntos respectivos de las rectas $BC$, $CA$, $AB$, tales que $PA'$ es perpendicular a $BC$, $PB'$ es perpendicular a $CA$ y $PC'$ es perpendicular a $AB$. Demuestra que $PA'$ y $B'C'$ se cortan sobre la m... | [
"Solution:\n\nSea $E$ el punto de intersección de $PA'$ y $B'C'$. Si $P$ se mueve sobre la bisectriz $AI$ ($I$ es el incentro), la figura $PB'C'E$ es homotética de sí misma con respecto al punto $A$. Luego $E$ describe una recta que pasa por $A$. La bisectriz $AI$ corta a la circunferencia circunscrita a $ABC$ en $... | Spain | XLVI Olimpiada Matemática Española Fase nacional | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler ... | null | proof only | null | |
019f | A number $N$, written in decimal notation, consists of $2011$ digits. All the digits are $1$, except the middle digit. If $N$ is divisible by $13$, find the middle digit. | [
"Since $1001$ is divisible by $13$, so is $111 \\times 1001 = 111111$. Noting that $2011 = 6 \\times 334 + 7$, by taking off blocks of $111111$ from $N$ we deduce that $111X111$ is divisible by $13$.\n\nNow reduce the number further by subtracting multiples of $1001$, obtaining multiples of $13$ at every step:\n$$1... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2 | |
0f1n | Problem:
Three flies crawl along the perimeter of a triangle. At least one fly makes a complete circuit of the perimeter. For the entire period the center of mass of the flies remains fixed. Show that it must be at the centroid of the triangle. [You may not assume, without proof, that the flies have the same mass, or ... | [] | Soviet Union | ASU | [
"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 > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordi... | null | proof only | null | |
0kqs | Problem:
Show that $a^{1729} \equiv a \pmod{1729}$ for all positive integers $a$. | [
"Solution:\n\nSince $1729 = 7 \\cdot 13 \\cdot 19$, by the Chinese Remainder Theorem it suffices to show that $a^{1729} \\equiv a$ in $\\bmod\\ 7$, $\\bmod\\ 13$, and $\\bmod\\ 19$. This follows from Fermat's Little Theorem, since\n$$\n\\begin{aligned}\n& a^{1729} \\equiv \\left(a^6\\right)^{288} \\cdot a \\equiv a... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
066v | Let $n$ be a positive integer. Two players, Alice and Bob, are playing the following game:
* Alice chooses $n$ real numbers, not necessarily distinct
* Alice writes all pairwise sums on a sheet of paper and gives it to Bob (there are $\frac{n(n-1)}{2}$ such sums, not necessarily distinct)
* Bob wins if he finds correct... | [] | Greece | Junior Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a. Yes
b. No
c. No | |
05j8 | Problem:
Dans un pays, il y a $n$ villes. Deux villes quelconques sont toujours reliées soit par une autoroute, soit par une ligne de train. Montrer qu'un des deux moyens de transport permet de relier n'importe quelle ville à n'importe quelle autre. | [
"Solution:\n\nOn raisonne par récurrence sur $n$ : pour $n=1$ ou $n=2$, le résultat est immédiat. Supposons qu'on l'a montré au rang $n$, et considérons $n+1$ villes :\nOn isole une des villes, appelons-la par exemple Paris. Alors un des deux moyens de transports permet de relier entre elles toutes les villes sauf ... | France | OFM 2013-2014 Envoi 2 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
082p | Problem:
Il piccolo Marco sale e scende da un piano all'altro la scala mobile di un centro commerciale, uno scalino alla volta. Se procede nel senso di marcia della scala a velocità costante rispetto ad essa (cioè l'intervallo di tempo fra un passo e l'altro è costante), calpesta 15 gradini, se procede in senso contra... | [
"Solution:\n\nLa risposta è 21. Indichiamo con $V_{s}$ la velocità della scala, in scalini per unità di tempo, e con $V_{m}$ la velocità di Marco, nella medesima unità di misura. Se $T_{1}$ e $T_{2}$ sono, rispettivamente, i tempi impiegati da Marco per percorrere la scala nel verso concorde e nel verso discorde, a... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 21 | |
0irw | Prove that for no integer $n$ is $n^7 + 7$ a perfect square. | [
"Assume on the contrary that $n^7 + 7 = m^2$ for some integer $m$.\nFor $n$ even, $n^7 + 7 \\equiv 3 \\pmod 4$, so it cannot be a perfect square. For $n \\equiv 3 \\pmod 4$, $n^7 + 7 \\equiv 2 \\pmod 4$, so again it cannot be a perfect square. We must have $n \\equiv 1 \\pmod 4$ and $m$ even.\n\nNote that $n^7 + 2^... | United States | Team Selection Test | [
"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 only | null | |
06s2 | Let $ABCD$ be a convex quadrilateral with non-parallel sides $BC$ and $AD$. Assume that there is a point $E$ on the side $BC$ such that the quadrilaterals $ABED$ and $AECD$ are circumscribed. Prove that there is a point $F$ on the side $AD$ such that the quadrilaterals $ABCF$ and $BCDF$ are circumscribed if and only if... | [
"Let $\\omega_1$ and $\\omega_2$ be the incircles and $O_1$ and $O_2$ the incenters of the quadrilaterals $ABED$ and $AECD$ respectively. A point $F$ with the stated property exists only if $\\omega_1$ and $\\omega_2$ are also the incircles of the quadrilaterals $ABCF$ and $BCDF$.\n\n\n\nLe... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterials",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l5f | Let $a_1, a_2, \dots$ and $b_1, b_2, \dots$ be sequences of real numbers for which $a_1 > b_1$ and
$$
a_{n+1} = a_n^2 - 2b_n
$$
$$
b_{n+1} = b_n^2 - 2a_n
$$
for all positive integers $n$. Prove that $a_1, a_2, \dots$ is eventually increasing (that is, there exists a positive integer $N$ for which $a_k < a_{k+1}$ for al... | [] | United States | TST2025 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0gur | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x, y$,
$$f(x+y)^3 = (x+2y)f(x^2) + f(f(y))(x^2+3xy+y^2).$$ | [
"Answer: $f(x) = 0$ and $f(x) = x$.\n\n$f(x + y)^3 = (x + 2y)f(x^2) + f(f(y))(x^2 + 3xy + y^2)$.\n\nBy putting $(x, y) = (0, 0)$ to the original equation we get $f(0) = 0$.\n\nBy putting $(x, y) = (x, 0)$ to the original equation we get $f(x)^3 = x f(x^2)$.\n\nBy putting $(x, y) = (0, x)$ to the original equation w... | Turkey | Team Selection Test for IMO 2024 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 and f(x) = x | |
0360 | Problem:
Find all values of the real parameter $a$ for which the equations $x^{2}-(2 a+1) x+a=0$ and $x^{2}+(a-4) x+a-1=0$ have real roots $x_{1}, x_{2}$ and $x_{3}, x_{4}$, respectively, such that
$$
\frac{x_{1}}{x_{3}}+\frac{x_{4}}{x_{2}}=\frac{x_{1} x_{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right)}{a}
$$ | [
"Solution:\nFor $a \\neq 0, a \\neq 1$, the given equality is equivalent to\n$$\n\\begin{aligned}\n& a\\left(x_{1} x_{2}+x_{3} x_{4}\\right)=x_{1} x_{2} x_{3} x_{4}\\left(x_{1}+x_{2}+x_{3}+x_{4}\\right) \\\\\n& \\Longleftrightarrow 2 a-1=(a-1)(a+5) \\Longleftrightarrow a^{2}+2 a-4=0 \\\\\n& \\Longleftrightarrow a_{... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = -1 ± sqrt(5) | |
04ab | Prove that $7 \mid 2^{n+2} + 3^{2n+1}$ for all $n \in \mathbb{N}$. | [] | Croatia | Hrvatska 2011 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
00jr | For which number from $2000$ through $2100$ is the probability that a randomly chosen divisor will not be greater than $45$ the largest? (Note: The probability is equal to the number of divisors not greater than $45$ divided by the total number of divisors.) | [
"We first note that $45^2 = 2025$. For any number $n$, the number of divisors less than $\\sqrt{n}$ is certainly equal to the number of divisors greater than $\\sqrt{n}$, since $0 < t < \\sqrt{n}$ implies $\\frac{n}{t} > \\sqrt{n}$ and $t|n$ implies $\\frac{n}{t}|n$ (and vice versa).\n\nFor all numbers from $2000$ ... | Austria | AustriaMO2013 | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2025 | |
03cq | An equilateral triangle $T$ with side length $2022$ is coloured in white and partitioned into equilateral triangles with side length $1$ (called *cells*) through lines, parallel to the sides of $T$. Two cells are called *adjacent* if they have at least one vertex in common. Ivan colors some of the cells in black. Witho... | [
"For simplicity we write $A$ instead of Ivan and $B$ instead of Peter. The crucial idea is in the following\n\n*Lemma.* In the graph $G$ all vertices are initially colored white. Let $A$ colors some of the vertices in black and $B$ asks $A$ about the parity of the number of blacks in a set $S$ from vertices, and wi... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 12120 | |
00ju | Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the conditions $f(0) = 0$ and
$$
f(x^k y^k) = xy f(x) f(y) \quad \text{for all } x, y \neq 0.
$$ | [
"Putting $x = y = 1$ in (2) yields $f(1) = f(1)^2$, that is $f(1) \\in \\{0, 1\\}$.\n\n**Case 1:** $f(1) = 0$\n(a) If $k = 0$, (2) becomes for $x = y = t$, $t \\neq 0$: $f(1) = t^2 f(t)^2$, whence $f(t) = 0$, $t \\neq 0$. Let $k \\neq 0$. Then $y = 1$ and $x = t$ in (2) yields $f(t^k) = t f(t) f(1) = 0$. Letting $x... | Austria | AustriaMO2013 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All solutions are:
1) f(x) = 0 for all real x (for any integer k).
2) If k is an even integer, then f(0) = 0 and f(x) = x^{1/(k-1)} for x ≠ 0. | |
084z | Problem:
Quanti sono i numeri di cinque cifre (cioè fra $10000$ e $99999$) che non contengono zeri e sono multipli di $12$? | [
"Solution:\n\nLa risposta è $4374$. Per determinare quanti sono esattamente i numeri da contare, ricordiamo innanzitutto che un numero è divisibile per $12$ se e solo se è divisibile per $4$ e per $3$, e i criteri di divisibilità ci dicono che:\n- Le ultime due cifre del numero devono formare un multiplo di $4$.\n-... | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 4374 | |
0433 | Given quintic polynomial $f(x)$ with its leading coefficient being 1, it satisfies $f(n) = 8n$, $n = 1, 2, \dots, 5$. Then the coefficient of the term of degree 1 of $f(x)$ is ______. | [
"Let $f(x) = g(x) - 8x$, and then $g(x)$ is also a quintic polynomial with its leading coefficient being 1. And there is\n$$\ng(n) = f(n) - 8n = 0, \\quad n = 1, 2, \\dots, 5.\n$$\nHence, $g(x)$ has 5 real roots, namely, $1, 2, \\dots, 5$. Therefore,\n$$\ng(x) = (x-1)(x-2)\\cdots(x-5),\n$$\nand thus,\n$$\nf(x) = (x... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 282 | |
0a3v | Let $\triangle ABC$ be an acute triangle with $|AB| > |AC|$ and let $\omega$ be the circumcircle of $\triangle ABC$ with centre $O$. The altitude from $A$ intersects $BC$ in $D$ and intersects $\omega$ a second time in $P$. Let $H$ be the orthocentre of $\triangle ABC$ and let $K$ be the point on the line segment $BC$ ... | [
"Let $M$ be the circumcentre of $\\triangle PKH$. Note that the reflection of $H$ in $BC$ lies on the circumcircle $\\omega$. (One way to see this is to note that $\\angle BHC = \\angle BHD + \\angle DHC = \\angle BCA + \\angle ABC$.) Therefore this reflection is the point $P$. So the reflection in $BC$ transforms ... | Netherlands | IMO Team Selection Test 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Cons... | null | proof only | null | |
0fjm | Problem:
Ensartamos $2n$ bolas blancas y $2n$ bolas negras formando una cadena abierta. Demuestra que, se haga en el orden en que se haga, siempre es posible cortar un segmento de cadena que contenga exactamente $n$ bolas blancas y $n$ bolas negras. | [
"Solution:\n\nTenemos la cadena con el total de $4n$ bolas, $2n$ blancas y $2n$ negras. Cogemos un grupo de un extremo con $2n$ bolas; este grupo tendrá $x$ bolas negras e $y$ bolas blancas, de forma que la diferencia es $x-y=2k$ para $k \\in \\{-n,-n+1, \\ldots, 1,0,1, \\ldots, n-1, n\\}$.\n\nVamos moviéndonos de ... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
03d3 | For a positive integer $n > 3$ the set $A$ of sequences of 0 and 1 with length $n+1$ is called *good* if any sequence of 0 and 1 of length $n$ can be obtained from a sequence from $A$ by deleting one term. Let $a_n$ be the minimum number of elements in a good set. Prove that: $\frac{2^n + 2n - 7}{n} \le a_n \le 2^{n-1}... | [
"Let $B$ be the set of all sequences of 0 and 1 of length $n-1$, thus $|B| = 2^{n-1}$. It is clear that the set $A$ obtained by adding 01 at the end of any sequence from $B$ is good. Therefore $a_n \\le 2^{n-1}$.\n\nA sequence of length $n$ with all zeroes can be obtained either by the sequence with all zeroes of l... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ezx | Problem:
The projections of a body on two planes are circles. Show that the circles have the same radius. | [] | Soviet Union | ASU | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0aqm | Problem:
Let $y = \left(1 + e^{x}\right)\left(e^{x} - 6\right)^{-1}$. If the values of $x$ run through all real numbers, determine the values of $y$. | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, -1/6) ∪ (1, ∞) | |
0gmg | Graph Airlines (GA) operates flights between some of the cities of the Republic of Graphia. There are GA flights between each city and at least three different cities and it is possible to travel from any city to any other city in the Republic of Graphia using GA flights. GA decides to discontinue some of its flights. ... | [] | Turkey | X. NATIONAL MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Graph Theory"
] | English | proof only | null | |
09d0 | $a + b + c + d - 3 = ab$ ба $a + b + c + d - 3 = cd$ похцлийг хангах бүх ээрэг бүхэл ($a, b, c, d$) дөрөвтүүдийг ол. | [
"Гишүүнчлэн нэмбэл\n$$\nab + cd = 2(a + b + c + d) - 6 \\text{ болно.} \\Rightarrow (a - 2)(b - 2) + (c - 2)(d - 2) = 2 \\text{ болно.} a, b, c, d\\text{-ийн хамгийн багыг нь } a \\text{ гэвэл } -1 \\leq a - 2 \\leq 1 \\text{ болох учир дараах тохиолдлуудыг авч үзье.}\n$$\n$$\n1) a - 2 = 1 \\text{ бол } b - 2 = c -... | Mongolia | ММО-48 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Mongolian | proof and answer | All positive integer solutions are: (1,1,1,1), (3,3,3,3), (2,6,3,4), (2,6,4,3), (6,2,3,4), (6,2,4,3), (3,4,2,6), (4,3,2,6), (3,4,6,2), (4,3,6,2). | |
031g | Problem:
Consider the functions
$$
f(x) = \frac{\cos^2 x}{1 + \cos x + \cos^2 x} \text{ and } g(x) = k \tan x + (1-k) \sin x - x
$$
where $k$ is a real number.
a) Prove that $g'(x) = \frac{(1-\cos x)(k-f(x))}{f(x)}$.
b) Find the image of $f(x)$ if $x \in \left[0 ; \frac{\pi}{2}\right)$.
c) Find all $k$ such that $g... | [
"Solution:\n\na) We have\n$$\n\\begin{aligned}\ng'(x) & = \\frac{(1-\\cos x)\\left[k\\left(1+\\cos x+\\cos^2 x\\right)-\\cos^2 x\\right]}{\\cos^2 x} \\\\\n& = \\frac{(1-\\cos x)\\left(1+\\cos x+\\cos^2 x\\right)}{\\cos^2 x}[k-f(x)] \\\\\n& = \\frac{(1-\\cos x)[k-f(x)]}{f(x)}\n\\end{aligned}\n$$\n\nb) Set $\\cos x =... | Bulgaria | Bulgarian Mathematical Competitions | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Trigonometric functions"
] | null | proof and answer | Range of the first function on the given interval: (0, 1/3]; Parameter values: k ∈ [1/3, ∞). | |
0f30 | Problem:
Let $x_i$ be real numbers between $0$ and $1$. Show that
$$(x_1 + x_2 + \ldots + x_n + 1)^2 \geq 4(x_1^2 + x_2^2 + \ldots + x_n^2).$$ | [] | Soviet Union | ASU | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0fso | Problem:
a. Die positiven reellen Zahlen $x$, $y$ und $z$ haben Summe $1$. Zeige, dass gilt
$$
\sqrt{7x+3} + \sqrt{7y+3} + \sqrt{7z+3} \leq 7
$$
Kann die Zahl $7$ auf der rechten Seite durch eine kleinere Zahl ersetzt werden? | [
"Solution:\n\nNach AM-QM ist die linke Seite höchstens gleich\n$$\n3 \\cdot \\sqrt{\\frac{(7x+3)+(7y+3)+(7z+3)}{3}} = 3 \\cdot \\sqrt{16/3} = 4\\sqrt{3} < 7\n$$"
] | Switzerland | IMO - Selektion | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4*sqrt(3) | |
05r4 | Problem:
Soit $p \geqslant 5$ un nombre premier. Montrer qu'il existe un entier $n$ tel que pour tout $x \in \{n-1, n, n+1\}$, $p^{2} \nmid x^{p-1}-1$ et $p \nmid x$. | [
"Solution:\n\nLemme : entre $0$ et $p^{2}$ il y a exactement $p-1$ éléments dont la puissance $(p-1)$-ième est congrue à $1$ modulo $p^{2}$.\n\nPreuve : On prend $\\omega$ une racine primitive modulo $p^{2}$ (pour rappel $\\omega$ une racine primitive modulo $m$ signifie que les puissances de $\\omega$ corresponden... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
022t | Problem:
Num triângulo $ABC$, o ângulo $A\hat{BC}$ mede $20^\circ$ e o ângulo $A\hat{C}B$ mede $40^\circ$. Seja $E$ um ponto sobre $BC$ tal que $BE = BA$.
a. Mostre que o triângulo $CEA$ é isósceles.
b. Sabendo que o comprimento da bissetriz do ângulo $B\hat{A}C$ é $2$, determine $BC - AB$. | [
"Solution:\n\na.\nTemos $C\\hat{AB} = 180^\\circ - 20^\\circ - 40^\\circ = 120^\\circ$. Como o triângulo $ABE$ é isósceles, segue que\n$$\nA\\hat{E}B = E\\hat{A}B = \\frac{180^\\circ - 20^\\circ}{2} = 80^\\circ\n$$\nAssim, $CA\\hat{E} = 120^\\circ - 80^\\circ = 40^\\circ$ e o triângulo $ACE$ tem dois ângulos de $40... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2 | |
0b0t | Problem:
A geometric sequence with at least three terms and a rational common ratio has first term $32^{16}$ and last term $625^{30}$. If the product of all possible values of the second term of this sequence is $16^{x} \cdot 625^{y}$, where $x$ and $y$ are integers, what are the values of $x$ and $y$ ?
Answer: $x=21... | [
"Solution:\n\nLet $a$ be the common ratio of the geometric sequence and $k$ be the number of terms, then $32^{16} \\cdot a^{k-1}=625^{30}$ or $2^{80} \\cdot a^{k-1}=5^{120}$. Solving for $a$, we have\n$$\na=\\sqrt[k-1]{\\frac{5^{120}}{2^{80}}}=\\left(\\frac{5^{120}}{2^{80}}\\right)^{1 /(k-1)}\n$$\nFrom the equation... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | x=214, y=69 | |
0gc4 | 令 $a, b, c$ 為任意實數使得 $a + b + c = 0$. 試證:
$$
\frac{33a^2 - a}{33a^2 + 1} + \frac{33b^2 - b}{33b^2 + 1} + \frac{33c^2 - c}{33c^2 + 1} \ge 0.
$$ | [
"Notice the original inequality is equivalent to\n$$\n\\sum_{cyc} \\left( \\frac{33a^2 - a}{33a^2 + 1} + t \\right) \\ge 3t \\quad (1)\n$$\nWe want to try to make every term on the left hand side non negative, so we need to choose a $t$ such that the numerator of $\\left(\\frac{33a^2-a}{33a^2+1}+t\\right)$ is alway... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0056 | Se tiene un pentágono de papel, $ABCDE$, tal que
$$
AB = BC = 3 \text{ cm},\ CD = DE = 5 \text{ cm},\ EA = 4 \text{ cm};\ \angle ABC = 100^\circ,\ \angle CDE = 80^\circ.
$$
Hay que dividir el pentágono en cuatro triángulos, mediante tres cortes rectos, de manera que con los cuatro triángulos se arme un rectángulo, sin ... | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Español | proof only | null | |
01zx | Problem:
Find all quadruples $(a, b, p, n)$ of positive integers, such that $p$ is a prime and
$$
a^{3}+b^{3}=p^{n}
$$ | [
"Solution:\nLet $(a, b, p, n)$ be a solution. Note that we can write the given equation as\n$$\n(a+b)\\left(a^{2}-a b+b^{2}\\right)=p^{n}\n$$\nAs $a$ and $b$ are positive integers, we have $a+b \\geq 2$, so $p \\mid a+b$. Furthermore, $a^{2}-a b+b^{2}=(a-b)^{2}+a b$, so either $a=b=1$ or $a^{2}-a b+b^{2} \\geq 2$. ... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | All solutions are the three families: (2^k, 2^k, 2, 3k+1), (2·3^k, 3^k, 3, 3k+2), and (3^k, 2·3^k, 3, 3k+2) for all integers k ≥ 0. | |
0hxn | Problem:
The cost of 3 hamburgers, 5 milk shakes, and 1 order of fries at a certain fast food restaurant is $\$ 23.50$. At the same restaurant, the cost of 5 hamburgers, 9 milk shakes, and 1 order of fries is $\$ 39.50$. What is the cost of 2 hamburgers, 2 milk shakes, and 2 orders of fries at this restaurant? | [
"Solution:\nLet $H = $ hamburger, $M = $ milk shake, and $F = $ order of fries. Then $3H + 5M + F = \\$ 23.50$. Multiplying the equation by 2 yields $6H + 10M + 2F = \\$ 47$. Also, it is given that $5H + 9M + F = \\$ 39.50$. Then subtracting the following equations\n$$\n\\begin{aligned}\n& 6H + 10M + 2F = \\$ 47.00... | United States | HMMT 1998 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | $15 | |
0kxy | Problem:
Let $r_{k}$ denote the remainder when $\binom{127}{k}$ is divided by $8$. Compute $r_{1} + 2 r_{2} + 3 r_{3} + \cdots + 63 r_{63}$.
Proposed by: Rishabh Das | [
"Solution:\nLet $p_{k} = \\frac{128 - k}{k}$, so\n$$\n\\binom{127}{k} = p_{1} p_{2} \\cdots p_{k}\n$$\nNow, for $k \\leq 63$, unless $32 \\mid \\gcd(k, 128 - k) = \\gcd(k, 128)$, $p_{k} \\equiv -1 \\pmod{8}$. We have $p_{32} = \\frac{96}{32} = 3$. Thus, we have the following characterization:\n$$\nr_{k} = \\begin{c... | United States | HMMT November 2023 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 8096 | |
013p | Problem:
Let $f$ be a non-constant polynomial with integer coefficients. Prove that there is an integer $n$ such that $f(n)$ has at least 2004 distinct prime factors. | [
"Solution:\n\nSuppose the contrary. Choose an integer $n_{0}$ so that $f(n_{0})$ has the highest number of prime factors. By translating the polynomial we may assume $n_{0}=0$. Setting $k=f(0)$, we have $f(w k^{2}) \\equiv k \\pmod{k^{2}}$, or $f(w k^{2})=a k^{2}+k=(a k+1) k$. Since $\\gcd(a k+1, k)=1$ and $k$ alon... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
070z | Problem:
Given a pile of 2000 stones, two players take turns in taking stones from the pile. Each player must remove $1, 2, 3, 4$, or $5$ stones from the pile at each turn, but may not take the same number as his opponent took on his last move. The player who takes the last stone wins. Does the first or second player ... | [
"Solution:\n\nThe first player has a winning strategy. He takes $4$ on his first move leaving $7 \\bmod 13$ ($2000 = 153 \\times 13 + 7 + 4$). Now we claim that the first player can always leave: (1) $0 \\bmod 13$, (2) $3 \\bmod 13$ by taking away $3$, (3) $5 \\bmod 13$ by taking away $5$, or (4) $7 \\bmod 13$, and... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | First player | |
0ior | Problem:
g is a twice differentiable function over the positive reals such that
$$
\begin{aligned}
g(x)+2 x^{3} g^{\prime}(x)+x^{4} g^{\prime \prime}(x) & =0 \quad \text{ for all positive reals } x . \\
\lim _{x \rightarrow \infty} x g(x) & =1
\end{aligned}
$$
Find the real number $\alpha>1$ such that $g(\alpha)=1 / 2... | [
"Solution:\n\nAnswer: $\\frac{6}{\\pi}$.\n\nIn the first equation, we can convert the expression $2 x^{3} g^{\\prime}(x)+x^{4} g^{\\prime \\prime}(x)$ into the derivative of a product, and in fact a second derivative, by writing $y=1 / x$. Specifically,\n$$\n\\begin{aligned}\n0=g(x)+2 x^{3} g^{\\prime}(x)+x^{4} g^{... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Equations > ODEs",
"Precalculus > Limits",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 6/pi | |
09dg | $a, b, c$ эерэг тоонууд бол
$$
\sqrt{a^3 - ab + b^2 + \sqrt{b^2 - bc + c^2}} \ge \sqrt{a^2 + ac + c^2}
$$
гэж батал. | [
"$a, b, c$ зэрэг бодит тоонууд тул $a, b, c$-с урттай хэрчмүүдийг байгуулж чадна. Иймд $O$ дээр эхлэлтэй\n$$\n\\begin{aligned}\n|OA| &= a; \\quad |OB| = b; \\quad |OC| = c \\text{ байх хэрчмүүдийг авч үзье. Энд} \\\\\n\\angle AOB &= 60^\\circ; \\quad \\angle BOC = 60^\\circ \\text{ байхаар авбал } \\angle AOC = 120... | Mongolia | ММО-48 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | Mongolian | proof only | null | |
0dg4 | One cuts a grid of size $8 \times 8$ by a straight line. Find the maximal possible number of cells that are cut by the line. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 15 | |
0jhy | Problem:
Let $b(n)$ be the number of digits in the base $-4$ representation of $n$. Evaluate $\sum_{i=1}^{2013} b(i)$. | [
"Solution:\n\nWe have the following:\n\n- $b(n) = 1$ for $n$ between $1$ and $3$.\n- $b(n) = 3$ for $n$ between $4^{2} - 3 \\cdot 4 = 4$ and $3 \\cdot 4^{2} + 3 = 51$. (Since $a \\cdot 4^{2} - b \\cdot 4 + c$ takes on $3 \\cdot 4 \\cdot 4$ distinct values over $1 \\leq a \\leq 3$, $0 \\leq b \\leq 3$, $0 \\leq c \\... | United States | HMMT November 2013 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 12345 | |
0ct5 | Peter chose 10 consecutive positive integers. He wrote down each of the chosen numbers either in red or in blue (both colors are present). Is it possible that the sum of the l.c.m. of the red numbers and the l.c.m. of the blue numbers has a decimal representation ending up with 2016? (O. Dmitriev, R. Zhenodarov)
Петя ... | [
"No.\n\nAmong Peter's numbers, either there is exactly one number divisible by $8$ (then the sum of the l.c.m.'s is not divisible by $8$), or there are exactly two numbers divisible by $8$ (then exactly one of them is divisible by $16$, and the sum is not divisible by $16$). In any case, $16$ does not divide the su... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English; Russian | proof and answer | No | |
05qb | Problem:
Déterminer la valeur maximale de $\sqrt{x}+\sqrt{2 y+2}+\sqrt{3 z+6}$ lorsque $x, y, z$ sont des nombres réels strictement positifs vérifiant $x+y+z=3$. | [
"Solution:\n\nD'après l'inégalité de Cauchy-Schwarz, on a\n$$\n\\sqrt{x} \\sqrt{1}+\\sqrt{y+1} \\sqrt{2}+\\sqrt{z+2} \\sqrt{3} \\leqslant \\sqrt{x+y+1+z+2} \\sqrt{1+2+3}=6\n$$\nL'égalité est atteinte lorsque $x=y=z=1$. Donc la valeur maximale est $6$."
] | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 6 | |
0kqc | Problem:
Consider the set $S$ of all complex numbers $z$ with nonnegative real and imaginary part such that
$$
|z^{2}+2| \leq |z|
$$
Across all $z \in S$, compute the minimum possible value of $\tan \theta$, where $\theta$ is the angle formed between $z$ and the real axis. | [
"Solution:\n\nLet $z = a + b i$. Then,\n$$\nz^{2} + 2 = (a^{2} - b^{2} + 2) + 2 a b \\cdot i.\n$$\nRecall the identity $(a^{2} - b^{2})^{2} + (2 a b)^{2} = (a^{2} + b^{2})^{2}$, so we have\n$$\n|z^{2} + 2|^{2} = (a^{2} + b^{2})^{2} + 4(a^{2} - b^{2}) + 4\n$$\nThus, $z \\in S$ if and only if $(a^{2} + b^{2})^{2} + 4... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | sqrt(7) | |
0f20 | Problem:
$n$ rectangles are drawn on a rectangular sheet of paper. Each rectangle has its sides parallel to the sides of the paper. No pair of rectangles has an interior point in common. If the rectangles were removed show that the rest of the sheet would be in at most $n + 1$ parts. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | proof only | null | |
03ja | Problem:
In the diagram line segments $AB$ and $CD$ are of length $1$ while angles $ABC$ and $CBD$ are $90^{\circ}$ and $30^{\circ}$ respectively. Find $AC$.
 | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | null | |
0kjc | For each integer $n \ge 2$, let $S_n$ be the sum of all products $jk$, where $j$ and $k$ are integers and $1 \le j < k \le n$. What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by 3?
(A) 196 (B) 197 (C) 198 (D) 199 (E) 200 | [
"$$\nU_n = (1 + 2 + 3 + \\dots + (n-1)) \\cdot n = \\frac{(n-1)n^2}{2}.\n$$\nThen $S_n = S_{n-1} + U_n$ for $n \\ge 3$. Note that $U_n$ is divisible by 3 if $n \\equiv 0 \\text{ or } 1 \\pmod 3$; and if $n \\equiv 2 \\pmod 3$, then $(n-1)n^2 \\equiv 1 \\pmod 3$ and is even, so $U_n \\equiv 2 \\pmod 3$. Hence $S_{n+... | United States | Fall 2021 AMC 10 B | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | MCQ | B | |
07xb | Let $a_0, a_1, a_2, \dots$ be an arithmetic sequence of positive integers. A hexagon is called equiangular if all its internal angles have the same measure. Determine with proof for each $i \ge 0$ the number of incongruent equiangular hexagons the side lengths of which are the numbers $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a... | [
"If $d = a_{i+1} - a_i$ is the constant difference of the given arithmetic sequence, we have $a_i = a_0 + d i$ for all $i \\ge 0$. Because we are given $a_i > 0$ for all $i$, it follows that $a_0 > 0$ and $d \\ge 0$.\n\nIf $d = 0$, the sequence is constant and any equiangular hexagon with side lengths $a_i, a_{i+1}... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | If the common difference is zero, the number is 1; otherwise, the number is 2. |
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