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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0atj | Problem:
Let $x$ be a real number so that $x + \frac{1}{x} = 3$. Find the last two digits of $x^{2^{2013}} + \frac{1}{x^{2^{2013}}}$. | [
"Solution:\n\nLet $x + \\frac{1}{x} = 3$.\n\nLet $a_n = x^n + \\frac{1}{x^n}$ for $n \\geq 0$.\n\nWe have the recurrence:\n$$\na_{n} = (x^n + x^{-n}) = (x + x^{-1}) a_{n-1} - a_{n-2}\n$$\nSo,\n$$\na_n = 3 a_{n-1} - a_{n-2}\n$$\nwith $a_0 = 2$, $a_1 = 3$.\n\nLet us compute $a_2$:\n$$\na_2 = 3 a_1 - a_0 = 3 \\times 3... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 07 | |
0453 | Let $\Gamma_2$ be a circle contained in the interior of another circle $\Gamma_1$ on the plane. Prove that there exists a point $P$ on the plane satisfying the following conditions: if $\ell$ is a line that does not contain $P$, that intersects $\Gamma_1$ at two different points $A, B$, and that intersects $\Gamma_2$ a... | [
"\n\n**Proof.** Denote the centers of the two circles by $O_1, O_2$, and the radii by $r_1, r_2$, respectively, where $r_1 > r_2$. We first show that there are two points $P, Q$ on the ray $O_1O_2$ such that $O_1P \\cdot O_1Q = r_1^2$ and $O_2P \\cdot O_2Q = r_2^2$.\nOne can choose a point ... | China | 2022 China Team Selection Test for IMO | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
07kj | Find, with proof, all triples of integers $(a, b, c)$ such that $a$, $b$ and $c$ are the lengths of the sides of a right angled triangle whose area is $a + b + c$. | [
"Assume for the moment that $a \\leq b \\leq c$. So\n$$\na^2 + b^2 = c^2. \\quad (1)\n$$\nThe condition on the area entails $ab = 2(a + b + c)$ which implies that\n$$\n4c^2 = (ab)^2 - 4ab(a + b) + 4(a^2 + b^2 + 2ab). \\quad (2)\n$$\nEquations (1) and (2) imply that\n$$\nab - 4(a + b) + 8 = 0\n$$\nor\n$$\nab - 4(a +... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (5, 12, 13) and (6, 8, 10) (up to permutation) | |
0h82 | Determine all positive integers $n$ such that number $N$ is also integer, where
$$
N = \frac{2 \cdot 1}{\sqrt{1^2 + 1 + 4 + \sqrt{1^2 - 1 + 4}}} + \frac{2 \cdot 2}{\sqrt{2^2 + 2 + 4 + \sqrt{2^2 - 2 + 4}}} + \dots + \frac{2n}{\sqrt{n^2 + n + 4 + \sqrt{n^2 - n + 4}}}
$$
(Anikushyn A., Rublyov B.) | [
"Let us note that\n$$\n\\frac{2k}{\\sqrt{k^2 + k + 4} + \\sqrt{k^2 - k + 4}} = \\frac{(\\sqrt{k^2 + k + 4} - \\sqrt{k^2 - k + 4})2k}{(k^2 + k + 4) - (k^2 - k + 4)} = \\\\\n= \\sqrt{k^2 + k + 4} - \\sqrt{k^2 - k + 4} = \\sqrt{k^2 + k + 4} - \\sqrt{(k-1)^2 + (k-1) + 4}.\n$$\nAnd thus\n$$\nN = (\\sqrt{6} - \\sqrt{4}) ... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3 | |
029l | Problem:
Construindo quadrados perfeitos - Observe as seguintes igualdades:
Será que isso é sempre verdadeiro? Isto é: o produto de quatro números inteiros consecutivos, mais $1$, é sempre um quadrado perfeito? | [
"Solution:\n\nSim, será sempre um quadrado perfeito. De fato, se $n-1$, $n$, $n+1$ e $n+2$ são quatro inteiros consecutivos, então seu produto mais $1$ é dado por:\n$$\n\\begin{aligned}\n(n-1) n(n+1)(n+2)+1 & = n\\left(n^{2}-1\\right)(n+2)+1 \\\\\n& = n\\left(n^{3}+2 n^{2}-n-2\\right)+1 \\\\\n& = n^{4}+2 n^{3}-n^{2... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | null | proof only | null | |
02g7 | Line *r* passes through the corner $A$ of a sheet of paper and makes an angle $\alpha$ with the horizontal border, as shown in figure 1. In order to divide $\alpha$ into three equal parts we proceed as follows:
a) initially we mark two points $B$ and $C$ on the vertical border such that $AB = BC$; through $B$ we draw ... | [
"Let $P$ and $X$ be the points determined by the crease in the lower horizontal border of the sheet and on the line $s$, respectively. Let $\\beta = \\angle PAA'$. Since $AP = AP'$, $\\angle PA'A = \\angle AA'B = \\beta$, thus $\\angle PA'X = 2\\beta$. Moreover $\\triangle APX \\cong \\triangle A'PX$, so that $\\an... | Brazil | XXII OBM | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0esv | Let $\triangle ABC$ be a triangle with circumradius $R$, and let $\ell_A, \ell_B, \ell_C$ be the altitudes through $A, B, C$ respectively. The altitudes meet at $H$. Let $P$ be an arbitrary point in the same plane as $ABC$. The feet of the perpendicular lines through $P$ onto $\ell_A, \ell_B, \ell_C$ are $D, E, F$ resp... | [
"\n\nNote that $\\angle PDH = \\angle PFH = \\angle PEH = 90^\\circ$ by construction, so by Thales's Theorem, $D, E$ and $F$ lie on a circle whose diameter is $PH$. Therefore, the angle between lines $DE$ and $FE$ is the same as the angle between lines $DH$ and $FH$ (angles subtended by the... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g20 | Problem:
Soit $ABC$ un triangle aigu et $O$ le centre de son cercle circonscrit. La droite $OA$ coupe la hauteur $h_{b}$ en $P$ et la hauteur $h_{c}$ en $Q$. Soit $H$ l'orthocentre du triangle $ABC$. Prouver que le centre du cercle circonscrit au triangle $PQH$ est sur la médiane du triangle $ABC$ passant par $A$. | [
"Solution:\n\nOn commence par une petite chasse aux angles. Soient $\\alpha, \\beta$ et $\\gamma$ les trois angles du triangle $\\triangle ABC$. Observez que $\\angle HPQ = \\pi / 2 - \\angle PAC$ parce que $PH$ est perpendiculaire à $AC$. De plus, comme $OA$ est un diamètre du cercle circonscrit à $\\triangle ABC$... | Switzerland | SMO-Selektion | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
02q0 | Problem:
Existem bolas azuis e bolas vermelhas em uma caixa. A probabilidade de sortear duas bolas de cores diferentes, ao retirar duas bolas ao acaso, é $1/2$. Prove que o número de bolas na caixa é um quadrado perfeito. | [
"Solution:\n\nSuponha que existam $a$ bolas azuis e $v$ bolas vermelhas na caixa.\n\n(1) O número de modos de escolher duas bolas de cores diferentes é $a v$.\n\n(2) O número de modos de escolher duas bolas quaisquer é $\\binom{a+v}{2}$.\n\n(3) De (1) e (2), a probabilidade de sortear duas bolas de cores diferentes... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0f9q | Problem:
A graph has 30 points and each point has 6 edges. Find the total number of triples such that each pair of points is joined or each pair of points is not joined. | [
"Solution:\nThere are $30 \\times 29 \\times 28 / 6 = 4060$ triples in all. Let $m$ be the number of triples with $0$ or $3$ edges, and let $n$ be the number of triples with $1$ or $2$ edges. So $m + n = 4060$.\n\nEach point is joined to $6$ others, so it is in $6 \\times 5 / 2 = 15$ triples where it is joined to b... | Soviet Union | 24th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1990 | |
0jg3 | Problem:
How many orderings $\left(a_{1}, \ldots, a_{8}\right)$ of $(1,2, \ldots, 8)$ exist such that $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}-a_{8}=0$? | [
"Solution:\nWe can divide the numbers up based on whether they have a $+$ or $-$ before them. Both the numbers following $+$'s and $-$'s must add up to $18$. Without loss of generality, we can assume the $+$'s contain the number $1$ (and add a factor of $2$ at the end to account for this). The possible $4$-element ... | United States | HMMT | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 4608 | |
06ko | For which natural number $n$ is it possible to place natural numbers from $1$ to $3n$ on the edges of a right $n$-angled prism (on each edge there is exactly one number placed and each one is used exactly 1 time) in such a way that the sum of all the numbers that surround each face is the same? | [
"The only possible $n$ is $4$.\nWe call those edges which are the sides of a base the *base edges*, and call the remaining edges the *lateral edges*. Let $s$ be the sum of all numbers surrounding each face, and let $x$ be the sum of all numbers on the lateral edges.\n\nFirstly, we consider the sum of all numbers on... | Hong Kong | null | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 4 | |
0f2g | Problem:
Given a polynomial $x^{10} + a_9 x^9 + \ldots + a_1 x + 1$. Two players alternately choose one of the coefficients $a_1$ to $a_9$ (which has not been chosen before) and assign a real value to it. The first player wins iff the resulting polynomial has no real roots. Who wins? | [] | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | Second player | |
0ku4 | Problem:
Let $\zeta = e^{2 \pi i / 99}$ and $\omega = e^{2 \pi i / 101}$. The polynomial
$$
x^{9999} + a_{9998} x^{9998} + \cdots + a_{1} x + a_{0}
$$
has roots $\zeta^{m} + \omega^{n}$ for all pairs of integers $(m, n)$ with $0 \leq m < 99$ and $0 \leq n < 101$. Compute $a_{9799} + a_{9800} + \cdots + a_{9998}$. | [
"Solution:\nLet $b_{k} := a_{9999-k}$ for sake of brevity, so we wish to compute $b_{1} + b_{2} + \\cdots + b_{200}$. Let $p_{k}$ be the sum of the $k$-th powers of $\\zeta^{m} + \\omega^{n}$ over all ordered pairs $(m, n)$ with $0 \\leq m < 99$ and $0 \\leq n < 101$. Recall that Newton's sums tells us that\n$$\n\\... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 14849 - (9999/200) * C(200, 99) | |
0l8b | The parabola with equation $y = x^2 - 4$ is rotated $60^\circ$ counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has $y$-coordinate $\frac{a - \sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers, and $a$ and $c$ are relatively prime.... | [] | United States | 2025 AIME I | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 62 | |
0ix6 | Find all pairs of positive integers $(m, n)$ such that $mn - 1$ divides $(n^2 - n + 1)^2$. | [
"**Solution 1** (Based on the work of John Berman). The answer is $(m, n) = (2, 2)$ or $(m, n) = (t^2 \\pm 2t + 2, t^2 + 1)$ for some $t \\ge 0$.\nWe first check that these pairs work: $2 \\cdot 2 - 1 = 3 \\mid 3^2 = (2^2 - 2 + 1)^2$ and\n$$\n(t^2 \\pm 2t + 2)(t^2 + 1) - 1 = (t^2 \\pm t + 1)^2 \\mid (t^4 + t^2 + 1)... | United States | Team Selection Test 2009 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | (m, n) = (2, 2) or (m, n) = (t^2 ± 2t + 2, t^2 + 1) for any integer t ≥ 0 | |
0iuf | Problem:
In general, if there are $d$ doors in every room (but still only 1 correct door) and $r$ rooms, the last of which leads into Bowser's level, what is the expected number of doors through which Mario will pass before he reaches Bowser's level? | [
"Solution:\n\nAnswer: $\\frac{d\\left(d^{r}-1\\right)}{d-1}$\n\nLet $E_{i}$ be the expected number of doors through which Mario will pass in the future if he is currently in room $i$ for $i=1,2, \\ldots, r+1$ (we will set $E_{r+1}=0$). We claim that\n$$\nE_{i} = 1 + \\frac{d-1}{d} E_{1} + \\frac{1}{d} E_{i+1}.\n$$\... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | d(d^r - 1)/(d - 1) | |
06bp | For a positive integer $n$, let $f(n)$ be the largest $k$ such that $2^k$ divides $n$ and $g(n)$ be the sum of the digits of the binary representation of $n$. Prove that for any positive integers $n$,
(i) $f(n!) = n - g(n)$;
(ii) $4$ divides $\binom{2n}{n} = \frac{(2n)!}{n!n!}$ if and only if $n$ is not a power of $2$. | [
"(i) This is an alternative form of Legendre's formula. Here is a simple proof of the result. Let $n = \\overline{a_s a_{s-1} \\cdots a_0}$ be the binary representation of $n$. Then we have\n$$\n\\begin{aligned}\nf(n!) &= \\sum_{j=1}^{s} \\left\\lfloor \\frac{n}{2^j} \\right\\rfloor = \\sum_{j=1}^{s} \\overline{a_s... | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0e3i | Ten pirates find a chest filled with golden and silver coins. There are twice as many silver coins in the chest as there are golden. They divide the golden coins in such a way that the difference of the numbers of coins given to any two of the pirates is not divisible by $10$. Prove that they cannot divide the silver c... | [
"Let $a_1$ denote the number of golden coins given to the first pirate, $a_2$ the number of golden coins given to the second pirate, and so on. Since $10 \\nmid a_i - a_j$ for all $i \\ne j$, the numbers $a_1, a_2, \\dots, a_{10}$ must give different remainders when divided by $10$. There are only $10$ possible rem... | Slovenia | National Math Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
08si | 2008 boys and 2008 girls decided to get together and play the game of exchanging presents. Each participating boy was asked to bring a bouquet of flowers and each participating girl was asked to bring a bar of chocolate to the get-together. When all the participants showed up, they were lined up in some way and were se... | [
"$[2^{251} + 2^{502} + 2^{1004} + 2^{2008}]$\n\nLet us call the situation the [initial state] if every participating boy has a bouquet of flowers and every participating girl has a chocolate bar, and call the situation the [good state] if every participating boy has a chocolate bar and every participating girl has ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2^{251} + 2^{502} + 2^{1004} + 2^{2008} | |
08fz | Problem:
Data una tabella $2n \times 2n$, un sottoinsieme $S$ delle sue caselle è detto labirintico se ha le seguenti proprietà:
- da ogni casella di $S$ è possibile spostarsi in ogni altra casella di $S$ muovendosi solo da caselle di $S$ a caselle di $S$ adiacenti (in orizzontale o verticale, cioè che abbiano un lato... | [
"Solution:\n\na. Innanzitutto osserviamo che un quadrato $2 \\times 2$ all'interno della tabella può contenere al più 3 caselle appartenenti all'insieme $S$: infatti, se per assurdo ci fosse un quadrato $2 \\times 2$ con tutte e 4 le caselle appartenenti all'insieme $S$, allora si potrebbe trovare un percorso, tutt... | Italy | Olimpiadi di Matematica - Febbraio | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | proof only | a) At most 3·n^2 cells. b) An explicit 12-cell labyrinthine set exists on the 4×4 grid (example provided in the solution). c) The perimeter equals 2k + 2 for k cells. d) No labyrinthine set of 300 cells exists when n = 10 (on the 20×20 grid). | |
0bv8 | How many of the first 2017 positive integers can be uniquely represented as $2^a + 2^b + 2^c$, with $a$, $b$, $c$ non-negative integers? (Two representations that only differ by the order of the terms are considered identical.) | [
"If a number can be represented as a sum of three, not necessarily distinct, powers of $2$, regrouping the equal terms (if such terms exist), one obtains a sum of at most three *distinct* powers of $2$, hence the base $2$ representation of such a number has at most three digits equal to $1$. Convenient numbers are ... | Romania | Eleventh STARS OF MATHEMATICS Competition | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 184 | |
08l5 | Problem:
If the real numbers $a, b, c, d$ are such that $0 < a, b, c, d < 1$, show that
$$
1 + a b + b c + c d + d a + a c + b d > a + b + c + d
$$ | [
"Solution:\nIf $1 \\geq a + b + c$ then we write the given inequality equivalently as\n$$\n\\begin{gathered}\n1 - (a + b + c) + d[(a + b + c) - 1] + a b + b c + c a > 0 \\\\\n\\Leftrightarrow [1 - (a + b + c)](1 - d) + a b + b c + c a > 0\n\\end{gathered}\n$$\nwhich is of course true.\n\nIf instead $a + b + c > 1$,... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0d5x | How many sequences of integers
$$
1 \leq a_{1} \leq a_{2} \leq \ldots \leq a_{11} \leq 2015
$$
that satisfy $a_{i} \equiv i^{2} (\bmod\ 12)$ for all $1 \leq i \leq 11$ are there? | [
"Let $r_{i}$ be the remainder when $i^{2}$ is divided by $12$, and $a_{i} = 12 k_{i} + r_{i}$ for some nonnegative integer $k_{i}$, when $1 \\leq i \\leq 11$. From the following table showing the values of $r_{i}$\n\n| $i$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |\n| :---: | :---: | :---: | :---: | :---: | :-... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English, Arabic | proof and answer | C(173, 11) | |
031p | Problem:
Consider the sequence
$$
y_{1} = y_{2} = 1, \quad y_{n+2} = (4k - 5) y_{n+1} - y_{n} + 4 - 2k, \quad n \geq 1
$$
Find all integers $k$ such that any term of the sequence is a perfect square. | [
"Solution:\nLet $k$ have the given property. We have that $y_{3} = 2k - 2 = 4a^{2}$ ($a \\geq 0$), i.e., $k = 2a^{2} + 1$. Further, $y_{4} = 8k^{2} - 20k + 13$ and $y_{5} = 32k^{3} - 120k^{2} + 148k - 59 = 256a^{6} - 96a^{4} + 8a^{2} + 1$.\n\nIf $a = 0$ we get that $k = 1$ and the given sequence is $1, 1, 0, 1, 1, ... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | k = 1 and k = 3 | |
0dh8 | The sequence $(a_k)$ is given by $a_1 = \frac{1}{2}$ and $a_{n+1} = 1 - a_1 a_2 \cdots a_n$ for every $n \ge 1$. Prove that $a_{100} > 0.99$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
010x | Problem:
Determine all positive integers $n$ with the property that the third root of $n$ is obtained by removing the last three decimal digits of $n$. | [
"Solution:\n32768 is the only such integer.\nIf $n = m^{3}$ is a solution, then $m$ satisfies $1000 m \\leqslant m^{3} < 1000(m+1)$. From the first inequality, we get $m^{2} \\geqslant 1000$, or $m \\geqslant 32$. By the second inequality, we then have\n$$\nm^{2} < 1000 \\cdot \\frac{m+1}{m} \\leqslant 1000 \\cdot ... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 32768 | |
0hlk | Problem:
A car is moving at a constant speed. Every 15 minutes it makes a $90^{\circ}$ turn to either left or right. If the car has started the trip at some point $A$, prove that it can return to $A$ only after an integer number of hours. | [
"Solution:\nIt is important to notice that an hour consists of 60 minutes and $60 / 15 = 4$. That's why it is common to say that 15 minutes is a \"quarter of an hour\". We may assume that the car is traveling in a square grid where the size of each square is the distance that the car travels in 15 minutes. Assume f... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
01mo | Prove that if positive numbers $a, b, x, y$ satisfy the inequality $ab \ge xa + yb$, then they satisfy the inequality $\sqrt{a+b} \ge \sqrt{x} + \sqrt{y}$. | [] | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0d63 | Define the sequence $a_{1}, a_{2}, \ldots$ as follows: $a_{1}=1$, and for every $n \geq 2$, $a_{n}=n-2$ if $a_{n-1}=0$ and $a_{n}=a_{n-1}-1$, otherwise. Find the number of $1 \leq k \leq 2016$ such that there are non-negative integers $r, s$ and a positive integer $n$ satisfying $k=r+s$ and $a_{n+r}=a_{n}+s$. | [
"Let $N=n+r$ and $M=n$, then\n$$\nr=N-M,\\ s=a_{N}-a_{M}\\ \\text{ and }\\ k=r+s=\\left(a_{N}+N\\right)-\\left(a_{M}+M\\right) .\n$$\nWe need to find the number of possible values of $\\left(a_{N}+N\\right)-\\left(a_{M}+M\\right)$, where $N \\geq M$ and $a_{N} \\geq a_{M}$.\nIt is easy to see by induction that $a_{... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 51 | |
0j72 | Problem:
Let $S$ be a set of consecutive positive integers such that for any integer $n$ in $S$, the sum of the digits of $n$ is not a multiple of 11. Determine the largest possible number of elements of $S$. | [
"Solution:\nAnswer: 38\n\nWe claim that the answer is 38. This can be achieved by taking the smallest integer in the set to be 999981. Then, our sums of digits of the integers in the set are\n$$\n45, \\ldots, 53, 45, \\ldots, 54, 1, \\ldots, 10, 2, \\ldots, 10\n$$\nnone of which are divisible by 11.\n\nSuppose now ... | United States | Harvard-MIT November Tournament | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 38 | |
01bt | The sum of angles $A$ and $C$ of a convex quadrilateral $ABCD$ is less than $180^\circ$. Prove that
$$
AB \cdot CD + AD \cdot BC < AC(AB + AD).
$$ | [
"Let $s$ be a circumcircle $ABD$. Then the point $C$ is outside this circle but inside the angle $BAD$.\n\nApply the inversion with the center $A$ and radius $1$. This inversion maps the circle $s$ to the line $s' = B'D'$, where $B'$ and $D'$ are images of $B$ and $D$. The point $C$ goes to the point $C'$ inside th... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
0cf0 | Given a non-negative integer $n$, let $\sum_{k=0}^{2n} a_k X^k$ be the standard power expansion of the polynomial $\sum_{k=0}^{n} \binom{n}{k}^2 (X+1)^{2k} (X-1)^{2(n-k)}$. The coefficients $a_{2k+1}$ all vanish since the polynomial is invariant under the change $X \mapsto -X$. Show that the $a_{2k}$ are all positive. | [
"Leaving aside the trivial cases $n=0$ and $n=1$, assume $n \\ge 2$. The conclusion follows from the fact that the polynomial under consideration,\n\n$$\nf = \\sum_{k=0}^{n} \\binom{n}{k}^2 (X+1)^{2k} (X-1)^{2(n-k)},\n$$\nis expressible as a sum of $n+1$ polynomials of the form $(b_k X^2 + c_k)^n$, where the $b_k$ ... | Romania | Nineteenth IMAR Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0dlg | There are $30$ students in a mathematics competition. We wish to give them a total of $N$ candies so that the following hold:
(i) Each student gets at least one candy.
(ii) Any student with larger score gets more candies than any student with lower score.
(iii) Any two students with the same score get the same amount o... | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 480 | |
0hrg | Problem:
A $23 \times 23$ square is divided into smaller squares of dimensions $1 \times 1$, $2 \times 2$, and $3 \times 3$. What is the minimum possible number of $1 \times 1$ squares? | [
"Solution:\n\nColor the rows of the square black and white alternately, so the top and bottom rows are black. Then each $2 \\times 2$ tile covers two cells of each color, and each $3 \\times 3$ tile covers six of one color and three of the other. In particular, if only $2 \\times 2$ and $3 \\times 3$ tiles are used... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1 | |
08bo | Problem:
Cecilia ha un dado a sei facce (numerate da 1 a 6) e 4 colori a disposizione. In quanti modi può colorare le sei facce del dado usando in totale almeno tre colori diversi e facendo in modo che facce opposte siano di colori diversi?
(A) $4^{3} \cdot 3^{3}$
(B) $3^{6}-2^{6}$
(C) $2^{6} \cdot 3^{2}$
(D) $2^{4} ... | [
"Solution:\n\nLa risposta è (D). Contiamo innanzitutto in quanti modi possiamo colorare il cubo affinché facce opposte abbiano colori diversi. Abbiamo $4 \\cdot 3$ scelte per le facce 1 e 6, e altrettante per ciascuna delle coppie di facce 2 e 5 e 3 e 4. Complessivamente abbiamo $4^{3} \\cdot 3^{3}$ colorazioni.\n\... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | D | |
0exf | Problem:
Prove that the sum of the lengths of the edges of a polyhedron is at least 3 times the greatest distance between two points of the polyhedron. | [
"Solution:\n\nIf $A$ and $B$ are at the greatest distance, then they must be vertices. For suppose $A$ is not a vertex. Then there is a segment $XY$ entirely contained in the polyhedron with $A$ as an interior point. But now at least one of angles $BAX$, $BAY$ must be at least $90^{\\circ}$. Suppose it is $BAX$. Th... | Soviet Union | 5th ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
0ipj | Problem:
Al has a rectangle of integer side lengths $a$ and $b$, and area $1000$. What is the smallest perimeter it could have? | [
"Solution:\n\nAnswer: $130$\n\nTo minimize the sum of the side lengths, we need to keep the height and width as close as possible, because the square has the smallest perimeter of all rectangles with a fixed area. So, $40$ and $25$ multiply to $1000$ and are as close as possible—the $40 \\times 25$ rectangle has pe... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | final answer only | 130 | |
078g | Let $ABC$ be a triangle with circumcentre $O$ and centroid $G$. Let $M$ be the midpoint of $BC$ and $N$ be the reflection of $M$ across $O$. Prove that $NO = NA$ iff $\angle AOG = 90^\circ$. | [
"Let $H$ be the orthocenter of $\\triangle ABC$ and let $X$ be the midpoint of $AH$. Then we know that $AXON$ is a parallelogram.\nNow, observe that $\\angle AOH = \\angle AOG$. Now,\n$$\nNO = NA \\iff XA = XO \\iff \\angle AOH = 90^\\circ\n$$\nThus, we are done. ☐",
"Let $H$ be the orthocenter of $\\triangle ABC... | India | EGMO TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ayq | Problem:
Evaluate the expression $\left(1+\tan 7.5^\circ\right)\left(1+\tan 18^\circ\right)\left(1+\tan 27^\circ\right)\left(1+\tan 37.5^\circ\right)$. | [] | Philippines | 21st PMO Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 4 | |
0g7w | 設 $n$ 是大於或等於 1 的整數。在集合 $\{1, 2, \dots, n\}$, 最多可以選出幾對數字, 滿足其中每一對都是相異的兩個數, 每一對中的兩數之和與其他對的和皆不同, 而且每一對的和都不超過 $n$? | [
"考慮集合 $\\{1, 2, 3, \\dots, n\\}$ 中的 $x$ 對這樣的數字。這些 $2x$ 個數字的總和 $S$ 至少是 $1+2+\\cdots+2x$, 因為這些選出來的數字兩兩互異。另一方面 $S \\le n+(n-1)+\\cdots+(n-x+1)$, 因為每對數字的和均不相同, 並且都不超過 $n$。於是給出下列的不等式\n$$\n\\frac{2x(2x+1)}{2} \\le nx - \\frac{x(x-1)}{2},\n$$\n所以推得 $x \\le \\frac{2n-1}{5}$。所以最多可以有 $\\lfloor \\frac{2n-1}{5} \\rfloor$ 對滿足題設... | Taiwan | 二〇一三數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | floor((2n-1)/5) | |
0i2u | Let $S$ be a set of integers (not necessarily positive) such that
a. there exist $a, b \in S$ with $\gcd(a, b) = \gcd(a - 2, b - 2) = 1$;
b. if $x$ and $y$ are elements of $S$ (possibly equal), then $x^2 - y$ also belongs to $S$.
Prove that $S$ is the set of all integers. | [
"In the solution below we use the expression $S$ is stable under $x \\mapsto f(x)$ to mean that if $t$ belongs to $S$, then $f(t)$ also belongs to $S$. If $c, d \\in S$, then by condition (b), $S$ is stable under $x \\mapsto c^2 - x$ and $x \\mapsto d^2 - x$. Hence, it is stable under $x \\mapsto c^2 - (d^2 - x) = ... | United States | USA IMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof only | null | |
0htc | Problem:
Prove that there exists an infinite set $S$ of positive integers with the property that if we take any finite subset $T$ of $S$, the sum of the elements of $T$ is not a perfect $k$th power for any $k \geq 2$. | [
"Solution:\n\n$$\nS = \\left\\{2, 2^{2} \\cdot 3, 2^{2} \\cdot 3^{2} \\cdot 5, 2^{2} \\cdot 3^{2} \\cdot 5^{2} \\cdot 7, 2^{2} \\cdot 3^{2} \\cdot 5^{2} \\cdot 7^{2} \\cdot 11, \\ldots \\right\\}.\n$$\nLet $T \\subseteq S$ be a finite nonempty set, and $m$ its smallest element. Let $p$ be the prime factor which app... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0ikb | Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ such that $AP = AB$ and $AQ = AC$ and $\angle BAP = \angle CAQ$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \perp PQ$. | [
"**Note:** We present five different approaches. The first three synthetic solutions are all based on the following simple observation.\nWe first note that $APBR$ and $AQCR$ are cyclic quadrilaterals. It is easy to see that triangles $APC$ and $ABQ$ are congruent to each other, implying that $\\angle APR = \\angle ... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transfor... | null | proof only | null | |
01ro | Find all positive integers $n$ such that
$$
n = q(q^2 - q - 1) = r(2r + 1)
$$
for some primes $q$ and $r$. | [
"(Solution by V. Vityaz.) First, $q \\neq r$ otherwise from\n$$\nq(q^2 - q - 1) = r(2r + 3) \\quad (1)\n$$\nwe would have $r^2 - r - 1 = 2r + 3$ which gives $r = 4$, which is not prime.\n\nHence (1) implies $q^2 - q - 1 = kr$, so $q^2 - q - 1 = kr$, $2r + 3 = kq$. Eliminating $r$ we obtain\n$$\n2q^2 - (2 + k^2)q + ... | Belarus | SELECTION and TRAINING SESSION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 2015 | |
0esd | Determine all pairs $(P, d)$ of a polynomial $P$ with integer coefficients and an integer $d$ such that the equation $P(x) - P(y) = d$, where $x$ and $y$ are integers and $x \neq y$, has infinitely many solutions. | [
"Note first that $x - y$ divides $P(x) - P(y)$. So if $d \\neq 0$, there are only finitely many possibilities for $x - y$. For one of these possible values, let us denote it by $a$, there must be\n\ninfinitely many pairs $(x, y)$ such that $x - y = a$ and $P(x) - P(y) = d$. Note that\n$$\nP(x) - P(x - a) - d\n$$\ni... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All such pairs are exactly:
1) For any integer d, any linear polynomial with integer coefficients P(x) = (d/a) x + b, where a is a nonzero integer divisor of d and b is any integer (including the case d = 0, which yields constant polynomials).
2) For d = 0, any polynomial that is even after a shift, i.e., P(x) = R((x −... | |
08tb | On each of 9 balls a distinct number chosen from 1 through 9 is marked. How many distinct ways of choosing some balls from this collection of 9 balls are there if the choice has to satisfy the following condition? Selecting no ball from the collection should be considered as 1 possibility.
**Condition:** By selecting s... | [
"If no balls are selected, the condition is satisfied by agreement.\n\nIn all other cases, let us denote by $M$ and $m$ the maximum and the minimum number, respectively, among the numbers marked on the selected balls. When $M - m$ is fixed, there are exactly $9 - (M - m)$ ways of choosing the pair $(m, M)$.\n\nFor ... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 256 | |
0im5 | Let $n$ be a positive integer and let $a_1 \le a_2 \le \dots \le a_n$ and $b_1 \le b_2 \le \dots \le b_n$ be two nondecreasing sequences of real numbers such that
$$
a_1 + \dots + a_i \le b_1 + \dots + b_i \quad \text{for every } i = 1, \dots, n-1
$$
and
$$
a_1 + \dots + a_n = b_1 + \dots + b_n.
$$
Suppose that for an... | [
"**First Solution:** Put $s_n = a_1 + \\cdots + a_n = b_1 + \\cdots + b_n$. Then\n$$\n\\begin{aligned}\n2 \\sum_{i=1}^{n-1} (a_1 + \\cdots + a_i) &= 2(n-1)a_1 + 2(n-2)a_2 + \\cdots + 2(1)a_{n-1} \\\\\n&= (n-1)a_1 + (n-3)a_2 + \\cdots + (1-n)a_n + (n-1)s_n \\\\\n&= (n-1)s_n + \\sum_{1 \\le i < j \\le n} (a_i - a_j)\... | United States | Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0c71 | On the board are written initially three consecutive positive integers, $n-1$, $n$, $n+1$. A move consists of choosing two numbers written on the board $a$ and $b$, and replacing them with $2a-b$ and $2b-a$. For what values of $n$ is it possible to obtain, after a succession of such moves, that two of the numbers writt... | [
"We prove that we can obtain two 0-s on the board if and only if $n$ is a power of 3.\n\nAs the sum of the numbers written on the board stays the same, we must obtain on the board the numbers $(0, 0, 3n)$. If $p \\neq 3$ is a prime divisor of $n$, then, in the final configuration, all the numbers are divisible by $... | Romania | Stars of Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | n is a power of 3 (n = 3^k for some nonnegative integer k) | |
023w | Problem:
Aonde está o erro? - Seja $x$ solução de $x^{2}+x+1=0$. Então $x \neq 0$ e por isso podemos dividir ambos os membros da equação por $x$, obtendo $x+1+\frac{1}{x}=0$. Da equação temos que $x+1=-x^{2}$, logo $-x^{2}+\frac{1}{x}=0$, isto é: $x^{2}=1 / x$ ou ainda $x^{3}=1$ e $x=1$. Substituindo $x=1$ na equação ... | [
"Solution:\n\nAonde está o erro? - Esse deixamos para os alunos!"
] | Brazil | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
058s | Juku has the first 100 volumes of the Harrie Totter book series at his home. For every $i$ and $j$, where $1 \le i < j \le 100$, call the pair $(i, j)$ *reversed* if volume No $j$ is before volume No $i$ on Juku's shelf. Juku wants to arrange all volumes of the series to one row on his shelf in such a way that there do... | [
"Let all 100 volumes be placed to a shelf $a$ in some order. For every $i = 1, 2, ..., 100$, let $a_i$ be the number of the volume that occurs as the $i$th in this order. In the order the volumes occur on shelf $a$, we start relocating the volumes to two new shelves $b$ and $c$. We place a volume to the end of shel... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | 2500 | |
0881 | Problem:
Sia $ABC$ un triangolo equilatero, indichiamo con $D, E, F$ i punti medi dei lati. Quanti triangoli non degeneri e non congruenti fra loro si possono ottenere scegliendo 3 dei punti $A, B, C, D, E, F$?
Quante sono le coppie ordinate $(x, y)$ di interi relativi che verificano l'equazione $y^{4}-8 y^{2}+7=8 x^... | [
"Solution:\n\nLa risposta è 4. A meno di isometrie è possibile determinare un triangolo ottusangolo, un triangolo rettangolo e due diversi triangoli equilateri."
] | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Diophantine Equations > Techniques: modulo, s... | null | proof and answer | First: 4; Second: 4 | |
0g0s | Problem:
Soit $ABC$ un triangle avec $AB < AC$. La bissectrice de $\angle BAC$ coupe le côté $BC$ en $D$. Soit $k$ le cercle qui passe par $D$ et qui est tangent aux segments $AC$ et $AB$ en $E$, respectivement $F$. Soit $G$ le deuxième point d'intersection de $k$ avec $BC$. Soit $S$ le point d'intersection de $EG$ et... | [
"Solution:\n\n\n\nComme les points $E$ et $F$ sont symétriques par rapport à la droite $AD$,\n$$\n\\angle AED = \\angle DFA\n$$\nPar la théorème de l'angle tangent,\n$$\n\\angle EGD = \\angle DEC\n$$\nOr\n$$\n\\angle DEC = 180^\\circ - \\angle AED = 180^\\circ - \\angle DFA = \\angle BFD\n$... | Switzerland | SMO - Vorrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hu0 | Problem:
A society has $100$ members and every two members are either friends or enemies. Prove that there are two persons from the society that have an even number of common enemies. | [
"Solution:\n\nAssume that for each two members the set of common enemies has an odd number of elements. Then the set of their common friends has also an odd number of elements. Fix a person $a$. Let $F=\\{f_{1}, \\ldots, f_{k}\\}$ be the set of friends of the member $a$, and $E=\\{e_{1}, \\ldots, e_{99-k}\\}$ the s... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
00gq | Let $\Gamma$ be the circumcircle of a triangle $A B C$. A circle passing through points $A$ and $C$ meets the sides $B C$ and $B A$ at $D$ and $E$, respectively. The lines $A D$ and $C E$ meet $\Gamma$ again at $G$ and $H$, respectively. The tangent lines of $\Gamma$ at $A$ and $C$ meet the line $D E$ at $L$ and $M$, r... | [
"Let $M G$ meet $\\Gamma$ at $P$. Since $\\angle M C D = \\angle C A E$ and $\\angle M D C = \\angle C A E$, we have $M C = M D$. Thus\n$$\nM D^{2} = M C^{2} = M G \\cdot M P\n$$\nand hence $M D$ is tangent to the circumcircle of $\\triangle D G P$. Therefore $\\angle D G P = \\angle E D P$.\nLet $\\Gamma'$ be the ... | Asia Pacific Mathematics Olympiad (APMO) | XX Asian Pacific Mathematics Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k66 | Problem:
Fred the Four-Dimensional Fluffy Sheep is walking in 4-dimensional space. He starts at the origin. Each minute, he walks from his current position $(a_{1}, a_{2}, a_{3}, a_{4})$ to some position $(x_{1}, x_{2}, x_{3}, x_{4})$ with integer coordinates satisfying
$$(x_{1}-a_{1})^{2}+(x_{2}-a_{2})^{2}+(x_{3}-a_{... | [
"Solution:\n\nThe possible moves correspond to the vectors $\\pm\\langle 2,0,0,0\\rangle$, $\\pm\\langle 1,1,1,-1\\rangle$, and their permutations. It's not hard to see that these vectors form the vertices of a 4-dimensional hypercube, which motivates the change of coordinates\n$$\n(x_{1}, x_{2}, x_{3}, x_{4}) \\Ri... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | C(40,10) * C(40,20)^3 | |
00x1 | Problem:
Let $p$ and $q$ be two consecutive odd prime numbers. Prove that $p+q$ is a product of at least three positive integers greater than 1 (not necessarily different). | [
"Solution:\nSince $q-p=2k$ is even, we have $p+q=2(p+k)$. It is clear that $p < p+k < p+2k = q$. Therefore $p+k$ is not prime and, consequently, is a product of two positive integers greater than 1."
] | Baltic Way | Baltic Way 1992 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0emy | Each positive integer is coloured red or blue. A function $f$ from the set of positive integers to itself has the following two properties:
(a) if $x \le y$, then $f(x) \le f(y)$; and
(b) if $x, y$ and $z$ are (not necessarily distinct) positive integers of the same colour and $x + y = z$, then $f(x) + f(y) = f(z)$.
Pr... | [
"For integers $x$ and $y$, by a segment $[x, y]$ we mean the set of all integers $t$ such that $x \\le t \\le y$; the length of this segment is $y - x$.\nIf, for every two positive integers $x$ and $y$ of the same colour we have $\\frac{f(x)}{x} = \\frac{f(y)}{y}$, then one can choose $a = \\max\\left\\{\\frac{f(r)... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0g87 | 設圓 $O_1, O_2$ 的半徑分別為 $R_1, R_2$, 且此兩圓交於 $A, D$ 兩點。過 $D$ 作一直線 $L$, 設 $L$ 分別再交圓 $O_1, O_2$ 於 $B, C$ 兩點。現在讓兩圓圓心的距離可以變動, 直線 $L$ 也可以變動。當 $\triangle ABC$ 的面積達到最大時, 求 $AD$ 的長度。
Let $O_1, O_2$ be circles with radius $R_1$ and $R_2$, respectively. Let the two circles intersect each other at $A$ and $D$. Let $L$ be a straight l... | [
"觀察當 $AD$ 為某一定值時，$\\angle B, \\angle C$ (或其補角，下同) 為一定角，因此 $\\angle A$ 為一定角。因\n$$\n\\begin{aligned}\n|\\triangle ABC| &= \\frac{1}{2} AB \\cdot AC \\sin \\angle A \\\\\n&= 2R_1 R_2 \\sin \\angle ADB \\sin \\angle ADC \\sin \\angle A \\\\\n&= 2R_1 R_2 \\sin^2 \\angle ADB \\sin \\angle A.\n\\end{aligned}\n$$\n當 $\\ang... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2 / sqrt(1/R1^2 + 1/R2^2) | |
07mw | Each of $117$ spies, operating in a certain country, is to assign himself to one of three missions, such that every mission has at least one spy assigned. At this point, no two spies can communicate. Headquarters will then sequentially issue a number of passwords, each of which allows a single pair of spies to communic... | [
"We represent the given information by a graph the vertices of which correspond to the $117$ spies. Two vertices are connected iff the corresponding spies share a password which allows them to communicate directly. We need to ensure that at least one of the sub-graphs formed by spies sharing the same mission is con... | Ireland | Ireland | [
"Discrete Mathematics > Graph Theory",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Assign the spies evenly among the three missions: 39, 39, 39. The minimal guaranteed number of passwords is 2110. | |
0f4e | Problem:
$ABC$, $CDE$, $EFG$ are equilateral triangles (not necessarily the same size). The vertices are counter-clockwise in each case. $A$, $D$, $G$ are collinear and $AD = DG$. Show that $BFD$ is equilateral. | [] | Soviet Union | 15th ASU | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02dm | Show how to construct a line segment length $\sqrt[4]{a^4 + b^4}$ given segments of length $a$ and $b$. | [
"We show first how to get the square and the square root. Take $ABC$ with $\\angle A = 90^\\circ$, altitude $AD$ of length $a$ and $BD = 1$. Then by similar triangles $\\frac{BC}{AB} = \\frac{AB}{BD} \\iff BC = \\frac{AB^2}{BD} = AB^2 = a^2 + 1$. Hence $CD = a^2$.\n\n\n\nConversely, we can ... | Brazil | IV OBM | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
068r | Determine all positive integers $a$, $b$, $p$, where $p$ is prime, satisfying the equation: $$\frac{1}{p} = \frac{1}{a^2} + \frac{1}{b^2}.$$ | [
"The given equation can be written as: $p a^2 + b^2 = a^2 b^2$ (1)\nSince $p$ is prime, from (1) we have: $p|a$ or $p|b$. We suppose that $p|a$, and hence $a = p a_1$, $a_1 \\in \\mathbb{N}^*$. Moreover, we have that\n$$\n\\begin{aligned}\n\\frac{1}{p} > \\frac{1}{b^2} &\\Rightarrow b^2 > p \\Rightarrow b^2 \\ge p+... | Greece | 34th Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (a, b, p) = (2, 2, 2) | |
04nq | Let $n$ be a positive integer. Prove that
$$
\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{3n+1} > 1.
$$ | [
"Let us consider the sum:\n$$\nS = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{3n+1}\n$$\nThere are $(3n+1) - (n+1) + 1 = 2n+1$ terms in the sum.\n\nWe will compare this sum to the integral:\n$$\n\\int_{n+1}^{3n+2} \\frac{1}{x} \\, dx = \\ln(3n+2) - \\ln(n+1) = \\ln\\left(\\frac{3n+2}{n+1}\\right)\n$$\nNot... | Croatia | Croatia_2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0d14 | Determine all positive integers $n \ge 2$ for which the following statement is true:
Given any $n$ distinct points on the plane such that the distance between each pair of points is distinct, there exists a pair of points $A$, $B$ for which the difference between the number of points lying on either side of the perpend... | [
"The statement trivially holds for $n = 2$ and $n = 3$. We will show that it does not hold for any $n \\ge 4$.\n\nFirst suppose that $n$ is even. Setup a coordinate, and place a point at $(0,0)$. Place $n/2 - 1$ points on the negative $x$-axis and $n/2 - 1$ points on the positive $x$-axis at $(-1/2^i, 0)$ and $(2^i... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | n = 2 and n = 3 | |
03x0 | Let $O, I$ be the circumcenter and incenter of $\triangle ABC$. Prove that, for an arbitrary point $D$ on the circle $O$, one can construct a triangle $DEF$, such that $O, I$ are the circumcenter and incenter of $\triangle DEF$. (Posed by Tao Pingsheng)
 | [
"As shown in Fig. 2, let $OI = d$, $R$, $r$ be the circumradius and incircle radius of $\\triangle ABC$. The point $K$ is the intersection of $AI$ and the circle $O$; then\n$$\nKI = KB = 2R \\sin \\frac{\\angle BAC}{2},\n$$\n$$\nAI = \\frac{r}{\\sin \\frac{\\angle BAC}{2}}.\n$$\n\nFig. 2\nL... | China | China Southeastern Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theor... | English | proof only | null | |
0e1h | Problem:
Tabela velikosti $5 \times 7$ ima eno polje rdeče, ostala polja pa bela. Na to tabelo postavljamo ploščice oblike
(ploščico lahko zrcalimo ali zasukamo),
tako da cele ležijo na tabeli. Pri tem se ploščice na rdečem polju lahko prekrivajo, na belih pa ne. Določi vse možne položaje rde... | [
"Solution:\n\nDenimo, da smo v celoti uspeli prekriti tabelo. Vsaka ploščica, ki prekriva rdeče polje, prekriva vsaj še eno belo polje, ki leži levo, desno, nad ali pod rdečim poljem. V primeru, da bi rdeče polje prekrivalo 5 ali več ploščic, bi se vsaj dve izmed njih prekrivali na belem polju. Torej se na rdečem p... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0f53 | Problem:
A $4 \times 4$ array of unit cells is made up of a grid of total length $40$. Can we divide the grid into $8$ paths of length $5$? Into $5$ paths of length $8$? | [] | Soviet Union | 17th ASU | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes for eight paths of length five; No for five paths of length eight. | |
0jil | Let $ABC$ be a triangle and $D, E, F$ be the midpoints of arcs $BC, CA, AB$ on the circumcircle. Line $l_a$ passes through the feet of the perpendiculars from $A$ to $DB$ and $DC$. Line $m_a$ passes through the feet of the perpendiculars from $D$ to $AB$ and $AC$. Let $A_1$ denote the intersection of lines $l_a$ and $m... | [
"We prove the following stronger statement: $A_1, B_1, C_1$ are midpoints of segments $HD, DE, HF$, respectively, where $H$ is the orthocenter of triangle $ABC$. Therefore, there is a dilation centered at $H$ with magnitude 2 sending triangle $A_1B_1C_1$ to $DEF$. (For readers familiar with the nine-point circle of... | United States | IMO Team Selection Team Selection Test | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations... | null | proof only | null | |
0dz6 | Let $n = (p^2 - 1)(p^2 - 4) + 9$. Assuming $p$ is a prime number, what is the least possible sum of the digits of $n$? Find all prime numbers $p$ for which this value is attained. | [
"Let us find the first few $n$. When $p=2$ we have $n=9$, when $p=3$ we have $n=49$ and when $p=5$ we have $n=513$. Now, let $p > 5$. Rewrite $n$ as $n = (p-2)(p-1)(p+1)(p+2)+9$. Since $(p-2), (p-1), p, (p+1), (p+2)$ are five consecutive positive integers, at least one of them is divisible by $5$. Since $p > 5$, $p... | Slovenia | Slovenija 2008 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Least sum of digits is 9; attained for p = 2 and p = 5. | |
04gd | In each vertex of a regular $50$-gon one of the numbers $1$ or $2$ is written. No three consecutive vertices have the same number written in them. The number $1$ is written twenty times and the number $2$ thirty times in total. For each vertex the product of the numbers written in two adjacent vertices and the number i... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 180 | |
0jpx | Problem:
Consider a $4 \times 4$ grid of squares. Aziraphale and Crowley play a game on this grid, alternating turns, with Aziraphale going first. On Aziraphale's turn, he may color any uncolored square red, and on Crowley's turn, he may color any uncolored square blue. The game ends when all the squares are colored, ... | [
"Solution:\n\nWe claim that the answer is $6$.\n\nOn Aziraphale's first two turns, it is always possible for him to take $2$ adjacent squares from the central four; without loss of generality, suppose they are the squares at $(1,1)$ and $(1,2)$. If allowed, Aziraphale's next turn will be to take one of the remainin... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 6 | |
09qu | Problem:
Vind alle drietallen $(a, b, c)$ van positieve gehele getallen met $a+b+c=10$ zodat er $a$ rode, $b$ blauwe en $c$ groene punten (allemaal verschillend) in het vlak bestaan met de volgende eigenschappen:
- voor elk rood punt en elk blauw punt bekijken we de afstand tussen deze twee punten; de som van al deze ... | [
"Solution:\n\nWe maken driehoeken bestaande uit een blauw, een rood en een groen punt. Deze driehoeken mogen ook gedegenereerd zijn. In elk van de driehoeken geldt de (nietstrikte) driehoeksongelijkheid: de afstand tussen het blauwe en het rode punt is hoogstens gelijk aan de som van de afstanden tussen het blauwe ... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | (8, 1, 1) | |
0ay4 | Problem:
Two semicircles, each with radius $\sqrt{2}$, are tangent to each other, as shown in the figure below. If $AB \parallel CD$, determine the length of segment $AD$.
 | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2√2 | |
09t9 | Problem:
Zij $ABC$ een driehoek met $\angle A = 90^{\circ}$ en zij $D$ het voetpunt van de hoogtelijn vanuit $A$. De middens van $AD$ en $AC$ noemen we respectievelijk $E$ en $F$. Zij $M$ het middelpunt van de omschreven cirkel van $\triangle BEF$. Bewijs dat $AC \parallel BM$. | [
"Solution:\n\nVanwege de rechte hoeken bij $A$ en $D$ geldt $\\triangle ADB \\sim \\triangle CAB$ (hh). Hieruit volgt $\\frac{|AD|}{|AB|} = \\frac{|CA|}{|CB|}$. Omdat $|AE| = \\frac{1}{2}|AD|$ en $|CF| = \\frac{1}{2}|CA|$ geldt ook $\\frac{|AE|}{|AB|} = \\frac{|CF|}{|CB|}$. Uit de vorige gelijkvormigheid volgt ook ... | Netherlands | Selectietoets | [
"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 | |
0l8h | Let $n$ and $k$ be positive integers with $k < n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0, a_1, \dots, a_k$ such that the polynomial $a_kx^k + \dots + a_1x + a_0$ divides $P(x)$, the product $a_0a_1\dots a_k$ ... | [] | United States | USAMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
060q | Problem:
Dans une école il y a $n$ cours et $n$ élèves. Les élèves sont inscrits dans plusieurs cours de sorte que deux élèves différents n'ont jamais exactement les mêmes cours. Prouver qu'on peut supprimer un cours de sorte qu'aucune paire d'élèves ne se retrouve avec exactement les mêmes cours. | [
"Solution:\n\nOn va montrer par récurrence forte sur $n$ la propriété plus forte suivante :\n$$\n\\mathcal{P}_{n} : \\text{Pour tout } m \\geqslant n, \\text{ dans une classe à } n \\text{ élèves et } m \\text{ cours telle que deux élèves n'ont jamais exactement les mêmes cours, alors il est possible de retirer un ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
01mq | Prove that if positive numbers $a$, $b$, $c$, $k$, $l$, $m$ satisfy the inequalities $abc \ge ka + lb + mc$, then they satisfy the inequality
$$
a + b + c \ge \sqrt{3}(\sqrt{k} + \sqrt{l} + \sqrt{m}).
$$ | [] | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0b0k | Problem:
Let $a_{1}, a_{2}, \ldots$ be a sequence of integers defined by $a_{1}=3$, $a_{2}=3$, and
$$
a_{n+2}=a_{n+1} a_{n}-a_{n+1}-a_{n}+2
$$
for all $n \geq 1$. Find the remainder when $a_{2020}$ is divided by $22$. | [
"Solution:\n\nLet $\\{F_{n}\\}_{n=1}^{\\infty}=\\{1,1,2,3,5,8, \\ldots\\}$ be the sequence of Fibonacci numbers. We first claim that $a_{n}=2^{F_{n}}+1$ for all $n \\in \\mathbb{N}$. Clearly, this is true for $n=1,2$. Let $k \\in \\mathbb{N}$ and suppose that the claim is true for $n=k$ and for $n=k+1$. Then\n$$\n\... | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 11 | |
09ha | Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$
(a-b)f(a+b) + (b-c)f(b+c) + (c-a)f(c+a) = 0
$$
for all $a, b, c \in \mathbb{R}$. | [
"Answer: $f(x) = Ax + B$ for real numbers $A$ and $B$.\nFor $x \\in \\mathbb{R}$, let $a = \\frac{x-1}{2}$, $b = \\frac{x+1}{2}$ and $c = \\frac{1-x}{2}$. Then we have\n$$\n\\begin{cases}\na+b=x \\\\\nb+c=1 \\\\\nc+a=0\n\\end{cases}\n$$\nand\n$$\n\\begin{cases}\na-b=-1 \\\\\nb-c=x \\\\\nc-a=1-x\n\\end{cases}\n$$\n,... | Mongolia | Mongolian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = A x + B for real numbers A and B | |
0drs | Let $n$ be a positive integer. Determine the minimum number of lines that can be drawn on the plane so that they intersect in exactly $n$ distinct points. | [
"Let $m$ be the integer so that $\\left(\\frac{m}{2}\\right) < n \\le \\left(\\frac{m+1}{2}\\right)$. Then since $m$ lines intersect in at most $\\left(\\frac{m}{2}\\right)$ points, we have $n > m$. We shall show that there exist $m + 1$ lines that intersect in exactly $n$ points. Let $p = n - \\left(\\frac{m}{2}\\... | Singapore | Singapur | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | the smallest integer k such that k(k−1)/2 ≥ n | |
0bjx | Find all positive integers $n$ so that $17^n + 9^{n^2} = 23^n + 3^{n^2}$. | [
"Clearly $n=0$ and $n=1$ are solutions.\n\nIf $n \\ge 2$, then $n^2 \\ge 2n$, hence $9^{n^2} - 3^{n^2} = 3^{n^2}(3^{n^2} - 1) \\ge 3^{2n}(3^{2n} - 1) = 81^n - 9^n$. Since $81^n - 9^n > 23^n - 17^n$, the equation does not have solutions $n \\ge 2$."
] | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 1 | |
0dja | Let $ABC$ be an acute, non-isosceles triangle inscribed in a circle $(O)$ and $M$ is the midpoint of $BC$. Take $D$ on the segment $BC$ and the circle $(ABD)$ meets $AC$ again at $E$, the circle $(ACD)$ meets $AB$ again at $F$. Lines $BE$, $CF$ meet at $K$ and lines $AK$, $EF$ meet at $R$. Suppose that there exist the ... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
02jp | Problem:
As vizinhas Elza, Sueli, Patrícia, Heloísa e Cláudia chegam juntas do trabalho e começam a subir as escadas do prédio de 5 andares onde moram. Cada uma mora num andar diferente. Heloísa chega a seu andar depois de Elza, mas antes de Cláudia. Quando Sueli chega ao seu andar, Heloísa ainda tem 2 andares para su... | [
"Solution:\n\nVejamos as informações dadas no enunciado:\n\"Heloísa chega a seu andar depois de Elza, mas antes de Cláudia\".\n$$\n\\Rightarrow \\text{Heloísa mora acima de Elza e abaixo de Cláudia.}\n$$\n\"Quando Sueli chega ao seu andar, Heloísa ainda tem 2 andares para subir, e o mesmo ocorre a Patrícia quando E... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | null | proof and answer | Elza on the 1st floor, Sueli on the 2nd floor, Patrícia on the 3rd floor, Heloísa on the 4th floor, and Cláudia on the 5th floor. | |
0csj | In a convex $n$-gon, several diagonals are drawn. A drawn diagonal is good if it intersects (by an interior point) with exactly one of other drawn diagonals. Find the maximal possible number of good diagonals. (S. Berlov) | [
"**Ответ.** $n - 2$ при чётных $n$, $n - 3$ при нечётных $n$.\n\nМы будем пользоваться следующей известной леммой.\n\n**Лемма.** В выпуклом $n$-угольнике нельзя провести более $n - 3$ диагоналей, не имеющих общих внутренних точек.\n\nСначала докажем индукцией по $n$, что количество хороших диагоналей не превосходит... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n − 2 for even n; n − 3 for odd n | |
0cif | Consider a set $M$ of real numbers satisfying the following properties:
a) $(1 + \sqrt{2}) \in M;$
b) if $x, y \in M$, then $x \cdot y \in M$ and $(6x - y) \in M$;
c) if $(6x + y) \in M$, then $x \in M$ or $y \in M$.
Prove that the numbers $3 + 4\sqrt{2}$, $\sqrt{2} - 1$, and $2025$ are elements of the set $M$. | [] | Romania | 75th NMO | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof only | null | |
0kaz | Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on line $BC$ satisfying $\angle AID = 90^\circ$. Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to $\overline{BC}$ at point $A_1$. Define points $B_1$ on $\overline{CA}$ and $C_1$ on $\overline{AB}$ analogously, using the excircles... | [
"**First solution using spiral similarity (Ankan Bhattacharya)** First, we prove the part of the problem which does not depend on the condition $AB_1A_1C_1$ is cyclic.\n\n**Lemma**\nLet $ABC$ be a triangle and define $I, D, B_1, C_1$ as in the problem. Moreover, let $M$ denote the midpoint of $\\overline{AD}$. Then... | United States | USA IMO TST | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Mis... | null | proof only | null | |
0e27 | For which positive integers $n$ does there exist a multiple of $7$, such that the sum of its digits is equal to $n$? | [
"Any number with the sum of the digits equal to $1$ is a power of $10$, so it cannot be a multiple of $7$. Let us try and find a multiple of $7$ such that the sum of its digits will be equal to $2$. This number must have two digits equal to $1$. We check the first few positive integers with this property. The numbe... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All positive integers greater than 1 | |
08gc | Problem:
La successione $a_{n}$ è costruita nel modo seguente: $a_{1}, a_{2}$ sono interi compresi fra 1 e 9 (estremi inclusi); per $n \geq 3$, se la somma fra $a_{n-1}$ e $a_{n-2}$ consta di una sola cifra, allora tale somma è il valore di $a_{n}$; se invece $a_{n-1}+a_{n-2}$ ha più di una cifra, la somma delle sue c... | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. Osserviamo che per $n \\geq 3, a_{n}$ è il numero in $\\{1, \\ldots, 9\\}$ dato dal resto della divisione per 9 di $a_{n-2}+a_{n-1}$ (dove 9 rappresenta resto 0). In particolare, $a_{n-2}$ è il resto (come sopra, preso in $\\{1, \\ldots, 9\\}$ ) nella divisione per 9 d... | Italy | Italian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | MCQ | C | |
0gbx | 令 $\mathbb{Z}$ 表示所有整數之集合。試求所有的函數 $f: \mathbb{Z} \to \mathbb{Z}$ 使得 $f(0) = 0$ 且對任意的整數 $x, y,$
$$
f(x + f(y))f(y + f(x)) = (2x + f(y - x))(2y + f(x - y)) \text{ 成立。}
$$ | [
"(1) $f(f(x)) = 2x + f(-x), \\forall x \\in \\mathbb{Z}$\n注意到, 若 $a \\in \\mathbb{Z}$ 使得 $f(a) = 0$, 那麼在條件中代入 $x, y = a$,\n$$\n4a^2 = f(a + f(a))^2 = f(a)^2 = 0.\n$$\n可知 $a = 0$. 現在改代入 $y = 0$ 得\n$$\nf(x)f(f(x)) = f(x)(2x + f(-x)).\n$$\n所以如果 $x \\neq 0$, 結論成立。另一方面, 當 $x = 0$ 時, 顯然也對。\n\n(2) $f(x + f(x)) = 2x, \\for... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 獨立研究(二) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | f(x) = x for all integers x; and f(x) = -2x for all integers x | |
015a | About a monic polynomial $p(x)$ of degree $n \ge 2$ is known that all its complex roots $\alpha$ are real and satisfy $\alpha \le 1$ and that $p(2) = 3^n$. Which values can $p(1)$ have?
(A monic polynomial $p(x)$ of degree $n$ is one whose coefficient of $x^n$ is equal to one.) | [
"Let $\\prod_{i=1}^{n}(x - \\alpha_i)$ be the factorisation of $p(x)$. Then, $\\alpha_i \\le 1$, $i = 1 \\ldots n$, whence $p(1) \\ge 0$. By the AG theorem,\n$$\n3^n = p(2) = \\prod_{i=1}^{n}(2 - \\alpha_i) = \\prod_{i=1}^{n}(1 + (1 - \\alpha_i)) = \\sum_{k=0}^{n} \\sum_{1 \\le i_1 < \\dots < i_k \\le n} \\prod_{j=... | Baltic Way | Baltic Way SHL | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | [0, 2^n] | |
04hz | The first $n$ positive integers are written on a board ($n \ge 3$). Ante repeats the following procedure: first he chooses two numbers on the board, and then he increases them both by the same arbitrary positive integer.
Determine all positive integers $n$ such that Ante can, by repeating this procedure, achieve that a... | [
"Assume $n = 4k$. Then Ante can achieve that all numbers on the board are equal in the following way: he will increase by $1$ the numbers $1$ and $3$, $5$ and $7$, ..., $4k-3$ and $4k-1$. By doing that, he gets that the numbers on the board are all even numbers smaller than or equal to $n$, and each is written twic... | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All positive integers n that are not congruent to 2 modulo 4; equivalently, all odd n and all n divisible by 4. | |
06v5 | Let $n \geqslant 2$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{n}$ be real numbers such that
$$
a_{1}+a_{2}+\cdots+a_{n}=0 .
$$
Define the set $A$ by
$$
A=\left\{(i, j)\left|1 \leqslant i<j \leqslant n,\left|a_{i}-a_{j}\right| \geqslant 1\right\} .\right.
$$
Prove that, if $A$ is not empty, then
$$
\sum_{(i, j... | [
"Define sets $B$ and $C$ by\n$$\n\\begin{aligned}\n& B=\\left\\{(i, j)\\left|1 \\leqslant i, j \\leqslant n,\\left|a_{i}-a_{j}\\right| \\geqslant 1\\right\\},\\right. \\\\\n& C=\\left\\{(i, j)\\left|1 \\leqslant i, j \\leqslant n,\\left|a_{i}-a_{j}\\right|<1\\right\\} .\\right.\n\\end{aligned}\n$$\nWe have\n$$\n\\b... | IMO | IMO 2019 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0gt0 | Find the largest value of $x - y$, where $x$, $y$, $z$ are real numbers satisfying the following two conditions
$$
x + y + z = 2, \quad xy + yz + zx = 1.
$$ | [
"Answer: $\\frac{2\\sqrt{3}}{3}$.\nThe equality holds at $x = \\frac{2+\\sqrt{3}}{3}$, $y = \\frac{2-\\sqrt{3}}{3}$, $z = \\frac{2}{3}$.\nLet us show that $x - y \\le \\frac{2\\sqrt{3}}{3}$.\n\n*Solution 1:* Problem conditions yield\n$$\nx + y = 2 - z, \\quad xy = 1 - z(x + y) = 1 - z(2 - z) = (z - 1)^2\n$$\nTheref... | Turkey | Junior Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 2\sqrt{3}/3 | |
0112 | Problem:
What is the least number of moves it takes a knight to get from one corner of an $n \times n$ chessboard, where $n \geqslant 4$, to the diagonally opposite corner? | [
"Solution:\n\nAnswer: $2 \\cdot \\left\\lfloor \\dfrac{n+1}{3} \\right\\rfloor$.\n\nLabel the squares by pairs of integers $(x, y)$, $x, y = 1, \\ldots, n$, and consider a sequence of moves that takes the knight from square $(1,1)$ to square $(n, n)$.\n\nThe total increment of $x+y$ is $2(n-1)$, and the maximal inc... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2 * floor((n+1)/3) | |
09qs | Problem:
Gegeven is een trapezium $A B C D$ met $B C \| A D$. Neem aan dat de bissectrices van de hoeken $B A D$ en $C D A$ elkaar snijden op de middelloodlijn van lijnstuk $B C$. Bewijs dat $|A B|=|C D|$ of $|A B|+|C D|=|A D|$. | [
"Solution:\n\nOplossing I. Zij $M$ het midden van $B C$ en zij $P$ het snijpunt van de middelloodlijn van $B C$ met $A D$. Noem $K$ het snijpunt van $M P$ en de twee bissectrices. Laat $L$ en $N$ de voetpunten zijn van $K$ op respectievelijk zijden $A B$ en $D C$. Omdat $A K$ en $D K$ bissectrices zijn, geldt $|K L... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jfc | Problem:
Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be a convex hexagon such that $A_{i} A_{i+2} \parallel A_{i+3} A_{i+5}$ for $i=1,2,3$ (we take $A_{i+6}=A_{i}$ for each $i$). Segment $A_{i} A_{i+2}$ intersects segment $A_{i+1} A_{i+3}$ at $B_{i}$, for $1 \leq i \leq 6$, as shown. Furthermore, suppose that $\triangle ... | [
"Solution:\nBecause $B_{6} A_{3} B_{3} A_{6}$ and $B_{1} A_{4} B_{4} A_{1}$ are parallelograms, $B_{6} A_{3}=A_{6} B_{3}$ and $A_{1} B_{1}=A_{4} B_{4}$. By the congruence of the large triangles $A_{1} A_{3} A_{5}$ and $A_{2} A_{4} A_{6}$, $A_{1} A_{3}=A_{4} A_{6}$. Thus, $B_{6} A_{3}+A_{1} B_{1}-A_{1} A_{3}=A_{6} B... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 22 | |
01qn | Find all functions $f: \mathbb{N} \to \mathbb{N}$, such that $mf(m) + n$ is divisible by $m^2 + f(n)$ for all $m, n \in \mathbb{N}$. | [
"### 2. See IMO-2013 Shortlist, Problem N1."
] | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | f(n) = n for all n in the positive integers | |
0981 | Problem:
Fie funcția $f:\left[\pi ; \frac{3 \pi}{2}\right] \rightarrow \mathbb{R}$, $f(x)=\frac{3 \cos (2 x)-4 \sin (2 x)+\cos x-3 \sin x}{(\sin x+3 \cos x)^{2022}}$. Determinați primitiva $F:\left[\pi ; \frac{3 \pi}{2}\right] \rightarrow \mathbb{R}$ a funcției $f$, pentru care $F\left(\frac{3 \pi}{2}\right)=\frac{1}{... | [
"Solution:\n\n$$\n\\begin{aligned}\n& \\int \\frac{3 \\cos (2 x)-4 \\sin (2 x)+\\cos x-3 \\sin x}{(\\sin x+3 \\cos x)^{2022}} \\, d x= \\\\\n& =\\int \\frac{3 \\cos ^{2} x-3 \\sin ^{2} x-8 \\sin x \\cos x+\\cos x-3 \\sin x}{(\\sin x+3 \\cos x)^{2022}} \\, d x= \\\\\n& =\\int \\frac{(\\sin x+3 \\cos x)(\\cos x-3 \\s... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions"
] | null | final answer only | F(x) = -1/(2020 (sin x + 3 cos x)^{2020}) - 1/(2021 (sin x + 3 cos x)^{2021}) + 1/2020 | |
0a58 | Problem:
Let $ABCD$ be a convex quadrilateral such that $AB + BC = 2021$ and $AD = CD$. We are also given that
$$\angle ABC = \angle CDA = 90^{\circ}$$
Determine the length of the diagonal $BD$. | [
"Solution:\nSince $AD = DC$ and $\\angle ADC = 90^{\\circ}$, we can fit four copies of quadrilateral $ABCD$ around vertex $D$ as shown in the diagram.\n\n\n\nThe outer shape is a quadrilateral because $\\angle DAB + \\angle BCD = 180^{\\circ}$. Moreover it is a rectangle because $\\angle AB... | New Zealand | NZMO Round Two | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2021/√2 |
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