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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
044r | Let $a_1, a_2, \dots, a_{21}$ be a permutation of $1, 2, \dots, 21$, satisfying
$$
|a_{20} - a_{21}| \geq |a_{19} - a_{21}| \geq |a_{18} - a_{21}| \geq \dots \geq |a_1 - a_{21}|.
$$
The number of such permutations is ________. | [
"For a given $k \\in \\{1, 2, \\dots, 21\\}$, consider the number of permutations $N_k$ that satisfy the conditions such that $a_{21} = k$.\nWhen $k \\in \\{1, 2, \\dots, 11\\}$, for $i = 1, 2, \\dots, k-1$, there exist $a_{2i-1}, a_{2i}$ that are permutations of $k-i, k+i$ (if $k=1$, there exists no such $i$), and... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 3070 | |
0jmu | Problem:
In equilateral triangle $ABC$ with side length $2$, let the parabola with focus $A$ and directrix $BC$ intersect sides $AB$ and $AC$ at $A_{1}$ and $A_{2}$, respectively. Similarly, let the parabola with focus $B$ and directrix $CA$ intersect sides $BC$ and $BA$ at $B_{1}$ and $B_{2}$, respectively. Finally, ... | [
"Solution:\n\nAnswer: $\\quad 66-36 \\sqrt{3}$\n\nSince everything is equilateral it's easy to find the side length of the wanted triangle. By symmetry, it's just $AA_{1} + 2A_{1}B_{2} = 3AA_{1} - AB$. Using the definition of a parabola, $AA_{1} = \\frac{\\sqrt{3}}{2} A_{1}B$ so some calculation gives a side length... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 66-36 \sqrt{3} | |
0bps | Let $A$ and $B$ be two $3 \times 3$ matrices with real entries, so that $AB^2A = BA^2B$.
a) Prove that there exists $\alpha, \beta \in \mathbb{R}$ so that
$$
\det[(AB)^2 + (BA)^2] = [\det(A)\det(B) - \alpha]^2 + [\det(A)\det(B) - \beta]^2.
$$ | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
099x | $G$ is a given simple graph. If any $1100$ of the edges of the graph can be represented by $45$ of its vertices, then prove that there exists $45$ vertices that can represent all of its edges ($v \in e$, then $v$ vertices represents edge $e$).
(proposed by B. Batbayasgalan) | [
"Suppose that such $45$ vertices do not exist. If removal of any one of the edges does not interfere with this condition, then remove this edge. Repeating this action until there are no longer any edges to be removed results in the graph such that after a removal of any arbitrary edge, $45$ vertices that represent ... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0eom | I buy a number of chocolates which cost R25 each and cool-drinks which cost R9 each. I buy more chocolates than cool-drinks. How many cool-drinks do I buy if I pay R839? | [
"If I buy $x$ chocolate bars and $y$ cool-drinks, then $25x + 9y = 839$ with $x > y$. If we put $x = y$ as a first guess, then $x = y = 839/34 \\approx 24$, so we can write $x = 24 + a$ and $y = 24 - b$. The equation becomes $25a - 9b = 23$, which by easy trial and error (guess and check) has a solution $a = 2$ and... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 21 | |
0kp2 | Let $x$, $y$, and $z$ be positive real numbers satisfying the system of equations
$$
\begin{aligned}
\sqrt{2x - xy} + \sqrt{2y - xy} &= 1 \\
\sqrt{2y - yz} + \sqrt{2z - yz} &= \sqrt{2} \\
\sqrt{2z - zx} + \sqrt{2x - zx} &= \sqrt{3}.
\end{aligned}
$$
Then $[(1-x)(1-y)(1-z)]^2$ can be written as $\frac{m}{n}$, where $m$ ... | [] | United States | 2022 AIME I | [
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | 33 | |
02b1 | Problem:
Determine 4 números distintos $a_{1}, a_{2}, a_{3}$ e $a_{4}$ que sejam termos consecutivos de uma progressão aritmética e que os números $a_{1}, a_{3}$ e $a_{4}$ formem uma progressão geométrica. | [
"Solution:\n\nOs 4 termos de uma progressão aritmética de razão $r$ podem ser escritos como:\n$$\nx-2r,\\ x-r,\\ x,\\ x+r\n$$\nLogo, os 3 termos da progressão geométrica de razão $q$ serão\n$$\nx-2r,\\ x,\\ x+r\n$$\nonde\n$$\nx = (x-2r)q \\text{ e } x+r = xq\n$$\nDaí segue que:\n$$\nx = xq - 2rq \\Rightarrow x = x ... | Brazil | Nível 3 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | All solutions are given by 2x, 3x/2, x, x/2 with x ≠ 0 (e.g., 2, 3/2, 1, 1/2). | |
02w0 | Problem:
Seja $n$ um inteiro positivo. Se a equação $2x + 2y + z = n$ tem 28 soluções em inteiros positivos $x$, $y$ e $z$, determine os possíveis valores de $n$. | [
"Solution:\n\nPerceba inicialmente que se $x$ e $y$ estão definidos, só existe uma possível escolha para $z$ e, além disso, $z$ e $n$ possuem a mesma paridade. Consideremos os seguintes casos:\n\ni) O número $n$ é par, ou seja, $n = 2i$. Assim, devemos ter $z = 2j$ e\n$$\n\\begin{aligned}\n2x + 2y + z &= n \\\\\nx ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 17, 18 | |
0arw | Problem:
Find the polynomial expression in $Z = x - \frac{1}{x}$ of $x^{5} - \frac{1}{x^{5}}$. | [
"Solution:\n$x^{5} - \\frac{1}{x^{5}} = \\left(x - \\frac{1}{x}\\right)\\left(x^{4} + x^{2} + 1 + \\frac{1}{x^{2}} + \\frac{1}{x^{4}}\\right) = \\left(x - \\frac{1}{x}\\right)\\left(x^{2} + \\frac{1}{x^{2}} + x^{4} + \\frac{1}{x^{4}} + 1\\right)$.\n\nNow $x^{2} + \\frac{1}{x^{2}} = \\left(x - \\frac{1}{x}\\right)^{... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | Z^5 + 5Z^3 + 5Z | |
03zb | Let $a_1, a_2, \dots$ be a permutation of all positive integers. Prove that there exist infinite positive integers $i$'s, such that $(a_i, a_{i+1}) \le \frac{3}{4}i$. (posed by Chen Yonggao) | [
"We prove this problem by contradiction. If the conclusion of the problem is not true, then there exists $i_0$, and we have $(a_i, a_{i+1}) > \\frac{3}{4}i$ for $i \\ge i_0$.\n\nTake a positive number $M > i_0$, so if $i \\ge 4M$, then $(a_i, a_{i+1}) > \\frac{3}{4}i \\ge 3M$.\n\nSo, if $i \\ge 4M$, $a_i \\ge (a_i,... | China | China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0b1e | Problem:
Find the largest three-digit integer for which the product of its digits is 3 times the sum of its digits. | [] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 951 | |
005c | Sea $n > 2$ un número natural. Un subconjunto $A$ de $\mathbb{R}$ se dice $n$-pequeño si existen $n$ números reales $t_1, t_2, \ldots, t_n$ tales que los conjuntos $t_1 + A, t_2 + A, \ldots, t_n + A$ sean disjuntos dos a dos. Demuestre que $\mathbb{R}$ no puede ser representado como unión de $n$ conjuntos $n$-pequeños.... | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof only | null | |
056f | Let $M$ be the intersection of the diagonals of a cyclic quadrilateral $ABCD$. Find the length of $AD$, if it is known that $AB = 2$ mm, $BC = 5$ mm, $AM = 4$ mm, and $\frac{CD}{CM} = 0.6$. | [
"Opposite angles $AMB$ and $DMC$ equal. Also notice that $\\angle ABM = \\angle ABD = \\angle ACD = \\angle MCD$, as $ABD$ and $ACD$ are subtended to the same arc. Therefore the triangles $AMB$ and $DMC$ are similar. Hence $\\frac{BA}{BM} = \\frac{CD}{CM}$, from which $BM = BA \\cdot \\frac{CM}{CD}$. Analogously th... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 6 mm | |
0aar | In the square $ABCD$ with side length $4\,\mathrm{cm}$, from each vertex arcs are circumscribed with radii that equal half of the side length of the square, like in the picture. Calculate the perimeter and area of the marked part of the square.
 | [
"The four arcs form a circumference with radius $r = 2\\,\\mathrm{cm}$. We have that $L = 2r\\pi = 4\\pi\\,\\mathrm{cm}$ and\n$$\nP = 4^2 - r^2\\pi = 16 - 4\\pi = 4(4 - \\pi)\\,\\mathrm{cm}^2\n$$"
] | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Perimeter: 4π cm; Area: 16 − 4π cm² | |
0ihv | Problem:
Fifteen freshmen are sitting in a circle around a table, but the course assistant (who remains standing) has made only six copies of today's handout. No freshman should get more than one handout, and any freshman who does not get one should be able to read a neighbor's. If the freshmen are distinguishable but... | [
"Solution:\n\nSuppose that you are one of the freshmen; then there's a $6 / 15$ chance that you'll get one of the handouts. We may ask, given that you do get a handout, how many ways are there to distribute the rest? We need only multiply the answer to that question by $15 / 6$ to answer the original question.\n\nG... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 125 | |
09xo | We consider security codes consisting of four digits. We say that one code *dominates* another code if each digit of the first code is at least as large as the corresponding digit in the second code. For example, $4961$ dominates $0761$, because $4 \ge 0$, $9 \ge 7$, $6 \ge 6$, and $1 \ge 1$. We would like to assign a ... | [
"$37$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 37 | |
050h | Prove that if $n$ and $k$ are positive integers such that $1 < k < n-1$, then the binomial coefficient $\binom{n}{k}$ is divisible by at least two different primes. | [
"Assume w.l.o.g. that $n \\ge 2k$ (if $n < 2k$, then interchange the roles of $k$ and $n-k$). Let $p$ be an arbitrary prime number. Consider the numbers that remain into the numerator of the expression\n$$\n\\binom{n}{k} = \\frac{n \\cdot (n-1) \\cdots (n-k+1)}{k \\cdot (k-1) \\cdots 1}\n$$\nafter reducing all fact... | Estonia | IMO Team Selection Contest | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0kpj | Problem:
On Semi-Predictable Island, everyone is either a liar (who always lies), a truth-teller (who always tells the truth), or a spy (who could do either). Aerith encounters three people and knows that one is a liar, one a truth-teller, and one a spy. She can ask two yes-or-no questions, and all three of them will ... | [
"Solution:\n\nFirst she asks the three people, \"Are you a spy?\" The truth-teller will say \"no,\" the liar will say \"yes,\" and the spy could say either. Either way, two people will give the same answer and the third will give a different answer. She can now tell the identity of that third person: if they said \... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Logic"
] | null | proof only | null | |
0991 | Problem:
Fie $x$, $y$, $z$ numere reale pozitive. Arătați că
$$
\frac{(x+y)(x+z)(y+z)}{4 x y z} \geq \frac{x+z}{y+z}+\frac{y+z}{x+z}
$$ | [
"Solution:\nÎmpărțim inegalitatea din enunț la produsul pozitiv $(x+z)(y+z)$ obținem inegalitatea\n$$\n\\frac{x+y}{4 x y z} \\geq \\frac{1}{(y+z)^{2}}+\\frac{1}{(x+z)^{2}}\n$$\nechivalentă celei din enunț. Dar\n$$\n\\frac{x+y}{4 x y z}=\\frac{x}{4 x y z}+\\frac{y}{4 x y z}=\\frac{1}{4 y z}+\\frac{1}{4 x z}\n$$\nDin... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0l65 | Problem:
Ben has 16 balls labeled $1$, $2$, $3$, $\ldots$, $16$, as well as 4 indistinguishable boxes. Two balls are neighbors if their labels differ by $1$. Compute the number of ways for him to put 4 balls in each box such that each ball is in the same box as at least one of its neighbors. (The order in which the ba... | [
"Solution:\n\nEach box must either contain a single group of four consecutive balls (e.g. $5$, $6$, $7$, $8$) or two groups of two consecutive balls (e.g. $5$, $6$, $9$, $10$). Since all groups have even lengths, this means that $1$ and $2$ are in the same group, $3$ and $4$ are in the same group, and so on. We can... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 105 | |
00go | Let $ABC$ be a triangle with $\angle A < 60^{\circ}$. Let $X$ and $Y$ be the points on the sides $AB$ and $AC$, respectively, such that $CA + AX = CB + BX$ and $BA + AY = BC + CY$. Let $P$ be the point in the plane such that the lines $PX$ and $PY$ are perpendicular to $AB$ and $AC$, respectively. Prove that $\angle BP... | [
"Let $I$ be the incenter of $\\triangle ABC$, and let the feet of the perpendiculars from $I$ to $AB$ and to $AC$ be $D$ and $E$, respectively. (Without loss of generality, we may assume that $AC$ is the longest side. Then $X$ lies on the line segment $AD$. Although $P$ may or may not lie inside $\\triangle ABC$, t... | Asia Pacific Mathematics Olympiad (APMO) | XX Asian Pacific Mathematics 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 | |
0al9 | Problem:
Find the area of the region bounded by the graph of $2x^{2} - 4x - xy + 2y = 0$ and the $x$-axis.
(a) 9
(b) 12
(c) 4
(d) 6 | [] | Philippines | Qualifying Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | c | |
0aj2 | Prove that for positive real numbers $a, b, c$ the following inequality holds:
$$
(16a^2 + 8b + 17)(16b^2 + 8c + 17)(16c^2 + 8a + 17) \geq 2^{12}(a+1)(b+1)(c+1).
$$
When does equality hold? | [
"By twice using the inequality between the arithmetical mean and geometrical mean we get\n$$\n\\begin{aligned}\n(16a^2 + 8b + 17) &= (16a^2 + 1 + 8b + 16) \\ge 8a + 8b + 16 = 8(a+b+2) = 8(a+1+b+1) \\ge \\\\\n&\\ge 8 \\cdot 2\\sqrt{(a+1)(b+1)} = 2^4\\sqrt{(a+1)(b+1)}.\n\\end{aligned} \\quad (1)\n$$\nAnalogously we h... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds when a = b = c = 1/4. | |
092y | Problem:
Let $ABCDE$ be a convex pentagon. Let $P$ be the intersection of the lines $CE$ and $BD$. Assume that $\angle PAD = \angle ACB$ and $\angle CAP = \angle EDA$. Prove that the circumcentres of the triangles $ABC$ and $ADE$ are collinear with $P$. | [
"Solution:\n\nSimple angle chasing gives us:\n$$\n\\begin{gathered}\n\\angle BCD + \\angle EDC = \\angle ACB + \\angle ACD + \\angle EDA + \\angle ADC = \\\\\n= \\angle PAD + \\angle ACD + \\angle CAP + \\angle ADC = 180^\\circ,\n\\end{gathered}\n$$\nso $BC \\parallel DE$. Therefore there exists a homothety $H$ cen... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0eec | Find all integers $a$, $b$, $c$, and $d$ that solve the system of equations
$$
a^2 + b^2 + c^2 = d + 13,
$$
$$
a + 2b + 3c = \frac{d}{2} + 13.
$$ | [
"From the second equation we express $d = 2a + 4b + 6c - 26$ and insert it into the first equation to get\n$$\na^2 + b^2 + c^2 = 2a + 4b + 6c - 13.\n$$\nMoving all the terms to the left we have $a^2 + b^2 + c^2 - 2a - 4b - 6c + 13 = 0$, which we now rewrite as the sum of perfect squares\n$$\n(a - 1)^2 + (b - 2)^2 +... | Slovenia | Slovenija 2016 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (0, 2, 3, 0), (2, 2, 3, 4), (1, 1, 3, -2), (1, 3, 3, 6), (1, 2, 2, -4), (1, 2, 4, 8) | |
0ixm | Problem:
The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placed on a table? | [
"Solution:\nThe major diagonal has a length of $\\sqrt{3}$. The volume of the pyramid is $1/6$, and so its height $h$ satisfies\n$$\n\\frac{1}{3} \\cdot h \\cdot \\frac{\\sqrt{3}}{4}(\\sqrt{2})^{2} = 1/6\n$$\nsince the freshly cut face is an equilateral triangle of side length $\\sqrt{2}$. Thus $h = \\sqrt{3}/3$, a... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | √3/3 | |
07xx | Consider triangle $ABC$ with $|AB| < |AC| < |BC|$, and let $I$ be its incentre. The incircle of $\triangle ABC$ touches sides $BC$ and $AC$ at points $D$ and $E$, respectively. Let $R$ denote the midpoint of side $AC$. The point $N$ lies on the segment $DE$ such that $|RE| = |RN|$. Let $Q$ be the midpoint of segment $N... | [
"Note that $|CD| = |CE|$ and $|RN| = |RE|$. Hence, triangles $\\triangle ENR$ and $\\triangle EDC$ which share an angle at $E$, are both isosceles and hence similar. In particular, $RN \\parallel CB$ and\n$$\n\\angle CDE = \\angle CED = \\angle ENR = 90^\\circ - \\frac{1}{2}\\angle C.\n$$\n\n![](attached_image_1.pn... | Ireland | IRL_ABooklet_2025 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic qu... | null | proof only | null | |
01b5 | Let $f: \mathbb{R} \to \mathbb{R}$ be a strictly increasing function, and $f(x) > x$ for every $x$. Assume that
$$ f(x) + f^{-1}(x) = 2x $$
for all $x \in \mathbb{R}$. Show that $f(x) = x + f(0)$ for all $x \in \mathbb{R}$. | [
"Set $g(x) = f(x) - x$. Then $g(x) > 0$ for all $x$. Let $g(x) = c$ for some $x$. So $f(x) = x + c$. Then $x = f^{-1}(x + c)$, $2x + 2c = f(x + c) + f^{-1}(x + c) = f(x + c) + x$, and $g(x + c) = f(x + c) - (x + c) = c$. By induction, then $g(x + kc) = c$ for $k \\in \\mathbb{N}$.\n\nWe show that $g$ only assumes o... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
0ami | Problem:
The number $x$ is chosen randomly from the interval $(0,1]$. Define $y=\left\lceil\log_{4} x\right\rceil$. Find the sum of the lengths of all subintervals of $(0,1]$ for which $y$ is odd. For any real number $a,\lceil a\rceil$ is defined as the smallest integer not less than $a$. | [] | Philippines | AREA STAGE | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 1/5 | |
07tk | Solve $\sqrt{x + 5 - 4\sqrt{x + 1}} + \sqrt{x + 17 - 8\sqrt{x + 1}} = 2$ for all real $x$. | [
"It is necessary that $x \\ge -1$. Let $u = \\sqrt{x+1}$. Then\n$$\n\\begin{align*}\n& \\sqrt{x+5-4\\sqrt{x+1}} + \\sqrt{x+17-8\\sqrt{x+1}} \\\\\n&= \\sqrt{u^2+4-4u} + \\sqrt{u^2+16-8u} \\\\\n&= \\sqrt{(u-2)^2} + \\sqrt{(u-4)^2} \\\\\n&= |u-2| + |u-4|.\n\\end{align*}\n$$\nThus the equation becomes\n$|u - 2| + |u - ... | Ireland | IRL_ABooklet | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x in [3, 15] | |
0gyy | Numbers from $1$ to $2007$ are arbitrarily written down in strip cells. Two players by turns mark cells of this strip. That player after whose move there are such two natural numbers $m < n$ loses that the sum of numbers of all noted cells from $m$ to $n$ divides by $2008$. Prove that there is at least one thousand ini... | [
"Let in a cell with number $n$ number $a_n$ is written down. We will consider such sequence $(a_n)$: $a_{2k-1} = 2k-1$, $k = \\overline{1,1004}$, $a_{2k} = 2008 - 2k$, $k = \\overline{1,1003}$. We will prove that such placing approaches us. Let $S_n = a_1 + a_2 + \\dots + a_n$. Then $S_{2k} = 2007k$, $k = \\overlin... | Ukraine | The Problems of Ukrainian Authors | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
077c | Problem:
Let $X=\{0,1,2,3,4,5,6,7,8,9\}$. Let $S \subseteq X$ be such that any nonnegative integer $n$ can be written as $p+q$ where the nonnegative integers $p, q$ have all their digits in $S$. Find the smallest possible number of elements in $S$. | [
"Solution:\n\nWe show that 5 numbers will suffice. Take $S=\\{0,1,3,4,6\\}$. Observe the following splitting:\n\n| $n$ | $a$ | $b$ |\n| :--- | :--- | :--- |\n| 0 | 0 | 0 |\n| 1 | 0 | 1 |\n| 2 | 1 | 1 |\n| 3 | 0 | 3 |\n| 4 | 1 | 3 |\n| 5 | 1 | 4 |\n| 6 | 3 | 3 |\n| 7 | 3 | 4 |\n| 8 | 4 | 4 |\n| 9 | 3 | 6 |\n\nThus e... | India | INMO | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 5 | |
0af4 | Имаме две свеќи со различни должини и дебелини. Подолгата и потенка свеќа целосно изгорува за $3{,}5$ часа, а пократката и подебела свеќа за $5$ часа. Свеќите биле запалени истовремено, а после $2$ часа горење нивните должини биле еднакви. За колку проценти потенката свеќа е подолга од подебелата? | [
"За еден час изгоруваат $\\frac{2}{7}$ од првата (подолгата и потенка) свеќа, а $\\frac{1}{5}$ од втората (пократката и подебела) свеќа. По два часа изгореле $\\frac{4}{7}$ од првата и $\\frac{2}{5}$ од втората свеќа. Значи останале $\\frac{3}{7}$ од првата и $\\frac{3}{5}$ од втората свеќа. Бидејќи тие големини се... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | final answer only | 40% | |
06w7 | Consider any rectangular table having finitely many rows and columns, with a real number $a(r, c)$ in the cell in row $r$ and column $c$. A pair ($R, C$), where $R$ is a set of rows and $C$ a set of columns, is called a saddle pair if the following two conditions are satisfied:
(i) For each row $r'$, there is $r \in R$... | [
"Solution 1. We say that a pair ($R', C'$) of nonempty sets is a subpair of a pair ($R, C$) if $R' \\subseteq R$ and $C' \\subseteq C$. The subpair is proper if at least one of the inclusions is strict.\nLet ($R_1, C_1$) and ($R_2, C_2$) be two saddle pairs with $|R_1| > |R_2|$. We will find a saddle subpair ($R', ... | IMO | IMO 2020 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ho5 | Problem:
On the sides of a convex quadrilateral $A B C D$, construct squares externally. Prove that the quadrilateral with vertices at the centers of the squares has equal and perpendicular diagonals. | [
"Solution:\n\nFirst, a lemma: If squares constructed on $X Y$ and $Y Z$ of triangle $X Y Z$ have respective centers $U, V$, then a rotation through angle $\\pi / 2$ about the midpoint $M$ of $Z X$ carries $U$ into $V$.\n\nThe proof is as follows: Rotating quadrilateral $M X U Y$ by an angle $\\pi$ about $M$ gives u... | United States | Berkeley Math Circle Take-Home Contest #6 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a43 | Problem:
Zij $ABCD$ een parallellogram en zij $M$ het snijpunt van de diagonalen. De omgeschreven cirkel van $\triangle ABM$ snijdt het lijnstuk $AD$ in $E \neq A$ en de omgeschreven cirkel van $\triangle EMD$ snijdt het lijnstuk $BE$ in het punt $F \neq E$.
Bewijs dat $\angle ACB = \angle DCF$.
![](attached_image_1.... | [
"Solution:\n\nWe laten eerst zien dat $CBFD$ een koordenvierhoek is. Er geldt\n\\[\n\\begin{align*}\n\\angle BCD &= \\angle BAD = \\angle BAE \\quad (\\text{parallellogram}) \\\\\n&= 180^\\circ - \\angle EMB \\quad (\\text{koordenvierhoekstelling in } EABM) \\\\\n&= \\angle EMD \\quad (\\text{gestrekte hoek}) \\\\\... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
045t | Prove that there exist two positive real numbers $C$ and $\alpha > \frac{1}{2}$, such that, for any positive integer $n$, there is a subset $A$ of $\{1, 2, \dots, n\}$ with $|A| \ge Cn^\alpha$, such that the difference between any pair of different numbers in $A$ is not a perfect square. | [
"**Proof.** For $n \\ge 25$, write $5^{2t} \\le n < 5^{2t+2}$ for $t \\in \\mathbb{N}^*$. Take the set\n$$\nA = \\{(\\alpha_{2t}, \\dots, \\alpha_1)_5 \\mid \\alpha_{2i} \\in \\{0, 1, 2, 3, 4\\} \\text{ and } \\alpha_{2i-1} \\in \\{1, 3\\} \\text{ for } i = 1, \\dots, t\\},\n$$\nwhere $(\\alpha_{2t}, \\dots, \\alph... | China | 2022 China Team Selection Test for IMO | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0glh | There are $2016$ real numbers written on the blackboard. In each step, we choose two numbers, erase them, and replace each of them by their product. Determine whether it is possible to obtain $2016$ equal numbers on the blackboard after a finite number of steps.
(the 15th Czech-Polish-Slovak Mathematics Competition) | [
"We shall prove that it is possible to obtain equal numbers after some finite steps by induction with respect to $n$. The claim is trivial for $n = 2$ (we can get the desired $2$-tuple after a single step $(a, b) \\to (ab, ab)$) and $n = 4$ (we can follow the scheme $(\\underline{a}, \\underline{b}, c, d) \\to (ab,... | Thailand | The first T3MO | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Yes | |
02bm | Problem:
Um número de 3 algarismos e seu sêxtuplo são formados pelos mesmos algarismos. A soma dos algarismos desse número é 17 e a de seu sêxtuplo é 21. Qual é esse número? Existe mais do que um? | [
"Solution:\n\n746 (solução única?)"
] | Brazil | Desafios | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 746; unique | |
0dkj | Let positive integer $n$ is given. The cube of size $2n + 1$ consists of $(2n + 1)^3$ unit cubes. Each cube is either colored green or orange. It is known that within any 8 unit cubes that form a cube $2 \times 2 \times 2$, there are at most 4 green cubes. Find the maximum number of green cubes. | [
"(Solution of Yousif Alkhalawi, IMO 2025 team's candidate)\nSince each cube $2 \\times 2 \\times 2$, there are at least 4 orange unit cubes (here we call it by square) so we want to minimize the number of orange squares.\n\n**Construction:** Let number the layer $n \\times n$ of cubes from $1 \\to n$ (bottom to top... | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 4n^3 + 9n^2 + 6n + 1 | |
0g8r | 瘋狂科學家無意間在他的實驗室發現了一種叫瓜克的新粒子。兩個瓜克可以組成一個瓜克對；一個瓜克可以同時和很多瓜克組成瓜克對。此外，他發現他可以進行以下兩種動作：
(1) 如果有一個瓜克和奇數個其他瓜克組成瓜克對，則他可以將這個瓜克消滅。
(2) 將整個實驗室裡的瓜克複製，也就是說，每個瓜克 $I$ 都會複製出一個瓜克 $I'$. 新的瓜克 $I'$ 和 $J'$ 組成瓜克對若且唯若舊的瓜克 $I$ 和 $J$ 組成瓜克對；此外，$I'$ 會與 $I$ 組成瓜克對。除以上瓜克對外，複製不會製造其他額外的瓜克對。
試證：科學家可以經由一系列的動作，讓實驗室裡最後只剩下一群瓜克，它們兩兩之間不成瓜克對。
A crazy physicist ... | [
"以瓜克為頂點作圖，並將組成瓜克對的兩瓜克間連線。此外，我們將定頂點著色。我們稱一個圖 $G$ 是好圖，若且唯若 $G$ 中相連的兩頂點都是不同顏色。\n\n1. **引理.** 給定 $n$ 色好圖，必可透過一系列動作，得到一 $n-1$ 色好圖（其中 $n > 1$）\n**Proof.** 首先，我們重複執行動作 (1)，直到所有端點都是偶數邊。這個新圖當然是好圖。\n接著執行動作 (2)；若原瓜克 $I$ 是第 $k$ 個顏色，則我們將複製出來的 $I'$ 塗上第 $k+1 \\pmod n$ 個顏色。易知這仍然是個好圖。\n現在，這個新圖的所有端點都是奇數邊，因此我們可以對所有塗上第 $n$ 個顏色的瓜克進行動作 ... | Taiwan | 國際數學奧林匹亞競賽第二階段選訓營 模擬競賽(二) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
03xh | As shown below, two circles $\Gamma_1$, $\Gamma_2$ intersect at points $A$, $B$, one line passing through $B$ intersects $\Gamma_1$, $\Gamma_2$ at points $C$, $D$, another line passing through $B$ intersects $\Gamma_1$, $\Gamma_2$ at points $E$, $F$, and line $CF$ intersects $\Gamma_1$, $\Gamma_2$ at points $P$, $Q$, r... | [
"**Proof** Draw lines $AC$, $AD$, $AE$, $AF$, $DF$. From $\\angle ADB = \\angle AFB$, $\\angle ACB = \\angle AEF$ and the assumption $CD = EF$; one obtains $\\triangle ACD \\cong \\triangle AEF$. So we have $AD = AF$, $\\angle ADC = \\angle AFE$ and $\\angle ADF = \\angle AFD$. Then $\\angle ABC = \\angle AFD = \\a... | China | Chinese Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0dze | Problem:
Rešitvi enačbe $\frac{\log \left(35-x^{3}\right)}{\log (5-x)}=3$ sta dolžini katet pravokotnega trikotnika. Izračunaj polmer kroga, ki je temu trikotniku očrtan. | [
"Solution:\n\nEnačbo množimo z $\\log (5-x)$, antilogaritmiramo in uredimo v kvadratno $15 x^{2}-75 x+90=0$. Rešitvi kvadratne enačbe sta $x_{1}=2$, $x_{2}=3$. Rešitvi enačbe sta dolžini katet pravokotnega trikotnika. Izračunamo dolžino hipotenuze. Polovična vrednost dolžine hipotenuze je enaka polmeru $R$ temu tri... | Slovenia | Državno tekmovanje | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | sqrt(13)/2 | |
0gvi | Find all non-decreasing and all non-increasing functions $f : [0; +\infty) \to \mathbf{R}$ such that for all $x, y \ge 0$ the equality
$$
f(x+y) - f(x) - f(y) = f(xy+1) - f(xy) - f(1)
$$
holds, and additionally $f(3) + 3f(1) = 3f(2) + f(0)$. | [
"**Лема.** Якщо функція $F:(0;+\\infty) \\rightarrow \\mathbf{R}$ задовольняє на проміжку $(0;+\\infty)$ **функціональне рівняння Коші**\n$$\nF(u + v) = F(u) + F(v)\n$$\nі є обмеженою на інтервалі $(0;1)$ (тобто, існує така стала $M$, що $|F(u)| \\le M$ для всіх $u \\in (0;1)$), то $F(u) = F(1)u$ для всіх $u \\in (... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All solutions are quadratic polynomials f(u) = A u^2 + B u + C for u ≥ 0. The functional equation holds for any real A, B, C, and the extra condition forces f(0) = C. Monotonicity constraints: nondecreasing if and only if A ≥ 0 and B ≥ 0; nonincreasing if and only if A ≤ 0 and B ≤ 0. | |
0a1u | Problem:
Vind de kleinst mogelijke waarde van
$$
x y + y z + z x + \frac{1}{x} + \frac{2}{y} + \frac{5}{z},
$$
voor positieve reële getallen $x$, $y$ en $z$. | [
"Solution:\n\nAntwoord: de kleinst mogelijke waarde is $3 \\sqrt[3]{36}$.\nMet de ongelijkheid van het rekenkundig-meetkundig gemiddelde vinden we\n$$\n\\begin{aligned}\nx y + \\frac{1}{3 x} + \\frac{1}{2 y} & \\geq 3 \\sqrt[3]{x y \\frac{1}{3 x} \\frac{1}{2 y}} = 3 \\sqrt[3]{\\frac{1}{6}} \\\\\ny z + \\frac{3}{2 y... | Netherlands | IMO-selectietoets II | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3 \sqrt[3]{36} | |
0k7m | Problem:
In isosceles $\triangle ABC$, $AB = AC$ and $P$ is a point on side $BC$. If $\angle BAP = 2 \angle CAP$, $BP = \sqrt{3}$, and $CP = 1$, compute $AP$. | [
"Solution:\n\nLet $\\angle CAP = \\alpha$. By the Law of Sines,\n$$\n\\frac{\\sqrt{3}}{\\sin 2\\alpha} = \\frac{1}{\\sin \\alpha}\n$$\nwhich rearranges to\n$$\n\\cos \\alpha = \\frac{\\sqrt{3}}{2} \\Rightarrow \\alpha = \\frac{\\pi}{6}.\n$$\nThis implies that $\\angle BAC = \\frac{\\pi}{2}$.\n\nBy the Pythagorean T... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(2) | |
02ky | Problem:
Na campanha "Vamos ao teatro", 5 ingressos podem ser adquiridos pelo preço usual de 3 ingressos. Mário comprou 5 ingressos nessa campanha. A economia que Mário fez representa que percentual sobre o preço usual dos ingressos?
(a) $20\%$
(b) $33 \frac{1}{3} \%$
(c) $40\%$
(d) $60\%$
(e) $66 \frac{2}{3} \%$ | [
"Solution:\n\nMário pagou 3 e levou 5, logo ele pagou apenas $\\frac{3}{5}$ do preço usual e portanto, economizou $\\frac{2}{5}$. Como $\\frac{2}{5}=\\frac{40}{100}$, a economia foi de $40\\%$. A opção correta é (c)."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | (c) | |
0adn | Let $A_1B_1C_1$ and $A_2B_2C_2$ are given triangles. Let $T_1$ and $T_2$ are their centers of mass correspondently. Prove that $3\overline{T_1T_2} = \overline{A_1A_2} + \overline{B_1B_2} + \overline{C_1C_2}$. | [
"We have $\\overline{A_1A_2} = \\overline{A_1T_1} + \\overline{T_1A_2} + \\overline{T_2A_2}$, $\\overline{B_1B_2} = \\overline{B_1T_1} + \\overline{T_1B_2} + \\overline{T_2B_2}$, $\\overline{C_1C_2} = \\overline{C_1T_1} + \\overline{T_1C_2} + \\overline{T_2C_2}$. If we sum the last three equalities we obtain\n$$\n\... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
01m1 | Three distinct real numbers satisfy the following condition: the square of any of them is $1$ greater than the product of the remaining two.
Find all possible values of the sum of pairwise products of these numbers. | [
"Let the three distinct real numbers be $a$, $b$, and $c$.\n\nThe condition says: the square of any of them is $1$ greater than the product of the other two.\nSo:\n\n$a^2 = bc + 1$\n$b^2 = ca + 1$\n$c^2 = ab + 1$\n\nAdd all three equations:\n$a^2 + b^2 + c^2 = ab + bc + ca + 3$\n\nRecall that $(a + b + c)^2 = a^2 +... | Belarus | 61st Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | -1 | |
0cae | Problem:
Arătaţi că, pentru orice număr natural $n \geqslant 2$, există un multiplu $m$ al său, nenul, cu următoarele proprietăţi:
a) $m < n^{4}$;
b) în scrierea lui $m$ în baza 10 se folosesc cel mult patru cifre distincte. | [] | Romania | Al doilea test de selecţie pentru OBMJ | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0dxh | Problem:
Poišči vse pare naravnih števil $m$ in $n$, za katere ima kvadratna enačba
$$
2007 x^{2} + m n x + n = 0
$$
samo eno rešitev. Za vsak tak par rešitev tudi zapiši. | [
"Solution:\n\nEnačba ima dvojno ničlo natanko tedaj, ko je diskriminanta enaka $0$, torej\n$$\n0 = (m n)^{2} - 4 \\cdot n \\cdot 2007 = n (m^{2} n - 4 \\cdot 2007)\n$$\nOd tod sledi $m^{2} n = 4 \\cdot 2007 = 2^{2} \\cdot 3^{2} \\cdot 223$. Ker je $223$ praštevilo, mora biti delitelj števila $n$. Pišimo $n = 223 n'... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (m,n) = (1, 8028) with root -2; (2, 2007) with root -1; (3, 892) with root -2/3; (6, 223) with root -1/3. | |
0clf | Find all matrices $A$, $B$, $C \in \mathcal{M}_2(\mathbb{R})$ such that $A = BC - CB$, $B = CA - AC$, and $C = AB - BA$. | [
"If one of the matrices $A$, $B$, $C$ is null, then all the matrices are null. Suppose that there are three matrices $A$, $B$, $C \\in \\mathcal{M}_2(\\mathbb{R}) \\setminus \\{O_2\\}$ which satisfy the equations from the statement. We have $\\text{tr}(A) = \\text{tr}(BC - CB) = 0$. Similarly, $\\text{tr}(B) = \\te... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof and answer | A = B = C = O_2 | |
0k73 | Problem:
Meghana writes two (not necessarily distinct) primes $q$ and $r$ in base 10 next to each other on a blackboard, resulting in the concatenation of $q$ and $r$ (for example, if $q=13$ and $r=5$, the number on the blackboard is now 135). She notices that three more than the resulting number is the square of a pr... | [
"Solution:\n\nTrying $p=2$, we see that $p^{2}-3=1$ is not the concatenation of two primes, so $p$ must be odd. Then $p^{2}-3$ is even. Since $r$ is prime and determines the units digit of the concatenation of $q$ and $r$, $r$ must be $2$. Then $p^{2}$ will have units digit $5$, which means that $p$ will have units... | United States | HMMT November 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 5 | |
00sy | Let $G$, $H$ be the centroid and orthocentre of $\triangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot ABC$ again at $L \neq A$. The tangent to $\omega$ at $L$ intersects $\odot ABC$ at $K \neq L$.
Given that $AG = GH$, prove $\angle HKG = 90^\... | [
"Let $L'$ be the midpoint of $AH$. Then we claim $L'$ lies on $\\odot ABC$.\n\nIndeed, let $D$ be the foot of the $A$-altitude on $BC$. Then:\n$$\nAG = GH \\Rightarrow \\angle GL'A = 90^\\circ \\Rightarrow GL' \\parallel BC \\Rightarrow DL' = \\frac{AL'}{2} = \\frac{HL'}{2} \\Rightarrow DL'... | Balkan Mathematical Olympiad | BMO Short List | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0566 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ which for any real numbers $x$ and $y$ satisfy $$(f(x+y))^2 = x f(x) + 2 f(xy) + (f(y))^2.$$ | [
"*Answer:* $f(x) = 0$ and $f(x) = x$.\n\nSubstituting $x = y = 0$ to the equation, we get $(f(0))^2 = 0 + 2f(0) + (f(0))^2$. Simplifying this gives $f(0) = 0$.\n\nSubstituting $y = 0$ to the original equation, we get the equation $(f(x))^2 = x f(x) + 2 f(0) + (f(0))^2$ which must be satisfied for all real $x$. Sinc... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
01zf | Given $n \ge 2$ different integers greater than $-10$. It turned out that among them the amount of odd numbers equals to the largest even number, and the amount of even numbers equals to the largest odd number.
a) Find the smallest possible value of $n$.
b) Find the greatest possible value of $n$. | [
"a) The largest odd and even numbers are positive integers, since there are numbers of both parities. This means that the largest even number is not less than two, the largest even number is not less than one, and the total number of numbers is not less than three. Note that $n = 3$ is possible if given numbers are... | Belarus | Belarus2022 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | a) 3; b) 19 | |
07xs | Find the smallest prime number which divides $n^2 - 3n + 13$ for some integer $n$. | [
"We define $f(x) = x^2 - 3x + 13$ and note that $f(x) = (x-3)x + 13 = 13 - x(3-x)$ from which we easily discover that $f(3-x) = f(x)$ for all $x$. Calculating $f(0) = 13$ and $f(1) = 11$, we see that the smallest prime $p$ that divides $f(n)$ for some $n$ is at most $11$. We may finish in two different ways.\n\n**W... | Ireland | IRL_ABooklet_2025 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 11 | |
0fs5 | Problem:
Betrachte eine Tabelle mit $m$ Zeilen und $n$ Spalten. Auf wieviele Arten kann diese Tabelle mit lauter Nullen und Einsen ausgefüllt werden, sodass in jeder Zeile und jeder Spalte eine gerade Anzahl Einsen stehen? | [
"Solution:\n\nNenne eine Tabelle, die die Forderungen der Aufgabe erfüllt, zulässig. Wir zeigen zuerst, dass eine zulässige Tabelle durch die Einträge in der oberen linken $m-1 \\times n-1$ Tabelle bereits eindeutig bestimmt ist. Denn die Einträge in der letzten Spalte, ausgenommen das Feld unten rechts, müssen 0 o... | Switzerland | Vorselektionsprüfung | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 2^{(m-1)(n-1)} | |
04pe | Calculate
$$
\frac{\tan 192^\circ + \tan 48^\circ}{1 + \tan 168^\circ \cdot \tan 408^\circ}
$$ | [] | Croatia | Croatian Mathematical Society Competitions | [
"Precalculus > Trigonometric functions"
] | English | final answer only | sqrt(3) | |
09ge | Let $p$ be an odd prime. Show that there is a positive integer $n$ such that $n^{n^n} + n^n + 1$ is divisible by $p$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0h8l | Find possible positive solutions $(x, y, z)$ for the given system of equations:
$$
\begin{cases}
x + y^2 = 2z^3, \\
y + z^2 = 2x^3, \\
z + x^2 = 2y^3.
\end{cases}
$$ | [
"Without loss of generality let $z \\ge \\max\\{x, y\\}$. Consider the following cases.\n\nCase 1. $z > 1$. Then due to the first equation $x + y^2 \\le z + z^2 < 2z^3$ – contradiction.\n\nCase 2. $z = 1$. Due to the first equation, $2 = x + y^2$.\nSince $\\max\\{x, y\\} \\le 1$, it is possible only if $x = y = 1$.... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (1, 1, 1) | |
0aee | а) За кои природни броеви $n$, постојат природни броеви $x$ и $y$ за кои важи
$$
\text{НЗС}(x, y) = n!; \quad \text{НЗД}(x, y) = 2009
$$
б) Одреди го бројот на парови $(x, y)$ за кои важи
$$
\text{НЗС}(x, y) = 4!; \text{НЗД}(x, y) = 2009; x \le y
$$ | [
"а) Од $\\text{НЗД}(x, y) = 2009$ следува дека $x = 2009a$ и $y = 2009b$, каде што $a$ и $b$ се природни броеви така што $\\text{НЗД}(a,b)=1$.\nОд $\\text{НЗС}(x, y) = n!$ следува $2009ab = n!$, односно $7^2 \\cdot 4! ab = n!$. Од последново следува $n \\ge 4!$. Условот е и доволен бидејќи ако $n \\ge 4!$, за $x = ... | North Macedonia | Републички натпревар по математика за средно образование | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | Macedonian, English | proof and answer | a) n ≥ 4!; b) 2048 | |
08bj | Problem:
Siano $m, n$ due interi maggiori o uguali a $2$. Di una tabella a $m$ righe e $n$ colonne si sa che ogni casella contiene o il numero $1$ o il numero $-1$, e che la somma totale di tutte le caselle è maggiore o uguale a zero. Genoveffa considera i percorsi che uniscono una casella della prima colonna (a sua s... | [
"Solution:\n\na. Supponiamo che la tabella contenga $1$ e $-1$ disposti \"a scacchiera\" (in modo che le caselle adiacenti in orizzontale e verticale a una casella contenente $1$ contengano $-1$, e viceversa), in modo che la casella in alto a sinistra contenga il numero $1$ (questo garantisce che la somma dei numer... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
00ae | Given 100 infinitely large boxes with markers in them, the following procedure is carried out. At step 1 one adds one marker in every box. At step 2 one marker is added in every box containing an even number of markers. At step 3 one marker is added in every box in which the number of markers is divisible by 3, and so ... | [
"The answer is *no*. Regardless of the initial distribution all boxes will contain the same number of markers after finitely many steps. Moreover this is true for any number of boxes.\n\nDenote by $x_n$ the number of markers in a certain box before step $n$, $n = 1, 2, \\dots$. Suppose that $x_n = n$ for some $n$. ... | Argentina | Argentine National Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | no | |
0kpt | Problem:
Does there exist a regular pentagon whose vertices lie on edges of a cube? | [
"Solution:\n\nIf two of the sides of the pentagon lie on the same face of the cube, then we have that all sides of the pentagon lie on this face, which would mean that one could choose five points on the boundary of the unit square that forms a pentagon. This is impossible by the following argument. By pigeonhole, ... | United States | HMIC | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > An... | null | proof only | null | |
0gk5 | Let $A$, $B$, $C$ be three distinct points on a unit circle. Let $G$ and $H$ be the centroid and the orthocenter of the triangle $ABC$, respectively. Let $F$ be the midpoint of the segment $GH$. Evaluate $|AF|^2 + |BF|^2 + |CF|^2$. | [
"Define a coordinate system with the origin at the center of the circle. We can see that $\\vec{H} = \\vec{A} + \\vec{B} + \\vec{C}$ and $\\vec{G} = \\frac{1}{3}(\\vec{A} + \\vec{B} + \\vec{C})$.\nThus, $\\vec{F} = \\frac{\\vec{G} + \\vec{H}}{2} = \\frac{2}{3}(\\vec{A} + \\vec{B} + \\vec{C})$. We now have\n$$\n\\be... | Thailand | Thai Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | 3 | |
0ilo | Problem:
The polynomial $f(x) = x^{2007} + 17 x^{2006} + 1$ has distinct zeroes $r_1, \ldots, r_{2007}$. A polynomial $P$ of degree $2007$ has the property that $P\left(r_j + \frac{1}{r_j}\right) = 0$ for $j = 1, \ldots, 2007$. Determine the value of $\frac{P(1)}{P(-1)}$. | [
"Solution:\n\nAnswer: $\\mathbf{289}$.\n\nFor some constant $k$, we have\n$$\nP(z) = k \\prod_{j=1}^{2007} \\left(z - \\left(r_j + \\frac{1}{r_j}\\right)\\right)\n$$\nNow writing $\\omega^3 = 1$ with $\\omega \\neq 1$, we have $\\omega^2 + \\omega = -1$. Then\n$$\n\\begin{gathered}\n\\frac{P(1)}{P(-1)} = \\frac{k \... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 289/259 | |
0f8e | Problem:
What is the smallest $n$ for which there is a solution to
$$\sin x_{1} + \sin x_{2} + \ldots + \sin x_{n} = 0,$$
$$\sin x_{1} + 2 \sin x_{2} + \ldots + n \sin x_{n} = 100?$$ | [
"Solution:\n\nPut $x_{1} = x_{2} = \\ldots = x_{10} = 3\\pi /2$, $x_{11} = x_{12} = \\ldots = x_{20} = \\pi /2$. Then\n$$\nsin x_{1} + \\sin x_{2} + \\ldots + \\sin x_{20} = (-1 - 1 - 1 - \\ldots - 1) + (1 + 1 + \\ldots + 1) = 0,\n$$\nand\n$$\nsin x_{1} + 2 \\sin x_{2} + \\ldots + 20 \\sin x_{20} = - (1 + 2 + \\ldo... | Soviet Union | 22nd ASU | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 20 | |
00z5 | Problem:
Let $\alpha$ be the angle between two lines containing the diagonals of a regular 1996-gon, and let $\beta \neq 0$ be another such angle. Prove that $\alpha / \beta$ is a rational number. | [
"Solution:\n\nLet $O$ be the circumcentre of the 1996-gon. Consider two diagonals $AB$ and $CD$. There is a rotation around $O$ that takes the point $C$ to $A$ and $D$ to a point $D'$. Clearly the angle of this rotation is a multiple of $2\\varphi = 2\\pi / 1996$.\n\nThe angle $BAD'$ is the inscribed angle on the a... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06hb | Let $a = BC$, $b = CA$ and $c = AB$ be respectively the lengths of a triangle $ABC$; $i_a, i_b, i_c$ be respectively the lengths of the angle bisectors from $A, B$ and $C$. Let $R$ be the circumradius of the triangle. Prove that
$$
ai_a + bi_b + ci_c < 9R^2.
$$ | [
"Let $D$ be the intersection point of the internal angle bisector of $\\angle BAC$ and $BC$.\nBy the angle bisector theorem, we have $\\frac{BD}{CD} = \\frac{c}{b}$, hence\n$$\nBD = \\frac{ac}{b+c} \\quad \\text{and} \\quad CD = \\frac{ab}{b+c}.\n$$\nBy Stewart's theorem, we have\n$$\ni_a^2 = \\frac{1}{a} \\left( b... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
056p | Let a simple polynomial function be a polynomial function $P(x)$ whose coefficients belong to the set $\{-1, 0, 1\}$. Let $n$ be a positive integer, $n > 1$. Find the smallest possible number of non-zero coefficients in a simple polynomial function of $n$th order whose values at all integral arguments are divisible by ... | [
"A single non-zero coefficient is not sufficient for any $n > 1$ as the only simple polynomial functions with a single non-zero coefficient are $P(x) = x^n$ and $P(x) = -x^n$ but in both cases $n \\nmid P(1)$. Let us show that the values of the polynomial function $P_n(x) = x^n - x^{n-\\varphi(n)}$ at all integral ... | Estonia | IMO Team Selection Contest | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 2 | |
014m | Problem:
A 12-digit positive integer consisting only of digits $1$, $5$ and $9$ is divisible by $37$. Prove that the sum of its digits is not equal to $76$. | [
"Solution:\n\nLet $N$ be the initial number. Assume that its digit sum is equal to $76$.\n\nThe key observation is that $3 \\cdot 37 = 111$, and therefore $27 \\cdot 37 = 999$. Thus we have a divisibility test similar to the one for divisibility by $9$: for $x = a_n 10^{3n} + a_{n-1} 10^{3(n-1)} + \\cdots + a_1 10^... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
03le | Problem:
Let $S$ be a set of $n$ points in the plane such that any two points of $S$ are at least $1$ unit apart. Prove there is a subset $T$ of $S$ with at least $n / 7$ points such that any two points of $T$ are at least $\sqrt{3}$ units apart. | [
"Solution:\nWe will construct the set $T$ in the following way: Assume the points of $S$ are in the $xy$-plane and let $P$ be a point in $S$ with maximum $y$-coordinate. This point $P$ will be a member of the set $T$ and now, from $S$, we will remove $P$ and all points in $S$ which are less than $\\sqrt{3}$ units f... | Canada | 2003 CMO | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
08n7 | Problem:
Solve in positive integers the equation $1005^{x} + 2011^{y} = 1006^{z}$. | [
"Solution:\nWe have $1006^{z} > 2011^{y} > 2011$, hence $z \\geq 2$. Then $1005^{x} + 2011^{y} \\equiv 0 \\pmod{4}$.\nBut $1005^{x} \\equiv 1 \\pmod{4}$, so $2011^{y} \\equiv -1 \\pmod{4} \\Rightarrow y$ is odd, i.e. $2011^{y} \\equiv -1 \\pmod{1006}$.\nSince $1005^{x} + 2011^{y} \\equiv 0 \\pmod{1006}$, we get $10... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | (2, 1, 2) | |
09dj | Гэр бүлийн $n$ ($n \ge 3$) хосыг дугуй ширээ тойруулан суулгахад эрэгтэйчүүд ба эмэгтэйчүүд нь сөөлжлөн суусан ба нэг бүлийн ямарч хос зэрэгцэж суугаагүй байх боломжийн тоог ол (сандлууд дугаартай гэж ойлгоно). | [
"**ДБ-А1:** (П.Энхболын ХІІ уралдаан х.30.)"
] | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | Mongolian | proof and answer | 2n · a(n), where a(n) is the ménage number (the number of alternating round-table seatings of n couples up to rotation with no spouses adjacent). Equivalently, 2 · n! · W_n, where W_n is the number of permutations π of {1,…,n} with π(i) not equal to i or i−1 modulo n. | |
00yg | Problem:
Find the largest value of the expression
$$
x y + x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}} - \sqrt{(1 - x^{2})(1 - y^{2})}
$$ | [
"Solution:\nThe expression is well-defined only for $|x|, |y| \\leq 1$ and we can assume that $x, y \\geq 0$. Let $x = \\cos \\alpha$ and $y = \\cos \\beta$ for some $0 \\leq \\alpha, \\beta \\leq \\frac{\\pi}{2}$. This reduces the expression to\n$$\n\\cos \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta + \\cos \... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(2) | |
05j7 | Problem:
Soit $ABCD$ un quadrilatère convexe. Soit $M$ l'intersection entre les bissectrices intérieures des angles $B$ et $C$, et $N$ l'intersection entre les bissectrices intérieures des angles $A$ et $D$. Montrer que les droites $AB$, $CD$ et $MN$ sont concourantes. | [
"Solution:\n\nQuitte à échanger les rôles de $A$ et de $B$ et les rôles de $C$ et de $D$, on peut supposer que $A$ et $D$ se trouvent sur les segments $[EB]$ et $[EC]$ respectivement.\n\n\n\nNotons $E$ le point d'intersection entre $(AB)$ et $(CD)$. Alors $M$ est l'intersection des bissectr... | France | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08w1 | For each positive integer $k$, denote by $S(k)$ the sum of the digits of $k$. How many positive integers $n$ less than or equal to $999$ are there for which $\frac{S(n)}{S(n+1)}$ is an integer? | [
"A positive integer $n$ less than or equal to $999$ can be represented as $10^2a + 10b + c$ where $a, b, c$ are integers satisfying $0 \\le a, b, c \\le 9$ and $a + b + c \\ne 0$. Then we have $S(n) = a + b + c$ and\n$$\nS(n+1) = \\begin{cases} a + b + c + 1, & \\text{if } c < 9 \\cdots (\\text{i}), \\\\ a + b + 1,... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 17 | |
0fr6 | Problem:
Sea $n$ un entero positivo. En una cuadrícula de tamaño $n \times n$, algunas casillas tienen un espejo de doble cara a lo largo de una de sus diagonales. En el exterior de cada casilla de los lados izquierdo y derecho de la cuadrícula se encuentra un puntero láser, que apunta horizontalmente hacia la cuadríc... | [
"Solution:\n\nConsideremos la unión $S$ de las líneas de centros de cada fila y columna. Como cada espejo forma un ángulo de $45$ grados con las direcciones de la cuadrícula,\n\n\nFigura 2: Un ejemplo de una configuración para el Problema 3, donde se muestran los recorridos de dos láseres.\... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0i16 | Problem:
In the UC Berkeley bureaucratic hierarchy, certain administrators report to certain other administrators. It so happens that if $A$ reports to $B$ and $B$ reports to $C$, then $C$ reports to $A$. Also, administrators do not report to themselves. Prove that all the administrators can be divided into three disj... | [
"Solution:\n\nForm a graph whose vertices represent the administrators, with an edge between $A$ and $B$ if one of $A, B$ reports to the other. We will temporarily assume that the graph is connected. Consider any walk on this graph (i.e. a finite sequence of vertices, any two consecutive members of which are connec... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0f5f | Problem:
$x$ is a real. The decimal representation of $x$ includes all the digits at least once. Let $f(n)$ be the number of distinct $n$-digit segments in the representation. Show that if for some $n$ we have $f(n) \leq n + 8$, then $x$ is rational. | [] | Soviet Union | 17th ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
0g76 | 由正整數 $37$ 開始，依序在各項的前方加一數字 $5$，形成下面的數列：
$37$, $537$, $5537$, $55537$, $555537$, ...
請問此數列中有多少項是質數？ | [
"此數列只有第 $1$ 項是質數，其他項均為合數。\n\n將此數列的第 $n$ 項記為 $a_n$。由數學歸納法可知下列事實：\n\n- $a_1$ 被 $37$ 整除。$a_{n+3} = 555 \\cdot 10^{n+1} + a_n$，而 $555 = 3 \\cdot 5 \\cdot 37$。所以 $a_1, a_4, a_7, \\dots$ 均為 $37$ 的倍數。\n\n- $a_2 = 537$ 被 $3$ 整除。$a_{n+3} = 555 \\cdot 10^{n+1} + a_n$，而 $555 = 3 \\cdot 5 \\cdot 37$。所以 $a_2, a_5, a_8, \\dots$ ... | Taiwan | 二0一三數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1 | |
02oy | Problem:
Observe que
$$
\begin{gathered}
1^{2}+2^{2}+(1 \times 2)^{2}=3^{2} \\
2^{2}+3^{2}+(2 \times 3)^{2}=7^{2} \\
3^{2}+4^{2}+(3 \times 4)^{2}=13^{2}
\end{gathered}
$$
Prove que se $a$ e $b$ são inteiros consecutivos então o número
$$
a^{2}+b^{2}+(a b)^{2}
$$
é um quadrado perfeito. | [
"Solution:\n\nSuponha, sem perda de generalidade, que $b>a$, isto é, $b-a=1$. Então\n$$\n\\begin{gathered}\n(b-a)^{2}=1^{2} \\\\\nb^{2}-2 a b+a^{2}=1 \\\\\na^{2}+b^{2}=2 a b+1\n\\end{gathered}\n$$\nSomando $(a b)^{2}$ em cada lado da igualdade, temos\n$a^{2}+b^{2}+(a b)^{2}=(2 a b+1)+(a b)^{2}=(a b)^{2}+2(a b) \\cd... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
005r | Dos personas participan en un juego donde hay fichas negras, fichas blancas y dos cajas. El primer jugador pone varias de sus fichas en una de las cajas, y otras varias en otra caja. Está permitido no poner ninguna ficha en una de las cajas y, además, no es obligatorio poner todas las fichas disponibles en las cajas. A... | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Spanish | proof and answer | Exactly when a + b is divisible by 3 and both colors are initially present; more precisely, the first player can force the goal if and only if a + b ≡ 0 (mod 3) and ab > 0, with the trivial case (a, b) = (0, 0) also winning. | |
0gw6 | Find all positive integers $a$, $100 \le a \le 999$, such that the decimal values of $a^2$ and $(3a - 2)^2$ have the same three-digits endings. | [
"Відповідь: $251$, $313$, $501$, $563$, $751$, $813$.\n\nМаємо: $(3a-2)^2 - a^2 = 1000k$.\nТобто, $(2a-1)(a-1) = 250k = 2 \\cdot 5^3 \\cdot k$.\nЧисла $2a-1$ і $a-1$ взаємно прості, причому перше з них непарне. Отже, друге число парне, і, відповідно, число $a$ непарне. Звідси випливає, що $a-1=250$, або ж $2a-1=125... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [251, 313, 501, 563, 751, 813] | |
0jr5 | Problem:
Find the number of ordered pairs of integers $(a, b) \in \{1,2, \ldots, 35\}^2$ (not necessarily distinct) such that $a x+b$ is a "quadratic residue modulo $x^2+1$ and 35", i.e. there exists a polynomial $f(x)$ with integer coefficients such that either of the following equivalent conditions holds:
- there exi... | [
"Solution:\nAnswer: $225$\n\nBy the Chinese remainder theorem, we want the product of the answers modulo $5$ and modulo $7$ (i.e. when $35$ is replaced by $5$ and $7$, respectively).\n\nFirst we do the modulo $7$ case. Since $x^2+1$ is irreducible modulo $7$ (or more conceptually, in $\\mathbb{F}_7[x]$), exactly ha... | United States | HMMT February 2015 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof and answer | 225 | |
0jom | Problem:
Let $S=\{1,2,4,8,16,32,64,128,256\}$. A subset $P$ of $S$ is called squarely if it is nonempty and the sum of its elements is a perfect square. A squarely set $Q$ is called super squarely if it is not a proper subset of any squarely set. Find the number of super squarely sets.
(A set $A$ is said to be a proper... | [
"Solution:\nAnswer: 5 Clearly we may biject squarely sets with binary representations of perfect squares between 1 and $2^{0}+\\cdots+2^{8}=2^{9}-1=511$, so there are 22 squarely sets, corresponding to $n^{2}$ for $n=1,2, \\ldots, 22$. For convenience, we say $N$ is (super) squarely if and only if the set correspon... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 5 | |
06lx | Let $\triangle ABC$ be an acute triangle with circumcircle $\Gamma$, and let $P$ be the midpoint of the minor arc $BC$ of $\Gamma$. Let $AP$ meet $BC$ at $D$, and let $M$ be the midpoint of $AB$. Also, let $E$ be the point such that $AE \perp AB$ and $BE \perp MP$. Prove that $AE = DE$. | [
"Since $P$ is the midpoint of the minor arc $BC$, $AD$ is the internal angle bisector of $\\angle BAC$.\nLet $E'$ be the intersection of the perpendicular bisector of $AD$ and the line through $A$ perpendicular to $AB$. Using the method of false position, it suffices to show $BE' \\perp MP$, and then conclude $E' =... | Hong Kong | Year 2021 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fbq | Problem:
Un cuadrilátero convexo tiene la propiedad que cada una de sus dos diagonales biseca su área. Demuestra que este cuadrilátero es un paralelogramo. | [
"Solution:\n\n\n\nSea $ABCD$ el cuadrilátero dado. Es sabido que al trazar paralelas a cada diagonal por los extremos de la otra se forma un paralelogramo ($XYZT$ en la figura) de área doble de la del cuadrilátero de partida.\n\nSi $AC$ biseca a $ABCD$ también biseca a $XYZT$, pero siendo $... | Spain | null | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0bf1 | Let $n$ be an integer greater than $1$. The set $S$ of all diagonals of a $(4n-1)$-gon is partitioned into $k$ sets, $S_1, \dots, S_k$, so that, for every pair of distinct indices $i$ and $j$, some diagonal in $S_i$ crosses some diagonal in $S_j$; that is, the two diagonals share an interior point. Determine the larges... | [
"The required maximum is $k = (n-1)(4n-1)$. Clearly, $|S| = 2(n-1)(4n-1)$. To begin, we show that $k \\le (n-1)(4n-1)$. Otherwise, some $S_i$ is a singleton set, say $S_i = \\{\\delta\\}$. Let $m$ be the number of vertices on one side of $\\delta$, so the number of vertices on the other side is $4n - m - 3$, and th... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | (n-1)(4n-1) | |
04io | Let $ABCD$ be a quadrilateral such that $|AB| = 6$, $|BC| = 9$, $|CD| = 18$ and $|AD| = 5$ hold. Determine the length of the diagonal $AC$ if it is known that it is a positive integer.
(Andrea Aglić-Aljinović) | [
"**2.5.** Let $a_n$ be the number the grasshopper is located at after the $n^{th}$ jump, i.e.\n$$\na_1 = 1, \\quad a_n = 1 + k + \\dots + k^{n-1}, \\quad n \\ge 2.\n$$\nWe are looking for all numbers $k$ such that $2015 \\nmid a_n$ for all $n = 1, \\dots, 2015$.\nSuppose that $M(k, 2015) = d > 1$. Then every $a_n$ ... | Croatia | First round – City competition | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof and answer | 14 | |
0d12 | Let $\mathbb{Q}$ be the set of rational numbers. Find all functions $f: \mathbb{Q} \to \mathbb{Q}$ such that for all rational numbers $x, y$,
$$
f(f(x) + x f(y)) = x + f(x)y.
$$ | [
"**Solution 1.** We show $f(x) = x$ is the only solution. It is easy to check that it works.\n\nPut in $y = 0$ to obtain that $f$ is surjective. Let $c$ be a real with $f(c) \\neq 0$ and suppose $f(a) = f(b)$. Then\n$$\na = \\frac{f(f(c) + c f(a)) - c}{f(c)} = \\frac{f(f(c) + c f(b)) - c}{f(c)} = b,\n$$\nso $f$ is ... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = x for all rational x | |
09gs | $2 \times n$ хэмжээтэй хүснэгтийн зарим нүдийг будахад, будагдсан аль ч хоёр нүд нь хөрш биш байвал уг будалтыг “зөв будалт” гэж нэрлэе. Тэгш тооны нүдийг будсан зөв будалтын тоо ба сондгой тооны нүдийг будсан зөв будалтын тооны ялгаврыг ол. (Ерөнхий талтай нүднүүдийг хөрш нүднүүд гэж нэрлэнэ) | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Generating functions"
] | English | proof and answer | (-1)^{ceil(n/2)} | |
01wv | $1019$ stones are placed into two non-empty boxes. Each second Alex chooses a box with an even amount of stones and shifts half of these stones into another box.
Prove that for each $k$, $1 \le k \le 1018$, at some moment there will be a box with exactly $k$ stones. | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
04xp | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x + f(y)) - f(x) = (x + f(y))^4 - x^4
$$
for all $x, y \in \mathbb{R}$. | [
"We rewrite the given equation into the equivalent form\n$$\nf(x + f(y)) = (x + f(y))^4 - x^4 + f(x). \\quad (1)\n$$\nSetting $x = -f(z)$, $y = z$ in (1) we obtain\n$$\nf(0) = -(f(z))^4 + f(-f(z)) \\quad \\text{for all } z \\in \\mathbb{R}. \\quad (2)\n$$\nNow, setting $x = -f(z)$ in (1) and using (2) we get\n$$\nf... | Czech-Polish-Slovak Mathematical Match | 12th Czech-Polish-Slovak Mathematics Competition | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All solutions are f(x) = x^4 + k for any real constant k, and the zero function f(x) = 0. | |
0etl | Show that there are infinitely many polynomials $P$ with real coefficients such that if $x$, $y$, and $z$ are real numbers such that $x^2 + y^2 + z^2 + 2xyz = 1$, then
$$
P(x)^2 + P(y)^2 + P(z)^2 + 2P(x)P(y)P(z) = 1.
$$ | [
"Let us call a triple $(x, y, z)$ of real numbers a *-triple if it satisfies\n$$\nx^2 + y^2 + z^2 + 2xyz = 1.\n$$\nLet us call a polynomial $p(x)$ with real coefficients a *-polynomial if $(x, y, z)$ a *-triple implies $(p(x), p(y), p(z))$ a *-triple.\nWe first investigate polynomials of degree at most 1. Hence, as... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
0f9d | Problem:
Find all integers $n$ such that $\left[ \frac{n}{1!} \right] + \left[ \frac{n}{2!} \right] + \ldots + \left[ \frac{n}{10!} \right] = 1001$. | [] | Soviet Union | 24th ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 584 | |
04tf | In how many ways can you partition the set $\{1, 2, \ldots, 12\}$ into six mutually disjoint two-element sets in such a way that the two elements in any set are coprime? | [
"No two even numbers can be in the same set (pair). Let us call partitions of $\\{1, 2, \\ldots, 12\\}$ with this property, that is one even and one odd number in each pair, even-odd partitions. The only further limitations are, that $6$ nor $12$ cannot be paired with $3$ or $9$, and $10$ cannot be paired with $5$.... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 252 | |
01aa | Three line segments, all of length $1$, form a connected figure on the plane. Any point that is common to two of these line segments is an endpoint of both segments. Find the maximum area of the convex hull of the figure. | [
"**Answer:** $\\frac{3}{4}\\sqrt{3}$.\n\nClearly all vertices of the convex hull are some endpoints of the line segments. As the figure is connected, there are at most $4$ different locations of the endpoints of line segments. Hence the convex hull is either a quadrilateral or a triangle. We can assume that there a... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > P... | null | proof and answer | 3/4*sqrt(3) | |
0316 | Problem:
Let $k_{1}$ and $k_{2}$ be circles with centers $O_{1}$ and $O_{2}$, $O_{1}O_{2}=25$, and radii $R_{1}=4$ and $R_{2}=16$, respectively. Consider a circle $k$ such that $k_{1}$ is internally tangent to $k$ at a point $A$, and $k_{2}$ is externally tangent to $k$ at a point $B$.
a) Prove that the segment $AB$ ... | [
"Solution:\n\na) We shall prove that the position of the point $S = O_{1}O_{2} \\cap AB$ does not depend on $k$. Let $O_{3}$ be the center of $k$. It follows by the Menelaus theorem for $\\triangle O_{1}O_{2}O_{3}$ and the line $AB$ that\n$$\n\\frac{O_{3}B}{BO_{2}} \\cdot \\frac{O_{2}S}{SO_{1}} \\cdot \\frac{O_{1}A... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-G... | null | proof and answer | 12 |
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