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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kwe | Problem:
A standard $n$-sided die has $n$ sides labeled $1$ to $n$. Luis, Luke, and Sean play a game in which they roll a fair standard $4$-sided die, a fair standard $6$-sided die, and a fair standard $8$-sided die, respectively. They lose the game if Luis's roll is less than Luke's roll, and Luke's roll is less than ... | [
"Solution:\nWe perform casework on Luke's roll. If Luke rolls $n$, with $2 \\leq n \\leq 5$, then the probability Luis rolls less than Luke is $\\frac{n-1}{4}$, and the probability Sean rolls more than Luke is $\\frac{8-n}{8}$. If Luke rolls $6$ then Luis will definitely roll less than Luke, and Sean rolls more tha... | United States | HMMT February 2023 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1/4 | |
02x4 | Problem:
a) Encontre a soma dos dígitos de $99 \cdot 33$.
b) Usando $999999=10^{6}-1$, encontre a soma dos dígitos de $999999 \cdot 333333$.
c) Encontre a soma dos dígitos de $999\ldots999 \cdot 333\ldots333$, em que cada dígito nos fatores anteriores aparece 2018 vezes. | [
"Solution:\na) Como $99 \\cdot 33 = 3267$, a soma procurada é $3+2+6+7=18$.\n\nb) Temos que:\n$$\n\\begin{aligned}\n999999 \\cdot 333333 &= (10^{6}-1) \\cdot 333333 \\\\\n&= 333333 \\cdot 10^{6} - 333333 \\\\\n&= 333332666667\n\\end{aligned}\n$$\nNote que os 12 algarismos podem ser agrupados em 6 pares de números c... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 18; b) 54; c) 18162 | |
0gj5 | 定義費氏數: $F_0 = 0, F_1 = 1, F_2 = 1$, 且 $F_{n+2} = F_{n+1} + F_n$ 對所有正整數 $n$ 均成立。
證明存在一個正整數 $N$ 滿足下列性質：對於任意非常數的、長度為 $N$ 的正整數等差數列, 其中必有一項, 當其寫成若干個費氏數之和時, 至少需要 2024 個費氏數 (費氏數可重複)。
Define the Fibonacci numbers $F_n$ as follows: $F_0 = 0, F_1 = 1, F_2 = 1; F_{n+2} = F_{n+1} + F_n$ for all positive integers $n$.
Prove that t... | [
"以下約定：將正整數寫成費氏數之和時，使用的費氏數要最大。例如：18 雖然可以寫成 $8+5+5$，但我們寫成 $13+5$。\n令 $k$ 為正整數, $\\mathcal{F}$ 為費氏數形成的集合。定義集合加法 $A+B := \\{a+b: a \\in A, b \\in B\\}$。令 $N(k)$ 為集合 $\\underbrace{\\mathcal{F} + \\dots + \\mathcal{F}}_{k}$ 中最長的等差數列的長度。\n我們將用數學歸納法證明 $N(k)$ 皆存在且有限，則取 $N = N(2023) + 1$ 即為所求。\n\nBase case: $k=1$ 時, $N(1) \\... | Taiwan | IMO 2J, Independent Study 1 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | Chinese; English | proof only | null | |
02dh | The centers of the faces of a cube form a regular octahedron of volume $V$. Through each vertex of the cube we may take the plane perpendicular to the long diagonal from the vertex. These planes also form a regular octahedron. Show that its volume is $27V$. | [
"Let the cube have side $k$. $A$, $B$ are two adjacent vertices of the small octahedron, and $AX = BX = k/2$ and $\\angle AXB = 90^\\circ$, so $AB = k/\\sqrt{2}$.\n\n\n\nThe large octahedron has the vertices of the cube at the center of its faces.\nThe line joining the centers of $OPS$ and ... | Brazil | III OBM | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | English | proof only | null | |
0fhy | Problem:
Se consideran conjuntos $A$ de cien números naturales distintos, que tengan la propiedad de que si $a, b, c$ son elementos cualesquiera (iguales o distintos) de $A$, existe un triángulo no obtusángulo cuyos lados miden $a, b$ y $c$ unidades.
Se denomina $S(A)$ a la suma de los perímetros considerados en la de... | [
"Solution:\n\nSea $A$ un conjunto que cumpla las condiciones del enunciado. Queremos calcular la suma de los perímetros de los triángulos, y lo haremos por dos procedimientos distintos.\n\nPrimer procedimiento\n\nSea $k \\in A$ y contemos el número de triángulos que tienen lado $k$ y su contribución a la suma de pe... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 149121450 | |
08p8 | Problem:
Consider any four pairwise distinct real numbers and write one of these numbers in each cell of a $5 \times 5$ array so that each number occurs exactly once in every $2 \times 2$ subarray. The sum over all entries of the array is called the total sum of that array. Determine the maximum number of distinct tot... | [
"Solution:\n\nWe will prove that the maximum number of total sums is $60$.\n\nThe proof is based on the following claim.\n\nClaim. Either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.\n\nProof of the Claim. Indeed, let $R$ be a row containing at least three of the... | JBMO | Junior Balkan Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 60 | |
0e4m | Let $O_1$ be the centre of the circle $K_1$. Let $K_2$ be the circle centered at $O_2$ and passing through $O_1$. The circles $K_1$ and $K_2$ meet at $A$ and $B$. The circle $K_1$ intersects the segment $O_1O_2$ at $C$. The line $BC$ intersects the circle $K_2$ at $B$ and $D$. The line $AD$ intersects the circle $K_1$ ... | [
"Write $\\angle ABD = \\alpha$. Since $AO_1BD$ is a cyclic quadrilateral, we have $\\angle AO_1D = \\angle ABD = \\alpha$. The central angle $\\angle AO_1C$ in the circle $K_1$ is twice the size of the inscribed angle $\\angle ABC$, so $\\angle AO_1C = 2\\alpha$.\n\n\n\n\\[\n\\begin{aligned... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f4p | Problem:
What is the largest subset of $\{1, 2, \ldots , 1982\}$ with the property that no element is the product of two other distinct elements. | [] | Soviet Union | 16th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1939; for example, the set {1} together with all integers from 45 to 1982 | |
0idw | Problem:
Let $x$ be a real number such that $x^{3} + 4x = 8$. Determine the value of $x^{7} + 64x^{2}$. | [
"Solution:\nFor any integer $n \\geq 0$, the given implies $x^{n+3} = -4x^{n+1} + 8x^{n}$, so we can rewrite any such power of $x$ in terms of lower powers. Carrying out this process iteratively gives\n$$\n\\begin{aligned}\nx^{7} & = -4x^{5} + 8x^{4} \\\\\n& = 8x^{4} + 16x^{3} - 32x^{2} \\\\\n& = 16x^{3} - 64x^{2} ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 128 | |
0a4x | Problem:
What is the maximum integer $n$ such that $\frac{50!}{2^n}$ is an integer? | [
"Solution:\n$50! = 1 \\times 2 \\times 3 \\times \\dots \\times 50$. Of the numbers up to $50$, we need to find how many of them are divisible by $2^k$ for each $k = 1, 2, 3, 4, 5, 6$. There are $\\left\\lfloor \\frac{50}{2^k} \\right\\rfloor$ numbers which are divisible by $2^k$ and there are $\\left\\lfloor \\fra... | New Zealand | NZMO Round One | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | final answer only | 47 | |
00dd | Four teams $A$, $B$, $C$, $D$ play a football tournament in which every team faces each of the other teams exactly twice (for a total of 12 matches). In each match, if the teams draw against each other, both of them earn 1 point; otherwise, the winner earns 3 points and the loser earns no points.
Given that at the end... | [
"Let $d$ be the number of draws in the tournament. Then the total number of points awarded equals $2d + 3(12 - d) = 36 - d$, because every draw awards 2 points in total and every non-draw awards 3 points in total. Since teams $A$, $B$, $C$ combined have 24 points, the score of team $D$ equals $12 - d$.\n\nIf $D$ ha... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 3, 4, 6, 7, 8, 9 | |
03bf | Let $A$ be a set with $n \ge 2$ elements and let $f_1, \dots, f_k : A \to \mathbb{R}$ be functions. Prove that there exist functions $g_1, \dots, g_{n-1} : A \to \mathbb{R}$ such that
$$
\max_{1 \le i \le k} |f_i(x) - f_i(y)| = \max_{1 \le i \le n-1} |g_i(x) - g_i(y)|
$$
for any $x, y \in A$. | [
"Let $A = \\{a_1, a_2, \\dots, a_n\\}$ and set\n$$\nd(x, y) = \\max_{1 \\le i \\le k} |f_i(x) - f_i(y)|, \\quad g_i(x) = d(x, a_i) - d(a_i, a_i), \\quad 1 \\le i \\le n-1\n$$\nand $d'(x, y) = \\max_{1 \\le i \\le n-1} |g_i(x) - g_i(y)|$. Since\n$$\n|g_i(x) - g_i(y)| = |d(x, a_i) - d(y, a_i)| \\le d(x, y) \\quad (\\... | Bulgaria | Bulgaria | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Other"
] | English | proof only | null | |
0aaf | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(f(x) f(1-x))=f(x) \quad \text{ and } \quad f(f(x))=1-f(x)
$$
for all real $x$. | [
"Solution:\nNotice that $f(f(f(x)))=1-f(f(x))=f(x)$. This is equation 3.\nBy substituting $f(x)$ for $x$ in the first equation we get:\n$$\nf(f(f(x)) f(1-f(x)))=f(f(x))\n$$\nAgain we substitute $f(x)$ for $x$ above:\n$$\nf(f(f(x)) f(f(f(x))))=f(f(f(x)))\n$$\nEquation 3 applied on both sides gives us:\n$$\nf(x)=f(f(... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/2 | |
083f | Problem:
Per quanti valori interi di $a$ l'equazione
$$
x^{3} + (a-4)x^{2} + (a+4)x + 9 = 0
$$
ha esattamente due soluzioni intere distinte?
(A) 1
(B) 2
(C) 3
(D) 4
(E) più di 4. | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | MCQ | C | |
09n2 | Let $a$, $b$, $c$ be real numbers with sum $a + b + c = 3$. Prove that
$$
\frac{a^2 + b^2 + c^2}{2} + \frac{1}{a^2 + 1} + \frac{1}{b^2 + 1} + \frac{1}{c^2 + 1} \geq 3.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
06b4 | (i) If $x$, $y$ are positive real numbers, prove that:
$$
\frac{4}{x+y} \le \frac{1}{x} + \frac{1}{y}.
$$
(ii) If $\alpha$, $\beta$, $\gamma$, $\delta$ are positive real numbers, prove that
$$
\frac{2}{(\alpha + \beta)(\gamma + \delta) + (\beta + \gamma)(\alpha + \delta)} \le \frac{1}{(\alpha + \gamma)(\beta + \delta)... | [
"(i) Since $x$, $y$ are positive real numbers, we have:\n$$\n\\frac{4}{x+y} \\le \\frac{1}{x} + \\frac{1}{y} \\Leftrightarrow \\frac{4}{x+y} \\le \\frac{x+y}{xy} \\Leftrightarrow (x+y)^2 \\ge 4xy \\Leftrightarrow (x-y)^2 \\ge 0.\n$$\n\n(ii)\nLet\n$$\nx = (\\alpha + \\gamma)(\\beta + \\delta) + 4\\alpha\\gamma, \\qu... | Greece | Selection Examination | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Equality holds if and only if alpha times gamma equals beta times delta. | |
06ae | Determine all integers $n \ge 1$, for which there exist at least one pair of positive integers $(a, b)$ such that:
(i) The integer $a^2 + b + 3$ is not divisible by the cube of a prime number.
(ii) $\frac{ab+3b+8}{a^2+b+3} = n$ | [] | Greece | Selection examinations | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 2 | |
0h1w | Find all prime numbers $p$, $q$, $r$, such that:
$$
p(p+1) + q(q+1) = r(r+1).
$$ | [
"WLOG, $p \\le q$, clearly $r > p$. We can write:\n$$\np(p+1) = (r-q)(r+q+1). \\quad (1)\n$$\nSince $p$ is prime, then either $r-q$ or $r+q+1$ is divisible by $p$. If $p$ divides $r-q$, then $r-q \\ge p$,\n$$\np(p+1) \\le (r-q)(r-q+1) < (r-q)(r+q+1),\n$$\nwhich contradicts to (1).\n\nLet suppose that $r+q+1$ is div... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p = 2, q = 2, r = 3 | |
0axw | Problem:
Given $O(0,0)$ and $Q(1,2)$, consider the set of points $P$ for which $O P: P Q=5: 4$. It is known that this set is a circle. Find its center. | [
"Solution:\n\nSuppose we have $P(x, y)$. This implies in particular that $4 O P=5 P Q$, or $16 O P^{2}=25 P Q^{2}$. Hence, we have\n$$\n16\\left(x^{2}+y^{2}\\right)=25(x-1)^{2}+25(y-2)^{2}\n$$\nThis simplifies to\n$$\n9 x^{2}-50 x+25+9 y^{2}-100 y+100=0\n$$\nand, completing squares,\n$$\n\\left(3 x-\\frac{25}{3}\\r... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | (25/9, 50/9) | |
00sd | Let $a_0$ be an arbitrary positive integer. Let $\{a_n\}$ be an infinite sequence of positive integers such that for every positive integer $n$ the term $a_n$ is the smallest positive integer such that $a_0 + a_1 + \dots + a_n$ is divisible by $n$. Prove that there is a positive integer $N$ such that $a_{n+1} = a_n$ fo... | [
"Define $b_n = \\frac{a_0+a_1+\\dots+a_n}{n}$ for every positive integer $n$. According to condition, $b_n$ is a positive integer for every positive integer $n$.\nSince $a_{n+1}$ is the smallest positive integer such that $\\frac{a_0+a_1+\\dots+a_n}{n+1}$ is a positive integer and\n$$\n\\frac{a_0 + a_1 + \\dots + a... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0ipi | Problem:
Let $f(r) = \sum_{j=2}^{2008} \frac{1}{j^{r}} = \frac{1}{2^{r}} + \frac{1}{3^{r}} + \cdots + \frac{1}{2008^{r}}$. Find $\sum_{k=2}^{\infty} f(k)$. | [
"Solution:\nAnswer: $\\frac{2007}{2008}$\nWe change the order of summation:\n$$\n\\sum_{k=2}^{\\infty} \\sum_{j=2}^{2008} \\frac{1}{j^{k}} = \\sum_{j=2}^{2008} \\sum_{k=2}^{\\infty} \\frac{1}{j^{k}} = \\sum_{j=2}^{2008} \\frac{1}{j^{2}\\left(1-\\frac{1}{j}\\right)} = \\sum_{j=2}^{2008} \\frac{1}{j(j-1)} = \\sum_{j=... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2007/2008 | |
09xn | Jesse and Tjeerd are playing a game. Jesse has $n \ge 2$ stones. There are two boxes: in the black box there is space for half of the stones (rounded down) and in the white box there is space for half of the stones (rounded up). Jesse and Tjeerd alternate turns, with Jesse as first player. In his turn, Jesse takes one ... | [
"We will show that the capacity of the two boxes does not matter, as long as the total capacity is $n$ (and at least 1 for each box). Jesse can always win this game, and can do that by first playing the power $2^0 = 1$ of two, and then in each following turn the next power of two that is smaller or greater. That me... | Netherlands | BxMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Jesse always wins for every n ≥ 2. | |
011n | Problem:
Fourteen friends met at a party. One of them, Fredek, wanted to go to bed early. He said goodbye to 10 of his friends, forgot about the remaining 3, and went to bed. After a while he returned to the party, said goodbye to 10 of his friends (not necessarily the same as before), and went to bed. Later Fredek ca... | [
"Solution:\n\nFredek returned at least $32$ times.\n\nAssume Fredek returned $k$ times, i.e. he was saying goodbye $k+1$ times to his friends. There exists a friend of Fredek, call him $X_{13}$, about whom Fredek forgot $k$ times in a row, starting from the very first time—otherwise Fredek would have come back less... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 32 | |
0b90 | Given positive integers $m \ge 2$, $n \ge 2$, consider the set
$$
S(m, n) = \left\{ \mathbf{x} = (x_1, x_2, \dots, x_m) ; \mathbf{x} \in \mathbb{Z}_+^m, \sum_{k=1}^m x_k = n \right\}.
$$
Determine $N(m, n)$, the largest cardinality $|X|$ of a subset $X \subseteq S(m, n)$ with
$$
|X| = |\{x_k ; \mathbf{x} \in X\}|, \tex... | [
"Consider such a set $X \\subseteq S(m, n)$. Clearly, $S(m, n)$ and thus a fortiori $X$ are finite sets. Then\n$$\n\\sum_{\\mathbf{x} \\in X} x_k \\ge \\sum_{i=1}^{|X|} (i-1) = |X|(|X| - 1)/2, \\text{ for all } k = 1, 2, \\dots, m,\n$$\nand so\n$$\nn|X| = \\sum_{\\mathbf{x} \\in X} \\sum_{k=1}^{m} x_k = \\sum_{k=1}... | Romania | Local Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | floor(2n/m) + 1 | |
08e0 | Problem:
Alberto e Barbara scrivono dei numeri alla lavagna. Parte Alberto e scrive il numero reale $x$. Poi Barbara scrive il numero $1$. I due poi si alternano, e ad ogni turno scrivono un numero. Alberto nel suo turno moltiplica l'ultimo numero scritto per $x^{2}$ e scrive il risultato. Barbara nel suo turno somma ... | [
"Solution:\n\nLa risposta è (E). Possiamo procedere per esclusione. La (A) è falsa perché se $x>0$ allora ad ogni turno Alberto e Barbara moltiplicano e sommano numeri positivi a quelli già presenti sulla lavagna. Dato che i primi due numeri sono $x$ e $1$, allora non c'è nessun numero negativo sulla lavagna. (B) e... | Italy | Olimpiadi della Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | E | |
0erg | A palindrome is a whole number that reads the same forwards and backwards (e.g. $47274$). How many $3$-digit palindromes are there?
(A) $95$ (B) $90$ (C) $85$ (D) $80$ (E) $70$ | [
"There are $9$ ways to select the nonzero hundreds' digit, then $1$ way to select the units' digit (same as the hundreds' digit), and then $10$ ways to select the tens' digit. Hence there are $9 \\times 1 \\times 10 = 90$ three-digit palindromes."
] | South Africa | South African Mathematics Olympiad First Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | MCQ | B | |
0738 | Problem:
Let $\sigma=\left(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\right)$ be a permutation of $(1,2,3, \ldots, n)$. A pair $\left(a_{i}, a_{j}\right)$ is said to correspond to an inversion of $\sigma$, if $i<j$ but $a_{i}>a_{j}$. (Example: In the permutation $(2,4,5,3,1)$, there are 6 inversions corresponding to the pairs ... | [
"Solution:\nIn a permutation of $(1,2,3, \\ldots, n)$, two inversions can occur in only one of the following two ways:\n(A) Two disjoint consecutive pairs are interchanged:\n$$\n\\begin{aligned}\n& (1,2,3, \\ldots, j-1, j, j+1, \\ldots, k-1, k, k+1, k+2, \\ldots, n) \\\\\n& \\quad \\longrightarrow (1,2, \\ldots, j-... | India | INMO | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | ((n+1)(n-2))/2 | |
0f58 | Problem:
A point is chosen on each of the three sides of a triangle and joined to the opposite vertex. The resulting lines divide the triangle into four triangles and three quadrilaterals. The four triangles all have area $A$. Show that the three quadrilaterals have equal area. What is it (in terms of $A$)? | [] | Soviet Union | 17th ASU | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Each quadrilateral has area 2A. | |
01v0 | Given the parallelogram $ABCD$. The circle $S_1$ passes through the vertex $C$ and touches the sides $BA$ and $AD$ at points $P_1$ and $Q_1$ respectively. The circle $S_2$ passes through the vertex $B$ and touches the sides $DC$ and $AD$ at points $P_2$ and $Q_2$ respectively. Let $d_1, d_2$ be the distances from $C$ a... | [
"**Answer : 1.**\nWe prove that $d_1 = d_2$. Let $\\angle A = \\angle C = \\alpha$, $\\angle B = \\angle D = -\\alpha$. Choose points $F$ and $T$ on $P_1Q_1$ so that $\\angle P_1FC = \\angle Q_1TC = \\angle D$. Then $CF = CT$. Further, $\\triangle P_1BC \\sim \\triangle Q_1TC$, thus we have $\\frac{BC}{TC} = \\frac... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof and answer | 1 | |
05ib | Problem:
Peut-on numéroter les arêtes d'un cube de 1 à 12 en sorte que la somme des nombres sur les arêtes entrant dans un sommet du cube soit la même pour tous les sommets? | [
"Solution:\n\nNous allons prouver par l'absurde que la réponse à la question est négative.\nSupposons qu'une telle numérotation existe et notons $k$ la somme dans chaque sommet. Sur chaque arête, écrivons son numéro deux fois : une fois à un bout de l'arête et l'autre fois à l'autre bout. Maintenant, calculons la s... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No | |
0afw | **Дадени се пет кружници. Ако избереме било кои четири кружници од дадените пет, тие имаат заедничка точка. Докажи дека постои точка која е заедничка за сите пет кружници.** | [
"Нека $k_1, k_2, k_3, k_4$ и $k_5$ се кружинците. Нека $k_1, k_2, k_3$ и $k_4$ поминуваат низ $A$, $k_1, k_2, k_4$ и $k_5$ поминуваат низ $B$ и $k_1, k_2, k_3$ и $k_5$ поминуваат низ $C$. Но тогаш $k_1$ и $k_2$ поминуваат низ $A$, $B$ и $C$ и бидејќи сите се различни кружници мора две од точките $A, B$ и $C$ да се ... | North Macedonia | Републички натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Circles"
] | Macedonian, English | proof only | null | |
0fex | Problem:
Los números naturales $22$, $23$, y $24$ tienen la siguiente propiedad: los exponentes de los factores primos de su descomposición son todos impares:
$$
22=2^{1} \cdot 11^{1} ; \quad 23=23^{1} ; \quad 24=2^{3} \cdot 3^{1}
$$
¿Cuál es el mayor número de naturales consecutivos que pueden tener esa propiedad?. R... | [
"Solution:\n\nVamos a demostrar que es imposible encontrar $8$ números consecutivos con esta propiedad. Supongamos, para razonar por reducción al absurdo, que tales $8$ números consecutivos existen. Uno de ellos, al que llamaremos $n$, es divisible por $8$.\n\nEntre los $8$ números deberá estar, o bien $n+4$, o bie... | Spain | TANDA II | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 7 | |
02zu | Problem:
Na figura a seguir, $\angle C A B = 2 \cdot \angle C B A$, $A D$ é uma altura e $M$ é o ponto médio de $A B$. Se $A C = 2~\mathrm{cm}$, encontre o comprimento do segmento $D M$.
 | [
"Solution:\n\nSeja $K$ o ponto médio de $A C$. Como o círculo de centro $K$ e diâmetro $A C$ passa por $D$, segue que $C K = A K = D K = 1~\\mathrm{cm}$. Dessa última igualdade, decorre que o triângulo $A D K$ é isósceles e assim $\\angle K D A = \\angle K A D$. O segmento $K M$ é base média do triângulo $A B C$, i... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1 cm | |
03vs | Find all triples $(p, q, n)$ such that
$$
q^{n+2} \equiv 3^{n+2} \pmod{p^n}, \quad p^{n+2} \equiv 3^{n+2} \pmod{q^n}
$$
where $p$, $q$ are positive odd primes and $n > 1$ is an integer. | [
"It is easy to check that $(3, 3, n)$ ($n = 2, 3, \\dots$) satisfy both equations. Now let $(p, q, n)$ be another triple satisfying the condition. Then we must have $p \\ne q$, $p \\ne 3$, $q \\ne 3$. We may assume that $q > p \\ge 5$.\n\nIf $n = 2$, then $q^2 \\mid p^4 - 3^4$, or $q^2 \\mid (p^2 - 3^2)(p^2 + 3^2)$... | China | China Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order... | English | proof and answer | p = 3, q = 3, and n ≥ 2 | |
0j6c | Problem:
Let $ABC$ be a triangle with $AB = 23$, $BC = 24$, and $CA = 27$. Let $D$ be the point on segment $AC$ such that the incircles of triangles $BAD$ and $BCD$ are tangent. Determine the ratio $CD / DA$. | [
"Solution:\nAnswer: $\\frac{14}{13}$\nLet $X, Z, E$ be the points of tangency of the incircle of $ABD$ to $AB$, $BD$, $DA$ respectively. Let $Y, Z, F$ be the points of tangency of the incircle of $CBD$ to $CB$, $BD$, $DC$ respectively. We note that\n$$\nCB + BD + DC = CY + YB + BZ + ZD + DF + FC = 2(CY) + 2(BY) + 2... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 14/13 | |
02g5 | There are $n$ football teams in *Tumbólia*.
A championship is to be organized in which each team plays against every other exactly once. Every match takes place on Sundays and one team must not play more than once in the same day.
Find the least integer $m$ for which it's possible to set up a championship lasting $m$ S... | [
"First case: $n$ is even.\nLet $T_1, T_2, \\dots, T_n$ be the teams. Each team shall play $n-1$ times on at least $n-1$ Sundays. Let us show that $n-1$ Sundays are sufficient to set up the championship.\n\nTeams $T_i$ and $T_j$ play on the $d_{ij}$-th Sunday. Define $d_{ij}$ by the following rules:\n$$\nd_{ij} = \\... | Brazil | XXI OBM | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n-1 if n is even; n if n is odd | |
0d9l | Let $n$ be an odd positive integer with $n>1$ and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers such that $\operatorname{gcd}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=1$. Let $d=\operatorname{gcd}\left(a_{1}^{n}+a_{1} a_{2} \cdots a_{n}, a_{2}^{n}+a_{1} \cdot a_{2} \cdots a_{n}, \ldots, a_{n}^{n}+a_{1} \cdot a_... | [
"Let $p$ be a prime with $p \\mid d$. We know that $p \\mid a_{i}^{n}+a_{1} \\cdot a_{2} \\cdots a_{n}$ for all $i$. If $\\exists i$ such that $p \\mid a_{i}$, then $p \\mid a_{1} \\cdot a_{2} \\cdots a_{n}$. Since $p \\mid a_{j}^{n}+a_{1} \\cdot a_{2} \\cdots a_{n}$ for all $j$, we find $p \\mid a_{j}$ for all $j$... | Saudi Arabia | Team selection tests for GMO 2018 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | d = 1 or d = 2 | |
0byy | In the Cartesian plane, let $\gamma_1$ and $\gamma_2$ be the graphs of the quadratic functions $f_1(x) = p_1x^2+q_1x+r_1$ and $f_2(x) = p_2x^2+q_2x+r_2$, where $p_1 > 0 > p_2$. The graphs $\gamma_1$ and $\gamma_2$ cross at distinct points $A$ and $B$. The four tangents to $\gamma_1$ and $\gamma_2$ at $A$ and $B$ form a... | [
"Let the tangents to $\\gamma_i$ at $A$ and $B$ meet at $C_i$. Since $f_1(x)$ is convex and $f_2(x)$ is concave, the convex quadrangle formed by the four tangents is exactly $AC_1BC_2$.\n\n\n\n**Lemma.** If $CA$ and $CB$ are the tangents drawn from a point $C$ to the graph $\\gamma$ of a qu... | Romania | THE Tenth ROMANIAN MASTER OF MATHEMATICS | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic E... | English | proof only | null | |
03ru | Determine all positive real numbers $a$ such that there exists a positive integer $n$ and sets $A_1, A_2, \dots, A_n$ satisfying the following conditions:
(1) every set $A_i$ has infinitely many elements;
(2) every pair of distinct sets $A_i$ and $A_j$ do not share any common element;
(3) the union of sets $A_1, A_2, \... | [
"The answer of the problem is the set of all positive real numbers less than $2$. We consider two cases.\n\n**Case I** We assume that $0 < a < 2$. Then there is a positive $n$ such that $2^{n-1} > a^n$. We define $A_n = \\{m \\mid m \\text{ is a multiple of } 2^{n-1}\\}$ and\n$$\nA_i = \\{2^{i-1}m \\mid m \\text{ i... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 0 < a < 2 | |
0kye | Problem:
In an empty $100 \times 100$ grid, 300 cells are colored blue, 3 in each row and each column. Compute the largest positive integer $k$ such that you can always recolor $k$ of these blue cells red so that no contiguous $2 \times 2$ square has four red cells. | [
"Solution:\n\nWe first prove the lower bound. We can recolor all three blue cells in odd rows, and the first and third blue cells in even rows.\nA $2 \\times 2$ square must cover two adjacent cells in an even row. Therefore, it's impossible for there to exist a red $2 \\times 2$ square in our recoloring, because co... | United States | HMIC 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 250 | |
0ch0 | Let $ABC$ be a triangle with $AB < AC$ and $\omega$ be its circumcircle. The tangent at $A$ to the circle $\omega$ intersects line $BC$ at $D$, and the line through $B$ parallel to $AD$ meets again the circle $\omega$ at $E$. Line $DE$ intersects $AB$ at $F$ and the circle $\omega$ for the second time at $G$. On the li... | [
"$$\n\\angle TAB = \\angle BCE = 180^\\{\\circ\\} - \\angle BGE = 180^\\{\\circ\\} - \\angle BGF = \\angle BNF = \\angle BNT,\n$$\nso ANBT is also cyclic.\n\nUsing the fact that $AD \\parallel BE$, it follows that $\\angle ATS = \\angle ABN = \\angle BAD = \\angle FAS$, so line AB is the tangent at A to the circumc... | Romania | 74th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fu3 | Problem:
Bestimme alle Mengen $M$ natürlicher Zahlen, sodass für je zwei (nicht notwendig verschiedene) Elemente $a, b$ aus $M$ auch
$$
\frac{a+b}{\operatorname{ggT}(a, b)}
$$
in $M$ liegt. | [
"Solution:\n\nWir setzen zur Abkürzung $f(a, b)=(a+b) / \\operatorname{ggT}(a, b)$. Die Lösung verläuft in mehreren Schritten.\n\n(i) Sei $a \\in M$ ein beliebiges Element. Dann ist $f(a, a)=2 \\in M$. Beachte, dass $M=\\{2\\}$ eine Lösung ist.\n\n(ii) Nehme an, $M$ enthält noch eine weitere gerade Zahl. Sei $2 m$ ... | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | {2}, ℕ \ {1}, ℕ | |
00lc | Consider arrangements of the numbers $1$ through $64$ on the squares of an $8 \times 8$ chess board, where each square contains exactly one number and each number appears exactly once.
A number in such an arrangement is called super-plus-good, if it is the largest number in its row and at the same time the smallest num... | [
"a. This is wrong. For example, one might place the numbers from $1$ to $8$ along the main diagonal and the numbers from $57$ to $64$ along the secondary diagonal:\n\n| 1 | 9 | 10 | 11 | 12 | 13 | 14 | 57 |\n|----|----|----|----|----|----|----|----|\n| 15 | 2 | 16 | 17 | 18 | 19 | 58 | 20 |\n| 21 | 22 | 3 | 23 ... | Austria | National Competition | [
"Discrete Mathematics > Other",
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
00uq | Let $ABC$ be an acute triangle ($AB < BC < AC$) with circumcircle $\Gamma$. Assume there exists $X \in AC$ satisfying $AB = BX$ and $AX = BC$. Points $D, E \in \Gamma$ are taken such that $\angle ADB < 90^\circ$, $DA = DB$ and $BC = CE$. Let $P$ be the intersection point of $AE$ with the tangent line to $\Gamma$ at $B$... | [
"First, it is easy to check that $B$ lies between $A$ and $Q$ and $A$ lies between $P$ and $E$ using $AB < BC < AC$. Let $T$ and $S$ be the intersection points of the lines $BP$ and $AP$ with $QC$, respectively. Since $CE = CB$, we have $\\angle EAC = \\angle EBC = \\angle BEC = \\angle A$, which gives $\\angle BAS... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04oy | Find all complex numbers $a$ such that all coefficients of
$$
P(x) = (x - a)(x - a^2)(x - a^3)
$$
are real numbers.
(Matko Ljulj) | [
"Obviously, all real numbers satisfy the given condition.\nLet us, from now on, assume that $a$ is not a real number.\nSince $P$ is a polynomial of degree 3, it must have at least one real root. We also know that if $z$ is a root of $P$, then $\\bar{z}$ is a root of $P$ as well. Hence $\\bar{a} = a^2$ or $\\bar{a} ... | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All real numbers together with i, -i, (-1 + i√3)/2, (-1 - i√3)/2 | |
04ih | Determine all quadruples $(a, b, c, d)$ of positive integers such that
$$a^3 = b^2, \quad c^5 = d^4 \quad \text{and} \quad a - c = 9.$$ | [] | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (25, 125, 16, 32) | |
05wn | Problem:
Soit $\left(u_{n}\right)_{n \in \mathbb{N}}$ une suite périodique de période 2022 et de période 5. Montrer que $\left(u_{n}\right)$ est constante.
Une suite $\left(u_{n}\right)$ est dite périodique de période $t$ si, pour tout entier naturel $n$, $u_{n+t}=u_{n}$. | [
"Solution:\n\nCommençons par remarquer que si une suite $\\left(u_{n}\\right)$ est $t$-périodique, alors on a pour tout $n$, $u_{n}=u_{n+t}=u_{n+2t}$ par exemple. Ainsi, on peut montrer par récurrence que pour tous les entiers naturels $k$ et $n$, $u_{n+kt}=u_{n}$ : la suite est $kt$-périodique pour tout $k$.\n\nSo... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0hhs | Let's call a pair of natural numbers $\overline{a_1 a_2 \dots a_k}$ and $\overline{b_1 b_2 \dots b_k}$ $k$-similar if all the digits $a_1, a_2, \dots, a_k, b_1, b_2, \dots, b_k$ are pairwise distinct and there exist distinct natural numbers $m, n$ such that the following equation holds:
$$
a_1^m + a_2^m + \dots + a_k^m... | [
"Let us consider the numbers $1234$ and $6789$. For them, we see that the following equation holds:\n$$\n6^1 + 7^1 + 8^1 + 9^1 = 30 = 1^2 + 2^2 + 3^2 + 4^2.\n$$\nThus, this pair of numbers is $4$-similar.\n\nObviously, for $k > 5$ there are no $k$-similar numbers, because there are only $10$ different digits. Let u... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 4 | |
0bco | Problem:
Numerele întregi pozitive $p$ şi $q$ sunt prime şi satisfac
$$
\frac{p}{p+1} + \frac{q+1}{q} = \frac{2n}{n+2}
$$
pentru un anume număr întreg pozitiv $n$. Determinaţi toate valorile posible ale diferenţei $q-p$. | [] | Romania | Olimpiada europeana de matematica a fetelor | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [2, 3, 5] | |
07ii | Suppose $2n$ subsets of the set $\{0, 1, \dots, 2n-1\}$ are chosen such that the sum of the cardinals (i.e., total number of elements) of these subsets equals $\binom{2n+1}{2}$. Prove that we can choose exactly one element from each subset such that the sum of the chosen numbers equals $\binom{2n}{2}$. | [
"We prove the statement using the following lemmas.\n\n**Lemma 1.** If $|A_i| + |A_j| = 2n + 1 - r$ for a non-negative $r$, then there exist $a_i \\in A_i, a_j \\in A_j$ such that $2n - 1 - r \\le a_i + a_j \\le 2n - 1 + r$.\n*Proof.* Assume the contrary. Let $A_i = \\{x_1 \\le x_2 \\le \\cdots \\le x_{|A_i|}\\}$ a... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ik4 | Problem:
Find the number of solutions in positive integers $\left(k ; a_{1}, a_{2}, \ldots, a_{k} ; b_{1}, b_{2}, \ldots, b_{k}\right)$ to the equation
$$
a_{1}\left(b_{1}\right)+a_{2}\left(b_{1}+b_{2}\right)+\cdots+a_{k}\left(b_{1}+b_{2}+\cdots+b_{k}\right)=7 .
$$ | [
"Solution:\nLet $k, a_{1}, \\ldots, a_{k}, b_{1}, \\ldots, b_{k}$ be a solution. Then $b_{1}, b_{1}+b_{2}, \\ldots, b_{1}+\\cdots+b_{k}$ is just some increasing sequence of positive integers. Considering the $a_{i}$ as multiplicities, the $a_{i}$'s and $b_{i}$'s uniquely determine a partition of $7$. Likewise, we c... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 15 | |
0hsw | Problem:
Suppose you have only an unmarked straightedge (no compass), and you are given a line segment $A B$ with midpoint $O$ and a point $P$ not on line $A B$.
a. Construct a line through $P$ parallel to $A B$.
b. If you are also given the circle with center $O$ and radius $O A$ and $P$ does not lie on the circle,... | [
"Solution:\n\na.\nExtend line $A P$ to some point $Q$ on the opposite side of $P$ from $A$. Let $R$ be the intersection of lines $Q O$ and $P B$, and let $S$ be the intersection of lines $A R$ and $Q B$, as shown below.\n\n\n\nBy Ceva's theorem,\n$$\n\\frac{Q P}{P A} \\cdot \\frac{A O}{O B}... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometr... | null | proof only | null | |
0aea | Реши ја равенката
$$
4^{\log_{10} x} - 32 + x^{\log_{10} 4} = 0.
$$ | [
"Јасно е дека равенката е определена за $x > 0$. За $x > 0$, $\\log_{10} x$ можеме да го запишеме во облик\n$$\n\\log_{10} x = \\frac{\\log_4 x}{\\log_4 10} = (\\log_4 x)(\\log_{10} 4),\n$$\nод каде добиваме\n$$\n4^{\\log_{10} x} = 4^{(\\log_4 x)(\\log_{10} 4)} = (4^{\\log_4 x})^{\\log_{10} 4} = x^{\\log_{10} 4}.\n... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | Macedonian, English | proof and answer | 100 | |
01zd | Positive integers $a$ and $b$ satisfy the equality $a + \tau(a) = b^2 + 2$, where by $\tau(n)$ we denote the number of all divisors of a positive integer $n$, including $1$ and $n$.
Prove that the sum $a + b$ is even. | [
"$$\nc^2 + \\tau(c^2) - 2 = b^2 \\ge (c+1)^2 = c^2 + 2c + 1,\n$$\nwhich means that $\\tau(c^2) > 2c$. All divisors of $c^2$, except $c$, split into pairs of the form $(d, c^2/d)$, where $d < c$ and $c^2/d > c$, so $\\tau(c^2) \\le 1 + 2(c-1) = 2c - 1$, which contradicts the previously obtained inequality $\\tau(c^2... | Belarus | Belarus2022 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
03re | Let $\theta_i \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $i = 1, 2, 3, 4$. Prove that there exists $x \in \mathbb{R}$ such that the following two inequalities
$$
\cos^2\theta_1 \cos^2\theta_2 - (\sin\theta_1 \sin\theta_2 - x)^2 \ge 0, \quad ①
$$
$$
\cos^2\theta_3 \cos^2\theta_4 - (\sin\theta_3 \sin\theta_4 - x)^2 \ge 0, \qu... | [
"**Proof** Clearly, ① and ② are equivalent to\n$$\n\\sin \\theta_1 \\sin \\theta_2 - \\cos \\theta_1 \\cos \\theta_2 \\le x \\le \\sin \\theta_1 \\sin \\theta_2 + \\cos \\theta_1 \\cos \\theta_2, \\quad ④\n$$\n$$\n\\sin \\theta_3 \\sin \\theta_4 - \\cos \\theta_3 \\cos \\theta_4 \\le x \\le \\sin \\theta_3 \\sin \\... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0gke | A convex polygon of $n$ sides is divided into triangles by diagonals which do not intersect inside the polygon. The triangles are painted black and white so that any two triangles with a common side are painted in different colors. For each $n$, determine the maximal difference between the number of black and the numbe... | [
"Construct a graph by representing each triangle by a vertex of a graph. Two vertices are joined by an edge if their corresponding triangles share a common side. This graph is connected and contains no cycle and hence is a tree. Since it has $n-2$ vertices, it has $n-3$ edges. Moreover, each vertex has degree at mo... | Thailand | Thai Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | If n is congruent to 0 or 2 modulo 3, the maximum difference is floor(n/3). If n is congruent to 1 modulo 3, the maximum difference is floor(n/3) − 1. | |
03ll | Problem:
Let $S$ be a set of $n \geq 3$ points in the interior of a circle.
a) Show that there are three distinct points $a, b, c \in S$ and three distinct points $A, B, C$ on the circle such that $a$ is (strictly) closer to $A$ than any other point in $S$, $b$ is closer to $B$ than any other point in $S$ and $c$ is ... | [
"Solution:\n\na) Let $H$ be the smallest convex set of points in the plane which contains $S$. Take 3 points $a, b, c \\in S$ which lie on the boundary of $H$. (There must always be at least 3 (but not necessarily 4) such points.)\n\nSince $a$ lies on the boundary of the convex region $H$, we can construct a chord ... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0ggl | 平面上有兩個三角形 $ABC$ 與 $A'B'C'$。已知三角形 $ABC$ 的各邊長不小於 $a$, 且三角形 $A'B'C'$ 的各邊長不小於 $a'$。證明總是能夠從這兩個三角形中各選出一頂點, 使得它們的距離不小於 $\sqrt{\frac{a^2+a'^2}{3}}$。 | [
"令點 $G, G'$ 分別為 $\\triangle ABC$ 與 $\\triangle A'B'C'$ 的重心。由重心公式知，對於平面上的任一點 $P$，均有\n$$\nPA^2 + PB^2 + PC^2 = 3PG^2 + \\frac{AB^2 + BC^2 + CA^2}{3}.\n$$\n分別令 $P$ 為 $A', B', C'$ 三點，所得的三條式子加總，再使用上述公式在三角形 $\\triangle A'B'C'$ 及 $P=G$ 上，得\n$$\n\\begin{aligned}\nA'A^2 + A'B^2 + A'C^2 + B'A^2 + B'B^2 + B'C^2 + C'A^2 + C'B^... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 獨立研究(一) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | Chinese; English | proof only | null | |
04ze | For arbitrary positive integers $a, b$, denote $a \ominus b = \frac{a-b}{\text{gcd}(a,b)}$.
Let $n$ be a positive integer. Prove that the following conditions are equivalent:
(i) $\text{gcd}(n, n \ominus m) = 1$ for every positive integer $m < n$;
(ii) $n = p^k$ where $p$ is a prime number and $k$ is a non-negative int... | [
"Note at first that $da \\ominus db = a \\ominus b$ for all positive integers $a, b$, and $d$. Indeed,\n$$\nda \\ominus db = \\frac{da - db}{\\text{gcd}(da, db)} = \\frac{d \\cdot (a-b)}{d \\cdot \\text{gcd}(a,b)} = \\frac{a-b}{\\text{gcd}(a,b)} = a \\ominus b.\n$$\nShow now that if $n$ is a prime power and $m < n$... | Estonia | Estonija 2010 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
03t4 | Let $S$ be the set of all those $2007$-place decimal integers $\overline{2a_1a_2a_3\cdots a_{2006}}$ which contain odd number of digit '9' in each sequence $a_1, a_2, a_3, \cdots, a_{2006}$. The cardinal number of $S$ is | [
"Define $A$ as the number of the elements in $S$, we have\n$$\nA = \\binom{2006}{1} 9^{2005} + \\binom{2006}{3} 9^{2003} + \\dots + \\binom{2006}{2005} 9.\n$$\nOn the other hand,\n$$\n(9+1)^{2006} = \\sum_{k=0}^{2006} \\binom{2006}{k} 9^{2006-k}\n$$\nand\n$$\n(9-1)^{2006} = \\sum_{k=0}^{2006} \\binom{2006}{k} (-1)^... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1/2*(10^2006 - 8^2006) | |
0h9e | Given a foundation that is in form of a square $4 \times 4$, that is divided into smaller $1 \times 1$ squares. There is a gap of length $1$ between any two adjacent squares. The foundation is covered with several layers of bricks of size $2 \times 1$. Every layer consists of $8$ bricks and each brick fully covers exac... | [
"Consider a square $A$ of size $1 \\times 1$, that does not touch the borders of $4 \\times 4$. All $4$ sides of it have to be covered by bricks. Moreover, these bricks have to be different for different\n\n**Fig. 16**\nsides, since one half of each brick covers the square $A$. Thus, there ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 4 | |
0j01 | Problem:
A 5-dimensional ant starts at one vertex of a 5-dimensional hypercube of side length $1$. A move is when the ant travels from one vertex to another vertex at a distance of $\sqrt{2}$ away. How many ways can the ant make $5$ moves and end up on the same vertex it started at? | [
"Solution:\n\nAnswer: $6240$\n\nWe let the cube lie in $\\mathbb{R}^5$ with each corner with coordinates $1$ or $0$. Assume the ant starts at $(0,0,0,0,0)$. Every move the ant adds or subtracts $1$ to two of the places. Note that this means the ant can only land on a vertex with the sum of its coordinates an even n... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Other"
] | null | final answer only | 6240 | |
02co | Problem:
Um terreno compartilhado - Três amigas compraram um terreno quadrado e querem reparti-lo como indicado na figura, por que em $A$ se encontra uma fonte de água. Elas querem também que as áreas das três partes sejam iguais. Onde devem estar os pontos $M$ (sobre $BC$) e $N$ (sobre $CD$)?
![](attached_image_1.pn... | [
"Solution:\n\nComo as áreas de $\\triangle ABM$ e $\\triangle ADN$ são iguais e $AB = AD$, temos então\n$$\n\\frac{BM \\times AB}{2} = \\frac{ND \\times AD}{2} \\Longrightarrow BM = DN\n$$\n\n\n\nAssim, a figura $AMCN$ é simétrica com respeito à diagonal $AC$. Portanto, a área do $\\triangl... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | BM = DN = 2/3 of the side length (equivalently, MC = CN = 1/3 of the side length). | |
07w5 | Suppose $Q = ABCD$ is a cyclic quadrilateral. Prove that its longest diagonal doesn't exceed $\sqrt{2}$ times its longest side, with equality holding iff $Q$ is a square. | [
"Let $a = |AB|$, $b = |BC|$, $c = |CD|$, $d = |DA|$ be the side lengths of $Q$, and $f = |AC|$, $g = |BD|$ the lengths of its diagonals. It's well-known and easy to prove that\n$$\nf^2 = \\frac{ab(c^2 + d^2) + cd(a^2 + b^2)}{ab + cd},\n$$\nand\n$$\ng^2 = \\frac{ad(b^2 + c^2) + bc(a^2 + d^2)}{ad + bc}.\n$$\nFor inst... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0g5r | 已知圓 $O_1$ 與圓 $O_2$ 外切於點 $T$, 一直線與圓 $O_2$ 相切於點 $X$, 與圓 $O_1$ 交於點 $A, B$ 且點 $B$ 在線段 $AX$ 的內部, 直線 $XT$ 與圓 $O_1$ 交於另一點 $S$。$C$ 是不包含點 $A, B$ 的 $TS$ 上的一點, 過點 $C$ 作圓 $O_2$ 的切線, 切點為 $Y$ 且線段 $CY$ 與線段 $ST$ 不相交, 直線 $SC$ 與 $XY$ 交於點 $I$。試證：
(1) $C, T, I, Y$ 四點共圓；
(2) $I$ 是 $\triangle ABC$ 的 $\angle A$ 的旁切圓的圓心。 | [
"(1) 作輔助線如圖所示。因為弧 $ST = $ 弧 $XT$，則\n$$\n\\angle BXT = \\frac{\\text{弧}XT^{\\circ}}{2} = \\frac{\\text{弧}ST^{\\circ}}{2} = \\angle TAS.\n$$\n從而, $\\triangle SAT \\sim \\triangle SXA$。故 $\\angle XAS = \\angle ATS$。即弧 $BS = $ 弧 $AS$。\n所以 $S$ 是弧 $AB$ 的中點。因為\n$$\n\\angle TCI = \\angle TAS = \\angle BXT = \\angle TYX.\n$... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,獨立研究(二) | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 ... | null | proof only | null | |
06uw | Let $a$ and $b$ be distinct positive integers. The following infinite process takes place on an initially empty board.
(i) If there is at least a pair of equal numbers on the board, we choose such a pair and increase one of its components by $a$ and the other by $b$.
(ii) If no such pair exists, we write down two times... | [
"We may assume $\\gcd(a, b)=1$; otherwise we work in the same way with multiples of $d=\\gcd(a, b)$.\nSuppose that after $N$ moves of type (ii) and some moves of type (i) we have to add two new zeros. For each integer $k$, denote by $f(k)$ the number of times that the number $k$ appeared on the board up to this mom... | IMO | IMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequ... | null | proof only | null | |
0aiu | We are given an $n \times n$ board. Rows are labeled with numbers $1$ to $n$ downwards and columns are labeled with numbers $1$ to $n$ from left to right. On each field we write the number $x^2 + y^2$ where $(x, y)$ are its coordinates. We are given a figure and can initially place it on any field. In every step we can... | [
"a) The answer is NO.\n\n| | 1 | 2 | 3 | 4 |\n|---|---|---|---|---|\n| 1 | 2 | 5 | 10 | 17 |\n| 2 | 5 | 8 | 13 | 20 |\n| 3 | 10 | 13 | 18 | 25 |\n| 4 | 17 | 20 | 25 | 36 |\n\n| 1 | 2 | 3 | 4 |\n|---|---|---|---|\n| 2 | 1 | 2 | 1 |\n| 1 | 0 | 1 | 0 |\n| 2 | 1 | 2 | 1 |\n| 1 | 0 | 1 | 0 |\n\nOn the left we have the... | North Macedonia | European Mathematical Cup | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | a) No; b) Yes | |
0eqy | $\left(\frac{1}{8}\right)^2 \div \left(\frac{1}{2}\right)^8$ is equal to
(A) 2
(B) 4
(C) 8
(D) 16
(E) 32 | [
"$\\frac{\\left(\\frac{1}{8}\\right)^2}{\\left(\\frac{1}{2}\\right)^8} = \\frac{\\frac{1}{64}}{\\frac{1}{256}} = \\frac{1}{64} \\times \\frac{256}{1} = 4$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | B | |
0fva | Problem:
Sei $ABC$ ein gleichseitiges Dreieck und sei $D$ ein innerer Punkt der Seite $BC$. Ein Kreis berühre $BC$ in $D$ und scheide die Seiten $AB$ und $AC$ in den inneren Punkten $M, N$ und $P, Q$. Beweise, dass gilt
$$
|BD| + |AM| + |AN| = |CD| + |AP| + |AQ|
$$ | [
"Solution:\n\nAnwenden des Potenzsatzes bei den Punkten $A$, $B$, $C$ ergibt der Reihe nach\n$$\n\\begin{aligned}\n|AM| \\cdot |AN| &= |AP| \\cdot |AQ| \\\\\n|BM| \\cdot |BN| &= |BD|^{2} \\\\\n|CP| \\cdot |CQ| &= |CD|^{2}\n\\end{aligned}\n$$\nWir bezeichnen die Seitenlänge des gleichseitigen Dreiecks $ABC$ mit $s$.... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0dqx | For each $i = 1, 2, \dots, N$, let $a_i, b_i, c_i$ be integers such that at least one of them is odd. Show that one can find integers $x, y, z$ such that $xa_i + yb_i + zc_i$ is odd for at least $4N/7$ different values of $i$. | [
"Consider all the 7 triples $(x, y, z)$, where $x, y, z$ are either $0$ or $1$ but not all $0$. For each $i$, at least one of the numbers $a_i, b_i, c_i$ is odd. Thus among the 7 sums $xa_i + yb_i + zc_i$, 3 are even and 4 are odd. Thus there are altogether $4N$ odd sums. Thus there is choice of $(x, y, z)$ for whi... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
05fn | Problem:
Trouver tous les entiers $n \geqslant 1$ pour lesquels la fonction :
$$
x \longmapsto x^{x}
$$
prenne toutes les valeurs possibles modulo $n$ lorsque $x$ parcourt $\llbracket 0, n-1 \rrbracket$.
On dit que c'est une surjection dans $\mathbb{Z} / n \mathbb{Z}$. | [
"Solution:\n\nNotons déjà que si $n$ vérifie la propriété, alors tout diviseur $d$ de $n$ la vérifie également. Posons $f$ l'application qui à $x$ dans $\\mathbb{N}^{*}$ associe $x^{x}$ dans $\\mathbb{Z} / n \\mathbb{Z}$.\n\nPour $n = p^{2}$ avec $p$ premier, notons que $p$ n'est pas dans l'image de $f$. En effet, ... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Div... | null | proof and answer | Exactly those n that are squarefree and whose distinct prime factors p and q satisfy that no prime p divides q minus one; equivalently, n is a product of distinct primes with the property that for any two distinct prime factors p_i and p_j, p_i does not divide p_j − 1. (This includes n = 1.) | |
0a9g | Problem:
Laura has $2010$ lamps connected with $2010$ buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the b... | [
"Solution:\n\na. Let us say that two lamps are separated, if one of the lamps is turned on while the other lamp remains off. Laura can find out which lamps belong to the buttons if every two lamps are separated. Let Richard choose two arbitrary lamps. To begin with, he turns both lamps on and then varies all the ot... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 24 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a) 2^2009 + 1; b) 11 | |
016a | Suppose that for a prime number $p$ and integers $a$, $b$, $c$ the following holds:
$$
6 \mid p + 1, \quad p \mid a + b + c, \quad p \mid a^4 + b^4 + c^4.
$$
Prove that $p \mid a$, $b$, $c$. | [
"Observe that $p \\mid a^4 + b^4 + (-a-b)^4 = 2(a^2 + ab + b^2)^2$. Thus, since $p \\ne 2$, we have $p \\mid a^2 + ab + b^2$ which implies that $p \\mid (a^2 + ab + b^2)(a - b) = a^3 - b^3$. From this fact and Fermat's little theorem we obtain the following congruences, with $p = 6n - 1$:\n$$\nb \\equiv b^p \\equiv... | Baltic Way | Baltic Way SHL | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0hxz | Problem:
Let $f(x) = 1 + \frac{x}{2} + \frac{x^{2}}{4} + \frac{x^{3}}{8} + \ldots$, for $-1 \leq x \leq 1$. Find $\sqrt{e^{\int_{0}^{1} f(x) d x}}$. | [
"Solution:\nObserve that $f(x)$ is merely an infinite geometric series. Thus $f(x) = \\frac{1}{1 - \\frac{x}{2}} = \\frac{2}{2 - x}$.\n\nThen\n$$\n\\int_{0}^{1} \\frac{2}{2 - x} \\, dx = 2 \\int_{0}^{1} \\frac{1}{2 - x} \\, dx = 2 [ -\\ln|2 - x| ]_{0}^{1} = 2 ( -\\ln(1) + \\ln(2) ) = 2 \\ln 2.\n$$\n\nThen\n$$\n\\sq... | United States | HMMT | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 2 | |
0eog | A hiker walks $1$ km East, then $2$ km North, then $3$ km West, then $4$ km South, then $5$ km East and finally $6$ km North. The hiker's straight-line distance in km from the starting point is
(A) $4$ (B) $5$ (C) $6$ (D) $7$ (E) $8$ | [
"Horizontal distance travelled:\n$$\n1 - 3 + 5 = 3 \\text{ (i.e. 3 km East)}\n$$\nVertical distance travelled:\n$$\n2 - 4 + 6 = 4 \\text{ (i.e. 4 km North)}\n$$\nBy Pythagoras, the straight-line distance from the starting point is thus $5$ km."
] | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | MCQ | B | |
0g5s | $ABCD$ 為一凸四邊形。點 $E, F, G, H$ 分別在線段 $AB, BC, CD, DA$ 上，使得延長後的直線 $AB, FH, CD$ 三線共點，直線 $BC, EG, AD$ 亦三線共點。令 $O$ 為 $EG$ 和 $FH$ 的交點。考慮以下四個四邊形：
$AHOE, BEOF, CFOG$ 及 $DGOH$。
試證明：若這四個四邊形的其中三個是圓外切四邊形，則第四個必定也是。
註：一個四邊形稱為圓外切四邊形的條件是存在一個圓在四邊形內使得該圓與四條邊都相切。 | [
"令 $X$ 為 $AB, FH, CD$ 的交點，$Y$ 為 $BC, EG, AD$ 的交點。令 $L_1, L_2$ 分別為 $\\angle AXO, \\angle OXC$ 的角平分線，且令 $K_1, K_2$ 分別為 $\\angle AYO, \\angle OYC$ 的角平分線。\n顯然四個四邊形的內切圓（如果存在的話）的圓心必須分別落在 $L_i, K_j$ 的交點上 $(i, j = 1, 2)$。令 $P_{ij}$ 表示 $L_i$ 和 $K_j$ 的交點。我們令\n$$\n\\alpha_1 = \\frac{1}{2} \\angle AXO, \\quad \\alpha_2 = \\fra... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
05tk | Problem:
Dans 5 boîtes se trouvent respectivement $402$, $403$, $404$, $405$ et $406$ pierres. La seule opération autorisée est de prendre $4$ pierres dans un tas ayant au moins $4$ pierres et d'en mettre une dans chacun des autres tas. Quel est le plus grand nombre de pierres qu'il est possible d'avoir dans un seul t... | [
"Solution:\n\nL'énoncé présente une suite d'opérations et décrit une certaine situation initiale. Une première idée à essayer dans ce cas est de chercher un invariant.\n\nUne deuxième idée est de tester l'énoncé pour des valeurs plus petites, par exemple pour des tas de taille $0,1,2,3$ et $4$. Après plusieurs essa... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 2014 | |
09z0 | Sofie and her grandmother both have their birthday on 1 January. In six consecutive years, the age of grandma is an integer multiple of the age of her granddaughter Sofie. In the seventh year this is not the case. A few years later the age of grandma is again an integer multiple of the age of Sofie.
How old can grandma... | [] | Netherlands | First Round | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
0efn | Problem:
Za koliko \% se spremeni vrednost ulomka, če števec povečamo za $20 \%$, imenovalec pa zmanjšamo za $20 \%$?
(A) $0 \%$
(B) $20 \%$
(C) $40 \%$
(D) $50 \%$
(E) $60 \%$ | [
"Solution:\n\n$a + 0,2a = 1,2a$, $b - 0,2b = 0,8b$ in $\\frac{1,2a}{0,8b} = 1,5 \\frac{a}{b}$. Vrednost ulomka se poveča za $50 \\%$. Pravilen odgovor je (D)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | D | |
04cc | There are some integers written on the blackboard. In each step two numbers $a$ and $b$ are chosen and replaced with the numbers $3a - b$ and $13a - 3b$. If there are numbers $1$, $2$, $3$, $4$, $\ldots$, $2011$, $2012$ on the blackboard at the beginning, is it possible to get numbers $2$, $4$, $6$, $8$, $\ldots$, $402... | [
"In each step the sum of all numbers on the blackboard changes by\n$$\n((3a - b) + (13a - 3b)) - (a + b) = 15a - 5b = 5(3a - b).\n$$\nThis difference is divisible by $5$, i.e. the sum of all numbers written on the board in each step gives the same remainder when divided by $5$.\nAt the beginning, the sum of all num... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | No | |
0dj9 | Let $ABC$ be a triangle with circumcircle $(O)$ with orthocenter $H$. Take $X \in AB$, $Y \in AC$ such that $XB = XH$, $YC = YH$. Prove that $O$ belongs to the Euler line of triangle $AXY$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kjf | Azar and Carl play a game of tic-tac-toe. Azar places an $X$ in one of the boxes in a 3-by-3 array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all 9 boxes are filled or one of the players has 3 of their symbols in a row—h... | [
"Solution:\n\n**Answer (D):** For Carl to win at his third turn, his 3 $O$s must lie in one of the 8 winning configurations and Azar's 3 $X$s must not (because that would have resulted in her winning after her third move). There are 6 vertical or horizontal rows for Carl's $O$s, and in each case there are $\\binom{... | United States | AMC 12 A | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | D | |
0ejn | Problem:
Poišči vsa realna števila $x$, za katera velja $\left(x^{2}-7 x+11\right)^{x^{2}-13 x+42}=1$. | [
"Solution:\nVrednost potence $a^{b}$ je enaka $1$ le v primeru, ko je $a=1$, $a=-1$ in $b$ sodo celo število ali $b=0$ in $a \\neq 0$.\n\nČe je $x^{2}-7 x+11=1$, sledi $x^{2}-7 x+10=0$ oziroma $(x-2)(x-5)=0$. Od tod dobimo rešitvi $x=2$ in $x=5$.\n\nČe je $x^{2}-7 x+11=-1$ oziroma $x^{2}-7 x+12=0$, sledi $(x-3)(x-4... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2, 3, 4, 5, 6, 7 | |
0jcf | Problem:
Find the number of ordered $2012$-tuples of integers $(x_{1}, x_{2}, \ldots, x_{2012})$, with each integer between $0$ and $2011$ inclusive, such that the sum $x_{1} + 2x_{2} + 3x_{3} + \cdots + 2012x_{2012}$ is divisible by $2012$. | [
"Solution:\n\nWe claim that for any choice of $x_{2}, x_{3}, \\ldots, x_{2012}$, there is exactly one possible value of $x_{1}$ satisfying the condition. We have\n$$\nx_{1} + 2x_{2} + \\ldots + 2012x_{2012} \\equiv 0 \\pmod{2012}\n$$\nor\n$$\nx_{1} \\equiv -\\left(2x_{2} + 3x_{3} + \\cdots + 2012x_{2012}\\right) \\... | United States | HMMT November 2012 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2012^2011 | |
08q5 | Problem:
Let $ABC$ be a triangle and let $\omega$ be its circumcircle. Let $\ell_{B}$ and $\ell_{C}$ be two parallel lines passing through $B$ and $C$ respectively. The lines $\ell_{B}$ and $\ell_{C}$ intersect with $\omega$ for the second time at the points $D$ and $E$ respectively, with $D$ belonging on the arc $AB$... | [
"Solution:\n\nWe write $\\omega_{1}, \\omega_{2}$ and $\\omega'$ for the circumcircles of $AGD, AEF$ and $OO_{1}O_{2}$ respectively. Since $O_{1}$ and $O_{2}$ are the centers of $\\omega_{1}$ and $\\omega_{2}$, and because $DG$ and $EF$ are parallel, we get that\n$$\n\\angle GAO_{1} = 90^{\\circ} - \\frac{\\angle G... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
05xs | Problem:
Trouver tous les réels $x, y$ tels que
$$
x(x-2y) + y(2y-4) + 4 = 0
$$ | [
"Solution:\nDéveloppons l'équation pour obtenir\n$$\nx^2 - 2xy + 2y^2 - 4y + 4 = 0.\n$$\nLes termes $x^2 - 2xy$ nous font penser à l'identité remarquable $x^2 - 2xy + y^2 = (x-y)^2$. On sépare alors le $2y^2$ en $y^2 + y^2$ pour écrire\n$$\n(x-y)^2 + y^2 - 4y + 4 = 0\n$$\nMais on reconnait alors une deuxième identi... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = 2, y = 2 | |
09tj | Problem:
Zij $\triangle ABC$ een driehoek waarvan de zijdelengtes positieve gehele getallen zijn die paarsgewijs relatief priem zijn. De raaklijn in $A$ aan de omgeschreven cirkel snijdt de lijn $BC$ in $D$. Bewijs dat $|BD|$ geen geheel getal is.
 | [
"Solution:\n\nEr zijn twee configuraties mogelijk. Neem zonder verlies van algemeenheid aan dat $B$ tussen $D$ en $C$ ligt. Schrijf $a=|BC|$, $b=|CA|$, $c=|AB|$, $x=|BD|$ en $y=|AD|$. Vanwege de raaklijnomtrekshoekstelling geldt $\\angle BAD=\\angle ACB=\\angle ACD$, dus $\\triangle ABD \\sim \\triangle CAD$ (hh), ... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0daa | The partition of $2n$ positive integers into $n$ pairs is called square-free if the product of numbers in each pair is not a perfect square. Prove that if for $2n$ distinct positive integers, there exists one square-free partition, then there exists at least $n!$ square-free partitions. | [
"For convenience, let us call integers that are not perfect squares \"good\" and integers that are perfect squares \"bad\". We shall prove the given estimation by induction.\n\nIndeed,\n\nFor $n=1$ we only have 1 square-free partition.\n\nFor $n=2$ we have four numbers $x_{1}, x_{2}, y_{1}, y_{2}$ with $x_{1}y_{1}$... | Saudi Arabia | Team selection tests for BMO 2018 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
04g0 | Find all pairs of real numbers $(x, y)$ satisfying the following system:
$$
x + y^2 = y^3,
$$
$$
y + x^2 = x^3.
$$ | [
"Subtracting the given equations, we obtain $y - x + x^2 - y^2 = x^3 - y^3$, which is equivalent to\n$$\n(x - y)(-1 + x + y) = (x - y)(x^2 + xy + y^2),\n$$\nimplying that either $x - y = 0$ or $x + y - 1 = x^2 + xy + y^2$.\n\nIn the latter case, $x + y - 1 = x^2 + xy + y^2$ is equivalent to\n$$\ny^2 + (x-1)y + x^2 ... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (0, 0), ((1 + sqrt(5)) / 2, (1 + sqrt(5)) / 2), ((1 - sqrt(5)) / 2, (1 - sqrt(5)) / 2) | |
0d6t | Let $ABC$ be a triangle inscribed in the circle $(O)$ and $P$ is a point inside the triangle $ABC$. Let $D$ be a point on $(O)$ such that $AD \perp AP$. The line $CD$ cuts the perpendicular bisector of $BC$ at $M$. The line $AD$ cuts the line passing through $B$ and is perpendicular to $BP$ at $Q$. Let $N$ be the refle... | [
"\nLet $K, L$ be the midpoints of $PQ, PN$.\nSince $PBQA$ is cyclic\n$$\n\\angle KPB = \\angle BAQ = \\angle DCB.\n$$\nHence the isosceles triangles $KPB$ and $MCB$ are similar. The line $BM$ cuts $PQ$ at $R$. We have\n$$\n\\angle LMC = \\angle LMB - \\angle BMC = \\angle MRQ - \\angle BKP ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0f3a | Problem:
a and b are real numbers. Find real $x$ and $y$ satisfying:
$$(x - y)(x^2 - y^2)^{1/2} = a(1 - x^2 + y^2)^{1/2}$$
and
$$(y - x)(x^2 - y^2)^{1/2} = b(1 - x^2 + y^2)^{1/2}.$$ | [] | Soviet Union | ASU | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Solutions exist if and only if b = −a.
Case 1: a = b = 0. Then any real x, y with x^2 = y^2 (i.e., x = y or x = −y) satisfy the equations.
Case 2: a ≠ 0 and b = −a. Let u be any nonzero real with the same sign as a. Define v = a^2 / (u (u^2 + a^2)). Then set
x = (u + v) / 2, y = (v − u) / 2.
These give all real solut... | |
02wb | Problem:
As somas das três colunas e das três linhas da tabela são iguais.
| 4 | 9 | 2 |
| :--- | :--- | :--- |
| 8 | 1 | 6 |
| 3 | 5 | 7 |
Qual é o menor número de casas da tabela, que devem ser alteradas, para que todas as novas seis somas sejam diferentes entre si? | [
"Solution:\n\nSe três ou menos casas são alteradas, ou existirão duas filas sem usarem casas alteradas ou uma casa é a única alterada em sua linha e coluna. No primeiro caso, essas duas filas sem casas alteradas possuem a mesma soma. No segundo caso, se apenas uma casa é a única alterada na sua linha e coluna, elas... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
0gs0 | Let $(a_n)_{n=1}^\infty$ be a sequence of integers with $a_1 = 1$, $a_2 = 2$ and
$$
a_{n+2} = a_{n+1}^2 + (n+2)a_{n+1} - a_n^2 - n a_n
$$
for all $n \ge 1$.
a) Show that there exist infinitely many prime numbers dividing at least one term of this sequence.
b) Find three different prime numbers not dividing any term o... | [
"a.\nBy putting $b_1 = 1$ and $b_{n+1} = a_n^2 + n a_n$ for all $n \\ge 1$ we get\n$$\na_{n+2} - a_{n+1} = a_{n+1}^2 + (n+1)a_{n+1} - a_n^2 - n a_n = b_{n+2} - b_{n+1}.\n$$\nWe also have $a_1 = b_1 = 1$, $a_2 = b_2 = 2$. Hence, we conclude that $a_n = b_n$ for all $n \\ge 1$. Then we get $a_{n+1} = b_{n+1} = a_n(a_... | Turkey | Team Selection Test for IMO 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 3, 5, 19 | |
08n1 | Problem:
Decide if it is possible to consider $2011$ points in a plane such that the distance between every two of these points is different from $1$ and each unit circle centered at one of these points leaves exactly $1005$ points outside the circle. | [
"Solution:\n\nNO. If such a configuration existed, the number of segments starting from each of the $2011$ points towards the other one and having length less than $1$ would be $1005$. Since each segment is counted twice, their total number would be $1005 \\cdot 2011 / 2$ which is not an integer, contradiction!"
] | JBMO | JBMO Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No | |
0c6p | We consider $D$, the midpoint of the baseline $[BC]$ of the isosceles triangle $ABC$, in which $m(\angle BAC) < 90^\circ$. On the perpendicular line in $B$ on the line $BC$, one takes the point $E$, such that $\angle EAB \equiv \angle BAC$, and on the parallel through $C$ to the line $AB$, one takes the point $F$, such... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
09mr | Find all non-negative integer solutions $(k, m, n)$ of the equation
$$
211 \cdot 3^k + 5^m = 2024^n.
$$
(Ganbat Atarsaikhan) | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | (k, m, n) = (2, 3, 1) | |
00l2 | Let $f: \mathbb{Z}_{>0} \to \mathbb{Z}$ be a function with the following properties:
(i) $f(1) = 0$,
(ii) $f(p) = 1$ for all prime numbers $p$,
(iii) $f(xy) = y f(x) + x f(y)$ for all $x, y$ in $\mathbb{Z}_{>0}$.
Determine the smallest integer $n \ge 2015$ that satisfies $f(n) = n$. | [
"We claim that\n$$\nf(q_1 \\cdots q_s) = q_1 \\cdots q_s \\left( \\frac{1}{q_1} + \\cdots + \\frac{1}{q_s} \\right) \\quad (1)\n$$\nholds for (not necessarily distinct) prime numbers $q_1, \\dots, q_s$.\nWe prove the claim by induction on $s$. For $s=0$, the claim reduces to $f(1) = 0$, which is true by assumption.... | Austria | 46th Austrian Mathematical Olympiad National Competition (Final Round, part 2) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 3125 | |
03ul | Suppose that the sequence of positive numbers $x_1, x_2, \ldots, x_n, \ldots$ satisfies $(8x_2 - 7x_1)x_1^7 = 8$ and
$$
x_{k+1}x_{k-1} - x_k^2 = \frac{x_{k-1}^8 - x_k^8}{(x_kx_{k-1})^7}, \quad k \ge 2.
$$
Find positive real number $a$ such that when $x_1 > a$ one has $x_1 > x_2 > \cdots > x_n > \cdots$, and when $0 < x... | [
"By $x_{k+1}x_{k-1} - x_k^2 = \\frac{x_{k-1}^8 - x_k^8}{(x_kx_{k-1})^7}$, we have\n$$\n\\frac{x_{k+1}}{x_k} - \\frac{x_k}{x_{k-1}} = \\frac{1}{x_k^8} - \\frac{1}{x_{k-1}^8},\n$$\ni.e.\n$$\n\\frac{x_{k+1}}{x_k} - \\frac{1}{x_k^8} = \\frac{x_k}{x_{k-1}} - \\frac{1}{x_{k-1}^8} = \\cdots = \\frac{x_2}{x_1} - \\frac{1}{... | China | China Girls' Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | a = 8^{1/8} | |
0l1w | Problem:
Let $S$ be a set of nonnegative integers such that
- there exist two elements $a$ and $b$ in $S$ such that $a, b > 1$ and $\operatorname{gcd}(a, b) = 1$; and
- for any (not necessarily distinct) element $x$ and nonzero element $y$ in $S$, both $x y$ and the remainder when $x$ is divided by $y$ are in $S$.
Prov... | [
"Solution:\nAssume $a < b$. Note that we can get $1 \\in S$ via the Euclidean algorithm, and $0 \\in S$ from $a \\bmod 1$. Suppose $(a, b) \\neq (2, 3)$. We will show that there exists $c, d \\in S$ with $1 < c \\leq a$ and $1 < d \\leq b$, with $\\operatorname{gcd}(c, d) = 1$ and at least one of $c \\neq a$ and $d... | United States | HMIC 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis,... | null | proof only | null | |
041m | Suppose there are 101 persons sitting around a round table in an arbitrary order. The $k$th person possesses $k$ pieces of cards, $k = 1, \dots, 101$. We call it a *transition* if one transits one of his cards to one of his adjacent persons. Find the minimum positive number $k$, such that whatever the order of the seat... | [
"The answer is $k = 42\\,925$.\n\nLet the circumference of the table be 101, and the distance between two adjacent persons be 1. In the following, we consider the least transition times.\n\nDenote the person who initially possesses $i$ cards by $[i - 51]$. So, if $p > 0$, then we can think of person $[p]$ as the so... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 42925 |
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