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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
00yv | Problem:
The following construction is used for training astronauts: A circle $C_{2}$ of radius $2R$ rolls along the inside of another, fixed circle $C_{1}$ of radius $nR$, where $n$ is an integer greater than $2$. The astronaut is fastened to a third circle $C_{3}$ of radius $R$ which rolls along the inside of circle... | [
"Solution:\n\nConsider a circle $C_{4}$ with radius $R$ that rolls inside $C_{2}$ in such a way that the two circles always touch in the point opposite to the touching point of $C_{2}$ and $C_{3}$. Then the circles $C_{3}$ and $C_{4}$ follow each other and make the same number of revolutions, and so we will assume ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | n - 1 | |
06j4 | In $\triangle ABC$, $\angle A = 54^\circ$ and $\angle C = 24^\circ$. $P$ is a point on $AC$ such that $AP = BC$. Find $\angle CBP$. | [
"As shown in the figure, let $D$ be the point for which $ABPD$ is an isosceles trapezium with $AB \\parallel DP$. Let also $E$ be the point for which $APED$ is a parallelogram, and $F$ be the point such that $A$ and $F$ lie on different sides of $BC$ and for which $CBF$ is an equilateral triangle.\n\nNote that by o... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 6° | |
00pr | Let the sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ satisfy $a_0 = b_0 = 1$, $a_n = 9a_{n-1} - 2b_{n-1}$ and $b_n = 2a_{n-1} + 4b_{n-1}$ for all positive integers $n$. Let $c_n = a_n + b_n$ for all positive integers $n$. Prove that there do not exist positive integers $k, r, m$ such that $c_r^2 = c_k c_m$. | [
"**Solution 1.** Multiplying $b_n = 2a_{n-1} + 4b_{n-1}$ by $t \\in \\mathbb{R}$ and adding $a_n = 9a_{n-1} - 2b_{n-1}$ we have\n$$\na_n + t b_n = (9 + 2t)a_{n-1} + (-2 + 4t)b_{n-1}.\n$$\nSelecting $t$ such that $9 + 2t = (-2 + 4t)/t$, that is, $t = -1/2$ or $t = -2$; this becomes\n$$\na_n + t b_n = (9 + 2t)(a_{n-1... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Other"
] | English | proof only | null | |
0h2x | 1. In the plane 5 circles are given such that no three of them have a common point. Can it happen that they have exactly:
a) 12;
b) 24 different intersection points? | [
"a) Yes, it is possible. It is enough to take 4 circles, each pair of which intersects in two points, and a fifth circle such that it does not intersect the others.\n\nb) No, it is not possible. The first circle can intersect the others in at most 8 points, the second adds at most 6 intersection points, and so on. ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) Yes. b) No. | |
0a62 | Problem:
Find all triples $(a, b, n)$ of positive integers such that $a$ and $b$ are both divisors of $n$, and $a + b = \frac{n}{2}$. | [
"Solution:\nSince $a$ and $b$ are both factors of $n$, we can find positive integers $x$ and $y$ such that $a = \\frac{n}{x}$ and $b = \\frac{n}{y}$. Then $\\frac{n}{x} + \\frac{n}{y} = \\frac{n}{2}$ so\n$$\n\\frac{1}{x} + \\frac{1}{y} = \\frac{1}{2}.\n$$\nWithout loss of generality assume $a \\leqslant b$. So $x \... | New Zealand | NZMO Round One | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (k, 2k, 6k), (2k, k, 6k), and (k, k, 4k) for any positive integer k | |
0k4k | Problem:
Anders is solving a math problem, and he encounters the expression $\sqrt{15!}$. He attempts to simplify this radical by expressing it as $a \sqrt{b}$ where $a$ and $b$ are positive integers. The sum of all possible distinct values of $a b$ can be expressed in the form $q \cdot 15!$ for some rational number $... | [
"Solution:\n\nNote that $15! = 2^{11} \\cdot 3^{6} \\cdot 5^{3} \\cdot 7^{2} \\cdot 11^{1} \\cdot 13^{1}$. The possible $a$ are thus precisely the factors of $2^{5} \\cdot 3^{3} \\cdot 5^{1} \\cdot 7^{1} = 30240$. Since $\\frac{a b}{15!} = \\frac{a b}{a^{2} b} = \\frac{1}{a}$, we have\n$$\n\\begin{aligned}\nq & = \... | United States | HMMT November 2018 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | null | proof and answer | 4 | |
071h | Problem:
$\mathrm{ABCD}$ is a square. $\mathrm{P}, \mathrm{Q}$ are points on the sides $\mathrm{BC}, \mathrm{CD}$ respectively, distinct from the endpoints such that $\mathrm{BP}=\mathrm{CQ}$. $\mathrm{X}, \mathrm{Y}$ are points on $\mathrm{AP}, \mathrm{AQ}$ respectively. Show that there is a triangle with side length... | [
"Solution:\n\n\n\nWe have $DY < BY \\leq BX + XY$ (this is almost obvious, but to prove formally use the cosine formula for $BAY$ and $DAY$ and notice that $\\angle BAY > \\angle DAY$). Similarly, $BX < DX \\leq DY + YX$. So it remains to show that $XY < BX + DY$.\n\nTake $Q'$ on the extens... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07pn | Suppose $x$, $y$ are non-negative real numbers such that $x + y \le 1$. Prove that
$$
8xy \le 5x(1-x) + 5y(1-y),
$$
and determine the cases of equality. | [
"Let $z = x + y$. Then $0 \\le z \\le 1$, whence $z(1-z) \\ge 0$. Now\n$$\n\\begin{aligned}\n5x(1-x) + 5y(1-y) - 8xy &= 5(x+y) - 5(x^2+y^2) - 8xy \\\\\n&= 5z - 5(x^2+2xy+y^2) + 2xy \\\\\n&= 5z - 5z^2 + 2xy \\\\\n&= 5z(1-z) + 2xy \\\\\n&\\ge 0,\n\\end{aligned}\n$$\n\n\nSolution 2:\n\nFrom $x+y \\le 1$ we obtain $y \... | Ireland | Ireland | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Equality holds exactly at (x, y) = (0, 0), (1, 0), or (0, 1). | |
0gc5 | 設三角形 $ABC$ 的內心、重心、外心分別為點 $I, G, O$。令點 $X, Y, Z$ 分別落在射線 $BC, CA, AB$ 上, 並且滿足 $BX = CY = AZ$。設點 $F$ 為三角形 $XYZ$ 的重心。
試證: 直線 $FG$ 與 $IO$ 垂直。
設三角形 ABC 的內心、重心、外心分別為點 I, G, O. 令點 X, Y, Z 分別落在射線 BC, CA, AB 上, 並且滿足 $BX = CY = AZ$. 設點 F 為三角形 XYZ 的重心。
試證：直線 FG 與 IO 垂直。 | [
"先來證明一個引理。\n\n**Lemma 1.** 設 $I, O$ 分別為 $\\triangle ABC$ 的內心與外心。設點 $E, F$ 分別在射線 $CA, BA$ 上, 並滿足 $BF = BC = CE$。則 $EF \\perp IO$。\n\n**引理證明.** 設 $M$ 為 $\\triangle ABC$ 的外接圓上包含 $A$ 的 $BC$ 弧中點, 且令 $X, Y, Z$ 分別為三角形 $AEF, CIA, AIB$ 的外心。因為 $MB = MC$, $BF = CE$ 且 $\\angle MCE = \\angle MBF$, 所以 $\\triangle MBF$ 與 $\\trian... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05y5 | Problem:
Trouver tous les entiers $n \geqslant 1$ pour lesquels il existe un multiple de 222 dont la somme des carrés des chiffres est égale à $n$. | [
"Solution:\n\nPour tout entier $k$, on note $\\mathrm{S}_{2}(k)$ la somme des carrés des chiffres de $k$. Soit $\\mathcal{E}$ l'ensemble des entiers $n$ recherchés, c'est-à-dire\n$$\n\\mathcal{E}=\\left\\{\\mathrm{S}_{2}(222 k): k \\in \\mathbb{Z}_{\\geqslant 1}\\right\\} .\n$$\n\nOn remarque tout d'abord que $\\ma... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n = 3, n = 6, or n ≥ 8 | |
0377 | Problem:
Consider the excircles of a triangle $ABC$ tangent to the sides $AB$ and $AC$. Denote by $M$, $N$ and $P$ the tangent points of the first circle to the side $AB$ and the extensions of the sides $BC$ and $CA$ and by $S$, $Q$ and $R$ the tangent points of the second circle to the side $AC$ and the extensions of... | [
"Solution:\n\nWe shall use the standard notations. We first prove that $YX \\perp BC$. In $\\triangle BQR$ we have $\\angle BRY = 90^\\circ - \\frac{\\beta}{2}$ and since $\\angle MNB = \\frac{\\beta}{2}$ it follows that $NX \\perp RY$. Analogously $RX \\perp NY$. This means that $X$ is the orthocenter of $\\triang... | Bulgaria | Spring Mathematical Competition | [
"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 > Miscellaneous > Angle chasin... | null | proof only | null | |
09yq | Determine all positive integers $n \ge 2$ which have a positive divisor $m \mid n$ satisfying
$$n = d^3 + m^3,$$
where $d$ is the smallest divisor of $n$ which is greater than $1$. | [
"The smallest divisor of $n$ greater than $1$ is the smallest prime divisor of $n$, hence $d$ is prime. Moreover, we have $d \\mid n$, hence $d \\mid d^3 + m^3$, and $d \\mid m^3$. This yields that $m > 1$. On the other hand we have $m \\mid n$, hence $m \\mid d^3 + m^3$, and $m \\mid d^3$. Because $d$ is prime and... | Netherlands | IMO Team Selection Test 1 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 16, 72, 520 | |
08o2 | Problem:
Find all positive integers $n$ for which $1^{3} + 2^{3} + \cdots + 16^{3} + 17^{n}$ is a perfect square. | [
"Solution:\nWe have $1^{3} + 2^{3} + \\cdots + 16^{3} = (1 + 2 + \\cdots + 16)^{2} = 136^{2}$.\n\nSo, $1^{3} + 2^{3} + \\cdots + 16^{3} + 17^{n} = 136^{2} + 17^{n}$.\n\nWe want $136^{2} + 17^{n}$ to be a perfect square.\n\nLet $136^{2} + 17^{n} = k^{2}$ for some integer $k$.\n\nThen $k^{2} - 136^{2} = 17^{n}$, so $... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | n = 3 | |
05qy | Problem:
On considère un tableau de taille $2018 \times 2018$ dont chaque case contient un entier naturel non nul. Noémie modifie ces entiers à sa guise, en appliquant les opérations suivantes :
$\triangleright$ choisir une ligne puis multiplier par 2 tous les entiers contenus dans cette ligne;
$\triangleright$ choisi... | [
"Solution:\n\nOn va dire qu'une colonne est positive si elle ne contient que des entiers naturels non nuls, et est nulle si elle ne contient que des 0. Remarquons que, chaque fois que Noémie double les entiers contenus sur une ligne, les colonnes positives restent positives et les colonnes nulles restent nulles. On... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0cft | Let $a$ and $b$ be two integers. Prove that the number:
$$
\prod_{k=1}^{n-1} \left( a^2 + b^2 + 1 - 2a \cos \frac{2k\pi}{n} - 2b \sin \frac{2k\pi}{n} \right)
$$
is a positive integer and it can be written as the sum of two perfect squares. | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
03l3 | Problem:
A permutation of the integers $1901, 1902, \ldots, 2000$ is a sequence $a_{1}, a_{2}, \ldots, a_{100}$ in which each of those integers appears exactly once. Given such a permutation, we form the sequence of partial sums
$$
s_{1} = a_{1}, \quad s_{2} = a_{1} + a_{2}, s_{3} = a_{1} + a_{2} + a_{3}, \ldots, s_{1... | [
"Solution:\n\nLet $\\{1901, 1902, \\ldots, 2000\\} = R_{0} \\cup R_{1} \\cup R_{2}$ where each integer in $R_{i}$ is congruent to $i$ modulo $3$. We note that $|R_{0}| = |R_{1}| = 33$ and $|R_{2}| = 34$. Each permutation $S = (a_{1}, a_{2}, \\ldots, a_{100})$ can be uniquely specified by describing a sequence $S' =... | Canada | Canadian Mathematics Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 99! * 33! * 34! / 66! | |
047s | Let $n \ge 2$ be an integer. Two players, Alice and Bob, play the following game: Initially all $\binom{n}{2}$ edges of the complete graph $K_n$ are uncolored. They take turns coloring edges red, with Alice starting first. In each move, a player selects one or two uncolored edges to color red, with the constraint that ... | [
"We consider the graph $G$ formed by red edges. Initially, $G$ is an empty graph on $n$ vertices, and the condition requires that $G$ remains triangle-free throughout the game.\n\n**Step 1:** We show that if at Alice's turn, $G$ has 24 isolated vertices, then Alice can create a 5-cycle among these vertices in four ... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
04ml | There are $599$ yellow and $301$ blue balls. Can these balls be ordered in a sequence so that the number of balls between any two blue balls is different from $2$ and $5$? | [] | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | No | |
0998 | $\mathbb{N}$ натурал тоон олонлог. Аливаа $m, n \in \mathbb{N}$ тоонуудын хувьд
$$
(g(m) + n)(m + g(n))
$$
тоо бүтэн квадрат байх бүх $g: \mathbb{N} \to \mathbb{N}$ функцийг ол. | [
"$0 \\le c \\in \\mathbb{Z}$ бол $f(n) = n + c$ хэлбэрийн бүх функц бодлогын нөхцөлийг хангана:\n$$\n(f(m) + n)(f(n) + m) = (n + m + c)^2\n$$\n\nҮүнээс өөр функц байхгүйг харуулъя.\n\nЛемм. Ямар нэг анхны тоо $p$ ба натурал тоо $k, l$-ийн хувьд $p|f(k)-f(l)$ бол $p|k-l$ байна.\n\na) $p^2|f(k)-f(l)$ бол $f(l)=f(k)+p... | Mongolia | International Mathematical Olympiad 51 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order an... | Mongolian | proof and answer | g(n) = n + c for any fixed integer c ≥ 0 | |
08nv | Problem:
On a billiards table in the shape of a rectangle $ABCD$ with $AB = 2013$ and $AD = 1000$, a billiard ball is shot along the bisector of the angle $\angle BAD$. Assuming that the ball is reflected from the sides at the same angle it comes in, determine whether it will ever go to the corner $B$. | [
"Solution:\n\nThe ball travels a horizontal distance of $1000$ units between two bounces from the sides $AB$ and $CD$ as it always moves on a line making a $45^{\\circ}$ angle with the sides. Hence it is always at a distance of even number of units to the line $AD$ when it hits $AB$ or $CD$. Hence it can never hit ... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No, the ball will never reach corner B. | |
0hsm | Problem:
For the sequence of numbers $n_{1}, n_{2}, n_{3}, \ldots$, the relation $n_{i}=2 n_{i-1}+a$ holds for all $i>1$. If $n_{2}=5$ and $n_{8}=257$, what is $n_{5}$? | [
"Solution:\n\n33 ."
] | United States | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 33 | |
009m | Find the number of $2013$-digit numbers $d_1 d_2 \dots d_{2013}$ with odd digits $d_1, d_2, \dots, d_{2013}$ so that
$d_1 \cdot d_2 + d_3 \cdot d_4 + \dots + d_{1809} \cdot d_{1810} \equiv 1 \pmod{4}$, $d_{1810} \cdot d_{1811} + d_{1811} \cdot d_{1812} + \dots + d_{2012} \cdot d_{2013} \equiv 1 \pmod{4}$. | [
"We use the following observation: Any odd numbers $x_1, \\dots, x_k$ satisfy\n$$\nx_1 x_2 + x_2 x_3 + \\dots + x_{k-1} x_k + x_k x_1 \\equiv k \\pmod{4}. \\quad (*)\n$$\nNote that the sum in $(*)$ is cyclic, unlike the ones in the statement. To justify $(*)$ reduce the $x_i$ mod $4$; then they become $+1$'s or $-1... | Argentina | NATIONAL XXX OMA | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 6*5^2011 | |
0f6m | Problem:
Given a large sheet of squared paper, show that for $n > 12$ you can cut along the grid lines to get a rectangle of more than $n$ unit squares such that it is impossible to cut it along the grid lines to get a rectangle of $n$ unit squares from it. | [] | Soviet Union | 19th ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09a8 | Let $\gamma_1$ and $\gamma_2$ be externally tangent circles and $S$ be the point of tangency. Let $\omega$ be a circle that touches internally $\gamma_1$ and $\gamma_2$ at $P$ and $Q$ respectively. Denote by $R$ one of the intersection points of $\omega$ and the common tangent line of $\gamma_1$ and $\gamma_2$ that pas... | [
"Let $(AC) \\cap (BD) = M$. It suffices to prove that $M$ is on the radical axis of $\\gamma_1$ and $\\gamma_2$. It's equivalent to $MC \\cdot MA = MD \\cdot MB$. This is equivalent to $ACDB$ is a cyclic quadrilateral. Since $PT$ and $TQ$ are tangents to $\\omega$, $\\angle TPQ = \\angle TQP = \\alpha$. From this $... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b3x | Problem:
Let $\omega \neq -1$ be a complex root of $x^{3}+1=0$. What is the value of $1+2 \omega+3 \omega^{2}+4 \omega^{3}+5 \omega^{4}$?
(a) 3
(b) -4
(c) 5
(d) -6 | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | MCQ | d | |
0g3r | Problem:
Seien $n \geq 4$ und $k, d \geq 2$ natürliche Zahlen mit $k \cdot d \leq n$. Die $n$ Teilnehmenden der MathematikOlympiade sitzen um einen runden Tisch und warten auf Patrick. Als Patrick auftaucht, gefällt ihm die Situation gar nicht, da die Regeln des Social Distancing nicht eingehalten werden. Er wählt als... | [
"Solution:\n\nAntwort: $\\frac{n}{k}\\binom{n-k d+k-1}{k-1}$ oder dazu äquivalente Ausdrücke.\n\nLösung 1 (Surjektive $k$-zu-1 Abbildung):\nZuerst lasst uns die Personen am Tisch im Uhrzeigersinn von $1$ bis $n$ nummerieren, wobei die $1$ Person beliebig ausgewählt wurde. Wir werden nun zählen wie viele Kombination... | Switzerland | Zweite Runde 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | (n/k) * C(n - k*d + k - 1, k - 1) | |
0dee | Let $ABCD$ be a rectangle with $P$ lies on the segment $AC$. Denote $Q$ as a point on minor arc $PB$ of $(PAB)$ such that $QB = QC$. Denote $R$ as a point on minor arc $PD$ of $(PAD)$ such that $RC = RD$. The lines $CB$, $CD$ meet $(CQR)$ again at $M$, $N$ respectively. Prove that $BM = DN$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
08mk | Problem:
Let $ABC$ be an acute-angled triangle. A circle $\omega_{1}(O_{1}, R_{1})$ passes through points $B$ and $C$ and meets the sides $AB$ and $AC$ at points $D$ and $E$, respectively. Let $\omega_{2}(O_{2}, R_{2})$ be the circumcircle of the triangle $ADE$. Prove that $O_{1}O_{2}$ is equal to the circumradius of t... | [
"Solution:\nRecall that, in every triangle, the altitude and the diameter of the circumcircle drawn from the same vertex are isogonal. The proof offers no difficulty, being a simple angle chasing around the circumcircle of the triangle.\nLet $O$ be the circumcenter of the triangle $ABC$. From the above, one has $\\... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incente... | null | proof only | null | |
0itp | Problem:
Let $ABC$ be a triangle, and $I$ its incenter. Let the incircle of $ABC$ touch side $BC$ at $D$, and let lines $BI$ and $CI$ meet the circle with diameter $AI$ at points $P$ and $Q$, respectively. Given $BI = 6$, $CI = 5$, $DI = 3$, determine the value of $(DP / DQ)^2$. | [
"Solution:\n\nAnswer: $\\frac{75}{64}$ Same as Geometry Test problem 9."
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > An... | null | proof and answer | 75/64 | |
0jpb | Problem:
Let $p$ be a real number and $c \neq 0$ an integer such that
$$
c-0.1 < x^{p} \left( \frac{1-(1+x)^{10}}{1+(1+x)^{10}} \right) < c+0.1
$$
for all (positive) real numbers $x$ with $0 < x < 10^{-100}$. (The exact value $10^{-100}$ is not important. You could replace it with any "sufficiently small number".)
Find... | [
"Solution:\nAnswer: $(-1,-5)$ This is essentially a problem about limits, but phrased concretely in terms of \"small numbers\" (like 0.1 and $10^{-100}$).\n\nWe are essentially studying the rational function $f(x) := \\frac{1-(1+x)^{10}}{1+(1+x)^{10}} = \\frac{-10x + O(x^{2})}{2 + O(x)}$, where the \"big-O\" notati... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (-1, -5) | |
0l8y | The positive integer $m$ has a prime divisor greater than $\sqrt{2m} + 1$. Find the least positive integer $M$ such that there exists a set $T$ consisting of a finite number of distinct positive integers satisfying simultaneously the following conditions:
i) $m$ and $M$ are respectively the least and the greatest numb... | [] | Vietnam | CONTEST FOR THE SELECTION OF VIETNAMESE INTERNATIONAL MATHEMATICAL OLYMPIAD TEAM | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | m + p | |
0g29 | Problem:
Sei $n$ eine natürliche Zahl. Wir nennen eine Sequenz bestehend aus $3 n$ Buchstaben rumänisch, falls die Buchstaben $I, M$ und $O$ alle genau $n$ Mal vorkommen. Ein swap ist eine Vertauschung von zwei benachbarten Buchstaben. Zeige, dass für jede rumänische Sequenz $X$ eine rumänische Sequenz $Y$ existiert, ... | [
"Solution:\n\nWir nummerieren die Positionen der Buchstaben von 1 bis $3 n$ und berechnen für jede Sorte von Buchstaben $\\{I, M, O\\}$ die Differenz zwischen der Summe ihrer Positionen und der Summe der Positionen $1, \\ldots, n$ :\n$$\n\\begin{aligned}\nS_{I} & :=\\sum_{j=1}^{n} p_{j}(I)-\\sum_{j=1}^{n} j \\\\\nS... | Switzerland | SMO-Selektion | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
06gj | In a city the phone numbers should consist of exactly eight digits, and $0$ is not allowed as a digit in them (i.e., only $1, 2, 3, \ldots, 9$ may be used). It is required that every two phone numbers either be different in at least two places, or have digits separated by at least two units, in at least one of the eigh... | [
"At most $21523361$ phone numbers can be selected, and there is only $1$ way to select this amount of phone numbers.\n\nConsider the situation when the phone numbers have $n$ digits instead of $8$. We shall prove the following by induction: at most $\\frac{9^n + 1}{2}$ phone numbers can be selected, and there is a ... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Maximum number: 21523361; Number of ways: 1 | |
0abx | In a given rectangle $ABCD$ the length of $AB$ is two times bigger than the length of $BC$. On the side $CD$ a point $M$ is chosen such that the angle $AMD$ is equal to the angle $AMB$.
a) Determine the measure of the angle $AMD$.
b) If $\overline{DM} = 1$, determine the area of the rectangle $ABCD$? | [
"$AB$ and $CD$ are parallel so we get that $\\angle BAM = \\angle AMD$. Because $\\angle AMB = \\angle AMD$ we conclude that $\\angle BAM = \\angle AMB$. So we obtain that the triangle $AMB$ is isosceles and $\\overline{AB} = \\overline{MB}$.\n\nIn the right-angled triangle $BCM$ $BM$ is a hypotenuse and two times ... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠AMD = 75°, area = 14 + 8√3 | |
04ci | a) Let $x$ and $y$ be real numbers such that $x + y$, $x^2 + y^2$ and $x^4 + y^4$ are integers. Prove that the number $x^n + y^n$ is an integer for all $n \in \mathbb{N}$.
b) Find an example of real numbers $x$ and $y$ that are not integers, such that the numbers $x + y$, $x^2 + y^2$ and $x^4 + y^4$ are all integers.
... | [
"a) Assume that $a = x + y$, $b = x^2 + y^2$ and $c = x^4 + y^4$ are integers. Then the numbers $a^2 - b = 2xy$ and $b^2 - c = 2x^2 y^2$ are integers as well.\n\nSuppose that $xy$ is not an integer. Then $xy = \\frac{m}{2}$, where $m \\in \\mathbb{Z}$ is odd. But, then $2x^2 y^2 = \\frac{m^2}{2}$ is not an integer.... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | a) For all natural numbers n, x^n + y^n is an integer.
b) Example: x = sqrt(2), y = −sqrt(2).
c) Example: x = 1/sqrt(2), y = −1/sqrt(2). | |
00g6 | Let $k \geq 14$ be an integer, and let $p_{k}$ be the largest prime number which is strictly less than $k$. You may assume that $p_{k} \geq 3k/4$. Let $n$ be a composite integer. Prove:
a. if $n = 2p_{k}$, then $n$ does not divide $(n-k)!$;
b. if $n > 2p_{k}$, then $n$ divides $(n-k)!$. | [
"a.\nNote that $n-k = 2p_{k} - k < 2p_{k} - p_{k} = p_{k}$, so $p_{k} \\nmid (n-k)!$, so $2p_{k} \\nmid (n-k)!$.\n\nb.\nNote that $n > 2p_{k} \\geq 3k/2$ implies $k < 2n/3$, so $n-k > n/3$. So if we can find integers $a, b \\geq 3$ such that $n = ab$ and $a \\neq b$, then both $a$ and $b$ will appear separately in ... | Asia Pacific Mathematics Olympiad (APMO) | XV APMO | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0f2f | Problem:
Each vertex of a convex polyhedron has three edges. Each face is a cyclic polygon. Show that its vertices all lie on a sphere. | [] | Soviet Union | ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof only | null | |
04v8 | In the real numbers, solve the system of equations
$$
\sqrt{\sqrt{x} + 2} = y - 2,
$$
$$
\sqrt{\sqrt{y} + 2} = x - 2.
$$ | [
"Let $(x, y)$ be any solution of the given system. Since $\\sqrt{\\sqrt{x} + 2}$ is obviously positive, we have $y > 2$ by the first equation. Similarly, the second equation implies $x > 2$.\n\nNow we prove that the numbers $x$ and $y$ must be equal. We will use the observation that the function square root is incr... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (4, 4) | |
0fm9 | Problem:
Los puntos $A_{1}, A_{2}, \ldots, A_{2n+1}$ son los vértices de un polígono regular de $2n+1$ lados. Hallar el número de ternas $A_{i}, A_{j}, A_{k}$ tales que el triángulo $A_{i}A_{j}A_{k}$ es obtusángulo. | [
"Solution:\nAl ser $2n+1$ impar, no es posible construir triángulos rectángulos. Observemos que cualquier triángulo obtusángulo dejará el centro $O$ (su circuncentro) fuera de él. Si lo giramos en sentido directo o inverso alrededor de $O$ podemos conseguir que uno de sus vértices agudos esté en $A_{1}$. Los otros ... | Spain | Spain | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | (2n+1) * C(n, 2) | |
05b4 | Mari and Jüri play the following game on an infinite grid: They take turns with Mari starting, Mari colours one uncolored square red in each of her turns, Jüri colours one uncolored square blue in each of his turns. If the centers of any four squares of the same colour form the corners of some square, then this player ... | [
"We will first prove that Mari can win in 5 turns. We will denote squares in the grid by the coordinates of their centers, taking Mari's first square to be $(0,0)$. W.l.o.g., let Jüri's first turn be $(1,0)$, allowing him to also color squares with fractional coordinates in future turns (Fig. 7). Mari will color $(... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | Mari has a winning strategy; the least number of her turns needed to guarantee victory is 5. | |
0i9m | Problem:
Find a set $S$ of positive integers such that no two distinct subsets of $S$ have the same sum. Your score will be $\left\lfloor 20\left(2^{n} / r-2\right)\right\rfloor$, where $n$ is the number of elements in the set $S$, and $r$ is the largest element of $S$ (assuming, of course, that this number is nonnega... | [] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | An optimal choice is S = {1, 2, 4, ..., 2^{n-1}}, which yields 2^n / r = 2 and thus the maximal score of 0. | |
0edg | Problem:
Ploščina pravilnega šestkotnika je $96 \sqrt{3}~\mathrm{cm}^{2}$. Koliko je obseg tega šestkotnika?
(A) $48~\mathrm{cm}$
(B) $24~\mathrm{cm}$
(C) $96~\mathrm{cm}$
(D) $16~\mathrm{cm}$
(E) $20~\mathrm{cm}$ | [
"Solution:\n\nUporabimo formulo za ploščino pravilnega šestkotnika $\\frac{6 a^{2} \\sqrt{3}}{4} = 96 \\sqrt{3}$ ter izračunamo $a = 8~\\mathrm{cm}$. Obseg pravilnega šestkotnika je $6a = 48~\\mathrm{cm}$."
] | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | A | |
0ek7 | Problem:
Diagonali kvadratov $ABCD$ in $BEFG$ sta zaporedoma dolgi $8~\mathrm{cm}$ in $11~\mathrm{cm}$. Točka $P$ je presečišče diagonal kvadrata $BEFG$ (glej sliko). Koliko kvadratnih centimetrov je ploščina trikotnika $APF$?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
 | [
"Solution:\n\nStranici kvadratov sta dolgi $|AB| = \\frac{8}{\\sqrt{2}}~\\mathrm{cm}$ in $|BE| = \\frac{11}{\\sqrt{2}}~\\mathrm{cm}$. Ker je točka $P$ razpolovišče diagonale $BF$, je ploščina trikotnika $APF$ enaka ploščini trikotnika $ABP$, ta pa je enaka\n$$\n\\frac{|AB| \\cdot \\frac{|BG|}{2}}{2} = \\frac{\\frac... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
0lg8 | Problem:
Find all real $a$ for which there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x-f(y))=f(x)+a[y]$ for every real $x$ and $y$ ( $[y]$ denotes the integral part of $y$ ). | [
"Solution:\n\nFirst note that $a=0$ satisfies the problem condition (for example, the equation is satisfied by the function $f(x) \\equiv 0$).\n\nNow suppose $a \\neq 0$.\n\nLemma. $f(y)=f(z)$ if and only if $[y]=[z]$.\n\nSuppose $f(y)=f(z)$ for some $y, z$. Then the given equation implies $f(x)+a[y]=f(x-f(y))=f(x-... | Zhautykov Olympiad | IZhO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | { a = -n^2 : n ∈ ℤ } | |
036a | Problem:
Let $ABCD$ be a cyclic quadrilateral with circumcircle $k$. The rays $\overrightarrow{DA}$ and $\overrightarrow{CB}$ meet at point $N$ and the line $NT$ is tangent to $k$, $T \in k$. The diagonals $AC$ and $BD$ meet at the centroid $P$ of $\triangle NTD$. Find the ratio $NT : AP$. | [
"Solution:\n\nIt follows from the condition that $T$ lies on the $\\operatorname{arc} BC$. Let $M = NT \\cap DP$ be the midpoint of $NT$. Then we have\n$$\nMB \\cdot MD = MT^{2} = MN^{2}\n$$\nThus $MB : MN = MN : MD$ and it follows that $\\triangle NMB \\sim \\triangle DMN$. Hence\n$$\n\\angle MNB = \\angle MDN = \... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point... | null | proof and answer | 3 | |
0e9r | For a real number $x$ let $[x]$ denote the greatest integer not greater than $x$.
a. Prove that for all positive integers $a$, $b$ and $c$ we have
$$
\left\lfloor \frac{\left\lfloor \frac{c}{a} \right\rfloor}{b} \right\rfloor \le \left\lfloor \frac{c}{ab} \right\rfloor.
$$
b. Find an example showing that the above eq... | [
"a. The number $c$ can be written in the form $c = kab + r$, where $k$ is a non-negative integer and $r < ab$ is the remainder of $c$ when divided by $ab$. The number $r$ can be further written as $r = ma + n$, where $m$ is a non-negative integer and $n < a$ is the remainder of $r$ when divided by $a$. From this it... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Example: a = 2, b = 1/2, c = 1 | |
01tg | Solve the equation $p^3 - q^3 = pq^3 - 1$ in primes $p, q$. | [
"Answer: $q = 7$, $p = 19$.\n\nNote that $q \\le p$, otherwise LHS $< 0$, RHS $> 0$. We have\n$$\n\\begin{aligned}\np^3 - q^3 &= pq^3 - 1 \\Leftrightarrow p^3 + 1 = pq^3 + q^3 \\\\\n&\\Leftrightarrow (p+1)(p^2-p+1) = q^3(p+1) \\\\\n&\\Leftrightarrow p(p-1) = q^3 - 1 \\\\\n&\\Leftrightarrow p(p-1) = (q-1)(q^2+q+1).\... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p = 19, q = 7 | |
0e6a | Problem:
Naj bo $ABCDE$ tetiven petkotnik, v katerem je $|CD| = |DE|$. Diagonali $AD$ in $BE$ se sekata v točki $K$, diagonali $AC$ in $BD$ pa v točki $L$. Dokaži, da sta premici $KL$ in $EC$ vzporedni. | [
"Solution:\n\nPo izreku o obodnih kotih v krogu je $\\angle CED = \\angle CAD$ in $\\angle DCE = \\angle DBE$. Ker je $ECD$ enakokrak trikotnik z vrhom pri $D$, je $\\angle CED = \\angle DCE$.\n\nZaradi kolinearnosti točk $A$, $L$, $C$ ter zaradi kolinearnosti točk $A$, $K$, $D$ je tudi $\\angle LAK = \\angle LBK$ ... | Slovenia | 56. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hi9 | Oleksii placed positive integers in the cells of the chessboard of size $8 \times 8$. For each pair of adjacent cells, Fedir wrote down the product of the numbers in them and added all the obtained numbers. Oleksii wrote down the sum of the numbers in each pair of adjacent cells and multiplied all the obtained numbers.... | [
"Suppose that this could happen. Since Oleksii's product is odd, all the factors are odd, and therefore the sum of the numbers in any neighboring cells is odd. But then for any neighboring cells, the parity of the numbers in those cells is different, and therefore the product of those numbers is even, and therefore... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0lfw | Problem:
Does it exist? a polynomial $P(x)$ with integer coefficients, such that $P(1+\sqrt{3})=2+\sqrt{3}$ and $P(3+\sqrt{5})=3+\sqrt{5}$. | [
"Solution:\n\nThe answer is No. The polynomial $P(x)-x$ has root $3+\\sqrt{5}$, and since it has integer coefficients, it also has root $3-\\sqrt{5}$. The quadratic having these two roots is $x^{2}-6x+4$. Therefore $P(x)-x=(x^{2}-6x+4)Q(x)$ for some polynomial $Q(x)$ with integer coefficients. Plugging in $x=1+\\sq... | Zhautykov Olympiad | International Zhautykov Olympiad in Sciences | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Algebraic Number Theory > Algebraic numbers"
] | null | proof only | null | |
09wg | In the table below each of the three rows is a correct calculation (the symbol $\div$ denotes division). Also each of the three columns (read from top to bottom) is a correct calculation. However, the digits in the table have been replaced by letters. Different letters represent different digits and no digits are $0$.
... | [
"E) $9$"
] | Netherlands | First Round | [
"Discrete Mathematics > Logic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | E | |
0ac1 | Solve the equation $\frac{x-1}{x} + \frac{x-2}{x} + \dots + \frac{1}{x} = \frac{3x-20}{4}$ in $\mathbb{N}$. | [
"The given equation is equivalent to\n$$\n\\frac{1}{x}[1+2+3+\\dots+(x-3)+(x-2)+(x-1)] = \\frac{3x-20}{4}.\n$$\nBecause $1+2+3+\\dots+(x-3)+(x-2)+(x-1) = \\frac{(x-1)x}{2}$, we have\n$$\n\\frac{1}{x} \\frac{(x-1)x}{2} = \\frac{3x-20}{4}.\n$$\nNow $4x-4=6x-40$, $2x=36$, $x=18$."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 18 | |
0gvk | Find all positive integers $n$ satisfying the inequality
$$
\cos(\pi\sqrt{n^2+n}) \ge 0.
$$ | [
"Легко довести, що $n < \\sqrt{n^2 + n} < n + \\frac{1}{2}$ для всіх натуральних $n$. Отже, маємо, що $\\pi\\sqrt{n^2 + n} \\in (\\pi n; \\pi n + \\frac{\\pi}{2})$. Залишається тільки зауважити, що для $k \\in \\mathbb{N}$ на проміжках $(2\\pi k; \\frac{\\pi}{2} + 2\\pi k)$ косинус набуває додатні значення, а на пр... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | all even positive integers | |
0hn7 | Problem:
The circle $\omega$ passes through the vertices $A$ and $B$ of a unit square $A B C D$. It intersects $A D$ and $A C$ at $K$ and $M$ respectively. Find the length of the projection of $K M$ onto $A C$. | [
"Solution:\nLet $T$ be the point of intersection of $\\omega$ with $B C$. Then, as $\\angle A B T$ is a right angle, $A T$ is a diameter, and $\\angle A M T$ is also a right angle. Therefore the projections of $K M$ and $K T$ on $A C$ coincide. But the length of the projection of $K T$ is $\\frac{\\sqrt{2}}{2}$ bec... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(2)/2 | |
03sd | Let $n \ge 2$ be a positive integer and $a_1, a_2, \dots, a_n \in (0, 1)$. Find the maximum value of the sum
$$
\sum_{i=1}^{n} \sqrt[6]{a_i(1-a_{i+1})}
$$
where $a_{n+1} = a_1$. | [
"By the AM-GM Inequality, we deduce that\n$$\n\\begin{aligned}\n& \\sqrt[6]{a_i(1-a_{i+1})} \\\\\n&= 2^{\\frac{4}{6}} \\sqrt[6]{a_i(1-a_{i+1}) \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}} \\\\\n&\\le 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} \\cdot (a_i + 1 - a_{i+1} + 2) \\\\\n&= 2^{... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | n / sqrt[3]{2} | |
04fx | How many four-digit numbers whose digits are mutually different and from the set $\{0, 1, 2, 3, 4, 5\}$, and that are divisible by $5$, are there? (Hong Kong) | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics"
] | English | final answer only | 108 | |
0cj5 | Determine the smallest natural number $n \ge 3$ with the property that there exists a unique set of natural numbers $a_1 < a_2 < \dots < a_n \le 100$ which are directly proportional to $1, 2, \dots, n$. | [] | Romania | 75th NMO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 51 | |
08en | Problem:
Sia $ABC$ un triangolo e sia $I$ il centro della sua circonferenza inscritta. Sia $D$ il simmetrico di $I$ rispetto al lato $AB$, e sia $E$ il simmetrico di $I$ rispetto al lato $AC$.
Dimostrare che le circonferenze circoscritte ai triangoli $BID$ e $CIE$ sono tra di loro tangenti. | [
"Solution:\n\nSiano $X, Y, Z$ i punti di tangenza della circonferenza inscritta con i lati $AB, AC, BC$, rispettivamente. Dimostriamo che la retta $IZ$ è tangente ad entrambe le circonferenze.\n\nIndichiamo con $\\alpha, \\beta, \\gamma$ le ampiezze degli angoli in $A, B, C$, rispettivamente. Indichiamo con $T$ un ... | Italy | XXXVII Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0dsd | Given 7 distinct positive integers, prove that there is an infinite arithmetic progression of positive integers $a, a+d, a+2d, \dots$, with $a \le d$, that contains exactly 3 or 4 of the 7 given integers. | [
"Let the numbers be $X_1, X_2, \\dots, X_7$ in ascending order. Let $[a, d]$ denote the arithmetic progression (AP) with initial term $a$, common difference $d$ and $a \\le d$.\n\nWe first show that there is an AP $[a, d]$ that contains $X_i$, i.e. $X_i = a + k_i d$, $i = 1, \\dots, 5$. For example, we can take $a ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof only | null | |
052i | Find all positive real-valued solutions to
$$
\begin{cases} x - y + \frac{1}{z} = 2013, \\ y - z + \frac{1}{x} = 2013, \\ z - x + \frac{1}{y} = 2013. \end{cases}
$$ | [
"Suppose w.l.o.g. that $z \\ge x$ and $z \\ge y$. From the second equation $\\frac{1}{x} \\ge 2013$,\n\ntherefore $x \\le \\frac{1}{2013}$. From the third equation $\\frac{1}{y} \\le 2013$, due to which $y \\ge \\frac{1}{2013} \\ge x$.\nBut now from the first equation $\\frac{1}{z} \\ge 2013$, therefore $z \\le \\f... | Estonia | Open Contests | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = y = z = 1/2013 | |
014h | Problem:
Are there four distinct positive integers such that adding the product of any two of them to $2006$ yields a perfect square? | [
"Solution:\n\nSuppose there are such integers. Let us consider the situation modulo $4$. Then each square is $0$ or $1$. But $2006 \\equiv 2 \\pmod{4}$. So the product of each two supposed numbers must be $2 \\pmod{4}$ or $3 \\pmod{4}$. From this it follows that there are at least three odd numbers (because the pro... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No | |
0jun | Problem:
The corners of a fixed convex (but not necessarily regular) $n$-gon are labeled with distinct letters. If an observer stands at a point in the plane of the polygon, but outside the polygon, they see the letters in some order from left to right, and they spell a "word" (that is, a string of letters; it doesn't ... | [
"Solution:\nLet us call our original $n$ points $V_{1}, V_{2}, \\ldots, V_{n}$.\nIf $A, B$ are two points, then viewers on one side of line $\\overleftrightarrow{A B}$ see $A$ to the left of $B$, and viewers on the other side see $B$ to the left of $A$. Therefore, if we draw the $\\binom{n}{2}$ lines determined by ... | United States | BAMO-12 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof and answer | n(n-1)(n^2 - 5n + 18)/12 | |
0dyc | Problem:
Jaka si je zamislil trimestno število $x$, ki ima v zapisu različne neničelne števke. Nato je na list napisal vsa druga trimestna števila, ki jih je lahko zapisal s števkami števila $x$. Določi vsa možna števila $x$, če je vsota števil na listu enaka $3434$. | [
"Solution:\n\nOznačimo števke števila $x$ z $a, b$ in $c$, torej $x=\\overline{a b c}$. Vsa trimestna števila, sestavljena iz števk $a, b$ in $c$ so $\\overline{a b c}$, $\\overline{a c b}$, $\\overline{b a c}$, $\\overline{b c a}$, $\\overline{c a b}$, $\\overline{c b a}$, njihova vsota pa je $100(2a+2b+2c)+10(2a+... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 784 | |
0fn4 | $ABCD$ is a quadrilateral inscribed in a circle $\Gamma$. Lines $AB$, $DC$ meet in $E$; lines $BC$, $AD$ meet in $F$. Show that the circle of diameter $EF$ cuts the circle $\Gamma$ orthogonally. | [
"Let $O$ be the center of the circle $\\Gamma$ and $R$ its radius. Let $G$ be the other point of intersection of the circles $FDC$ and $BCE$. (See Figure 1).\n\n\n\n**Remark.** The following argument needs some minor changes if all the angles of $ABCD$ are acute, that is, if the center $O$ ... | Spain | Mediterranean Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
0lah | In the plane fix two points $A, B$ ($A \neq B$). Consider a point $C$ moving in the plane such that $\overrightarrow{ACB} = \alpha$, where $\alpha$ is a given angle ($0^\circ < \alpha < 180^\circ$). The circle with center in $I$, inscribed the triangle $ABC$ touches the sides $AB$, $BC$ and $CA$ in the points $D$, $E$ ... | [
"1/ Consider the triangle $AFM$, we have:\n$$\n\\begin{align*}\n\\widehat{AMF} &= 180^\\circ - (\\widehat{MFA} + \\widehat{FAM}) = 90^\\circ - (\\widehat{EFI} + \\widehat{FAM}) = 90^\\circ - (\\widehat{ECI} + \\widehat{FAM}) \\\\\n&= 90^\\circ - \\left(\\frac{C}{2} + \\frac{A}{2}\\right) = \\frac{B}{2} = \\widehat{... | Vietnam | Vijetnam 2009 | [
"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 | |
039q | In a tournament of beach volleyball with $n$ players and $n$ games any two players play in one and the same game at least one. Find the maximal value of $n$. | [
"The four players in a game form 6 pairs. Since any pair plays in at least one game, the number of all pairs $\\binom{n}{2} = \\frac{n(n-1)}{2}$ do not exceed 6 times the number of the games, i.e., $6n$. Hence $\\frac{n(n-1)}{2} \\le 6n$ which is equivalent to $n \\le 13$.\n\nFor $n = 13$ let $1, 2, \\dots, 13$ be ... | Bulgaria | Spring Mathematical Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 13 | |
0a48 | Problem:
Zij $\triangle ABC$ een scherphoekige driehoek met $|AB| > |AC|$, zij $\omega$ de omgeschreven cirkel van $\triangle ABC$ met middelpunt $O$. De hoogtelijn vanuit $A$ snijdt $BC$ in $D$ en snijdt $\omega$ een tweede keer in $P$. Definieer $H$ als het hoogtepunt van $\triangle ABC$ en zij $K$ het punt op het l... | [
"Solution:\n\nZij $O$ het middelpunt van $\\omega$. Wegens $|BD| = |KC|$ vallen de middelloodlijnen van $BC$ en $KD$ samen en in het bijzonder geldt dus dat $|OK| = |OD|$. Zij $T'$ de spiegeling van $K$ in $O$. Wegens Thales geldt dan dat $\\angle KDT' = 90^\\circ$, dus $T'$ ligt op $AD$.\n\nHet is een standaardpla... | Netherlands | IMO-selectietoets II | [
"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 > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > C... | null | proof only | null | |
06qz | Denote by $\mathbb{Q}^{+}$ the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ which satisfy the following equation for all $x, y \in \mathbb{Q}^{+}$:
$$
f\left(f(x)^{2} y\right)=x^{3} f(x y)
$$ | [
"By substituting $y=1$, we get\n$$\nf\\left(f(x)^{2}\\right)=x^{3} f(x)\n$$\nThen, whenever $f(x)=f(y)$, we have\n$$\nx^{3}=\\frac{f\\left(f(x)^{2}\\right)}{f(x)}=\\frac{f\\left(f(y)^{2}\\right)}{f(y)}=y^{3}\n$$\nwhich implies $x=y$, so the function $f$ is injective.\n\nNow replace $x$ by $x y$ in the previous equa... | IMO | 51st IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | f(x) = 1/x for all positive rational x | |
0auo | Problem:
The operations below can be applied on any expression of the form $a x^{2}+b x+c$.
(I) If $c \neq 0$, replace $a$ by $4 a-\frac{3}{c}$ and $c$ by $\frac{c}{4}$.
(II) If $a \neq 0$, replace $a$ by $-\frac{a}{2}$ and $c$ by $-2 c+\frac{3}{a}$.
(III$_{t}$) Replace $x$ by $x-t$, where $t$ is an integer. (Diffe... | [
"Solution:\n\nEach operation changes the discriminant $D$ of $a x^{2}+b x+c$ into\n\n(I) $D' = b^{2}-4\\left(4 a-\\frac{3}{c}\\right)\\left(\\frac{c}{4}\\right) = b^{2}-4 a c+3 = D+3$\n\n(II) $D' = b^{2}-4\\left(-\\frac{a}{2}\\right)\\left(-2 c+\\frac{3}{a}\\right) = b^{2}-4 a c+6 = D+6$\n\n(III) $D' = (b-2 a t)^{2... | Philippines | 18th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | a: Not possible; b: Not possible | |
0juk | Problem:
Find all triples of continuous functions $f, g, h$ from $\mathbb{R}$ to $\mathbb{R}$ such that $f(x+y) = g(x) + h(y)$ for all real numbers $x$ and $y$. | [
"Solution:\nThe answer is $f(x) = c x + a + b$, $g(x) = c x + a$, $h(x) = c x + b$, where $a$, $b$, $c$ are real numbers. Obviously these solutions work, so we wish to show they are the only ones.\n\nFirst, put $y = 0$ to get $f(x+0) = g(x) + h(0)$, so $g(x) = f(x) - h(0)$. Similarly, $h(y) = f(y) - g(0)$. Therefor... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All solutions are f(x) = c x + a + b, g(x) = c x + a, h(x) = c x + b for real constants a, b, c. | |
0a7j | Problem:
Peter has many squares of equal side. Some of the squares are black, some are white. Peter wants to assemble a big square, with side equal to $n$ sides of the small squares, so that the big square has no rectangle formed by the small squares such that all the squares in the vertices of the rectangle are of eq... | [
"Solution:\n\nWe show that Peter only can make a $4 \\times 4$ square. The construction is possible, if $n=4$:\n\n\n\nNow consider the case $n=5$. We may assume that at least $13$ of the $25$ squares are black. If five black squares are on one horizontal row, the remaining eight ones are di... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 6 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 4 | |
01z1 | Fix a positive integer $n$ and a finite graph with at least one edge; the end points of each edge are distinct, and any two vertices are joined by at most one edge. Vertices and edges are assigned (not necessarily distinct) numbers in the range $0$ through $n-1$, one number each. A vertex assignment and an edge assignm... | [
"An edge assignment compatible with a fixed vertex assignment will be referred to as a *solution* relative to that vertex assignment. A solution relative to the all-zero vertex assignment will be referred to as a *fundamental solution*; the all-zero edge assignment is a fundamental solution.\n\nFix a vertex assignm... | Belarus | SELECTION and TRAINING SESSION | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0j0f | Problem:
Given a permutation $\pi$ of the set $\{1,2, \ldots, 10\}$, define a rotated cycle as a set of three integers $i, j, k$ such that $i<j<k$ and $\pi(j)<\pi(k)<\pi(i)$. What is the total number of rotated cycles over all permutations $\pi$ of the set $\{1,2, \ldots, 10\}$? | [
"Solution:\n\nLet us consider a triple $(i, j, k)$ with $i<j<k$ and determine how many permutations rotate it. There are $\\binom{10}{3}$ choices for the values of $\\pi(i), \\pi(j), \\pi(k)$ and the choice of this set of three determines the values of $\\pi(i), \\pi(j), \\pi(k)$. The other $7$ values then have $7!... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 72576000 | |
056h | A plus or a minus sign is placed between every pair of consecutive digits in the sequence $0\ 1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9$.
a) Find the smallest positive odd number that cannot be equal to the value of the resulting expression.
b) Find the smallest positive even number that cannot be equal to the value of the resulting... | [
"Let the sum of the digits with a plus sign in front of them be $x$ and the absolute value of the sum of the digits with a minus sign in front of them be $y$. Then the value $v$ of the expression equals $x - y$. Also $x + y = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. Therefore $v = 45 - 2y$.\n\na) As $45 - 2y \\... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) 47; b) 2 | |
04yg | Let $a$, $b$, $c$, $d$ be nonnegative real numbers for which $a^2 + b^2 = ac + bd$ holds and $c$, $d$ are not both zero. Find maximum and minimum value of the expression
$$ \frac{ad + bc - cd}{c^2 + d^2}. $$ | [
"We will show that the maximum value is $\\frac{1}{2}$ and the minimum is $-\\frac{1}{2}$. For maximum, after some rearranging, we want to prove\n$$\n2(ad + bc - cd) \\leq c^2 + d^2,\n$$\nor\n$$\n2(ad + bc) \\leq (c + d)^2.\n$$\nAfter adding double the expression $ac + bd = a^2 + b^2$ to both sides of this inequali... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2025 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | maximum = 1/2, minimum = -1/2 | |
0bvx | Let $(a_n)_{n \ge 1}$ be the sequence given by $a_1 = 2$ and $a_{n+1} = 1 + \frac{1}{a_1 a_2 \dots a_n}$, $\forall n \ge 1$.
a) Find the general term of the sequence.
b) Put $S = \sum_{k=1}^{n} \frac{2}{(k^3 + k^2)a_{2k+1}}$ in a close form and show that $S < 1$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) a_n = 1 + 1/n. b) S = 1 − 1/(n+1)^2, hence S < 1. | |
0hfz | **Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that the following condition is fulfilled for arbitrary real numbers $x$ and $y$**
$$
f^2(x + y) = f^2(x) + 2f(xy) + f^2(y).
$$ | [
"Let us denote the equation given in the statement of a problem by $(*)$ and substitute $y = -x$ into it. We obtain\n$$\nf^2(x) + f^2(-x) + 2f(-x^2) = f^2(0) \\text{\\{ (**).\\}}\n$$\nSubstituting $x = x + y, y = -x$ into $(*)$, we get\n$$\nf^2(y) = f^2(x + y) + f^2(-x) + 2f(-x^2 - xy).\n$$\nAdding $(*)$ and $(**)$... | Ukraine | Problems from Ukrainian Authors | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0; f(x) = -2; f(x) = x; f(x) = x - 2 | |
00e9 | Let $n > d > 0$ be integers. Ana, Beto and Carlitos play *blind man's bluff* over an infinite grid. Initially, Ana and Carlitos are in cells at distance $n$, and there is a candy in a cell which is at distance $d$ from Carlitos. Carlitos is blindfolded and can only see his own cell, whereas Ana and Beto can see the who... | [
"Answer: $n = 2d + 2$.\n\nAbbreviate Ana, Beto, Carlitos and the candy by $A, B, C$ and $D$ respectively. We also write $d(X, Y)$ for the distance between $X$ and $Y$.\n\nWe claim that for $n \\le 2d + 1$, $C$ cannot ensure his victory. For the first movement there is no information. Assume without loss of generali... | Argentina | Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | n = 2d + 2 | |
09d4 | a, b, c нь тэгээс ялгаатай бүхэл тоонууд ба $a \neq c$. $a(c^2 + b^2) = c(a^2 + b^2)$ болно. Тэгвэл $a^2 + b^2 + c^2$ зохиомол тоо гэж батал. | [
"$a(c^2 + b^2) = c(a^2 + b^2) \\Leftrightarrow (a-c)(b^2 - ac) = 0$ ба $a \\neq c \\Rightarrow b^2 = ac$ болно. Иймд $a^2 + b^2 + c^2 = a^2 + ac + c^2 = a^2 + 2ac + c^2 - b^2 = (a + c - b)(a + c + b)$ болох ба $a^2 + b^2 + c^2 > 3$ юм.\n\nЭсрэгээс нь $a^2 + b^2 + c^2$ анхны тоо гэж үзье. Тэгвэл\n\n(1) $a + c - b = ... | Mongolia | ММО-48 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Mongolian | proof only | null | |
05ww | Problem:
Les nombres $0,1, \ldots, n$ sont écrits sur un tableau. À tout moment, Thanima peut effacer un nombre s'il est la moyenne arithmétique de deux nombres encore présents sur le tableau. L'objectif de Thanima est d'effacer le plus de nombre possible, et elle joue de façon optimale. En fonction de $n$, combien de... | [
"Solution:\n\nSi $n \\leqslant 2$, la réponse est $n$.\n\nSinon, à la fin du procédé, il existe toujours au moins deux nombres : on ne peut effacer ni $0$ ni $n$.\n\nSi $n=2^{k}$ est une puissance de $2$, il est possible de n'avoir que deux nombres sur le tableau à la fin : on commence par effacer tous les nombres ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | If n ≤ 2: n. If n is a power of 2 (and n ≥ 3): 2. Otherwise (n ≥ 3 and not a power of 2): 3. | |
0gg5 | 令 $\mathcal{X}$ 為正整數集 $\mathbb{N}$ 的所有非空子集 (不一定有限) 所組成的集合。試求所有函數 $f: \mathcal{X} \to \mathbb{R}^+$ 滿足以下性質：
(i) 若 $S \subseteq T$ 皆為 $\mathbb{N}$ 的非空子集，則 $f(T) \le f(S)$；
(ii) 對於所有 $S, T \in \mathcal{X}$,
$$
f(S) + f(T) \le f(S + T), \quad f(S)f(T) = f(S \cdot T),
$$
其中 $S + T = \{s + t \mid s \in S, t \in T\}, S \cdot ... | [
"$f(S) = (\\min S)^\\alpha, \\forall S \\in \\mathcal{X}$，其中 $\\alpha \\ge 1$。易知所有這類函數都是解。以下證明僅有這樣的函數滿足所有條件。\n\n因為 $\\{1\\} \\cdot \\{1\\} = \\{1\\}$, $\\mathbb{N} \\cdot \\mathbb{N} = \\mathbb{N}$, 所以\n$$\nf(\\{1\\})^2 = f(\\{1\\}), \\quad f(\\mathbb{N\\})^2 = f(\\mathbb{N\\}) \\implies f(\\{1\\}) = f(\\mathbb{N\\... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 國際競賽實作(二) | [
"Algebra > Algebraic Expressions > Functional Equations"
] | Chinese; English | proof and answer | All functions are f(S) = (min S)^alpha for all nonempty S, where alpha >= 1. | |
003v | En un pizarrón Daniel escribió, de arriba hacia abajo, una lista de números enteros positivos menores o iguales que $10$. Al lado de cada número de la lista de Daniel, Martín anotó la cantidad de veces que ese número figuraba en la lista de Daniel y así obtuvo una lista de la misma longitud.
Si se lee la lista de Mart... | [] | Argentina | XVII OLIMPIADA MATEMATICA DEL CONO SUR | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Functional equations"
] | Español | proof and answer | 39 | |
03gf | Problem:
Show that for any quadrilateral inscribed in a circle of radius $1$, the length of the shortest side is less than or equal to $\sqrt{2}$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0dvl | Problem:
Vsota in zmnožek 2 ulomkov sta celi števili. Eden izmed ulomkov ima imenovalec 2003. Dokaži, da sta tudi oba ulomka celi števili. | [
"Solution:\n\nNaj bosta ulomka $\\frac{r}{2003}$ in $\\frac{p}{q}$, pri čemer smemo predpostaviti, da je ulomek $\\frac{p}{q}$ okrajšan. Če zapišemo\n$$\n\\frac{r}{2003} + \\frac{p}{q} = a \\quad \\text{in} \\quad \\frac{r}{2003} \\cdot \\frac{p}{q} = b\n$$\nsta $a$ in $b$ celi števili. Izrazimo\n$$\nq \\cdot r + 2... | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0ife | Problem:
In a chess-playing club, some of the players take lessons from other players. It is possible (but not necessary) for two players both to take lessons from each other. It so happens that for any three distinct members of the club, $A$, $B$, and $C$, exactly one of the following three statements is true: $A$ ta... | [
"Solution:\n\nIf $P$, $Q$, $R$, $S$, and $T$ are any five distinct players, then consider all pairs $A$, $B \\in \\{P, Q, R, S, T\\}$ such that $A$ takes lessons from $B$. Each pair contributes to exactly three triples $(A, B, C)$ (one for each of the choices of $C$ distinct from $A$ and $B$); three triples $(C, A,... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
0ew3 | Problem:
Given any natural numbers $m$, $n$ and $k$. Prove that we can always find relatively prime natural numbers $r$ and $s$ such that $rm + sn$ is a multiple of $k$. | [
"Solution:\nLet $d = (m, n)$, the greatest common divisor of $m$ and $n$. Let $r = n/d$, $s = nhk - m/d$, where $h$ is any integer sufficiently large to ensure that $s > 0$. Now $rm + sn = mn/d + nnhk - mn/d = nnhk$, which is a multiple of $k$. If $e$ divides $r$, then it also divides $rdhk = nhk$. So if $e$ divide... | Soviet Union | 1st ASU | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
09rk | Problem:
Bepaal alle gehele getallen $n \geq 2$ waarvoor geldt dat
$$
i+j \equiv \binom{n}{i} + \binom{n}{j} \pmod{2}
$$
voor alle $i$ en $j$ met $0 \leq i \leq j \leq n$. | [
"Solution:\n\nWe laten eerst zien dat $n$ voldoet dan en slechts dan als $\\binom{n}{i} \\equiv i+1 \\pmod{2}$ voor alle $i$ met $0 \\leq i \\leq n$. Stel dat $\\binom{n}{i} \\equiv i+1 \\pmod{2}$ voor alle $i$, dan geldt $\\binom{n}{i} + \\binom{n}{j} \\equiv i+1 + j+1 \\equiv i+j \\pmod{2}$ voor alle $i$ en $j$. ... | Netherlands | MO-selectietoets | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 2^k - 2 for integers k ≥ 2 | |
05mp | Problem:
Soit $n$ un entier strictement positif tel que $n(n+2015)$ est le carré d'un entier.
a) Prouver que $n$ n'est pas un nombre premier.
b) Donner un exemple d'un tel entier $n$. | [
"Solution:\na) Supposons que $n$ est premier et qu'il existe un entier $m$ vérifiant $n(n+2015)=m^{2}$. Alors $n$ divise $m^{2}$, donc $n$ divise $m$. On peut donc écrire $m=n r$. Il vient $n(n+2015)=n^{2} r^{2}$, puis $n+2015=n r^{2}$. Par conséquent, $2015=n r^{2}-n=n\\left(r^{2}-1\\right)$ est divisible par $n$.... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n is not prime; an example is n = 1612. | |
0k8v | Problem:
Find all positive integers $n$ such that the unit segments of an $n \times n$ grid of unit squares can be partitioned into groups of three such that the segments of each group share a common vertex. | [
"Solution:\nAnswer: $n \\equiv 0,2(\\bmod 6)$\nWe first prove that $n \\equiv 0,2(\\bmod 6)$ is necessary for there to be such a partitioning. We break this down into proving that $n$ has to be even and that $n \\equiv 0,2(\\bmod 3)$.\nThe only way a segment on a side of the square can be part of such a T-shape is ... | United States | HMMT February 2019 Team Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n ≡ 0, 2 (mod 6) | |
073l | Let $ABC$ be a non-isosceles triangle, and let $\Gamma$ be its in-circle. Let $D$, $E$, $F$ be the points of contact of $\Gamma$ with the sides $BC$, $CA$, $AB$ respectively. Suppose $FD$, $DE$, $EF$ intersect $CA$, $AB$, $BC$ in $U$, $V$, $W$ respectively. If $L$, $M$, $N$ are respectively the mid-points of $DW$, $EU$... | [
"\n\nDraw a line through $N$ which is parallel to $ED$. This passes through the midpoints $P$ of $DF$ and $Q$ of $FE$. Similarly the line through $M$ parallel to $FD$ passes through the mid-point $Q$ of $FE$ and $R$ of $ED$; the line through $L$ parallel to $FE$ passes through the mid-point... | India | Indija TS 2008 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
04sf | Find the least real $m$ such that there exist reals $a$ and $b$ for which the inequality
$$
|x^2 + a x + b| \le m
$$
holds for all $x \in (0, 2)$. | [
"Notice that no negative number $m$ satisfies the problem evidently (absolute value is non-negative number).\n\nNow we interpret the problem geometrically. A graph of some function $y = x^2 + a x + b$ lies in a horizontal strip between lines $y = +m$ and $y = -m$ and in the interval $(0, 2)$. Our aim is to find the... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1/2 | |
024d | Problem:
Será verdade que $\frac{1}{4^{3}}+\frac{1}{5^{3}}+\frac{1}{6^{3}}<\frac{1}{12}$? | [
"Solution:\nSolução 1: Uma maneira de verificar essa desigualdade é comparando cada parcela desta soma, como segue. Comparando as frações $\\frac{1}{5}$, $\\frac{1}{6}$ e $\\frac{1}{3}$ com $\\frac{1}{4}$, obtemos\n$$\n\\begin{aligned}\n& \\frac{1}{5}<\\frac{1}{4}, \\text{ portanto, } \\frac{1}{5^{3}}=\\left(\\frac... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | Yes | |
0iep | Problem:
Let $S=\{s_{0}, \ldots, s_{n}\}$ be a finite set of integers, and define $S+k=\{s_{0}+k, \ldots, s_{n}+k\}$. We say that $S$ and $T$ are equivalent, written $S \sim T$, if $T=S+k$ for some $k$. Given a (possibly infinite) set of integers $A$, we say that $S$ tiles $A$ if $A$ can be partitioned into subsets eq... | [
"Solution:\n\nLet the difference between the smallest and largest element of $S$ be $a$. Then the set equivalent to $S$ that contains $b^{3}$ can only contain integers between $b^{3}-a$ and $b^{3}+a$, inclusive. But for sufficiently large $b$, $b^{3}$ is the only cube in this range, so $S$ can only have one element... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0hpk | Problem:
Define an $n$-staircase to be the union of all squares of an $n \times n$ grid lying on or below its main diagonal. How many ways are there to divide a 10-staircase into 10 rectangles, each having a side of length 1? (Reflections are not included.) | [
"Solution:\n\nA 10-staircase has 10 \"upper right corners\" $P$, each of which must be the upper right corner of some rectangle, and no two of which can belong to the same rectangle. It also has a single lower left corner $Q$ which must belong to the same rectangle as one of the ten points $P$. Since this rectangle... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 256 | |
05tm | Problem:
Soit $k$ et $n$ deux entiers naturels non nuls, tels que $k \leqslant 2^{n}$. Morgane a écrit, sur son cahier, l'ensemble des $n$-uplets formés de 0 et de 1 : il y en a $2^{n}$. On dit que deux $n$-uplets $\left(x_{1}, \ldots, x_{n}\right)$ et $\left(y_{1}, \ldots, y_{n}\right)$ sont voisins s'ils ont $n-1$ t... | [
"Solution:\n\nDans toute la suite, on assimile nos $n$-uplets à des vecteurs de $(\\mathbb{Z} / 2 \\mathbb{Z})^{n}$, c'est-à-dire des vecteurs dont les $n$ coordonnées sont des éléments de $\\mathbb{Z} / 2 \\mathbb{Z}$. Ainsi, on note :\n\n$\\triangleright$ $0$ le vecteur dont toutes les coordonnées sont nulles,\n\... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Exactly those k with 1 + ceil(n/2) ≤ k ≤ 2^n | |
06rt | In a $999 \times 999$ square table some cells are white and the remaining ones are red. Let $T$ be the number of triples $(C_{1}, C_{2}, C_{3})$ of cells, the first two in the same row and the last two in the same column, with $C_{1}$ and $C_{3}$ white and $C_{2}$ red. Find the maximum value $T$ can attain. | [
"We prove that in an $n \\times n$ square table there are at most $\\frac{4 n^{4}}{27}$ such triples.\nLet row $i$ and column $j$ contain $a_{i}$ and $b_{j}$ white cells respectively, and let $R$ be the set of red cells. For every red cell $(i, j)$ there are $a_{i} b_{j}$ admissible triples $(C_{1}, C_{2}, C_{3})$ ... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4*999^4/27 | |
01l2 | Prove that there exist infinitely many positive integers $n$ so that $n$ and the sum of its digits are perfect squares and
a) the decimal representation of $n$ contains at most one $0$;
b) the decimal representation of $n$ does not contain $0$. (T. Lasy) | [
"Note that the sum of the digits of this number is equal to\n$$\nS(X_n) = 1 \\cdot n + 5 \\cdot (n - 1) + 6 = 6n + 1.\n$$\nFurther,\n$$\n\\begin{aligned}\nX_n &= \\frac{10^{2n} - 1}{9} + \\frac{4 \\cdot (10^n - 1)}{9} + 1 \\\\\n&= \\frac{10^{2n} - 1 + 4 \\cdot (10^n - 1) + 9}{9} \\\\\n&= \\frac{10^{2n} + 4 \\cdot 1... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
02i4 | Problem:
Entre 1986 e 1989, a moeda do nosso país era o cruzado ($\mathrm{Cz\ \$}$). De lá para cá, tivemos o cruzado novo, o cruzeiro, o cruzeiro novo e, hoje, temos o real. Para comparar valores do tempo do cruzado e de hoje, os economistas calcularam que 1 real equivale a 2.750.000.000 cruzados.
Imagine que a moeda... | [
"Solution:\nO enunciado diz que 1 real $= 275 \\times 10^{7}$ cruzados. O salário de João é 640 reais, o que é equivalente a $640 \\times 275 \\times 10^{7} = 176.000 \\times 10^{7} = 176 \\times 10^{10}$ cruzados. O número de pilhas de 100 notas que se podem fazer com este número de notas de 1 cruzado é $\\frac{17... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
01e0 | Let $a, b, c, d$ be positive numbers such that $abcd = 1$. Prove the inequality
$$
\frac{1}{\sqrt{a + 2b + 3c + 10}} + \frac{1}{\sqrt{b + 2c + 3d + 10}} + \frac{1}{\sqrt{c + 2d + 3a + 10}} + \frac{1}{\sqrt{d + 2a + 3b + 10}} \le 1.
$$ | [
"Let $x, y, z, t$ be positive numbers such that $a = x^4, b = y^4, c = z^4, d = t^4$.\nBy AM-GM inequality $x^4 + y^4 + z^4 + 1 \\ge 4xyz$, $y^4 + z^4 + 1 + 1 \\ge 4yz$ and $z^4 + 1 + 1 + 1 \\ge 4z$.\nTherefore we have the following estimation for the first fraction\n$$\n\\frac{1}{\\sqrt{x^4 + 2y^4 + 3z^4 + 10}} \\... | Baltic Way | Baltic Way shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0cxf | 1) Prove that there is a triangle with side lengths
$$
\sqrt{a^{2}-a+1},\ \sqrt{a^{2}+a+1},\ \text{ and }\ \sqrt{4 a^{2}+3}.
$$
2) Prove that the area of this triangle does not depend on $a$. | [] | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null |
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