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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0cj9 | Let $ABC$ be a triangle and $M$ a point in its plane, distinct from $A$, $B$ and $C$. Let $N$, $P$ and $Q$ denote the symmetries of point $M$ with respect to sides $AB$, $BC$ and $AC$, respectively.
a) Prove that the points $N$, $P$ and $Q$ are collinear if and only if the point $M$ belongs to the circumcircle of tria... | [
"The midpoint $S$ of the segment $MN$ belongs to the line $AB$, therefore $\\frac{s-a}{b-a} \\in \\mathbb{R}$. The lines $MN$ and $AB$ are perpendicular, therefore $\\frac{n-m}{b-a} \\in i\\mathbb{R}$.\n\nWe assume, without loss of generality, that the origin is at the circumcenter of the triangle, and $|a| = |b| =... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle"
] | English | proof only | null | |
01v5 | For all pairs of $(m, n)$ positive integers that have the same number $k$ of divisors we define the operation $\circ$. Write all their divisors in an ascending order: $1 = m_1 < \dots < m_k = m$, $1 = n_1 < \dots < n_k = n$ and set
$$
m \circ n = m_1 \cdot n_1 + \dots + m_k \cdot n_k.
$$
Find all pairs of numbers $(m, ... | [
"From the definition of $\\circ$ it is clear that $m \\circ n \\ge 1 + n^2$, so $n \\le \\sqrt{496} < 23$. Note that the minimal number that has at least 7 divisors is 24, hence it is sufficient to consider only $k$ from 1 to 6.\n\nExactly two divisors have only prime numbers, so for $k=2$ the numbers $m$ and $n$ a... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (20, 18) | |
024z | Problem:
Alguns números reais estão escritos nas casas de um tabuleiro $n \times n$ de modo que a soma total dos números escritos é positiva. Mostre que existe alguma permutação das colunas do tabuleiro, de modo que a soma dos números escritos nas casas da diagonal principal do novo tabuleiro seja positiva. | [
"Solution:\n\nCrie um cilindro a partir do tabuleiro como indicado na figura abaixo. Esse cilindro pode ser decomposto em $n$ diagonais disjuntas que começam em um extremo do cilindro e terminam no outro. Como a soma de todos os números do tabuleiro é positiva, pelo menos uma das diagonais terá soma positiva. Ela c... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
06e0 | In a school there are $b$ teachers and $c$ students. Suppose that
(i) each teacher teaches exactly $k$ students; and
(ii) for each pair of distinct students, exactly $h$ teachers teach both of them.
Show that
$$
\frac{b}{h} = \frac{c(c-1)}{k(k-1)}
$$ | [
"We count the number of triples $(T, S_1, S_2)$ such that $T$ is a teacher teaching two distinct students $S_1$ and $S_2$. Note that $(T, S_1, S_2)$ is different from $(T, S_2, S_1)$.\n\nFor each of the $b$ teachers, there are $k$ choices for $S_1$ and $k-1$ choices for $S_2$. Therefore, there are $b k (k-1)$ such ... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
03o7 | Problem:
Determine all positive integers $a$, $b$, $c$, $p$ where $p$ and $p + 2$ are odd primes and
$$2^{a}p^{b} = (p + 2)^{c} - 1.$$ | [
"Solution:\nThe only solution is $(a, b, c, p) = (3, 1, 2, 3)$. First, factor the right hand side. This gives us\n$$2^{a}p^{b} = (p + 1)((p + 2)^{c - 1} + (p + 2)^{c - 2} + \\dots +(p + 2) + 1).$$\nSince $\\gcd (p, p + 1) = 1$ it must be the case that $p + 1 = 2^{x}$ for some positive integer $x \\leq a$ and so $p ... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof and answer | (a, b, c, p) = (3, 1, 2, 3) | |
0brv | If $a$, $b$ and $c$ are the length of the sides of a triangle, show that
$$
\frac{3}{2} \le \frac{b+c}{b+c+2a} + \frac{a+c}{a+c+2b} + \frac{a+b}{a+b+2c} < \frac{5}{3}.
$$ | [
"The inequality\n$$\n\\frac{3}{2} \\le \\frac{b+c}{b+c+2a} + \\frac{a+c}{a+c+2b} + \\frac{a+b}{a+b+2c}\n$$\nis Nesbitt's inequality for the triple $(b+c, a+c, a+b)$.\n\nTriangle's inequality yields $b+c > a \\Leftrightarrow 3a+3b+3c > 4a+2b+2c \\Leftrightarrow \\frac{2}{3(a+b+c)} < \\frac{1}{2a+b+c} \\Leftrightarro... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0kg0 | Find all pairs of positive integers $(a, b)$ satisfying the following conditions:
(i) $a$ divides $b^4 + 1$,
(ii) $b$ divides $a^4 + 1$,
(iii) $\lfloor\sqrt{a}\rfloor = \lfloor\sqrt{b}\rfloor$. | [
"The only solutions are $(1, 1)$, $(1, 2)$, and $(2, 1)$, which clearly work. Now we show there are no others.\n\nObviously, $\\gcd(a, b) = 1$, so the problem conditions imply\n$$\nab \\mid (a - b)^4 + 1\n$$\nsince each of $a$ and $b$ divide the right-hand side. We define\n$$\nk \\stackrel{\\text{def}}{=} \\frac{(b... | United States | USA TSTST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(1, 1), (1, 2), (2, 1)] | |
0hgt | Find the integer closest to the value of the expression:
$$
((7 + \sqrt{48})^{2023} + (7 - \sqrt{48})^{2023})^2 - ((7 + \sqrt{48})^{2023} - (7 - \sqrt{48})^{2023})^2.
$$ | [
"Let's transform the given expression as follows:\n$$\n((7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023})^2 - ((7 + \\sqrt{48})^{2023} - (7 - \\sqrt{48})^{2023})^2 = \\\\\n= ((7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023} + (7 + \\sqrt{48})^{2023} - (7 - \\sqrt{48})^{2023}) \\cdot \\\\\n\\cdot ((7 + \\sqrt{48})... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | final answer only | 4 | |
05y4 | Problem:
Soit $x$, $y$ et $z$ des réels strictement positifs tels que $x y + y z + z x = 3$.
Démontrer que
$$
\frac{x+3}{y+z} + \frac{y+3}{z+x} + \frac{z+3}{x+y} + 3 \geqslant 27 \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^{2}}{(x+y+z)^{3}}
$$ | [
"Solution:\n\nPosons $s = x + y + z$, et soit $L$ et $R$ les membres de gauche et de droite de notre inégalité. On peut réécrire $L$ comme\n$$\nL = \\frac{x+3}{y+z} + 1 + \\frac{y+3}{z+x} + 1 + \\frac{z+3}{x+y} + 1 = (s+3)\\left(\\frac{1}{y+z} + \\frac{1}{z+x} + \\frac{1}{x+y}\\right).\n$$\nOr, l'inégalité de Cauch... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0la2 | Find the number of solutions of the following system of equations:
$$
\begin{cases} x^2 + y^3 = 29 \\ \log_3 x \cdot \log_2 y = 1. \end{cases}
$$ | [] | Vietnam | Vijetnam 2008 | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities"
] | null | proof and answer | 2 | |
08nz | Problem:
Let $I$ be the incenter and $AB$ the shortest side of a triangle $ABC$. The circle with center $I$ and passing through $C$ intersects the ray $AB$ at the point $P$ and the ray $BA$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $ABC$ belonging to angle $A$ touches the side $BC$, and ... | [
"Solution:\nFirst we will show that points $P$ and $Q$ are not on the line segment $AB$.\nAssume that $Q$ is on the line segment $AB$. Since $CI = QI$ and $\\angle IBQ = \\angle IBC$, either the triangles $CBI$ and $QBI$ are congruent or $\\angle ICB + \\angle IQB = 180^\\circ$. In the first case, we have $BC = BQ$... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
04nt | Let $f: N \to N$ be a function such that
$$
f(ab) = f(a+b)
$$
for all positive integers $a \ge 4$ and $b \ge 4$.
Prove that $f(n) = f(8)$ for all positive integers $n \ge 8$. | [
"Let $n \\ge 8$ be a positive integer. The problem condition implies that\n$$\n\\begin{aligned}\nf(n) &= f(4 + (n-4)) = f(4(n-4)) = f(2(n-4) + 2(n-4)) \\\\\n&= f(4(n-4)(n-4)) = f(4(n-4) + (n-4)) \\\\\n&= f(5(n-4)) = f(5 + n - 4) \\\\\n&= f(n+1).\n\\end{aligned}\n$$\nTherefore, by the principle of mathematical induc... | Croatia | Croatia_2018 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0elt | Find all functions $f : \mathbb{N} \to \mathbb{N}$ satisfying $f(n+1)f(n+2) = f(n)^2$ for all $n \in \mathbb{N}$. | [
"Suppose that the function is non-constant. Then it will have a least element $f(k)$. Then $f(k+1) \\neq f(k)$, or the function would be constant, and by definition $f(k+1) > f(k)$. But $f(k+2) = \\frac{f(k)^2}{f(k+1)} = f(k)\\frac{f(k)}{f(k+1)} < f(k)$, which is a contradiction. So $f(n)$ constant is the only allo... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | All constant functions f(n) = c for all n, where c is any natural number. | |
08vp | Let $ABC$ be a right triangle with $\angle ABC = 90^\circ$. Points $P, Q, R$ lie on the sides $BC, CA, AB$, respectively, in such a way that conditions
$$
AQ : QC = 2 : 1, \quad AR = AQ, \quad QP = QR, \quad \angle PQR = 90^\circ
$$
are satisfied. Find the value of $AR$ if $CP = 1$. Here for a line segment $XY$ its len... | [
"Take a point $C'$ on the line $AR$ in such a way that the points $A, R, C'$ lie on the line in this order and $RC' = 1$ is satisfied. From the hypothesis of the problem, it then follows that $RC' = PC, RQ = PQ$ are satisfied, and furthermore, we obtain the fact that the triangles $C'RQ$ and $CPQ$ are congruent sin... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2√5 + 4 | |
03qb | We say a positive integer $n$ is "good" if there is a permutation $(a_1, a_2, \dots, a_n)$ of $1, 2, \dots, n$ such that $a_k + k$ is a perfect square for all $1 \le k \le n$. Determine all the good numbers in the set $\{11, 13, 15, 17, 19\}$. | [
"The good numbers are $13$, $15$, $17$ and $19$. However $11$ is not.\n\nNote that for $1 \\le k \\le 11$, $4 + k$ is a perfect square if and only if $k = 5$. Likewise, $11 + k$ is a perfect square if and only if $k = 5$. Hence $11$ is not good.\n\nNote that $13$ is good because\n\n$k$: 1 2 3 4 5 6 7 8 9 10 11 12 1... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | 13, 15, 17, 19 | |
0c8s | For every positive, odd integer $n$, prove that
$$
\left[ \frac{1}{2} + \sqrt{n + \frac{1}{2}} \right] = \left[ \frac{1}{2} + \sqrt{n + \frac{1}{2020}} \right],
$$
where $[a]$ denotes the integer part of the real number $a$. | [
"We will prove by contradiction that there is no integer between $\\frac{1}{2} + \\sqrt{n + \\frac{1}{2020}}$ and $\\frac{1}{2} + \\sqrt{n + \\frac{1}{2}}$.\n\nSuppose there exists $k \\ge 1$ such that $\\frac{1}{2} + \\sqrt{n + \\frac{1}{2}} \\ge k > \\frac{1}{2} + \\sqrt{n + \\frac{1}{2020}}$. We obtain $n \\ge k... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0fv0 | Problem:
Sei $A_{1} A_{2} \ldots A_{n}$ ein reguläres $n$-Eck. Die Punkte $B_{1}, \ldots, B_{n-1}$ sind wie folgt definiert:
- Für $i=1$ oder $i=n-1$ ist $B_{i}$ der Mittelpunkt der Seite $A_{i} A_{i+1}$;
- Für $i \neq 1, i \neq n-1$ sei $S$ der Schnittpunkt von $A_{1} A_{i+1}$ und $A_{n} A_{i}$. Der Punkt $B_{i}$ ist... | [
"Solution:\n\nWir beginnen mit folgendem\nLemma 2. Sei $A B C D$ ein gleichschenkliges Trapez, wobei $A B \\| C D$. Die Diagonalen $A C$ und $B D$ schneiden sich in $S$. Sei $M$ der Mittelpunkt von $B C$ und die Winkelhalbierende von $\\Varangle C S B$ schneide $B C$ in $N$. Dann gilt $\\Varangle A M D=\\Varangle A... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07is | Let $n$ be a given positive integer. Find the number of permutations $(a_1, \dots, a_n)$ of $1, 2, \dots, n$ such that for all $1 \le i \le n$ we have $a_i \mid 2i$. | [
"We shall prove that the answer is $2^{\\lfloor \\frac{n}{2} \\rfloor}$. We shall firstly prove the following lemma;\n\n**Lemma 1.** Let $i = 2^{\\nu_2(i)}m$, $\\gcd(2, m) = 1$ then $a_i = 2^j m$ for some $j \\in \\{0, 1, \\dots, 1+\\nu_2(i)\\}$.\n\n*Proof.* We shall prove this statement through induction on $m$. N... | Iran | 41th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2^{\lfloor n/2 \rfloor} | |
0l9h | Let $n$ be a positive integer. Prove that the number $2^n + 1$ has no prime divisor of the form $8k-1$, where $k$ is a positive integer. | [] | Vietnam | Vietnamese Team Selection Contest for the 44th IMO | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
0aq6 | Problem:
Let $ABCD$ be a square. Let $M$ be the midpoint of $\overline{DC}$, $N$ the midpoint of $\overline{AC}$, and $P$ the intersection of $\overline{BM}$ and $\overline{AC}$. What is the ratio of the area of $\triangle MNP$ to that of the square $ABCD$? | [
"Solution:\n\n$1:24$\n\nRefer to Figure 8. Notice that $\\triangle MNP \\sim \\triangle BCP$, so that\n$$\n\\frac{NP}{PC} = \\frac{MN}{BC} = \\frac{1}{2} \\quad \\text{and} \\quad \\frac{MP}{BP} = \\frac{MN}{BC} = \\frac{1}{2}\n$$\nRecall that the ratio of the areas of two triangles of equal altitudes is equal to t... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1:24 | |
0aws | Problem:
Let $L_{1}$ be the line with equation $6x - y + 6 = 0$. Let $P$ and $Q$ be the points of intersection of $L_{1}$ with the $x$-axis and $y$-axis, respectively. A line $L_{2}$ that passes through the point $(1, 0)$ intersects the $y$-axis and $L_{1}$ at $R$ and $S$, respectively. If $O$ denotes the origin and th... | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | y = -3x + 3 or y = -10x + 10 | |
0900 | Positive integers such that all the digits are prime numbers are called excellent numbers. Determine all the three-digit positive integers $n$ such that $n + 2024$ and $n - 34$ are both excellent numbers. There exist exactly two such positive integers $n$. | [
"$$\n\\boxed{309,\\ 311}\n$$\nLet $a$, $b$ and $c$ be each digit of $n$ in the hundreds, tens, and ones place respectively, then $n$ is described as $n = 100a + 10b + c$. Both $n + 2024$ and $n - 34$ have a prime number in the ones place, hence $c$ is $1$ or $9$.\n\nIn the case of $c = 1$, both integers\n$$\nn + 20... | Japan | Japan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 309, 311 | |
0gni | Find the minimum of
$$
\frac{1+a+b+c}{3+2a+b} - \frac{c}{b}
$$
where $a$, $b$, $c$ are real numbers such that all roots of the equation $x^3 - a x^2 + b x - c = 0$ are real positive numbers. | [
"We prove that the minimum is $\\frac{1}{3}$.\nLet $x_i > 0$ for $i = 1, 2, 3$ be the roots of the equation $x^3 - a x^2 + b x - c = 0$. By Vieta's theorem\n$$\nx_1 + x_2 + x_3 = a, \\quad x_1 x_2 + x_2 x_3 + x_1 x_3 = b, \\quad x_1 x_2 x_3 = c.\n$$\nThen\n$$\nA = \\frac{1 + a + b + c - c}{3 + 2a + b} = \\frac{1}{b... | Turkey | Team Selection Test for IMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 1/3 | |
0j3j | Problem:
A knight moves on a two-dimensional grid. From any square, it can move 2 units in one axis-parallel direction, then move 1 unit in an orthogonal direction, the way a regular knight moves in a game of chess. The knight starts at the origin. As it moves, it keeps track of a number $t$, which is initially $0$. Wh... | [
"Solution:\nFor convenience, we will refer to $(a, b)$ as $[a x + b]$, the function it represents. This will make it easier to follow the trajectory of $t$ over a given sequence of moves.\n\nSuppose we start at $[x+1]$ with $t=a$. Taking the path $[x+1] \\rightarrow [-x] \\rightarrow [x-1] \\rightarrow [-x] \\right... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0hne | Problem:
Let $ABC$ be an equilateral triangle, and let $P$ be a point on minor $\operatorname{arc} BC$ of the circumcircle of $ABC$. Prove that $PA = PB + PC$. | [
"Solution:\n\nExtend line $PC$ through $C$ to point $D$ such that $CD = BP$. Note that $\\angle ACD = \\pi - \\angle PCA = \\angle ABP$ (since quadrilateral $ABPC$ is cyclic), and $AC = AB$ since $\\triangle ABC$ is equilateral. Consequently, $\\triangle ACD \\cong \\triangle ABP$. In particular, we have $\\angle P... | United States | Berkeley Math Circle | [
"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"
] | null | proof only | null | |
0hbo | A right triangle $ABC$ is called *special*, if the lengths of its sides $AB$, $BC$ and $CA$ are integers, and there is a point $X$ on each side (different from the vertices of $\triangle ABC$), for which lengths of $AX$, $BX$ and $CX$ are integers. Find at least one special triangle.
**Fig. 1... | [
"Let triangle $\\triangle ABC$ have a right angle at $C$. Then for any triangle with even integer hypotenuse, the midpoint of hypotenuse is the point of interest (Fig. 11).\n\nLet $BC = an$, $AC = bn$, $AB = cn$, $n \\in \\mathbb{N}$ and $a^2 + b^2 = c^2$. Let $K \\in AC$, $N \\in BC$, $CN = bk$, $CK = ak$, $k \\in... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | A special triangle is one with side lengths 36, 48, and 60. | |
0ijc | Problem:
2006 vertices of a regular 2007-gon are red. The remaining vertex is green. Let $G$ be the total number of polygons whose one vertex is green and the others are red. Denote by $R$ the number of polygons whose all vertices are red. Which number is bigger, $R$ or $G$? Explain your answer. | [
"Solution:\n\nWe will prove that $G \\geq R$. For each polygon $\\mathcal{P}$ with all red vertices we can correspond a polygon with one green vertex (namely we can add the green vertex to the set of vertices of $\\mathcal{P}$). Thus $G \\geq R$. However, $G > R$ since the triangles with one green vertex can't be c... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | G > R | |
0k6h | Problem:
Five people are at a party. Each pair of them are friends, enemies, or frenemies (which is equivalent to being both friends and enemies). It is known that given any three people $A$, $B$, $C$:
- If $A$ and $B$ are friends and $B$ and $C$ are friends, then $A$ and $C$ are friends;
- If $A$ and $B$ are enemies a... | [
"Solution:\nIf $A$ and $B$ are frenemies, then regardless of whether another person $C$ is friends or enemies with $A$, $C$ will have to be frenemies with $B$ and vice versa. Therefore, if there is one pair of frenemies then all of them are frenemies with each other, and there is only one possibility.\n\nIf there a... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 17 | |
0hdn | Let $\triangle ABC$ be an acute and non-isosceles triangle. Its angle bisectors $AL_1$ and $BL_2$ intersect at the point $I$. Points $D$ and $E$ are chosen on the segments $AL_1$ and $BL_2$ in such a way that $\angle DBC = \frac{1}{2}\angle A$ and $\angle EAC = \frac{1}{2}\angle B$. The lines $AE$ and $BD$ intersect at... | [
"Denote $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$. Let $N$ be the midpoint of arc $ACB$ of the circumcircle of $\\triangle ABC$ (Fig. 35). We are going to show that $N$ belongs to the line $DE$.\n\n\n\nFirstly, observe that both points $D$ and $E$ lie inside $... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocent... | null | proof only | null | |
0329 | Problem:
Let $n$ be a positive integer. Find all positive integers $m$, for which there exists a polynomial $f(x) = a_{0} + a_{1} x + \cdots + a_{n} x^{n} \in \mathbb{Z}[x],\ a_{n} \neq 0$, such that $\gcd(a_{0}, a_{1}, \ldots, a_{n}, m) = 1$ and $f(k)$ divides $m$ for any integer $k$. | [
"Solution:\n1. We shall use the following well-known facts.\n\nLEMMA 1. For any integer $x$ and any positive integer $t$ the number $t!$ divides $(x+1)(x+2) \\ldots (x+t)$.\n\nProof. The statement is obvious for $x \\in \\{0, -1, -2, \\ldots, -t\\}$. We have to prove it for $x > 0$. Let $p$ be a prime divisor of $t... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All positive integers m that divide n! | |
082l | Problem:
Prendiamo un intero positivo $n$, facciamo la somma delle sue cifre e poi addizioniamo nuovamente le cifre di tale somma ottenendo un intero $S$. Qual è il più piccolo $n$ che permette di ottenere $S \geq 10 ?$ | [
"Solution:\n\nLa risposta è 199. Infatti se $n$ avesse 1 o 2 cifre, $n \\leq 99$, e allora la somma delle sue cifre sarebbe $\\leq 18$ e la somma delle cifre del numero pari alla somma delle sue cifre sarebbe $\\leq 9$, dunque $n$ ha almeno tre cifre. Osserviamo che per $n \\leq 198$, ancora la somma delle cifre de... | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO BIENNIO | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 199 | |
05gm | Problem:
Soit $n \geqslant 1$ un entier. On suppose qu'il existe exactement 2005 couples $(x, y)$ d'entiers naturels tels que $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$. Montrer que $n$ est le carré d'un entier.
N.B. Si $x \neq y$ alors $(x, y) \neq (y, x)$. | [
"Solution:\n\nOn remarque que $x > n$ et $y > n$.\nL'équation s'écrit $x y = n(x + y)$, ou encore $n^{2} = (x - n)(y - n)$. On en déduit qu'il y a exactement 2005 couples d'entiers naturels $(u, v)$ tels que $n^{2} = u v$.\nSi $n^{2} = u v$ alors $u$ est un diviseur de $n^{2}$. Réciproquement, si $u$ est un diviseu... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0b0a | Problem:
Find the largest real number $x$ such that $\sqrt[3]{x} + \sqrt[3]{4-x} = 1$. | [
"Solution:\nCubing both sides of the given equation yields\n$$\nx + 3 \\sqrt[3]{x(4-x)} (\\sqrt[3]{x} + \\sqrt[3]{4-x}) + 4 - x = 1,\n$$\nwhich then becomes\n$$\n4 + 3 \\sqrt[3]{x(4-x)} = 1\n$$\nor\n$$\n\\sqrt[3]{x(4-x)} = -1.\n$$\nCubing both sides of this equation gives\n$$\nx(4-x) = -1\n$$\nor\n$$\nx^2 - 4x = 1.... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2 + sqrt(5) | |
0jff | Let $N$ be the set of positive integers. Let $f: \mathbb{N} \to \mathbb{N}$ be a function satisfying the following two conditions:
a. $f(m)$ and $f(n)$ are relatively prime whenever $m$ and $n$ are relatively prime.
b. $n \le f(n) \le n + 2012$ for all $n$.
Prove that for any natural number $n$ and any prime $p$, if... | [
"Let $g(n)$ be the smallest prime factor of $f(n)$. (Since $f(n) \\ge n$, $f(n)$ must have a prime factor unless $n=1$ and $f(1)=1$. In this case, we define $g(1)=1$.) First, we show that for any prime $p$ and any $k \\ge 1$, $f(p^k)$ is a power of $g(p)$.\n\nSuppose for the sake of contradiction that $f(p^k)$ is n... | United States | TSTST | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
04m2 | In some archipelago, there are $2017$ islands labelled $1, 2, \ldots, 2017$. Two agencies, Red dragon and Blue eye, are planning to establish ship lines between certain islands. Exactly one agency will operate between each two islands, with a ship line from the island labelled with a smaller number to the island labell... | [
"Ship lines maintained by Red dragon we will call red lines, and those maintained by Blue eye blue lines.\n\nLet $a_n$ denote the number of good arrangements for an archipelago with $n$ islands. Consider an archipelago with $n + 1$ islands and notice island $1$. Let $A$ be the set of all islands which are connected... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2017! | |
0fpb | Sea $p$ un número primo positivo dado. Demostrar que existe un entero $\alpha$ tal que $\alpha(\alpha - 1) + 3$ es divisible por $p$ si y sólo si existe un entero $\beta$ tal que $\beta(\beta - 1) + 25$ es divisible por $p$. | [
"Sean $f(x) = x^2 - x + 3$, $g(y) = y^2 - y + 25$. Para los enteros $\\alpha$ y $\\beta$, $f(\\alpha)$ y $g(\\beta)$ son ambos enteros impares. Por tanto $p \\neq 2$. Con $\\alpha = 2$ y $\\beta = 3$, $f(\\alpha) = 9$ y $g(\\beta) = 27$, son múltiplos de $3$. De este modo para $p = 3$ se cumple el enunciado. Es cla... | Spain | LII Olimpiada Matemática Española | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Spanish | proof only | null | |
08tj | When you represent the following fraction in a decimal form, what is the integral part of the resulting expression?
$$
\frac{9 + 98 + 987 + 9876 + 98765 + 987654 + 9876543 + 98765432}{12345678 + 1234567 + 123456 + 12345 + 1234 + 123 + 12 + 1}
$$ | [
"[8].\nConsider pairs of numbers $(9, 1)$, $(98, 12)$, ..., $(98765432, 12345678)$, which are formed by pairing the first number of the numerator with the last number of the denominator, second number of the numerator with the next-to-the last number of the denominator and so on and the last number of the numerator... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 8 | |
010b | Problem:
Is it possible to cover a $13 \times 13$ chessboard with forty-two tiles of size $4 \times 1$ so that only the central square of the chessboard remains uncovered? (It is assumed that each tile covers exactly four squares of the chessboard, and the tiles do not overlap.) | [
"Solution:\n\nAnswer: no.\n\nLabel the horizontal rows by integers from $1$ to $13$. Assume that the tiling is possible, and let $a_{i}$ be the number of vertical tiles with their outer squares in rows $i$ and $i+3$. Then $b_{i} = a_{i} + a_{i-1} + a_{i-2} + a_{i-3}$ is the number of vertical tiles intersecting row... | Baltic Way | Baltic Way 1998 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | no | |
08tu | Suppose for a tetrahedron $OABC$, $OA = 3$, $OB = 4$, $OC = 5$, where by $XY$ we denote the length of the line segment $XY$. Suppose also that we have $\angle AOB = \angle AOC = 45^\circ$, and $\angle BOC = 60^\circ$. Determine the volume of the tetrahedron $OABC$. | [
"Take points $P$, $Q$, $R$ on the line segments $OA$, $OB$, $OC$, respectively, in such a way that the conditions $OP = 1$, $\\angle OPQ = \\angle OPR = 90^\\circ$ are satisfied. Since $\\angle POQ = 45^\\circ$, we have $PQ = 1$ and $OQ = \\sqrt{2}$. Similarly, we get $PR = 1$ and $OR = \\sqrt{2}$. From $\\angle RO... | Japan | Japan Mathematical Olympiad First Round | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 5 | |
04wy | Let $k$ be the circumcircle of a given convex quadrilateral $ABCD$ with the property that the half-lines $DA$ and $CB$ meet at a point $E$ for which $|CD|^2 = |AD| \cdot |ED|$ holds. Let us denote by $F$ ($F \neq A$) the point of intersection of the circle $k$ with the perpendicular to $ED$ at $A$. Prove that the segme... | [
"Clearly $DF$ is a diameter of $k$. First we show that under the given conditions the vertex $C$ cannot lie in the half-plane $DFA$.\nIf the vertices $B$, $C$ are points on the subarc $DA$ of the arc $DAF$ (Fig. 1) then the angles $DCB$ and $DBA$ are obtuse, hence $|DC| < |DB| < |DA| < |DE|$, which contradicts to t... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08xi | Consider an operation of putting an integer greater than or equal to $1$ and less than or equal to $6$ to each square in a grid of $6 \times 6$ squares. For a pair of integers $(i, j)$ with $1 \le i, j \le 6$, denote by $i \diamond j$ the integer put into the square located on $i$-th row and $j$-th column after an oper... | [
"First, we consider how we can obtain an operation that satisfies the two conditions of the problem. Let us begin with the following Lemma:\n**Lemma 1:** For every pair of integers $i$ and $j$ with $1 \\le i, j \\le 6$, $i \\diamond j$ is the only integer which appears in both $i$-th row and $j$-th column.\n**Proof... | Japan | Japan Mathematical Olympiad Initial Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Functional equations"
] | English | proof and answer | 122 | |
03dc | Let $n$ be a positive integer. We call a graph $G$ $n$-good if among any $n$ vertices of $G$ there exist two vertices connected by an edge. Find the least positive integer $N$ such that for any $n$-good connected graph $G$ with $N$ vertices there exists a cycle $C$ with the property:
After deleting all edges of the cyc... | [
"We prove first that $N \\ge 3n-2$. Consider a graph with $3n-3$ vertices consisting of $n-1$ triangles with vertices $(u_i, v_i, w_i)$ for $i=1, 2, \\dots, n-1$ and the path $v_1v_2v_3 \\dots v_{n-1}$. It is easy to see that the graph is $n$-good and by deleting any cycle the graph is not connected.\n\nLet $G$ be ... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3n-2 | |
04u7 | Let $x, y, z$ be real numbers such that
$$
\frac{1}{|x^2 + 2yz|}, \quad \frac{1}{|y^2 + 2zx|}, \quad \frac{1}{|z^2 + 2xy|}
$$
are side-lengths of a (non-degenerate) triangle. Find all possible values of $xy + yz + zx$.
(Michal Rolínek) | [
"If $x = y = z = t > 0$ then the three fractions are sides of an equilateral triangle and $xy + yz + zx = 3t^2$, hence $xy + yz + zx$ can attain all positive values. Similarly, for $x = y = t > 0$ and $z = -2t$ the three fractions are $\\frac{1}{3}t^{-2}$, $\\frac{1}{3}t^{-2}$, $\\frac{1}{6}t^{-2}$ which are positi... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | all real numbers except zero | |
0gee | 令 $\|x\|_* = (|x| + |x - 1| - 1)/2$。請決定所有的函數 $f: \mathbb{N} \to \mathbb{N}$ 使得
$$
f^{(\|f(x)-x\|_*)}(x) = x, \forall x \in \mathbb{N}.
$$
其中 $f^{(0)}(x) = x$, $f^{(n)}(x) = f(f^{(n-1)}(x))$, $\forall n \in \mathbb{N}$。
Let $\|x\|_* = (|x| + |x - 1| - 1)/2$. Find all $f : \mathbb{N} \to \mathbb{N}$ such that
$$
f^{(\|f... | [
"All the solutions are $f: \\mathbb{N} \\to \\mathbb{N}$ such that $f(x) \\in \\{x, x+1\\}$, $\\forall x \\in \\mathbb{N}$. Note that in this case, we have $\\|f(x) - x\\|_* = 0$ so they are clearly solutions.\n\nNow we show that these are all the solutions. In fact, if there exists an $x \\in \\mathbb{N}$ such tha... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 國際競賽實作 (二) | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | All functions f with f(x) in {x, x+1} for every natural number x. | |
01k5 | Let $r$ be some positive real number. It is known that for some positive integer $n$ the following condition holds: all positive real numbers $a_1, \dots, a_n$ satisfying the equality $a_1 + \dots + a_n = r\left(\frac{1}{a_1} + \dots + \frac{1}{a_n}\right)$, $a_i \ne \sqrt{r}$, $i = 1, \dots, n$, also satisfy the equal... | [
"Note that if $a_1, \\dots, a_n$ satisfy the equality\n$$\na_1 + \\dots + a_n = r \\left( \\frac{1}{a_1} + \\dots + \\frac{1}{a_n} \\right), \\quad (*)\n$$\nthen the numbers $b_1 = \\frac{r}{a_1}, \\dots, b_n = \\frac{r}{a_n}$ satisfy the equality\n$$\nb_1 + \\dots + b_n = r \\left( \\frac{1}{b_1} + \\dots + \\frac... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | n = 2 | |
074y | Problem:
Define a sequence $\langle a_{n}\rangle_{n \geq 0}$ by $a_{0}=0$, $a_{1}=1$ and
$$
a_{n}=2 a_{n-1}+a_{n-2}
$$
for $n \geq 2$.
a. For every $m>0$ and $0 \leq j \leq m$, prove that $2 a_{m}$ divides $a_{m+j}+(-1)^{j} a_{m-j}$.
b. Suppose $2^{k}$ divides $n$ for some natural numbers $n$ and $k$. Prove that $2^... | [
"Solution:\n\na.\nConsider $f(j)=a_{m+j}+(-1)^{j} a_{m-j}$, $0 \\leq j \\leq m$, where $m$ is a natural number. We observe that $f(0)=2 a_{m}$ is divisible by $2 a_{m}$. Similarly,\n$$\nf(1)=a_{m+1}-a_{m-1}=2 a_{m}\n$$\nis also divisible by $2 a_{m}$. Assume that $2 a_{m}$ divides $f(j)$ for all $0 \\leq j<l$, wher... | India | INMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0d8g | Solve the following equation in positive integers $x$, $y$:
$$
x^{2017}-1=(x-1)(y^{2015}-1).
$$ | [
"It is clear that $x=1$ is a solution of the problem (with $y$ any positive integer).\n\nWe now show that for $x>1$, the given equation has no solution. In fact, suppose that there are $x>1$ and $y$ are positive integers satisfying the equation. Then, one has $y>1$ and\n$$\nx^{2016}+x^{2015}+\\cdots+x+1 = y^{2015}-... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | x = 1 and y is any positive integer | |
0a8d | Problem:
In the triangle $ABC$, the bisector of angle $B$ meets $AC$ at $D$ and the bisector of angle $C$ meets $AB$ at $E$. The bisectors meet each other at $O$. Furthermore, $OD = OE$. Prove that either $ABC$ is isosceles or $\angle BAC = 60^{\circ}$. | [
"Solution:\n\n(See Figure 11.) Consider the triangles $AOE$ and $AOD$. They have two equal pairs of sides and the angles facing one of these pairs are equal. Then either $AOE$ and $AOD$ are congruent or $\\angle AEO = 180^{\\circ} - \\angle ADO$.\n\nIn the first case, $\\angle BEO = \\angle CDO$, and\n\n![](attache... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 14 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08b5 | Problem:
Determinare tutte le coppie di numeri interi $(a, b)$ che risolvono l'equazione $a^{3}+b^{3}+3 a b=1$. | [
"Solution:\n\nL'equazione è simmetrica in $a$ e $b$, quindi possiamo limitarci a considerare le soluzioni con $a \\geq b$. Non ci sono soluzioni con $a$ e $b$ entrambi positivi: infatti in tal caso avremmo $a^{3}+b^{3}+3 a b \\geq 1+1+3=5>1$, assurdo.\n\nSe almeno uno tra $a$ e $b$ è uguale a zero, diciamo $b$, all... | Italy | XXXI Olimpiade Italiana di Matematica | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | {(a,b) in Z^2 : a + b = 1} union {(-1, -1)} | |
0amx | Problem:
Find the maximum value of $y = (7 - x)^{4} (2 + x)^{5}$ when $x$ lies strictly between $-2$ and $7$.
(a) $7^{4} 2^{5}$
(b) $(4.5)^{4} (2.5)^{5}$
(c) $(2.5)^{9}$
(d) $(4.5)^{9}$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | MCQ | null | |
0jk9 | Problem:
Suppose that $(a_{1}, \ldots, a_{20})$ and $(b_{1}, \ldots, b_{20})$ are two sequences of integers such that the sequence $(a_{1}, \ldots, a_{20}, b_{1}, \ldots, b_{20})$ contains each of the numbers $1, \ldots, 40$ exactly once. What is the maximum possible value of the sum
$$
\sum_{i=1}^{20} \sum_{j=1}^{20}... | [
"Solution:\nLet $x_{k}$, for $1 \\leq k \\leq 40$, be the number of integers $i$ with $1 \\leq i \\leq 20$ such that $a_{i} \\geq k$. Let $y_{k}$, for $1 \\leq k \\leq 40$, be the number of integers $j$ with $1 \\leq j \\leq 20$ such that $b_{j} \\geq k$. It follows from the problem statement that $x_{k}+y_{k}$ is ... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 5530 | |
0brq | In every unit square of an $n \times n$ board a positive integer is written. A move consists of choosing a $2 \times 2$ square and adding 1 to exactly three of the four numbers written in the chosen square. We call a positive integer *good* if starting from any initial numbers, there is a sequence of moves that makes a... | [
"a. Let us notice that the sum of the 36 numbers written on the board is invariant modulo 3. When all the numbers on the board are equal, their sum is a multiple of 36, hence a multiple of 3. But this requires that the initial sum is also a multiple of 3. In conclusion, if the sum of the initial numbers is not a mu... | Romania | 67th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
02rg | Let $ABC$ be a triangle, $M$ be the midpoint of side $AC$ and $N$ be the midpoint of side $AB$. Let $r$ and $s$ be the reflections of lines $BM$ and $CN$ across line $BC$, respectively. Lines $r$ and $s$ meet line $MN$ at points $D$ and $E$, respectively. The circumcircles of $BDM$ and $CEN$ meet at points $X$ and $Y$,... | [
"\nFirst, notice that $MN \\parallel BC$, so $\\frac{WD}{DB} = \\frac{EW}{EC}$. Applying Ceva's theorem to triangle $BCW$ and cevians $WP$, $CD$ and $BE$ we have $\\frac{WD}{DB} \\cdot \\frac{BP}{PC} \\cdot \\frac{CE}{EW} = 1 \\iff BP = PC$, so $WZ$ meets $BC$ at its midpoint $P$.\n\nNow it... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0apv | Problem:
The graphs of $x^{2}+y=12$ and $x+y=12$ intersect at two points. What is the distance between these points? | [
"Solution:\nSolving the following system of equations using the elimination method:\n$$\n\\left\\{\\begin{array}{r}\nx^{2}+y=12 \\\\\nx+y=12\n\\end{array}\\right.\n$$\nwe get the ordered pairs $(0,12)$ and $(1,11)$. That is, the given graphs intersect at $(0,12)$ and $(1,11)$, whose distance is $\\sqrt{(0-1)^{2}+(1... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | sqrt(2) | |
0crr | Each of positive rational numbers $a$ and $b$ has the minimal period of length 30 in the decimal notation. A number $a-b$ has the minimal period of length 15 in a decimal notation. Find the least positive integer $k$ for which it may happen that the length of the minimal period of the number $a+kb$ in the decimal notat... | [
"Домножив, если нужно, числа $a$ и $b$ на подходящую степень десятки, мы можем считать, что десятичные записи чисел $a, b, a-b$ и $a+kb$ — чисто периодические (то есть периоды начинаются сразу после запятой).\n\nВоспользуемся следующим известным фактом: десятичная запись рационального числа $r$ — чисто периодическа... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6 | |
0ke4 | Problem:
Marisa has two identical cubical dice labeled with the numbers $\{1,2,3,4,5,6\}$. However, the two dice are not fair, meaning that they can land on each face with different probability. Marisa rolls the two dice and calculates their sum. Given that the sum is 2 with probability 0.04, and 12 with probability 0... | [
"Solution:\n\nLet $p_{i}$ be the probability that the die lands on the number $i$. The problem gives that $p_{1}^{2}=0.04$, $p_{6}^{2}=0.01$, so we have\n$$\np_{1}=0.2, \\quad p_{6}=0.1, \\quad p_{2}+p_{3}+p_{4}+p_{5}=0.7\n$$\nWe are asked to maximize\n$$\n2\\left(p_{1} p_{6}+p_{2} p_{5}+p_{3} p_{4}\\right)=2\\left... | United States | HMMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | final answer only | 28 | |
03pq | A convex hexagon is given in which any two opposite sides have the following property: the distance between their midpoints is $\frac{\sqrt{3}}{2}$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.
(A convex hexagon $ABCDEF$ has three pairs of opposite sides: $AB$ and $DE$, $BC$ and $E... | [
"**Proof I** We first prove the following lemma.\nLemma Consider a triangle $PQR$ with $\\angle QPR \\ge 60^\\circ$. Let $L$ be the midpoint of $QR$. Then $PL \\le \\frac{\\sqrt{3}}{2} QR$. The equality holds if and only if the triangle $PQR$ is equilateral.\nProof of the lemma\nLet $S$ be the point such that the t... | China | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0j2q | Problem:
Let $ABCD$ be an isosceles trapezoid such that $AB = 10$, $BC = 15$, $CD = 28$, and $DA = 15$. There is a point $E$ such that $\triangle AED$ and $\triangle AEB$ have the same area and such that $EC$ is minimal. Find $EC$. | [
"Solution:\n\nAnswer:\n\n| $\\frac{216}{\\sqrt{145}}$ |\n| :---: |\n\n\nThe locus of points $E$ such that $[AED] = [AEB]$ forms a line, since area is a linear function of the coordinates of $E$; setting the areas equal gives a linear equation in the coordinates $E$. Note that $A$ and $M$, t... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 216/√145 | |
03hw | Problem:
If $A, B, C, D$ are four points in space, such that
$$
\angle A B C=\angle B C D=\angle C D A=\angle D A B=\pi / 2
$$
prove that $A, B, C, D$ lie in a plane. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j38 | Problem:
You are standing in an infinitely long hallway with sides given by the lines $x=0$ and $x=6$. You start at $(3,0)$ and want to get to $(3,6)$. Furthermore, at each instant you want your distance to $(3,6)$ to either decrease or stay the same. What is the area of the set of points that you could pass through o... | [
"\n\nIf you draw concentric circles around the destination point, the condition is equivalent to the restriction that you must always go inwards towards the destination. In the diagram above, the regions through which you might pass are shaded.\nWe find the areas of regions A, B, and C sepa... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 9√3 + 21π/2 | |
01i6 | Assume that $ABCD$ is a cyclic quadrilateral with circumcircle $\Omega$. Assume lines $AB$ and $CD$ intersect at point $P$ and lines $AD$ and $BC$ intersect at $Q$. Let $\Gamma$ be the circumcircle of triangle $APQ$. Then $\Omega$ and $\Gamma$ intersect in two points, $A$ is one of them and $R$ is the other. Assume $C ... | [
"**Solution 1.** Let $X$ be a point on $BC$ such that $LX \\perp AB$, as seen in figure 18. It is enough to prove that\n$$\n\\frac{DP}{PL} = \\frac{DK}{KX}\n$$\nbecause then $PK \\parallel LX$ and $LX \\perp AB$.\nApplying Menelaos for triangle *BDL* and transversal *MPC* we get\n$$\n\\frac{DP}{PL} \\cdot \\frac{LM... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous... | null | proof only | null | |
013n | Problem:
A positive integer is written on each of the six faces of a cube. For each vertex of the cube we compute the product of the numbers on the three adjacent faces. The sum of these products is $1001$. What is the sum of the six numbers on the faces? | [
"Solution:\n\nLet the numbers on the faces be $a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$, placed so that $a_{1}$ and $a_{2}$ are on opposite faces etc. Then the sum of the eight products is equal to\n$$\n\\left(a_{1}+a_{2}\\right)\\left(b_{1}+b_{2}\\right)\\left(c_{1}+c_{2}\\right) = 1001 = 7 \\cdot 11 \\cdot 13.\n... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 31 | |
0azi | Problem:
Given the square $ABCD$ with side length of $2$, triangle $AEF$ is constructed so that $AE$ bisects side $BC$, and $AF$ bisects side $CD$. Moreover, $EF$ is parallel to the diagonal $BD$ and passes through $C$. Find the area of triangle $AEF$.
 | [
"Solution:\nRefer to the figure below.\n\n\n\nObserve that $\\triangle GEC \\cong \\triangle BIG \\cong \\triangle FCH \\cong \\triangle HJD$.\nMoreover, $BI = IJ = JD$. This is because $IO = \\frac{1}{2} BI$ and $OJ = \\frac{1}{2} JD$ (the points $I$ and $J$ are centroids).\nTherefore:\n$$... | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 8/3 | |
0abu | A passenger wanted to know the speed of the bus in which he was traveling, so he looked through the window and saw on a road sign (denoting the distance from the town from where he started his traveling) two-digit number. After one hour drive he saw on another road sign a three-digit number written with the same two di... | [
"On the first road sign the passenger saw the number $10x + y$, $0 < x \\le 9$, $0 \\le y \\le 9$. One hour later he saw the number $100y + x$, $y \\ne 0$. During that time the bus has driven $(100y + x) - (10x + y) = 9(11y - x)\\text{ km}$. The bus had constant speed so in the next two hours it has driven $2 \\cdo... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Speed: 45 km/h; road signs: 61, 106, 196 | |
026c | Problem:
Um trapézio - No trapézio da figura abaixo $AB$ é paralelo a $CD$, $AD = AB = BC = 1~\mathrm{cm}$ e $DC = 2~\mathrm{cm}$. Quanto mede o ângulo $C \hat{A} D$?
(A) $30^\circ$
(B) $45^\circ$
(C) $60^\circ$
(D) $90^\circ$
(E) $120^\circ$
 | [
"Solution:\n\nA resposta correta é (D).\nSeja $P$ o ponto médio do segmento $CD$ e traçemos os segmentos $AP$ e $BP$. Os três triângulos formados $\\triangle ADP$, $\\triangle ABP$ e $\\triangle BCP$ são equiláteros (por quê?). Então, os ângulos $D \\widehat{A} P = 60^\\circ = P \\widehat{A} B$. Como o segmento $AC... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | MCQ | D | |
0kyf | Problem:
Compute the sum of all integers $n$ such that $n^{2}-3000$ is a perfect square. | [
"Solution:\n\nIf $n^{2}-3000$ is a square, then $(-n)^{2}-3000$ is also a square, so the sum is 0."
] | United States | HMMT February 2024 Guts Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 0 | |
0e0b | Let $p$ be a prime number and let $a$, $b$ and $c$ be integers divisible by $p$, such that the polynomial $q(x) = x^3 + a x^2 + b x + c$ has at least two different integer roots. Show that $p^2$ divides $b$ and $p^3$ divides $c$. | [
"Let $y$ and $z$ be two different integer roots of $q$. Then $y^3 + a y^2 + b y + c = 0$ and $z^3 + a z^2 + b z + c = 0$. We know that $p$ divides $a$, $b$ and $c$. Since $y^3 = -c - b y - a y^2$, $p$ divides $y^3$. Similarly, we show that $p$ divides $z^3$. As $p$ is a prime, it must divide $y$ and $z$.\n\nBy subt... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0kza | Problem:
Mark has a cursed six-sided die that never rolls the same number twice in a row, and all other outcomes are equally likely. Compute the expected number of rolls it takes for Mark to roll every number at least once. | [
"Solution:\n\nSuppose Mark has already rolled $n$ unique numbers, where $1 \\leq n \\leq 5$. On the next roll, there are $5$ possible numbers he could get, with $6-n$ of them being new. Therefore, the probability of getting another unique number is $\\frac{6-n}{5}$, so the expected number of rolls before getting an... | United States | HMMT February 2024 Guts Round | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 149/12 | |
03l5 | Problem:
Let $ABCD$ be a convex quadrilateral with
$$
\begin{aligned}
\angle CBD &= 2 \angle ADB, \\
\angle ABD &= 2 \angle CDB \\
\text{and}\quad AB &= CB.
\end{aligned}
$$
Prove that $AD = CD$. | [
"Solution:\nExtend $DB$ to a point $P$ on the circle through $A$ and $C$ centered at $B$. Then $\\angle CPD = \\frac{1}{2} \\angle CBD = \\angle ADB$ and $\\angle APD = \\frac{1}{2} \\angle ABD = \\angle CDB$,\nso $APCD$ is a parallelogram. Now $PD$ bisects $AC$ so $BD$ is an angle bisector of isosceles triangle $A... | Canada | Canadian Mathematics Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0c5w | Determine the triples $(m, n, p)$, of integers not greater than $5$, such that $10$ is a divisor of $A = 2^m + 3^n + 5^p$. | [] | Romania | 70th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | All triples with m ≥ 1 and 0 ≤ n,p ≤ 5 satisfying:
- For p ∈ {1,2,3,4,5}:
(m,n,p) ∈ {(1,1,p), (1,5,p), (2,0,p), (2,4,p), (3,3,p), (4,2,p), (5,1,p), (5,5,p)}.
- For p = 0:
(m,n,p) ∈ {(1,3,0), (3,0,0), (3,4,0), (4,1,0), (4,5,0), (5,3,0)}. | |
071u | Problem:
Let $\mathbf{R}$ denote the set of all real numbers. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfying the condition
$$
f(x+y)=f(x) f(y) f(x y)
$$
for all $x, y$ in $\mathbf{R}$. | [
"Solution:\nPutting $x=0, y=0$, we get $f(0)=f(0)^3$ so that $f(0)=0, 1$ or $-1$.\nIf $f(0)=0$, then taking $y=0$ in the given equation, we obtain $f(x)=f(x) f(0)^2=0$ for all $x$.\n\nSuppose $f(0)=1$. Taking $y=-x$, we obtain\n$$\n1=f(0)=f(x-x)=f(x) f(-x) f\\left(-x^2\\right)\n$$\nThis shows that $f(x) \\neq 0$ fo... | India | INMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all real x; f(x) = 1 for all real x; f(x) = -1 for all real x | |
0bp4 | Problem:
Se consideră paralelogramul $ABCD$, ale cărui diagonale se intersectează în $O$. Bisectoarele unghiurilor $DAC$ şi $DBC$ se intersectează în $T$. Se ştie că $\overrightarrow{TD} + \overrightarrow{TC} = \overrightarrow{TO}$. Determinaţi măsurile unghiurilor triunghiului $ABT$. | [
"Solution:\n\nDin ipoteză rezultă că $DOCT$ este paralelogram.\n\nDin $AO \\parallel DT$ deducem $\\angle DTA \\equiv \\angle OAT \\equiv \\angle DAT$, deci $DA = DT$.\n\nAstfel $DA = DT = OC$; analog $BC = CT = OD$, de unde $BD = AC$. Astfel $ABCD$ este dreptunghi, $AOTD$ este romb, triunghiul $AOD$ este echilater... | Romania | Olimpiada Naţională de Matematică | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60°, 60°, 60° | |
0b6o | Let $a$, $b$, $c$ be given positive integers. Prove there exists some positive integer $N$ such that
$$
\begin{aligned}
a &\mid Nbc + b + c \\
b &\mid Nca + c + a
\end{aligned}
$$ | [
"The necessity of having $x$, $y$, $z$ be pairwise co-prime is proved by, say, assuming $\\gcd(x, y) > 1$.\nThen $a \\mid Nbc + b + c$ becomes $x \\mid dNyz + y + z$, and so we must have $\\gcd(x, y) \\mid z$, absurd, since under this assumption it then follows $\\gcd(x, y) \\mid \\gcd(x, y, z) = 1$.\n\nOn the othe... | Romania | 2010 Fourth STARS OF MATHEMATICS COMPETITION | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
05cr | In an acute triangle $ABC$, the extension of the altitude $AD$ over $D$ intersects the circumcircle at $E$. The midpoint of $CE$ is $F$. The circumcircles of $ABC$ and $DEF$ intersect at $G \neq E$. The foot of the altitude drawn from $A$ to $FG$ is $P$. Prove that $DA = DP$. | [
"We first show that $\\angle BGF = 90^\\circ$ (Fig. 17). For this we notice that $F$ as the midpoint of the hypothenuse in $CDE$ is also its circumcenter, so $FC = FD$. Together with $BEGC$ being cyclic, we get\n$$\n\\angle EGB = \\angle ECB = \\angle FCD = \\angle FDC.\n$$\nThus\n$$\n\\begin{align*}\n\\angle BGF &... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01re | Let $x = \sqrt{ab}$ and $y = \sqrt{\frac{a^2 + b^2}{2}}$.
Compare the arithmetic mean of positive numbers $a$ and $b$ with the arithmetic mean of $x$ and $y$. | [
"Answer: $\\frac{a+b}{2} \\ge \\frac{x+y}{2}$.\nWe prove that\n$$\n\\frac{a+b}{2} \\ge \\frac{x+y}{2}.\n$$\n(*)\nIndeed,\n$$\n\\frac{a+b}{2} \\ge \\frac{x+y}{2} \\Leftrightarrow a+b \\ge \\sqrt{ab} + \\sqrt{\\frac{a^2+b^2}{2}} \\Leftrightarrow\n$$\n$$\n\\Leftrightarrow (a+b)^2 \\ge ab + \\frac{a^2+b^2}{2} + 2\\sqrt... | Belarus | Final Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | (a + b)/2 ≥ (x + y)/2, with equality if and only if a = b | |
09ai | Let $a$, $b$, $c$ be distinct positive real numbers. Show that
$$
\frac{(b-c)^4}{(a-b)^2(a-c)^2} + \frac{(a-c)^4}{(a-b)^2(b-c)^2} + \frac{(a-b)^4}{(a-c)^2(b-c)^2} \ge \frac{33}{2}.
$$ | [
"Let us assume that $a > b > c$. $a - c = (a - b) + (b - c) \\ge 2\\sqrt{(a - b)(b - c)}$. Then\n$$\n\\frac{(a - c)^4}{(a - b)^2 (b - c)^2} \\ge 16\n$$\nholds. Now it suffices to prove that\n$$\n\\frac{(a-b)^4}{(b-c)^2} + \\frac{(b-c)^4}{(a-b)^2} \\geq \\frac{(c-a)^2}{2}\n$$\nBy Cauchy-Schwarz inequality\n$$\n\\fra... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0emf | For each positive integer $a$ we consider the sequence $(a_n)$ with $a_0 = a$ and $a_n = a_{n-1} + 40^n$ for $n > 0$. Prove that every such sequence contains infinitely many numbers that are divisible by $2009$. | [
"Since $\\gcd(40, 2009) = 1$, we have $40^{k \\cdot \\varphi(2009)} \\equiv 1 \\pmod{2009}$ for all natural numbers $k$. For $n > \\varphi(2009)$ the exponent $n!$ is certainly a multiple of $\\varphi(2009)$, and therefore $a_{n+1} \\equiv a_n + 1 \\pmod{2009}$. This means that all values modulo $2009$ are taken cy... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
094c | Problem:
Let $\mathbb{R}$ denote the set of all real numbers. For each pair $(\alpha, \beta)$ of nonnegative real numbers subject to $\alpha+\beta \geq 2$, determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$
f(x) f(y) \leq f(x y)+\alpha x+\beta y
$$
for all real numbers $x$ and $y$. | [
"Solution:\nWe know $f(x) f(y) \\leq f(x y)+\\alpha x+\\beta y$ and by exchanging $x$ and $y$ we get $f(x) f(y) \\leq f(x y)+\\beta x+\\alpha y$. Combining the two we get\n$$\nf(x) f(y) \\leq f(x y)+\\gamma x+\\gamma y\n$$\nwhere $\\gamma=\\frac{\\alpha+\\beta}{2}$. Notice that $\\gamma \\geq 1$.\n\nSetting $x=y=-1... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | The only solution occurs when alpha equals beta equals one, in which case f(x) equals x plus one. For all other parameter pairs with nonnegative values summing to at least two, there is no function satisfying the inequality. | |
0eyl | Problem:
Which is greater: $31^{11}$ or $17^{14}$? [No calculators allowed!] | [
"Solution:\n\n$17^{2} = 289 > 9 \\cdot 31$. So $17^{14} > 9^{7} 31^{7}$. But $3^{7} = 2187 > 31^{2}$. Hence $17^{14} > 31^{11}$."
] | Soviet Union | 2nd ASU | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 17^14 | |
01ys | The incircle of the right triangle $ABC$ is tangent to the hypotenuse $AB$ at point $P$ and is tangent to the legs $AC$ and $BC$ at points $Q$ and $R$ respectively. Points $C_1$ and $C_2$ are symmetric to $C$ with respect to the lines $PQ$ and $PR$.
Find the angle $C_1IC_2$ where $I$ is the incenter of the triangle $AB... | [
"Denote $PR \\cap CC_2 = X$. Let's do some angle-chasing.\n$$\n\\angle RCC_2 = 90^\\circ - \\angle CRX = 90^\\circ - \\angle PRB = \\frac{\\angle B}{2}.\n$$\nTherefore $RI = RC = RC_2$, i.e. $R$ is the center of the circumcircle of the triangle $ICC_2$. Hence\n\n$$\n\\angle RIC_2 = 90^\\circ - \\frac{\\angle IRC_2}... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 135° | |
08by | Problem:
Il ricco Creso compra 88 vasi identici. Il prezzo di ognuno di essi, espresso in dracme, è un numero intero (lo stesso per tutti gli 88 vasi). Sappiamo che Creso paga un totale di $a1211b$ dracme, dove $a, b$ sono cifre da determinare (e che possono essere distinte o meno). Quante dracme costa un singolo vaso... | [
"Solution:\n\nLa risposta è 1274. Notiamo che il numero $a1211b$ dev'essere divisibile per 88, quindi dev'essere divisibile sia per 11 che per 8. Sappiamo che un numero è divisibile per 8 se e soltanto se lo sono le sue ultime 3 cifre, quindi $11b$ dev'essere divisibile per 8. Notiamo che $8 \\cdot 14 = 112$ è l'un... | Italy | Gara di Febbraio | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1274 | |
0bro | Determine the planar finite configurations $C$ consisting of at least three points, satisfying the following condition: if $x$ and $y$ are distinct points of $C$, then at least one of the two equilateral triangles erected on the segment $xy$ has all three vertices in $C$. | [
"The required configurations consist of the three vertices of an equilateral triangle. Clearly, the three vertices of an equilateral triangle form a configuration satisfying the condition in the statement.\n\nTo prove the converse, let $a$ and $b$ be the end points of a diameter of $C$ and notice that exactly one o... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof only | null | |
0dd0 | All of the numbers $1, 2, 3, \ldots, 1000000$ are initially colored black. On each move it is possible to choose the number $x$ (among the colored numbers) and change the color of $x$ and of all of the numbers that are not co-prime with $x$ (black into white, white into black). Is it possible to color all of the number... | [
"The answer is YES. We will prove by induction that the procedure can be applied for any positive integer $n$.\n\nThe statement is true for $n=1$. Suppose that it is also true for $n=k-1 \\geq 1$, which means there exists a way to change every number not exceeding $k-1$ from black to white, which we call process $A... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Yes | |
0hfx | Let $ABC$ be a right triangle with hypothenuse $BC$ and altitude $AD$. Let's denote the midpoints of $AD$ and $AC$ by $E$ and $F$ correspondingly. Let point $M$ be the circumcenter of $\triangle BEF$. Prove that $AC||BM$. | [
"As $\\triangle ADB \\sim \\triangle CAB$, we get $\\frac{AD}{AB} = \\frac{CA}{CB}$. As $AE = \\frac{1}{2}AD$, $CF = \\frac{1}{2}CA$, we get $\\frac{AF}{AB} = \\frac{CF}{CB}$. From this similarity we get that (fig. 16):\n$$\n\\angle BAE = \\angle BAD = \\angle BCA = \\angle BCF\n$$\nThen by angle and the ratio of t... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01i8 | Show that no non-zero integers $a$, $b$, $x$, $y$ satisfy
$$
\begin{cases} ax - by = 256 \\ ay + bx = 1 \end{cases}
$$ | [
"If we use the Diophantine sum of squares equality\n$$\n(ax - by)^2 + (ay + bx)^2 = (a^2 + b^2)(x^2 + y^2)\n$$\nthen we can see that for a system\n$$\n\\begin{cases} ax - by = s \\\\ ay + bx = t \\end{cases}\n$$\nto have a solution in positive integers the number $s^2 + t^2$ must be a composite number.\nThe number ... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Other"
] | null | proof only | null | |
0gsk | Let $x$, $y$, $z$ be positive real numbers and $x \le 1$. Show that
$$
xy + y + 2z \ge 4\sqrt{xyz}
$$ | [
"By applying AM-GM inequality we get\n$$\nxy + y + 2z = xy + y + z + z \\geq 4\\sqrt[4]{xy^2z^2}\n$$\nSince $0 < x \\leq 1$ we have $x \\geq x^2$ and hence\n$$\n4\\sqrt[4]{xy^2z^2} \\geq 4\\sqrt[4]{x^2y^2z^2} = 4\\sqrt{xyz}\n$$\nand we are done.",
"Since $0 < x \\leq 1$ we have $y \\geq xy$ and hence\n$$\nxy + y ... | Turkey | 30th Junior Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
07e4 | Let $\mathcal{P}$ be a simple polygon (non-self-intersecting) with perimeter $36$ that lies in a circle with radius $1$ and does not pass through the center of it. Prove that there is either a radius of this circle that intersects $\mathcal{P}$ at least $6$ times, or there is a second circle which is concentric with th... | [
"Fix a radius of circle $C$, $OX$ and for each segment $AB$ from the perimeter of $P$ (don't include polygon vertices in this process), consider the projection of $AB$ to $OX$ ($A'B'$) and the projection of $AB$ to the perimeter of $C$ ($A''B''$).\n\n\n\nCall $A'B'$ the radius projection of... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0531 | In a scalene triangle one angle is exactly two times as big as another one and some angle in this triangle is $36°$. Find all possibilities, how big the angles of this triangle can be. | [
"Based on the initial conditions the angles of the triangle are $\\alpha$, $2\\alpha$ and $180° - 3\\alpha$ and they all have to be different. It remains to perform calculations for three cases: $\\alpha = 36°$, $2\\alpha = 36°$ and $180° - 3\\alpha = 36°$."
] | Estonia | Open Contests | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The angle sets are {18°, 36°, 126°} and {36°, 48°, 96°}. | |
0c8g | Let $x$, $y$, $z$ be real numbers such that:
$$
x^2 + 4y^2 + 9z^2 + 20 = 4x + 12y + 24z.
$$
Prove that $x \in [-1, 5]$, $y \in [0, 3]$ and $z \in \left[\frac{1}{3}, \frac{7}{3}\right]$. | [
"We rewrite the given equality as:\n$$\n(x-2)^2 + (2y-3)^2 + (3z-4)^2 = 9.\n$$\nThis implies $(x-2)^2 \\le 9$, $(2y-3)^2 \\le 9$ and $(3z-4)^2 \\le 9$, therefore $|x-2| \\le 3$, $|2y-3| \\le 3$, $|3z-4| \\le 3$ and the last inequalities are equivalent to the conclusion."
] | Romania | Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
089v | Problem:
Nel triangolo $ABC$ supponiamo di avere $a > b$, dove $a = BC$ e $b = AC$. Sia $M$ il punto medio di $AB$, e siano $\alpha$ e $\beta$ le circonferenze inscritte, rispettivamente, ai triangoli $ACM$ e $BCM$. Siano poi $A^{\prime}$ e $B^{\prime}$ i punti di tangenza di $\alpha$ e $\beta$ con $CM$. Dimostrare ch... | [
"Solution:\n\nChiamiamo $x = MA^{\\prime}$, $y = MB^{\\prime}$, $r = CA^{\\prime}$, $s = CB^{\\prime}$. È chiaro che\n$$\nA^{\\prime}B^{\\prime} = x - y = s - r.\n$$\nDenotiamo con $A_{2}$ ed $A_{3}$ i punti di tangenza di $AB$ e $AC$ con $\\alpha$, e con $B_{2}$ e $B_{3}$ i punti di tangenza di $AB$ e $BC$ con $\\... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | (a - b) / 2 | |
01r9 | Some family pairs are friends with each other. Every St. Valentine's Day each husband of these pairs presents some roses to each wife of these pairs (including his own wife). Any wife will be offended by her husband if the number of roses that she obtains from her husband is less than or equal to the number of roses th... | [
"Consider $n$ ($n \\ge 2$) pairs. Let $M$ denote the set $\\{1, 2, ..., n\\}$. Let $a_{ij}$ be the number of roses presented by the husband from the $i$-th pair to the wife from the $j$-th pair. By condition, the wife from the $j$-th pair will be offended if\n$$\na_{jj} \\le \\sum_{i \\in M \\setminus \\{j\\}} a_{j... | Belarus | Final Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
08ln | Problem:
The side lengths of a parallelogram are $a$, $b$ and diagonals have lengths $x$ and $y$. Knowing that $a b = \frac{x y}{2}$, show that
$$
a = \frac{x}{\sqrt{2}},\ b = \frac{y}{\sqrt{2}} \quad \text{or} \quad a = \frac{y}{\sqrt{2}},\ b = \frac{x}{\sqrt{2}}
$$ | [
"Solution:\nLet us consider a parallelogram $ABCD$, with $AB = a$, $BC = b$, $AC = x$, $BD = y$, $\\widehat{AOD} = \\theta$.\nFor the area of $ABCD$ we know $(ABCD) = ab \\sin A$.\nBut it is also true that $(ABCD) = 4(AOD) = 4 \\cdot \\frac{OA \\cdot OD}{2} \\sin \\theta = 2 OA \\cdot OD \\sin \\theta = 2 \\cdot \\... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a = x/√2, b = y/√2 or a = y/√2, b = x/√2 | |
04j9 | There are ten white, and one red, blue, green, yellow and purple card. White cards are identical. On exactly one side of each card is the sign $X$. In how many ways is it possible to put the cards one on another such that no two cards face each other with the side having the sign $X$? | [] | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 5765760 | |
0bvb | Find all integers $a, b > 1$, such that both $a^2 + 16\left[\frac{b}{2}\right]$ and $b^2 + 16\left[\frac{a}{2}\right]$ are perfect squares. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | [(3,3), (4,7), (7,4)] | |
0lf3 | Given a set $A = \{1, 2, \dots, 4044\}$. One colors 2022 numbers of them by white and the rest by black. For each $i \in A$, denote the weight of $i$ by sum of the amount of white numbers that are smaller than $i$ and the amount of black numbers that are larger than $i$. For every positive integer $m$, find all positiv... | [
"Call a natural number $i$ good if its weight is $m$. We will prove the following claim.\n\n**Claim.** Consider a positive integer $i \\le 4044$.\n\na. If there are more black numbers than white from $1$ to $i-1$, then there exists a black number $j$ such that the numbers of black and white numbers from $j+1$ to $i... | Vietnam | IMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Let m be a positive integer. The possible k are exactly the even integers in the following ranges:
- If m < 2022: k ∈ {2, 4, ..., 2(m+1)}.
- If 2022 ≤ m ≤ 4044: k ∈ {2, 4, ..., 2(4044 − m)}.
- If m > 4044: no positive k exist. | |
08zr | A regular hexagon is inscribed in a rectangle as shown in the figure. The areas of the shaded triangle and quadrilateral are $20$ and $23$ respectively. Find the area of the regular hexagon.
 | [
"222\nLet $XY$ represent the length of segment $XY$. In the figure, let $A$, $B$, $C$ and $D$ be the vertices of the rectangle, and $P$, $Q$, $R$, $S$, $T$, $U$ and $O$ the vertices and the center of the regular hexagon.\n\nSince $\\angle QAP = \\angle PQS = 90^\\circ$, we can conclude that... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 222 | |
03fs | Find all pairs of positive integers $(n, k)$ such that all sufficiently large odd positive integers $m$ are representable as
$$
m = a_1^{n_1^2} + a_2^{(n_1+1)^2} + \dots + a_k^{(n+k-1)^2} + a_{k+1}^{(n+k)^2}
$$
for some non-negative integers $a_1, a_2, \dots, a_{k+1}$. | [] | Bulgaria | 4 Bulgarian National Olympiad - Regional Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All pairs with n = 1 and any positive integer k. | |
09ej | Let $n$ be a natural number satisfying the condition $n > 1$, $n | (6^n + 7^n)$. Prove that $13 | n$. | [
"Since $2^n + 7^n \\equiv 1 \\pmod{p}$, $n$ must be odd. Let $p$ be the least prime divisor of $n$. Hence $(p, 6) = 1$ and there exists $x, y \\in \\mathbb{Z}$ such that $px + 6y = 1$. Therefore $6y \\equiv 1 \\pmod{p}$.\n\nLet us consider $a$ such that $a \\equiv 7y \\pmod{p}$. Then $a^n + 1 \\equiv (7y)^n + (6y)^... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
03bj | All points in the plane having integer coordinates are colored in three colors. Find the least positive integer $S$ having the following property: for arbitrary such covering there exists a triangle of area $S$ having all its vertices in one and the same color. | [
"Consider the following two colorings:\n\n(1) Point $(x, y)$ is in color $i$, $1-i-2$ if and only if $x \\equiv i \\pmod{2}$.\n\n(2) Point $(x, y)$ is in color $i$, $1 - i - 3$ if and only if $x \\equiv i \\pmod{3}$.\n\nIt is clear that if $S$ exists then $2S$ is an integer. Coloring (1) implies that $S$ could be $... | Bulgaria | Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 3 | |
0403 | Let $\odot I$ be the incircle of $\triangle ABC$. The circle $\odot I$ intersects sides $AB$, $BC$ and $CA$ at points $D$, $E$ and $F$, respectively. Line $EF$ intersects lines $AI$, $BI$ and $DI$ at points $M$, $N$ and $K$, respectively. Prove that $DM \cdot KE = DN \cdot KF$. (posed by Zhang Pengcheng) | [
"It is easy to see that points $I$, $D$, $E$ and $B$ are concyclic and\n$$\n\\angle AID = 90^\\circ - \\angle IAD, \\\\\n\\angle MED = \\angle FDA = 90^\\circ - \\angle IAD.\n$$\nSo $\\angle AID = \\angle MED$, thus points $I$, $D$, $E$ and $M$ are concyclic.\nHence, five points $I$, $D$, $B$, $E$, $M$ are concycli... | China | China Southeastern Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis ... | English | proof only | null |
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