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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0g0t | Problem:
Soit $A$ un ensemble fini de nombres naturels. Une partition de $A$ en deux sous-ensembles disjoints non-vides $A_{1}$ et $A_{2}$ est appelée démoniaque si le plus petit multiple commun des éléments de $A_{1}$ est égal au plus grand diviseur commun des éléments de $A_{2}$. Quel est le plus petit nombre d'élém... | [
"Solution:\n\nSoit $A=A_{1} \\cup A_{2}$ une partition démoniaque et soient $a=\\max \\left(A_{1}\\right), b=\\min \\left(A_{2}\\right)$. On a alors que le plus petit multiple commun des éléments de $A_{1}$ est plus grand ou égal à $a$ et le plus grand diviseur commun des éléments de $A_{2}$ est plus petit ou égal ... | Switzerland | IMO-Selektion | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3025 | |
08f4 | Problem:
I $\pi$ cchi sono animali che vivono in famiglie di 1, 2 o 3 individui. Originariamente fu importata in Italia una famiglia di $3 \pi$ cchi. Una famiglia di $n \pi$ chi si riproduce crescendo di $2n-2$ nuovi individui, formando un totale di $3n-2$, e dividendosi in nuove famiglie (non necessariamente due fami... | [
"Solution:\n\nLa risposta è (B). Il numero di $\\pi$ cchi dopo 7 nidiate può assumere qualsiasi valore dispari fra 7 e $2^{9}-1$, e nessun altro. La risposta alla domanda è quindi $\\frac{2^{9}-1-7}{2}+1=2^{8}-3=253$, in quanto questa è la quantità di numeri dispari compresi fra 7 e $2^{9}-1=511$.\n\nGiustifichiamo... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | B | |
05in | Problem:
Déterminer tous les polynômes $X^{n}+a_{1} X^{n-1}+\cdots+a_{n-1} X+a_{n}$, non constants et à coefficients entiers, dont les racines sont exactement les nombres $a_{1}, \ldots, a_{n-1}, a_{n}$ (avec multiplicité). | [
"Solution:\n\nSoit $P(X) = X^n + a_1 X^{n-1} + \\cdots + a_{n-1} X + a_n$ un polynôme à coefficients entiers.\nAlors $P(X)$ est une solution du problème ssi\n$$\nP(X)=\\prod_{i=1}^{n}\\left(X-a_{i}\\right)\n$$\nSupposons tout d'abord que $a_{n}=0$. Soit alors $i$ minimal tel que $a_{n-i} \\neq 0$. On a donc $P(X)=X... | France | Olympiades Françaises de Mathématiques - Test de Sélection | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All solutions are precisely the polynomials of the form X^m, for m ≥ 1, and X^m · (X^2 + X − 2), and X^m · (X^3 + X^2 − X − 1), for m ≥ 0. | |
0ei8 | Problem:
Kateri od spodaj navedenih izrazov je enakovreden izrazu
$$
\frac{16 a^{-2} b^{-1} c^{-4}}{2^{-1} a^{-3} b^{-2} c^{-3}} \cdot \left(a^{0} + (a b)^{0}\right)^{-1}
$$
?
(A) $64 a b c^{-1}$
(B) $16 a b c^{-1}$
(C) $\frac{32}{a b c}$
(D) $32 a b c^{-1}$
(E) $16 a b c^{-7}$ | [
"Solution:\n\nNajprej uredimo ulomek:\n$$\n\\frac{16 a^{-2} b^{-1} c^{-4}}{2^{-1} a^{-3} b^{-2} c^{-3}}\n$$\nŠtevec: $16 a^{-2} b^{-1} c^{-4}$\nImenovalec: $2^{-1} a^{-3} b^{-2} c^{-3}$\n\nDelimo potence z enakimi osnovami:\n$$\n= 16 \\cdot 2^{1} \\cdot a^{-2 - (-3)} \\cdot b^{-1 - (-2)} \\cdot c^{-4 - (-3)}\n$$\nK... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | B | |
0ci7 | An $n$-type tiles triangle, where $n \geq 2$, is formed by the cells of a $(2n + 1) \times (2n + 1)$ array which are situated below its diagonals. For instance, a 3-type tiles triangle is the following:
Determine the maximal length of a sequence with pairwise distinct cells in an $n$-type tile... | [
"We alternately color (as a chessboard) the cells of an $n$-type tiles triangle, as below:\n\nWe have $b_n = 1 + 2 + \\ldots + n = \\frac{n(n+1)}{2}$ black cells and $a_n = 1 + 2 + \\ldots + (n-1) = \\frac{n(n-1)}{2}$ white cells. In any sequence of cells with the required propriety, their ... | Romania | 74th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | n^2 - n + 1 | |
0aan | In a school with $1000$ students, each one is friendly with exactly $4$ others. A group of three students $\{A, B, C\}$ is said to be a *friendly triplet* if any two students in the group are friends. Determine the maximal possible number of friendly triplets. | [] | North Macedonia | Fourth Memorial Mathematical Contest | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2000 | |
0hzf | Problem:
A rectangle has sides of length $\sin x$ and $\cos x$ for some $x$. What is the largest possible area of such a rectangle? | [
"Solution:\nWe wish to maximize $\\sin x \\cdot \\cos x = \\frac{1}{2} \\sin 2x$. But $\\sin 2x \\leq 1$, with equality holding for $x = \\pi/4$, so the maximum is $\\frac{1}{2}$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1/2 | |
0dez | What is the maximum number of $2 \times 3 \times 3$ bricks that can be fit inside an $8 \times 8 \times 9$ box? | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 31 | |
08iz | Problem:
The triangle $ABC$ has the semiperimeter $p$, the circumradius $R$, the inradius $r$ and $l_a, l_b, l_c$ are the lengths of internal bisectors, drawn from the vertices $A, B$ and $C$ respectively. Prove the inequality
$$
l_a l_b + l_b l_c + l_c l_a \leq p \sqrt{3 r^2 + 12 R r}.
$$ | [] | JBMO | The second selection test for IMO 2003 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Ine... | null | proof only | null | |
0i5f | Problem:
Points $P$ and $Q$ are $3$ units apart. A circle centered at $P$ with a radius of $\sqrt{3}$ units intersects a circle centered at $Q$ with a radius of $3$ units at points $A$ and $B$. Find the area of quadrilateral $APBQ$. | [
"Solution:\n\nThe area is twice the area of triangle $APQ$, which is isosceles with side lengths $3, 3, \\sqrt{3}$. By Pythagoras, the altitude to the base has length $\\sqrt{3^{2} - (\\sqrt{3} / 2)^{2}} = \\sqrt{33} / 2$, so the triangle has area $\\frac{\\sqrt{99}}{4}$. Double this to get $\\frac{3 \\sqrt{11}}{2}... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 3√11/2 | |
0jlf | Let $ABCD$ be a cyclic quadrilateral, and let $E, F, G$, and $H$ be the midpoints of $AB, BC, CD$, and $DA$, respectively. Let $W, X, Y$, and $Z$ be the orthocenters of triangles $AHE, BEF, CFG$, and $DGH$, respectively. Prove that quadrilaterals $ABCD$ and $WXYZ$ have the same area.
(This problem was suggested by Zhon... | [
"**Lemma.** Let $ABCD$ be any quadrilateral with $E, F, G, H$ the midpoints of sides $AB, BC, CD, DA$, respectively. Then $[ABCD] = 2[EFGH]$.\n*Proof.* Let $P$ be the intersection of diagonals $AC$ and $BD$. $HE$ is the midline of triangle $ABD$ so $HE$ bisects segment $AP$. It follows that $2[HPE] = [AHPE]$. Simil... | United States | IMO Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
0cza | Let $x$, $y$ be distinct positive integers. Prove that the number
$$
\frac{(x+y)^2}{x^3 + x y^2 - x^2 y - y^3}
$$
is not an integer. | [
"Denote by $E(x, y)$ the expression in the problem. We have\n$$\nE(x, y) = \\frac{(x+y)^2}{x(x^2 + y^2) - y(x^2 + y^2)} = \\frac{(x+y)^2}{(x-y)(x^2 + y^2)} \\tag{1}\n$$\nand we can assume that $x > y$.\n\nIf $x-y = 1$, then $E(x, y) \\in \\mathbb{Z}$ if and only if $x^2 + y^2 \\mid 2 x y$. This is not possible beca... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0jis | Problem:
Let $z$ be a complex number and $k$ a positive integer such that $z^{k}$ is a positive real number other than $1$. Let $f(n)$ denote the real part of the complex number $z^{n}$. Assume the parabola $p(n) = a n^{2} + b n + c$ intersects $f(n)$ four times, at $n = 0, 1, 2, 3$. Assuming the smallest possible val... | [
"Solution:\n\nAnswer: $\\boxed{\\frac{1}{3}}$\n\nLet $r = |z|$, $\\theta = \\arg z$, and $C = \\frac{\\Re z}{|z|} = \\cos \\theta = \\cos \\frac{2\\pi j}{k}$ for some $j$ with $\\gcd(j, k) = 1$. The condition of the four consecutive points lying on a parabola is equivalent to having the finite difference\n$$\nf(3) ... | United States | HMMT November 2014 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | 1/3 | |
05xw | Problem:
Anna et Baptiste jouent au jeu suivant. Au début du jeu sont placées devant eux 2022 cases blanches, numérotées de 1 à 2022. Puis, chacun son tour, en commençant par Anna, ils choisissent une case blanche et la colorient de la couleur de leur choix : soit en rouge, soit en bleu. La partie s'arrête au bout de ... | [
"Solution:\n\nPour plus de simplicité, on dira que deux entiers $a$ et $b$ sont amis si $a = b \\pm 3$. On dira que $a$ est joli si $a = b-3$ et si $a$ et $b$ sont coloriés de la même couleur, et que $a$ est laid si $a = b-3$ et si $a$ et $b$ sont coloriés de couleurs différentes : Anna souhaite donc maximiser le n... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1008 | |
0ea6 | Problem:
Naj bo $AB$ najdaljša stranica trikotnika $ABC$. Z $M$ in $N$ označimo taki točki na stranici $AB$, da velja $|AM| = |AC|$ ter $|BN| = |BC|$. Razpolovišči daljic $MC$ in $NC$ naj bosta $P$ in $R$, trikotniku $ABC$ včrtana krožnica pa naj se stranic $BC$ in $AC$ dotika v točkah $D$ in $E$. Dokaži, da so točke ... | [
"Solution:\n\n\n\nNaj bo $I$ središče trikotniku $ABC$ včrtane krožnice. Ker je $AMC$ enakokrak trikotnik z vrhom pri $A$, je $AP$ višina na osnovnico in hkrati simetrala kota $\\angle MAC$. Zato $I$ leži na premici $AP$. Podobno $I$ leži tudi na premici $BR$. Sledi $\\angle CPI = \\angle C... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06y5 | Let $p \neq q$ be coprime positive integers. Determine all infinite sequences $a_{1}, a_{2}, \ldots$ of positive integers such that the following conditions hold for all $n \geqslant 1$:
$$
\begin{aligned}
& \max \left(a_{n}, a_{n+1}, \ldots, a_{n+p}\right)-\min \left(a_{n}, a_{n+1}, \ldots, a_{n+p}\right)=p \quad \tex... | [
"Answer: The only such sequences are $a_{n}=n+C$, where $C$ is a nonnegative integer.\n\n## Common remarks.\n- Denote by $a_{[i, j]}$ the subsequence $a_{i}, a_{i+1}, \\ldots, a_{j}$.\n- Without loss of generality, in each solution we suppose $p<q$. It can be convenient to treat the case where $p=1$ separately.\n- ... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | a_n = n + C for some nonnegative integer C | |
0d1m | Let $ABCD$ be a convex quadrilateral such that $AB = AC = BD$. The lines $AC$ and $BD$ meet at point $O$, the circles $ABC$ and $ADO$ meet again at point $P$, and the lines $AP$ and $BC$ meet at point $Q$. Show that $\overline{COQ} = \overline{DOQ}$. | [
"We shall prove that the circles $ADO$ and $BCO$ meet again at the incenter $I$ of the triangle $ABO$, so the line $IO$ is the radical axis of the circles $ADO$ and $BCO$. Noticing further that the lines $AP$ and $BC$ are the radical axes of the pairs of circles $(ABC, ADO)$ and $(ABC, BCO)$, respectively, it follo... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dr4 | Let $ABC$ be a triangle with integral side lengths such that $\angle A = 3\angle B$. Find the minimum value of its perimeter. | [
"Let the sides be $a, b, c$. From the sine rule, we have\n$$\n\\begin{aligned}\n\\frac{a}{b} &= \\frac{\\sin 3B}{\\sin B} = 4\\cos^2 B - 1 \\\\\n\\frac{c}{b} &= \\frac{\\sin C}{\\sin B} = \\frac{\\sin 4B}{\\sin B} = 8\\cos^3 B - 4\\cos B\n\\end{aligned}\n$$\nThus\n$$\n2 \\cos B = \\frac{a^2 + c^2 - b^2}{ac} \\in \\... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Number Theory > Diophantine Equations > Pell's equations"
] | null | proof and answer | 21 | |
00hr | Find all positive integers $k<202$ for which there exists a positive integer $n$ such that
$$
\left\{\frac{n}{202}\right\}+\left\{\frac{2 n}{202}\right\}+\cdots+\left\{\frac{k n}{202}\right\}=\frac{k}{2}
$$
where $\{x\}$ denote the fractional part of $x$.
Note: $\{x\}$ denotes the real number $k$ with $0 \leq k<1$ such... | [
"Denote the equation in the problem statement as (*), and note that it is equivalent to the condition that the average of the remainders when dividing $n, 2 n, \\ldots, k n$ by $202$ is $101$. Since $\\left\\{\\frac{i n}{202}\\right\\}$ is invariant in each residue class modulo $202$ for each $1 \\leq i \\leq k$, i... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | [1, 100, 101, 201] | |
0g23 | Problem:
Sei $D$ ein Punkt im Inneren eines spitzwinkligen Dreiecks $A B C$, sodass $\angle B A D = \angle D B C$ und $\angle D A C = \angle B C D$. Sei $P$ ein Punkt auf dem Umkreis des Dreiecks $A D B$. Nehme an, $P$ befinde sich ausserhalb des Dreiecks $A B C$. Eine Gerade durch $P$ schneide den Strahl $B A$ in $X$... | [
"Solution:\n\nDurch die Aufgabenstellung ist gegeben, dass $\\angle C A D = \\angle D C B$. Mit dem Tangentenwinkelsatz erhält man, dass $B C$ eine Tangente zum Kreis $k_{1}$ durch $A D C$ und $k_{2}$ durch $A D B$ ist. Ebenso erhält man, dass $B C$ eine Tangente zu dem Umkreis von $A D B$ ist. Wir definieren $Z$ a... | Switzerland | SMO - Finalrunde | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k6l | Problem:
Kelvin the frog lives in a pond with an infinite number of lily pads, numbered $0, 1, 2, 3$, and so forth. Kelvin starts on lily pad $0$ and jumps from pad to pad in the following manner: when on lily pad $i$, he will jump to lily pad $(i+k)$ with probability $\frac{1}{2^{k}}$ for $k > 0$. What is the probabi... | [
"Solution:\n\nSuppose we combine all of the lily pads with numbers greater than $2019$ into one lily pad labeled $\\infty$. Also, let Kelvin stop once he reaches one of these lily pads.\n\nNow at every leap, Kelvin has an equal chance of landing on $2019$ as landing on $\\infty$. Furthermore, Kelvin is guaranteed t... | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1/2 | |
06t1 | Let $ABC$ be a triangle. The points $K$, $L$, and $M$ lie on the segments $BC$, $CA$, and $AB$, respectively, such that the lines $AK$, $BL$, and $CM$ intersect in a common point. Prove that it is possible to choose two of the triangles $ALM$, $BMK$, and $CKL$ whose inradii sum up to at least the inradius of the triang... | [
"Denote\n$$\na = \\frac{BK}{KC}, \\quad b = \\frac{CL}{LA}, \\quad c = \\frac{AM}{MB}\n$$\nBy Ceva's theorem, $abc = 1$, so we may, without loss of generality, assume that $a \\geqslant 1$. Then at least one of the numbers $b$ or $c$ is not greater than $1$. Therefore at least one of the pairs $(a, b)$, $(b, c)$ ha... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscell... | null | proof only | null | |
0adt | **Точката $M$ е средина на страната $BC$ на квадратот $ABCD$.** Точката $S$ е во внатрешноста на квадратот и е еднаков оддалечена од точките $A$, $D$ и $M$. Должината на страната на квадратот е $a = 40 \text{ cm}$. Да се пресмета периметарот и плоштината на четириаголникот $ABMS$.
 | [
"Точката $S$ е еднакво оддалечена од точките $A$ и $D$, па според тоа припаѓа на симетралата на страната $DA$. Ако $N$ е средина на страната на $AD$, тогаш точките $N$, $S$ и $M$ се колинеарни и лежат на симетралата на $DA$. Отсечката $SN$ е висина во триаголникот $ASD$. Бидејќи $MN = 40 \\text{ cm}$ и $SM = x$, до... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | Macedonian, English | proof and answer | Perimeter = 110 cm; Area = 650 cm^2 | |
0ajr | A segment $AB$ and its midpoint $K$ are given. An arbitrary point $C$, different from $K$, is chosen on the perpendicular to $AB$ through $K$. Let $N$ be the intersection of $AC$ and the line passing through $B$ and the midpoint of the segment $CK$. Let $U$ be the intersection of $AB$ with the line that passes through ... | [
"Let $M$ be the midpoint of the segment $CK$. From Menelaus' theorem for the triangle $AKC$ and the line $BN$ we have\n$$\n\\frac{\\overline{CN}}{\\overline{NA}} \\cdot \\frac{\\overline{AB}}{\\overline{BK}} \\cdot \\frac{\\overline{KM}}{\\overline{MC}} = 1.\n$$\nFrom this we get $\\overline{NA} = 2\\overline{NC}$,... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"
] | English | proof only | null | |
06hl | How many triples of integers $(a, b, c)$ satisfy $2 \le a \le b \le c$ and $abc = 2013 \times 2014$? | [
"The prime factorisation of $2013 \\times 2014$ is $2 \\times 3 \\times 11 \\times 19 \\times 53 \\times 61$. The problem amounts to counting the number of ways of distributing these 6 primes into 3 groups such that each group contains at least one prime. For example, if the groups are $\\{2, 3, 11\\}$, $\\{19, 53\... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 90 | |
0g80 | 試找出所有的函數 $f: \mathbb{R} \to \mathbb{R}$ 滿足下列條件:
(1) 對於所有的 $x, y \in \mathbb{R}$, $f(1 + xy) - f(x + y) = f(x)f(y)$, 和
(2) $f(-1) \neq 0$. | [
"唯一解是 $f(x) = x - 1, x \\in \\mathbb{R}$.\n我們令 $g(x) = f(x) + 1$ 且證明 $g(x) = x$ 對於所有的實數 $x$. 則題目的條件可寫成\n$$\ng(1+xy)-g(x+y) = (g(x)-1)(g(y)-1) \\quad \\text{對於所有 } x,y \\in \\mathbb{R} \\text{ 且 } g(-1) \\neq 1. \\quad (1)\n$$\n記 $C = g(-1) - 1 \\neq 0$. 設 (1) 式中的 $y = -1$, 可得\n$$\ng(1-x) - g(x-1) = C(g(x) - 1). \\q... | Taiwan | 二〇一三數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | f(x) = x - 1 | |
0hcu | For any natural number $n \ge 3$ find integers $a_1 < a_2 < \dots < a_n$ such that the following equality holds:
$$
\frac{a_1}{a_1} + \frac{a_1}{a_2} + \frac{a_1}{a_3} + \dots + \frac{a_1}{a_n} = \frac{a_2}{a_1} + \frac{a_2}{a_2} + \frac{a_2}{a_3} + \dots + \frac{a_2}{a_n}. \qquad (\text{BogdanRublyov})
$$ | [
"Let us first find rational numbers satisfying the equality, and then multiply them by the least common multiplier of the denominators in order to obtain integers.\nOne can fix $b_1 = -2^{n-1}$, $b_2 = -2^{n-2}$, ..., $b_{n-1} = -2^1$. Let us rewrite the equation as follows:\n$$\n\\frac{b_1}{b_1} + \\frac{b_1}{b_2}... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | One valid construction is: for k from 1 to n−1, set a_k = −2^{n−k}(2^{n−1}−1), and set a_n = 2^{n−1}. | |
0ae1 | На табла се напишани броевите $1, 2, \ldots, 2009$. Се бришат неколку од нив и наместо нив на таблата се запишува остатокот на збирот на избришаните броеви при делење со $13$. После одреден број повторувања на оваа постапка на таблата останале само три броја од кои двата се $99$ и $999$. Да се определи третиот број кој... | [
"Нека третиот број е $x$. Јасно после секој чекор остатокот при делење на збирот на броевите на таблата со $13$ не се менува.\n\nБидејќи $1+2+3+\\ldots+2009 = \\frac{2009 \\cdot 2010}{2} = 1005 \\cdot 2009$, т.е. има остаток $2$ при делење со $13$.\n\nПонатаму $99+999+x$ треба да има остаток $2$ при делење со $13$.... | North Macedonia | ЈММО 2009 година | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | Macedonian, English | proof and answer | 9 | |
0hua | Problem:
Prove that every positive real number $x$ satisfies
$$
\sqrt{x^{2}-x+\frac{1}{2}} \geq \frac{1}{x+\frac{1}{x}}
$$ | [
"Solution:\nBy the QM-AM inequality,\n$$\n\\sqrt{x^{2}-x+\\frac{1}{2}} = \\sqrt{\\frac{x^{2} + (1-x)^{2}}{2}} \\geq \\frac{x + (1-x)}{2} = \\frac{1}{2}.\n$$\nNow, by AM-GM (since we know $x$ is positive),\n$$\n\\frac{x + \\frac{1}{x}}{2} \\geq \\sqrt{x \\cdot \\frac{1}{x}} \\Longrightarrow \\frac{1}{2} \\geq \\frac... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0kmq | Problem:
Suppose that $x$, $y$, and $z$ are complex numbers of equal magnitude that satisfy
$$
x + y + z = -\frac{\sqrt{3}}{2} - i \sqrt{5}
$$
and
$$
xyz = \sqrt{3} + i \sqrt{5}.
$$
If $x = x_1 + i x_2$, $y = y_1 + i y_2$, and $z = z_1 + i z_2$ for real $x_1$, $x_2$, $y_1$, $y_2$, $z_1$, and $z_2$, then
$$
\left(x_1 x_... | [
"Solution:\nFrom the conditions, it is clear that $a$, $b$, $c$ all have magnitude $\\sqrt{2}$. Conjugating the first equation gives $2\\left(\\frac{ab + bc + ca}{abc}\\right) = -\\frac{\\sqrt{3}}{2} + i \\sqrt{5}$, which means $ab + bc + ca = \\left(-\\frac{\\sqrt{3}}{4} + i \\frac{\\sqrt{5}}{2}\\right)(\\sqrt{3} ... | United States | HMMT Spring 2021 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | final answer only | 1516 | |
043z | The range of $y = \sin^2 x + \sqrt{3} \sin x \cos x$ when $\frac{\pi}{4} \le x \le \frac{\pi}{2}$ is ______. | [
"$$\n\\begin{align*}\ny &= \\sin^2 x + \\sqrt{3} \\sin x \\cos x \\\\\n&= \\frac{1 - \\cos 2x}{2} + \\frac{\\sqrt{3}}{2} \\sin 2x \\\\\n&= \\sin \\left( 2x - \\frac{\\pi}{6} \\right) + \\frac{1}{2}.\n\\end{align*}\n$$\nWhen $\\frac{\\pi}{4} \\le x \\le \\frac{\\pi}{2}$, there is $\\frac{\\pi}{3} \\le 2x - \\frac{\\... | China | China Mathematical Competition | [
"Precalculus > Trigonometric functions"
] | null | final answer only | [1, 3/2] | |
0dlj | Let $ABC$ be a triangle inscribed in circle $(O)$ with $\angle A = 45^\circ$. Two rays $BO$, $CO$ intersect $AC$, $AB$ at $E$, $F$ respectively. The circumcircles of triangles $BOC$ and $EOF$ intersect at $K$. Let $J$ be circumcenter of triangle $AEF$. Prove that $JK$ passes through the orthocenter of triangle $ABC$. | [
"We have\n$$\n\\angle BOC + \\angle EOF = 90^\\circ + 90^\\circ = 180^\\circ\n$$\nso according to the familiar property of isogonal conjugates in quadrilaterals, we see that there exists a point $O'$ which is the isogonal conjugate of $O$ in $BFEC$. On the other hand, $BH$, $BO$ and $CH$, $CO$ are isogonal pairs in... | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilate... | null | proof only | null | |
09c8 | $k \ne l \in \mathbb{N}$ ба $0 \ne a_i \in \mathbb{C},\ i = 1, n$ (зарим нь адил байж болно) тоонууд өгчээ. $\{a_1^k, \dots, a_n^k\} = \{a_1^l, \dots, a_n^l\}$ бол $a_i$ бүр нэгжийн язгуур гэж батал. | [
"Бодлогын нөхцөлөөс\n$$\n\\exists f \\text{ хнух (биекц)} : \\{1, 2, \\dots, n\\} \\rightleftharpoons [n] \\to [n], \\\\\n\\mathbb{Z}f(j) = m \\text{ бол } a_j^k = a_m^l\\ (1) \\text{ гэж гарна.}\n$$\n$$\n1, f(1), f^{(2)}(1) \\Leftrightarrow f(f(1)), \\dots\n$$\nдарааллыг авч үзье. $f$ төгсгөлөг олонлог дээрх биекц... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | Mongolian | proof only | null | |
086p | Problem:
Sull'isola che non c'è ci sono 2008 abitanti, divisi in tre clan: i furfanti che mentono sempre, i cavalieri che non mentono mai, i paggi che mentono un giorno sì e uno no. Lorenza, in visita per due giorni, li incontra tutti il primo giorno. Il primo dice: "c'è esattamente un furfante sull'isola"; il secondo... | [
"Solution:\n\nLa risposta è (B). Lorenza nota che sia il primo giorno che il secondo tutti e 2008 gli abitanti fanno affermazioni contrastanti, segno che almeno 2007 di loro stanno mentendo. Ovvero al più uno dice la verità il primo giorno, e al più uno il secondo. Da ciò si deduce che almeno 2006 hanno mentito ent... | Italy | Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Logic"
] | null | MCQ | B | |
0i6t | Problem:
Our next object up for bid is an arithmetic progression of primes. For example, the primes $3$, $5$, and $7$ form an arithmetic progression of length $3$. What is the largest possible length of an arithmetic progression formed of positive primes less than $1,000,000$? Be prepared to justify your answer. | [
"Solution:\n\n$12$. We can get $12$ with $110437$ and difference $13860$."
] | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | final answer only | 12 | |
01x7 | Two collections $(x_1, x_2, \dots, x_n)$ and $(y_1, y_2, \dots, y_n)$ of real numbers satisfy the inequality $x_i \ge y_j$ for all indices $i$ and $j$. Let $P = \max_{1 \le i \le n}(x_i - y_i)$ and $G = \max_{1 \le i \le n} x_i - \min_{1 \le j \le n} y_j$.
Prove the inequality $P \le G \le nP$. | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0j9m | Problem:
Let $\alpha$ and $\beta$ be reals. Find the least possible value of
$$
(2 \cos \alpha + 5 \sin \beta - 8)^2 + (2 \sin \alpha + 5 \cos \beta - 15)^2.
$$ | [
"Solution:\nLet the vector $\\vec{v} = (2 \\cos \\alpha, 2 \\sin \\alpha)$ and $\\vec{w} = (5 \\sin \\beta, 5 \\cos \\beta)$. The locus of ends of vectors expressible in the form $\\vec{v} + \\vec{w}$ are the points which are five units away from a point on the circle of radius two about the origin. The expression ... | United States | HMMT November 2012 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 100 | |
04pw | Let $ABC$ be an acute-angled triangle such that $|AB| > |AC|$, and let $D$, $E$ and $F$ be the feet of its altitudes from the vertices $A$, $B$ and $C$, respectively. The lines $EF$ and $BC$ intersect at the point $P$. The line passing through $D$ parallel to $EF$ intersects the lines $AC$ and $AB$ at the points $Q$ an... | [
"The triangle $ABC$ is acute-angled, so the points $D$, $E$ and $F$ lie on the sides $\\overline{BC}$, $\\overline{CA}$ and $\\overline{AB}$, respectively. Since $|AB| > |AC|$, we have $|BD| > |CD|$.\n\nFurthermore, from $\\angle ACB < \\angle ABC$ it follows that $C$ lies between $B$ and $P$. Also, $C$ lies betwee... | Croatia | Croatian Mathematical Society Competitions | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
07z3 | Problem:
Una sbarra è costituita da un cilindro interno di acciaio di raggio $1~\mathrm{cm}$ rivestito da uno strato omogeneo di gomma leggera di spessore costante. Sapendo che invertendo i materiali il peso della sbarra non cambierebbe, determinare lo spessore del rivestimento.
(A) $1~\mathrm{cm}$
(B) $\sqrt{2}~\mat... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Circles",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0ahh | Find all positive integers $x$, $y$, $z$ and $t$ such that
$$
2^x \cdot 3^y + 5^z = 7^t.
$$ | [
"Reducing modulo $3$ we get $5^z \\equiv 1$, therefore $z$ is even, $z = 2c$, $c \\in \\mathbb{N}$.\nNext we prove that $t$ is even.\nObviously, $t \\ge 2$. Let us suppose that $t$ is odd, say $t = 2d+1$, $d \\in \\mathbb{N}$. The equation becomes $2^x \\cdot 3^y + 25^c = 7 \\cdot 49^d$. If $x \\ge 2$, reducing mod... | North Macedonia | XVI-th Junior Balkan Mathematical Olympiad | [
"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",
"Number Theory > Residues and Primitive Roots >... | English | proof and answer | (x, y, z, t) = (3, 1, 2, 2) | |
0a8f | Problem:
Let $A$ be a finite collection of squares in the coordinate plane such that the vertices of all squares that belong to $A$ are $(m, n)$, $(m+1, n)$, $(m, n+1)$, and $(m+1, n+1)$ for some integers $m$ and $n$. Show that there exists a subcollection $B$ of $A$ such that $B$ contains at least $25\%$ of the squar... | [
"Solution:\n\nDivide the plane into two sets by painting the strips of squares parallel to the $y$ axis alternately red and green. Denote the sets of red and green squares by $R$ and $G$, respectively. Of the sets $A \\cap R$ and $A \\cap G$ at least one contains at least one half of the squares in $A$. Denote this... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 15 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0hbj | The number $2019$ is written on the board. Katia and Mykola are playing the following game: one by one (starting with Katia) they choose any divisor $d$ of the number $N$ written on the board and change the number on the board $N$ to the number $N - (2d - 1)$, if it is positive integer. Whoever writes number $1$ loses.... | [
"Firstly, we will show that the number on the board decreases with every turn. Clearly, it will be smaller and integer. It will be positive, since: $N = dD \\Rightarrow M = N - (2d - 1) = dD - 2d + 1 = d(D - 2) + 1 \\ge 1$, since if $d < N$ then $D \\ge 2$. Therefore, number $1$ will be written on the board after f... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | Katia wins; because the number decreases and parity alternates each turn, the number one is written on Mykola’s turn, so he loses. Katia can simply make any legal move each turn. | |
08ji | Problem:
The triangle $ABC$ is isosceles with $AB = BC$. The point $F$ on the side $[BC]$ and the point $D$ on the side $[AC]$ are the feet of the internal bisector drawn from $A$ and altitude drawn from $B$ respectively so that $AF = 2 BD$. Find the measure of the angle $ABC$. | [] | JBMO | The second selection test for JMBO 2003 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 108° | |
0fn5 | Sean $\Gamma$ una circunferencia de centro $O$, $AE$ un diámetro de $\Gamma$ y $B$ el punto medio de uno de los arcos $AE$ de $\Gamma$. El punto $D \neq E$ está sobre el segmento $OE$. El punto $C$ es tal que el cuadrilátero $ABCD$ es un paralelogramo con $AB$ paralelo a $CD$ y $BC$ paralelo a $AD$. Las rectas $EB$ y $... | [
"Nótese que $AB$ es la simétrica de $BE$ respecto de $OB$, y al ser $OB$ perpendicular a $AE$, luego a $BC$, la mediatriz de $BC$ es paralela a $OB$. Pero como la simetría respecto de la mediatriz de $BC$ transforma a $B$ en $C$, y transforma las rectas paralelas a $BE$ en rectas paralelas a $AB$, luego a $CD$, se ... | Spain | Olimpiada Iberoamericana de Matemáticas | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | Spanish | proof only | null | |
0dcg | There are 3 clubs $A, B, C$ with non-empty members. For any triplet of members $(a, b, c)$ with $a \in A, b \in B, c \in C$, two of them are friend and two of them are not friend (here the friend relationship is bidirectional). Prove that one of these statements must be true
1. There exist one student from $A$ that kn... | [
"We will prove the statement by induction on the maximum number of members in clubs $A, B, C$.\n\nFor $n=1$, each club has exactly one member and the statement is obviously true.\n\nSuppose that when the maximum numbers in three clubs is $n \\geq 1$, then one of three above conditions holds. Assume that $a \\in A$ ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0ewb | Problem:
Prove that there are no integers $a$, $b$, $c$, $d$ such that the polynomial $ax^3 + bx^2 + cx + d$ equals $1$ at $x = 19$ and $2$ at $x = 62$. | [
"Solution:\nIf there were such values, then subtract the equation with $x = 19$ from the equation with $x = 62$ to get:\n$$\na(62^3 - 19^3) + b(62^2 - 19^2) + c(62 - 19) = 1.\n$$\nBut the left hand side is divisible by $62 - 19 = 43$, contradiction."
] | Soviet Union | 2nd ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0giu | A finite collection of triangles on the plane are *intersecting* if, for any two triangles in the collection, the intersection of their interiors is not empty. Show that for any two finite collections of intersecting triangles, there exists a straight line that simultaneously intersects the interiors of all triangles i... | [
"取平面上任意點 $P$, 令 $\\ell_\\theta$ 為過 $P$ 且與 $x$ 軸正向夾 $\\theta$ 角的有向直線。注意到對於第 $i$ 個集合 ($i = 1,2$), 第 $i$ 個集合中的每個三角形在 $\\ell_\\theta$ 上的投影都是一個有限區間, 且由相交性知道這些區間兩兩相交。這意味著第 $i$ 個集合中所有三角形投影的交集也是一個有限區間 (可用凸性證明或數歸¹), 令 $d_i(\\theta)$ 為從 $P$ 到第 $i$ 個集合投影所成區間中點的有向距離 (依據 $\\ell_\\theta$ 的方向)。\n\n現在, 注意到 $d_i(\\theta)$ 為 $\\thet... | Taiwan | Taiwan Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | Chinese; English | proof only | null | |
0fsz | Problem:
Der Inkreis des Dreiecks $A B C$ berührt die Seiten $A B, B C$ und $C A$ in den Punkten $D, E$ und $F$. Sei $P$ ein Punkt im Innern von $A B C$, sodass der Inkreis von $A B P$ die Seite $A B$ ebenfalls in $D$ berührt und die Seiten $A P$ und $B P$ in den Punkten $Q$ und $R$. Zeige, dass die vier Punkte $E, F,... | [
"Solution:\n\nWir zeigen im Folgenden mit Winkeljagd, dass $\\angle F E R+\\angle F Q R=180^{\\circ}$ gilt.\nSei $I$ der Inkreismittelpunkt und $k_{1}$ der Inkreis von $\\triangle A B C$ und $k_{2}$ der Inkreis von $\\triangle A B P$. Die Geraden $A F$ und $A D$ liegen beide tangential an $k_{1}$, es gilt also $A F... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
025a | Problem:
Um time de futebol ganhou $60\%$ das 45 partidas já disputadas. Qual é o número mínimo de partidas que esse time ainda precisa vencer para atingir uma porcentagem de $75\%$ de vitórias? | [
"Solution:\n\nO time ganhou $60\\%$ das 45 já disputadas, ou seja, $45 \\times \\frac{60}{100} = 27$ partidas. Se ele ganhar mais $n$ partidas, a porcentagem de partidas ganhas será\n$$\n\\frac{\\text{nº de partidas ganhas}}{\\text{nº de partidas disputadas}} = \\frac{27 + n}{45 + n} = 75\\% = \\frac{75}{100} = \\f... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 27 | |
0h7w | The points $A_1$ and $C_1$ are chosen on the sides $BC$ and $AB$ of the triangle $ABC$ so that the segments $AA_1$ and $CC_1$ are equal and perpendicular. Prove that if $\angle ABC = 45^\circ$, then $AC = AA_1$.
(Andrei Gogolev) | [
"Firstly we will show that the triangle is acute. Indeed, suppose that $\\angle BAC \\ge 90^\\circ$. Then $\\angle ACB \\le 45^\\circ = \\angle ABC < \\angle AA_1C$. From $\\triangle AA_1C$ we get that $AC > AA_1$. From the other side\n\n$\\angle C_1AC \\ge 90^\\circ$, then in $\\triangle AC_1C$ the side $CC_1$ is ... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06z7 | Problem:
If $a$, $b$, $c$, are the sides of a triangle, show that
$$
\frac{a-b}{a+b} + \frac{b-c}{b+c} + \frac{c-a}{a+c} < \frac{1}{16}.
$$ | [
"Solution:\n\nPut $f(a, b, c) = \\frac{a-b}{a+b} + \\frac{b-c}{b+c} + \\frac{c-a}{a+c}$. Let $A, B, C$ be a permutation of $a, b, c$, with $A \\leq B \\leq C$. If $(A, B, C) = (b, a, c), (a, c, b)$ or $(c, b, a)$, then $f(a, b, c) = X$, where\n$$\nX = \\frac{B-A}{B+A} + \\frac{C-B}{C+B} - \\frac{C-A}{A+C}.\n$$\nIf ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
059m | The largest angle of triangle $ABC$ is located at its vertex $C$. Let $\rho$ be the circle with centre $A$ and radius $AC$, and let $\sigma$ be the circle with centre $B$ and radius $BC$. The circle $\sigma$ intersects the circumcircle of the triangle $ABC$ and the circle $\rho$ at points $D$ and $F$, respectively ($D ... | [
"Denote $\\angle BAC = \\alpha$ (Figures 7 and 8 present two possible cases). Note that $AC = AF$ and $BC = BF$ as they are radii of circles $\\rho$ and $\\sigma$, respectively. Thus the triangles $ABF$ and $ABC$ are equal by three equal sides. Hence $\\angle BAF = \\alpha$. From equal radii of $\\sigma$, we also o... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles"
] | English | proof only | null | |
0dz1 | For what positive integers $n > 1$ does the expression
$$
\frac{\log_{10} 2 \cdot \log_{10} 3 \cdots \log_{10} n}{10^{n-1}}
$$
attain the smallest possible value? Find this value. | [
"Let us compare the expressions $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} (n-1)}{10^{n-2}}$ and $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}$. The inequality\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} (n-1)}{10^{n-2}} \\ge \\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_... | Slovenia | Slovenija 2008 | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | n = 10^{10} - 1 and n = 10^{10}; minimal value = (\prod_{k=2}^{10^{10}} \log_{10} k) / 10^{10^{10}-1} | |
04if | Determine the number of positive integers smaller than $1\,000\,000$, that are also perfect squares and give a remainder $4$ when divided by $8$. (Ilko Brnetić) | [] | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 250 | |
07dl | Let $ABC$ be a triangle with $\hat{A} \ne 90^\circ$. Points $E, F$ are the foot of altitudes from $B, C$ to $AC, AB$ respectively. The angle bisector of $\hat{A}$ intersects $EF, BC$ at $M, N$ respectively. Perpendicular lines to $EF$ at $M$ and to $BC$ at $N$ intersect each other at $P$. Prove that $P$ lies on the med... | [
"Let $Q$ be the midpoint of $BC$, $A'$ be the reflection of $A$ into $O$, where $O$ is the circumcenter of triangle $ABC$, and $H$ be the orthocenter of $ABC$.\n\n\n\nQuadrilateral $BHCA'$ is a parallelogram, therefore $A'H$ passes through $Q$. It also intersects the circumcircle again at $... | Iran | Iranian Mathematical Olympiad | [
"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 > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0440 | Let $S(k)$ denote the sum of all digits of $k$ in base 10. Find all integers $n \ge 2$ and rational numbers $\beta \in (0,1)$, such that there exist $n$ distinct positive integers $a_1, a_2, \dots, a_n$ that satisfy: for any subset $I \subseteq \{1, 2, \dots, n\}$ with two or more elements,
$$ S\left(\sum_{i \in I} a_i... | [
"The desired are integers $n \\in \\{2, 3, \\dots, 10\\}$ and all rational numbers $\\beta \\in (0, 1)$.\n\nFirst, for integer $n \\in \\{2, 3, \\dots, 10\\}$ and rational $\\beta \\in (0, 1)$, construct $a_1, a_2, \\dots, a_n$ as follows. Let $c$ and $s$ be positive integers that will be determined later. For $k =... | China | China National Team Selection Test | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All integers n from 2 to 10 inclusive, with any rational beta in the open interval (0, 1). | |
070e | Problem:
$A_{1}, A_{2}, \ldots, A_{n}$ are points in the plane. A non-zero real number $k_{i}$ is assigned to each point, so that the square of the distance between $A_{i}$ and $A_{j}$ (for $i \neq j$) is $k_{i} + k_{j}$. Show that $n$ is at most $4$ and that if $n = 4$, then $1 / k_{1} + 1 / k_{2} + 1 / k_{3} + 1 / k... | [
"Solution:\n\nSuppose we have four points $A$, $B$, $C$, $D$ with associated numbers $a$, $b$, $c$, $d$. Then $AB^{2} = a + b$, $AC^{2} = a + c$, so $AB^{2} - AC^{2} = b - c$. Similarly, $DB^{2} - DC^{2} = b - c$, so $AB^{2} - AC^{2} = DB^{2} - DC^{2}$.\n\nLet $X$ be the foot of the perpendicular from $A$ to $BC$, ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
084m | Problem:
Quanti sono i polinomi $p(x)$ di secondo grado, a coefficienti interi e con 2 radici intere, tali che $p(8)=1$ ? (Nota: ricordiamo che i numeri interi possono essere positivi, negativi o nulli)
(A) 1
(B) 2
(C) 3
(D) un numero finito maggiore di 3
(E) infiniti. | [
"Solution:\n\nLa risposta è (C). Dette $m, n$ le due radici (eventualmente coincidenti) del polinomio, si ha $p(x) = a(x-m)(x-n)$, ove $a$ è il coefficiente di $x^{2}$, quindi intero anch'esso. Quindi $1 = p(8) = a(8-m)(8-n)$. Ma $1$ si può ottenere come prodotto di tre interi solo se sono tutti e tre $1$, oppure d... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | C | |
01v1 | Determine, whether there exists a function $f$ defined on the set of all positive real numbers and taking positive values such that
$$
f(x + y) \geq y f(x) + f(f(x))
$$
for all positive $x$ and $y$? | [
"Answer: such function doesn't exist.\nSuppose that such a function exists. Putting $x = 1$ in the inequality\n$$\nf(x+y) \\geq y f(x) + f(f(x)) \\quad (1)\n$$\nwe obtain $f(1+y) \\geq a y + b$ where $a = f(1) > 0$, $b = f(f(1)) > 0$. So\n$$\nf(z) \\geq a z + b - a \\quad (2)\n$$\nfor all $z > 1$. From (2) it follo... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | No such function exists. | |
0kt7 | Problem:
Let $P_{1} P_{2} \cdots P_{n}$ be a regular $n$-gon in the plane and $a_{1}, \ldots, a_{n}$ be nonnegative integers. It is possible to draw $m$ circles so that for each $1 \leq i \leq n$, there are exactly $a_{i}$ circles that contain $P_{i}$ on their interior. Find, with proof, the minimum possible value of ... | [
"Solution:\n\nFor convenience, we take all indices modulo $n$. Let $[n]$ be the set $\\{1,2, \\ldots, n\\}$. Also, let $M=\\max \\left(a_{1}, \\ldots, a_{n}\\right), d=\\frac{1}{2} \\sum_{i}\\left|a_{i}-a_{i+1}\\right|$, and $M^{\\prime}=\\max (M, d)$. We claim that $M^{\\prime}$ is the answer.\nLet $\\Omega$ be th... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | m = max( max_i a_i, (1/2) * sum_i |a_i - a_{i+1}| ), with indices taken cyclically | |
03ba | Let $n \in \mathbb{N}$. Find the number of all sequences $A_0, A_1, \dots, A_n$ of pairwise different points in the plane with non-negative integer first coordinates and second coordinates $0$ or $1$, such that $A_0 = (0,0)$ and
$$
|x_{A_i} - x_{A_{i-1}}| + |y_{A_i} - y_{A_{i-1}}| = 1
$$
for $1 \le i \le n$. | [
"Denote by $c_n$ the number of the sequences from the problem. Define the number $d_n$ in a similar way but replacing the given equality by\n$$\nx_{A_i} - x_{A_{i-1}} + |y_{A_i} - y_{A_{i-1}}| = 1, \\quad 1 \\le i \\le n;\n$$\ncall the respective sequences *right* and the other sequences *wrong*. It is not difficul... | Bulgaria | Bulgaria | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | c_n = 2 f_{n+1} - \varepsilon_n = \frac{2}{\sqrt{5}}\big(q^{n+1} - (1 - q)^{n+1}\big) - \varepsilon_n, where f_k is the Fibonacci sequence with f_1 = f_2 = 1, q = (1 + \sqrt{5})/2, and \varepsilon_n = (1 + (-1)^n)/2. | |
0dty | Let $a_1, a_2, \dots, a_n$ be positive numbers and $A$ their arithmetic mean. Prove that
$$
A^{n-1} \geq \frac{1}{n} \sum_{i=1}^{n} a_1 \cdots \hat{a}_i \cdots a_n, \quad \text{where } \hat{a}_i \text{ denotes the term } a_i \text{ is omitted.}
$$ | [
"We use induction on $n$. When $n=2$, it is an equality. Assume $a_1 \\le a_2 \\le \\dots \\le a_n$. Then $a_1 \\le A \\le a_n$. Let $A_o = (a_1 + \\dots + a_{n-1})/(n-1)$. Then $A = A_o + (a_n - A_o)/n$ and $a_1 \\le A_o \\le A_{n-1} \\le a_n$. Since $a_n - A_o \\ge 0$, we use binomial expansion:\n$$\nA^{n-1} = \\... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0b8c | In the plane are given $100$ points, such that no three of them are on the same line. The points are arranged in $10$ groups, any group containing at least $3$ points. Any two points in the same group are joined by a segment.
a) Determine which of the possible arrangements in $10$ such groups is the one giving the min... | [
"a) If the groups contain respectively $a_1, a_2, \\dots, a_{10}$ points, the number of triangles is\n$$\nN = \\binom{a_1}{3} + \\binom{a_2}{3} + \\dots + \\binom{a_{10}}{3}.\n$$\nWe claim this number is minimal when\n$$\na_1 = a_2 = \\dots = a_{10} = 10,\n$$\nthe minimum value being $10\\binom{10}{3}$. Indeed, if ... | Romania | Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | a) The minimal number of triangles is achieved when the ten groups are all of size ten, giving 10 × C(10, 3) = 1200 triangles. b) Such an arrangement exists: in each group of ten, split the points into two convex pentagons; color pentagon sides with one color, pentagon diagonals with a second color, and edges between t... | |
02w9 | Problem:
a) Determine a quantidade de algarismos do produto $111111 \cdot 1111111111$, em que o primeiro fator possui 6 algarismos e o segundo possui 10 algarismos.
b) Os números $2^{2016}$ e $5^{2016}$ são escritos um ao lado do outro para formar um único número $N$ que possui uma quantidade de algarismos que é a som... | [
"Solution:\na) Sejam $A$ e $B$ os números com 6 e 10 algarismos e começados por 1, respectivamente. Temos $10^{5}<A<2 \\cdot 10^{5}$ e $10^{9}<B<2 \\cdot 10^{9}$. Com isso, multiplicando as duas desigualdades anteriores, podemos escrever:\n$$\n10^{14}<A \\cdot B<4 \\cdot 10^{14}<10^{15}\n$$\nConcluímos assim que o ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | a) 15; b) 2017 | |
0e9p | Find all pairs of primes $p$ and $q$, such that $p+q$ and $p+4q$ are perfect squares. | [
"Let $p+q = x^2$ and $p+4q = y^2$ for some positive integers $x$ and $y$. Subtracting the equations we get $3q = y^2 - x^2 = (y-x)(y+x)$. Since $q$ is a prime and $x+y \\ge 2$ we have the following possibilities:\n\n1. $y - x = 1$ and $y + x = 3q$. From the first equation we get $y = x + 1$. Plugging this into the ... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (5, 11), (13, 3) | |
0c42 | Problem:
Fie $ABCD$ un patrulater inscriptibil. Paralela prin $A$ la $BD$ intersectează paralela prin $B$ la $AC$ în punctul $E$. Cercul circumscris triunghiului $ABE$ intersectează a doua oară dreptele $EC$ şi $ED$ în punctele $F$, respectiv $G$. Arătaţi că dreptele $AB$, $CD$ şi $FG$ sunt paralele sau concurente. | [
"Solution:\n\nCum $\\varangle ACB \\equiv \\varangle ADB$, punctele $C$ şi $D$ sunt fie ambele în interiorul cercului circumscris triunghiului $ABE$, fie ambele pe acest cerc, fie ambele în exteriorul cercului. Avem aşadar trei cazuri: $F \\in (EC)$ şi $G \\in (ED)$, sau $F = C, G = D$ (caz în care afirmaţia din en... | Romania | Al treilea test de selecţie pentru OBMJ | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
035s | Problem:
Let $a$, $b$ and $c$ be positive integers such that $ab$ divides $c\left(c^{2}-c+1\right)$ and $a+b$ is divisible by $c^{2}+1$. Prove that the sets $\{a, b\}$ and $\left\{c, c^{2}-c+1\right\}$ coincide. | [
"Solution:\n\nWe shall use the following lemma.\n\n**Lemma.** Let $x$, $y$ and $n$ be positive integers such that $\\frac{xy}{x+y} > n$. Then\n$$\n\\frac{xy}{x+y} \\geq n + \\frac{1}{n^{2} + 2n + 2}\n$$\nwith equality if and only if $\\{x, y\\} = \\left\\{n+1, n^{2}+n+1\\right\\}$.\n\nProof of the lemma. Since $xy ... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
04oo | Let $a$ and $b$ be positive real numbers such that
$$
\frac{a}{b} + \frac{b}{a} = 3 \quad \text{and} \quad \frac{a^2}{b} + \frac{b^2}{a} = 10.
$$
Determine $\frac{1}{a} + \frac{1}{b}$.
(Kristina Ana Škreb) | [
"Let $x = a/b$ and $y = b/a$. Since $a, b > 0$, $x, y > 0$ and $xy = 1$.\n\nWe are given:\n$$\n\\frac{a}{b} + \\frac{b}{a} = x + y = 3.\n$$\nSince $xy = 1$, $x$ and $y$ are the roots of $t^2 - 3t + 1 = 0$.\n\nNow,\n$$\n\\frac{a^2}{b} + \\frac{b^2}{a} = \\frac{a^2}{b} + \\frac{b^2}{a} = a \\cdot \\frac{a}{b} + b \\c... | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 1 | |
04rz | We are given a segment $AB$ in the plane. Consider a triangle $XYZ$ with the following properties: the vertex $X$ is an interior point of the segment $AB$, the triangles $XBY$ and $XZA$ are similar ($\triangle XBY \sim \triangle XZA$) and the points $A$, $B$, $Y$, $Z$ lie on a circle in this order. Find the locus of mi... | [
"Let $XYZ$ be a satisfactory triangle. Then the vertices $Y$ and $Z$ must lie in the same half-plane with the boundary line $AB$. Denote by $Y'$ the reflection of $Y$ through the line $AB$. Due to the presumed similarity, the angles $XAZ$ and $BYX$ are congruent (Fig. 1) and hence $|\\angle BAZ| = |\\angle BY'Z|$ a... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | The locus is the set of points inside the circle with diameter formed by the given segment and outside both Thales circles with diameters from each endpoint to the midpoint of the segment. Equivalently, it is the intersection of the interior of the circle with the given segment as diameter and the exteriors of the circ... | |
0bta | Consider the triangle $ABC$, with $m(\widehat{A}) > 60^\circ$ and $m(\widehat{C}) > 30^\circ$. In the half-plane determined by the line $BC$ not containing $A$, consider points $D$ and $E$ so that $m(\widehat{ABE}) = m(\widehat{CBD}) = 90^\circ$ and $m(\widehat{BAE}) = m(\widehat{BCD}) = 60^\circ$. Denote $F$ and $H$ t... | [
"a) The hypothesis yields $\\Delta ABE \\sim \\Delta CBD$, whence $\\frac{BE}{BD} = \\frac{AB}{CB}$, that is $\\frac{EB}{AB} = \\frac{BD}{BC}$. Since angles $ABC$ and $EBD$ have the same complement, $\\widehat{ABC} = \\widehat{EBD}$, hence $\\Delta EBD \\sim \\Delta ABC$ (S.A.S.).\n\nb) From $\\Delta EBD \\sim \\De... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
04cf | In triangle $ABC$ the side $BC$ is of length $6$, cosine of the angle $\angle ABC$ equals $ rac{4}{5}$, and the radius of the incircle is $1$. Determine the lengths of the sides $AB$ and $AC$ of the triangle. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | AB = 15/4, AC = 15/4 | |
09n3 | Let $ABC$ be a triangle such that $\angle ACB = 90^\circ$ and $\angle BAC = 36^\circ$. Let $AL$ be the angle bisector of the angle $\angle BAC$. The point $M$ is taken on $AL$ so that $AM = MB$. Prove that $AM = 2 CL$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0co6 | Say that a positive integer $b$ is *fortunate* if for every positive integer $a$, a number $b$ divides $a^2$ whenever $b^2$ divides $a^5$. Find the number of fortunate positive integers which are less than $2010$. (P. Kozhevnikov)
Натуральное число $b$ назовём удачным, если для любого натурального $a$ такого, что $a^5... | [
"Установим следующее описание удачных чисел.\n\n**Лемма.** Число $b$ является удачным тогда и только тогда, когда каждое простое число входит в разложение $b$ на простые множители с одним из следующих показателей: $0, 1, 2, 3, 4, 6, 8$.\n\n**Доказательство.** Назовем целое неотрицательное число $k$ счастливым, если... | Russia | Regional round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English; Russian | proof and answer | 1961 | |
01al | *A* and *B* are two convex polygons without common points. None of them is fully contained inside the other one. Prove that there exists such a line *l* that does not intersect each of the polygons and *A* and *B* lie on the different sides of *l*. | [
"Take a point *P* on $A$ and a point *Q* on $B$ such that $|PQ|$ is minimal possible. We will prove that the perpendicular bisector of $PQ$ fits the definition of $l$.\n\nDraw a circle $\\omega$ with center $P$ and radius $|PQ|$, and denote the perpendicular bisector of $PQ$ as $t$. Suppose that some segment $s$ of... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
09o3 | Let $K$ and $L$ be points on side $BC$, $M$ a point on side $AC$, and $N$ a point on side $AB$ of triangle $ABC$. These points are chosen such that $\triangle KMC \sim \triangle ABC$ and $\triangle LBN \sim \triangle ABC$. Suppose that the segments $NL$ and $KM$ intersect at point $P$ inside triangle $ABC$. The circumc... | [
"Since $\\triangle KMC \\sim \\triangle ABC$, we have $\\angle MKC = \\angle BAC$, so quadrilateral $AMKB$ is cyclic. Hence we have, $BC \\cdot KC = AC \\cdot MC$, and since $AMPX$ is cyclic, we have $AC \\cdot MC = XC \\cdot PC$.\n\nCombining the two, we get $XC \\cdot PC = BC \\cdot KC$, so quadrilateral $BXPK$ i... | Mongolia | MMO2025 Round 4 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0efa | Problem:
V preglednici so podane vrednosti kvadratne funkcije $f$. Koliko je $f(4)$?
| $x$ | -2 | 0 | 1 | 2 |
| :---: | :---: | :---: | :---: | :---: |
| $f(x)$ | 0 | $\frac{8}{3}$ | 3 | $\frac{8}{3}$ |
(A) $\frac{2}{3}$
(B) $\frac{5}{3}$
(C) -3
(D) 0
(E) Nič od navedenega | [
"Solution:\n\nEna ničla funkcije $f$ je $-2$, v točki $T(1,3)$ ima graf funkcije $f$ teme, torej je $4$ druga ničla funkcije $f$, zato je $f(4)=0$. Pravilen odgovor je (D)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | D | |
06ve | Let $n>1$ be an integer. Suppose we are given $2n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \ldots, A_{2n}$ in some order. We then consider the $2n$ angles $\angle A_{1}A_{2}A_{3}, \angle A_{2}A_{3}A_{4}, \ldots, \angle A_{2n-2}A_{2n-1}A_{2n}, \angle A_{2n... | [
"Let $\\ell$ be a line separating the points into two groups ($L$ and $R$) with $n$ points in each. Label the points $A_{1}, A_{2}, \\ldots, A_{2n}$ so that $L=\\{A_{1}, A_{3}, \\ldots, A_{2n-1}\\}$. We claim that this labelling works.\nTake a line $s = A_{2n}A_{1}$.\n(a) Rotate $s$ around $A_{1}$ until it passes t... | IMO | IMO 2019 Shortlisted Problems | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
018v | A circulator is an instrument which draws the circumcircle of three given points in the plane (if the points happen to be collinear, it draws the line through them). Is it possible to construct, only with the help of a circulator, the centre of a given circle? | [
"No, it is not: for suppose you have an algorithm that constructs the centre of the given circle $\\gamma$ such that each step of the algorithm consists in choosing three points and constructing their circumcircle. Consider some arbitrary circle $\\Omega$ different from $\\gamma$. Invert $\\gamma$ in $\\Omega$ to g... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0k9c | Problem:
Six unit disks $C_{1}, C_{2}, C_{3}, C_{4}, C_{5}, C_{6}$ are in the plane such that they don't intersect each other and $C_{i}$ is tangent to $C_{i+1}$ for $1 \leq i \leq 6$ (where $C_{7}=C_{1}$). Let $C$ be the smallest circle that contains all six disks. Let $r$ be the smallest possible radius of $C$, and ... | [
"Solution:\n\nThe minimal configuration occurs when the six circles are placed with their centers at the vertices of a regular hexagon of side length $2$. This gives a radius of $3$.\n\nThe maximal configuration occurs when four of the circles are placed at the vertices of a square of side length $2$. Letting these... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellan... | null | proof and answer | sqrt(3) - 1 | |
08rx | Determine ten's place of $11^{12^{13}}$. ($12^{13}$th power of $11$, not $13$th power of $11^{12}$.) | [
"For positive integer $n$,\n$$\n11^n = (10 + 1)^n = \\sum_{k=0}^{n} C_n^k 10^k = n \\cdot 10^n + n \\cdot 10^0 = n \\cdot 10^n + 1 \\pmod{100}.\n$$\nTherefore, the ten's place of $11^n$ is equal to the last digit of $n$. The last digit of $12^n$ for $n = 1, 2, \\dots$ is $2, 4, 8, 6, 2, 4, \\dots$, with period $4$,... | Japan | Japan 2007 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | final answer only | 2 | |
0hbe | Find all natural numbers $a$, $b$ and $c$, such that the number $2^{a!} + 2^{b!} + 2^{c!}$ is a cube of a natural number.
(Arseniy Nicolaev) | [
"It is clear that if $n \\ge 3$, $n! \\nmid 3$, that is $n! \\ne 3k$, for some natural number $k$. But then $2^{n!} = 2^{3k} = 8^k \\equiv 1 \\pmod{7}$. Analogously modulo 9: if $n \\ge 3$ we have that $n! \\nmid 6$, that is $n! \\ne 6l$, for some natural $l$. But then $2^{n!} = 2^{6l} = 64^l \\equiv 1 \\pmod{9}$.\... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All permutations of (1,1,2). | |
01ht | A directed graph $G$ has 2021 vertices located on a circle, 101 directed edges go from each vertex of $G$ to the next 101 vertices in the counter-clockwise direction. We wish to colour the edges in such a way that for any two vertices $u$ and $v$ one can choose a directed path from $u$ to $v$ in which no two edges have... | [
"Answer: $\\lceil \\frac{2021}{101} \\rceil = 21$ colours.\n\nLabel the vertices along the cycle: $v_0, v_1, \\dots, v_{2020}$. It is clear that the distance from $v_0$ to any of $v_1, \\dots, v_{101}$ equals 1; the distance from $v_0$ to any of $v_{102}, \\dots, v_{202}$ equals 2, etc. Therefore, the distance from... | Baltic Way | Baltic Way 2021 Shortlist | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 21 | |
0cy6 | Find all pairs $(x, y)$ of real numbers that satisfy the system of equations
$$
\begin{cases}
x^{4} + 2x^{3} - y = \sqrt{3} - \dfrac{1}{4} \\
y^{4} + 2y^{3} - x = -\sqrt{3} - \dfrac{1}{4}
\end{cases}
$$ | [] | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | (x, y) ≈ (0.3842, -1.3468) | |
0409 | Let $x$, $y$, $z \in [0, 1]$. Then the maximum value of $M = \sqrt{|x-y|} + \sqrt{|y-z|} + \sqrt{|z-x|}$ is ________. | [
"We may assume $0 \\le x \\le y \\le z \\le 1$. Then\n$$\nM = \\sqrt{y-x} + \\sqrt{z-y} + \\sqrt{z-x}.\n$$\nSince\n$$\n\\sqrt{y-x} + \\sqrt{z-y} \\le \\sqrt{2[(y-x)+(z-y)]} = \\sqrt{2(z-x)},\n$$\nwe have\n$$\nM \\le \\sqrt{2(z-x)} + \\sqrt{z-x} = (\\sqrt{2} + 1) \\sqrt{z-x} \\le \\sqrt{2} + 1.\n$$\nThe equality hol... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | sqrt(2) + 1 | |
049g | Given $\log_a x = 3$, $\log_{ab} x = 2$, determine $\log_b x$. | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 6 | |
0gfi | 給定正整數 $n \ge 3$，並在圓上標了 $n$ 個點。安娜與艾莎進行以下遊戲：從安娜開始，兩人輪流從還沒用過的點中選三個點畫三角形，並不得通過已經畫好的三角形的任一邊；首先不能畫滿足條件的三角形的人判輸，另一人判贏。請問誰有必勝策略？ (答案得與 $n$ 有關。) | [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 國際競賽實作(一) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Chinese; English | proof and answer | Anna wins if and only if floor(n/3) is odd, equivalently when n mod 6 is in {3, 4, 5}; otherwise Elsa wins. | |
0gv3 | There are 207 boxes on the table numbered $1, 2, \ldots, 207$. Asli puts one red ball to each of some 100 boxes and one white ball to each of the remaining 107 boxes. After that Zehra writes $N$ pairs $(i, j)$, $1 \le i < j \le 207$ to the board. Finally for each pair $(i, j)$ on the board Asli informs Zehra whether ba... | [
"**Answer: 205.**\n\nLet us show that Zehra can guarantee to determine the colours of all balls for $N = 205$. Zehra writes the pairs $(1,2), (1,3), \\dots, (1,206)$ on the board. Let us assign a number $i$ to the ball contained in the box number $i$. Suppose that out of balls numbered $2, 3, \\dots, 206$ the numbe... | Turkey | Team Selection Test for JBMO 2024 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 205 | |
0c26 | On the side $BC$ of a parallelogram $ABCD$ lies a given point $M$. Find a compass-and-straightedge construction of a point $N \in CD$, so that the straight line $BD$ separates triangle $AMN$ into two equivalent areas.
Cecilia Deaconescu | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02rl | In the following diagram, *ABCDE* is a regular pentagon and *MNP* is an equilateral triangle. Find $\angle CMD$.
 | [
"First notice that $BE \\parallel CD$, so $BEM$ is an equilateral triangle. So $BE = EM$ and $\\angle BEM = 60^\\circ$.\n\nSince $ABCDE$ is a regular pentagon, $\\angle BEC = 36^\\circ$, so $\\angle MEC = \\angle BEM + \\angle BEC = 60^\\circ + 36^\\circ = 96^\\circ$.\n\nWe also have $CE = BE = EM$, so triangle $CE... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 24° | |
05zm | Problem:
Soit $k$ un entier strictement positif et $P$ un point du plan. Déterminer le plus petit entier $n$ ayant la propriété suivante : on peut tracer $n$ droites ne passant pas par $P$ de sorte que toute demi-droite d'origine le point $P$ coupe au moins $k$ droites. | [
"Solution:\n\nRéponse : $n = 2k + 1$\n\nL'énoncé demande de déterminer le plus petit entier $n$ possédant une certaine propriété, il contient nécessairement deux parties. D'une part, on montre que si $n$ est inférieur à $2k$, il ne vérifie pas la propriété, d'autre part on montre que $2k+1$ vérifie la propriété.\n\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2k+1 | |
0h38 | A function $y = f(x)$ is such that its graph is symmetric to the graph of $y = x^2$ with respect to the point $(1; 1)$. Solve the equation $f(f(x)) = f(x)$. | [
"Let us find the explicit form of the function $y = f(x)$. Let $P(p; p^2)$ be an arbitrary point of the parabola $y = x^2$, and $Q(q; f(q))$ be the point on the graph $y = f(x)$, symmetric to $P$ with respect to the point $(1; 1)$. Then:\n$$\n\\frac{p + q}{2} = 1, \\quad \\frac{p^2 + f(q)}{2} = 1.\n$$\nEliminating ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 1, 2, 3 | |
0eo9 | I have R4. How many sweets can I buy if one sweet costs 40c? | [
"Since total cost = number of sweets × cost of each sweet, it follows that the number of sweets is $\\text{R}4.00 \\div 40\\text{c} = \\text{R}4.00 \\div \\text{R}0.40 = 4 \\div 0.4 = 10$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | final answer only | 10 | |
036o | Problem:
Let $b$ and $c$ be real numbers such that the equation $x^{2} + b x + c = 0$ has two distinct real roots $x_{1}$ and $x_{2}$ with $x_{1} = x_{2}^{2} + x_{2}$.
a) Find $b$ and $c$ if $b + c = 4$.
б) Find $b$ and $c$ if they are coprime integers. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Greatest common diviso... | null | proof and answer | a) b = -8, c = 12. b) (b, c) = (-3, 2) or (1, 0). | |
0bzd | Let $ABC$ be a triangle such that $AB \neq AC$, let $G$ be its centroid, and let $H$ be its orthocenter. Let $D$ be the orthogonal projection of $A$ on the line $BC$, and let $M$ be the midpoint of the side $BC$. The circle $ABC$ crosses the ray $MH$ emanating from $M$ at $P$, and the ray $DG$ emanating from $D$ at $Q$... | [
"*First solution.* We show that the angles $APD$ and $AQM$ are either equal or one is the supplement of the other.\nTo begin, recall that the reflection of $H$ across $M$ is the antipode of $A$ in the circle $ABC$, to infer that the angle $APM$ is right, so $A$, $D$, $M$, $P$ are concyclic. Hence the angles $APD$ a... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"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"
] | English | proof only | null | |
0867 | Problem:
Sia $AB$ una corda di una circonferenza e $P$ un punto interno ad $AB$ tale che $AP = 2PB$. Sia $DE$ la corda passante per $P$ e perpendicolare ad $AB$. Dimostrare che il punto medio $Q$ di $AP$ è l'ortocentro di $ADE$. | [
"Solution:\n\nSia $H$ il punto in cui la retta $EQ$ interseca $AD$; si deve dimostrare che l'angolo $A\\widehat{H}E$ è retto. Tracciamo il segmento $BE$. Il triangolo $BQE$ è isoscele perché l'altezza $EP$ è anche mediana; infatti $P$, piede dell'altezza $EP$, è punto medio di $BQ$ in quanto $PQ = \\frac{1}{2} AP =... | Italy | Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"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 | |
0bbj | Solve the equation $(a^x + b^x)^{2011} = (a^{2011} + b^{2011})^x$, where $a, b$ are positive reals. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 2011 | |
0c62 | Determine all functions $f$ from the set of non-negative integers to itself such that
$$
f(a + b) = f(a) + f(b) + f(c) + f(d),
$$
whenever $a$, $b$, $c$, $d$ are non-negative integers satisfying $2ab = c^2 + d^2$. | [
"The required functions are $f(n) = k n^2$, where $k$ is a non-negative integer — these clearly satisfy the condition in the statement.\n\nConversely, let $f$ be a function satisfying the condition in the statement. Setting $(a, b, c, d) = (n, n, n, n)$ in the functional relation yields $f(2n) = 4f(n)$ for all $n$.... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof and answer | All functions f(n) = k n^2 with k a nonnegative integer. | |
02vv | Problem:
Um quadrilátero é dito cíclico quando seus quatro vértices estão sobre uma mesma circunferência. Considere um quadrilátero $ABCD$ cíclico com diagonais $AC$ e $BD$ perpendiculares. Além disso, sejam $O$ o centro da circunferência que passa pelos vértices do quadrilátero e $P$ o ponto de encontro das diagonais... | [
"Solution:\n\na) Seja $\\angle BCA=\\theta$. Veja que $\\angle BCA=\\angle BDA=\\theta$, pois eles estão inscritos no mesmo arco $AB$. Considerando a soma dos ângulos do triângulo $CXP$, temos $\\angle CPX=90^{\\circ}-\\theta$ e, consequentemente, $\\angle BPX=\\angle BPC-\\angle CPX=\\theta$. Vale que $\\angle APM... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07l3 | Suppose *a*, *b*, *c* are positive real numbers. Prove that
$$
abc(a + b + c) \le a^4 + b^4 + c^4,
$$
with equality iff $a = b = c$. | [
"By the Arithmetic mean-Geometric mean inequality\n$$\n3\\sqrt[3]{a^4b^4c^4} \\le a^4 + b^4 + c^4,\n$$\nwith equality iff $a = b = c$. Thus\n$$\n3^{3/4}abc \\le (a^4 + b^4 + c^4)^{3/4},\n$$\nwith equality iff $a = b = c$. Also, two applications of the Cauchy-Schwarz inequality tell us that\n$$\n(a+b+c)^4 \\le (3(a^... | Ireland | Irska | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
06sg | Let $n$ be a positive integer, and let $A$ be a subset of $\{1, \ldots, n\}$. An $A$-partition of $n$ into $k$ parts is a representation of $n$ as a sum $n=a_{1}+\cdots+a_{k}$, where the parts $a_{1}, \ldots, a_{k}$ belong to $A$ and are not necessarily distinct. The number of different parts in such a partition is the... | [
"If there are no $A$-partitions of $n$, the result is vacuously true. Otherwise, let $k_{\\text{min}}$ be the minimum number of parts in an $A$-partition of $n$, and let $n=a_{1}+\\cdots+a_{k_{\\min}}$ be an optimal partition. Denote by $s$ the number of different parts in this partition, so we can write $S=\\left\... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null |
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