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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0jl3 | Problem:
Pick a subset of at least four of the following geometric theorems, order them from earliest to latest by publication date, and write down their labels (a single capital letter) in that order. If a theorem was discovered multiple times, use the publication date corresponding to the geometer for which the theor... | [
"Solution:\nAnswer: HMPCVSE The publication dates were as follows.\n- Heron: 60 AD, in his book Metrica.\n- Menelaus: We could not find the exact date the theorem was published in his book Spherics, but because Menelaus lived from 70 AD to around 130 AD, this is the correct placement.\n- Pascal: 1640 AD, when he wa... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | final answer only | HMPCVSE | |
0fxc | Problem:
Sei $n$ eine natürliche Zahl und seien $a, b$ zwei verschiedene ganze Zahlen mit folgender Eigenschaft: Für jede natürliche Zahl $m$ ist $a^{m}-b^{m}$ durch $n^{m}$ teilbar. Zeige, dass $a, b$ beide durch $n$ teilbar sind. | [
"Solution:\n\nOffenbar genügt es den Fall zu betrachten, wo $n=p^{k}$ eine Primpotenz ist. Sei zunächst $k=1$. Nehme an, $a$ und $b$ seien nicht durch $p$ teilbar, mit $m=1$ folgt dann $a \\equiv b \\not \\equiv 0$ $(\\bmod p)$. Wir zeigen, dass $a-b$ durch beliebig grosse $p$-Potenzen teilbar ist, im Widerspruch z... | Switzerland | IMO Selektion | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
013r | Problem:
Let $a_{0}$ be a positive integer. Define the sequence $a_{n}$, $n \geq 0$, as follows: If
$$
a_{n} = \sum_{i=0}^{j} c_{i} 10^{i}
$$
where $c_{i}$ are integers with $0 \leq c_{i} \leq 9$, then
$$
a_{n+1} = c_{0}^{2005} + c_{1}^{2005} + \cdots + c_{j}^{2005}.
$$
Is it possible to choose $a_{0}$ so that all the... | [
"Solution:\n\nIt is clear that there exists a smallest positive integer $k$ such that\n$$\n10^{k} > (k+1) \\cdot 9^{2005}.\n$$\nWe will show that there exists a positive integer $N$ such that $a_{n}$ consists of less than $k+1$ decimal digits for all $n \\geq N$. Let $a_{i}$ be a positive integer which consists of ... | Baltic Way | Baltic Way 2005 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
0cdr | The triangle $ABC$ has $\angle ABC = 90^\circ$ and $\angle BCA = 30^\circ$. Let $AD$ be the bisector of the angle $\angle BAC$, $D \in BC$, and $BE \perp AC$, $E \in AC$. Denote $M$ the intersection of the lines $AD$ and $BE$, and $P$ the midpoint of the segment $CM$. Prove that $AC = 4 \cdot DP$. | [
"The $30^\\circ$ angle theorem yields $AC = 2AB$. Since $\\angle CAD = \\angle ACD = 30^\\circ$, the triangle $ADC$ is isosceles, with $DA = DC$.\n\nFrom $\\angle ADB = \\angle MBD = 60^\\circ$ follows that the triangle $MBD$ is equilateral.\n\nThis shows that $MB = BD = \\frac{1}{2}AD$, hence $M$ is the midpoint o... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06v0 | Let $f:\{1,2,3, \ldots\} \rightarrow\{2,3, \ldots\}$ be a function such that $f(m+n) \mid f(m)+f(n)$ for all pairs $m, n$ of positive integers. Prove that there exists a positive integer $c>1$ which divides all values of $f$. | [
"For every positive integer $m$, define $S_{m}=\\{n: m \\mid f(n)\\}$.\n\nLemma. If the set $S_{m}$ is infinite, then $S_{m}=\\{d, 2 d, 3 d, \\ldots\\}=d \\cdot \\mathbb{Z}_{>0}$ for some positive integer $d$.\n\nProof. Let $d=\\min S_{m}$; the definition of $S_{m}$ yields $m \\mid f(d)$.\nWhenever $n \\in S_{m}$ a... | IMO | IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
050c | Prove that none of the integers that contain one $2$, one $1$, and all the rest zeros, can be expressed as a sum of two perfect squares or as a sum of two perfect cubes. | [
"All the numbers described in the problem are divisible by $3$ (as their sum of digits is divisible by $3$). Note that all perfect squares leave the remainder $0$ or $1$ when divided by $3$, and therefore, for the sum of the two perfect squares to be divisible by $3$, they both have to be divisible by $3$. Now, as ... | Estonia | Selected Problems from Open Contests | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0ifu | Problem:
A regular octahedron $A B C D E F$ is given such that $A D$, $B E$, and $C F$ are perpendicular. Let $G$, $H$, and $I$ lie on edges $A B$, $B C$, and $C A$ respectively such that $\frac{A G}{G B}=\frac{B H}{H C}=\frac{C I}{I A}=\rho$. For some choice of $\rho>1$, $G H$, $H I$, and $I G$ are three edges of a re... | [
"Solution:\nLet $J$ lie on edge $C E$ such that $\\frac{E J}{J C}=\\rho$. Then we must have that $H I J$ is another face of the icosahedron, so in particular, $H I=H J$. But since $B C$ and $C E$ are perpendicular, $H J=H C \\sqrt{2}$. By the Law of Cosines, $H I^{2}=H C^{2}+C I^{2}-2 H C \\cdot C I \\cos 60^{\\cir... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (1+sqrt(5))/2 | |
0ibe | Problem:
An $n$-string is a string of digits formed by writing the numbers $1, 2, \ldots, n$ in some order (in base ten). For example, one possible 10-string is
35728910461
What is the smallest $n > 1$ such that there exists a palindromic $n$-string? | [
"Solution:\nThe following is such a string for $n=19$ :\n$$\n9|18|7|16|5|14|3|12|1|10|11|2|13|4|15|6|17|8|19\n$$\nwhere the vertical bars indicate breaks between the numbers. On the other hand, to see that $n=19$ is the minimum, notice that only one digit can occur an odd number of times in a palindromic $n$-string... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 19 | |
0ce6 | Let $\Gamma$ be a connected graph on $r + g + b + 1$ vertices. The edges of $\Gamma$ bear three colours: red, green, and blue. It turns out that $\Gamma$ has a spanning tree with exactly $r$ red edges, a spanning tree with exactly $g$ green edges, and a spanning tree with exactly $b$ blue edges. Prove that $\Gamma$ has... | [
"Induct on $n = r + g + b$. The base case, $n = 1$, is clear.\n\nLet now $n > 1$. Let $V$ denote the vertex set of $\\Gamma$, and let $T_r, T_g$, and $T_b$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.\n\n**Case 1:** There exists a partition $V = A \... | Romania | THE Fifteenth ROMANIAN MASTER OF MATHEMATICS | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
04q7 | A positive integer $n$ is good if each side and diagonal of a regular $n$-gon can be coloured in some colour so that for each pair of vertices $A$ and $B$ there is exactly one vertex $C$, different from $A$ and $B$, such that the segments $\overline{AB}$, $\overline{BC}$ and $\overline{CA}$ have the same colour.
Which ... | [
"The numbers 8, 10, 11 and 12 are not good, while 7 and 9 are good.\n\nWe first show that even numbers are not good. Let us assume $n$ is a good number and consider a fixed vertex $A$ of a regular $n$-gon coloured in the required way. For each vertex $B$ different from $A$, there is a unique vertex $C$ such that th... | Croatia | Croatian Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 7 and 9 are good; 8, 10, 11, and 12 are not good. | |
02rh | Christine has a deck of cards numbered from $1$ to $25$. She asked her friend Dorothy to choose six cards from the deck. Christine wrote down the chosen numbers and put the cards back in the deck. She then asked Dorothy to choose six cards again and, again, she wrote down the chosen numbers.
a. The first six numbers t... | [
"a. The prime numbers between $1$ and $25$ are $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$. The ones of the form $4k + 1$, $k \\in \\mathbb{Z}$, are $5$, $13$ and $17$; the ones of the form $4k + 3$ are $3$, $7$, $11$, $19$ and $23$. Since there are supposed to be five prime numbers, the numbers are $3$, $7$, ... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | a) 3, 7, 11, 15, 19, 23; b) 24 | |
0l7s | The 27 cells of a $3 \times 9$ grid are filled in using the numbers 1 through 9 so that each row contains 9 different numbers, and each of the three $3 \times 3$ blocks heavily outlined in the example below contains 9 different numbers, as in the first three rows of a Sudoku puzzle.
The numb... | [] | United States | 2025 AIME I | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 81 | |
08kc | Problem:
Let $p_{1}, p_{2}, \ldots, p_{2005}$ be different prime numbers. Let $\mathrm{S}$ be a set of natural numbers whose elements have the property that their simple divisors are some of the numbers $p_{1}, p_{2}, \ldots, p_{2005}$ and the product of any two elements from $\mathrm{S}$ is not a perfect square.
What ... | [
"Solution:\nLet $a, b$ be two arbitrary numbers from $\\mathrm{S}$. They can be written as\n$$\na = p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{2005}^{a_{2005}} \\text{ and } b = p_{1}^{\\beta_{1}} p_{2}^{\\beta_{2}} \\cdots p_{2005}^{\\beta_{2005}}\n$$\nIn order for the product of the elements $a$ and $b$ to be a squar... | JBMO | OJBM | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 2^2005 | |
08nk | Problem:
Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Show that
$$
\frac{1}{a^{3}+b c}+\frac{1}{b^{3}+c a}+\frac{1}{c^{3}+a b} \leq \frac{(a b+b c+c a)^{2}}{6}
$$ | [
"Solution:\nBy the AM-GM inequality we have $a^{3}+b c \\geq 2 \\sqrt{a^{3} b c} = 2 \\sqrt{a^{2}(a b c)} = 2 a$ and\n$$\n\\frac{1}{a^{3}+b c} \\leq \\frac{1}{2 a}\n$$\nSimilarly, $\\frac{1}{b^{3}+c a} \\leq \\frac{1}{2 b}$, $\\frac{1}{c^{3}+a b} \\leq \\frac{1}{2 c}$, and then\n$$\n\\frac{1}{a^{3}+b c}+\\frac{1}{b... | JBMO | JBMO Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
05tu | Problem:
Soit $A B C D E$ un pentagone convexe tel que $\widehat{A B E}=\widehat{A C E}=\widehat{A D E}=90^{\circ}$ et $B C=C D$. Enfin, soit $K$ un point sur la demi-droite $[A B]$ tel que $A K=A D$, et soit $L$ un point sur la demi-droite $[E D)$ tel que $E L=B E$. Démontrer que les points $B, D, K$ et $L$ appartien... | [
"Solution:\n\nAu vu des angles droits dont on dispose en $B$, $C$ et $D$, les points $A$, $B$, $C$, $D$ et $E$ appartiennent à un même demi-cercle de diamètre $[A E]$. Comme $B C=C D$, le point $C$ est donc le pôle Sud issu de $A$ dans le triangle $A B D$. Ainsi, $(A C)$ est la bissectrice de $\\widehat{D A B}$, c'... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07ro | We say that a rectangle with side lengths $a$ and $b$ fits inside a rectangle with side lengths $c$ and $d$ if either $(a \le c)$ and $(b \le d)$ or $(a \le d)$ and $(b \le c)$. For instance, a rectangle with side lengths $1$ and $5$ fits inside another rectangle with side lengths $1$ and $5$, and also fits inside a re... | [
"We write $R \\le R'$ if $R$ fits inside $R'$. (Note that $\\le$ is a preorder: it is reflexive and transitive. It might not be a partial order because two different rectangles might have matching widths and lengths.)\nWe call an $n$-element subset $C$ of $S$ an $n$-chain if its elements can be listed in \"increasi... | Ireland | Irish | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0dc4 | Find all positive integers of form $\overline{abcd}$ such that
$$
\overline{abcd} = a^{a+b+c+d} - a^{-a+b-c+d} + a.
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2018 | |
0jmn | Problem:
Let $f$ be a function from the nonnegative integers to the positive reals such that $f(x+y) = f(x) \cdot f(y)$ holds for all nonnegative integers $x$ and $y$. If $f(19) = 524288 k$, find $f(4)$ in terms of $k$. | [
"Solution:\n\nAnswer: $16 k^{4 / 19}$\n\nThe given condition implies $f(m n) = f(m)^n$, so\n$$\nf(4)^{19} = f(4 \\cdot 19) = f(19 \\cdot 4) = f(19)^4\n$$\nand it follows that $f(4) = 16 k^{4 / 19}$."
] | United States | HMMT November 2014 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 16 k^{4 / 19} | |
0b4f | Problem:
Three circles $\Gamma_{1}$, $\Gamma_{2}$, $\Gamma_{3}$ are pairwise externally tangent, with $\Gamma_{1}$, the smallest, having radius $1$, and $\Gamma_{3}$, the largest, having radius $25$. Let $A$ be the point of tangency of $\Gamma_{1}$ and $\Gamma_{2}$, $B$ be the point of tangency of $\Gamma_{2}$ and $\G... | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 156 | |
00tr | Let $a$, $b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b] = 2021^c$. If $|a - b|$ is a prime number, prove that the number $(a + b)^2 + 4$ is composite. | [
"We write $p = |a - b|$ and assume for contradiction that $q = (a + b)^2 + 4$ is a prime number.\nSince $(a, b) \\mid [a, b]$, we have that $(a, b) \\mid 2021^c$. As $(a, b)$ also divides $p = |a - b|$, it follows that $(a, b) \\in \\{1, 43, 47\\}$. We will consider all 3 cases separately:\n\n(1) If $(a, b) = 1$, t... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Other"
] | null | proof only | null | |
0aez | Нека $\Delta ABC$ е рамностран. На страната $AB$ се избрани точки $C_1$ и $C_2$, на $AC$ се избрани точки $B_1$ и $B_2$ и на страната $BC$ се избрани точки $A_1$ и $A_2$, притоа важи: $\overline{A_1A_2} = \overline{B_1B_2} = \overline{C_1C_2}$. Нека пресечните точки на правите $A_2B_1$ и $B_2C_1$, $B_2C_1$ и $C_2A_1$, ... | [
"Нека триаголникот формиран од отсечките $B_1A_2$, $A_1C_2$ и $C_1B_2$ го означиме со $\\Delta A_3B_3C_3$. Нека $P$ е точка во внатрешноста на $\\Delta EFG$ т.ш. $C_1C_2PB_2$ е паралелограм.\nТогаш $\\Delta B_2PB_1$ е рамностран.\nПа $PA_1A_2B_1$ е паралелограм. Па од претходниве разгледувања имаме $PC_2 \\parallel... | North Macedonia | ЈММО 2009 година | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Macedonian, English | proof only | null | |
0965 | Problem:
Determinaţi toate funcţiile $f: \mathbb{R} \rightarrow \mathbb{R}$, care verifică simultan condiţiile:
1) $|f(x)| \geq 1$, oricare ar fi numărul real $x$.
2) $f(x+y)=\frac{f(x)+f(y)}{1+f(x) \cdot f(y)}$, oricare ar fi numerele reale $x$ şi $y$. | [
"Solution:\n\nÎntrucât relaţia a doua are loc pentru orice valori reale ale numerelor $x$ şi $y$, $f(x) \\cdot f(y) \\neq -1$, $(\\forall) x \\in \\mathbb{R}, (\\forall) y \\in \\mathbb{R}$.\n\nRelaţia 2) este echivalentă cu $f(x+y) \\cdot (1+f(x) \\cdot f(y)) = f(x) + f(y)$, $(\\forall) x \\in \\mathbb{R}, (\\fora... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = 1 for all real x or f(x) = -1 for all real x | |
0gug | Two distinct circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Let the lines $l_1$ and $l_2$ be the common tangent lines of these circles such that $l_1$ is tangent to $\omega_1$ at $A$ and $\omega_2$ at $C$ and $l_2$ is tangent to $\omega_1$ at $B$ and $\omega_2$ at $D$. Let $Z$ be the reflection of $... | [
"\n\nLet $\\ell_1 \\cap \\ell_2 = P$. Since $\\angle AXY = \\angle YAC$ and $\\angle CXY = \\angle YCA$ we have $\\angle AXC + \\angle AYC = 180^\\circ$. Therefore, by symmetry, we have $\\angle AZC = \\angle AYC = 180^\\circ - \\angle AXC$ hence $A, X, C, Z$ are concyclic.\n\nNow we will p... | Turkey | Team Selection Test for EGMO 2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01jt | Find all pairs of positive integers $a$ and $b$ ($a \le b$) such that
$$
ab = 300 + 7[a, b] + 5(a, b),
$$
where $(a, b)$ is the greatest common divisor and $[a, b]$ is the least common multiple of $a$ and $b$. | [
"Let $[a, b] = x$, and $(a, b) = y$. It is known (and easy to prove) that $ab = [a, b] \\cdot (a, b)$, so $ab = xy$. Then the initial equation can be rewritten in the form\n$$\nxy = 300 + 7x + 5y \\Leftrightarrow xy - 7x - 5y + 35 = 335 \\Leftrightarrow x(y - 7) - 5(y - 7) = 335 \\Leftrightarrow (x - 5)(y - 7) = 5 ... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis,... | English | proof and answer | [(12, 72), (24, 36)] | |
0dxt | Problem:
a) Pokaži, da je vrednost izraza $\sqrt{9} \cdot \sqrt[3]{64} - \sqrt[3]{27} \cdot \sqrt{4}$ naravno število.
b) Pokaži, da je za vsako naravno število $x$ število $64^{-1} \cdot 8^{2x+4} - 24 \cdot 64^{x} + 4 \cdot 32^{x} \cdot 2^{x+3}$ deljivo s številom $\sqrt{9} \cdot \sqrt[3]{64} - \sqrt[3]{27} \cdot \sq... | [
"Solution:\n\na.\nIzračunajmo vrednost izraza $B = \\sqrt{9} \\cdot \\sqrt[3]{64} - \\sqrt[3]{27} \\cdot \\sqrt{4}$:\n\n$\\sqrt{9} = 3$\n\n$\\sqrt[3]{64} = 4$\n\n$\\sqrt[3]{27} = 3$\n\n$\\sqrt{4} = 2$\n\nTorej:\n\n$$\nB = 3 \\cdot 4 - 3 \\cdot 2 = 12 - 6 = 6\n$$\n\nKer je $6$ naravno število, je vrednost izraza nar... | Slovenia | 7. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
06xo | The Imomi archipelago consists of $n \geqslant 2$ islands. Between each pair of distinct islands is a unique ferry line that runs in both directions, and each ferry line is operated by one of $k$ companies. It is known that if any one of the $k$ companies closes all its ferry lines, then it becomes impossible for a tra... | [
"Answer: The largest $k$ is $k=\\left\\lfloor\\log_{2} n\\right\\rfloor$.\n\nWe reformulate the problem using graph theory. We have a complete graph $K_{n}$ on $n$ nodes (corresponding to islands), and we want to colour the edges (corresponding to ferry lines) with $k$ colours (corresponding to companies), so that ... | IMO | International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | floor(log_2 n) | |
0g2r | Problem:
Déterminer toutes les suites périodiques $x_{1}, x_{2}, x_{3}, \ldots$ de nombres réels strictement positifs telles que pour tout $n \geq 1$
$$
x_{n+2}=\frac{1}{2}\left(\frac{1}{x_{n+1}}+x_{n}\right)
$$ | [
"Solution:\n\n$$\n2 x_{n+2} x_{n+1}=x_{n+1} x_{n}+1\n$$\nIl paraît à présent naturel d'introduire la nouvelle suite $y_{n}:=x_{n+1} x_{n}$ pour $n \\geq 1$. Si la suite $\\left\\{x_{n}\\right\\}$ est périodique, alors la suite $\\left\\{y_{n}\\right\\}$ l'est aussi. On a la relation $2 y_{n+1}=y_{n}+1$. Au feeling,... | Switzerland | SMO 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All solutions are 2-periodic sequences with consecutive terms reciprocal: x_{n+1} x_n = 1 for all n. Equivalently, for some a > 0, x_n = a if n is odd and x_n = 1/a if n is even (the constant sequence x_n = 1 is included as a = 1). | |
09bh | $P(n) = n^4 + 2n^3 - n^2 + 2n + 1$ олон гишүүнт өгөгдөв. $|P(n)|$ анхны тоо байх бүх бүхэл $n$-ийг ол. | [
"$$\n n^4 + 2n^3 - n^2 + 2n + 1 = (n^2 + a n + 1)(n^2 + b n + 1) = n^4 + (a + b) n^3 + (2 + a b) n^2 + (a + b) n + 1\n$$\nба\n$$\n\\begin{cases} a + b = 2 \\\\ 2 + a b = -1 \\end{cases} \\Rightarrow a = 3, b = -1\n$$\nболно. Иймд\n$$\nP(n) = (n^2 + 3n + 1)(n^2 - n + 1)\n$$\nболов. $|P(n)|$ анхны тоо болохын тулд да... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Mongolian | proof and answer | -3, -2, -1, 1 | |
06i4 | Let $m, n$ be distinct positive integers not exceeding $2013$ and $d$ be their greatest common divisor. Suppose $d^2$ divides $3(m-n)$. Find the greatest possible value of $d(m+n)$. | [
"The greatest possible value of $d(m+n)$ is $166635$.\nLet $m = da$ and $n = db$ so that $(a, b) = 1$. The condition becomes $d \\mid 3(a - b)$. WLOG assume $m > n$. Let $N = d(m + n) = d^2(a + b)$.\nIf $3 \\nmid d$, then $d \\mid a-b$. This implies $d \\le a-b$. Also, we have $a = \\frac{m}{d} \\le \\frac{2013}{d}... | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 166635 | |
0kyk | Problem:
A circle of area $1$ is cut by two distinct chords. Compute the maximum possible area of the smallest resulting piece.
Proposed by: Derek Liu, Luke Robitaille
Answer: $\square$ | [
"Solution:\nAt least $3$ pieces are formed, so one of them has area at most $\\frac{1}{3}$. This can be achieved with two parallel chords:\n\n"
] | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | 1/3 | |
05xz | Problem:
Pour un réel $x$, on note $\lfloor x\rfloor$ le plus grand entier relatif inférieur ou égal à $x$ (par exemple, $\lfloor 2,7\rfloor=2$, $\lfloor\pi\rfloor=3$ et $\lfloor-1,5\rfloor=-2$).
Soient $a, b$ deux réels tels que
$$
a+\lfloor a\rfloor=b+\lfloor b\rfloor .
$$
Montrer que $a=b$. | [
"Solution:\n\nSupposons par l'absurde $a \\neq b$. Par symétrie (c'est-à-dire que $a$ et $b$ jouent le même rôle), on peut supposer $a<b$. On a alors\n$$\n\\lfloor a\\rfloor \\leqslant a<b\n$$\ndonc $\\lfloor a\\rfloor$ est un entier relatif inférieur ou égal à $b$, et donc par définition on a $\\lfloor a\\rfloor \... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0g7l | 令 $f$ 和 $g$ 是兩個係數是整數的非零多項式, 且 $\deg f > \deg g$。假設有無窮多的質數 $p$ 使得多項式 $pf + g$ 有一個有理根, 證明 $f$ 也有一個有理根。 | [
"因為 $\\deg f > \\deg g$, 所以當 $x$ 足夠大時, 我們有 $|g(x)/f(x)| < 1$。即存在一個實數 $R$ 使得對所有的 $|x| > R$, 我們有 $|g(x)/f(x)| < 1$。對於所有這樣的 $x$ 及所有的質數 $p$, 我們有\n$$\n|pf(x) + g(x)| \\geq |f(x)| \\left( p - \\frac{|g(x)|}{|f(x)|} \\right) > 0.\n$$\n因此多項式 $pf + g$ 的所有的實數根都會落在 $[-R, R]$。\n\n令 $f(x) = a_n x^n + a_{n-1} x^{n-1} + \\cdots +... | Taiwan | 二〇一三數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0hz1 | Problem:
Matt has somewhere between 1000 and 2000 pieces of paper he's trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries $2,3,4,5,6,7$, and $8$ piles but ends up with one sheet left over each time. How many piles does he need? | [
"Solution:\n\nThe number of sheets will leave a remainder of $1$ when divided by the least common multiple of $2,3,4,5,6,7$, and $8$, which is $8 \\cdot 3 \\cdot 5 \\cdot 7 = 840$. Since the number of sheets is between $1000$ and $2000$, the only possibility is $1681$. The number of piles must be a divisor of $1681... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 41 | |
0274 | Problem:
Um semicírculo de diâmetro $E F$, situado no lado $B C$ do triângulo $A B C$, é tangente aos lados $A B$ e $A C$ em $Q$ e $P$, respectivamente. As retas $E P$ e $F Q$ se encontram em $H$. Mostre que $A H$ é a altura do triângulo.
 | [
"Solution:\n\nSejam $K$ o pé da perpendicular de $H$ ao segmento $B C$ e $O$ o centro do semicírculo. Suponha sem perda de generalidade que $K$ está no segmento $O C$.\n\n\n\nComo $E F$ é um diâmetro, segue que $\\angle E Q F = \\angle H K E = 90^\\circ$ e consequentemente $E Q H K$ é um qu... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kk9 | Mr. Lopez has a choice of two routes to get to work. Route A is 6 miles long, and his average speed along this route is 30 miles per hour. Route B is 5 miles long, and his average speed along this route is 40 miles per hour, except for a $\frac{1}{2}$-mile stretch in a school zone where his average speed is 20 miles pe... | [] | United States | AMC 12 A | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | B | |
0bke | In triangle $ABC$ points $M, N \in (AB)$, $P, Q \in (BC)$ and $S, R \in (AC)$ are taken such that $AM = CR$, $AN = CS$, $\overline{MQB} \equiv \overline{RQC}$ and $\overline{NPB} \equiv \overline{SPC}$.
Show that if $MQ + QR = NP + PS$, then triangle $ABC$ is isosceles. | [
"Denote $R'$ and $S'$ the reflection of $R$, respectively $S$, across the line $BC$. Then $RQ = R'Q$ and $\\overline{RQC} \\equiv \\overline{R'QC}$. Since $\\overline{MQB} \\equiv \\overline{RQC}$, it follows that $\\overline{MQB} \\equiv \\overline{R'QC}$, and because $B, Q, C$ are collinear, so are $M, Q, R'$. Th... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
0gjq | Suppose function $f: \mathbb{N} \to \mathbb{N}$ satisfies
$$
f^{bf(a)}(a+1) = (a+1)f(b)
$$
for all positive integers $a, b \in \mathbb{N}$, where $f^k(n) = f(f(\cdots f(n)\cdots))$ denotes the composition of $f$ with itself $k$ times. Prove that $f(n) = n+1$ for all $n \in \mathbb{N}$. | [
"**Step 1. ($f$ is injective)**\n**Claim 1.** For any $a \\ge 2$, the set $\\{f^n(a) \\mid n \\in \\mathbb{Z}_{\\ge 0}\\}$ is infinite.\n*Proof.* First, we have $f^{f(a)}(a+1) \\stackrel{P(a,1)}{=} (a+1)f(1)$. Varying $a$, we see that $f(\\mathbb{Z}_{\\ge 0})$ is infinite. Next, we have $f^{bf(a-1)}(a) \\stackrel{P... | Taiwan | IMO 3J, Independent Study 2 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | Chinese; English | proof only | null | |
093o | Problem:
Let $N$ be a positive integer such that the sum of the squares of all positive divisors of $N$ is equal to the product $N(N+3)$. Prove that there exist two indices $i$ and $j$ such that $N=F_{i} \cdot F_{j}$, where $\left(F_{n}\right)_{n=1}^{\infty}$ is the Fibonacci sequence defined by $F_{1}=F_{2}=1$ and $F... | [
"Solution:\n\nDenote by $1=d_{1}<d_{2}<\\cdots<d_{k}=N$ all positive divisors of the given positive integer $N$ that satisfies\n$$\nd_{1}^{2}+d_{2}^{2}+\\cdots+d_{k}^{2}=N(N+3), \\quad \\text{ i.e. } \\quad d_{2}^{2}+d_{3}^{2}+\\cdots+d_{k-1}^{2}=3 N-1\n$$\nNow $3 N-1>0$ implies that $k \\geq 3$. However, if $k=3$,... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"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",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power... | null | proof only | null | |
0bhp | If $n$ is a composite number, denote $D_n$ its largest proper divisor. A number $n$ will be called *squarish* if the number $D_n + D_{n+1}$ is a perfect square.
a) Show that $35$, $76$ and $755$ are squarish.
b) Show that there are infinitely many squarish numbers. | [
"a.\nLet us compute $D_n + D_{n+1}$ for $n = 35$, $76$, and $755$.\n\nFor $n = 35$:\n$35$ is composite. Its largest proper divisor is $D_{35} = 35/5 = 7$ (since $35 = 5 \\times 7$ and $7 < 35$).\n$36$ is composite. Its largest proper divisor is $D_{36} = 36/2 = 18$ (since $36 = 2 \\times 18$ and $18 < 36$).\nSo $D_... | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 35, 76, and 755 are squarish; there are infinitely many squarish numbers. | |
05ew | Problem:
Soit $ABC$ un triangle, $L$, $M$ et $N$ les milieux respectifs des côtés $[BC]$, $[CA]$ et $[AB]$. On suppose que $\widehat{ANC} = \widehat{ALB}$. Montrer que $\widehat{CAL} = \widehat{ABM}$. | [
"Solution:\n\n\n\nOn essaie dans un premier temps d'interpréter la condition de l'énoncé $\\widehat{ANC} = \\widehat{ALB}$. Pour cela on commence par introduire $G$ le centre de gravité du triangle $ABC$ (on rappelle que c'est aussi l'intersection des trois médianes $(AL)$, $(BM)$ et $(CN)$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
085j | Problem:
Sia $p(x)=x^{20}+a_{19} x^{19}+a_{18} x^{18}+\ldots+a_{1} x+a_{0}$ un polinomio, con gli $a_{i}$ interi. Sappiamo che, per tutti gli interi $k$ compresi tra 1 e 20, $p(k)=2 k$. Quali sono le ultime 3 cifre di $p(21)$ ? | [
"Solution:\n\nLa risposta è 042 . Sia $q(x)=p(x)-2 x$. Poiché $p(k)=2 k$ per $k=1, \\ldots, 20$, allora anche $q(k)=0$ per $k=1, \\ldots, 20$. In base al teorema di Ruffini, questo equivale a dire che il polinomio $q(x)$ è divisibile per $x-1, x-2, \\ldots, x-20$. Ma allora è divisibile anche per il loro prodotto, ... | Italy | Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 042 | |
0c9k | Problem:
Determinați funcțiile continue $f:[0,1] \longrightarrow [0, \infty)$ care verifică relația
$$
\int_{0}^{1} f(x) \, dx \cdot \int_{0}^{1} f^{2}(x) \, dx \cdots \int_{0}^{1} f^{2020}(x) \, dx = \left(\int_{0}^{1} f^{2021}(x) \, dx\right)^{1010}
$$ | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | All solutions are the constant functions f(x) = c with c ≥ 0. | |
0hr9 | Problem:
Let $P$ be a $2023$-sided polygon. All but one side has length $1$. What is the maximum possible area of $P$? | [
"Solution:\nFirst, we claim $P$ must be convex to maximize its area. If not, let $A$ and $B$ be consecutive vertices on the perimeter of its convex hull that aren't consecutive vertices of $P$. Reflecting the path between $A$ and $B$ over line $AB$ must increase the area of $P$ as the new shape strictly contains $P... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1011/2 * cot(pi/4044) | |
00ta | Find all functions $f : \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that
$$
f(xf(x + y)) = y f(x) + 1
$$
holds for all $x, y \in \mathbb{R}^{+}$. | [
"We will show that for every $x \\in \\mathbb{R}^{+}$, $f(x) = \\frac{1}{x}$ for every $x \\in \\mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.\n\nWe write $P(x, y)$ for the assertion that $f(xf(x + y)) = y f(x) + 1$.\n\nWe first show that $f$ is injective. So assume $f(x_1) = f(x_2)... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = 1/x for all x > 0 | |
0a81 | Problem:
Let $A$, $B$, $C$, and $D$ be four different points in the plane. Three of the line segments $AB$, $AC$, $AD$, $BC$, $BD$, and $CD$ have length $a$. The other three have length $b$, where $b > a$. Determine all possible values of the quotient $\frac{b}{a}$. | [
"Solution:\n\nIf the three segments of length $a$ share a common endpoint, say $A$, then the other three points are on a circle of radius $a$, centered at $A$, and they are the vertices of an equilateral triangle of side length $b$. But this means that $A$ is the center of the triangle $BCD$, and\n$$\n\\frac{b}{a} ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 11 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(3); (sqrt(5)+1)/2 | |
0fu6 | Problem:
$m \times n$ Punkte sind in einem quadratischen Gitter zu einem Rechteck angeordnet. Wieviele Möglichkeiten gibt es, diese Punkte rot oder weiss zu färben, sodass unter je vier Punkten, die Ecken eines Einheitsquadrates bilden, genau zwei weisse und zwei rote vorkommen? | [
"Solution:\n\nWir betrachten das Gitter so, dass es $m$ Zeilen und $n$ Spalten hat. Wir färben die Punkte zeilenweise ein und beginnen mit der ersten Zeile. Nehme an, wir hätten die $k$-te Zeile schon eingefärbt.\n\nWir nennen eine Färbung der $k+1$-ten Zeile zulässig, wenn die Bedingung der Aufgabe für die $k$-te ... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^m + 2^n - 2 | |
07dv | Mojtaba and Hooman are playing a game. Initially Mojtaba draws $2018$ vectors with zero sum. Then starting with Mojtaba, each player takes a vector and puts it on the plane. After the first move, the players must put the starting point of their vector on the end of the vector that the previous person drew, until a clos... | [
"Mojtaba has a winning strategy.\nLet him consider a large vector $\\vec{V}$ to the left direction and $2017$ small vectors facing towards right. Let these vectors be $\\vec{V} = (-V, 0)$, $\\vec{v}_1 = (x_1, y_1)$, $\\vec{v}_2 = (x_2, y_2)$, $\\dots$, $\\vec{v}_{2017} = (x_{2017}, y_{2017})$. So that we have\n$$\n... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | Mojtaba has a winning strategy. | |
0dtu | Let $I$ be the incenter of $\triangle ABC$ and $\omega$ be its incircle. Let $D$ be the intersection of $\omega$ with $BC$. Suppose $X$, $Y$ are points on $\omega$ such that lines $IX$, $IY$ and $AD$ are tangent to the circumcircle of $\triangle AXY$. Prove that line $XY$ bisects segment $BC$. | [
"\n\nLet $M$ be the midpoint of $BC$. Denote $D'$ and $E$ as the reflection of $D$ across $I$ and $M$ respectively. It is well known that $E$ is the intersection of the $A$-excircle with $BC$. Hence, the homothety centered at $A$ sending the incircle to the $A$-excircle also sends $D'$ to $... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
0k11 | Problem:
Square $CASH$ and regular pentagon $MONEY$ are both inscribed in a circle. Given that they do not share a vertex, how many intersections do these two polygons have? | [
"Solution:\n\nPentagon $MONEY$ divides the circumference into $5$ circular arcs, and each vertex of $CASH$ lies in a different arc. Then each side of $CASH$ will intersect two sides of $MONEY$, for a total of $8$ intersections."
] | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 8 | |
0dsf | Let $A$ and $B$ be two $n \times n$ square arrays. The cells of $A$ are labelled by the numbers from $1$ to $n^2$ from left to right starting from the top row; whereas the cells of $B$ are labelled by the numbers from $1$ to $n^2$ along rising north-easterly diagonals starting with the upper left-hand corner. Stack the... | [
"For each $i = 1, 2, \\dots, n^2$, define $f(i) = j$ if the cells containing the number $i$ in $A$ and the cell containing $j$ in $B$ overlap. The two cells are coloured red iff $i = j$. We claim that $f$ has a fixed point not in the set $\\{1, n^2, (n^2+1)/2\\}$ if and only if $n$ has at least two different prime ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | There is at least one nontrivial red cell if and only if n has at least two distinct prime factors (i.e., n is not a prime power). | |
0czd | Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers for which
$$
\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\ldots+\frac{1}{1+x_{n}}=1
$$
Prove that
$$
x_{1} x_{2} \ldots x_{n} \geq (n-1)^{n}
$$ | [
"Let $y_{k}=\\frac{1}{1+x_{k}},\\ k=1,2, \\ldots, n$. Then\n$$\nx_{k}=\\frac{1}{y_{k}}-1=\\frac{y_{1}+\\ldots+y_{n}}{y_{k}}-1=\\frac{S_{k}}{y_{k}}\n$$\nwhere\n$$\nS_{k}=x_{1}+\\ldots+x_{k-1}+x_{k+1}+\\ldots+x_{n}, \\quad k=1,2, \\ldots, n.\n$$\nApplying AM-GM inequality it follows\n$$\n\\frac{S_{k}}{n-1} \\geq \\sq... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
06bw | From a point $P$ outside a circle centred at $O$, draw the two tangents to the circle touching it at $A, B$. Let $M$ be a point on the segment $AB$ and let $C, D$ be points on the circle with midpoint $M$. Let the tangents to the circle at $C, D$ intersect at $Q$. Show that $OQ \perp PQ$. | [
"This is a simple corollary of Brokard's theorem. Alternatively, we provide an elementary proof as follows.\nNote that $O$, $M$, $Q$ are collinear since all of them lie on the perpendicular bisector of $CD$. By the property of tangents, we know that $Q$, $C$, $O$, $D$ are concyclic. This yields\n$$\nMQ \\times MO =... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kcm | Problem:
Some bishops and knights are placed on an infinite chessboard, where each square has side length $1$ unit. Suppose that the following conditions hold:
- For each bishop, there exists a knight on the same diagonal as that bishop (there may be another piece between the bishop and the knight).
- For each knight,... | [
"Solution:\n\nColor the chessboard with the usual chessboard coloring. Note that bishops can only attack knights on the same colored squares, while knights can only attack bishops on different colored squares. Let $B_{B}, W_{B}, B_{N}, W_{N}$ denote the number of bishops on black colored squares, bishops on white c... | United States | HMIC | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | all positive multiples of 4 | |
0l5l | Problem:
Mark writes the expression $\sqrt{abcd}$ on the board, where $abcd$ is a four-digit number and $a \neq 0$. Derek, a toddler, decides to move the $a$, changing Mark's expression to $a\sqrt{bc}$. Surprisingly, these two expressions are equal. Compute the only possible four-digit number $abcd$. | [
"Solution:\n\nLet $x = bcd$. Then, we rewrite the given condition $\\sqrt{abcd} = a\\sqrt{bc}$ as\n\n$$1000a + x = a^{2}x,$$\n\nwhich simplifies as\n\n$$(a^{2} - 1)x = 1000a.$$\n\nIn particular, $a^{2} - 1$ divides $1000a$. Since $\\gcd(a^{2} - 1, a) = 1$, it follows that $a^{2} - 1 \\mid 1000$. The only $a \\in \\... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 3375 | |
0j8z | Problem:
Let $A B C$ be a non-isosceles, non-right triangle, let $\omega$ be its circumcircle, and let $O$ be its circumcenter. Let $M$ be the midpoint of segment $B C$. Let the tangents to $\omega$ at $B$ and $C$ intersect at $X$. Prove that $\angle O A M = \angle O X A$. (Hint: use SAS similarity). | [
"Solution:\n\nNote that $\\triangle O M C \\sim \\triangle O C X$ since $\\angle O M C = \\angle O C X = \\frac{\\pi}{2}$. Hence $\\frac{O M}{O C} = \\frac{O C}{O X}$, or, equivalently, $\\frac{O M}{O A} = \\frac{O A}{O X}$. By SAS similarity, it follows that $\\triangle O A M \\sim \\trian... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 | |
0bc3 | We consider an $n \times n$ ($n \in \mathbb{N}$, $n \ge 2$) square divided into $n^2$ unit squares. Determine all the values of $k \in \mathbb{N}$ for which we can write a real number in each of the unit squares such that the sum of the $n^2$ numbers is a positive number, while the sum of the numbers from the unit squa... | [
"We will prove that the desired numbers $k$ are those that are not factors of $n$.\n\nIf $k \\mid n$, then we can tile the $n \\times n$ square with $k \\times k$ squares and the total sum should simultaneously be positive and negative, which is impossible.\n\nIf $k \\nmid n$, then $n = kq + r$, where $0 < r < k$. ... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All natural k that do not divide n. | |
030w | Problem:
Aflați câte numere naturale $k \in \{1,2,3, \ldots, 2022\}$ au proprietatea că, dacă pe un cerc se scriu 2022 numere reale astfel încât suma oricăror $k$ numere aflate pe poziții consecutive este egală cu 2022, atunci toate cele 2022 de numere sunt egale. | [
"Solution:\nFie $k \\in \\{1,2,3, \\ldots, 2022\\}$ un număr cu proprietatea că, date fiind numerele reale $x_{1}, x_{2}, \\ldots x_{2022}$ scrise pe un cerc, suma oricăror $k$ numere aflate pe poziții consecutive este egală cu 2022. Pentru orice $n \\in \\mathbb{N}$, definim $x_{n}=x_{r}$, unde $r \\in \\{1,2,3, \... | Brazil | Primul baraj de selecție pentru OBMJ | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 672 | |
070g | Problem:
$k \geq 1$ is a real number such that if $m$ is a multiple of $n$, then $[mk]$ is a multiple of $[nk]$. Show that $k$ is an integer. | [
"Solution:\n\nSuppose $k$ is not an integer. Take an integer $n$ such that $nk > 1$, but $nk$ is not an integer. Now take a positive integer $c$ such that $\\frac{1}{c+1} \\leq nk - [nk] < \\frac{1}{c}$. Then $1 \\leq (c+1)nk - (c+1)[nk] < 1 + \\frac{1}{c}$. Hence $[(c+1) n k] = (c+1)[n k] + 1$. Put $m = (c+1) n$. ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
07ge | Points $D$ and $E$ lie on the side $BC$ of an acute-angled triangle $ABC$ such that $\angle DAB = \angle EAC$. Let $\omega$ be a circle whose center lies on the circumcircle of the triangle $ABC$ ($\Gamma$) and is tangent to $AD$ at $A$.
Denote by $A'$ the reflection of point $A$ with respect to $BC$ and the intersect... | [
"Let $F$ be the second intersection point of $\\omega$ and $\\Gamma$, and $K'$ the second intersection of $\\omega$ and $BF$. We just need to show that $K'$ lies on $A'E$. Suppose that $O$ is the center of $\\omega$ and $P$ is the intersection point of $AD$ and $\\Gamma$. It is clear that $\\angle OAP = 90^\\circ$,... | Iran | 38th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l4q | Problem:
Equilateral triangles $\triangle ABC$ and $\triangle DEF$ are drawn such that points $B$, $E$, $F$, and $C$ lie on a line in this order, and point $D$ lies inside triangle $\triangle ABC$. If $BE = 14$, $EF = 15$, and $FC = 16$, compute $AD$. | [
"Solution:\n\n\n\nExtend $DE$ to meet $AC$ at $X$. Observe that $ABEX$ and $DFCX$ are isosceles trapezoids (both with base angles of $60^\\circ$), so we have\n\n$AX = BE = 14$\n\n$DX = FC = 16$\n\nand $\\angle AXD = 120^\\circ$\n\nBy Law of Cosines on $\\triangle ADX$, the answer is\n\n$$\n... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 26 | |
0kgs | Problem:
Let $S$ be a set of positive integers satisfying the following two conditions:
- For each positive integer $n$, at least one of $n, 2n, \ldots, 100n$ is in $S$.
- If $a_{1}, a_{2}, b_{1}, b_{2}$ are positive integers such that $\operatorname{gcd}\left(a_{1} a_{2}, b_{1} b_{2}\right)=1$ and $a_{1} b_{1}, a_{2} ... | [
"Solution:\nThe optimal value of $r$ is $\\frac{1}{252}$. This is attained by letting $S$ be the set of integers $n$ for which $\\nu_{2}(n) \\equiv 4 \\bmod 5$ and $\\nu_{3}(n) \\equiv 1 \\bmod 2$.\n\nLet $S$ be a set of positive integers satisfying the two conditions. For each prime $p$, let $A_{p}=\\left\\{\\nu_{... | United States | HMMT Spring 2021 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | null | proof and answer | 396 | |
04po | Find all complex numbers $a$ such that all coefficients of
$$
P(x) = (x - a)(x - a^2)(x - a^3)(x - a^4)
$$
are real numbers. | [
"Obviously, all real numbers satisfy the given condition.\nLet us, from now on, assume that $a$ is not a real number.\nIf $z$ is a root of $P$, then $\\bar{z}$ is a root of $P$ as well. Hence $\\bar{a} = a^2$, $\\bar{a} = a^3$ or $\\bar{a} = a^4$.\n\nIn the first case, we get $|a| = |a|^2$, and then $|a| = 1$ (sinc... | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | All real numbers, together with a in {i, -i} and a = cos(2kπ/5) + i sin(2kπ/5) for k = 1, 2, 3, 4. | |
0f6x | Problem:
There are 30 cups each containing milk. An elf is able to transfer milk from one cup to another so that the amount of milk in the two cups is equalised. Is there an initial distribution of milk so that the elf cannot equalise the amount in all the cups by a finite number of such transfers? | [] | Soviet Union | 20th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Matrices"
] | null | proof and answer | Yes | |
0axd | Problem:
Given that $100^{2}+1^{2}=65^{2}+76^{2}=p q$ for some primes $p$ and $q$. Find $p+q$. | [
"Solution:\nRecall the identity $\\left(a^{2}+b^{2}\\right)\\left(c^{2}+d^{2}\\right)=(a d+b c)^{2}+(a c-b d)^{2}=(a c+b d)^{2}+(a d-b c)^{2}$. This, and the form of the equation, plus matching parities (note that $a d+b c$ and $a d-b c$ must share the same parity) suggests, as one possible setup:\n$$\n\\begin{alig... | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 210 | |
0and | Problem:
If $z^{3} - 1 = 0$ and $z \neq 1$, find the value of $z + \frac{1}{z} + 4$. | [
"Solution:\nSince $z^{3} - 1 = 0$, $z^{3} = 1$. The solutions are the cube roots of unity: $z = 1$, $z = \\omega$, $z = \\omega^{2}$, where $\\omega = e^{2\\pi i/3}$ and $\\omega^{2} = e^{4\\pi i/3}$.\n\nGiven $z \\neq 1$, so $z = \\omega$ or $z = \\omega^{2}$.\n\nRecall that $\\omega + \\omega^{2} = -1$ and $\\ome... | Philippines | AREA STAGE | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 3 | |
00kk | Prove that
$$
(x^2 + y^2 z^2) \cdot (y^2 + x^2 z^2) \cdot (z^2 + x^2 y^2) \geq 8 x y^2 z^3
$$
holds for all integer values of $x$, $y$ and $z$. When does equality hold? | [
"We first note that the left side of the inequality is certainly non-negative for all values of $x$, $y$ and $z$. If any of the variables is equal to $0$, the right-hand side is equal to $0$, and the inequality certainly holds. If equality holds with any variable being equal to $0$, we can without loss of generalit... | Austria | Austria 2014 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Equality holds if and only if either at least two of the variables are zero, or all variables have absolute value one with the first and third variables having the same sign, namely (1, 1, 1), (1, −1, 1), (−1, 1, −1), (−1, −1, −1). | |
0ct2 | В государстве $n$ городов, и между каждыми двумя из них курсирует экспресс (в обе стороны). Для любого экспресса цены билетов «туда» и «обратно» равны, а для любых разных экспрессов эти цены различны. Докажите, что путешественник может выбрать начальный город, выехать из него и проехать последовательно на $n-1$ экспрес... | [
"**Первое решение.** Уберём все экспрессы, а затем начнём запускать их обратно по одному в порядке возрастания цены (т. е. первым запустим самый дешёвый, вторым — самый дешёвый из остальных, и т. д.). В каждый момент в каждом городе будем писать максимальное количество экспрессов, на которых можно последовательно п... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
039s | Find all integers $b$ and $c$ such that the equation $x^2 - b x + c = 0$ has two real solutions $x_1$ and $x_2$ with $x_1^2 + x_2^2 = 5$. | [] | Bulgaria | Fall Mathematical Competition | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | (-3, 2), (-1, -2), (1, -2), (3, 2) | |
0cs5 | Дан многочлен
$$
P(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \dots + a_1 x + a_0,
$$
у которого каждый коэффициент $a_i$ принадлежит отрезку $[100, 101]$.
При каком минимальном натуральном $n$ у такого многочлена может найтись действительный корень? | [
"**Ответ.** $n = 100$.\n\nНазовём многочлен, удовлетворяющий условию задачи, **красивым**. Многочлен\n$$\nP(x) = 100(x^{200} + x^{198} + \\dots + x^2 + 1) + 101(x^{199} + x^{197} + \\dots + x)\n$$\nкрасив и имеет корень $-1$. Значит, при $n = 100$ требуемое возможно.\n\nОсталось показать, что при $n < 100$ у красив... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n = 100 | |
071y | Problem:
Do there exist 100 lines in the plane, no three of them concurrent, such that they intersect exactly in 2002 points? | [
"Solution:\n\nAny set of 100 lines in the plane can be partitioned into a finite number of disjoint sets, say $A_{1}, A_{2}, A_{3}, \\ldots, A_{k}$, such that\n\n(i) Any two lines in each $A_{j}$ are parallel to each other, for $1 \\leq j \\leq k$ (provided, of course, $|A_{j}| \\geq 2$);\n\n(ii) for $j \\neq l$, t... | India | INMO | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
05no | Problem:
Soit $ABC$ un triangle non-isocèle en $A$. On note $D, E, F$ les points de tangence du cercle inscrit sur les côtés $(BC)$, $(AC)$ et $(AB)$, $I$ le centre du cercle inscrit de $ABC$. Soient $P$ et $Q$ les intersections de la droite $(EF)$ avec le cercle $\Omega$ circonscrit à $ABC$. Soient enfin $O_{1}$ et $... | [
"Solution:\n\n\n\nOn peut déjà remarquer que $O_{1}$ et $O_{2}$ sont les pôles sud de $C$ et de $B$ respectivement dans le triangle $ABC$. La droite $\\left(O_{1}O_{2}\\right)$ est la médiatrice de $[AI]$. Or les droites $(AI)$ et $(EF)$ sont perpendiculaires. On en déduit alors que les dro... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > ... | null | proof only | null | |
0ey1 | Problem:
In the acute-angled triangle $ABC$, $AH$ is the longest altitude ($H$ lies on $BC$), $M$ is the midpoint of $AC$, and $CD$ is an angle bisector (with $D$ on $AB$).
a. If $AH \leq BM$, prove that the angle $ABC \leq 60^\circ$.
b. If $AH = BM = CD$, prove that $ABC$ is equilateral. | [
"Solution:\nAs usual let $a$, $b$, $c$ be the lengths of $BC$, $CA$, $AB$ respectively and let $A$, $B$, $C$ denote the angles $BAC$, $ABC$, $BCA$ respectively. We use trigonometry and try to express the quantities of interest in terms of $a$, $b$ and $C$.\n\na.\nSince $AH$ is the longest altitude, $BC$ must be the... | Soviet Union | 1st ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0e7l | Let $n$ and $m$ be natural numbers. We want to color each cell of a $n \times m$ table either white or black. For what $n$ and $m$ can this be done so that each cell has an odd number of neighbouring cells that are of the same color? Two cells are called neighbouring if they have a common side. | [
"We shall prove that this can be done if and only if at least one of the numbers $n$ and $m$ is even.\n\nFirst observe the case when one of the numbers is even. Without loss of generality, suppose $m$ is even. We divide a table of size $n \\times m$ ($n$ rows, $m$ columns) into quadrilaterals of size $1 \\times 2$ ... | Slovenia | Selection Examinations for the IMO 2013 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | It is possible if and only if at least one of the two dimensions is even. | |
0gpv | a. Find all prime triples $(p, q, r)$ such that $3 \nmid p+q+r$ and both $p+q+r$, $pq+qr+rp+3$ are perfect squares.
b. Is there any prime triple $(p, q, r)$ such that $3 \nmid p+q+r$ and both $p+q+r$, $pq+qr+rp+3$ are perfect squares. | [
"a. The answer: permutations of $(2, 3, 11)$.\nLet $p+q+r = x^2$ and $pq+qr+rp+3 = y^2$ where $x$, $y$ are integers. Let us show that one of the primes $p$, $q$, $r$ is $2$. If all primes $p$, $q$, $r$ are odd all possibilities up to permutations are: $(p, q, r) = (1, 1, 1)$, $(1, 1, 3)$, $(1, 3, 3)$, $(3, 3, 3)$ (... | Turkey | Team Selection Test for JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a) All permutations of (2, 3, 11). b) Yes; for example, (2, 11, 23). | |
090n | Let $ABCD$ be a cyclic quadrilateral with an incircle of radius $6$. Let the extensions of sides $AB$ and $DC$ beyond $B$ and $C$, respectively, meet at $P$, and let the extensions of sides $AD$ and $BC$ beyond $D$ and $C$, respectively, meet at $Q$. The inradii of triangle $PBC$ and $QCD$ are $5$ and $3$, respectively... | [
"$\\boxed{\\frac{15}{11}}$\n\nLet the inscribed circle of quadrilateral $ABCD$ be tangent to sides $AB$, $BC$, $CD$, and $DA$ at points $S$, $T$, $U$, and $V$, respectively.\nSince $AS$ and $AV$ are tangent segments from $A$ to this incircle, we have $AS = AV$. Similarly, $BS = BT$, $CT = CU$ and $DU = DV$ hold. Th... | Japan | The 35th Japanese Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocente... | English | proof and answer | 15/11 | |
0gns | Let $Q^+$ denote the set of positive rational numbers, and $\mathbb{Z}$ denote the set of all integers. Find all functions $f: Q^+ \to \mathbb{Z}$ that satisfy the conditions $f(1/x) = f(x)$ and $(x+1)f(x-1) = x f(x)$ for all $x \in Q^+$ such that $x > 1$. | [
"Substituting $x = 2$ in the second equation gives $3f(1) = 2f(2)$. In particular, $f(1)$ is even. It follows by induction that, for $n \\in \\mathbb{Z}^+$, $f(n) = (n+1)f(1)/2$.\n\nNow we show by induction on $p+q$ that for all $p, q \\in \\mathbb{Z}^+$ and $(p, q) = 1$, $f(p/q) = (p+q)f(1)/2$. If $p > q$, $p/q \\... | Turkey | Team Selection Test for IMO 2009 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All functions of the form f(p/q) = (p + q) m for coprime positive integers p, q, where m is a positive integer. | |
037o | Problem:
An infinite sequence of digits is obtained by writing all positive integers one after another in increasing order. Find the least positive integer $k$ such that among the first $k$ digits of the above sequence every two nonzero digits appear different number of times. | [
"Solution:\nDenote by $M_{n}$ the set of all digits of the numbers $1,2, \\ldots, n$. First we find the least positive integer $n=\\overline{a_{1} a_{2} \\ldots a_{t}}$ such that every two nonzero digits appear different number of times in $M_{n}$. By adding zeros on the left we may assume that all numbers $1,2, \\... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 56784 | |
08kr | Problem:
Prove that $\frac{a^{2}-b c}{2 a^{2}+b c}+\frac{b^{2}-c a}{2 b^{2}+c a}+\frac{c^{2}-a b}{2 c^{2}+a b} \leq 0$ for any real positive numbers $a, b, c$. | [
"Solution:\n\nThe inequality rewrites as $\\sum \\frac{2 a^{2}+b c-3 b c}{2 a^{2}+b c} \\leq 0$, or $3-3 \\sum \\frac{b c}{2 a^{2}+b c} \\leq 0$ in other words $\\sum \\frac{b c}{2 a^{2}+b c} \\geq 1$.\n\nUsing Cauchy-Schwarz inequality we have\n$$\n\\sum \\frac{b c}{2 a^{2}+b c}=\\sum \\frac{b^{2} c^{2}}{2 a^{2} b... | JBMO | 2007 Shortlist JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0c9w | Problem:
Pentru $n \geq 2$ numere reale nenule $a_{1}, a_{2}, \ldots, a_{n}$, nu neapărat distincte, definim matricea $A=\left(a_{ij}\right)_{1 \leq i, j \leq n} \in M_{n}(\mathbb{R})$ prin $a_{ij}=\max \{a_{i}, a_{j}\}$, $\forall i, j \in\{1,2, \ldots, n\}$. Arătaţi că $\operatorname{rang}(A)=\operatorname{card}\left... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0k35 | Problem:
Let $n$ be a positive integer. Let there be $P_{n}$ ways for Pretty Penny to make exactly $n$ dollars out of quarters, dimes, nickels, and pennies. Also, let there be $B_{n}$ ways for Beautiful Bill to make exactly $n$ dollars out of one dollar bills, quarters, dimes, and nickels. As $n$ goes to infinity, the... | [
"Solution:\n\nLet $d_{x}$ be the number ways to make exactly $x$ cents using only dimes and nickels. It is easy to see that when $x$ is a multiple of $5$,\n$$\nd_{x} = \\left\\lfloor \\frac{x}{10} \\right\\rfloor + 1\n$$\nNow, let $c_{x}$ be the number of ways to make exactly $x$ cents using only quarters, dimes an... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 20 | |
0d6n | Let $S = x + y + z$ where $x, y, z$ are three nonzero real numbers satisfying the following system of inequalities:
$$
\left\{\begin{array}{rl}
x y z & > 1 \\
x + y + z & > \frac{1}{x} + \frac{1}{y} + \frac{1}{z}
\end{array} .\right.
$$
Prove that $S$ can take on any real values when $x, y, z$ vary. | [
"First, if $z = \\frac{1}{y}$, then the given system becomes\n$$\n\\left\\{\\begin{array}{c}\nx \\cdot y \\cdot \\frac{1}{y} > 1 \\\\\nx + y + \\frac{1}{y} > \\frac{1}{x} + \\frac{1}{y} + y\n\\end{array} \\Leftrightarrow x > 1 .\\right.\n$$\nSo, all triples $\\left(x, y, \\frac{1}{y}\\right)$, with $x > 1$ and $y \... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
07wu | Consider three points $A$, $B$, $C$ on a circle $\Gamma$, with $\angle BAC > 90^\circ$. Let $d$ denote the line tangent to $\Gamma$ at $A$. Points $M$ and $N$ are chosen on $d$ such that $\angle MBA = \angle ABC$ and $\angle NCA = \angle ACB$. The circumcircles of triangles $ABM$ and $ACN$ intersect at $A$ and $T$. Pro... | [
"a. As $BTAM$ is cyclic, we have $\\angle BTM = \\angle BAM$. By the Alternate Segment Theorem, $\\angle BAM = \\angle ACB$. Since $CTAN$ is cyclic, $\\angle ACB = \\angle ACN = \\angle ATN$, hence $\\angle BTM = \\angle ATN$. Similarly $\\angle ATM = \\angle CTN$. Adding up we get $\\angle BTA = \\angle CTA$, i.e.... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04zw | The size of the angle $ABC$, expressed in degrees, in a right triangle $ABC$ is an integer. It is known that for some positive integer $n$, one can choose points $K_0 = A$, $K_2, \dots, K_{2n}$ on the hypotenuse $AB$ and points $K_1 = C$, $K_3, \dots, K_{2n+1} = B$ on the leg $CB$ in such a way that each triangle $K_{i... | [
"Let $\\angle ABC = \\alpha$ (Fig. 12). Then the base angle of the last isosceles triangle $K_{2n-1}K_{2n}K_{2n+1}$ is $\\alpha$. The base angle of the second last isosceles triangle $K_{2n-2}K_{2n-1}K_{2n}$ has the size $180^\\circ - (180^\\circ - 2\\alpha) = 2\\alpha$. The base angle of the next triangle before i... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 30, 18, 10, 6, 2 | |
0djs | Find all function $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for $x, y, z > 0$ pairwise distinct then
$$
f(x)^2 - f(y)f(z) \le f(xy)f(y)f(z)[f(yz) - f(zx)].
$$ | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All constant functions f(x) = c for any positive constant c. | |
0edq | Problem:
Vrednost izraza $10^{2016} - 10^{15}$ je naravno število. Koliko je vsota števk tega naravnega števila?
(A) 1
(B) 17
(C) 2001
(D) 18000
(E) 18009 | [
"Solution:\n\nNaravno število, ki predstavlja vrednost izraza $10^{2016} - 10^{15}$ ima 2016 števk, zadnjih 15 je enakih 0, ostale pa so enake 9. Vsota števk tega števila je torej $(2016 - 15) \\times 9 = 18009$."
] | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | E | |
0b8k | Each of the small squares of a $50 \times 50$ table is coloured in red or blue. Initially all squares are red. A step means changing the colour of all squares on a row or on a column.
a) Prove that there exists no sequence of steps, such that at the end there are exactly $2011$ blue squares.
b) Describe a sequence of... | [
"Without loss of generality, we may consider that the rows or columns to be modified in a sequence of steps are consecutive, and that each column or row is modified only once.\n\nSuppose then that the first $x$ rows and the first $50-y$ columns have been modified. One gets a $x \\times y$ rectangle and a $(50-x) \\... | Romania | Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | It is impossible to reach exactly 2011 blue squares. To obtain exactly 2010 blue squares, flip the first 44 rows and the first 5 columns. | |
0ey7 | Problem:
Find all integers $x$, $y$ satisfying $x^{2} + x = y^{4} + y^{3} + y^{2} + y$. | [
"Solution:\nThe only solutions are $x, y = -1, 1; 0, -1; 0, 0; -6, 2; 5, 2$.\n\n$(y^{2} + y / 2 - 1 / 2)(y^{2} + y / 2 + 1 / 2) = y^{4} + y^{3} + \\frac{1}{4}y^{2} - \\frac{1}{4} < y^{4} + y^{3} + y^{2} + y$ except for $-1 \\leq y \\leq -1/3$.\n\nAlso $(y^{2} + y / 2)(y^{2} + y / 2 + 1) = y^{4} + y^{3} + \\frac{5}{... | Soviet Union | 1st ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (-1, 1), (0, -1), (0, 0), (-6, 2), (5, 2) | |
0e7j | Let $D$ be the midpoint of the side $AB$, let $E$ be the point of intersection of the side $BC$ and the bisector of the angle $\angle BAC$, and let $F$ be the orthogonal projection of the point $E$ onto the side $AB$ of the triangle $ABC$. Suppose $\angle CDA = \angle ACB$ and $|CE| = |BF|$. Find the sizes of the angle... | [
"The equality $\\angle CDA = \\angle ACB$ implies that the triangles $ADC$ and $ACB$ are similar, so\n$$\n\\frac{|AC|}{|AB|} = \\frac{|AD|}{|AC|} = \\frac{|AB|}{2|AC|}\n$$\nwhich implies $|AB| = |AC|\\sqrt{2}$. Since $AE$ is the bisector of the angle $\\angle BAC$, we get\n\n$$\n\\frac{|BE|... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Angle at C is 90 degrees, angles at A and B are 45 degrees each. | |
083d | Problem:
Osservando le temperature registrate a Cesenatico negli ultimi mesi di dicembre e gennaio, Stefano ha notato una strana coincidenza: in tutti i giorni di questo periodo (esclusi il primo e l'ultimo) la temperatura minima è stata la somma della temperatura minima del giorno precedente e del giorno successivo.
... | [
"Solution:\n\nLa temperatura minima registrata il 25 dicembre è stata di -3 gradi.\n\nPer dimostrarlo, indichiamo con $x$ la temperatura minima registrata il primo dicembre e con $y$ la temperatura minima registrata il 2 dicembre. Usando la relazione osservata da Stefano, si può ricavare la temperatura minima in un... | Italy | Olimpiade Italiana di Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | -3 | |
0ezy | Problem:
An integer is written at each vertex of a regular $n$-gon. A move is to find four adjacent vertices with numbers $a$, $b$, $c$, $d$ (in that order), so that $(a - d)(b - c) < 0$, and then to interchange $b$ and $c$. Show that only finitely many moves are possible. For example, a possible sequence of moves is ... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
07s5 | A non-degenerate triangle is formed by three points that are not collinear. A triangle is called *nice* if it is non-degenerate and its vertices have integer coordinates $(x, y)$ such that $0 \le x \le 2018$ and $0 \le y \le 2018$. Let $V_n$ denote the number of nice isosceles triangles for which the coordinates of eac... | [
"We let $E_n$ be the number of nice isosceles triangles for which the coordinates $(x, y)$ of all vertices satisfy $y \\le n$ and for at least one vertex we have $y = n$. These triangles are part of the count for $V_n$ but not for $V_{n-1}$. Therefore, we obtain\n$$\nV_n = V_{n-1} + E_n \\quad \\text{for } n \\ge 1... | Ireland | Irish | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasi... | null | proof only | null | |
025b | Problem:
Dois dados são lançados. Qual é a probabilidade de o produto dos números obtidos nos dois dados ser divisível por $6$? | [
"Solution:\n\nNa tabela seguinte, marcamos com $\\times$ os produtos que são divisíveis por $6$.\n\n| $\\times$ | 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | | | | | | $\\times$ |\n| 2 | | | $\\times$ | | | $\\times$ |\n| 3 | | $\\times$ | | $\\times$ | | $\... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 5/12 | |
08zn | Find all positive integers $n$ such that $\frac{\phi(n)^{d(n)} + 1}{n}$ is an integer and $\frac{n^{\phi(n)} - 1}{d(n)^5}$ is not. Here, $\phi(n)$ denotes the number of integers in $\{1, \dots, n\}$ coprime to $n$, and $d(n)$ denotes the number of positive divisors of $n$. | [
"Let $\\prod_{i=1}^{m} x_i$ denote the product of $m$ real numbers $x_1, x_2, \\dots, x_m$. For any positive integer $N$, let $\\text{ord}_2 N$ denote the maximum nonnegative integer $\\ell$ such that $2^\\ell$ divides $N$.\n\n$n=1$ does not meet the assumption since $\\frac{1^{\\phi(1)} - 1}{d(1)^5} = 0$ is an int... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibi... | English | proof and answer | n = 2 | |
0it3 | Problem:
A positive integer $N$ is called stable if it is possible to split the set of all positive divisors of $N$ (including $1$ and $N$) into two subsets that have no elements in common, which have the same sum. For example, $6$ is stable, because $1+2+3=6$ but $10$ is not stable. Is $2^{2008} \cdot 2008$ stable? | [
"Solution:\nYes. In general, let $N$ be a number of the form $N=2^{k} p$, where $p$ is an odd prime less than $2^{k+1}$. We will show that one can form an expression, obtained by adding and subtracting together all the divisors of $N$, which is equal to zero. First note that\n$$\n2^{k} p-2^{k-1} p-\\cdots-p=p .\n$$... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Yes | |
03w2 | Find the maximum and minimum of the function
$$
y = \sqrt{x+27} + \sqrt{13-x} + \sqrt{x}.
$$ | [
"The domain of $y$ is $x \\in [0, 13]$. We have\n$$\n\\begin{aligned}\ny &= \\sqrt{x+27} + \\sqrt{13-x} + \\sqrt{x} \\\\\n&= \\sqrt{x+27} + \\sqrt{13 + 2\\sqrt{x(13-x)}} \\\\\n&\\geq \\sqrt{27} + \\sqrt{13} = 3\\sqrt{3} + \\sqrt{13}.\n\\end{aligned}\n$$\nThe equality holds when $x = 0$. Therefore, the minimum of $y... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | minimum = 3*sqrt(3) + sqrt(13), maximum = 11 | |
0kxc | Problem:
Svitlana writes the number $147$ on a blackboard. Then, at any point, if the number on the blackboard is $n$, she can perform one of the following three operations:
- if $n$ is even, she can replace $n$ with $\frac{n}{2}$;
- if $n$ is odd, she can replace $n$ with $\frac{n+255}{2}$; and
- if $n \geq 64$, she... | [
"Solution:\n\nThe answer is $163=\\sum_{i=0}^{4}\\binom{8}{i}$. This is because we can obtain any integer less than $2^{8}$ with less than or equal to $4$ ones in its binary representation. Note that $147=2^{7}+2^{4}+2^{1}+2^{0}$.\n\nWe work in binary. Firstly, no operation can increase the number of ones in $n$'s ... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 163 | |
0fo0 | Prove that the equation $2^x + 21^x = y^3$ has no solutions in positive integers and find all the solutions in nonnegative integers of the equation $2^x+21^y = z^2$. | [
"It is obvious that $x \\ge 1$, $y$ is odd and $21^x \\equiv 0 \\pmod 7$. Since $2^3 \\equiv 1 \\pmod 7$ then we have $2^{3n} \\equiv 1 \\pmod 7$, $2^{3n+1} \\equiv 2 \\pmod 7$, and $2^{3n+2} \\equiv 4 \\pmod 7$. On the other hand, $y^3 \\equiv 0 \\pmod 7$ or $y^3 \\equiv \\pm 1 \\pmod 7$. So, on account of the pre... | Spain | International Mathematical Arhimede Contest | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials ... | Spanish | proof and answer | No positive integer solutions to 2^x + 21^x = y^3. The nonnegative integer solutions to 2^x + 21^y = z^2 are (x, y, z) = (3, 0, 3) and (2, 1, 5). | |
02cd | Problem:
Considere um grupo de 15 pessoas. É possível que cada uma delas conheça exatamente:
(a) 4 pessoas do grupo?
(b) 3 pessoas do grupo?
(Admita que se $A$ conhece $B$ então $B$ conhece $A$.) | [
"Solution:\n(a) É possível. Representamos as 15 pessoas por pontos, conforme o diagrama ao lado. Um arco entre dois pontos significa que as duas pessoas representadas se conhecem. Como cada ponto está ligado a dois pontos à esquerda e a dois pontos à direita, saem quatro arcos de cada ponto, o que significa que é p... | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | a: yes; b: no | |
0g8d | 令 $R$ 表示實數所成的集合。定義集合 $S = \{1, -1\}$ 與函數 $sign : R \to S$ 如下：
$$
sign(x) = \begin{cases} 1 & \text{if } x \ge 0; \\ -1 & \text{if } x < 0. \end{cases}
$$
給定奇數 $n$。試問是否存在 $n^2 + n$ 個實數 $a_{ij}, b_i \in S$ ($1 \le i, j \le n$), 使得對於任意 $n$ 個數 $x_1, \dots, x_n \in S$, 利用下式
$$
y_i = \operatorname{sign}\left(\sum_{j=1}^{n} a... | [
"解：觀察小情況（如 $n=3$）容易猜想\n$$\na_{ij} = (-1)^{i+j},\\ b_i = 1\n$$\n符合題目要求。因此接下來困難處即是要證明其符合題目要求。注意到 $z$ 與 $x_1x_2\\cdots x_n$ 亦只有 $1, -1$ 兩種可能的值。$x_1x_2\\cdots x_n = 1$ 若且唯若 $x_j$ 中有偶數個 $-1$; 另一方面, 因為 $b_i = 1$, 此時 $z$ 為\n$$\nz = \\operatorname{sign}\\left(\\sum_{i=1}^{n} y_i\\right) \\qquad (1)\n$$\n因此 $z = 1$ 若且唯若 $\\... | Taiwan | 2014 IMO Training Camp Stage 3 Mock Competition (1) | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | Yes; for odd n one can choose a_ij = (-1)^(i+j) and b_i = 1 to achieve z = x_1 x_2 ... x_n for all inputs. | |
0im0 | Problem:
Let $ABC$ be a triangle with $AB = 7$, $BC = 9$, and $CA = 4$. Let $D$ be the point such that $AB \parallel CD$ and $CA \parallel BD$. Let $R$ be a point within triangle $BCD$. Lines $\ell$ and $m$ going through $R$ are parallel to $CA$ and $AB$ respectively. Line $\ell$ meets $AB$ and $BC$ at $P$ and $P'$, r... | [
"Solution:\n\nAnswer: $180$. Let $R'$ denote the intersection of the lines through $Q'$ and $P'$ parallel to $\\ell$ and $m$ respectively. Then $[RP'Q'] = [R'P'Q']$. Triangles $BPP'$, $R'P'Q'$, and $CQQ'$ lie in $ABC$ without overlap, so that on the one hand, $S \\leq ABC$. On the other, this bound is realizable by... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 180 |
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