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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02m1 | An ant crawls in plane as follows: it initially crawls 1 cm in either direction. Then, after each step, it turns $60^\circ$ to the left or to the right and crawls 1 cm in the new direction.
Is it possible for it to return to its starting point in
(a) 2008 steps?
(b) 2009 steps? | [
"Let $O$ be the starting point of the ant. Then the ant will walk on the following hexagonal lattice:\n\nColor the vertices in the lattice alternatively in black and white, as shown in the diagram. Then each step leads the ant from a black point to a white point or vice-versa. Hence the ant... | Brazil | XXXI Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) Yes. b) No. | |
01hp | John has a string of paper where $n$ real numbers $a_i \in [0, 1]$, for all $i \in \{1, \dots, n\}$, are written in a row. Show that for any given $k < n$, he can cut the string of paper into $k$ pieces, between adjacent numbers, in such a way that sum of the numbers on each piece does not differ from another by more t... | [
"Denote the sums on each piece by\n$$\n\\begin{align*}\nS_1 &= a_1 + a_2 + \\dots + a_{m_1}, \\\\\nS_2 &= a_{m_1+1} + a_{m_1+2} + \\dots + a_{m_2}, \\\\\n\\vdots \\\\\nS_k &= a_{m_{k-1}+1} + \\dots + a_{m_k}.\n\\end{align*}\n$$\nBy abuse of notation $S_i$ will both denote the set of numbers enclosed by cuts and its... | Baltic Way | Baltic Way 2021 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0cqa | Let $\triangle ABC$ be a regular triangle. Point $T$ is chosen on its side $AC$, and points $M$ and $N$ are chosen respectively on arcs $AB$ and $BC$ of its circumcircle so that $MT \parallel BC$ and $NT \parallel AB$. Segments $AN$ and $MT$ intersect at $X$, while segments $CM$ and $NT$ intersect at $Y$. Prove that th... | [
"**Первое решение.** Пусть $\\ell$ — касательная к описанной окружности в точке $B$, а $P$ и $Q$ — точки пересечения $\\ell$ с лучами $TM$ и $TN$ соответственно. Обозначим через $K$ и $L$ соответственно точки, в которых лучи $TM$ и $TN$ пересекают стороны треугольника (см. рис. 12).\nЗаметим, что четырёхугольники $... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Russian | proof only | null | |
0la3 | Given an angle $xOy$ on a plane. Let $M$ and $N$ are variable points on the rays $Ox$ and $Oy$, respectively. Denote by $d$ the outer angle-bisector of the angle $xOy$ and let $I$ denote the intersection of $d$ and perpendicular bisector of the segment $MN$. Choose two points $P$ and $Q$ on $d$ such that $IP = IQ = IM ... | [] | Vietnam | Vijetnam 2008 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurren... | null | proof only | null | |
0821 | Problem:
Sia $P(X) = a_{n} X^{n} + a_{n-1} X^{n-1} + \ldots + a_{1} X + a_{0}$ un polinomio a coefficienti interi (cioè, i numeri $a_{n}, a_{n-1}, \ldots, a_{1}, a_{0}$ sono interi). Se $P(2000) = 2000$ e $P(2001) = 2001$, quanti fra i numeri $2000, 2001, 2002, 2003, 2004$ possono essere uguali a $P(2002)$?
(A) 1
(B)... | [
"Solution:\n\nLa risposta è (C). Osserviamo innanzitutto che i numeri $2001$ e $2003$ non sono possibili, in quanto sono numeri dispari. Infatti, poiché $P(2000) = a_{n} 2000^{n} + a_{n-1} 2000^{n-1} + \\cdots + a_{1} 2000 + a_{0} = 2000$ è un numero pari, $a_{0}$ deve essere un numero pari, e di conseguenza, calco... | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | C | |
05tp | Problem:
Un coloriage des entiers $\{1,2, \cdots, 2020\}$ en bleu et rouge est dit agréable s'il n'existe pas deux entiers distincts dans $\{1,2, \cdots, 2020\}$ de même couleur dont la somme est une puissance de 2. Combien de tels coloriages existent-ils? | [
"Solution:\n\nUne première idée est de tester l'énoncé pour des valeurs plus petites, par exemple pour un coloriage des entiers de $1$ à $7$.\n\nColorions dans l'ordre les nombres : pour colorier $1$, on n'a pas de contrainte apparente, idem pour colorier $2$. Par contre, comme $3+1=4$, la couleur de $3$ est imposé... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^11 | |
0d5w | There are 22 chairs in a round table. Find the minimum $n$ such that for any group of $n$ people sitting in the table, we always can find two people with exactly 2 or 8 chairs between them. | [
"We consider the graph with 22 vertices $0, 1, \\ldots, 21$. Two vertices $x$ and $y$ are connected by an edge if $x - y \\equiv \\pm 3, \\pm 9 \\pmod{22}$. We consider only $\\pm 3$ modulo 22, then we have a cycle through all vertices. If we choose any $n \\geq 12$ vertices, there are two vertices (corresponding t... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Arabic | proof and answer | 12 | |
0i15 | Problem:
Prove that there exists a set $S$ of $3^{1000}$ points in the plane such that for each point $P$ in $S$, there are at least 2000 points in $S$ whose distance to $P$ is exactly 1 inch. | [
"Solution:\nLet us define a $k$-configuration to be a finite set of points on the plane such that for each point $P$ in the set, there are at least $k$ points of the set 1 inch from $P$. Then the problem is asking us to show that there is a 2000-configuration with $3^{1000}$ points.\n\nNotice that an equilateral tr... | United States | 2nd Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0819 | Problem:
In un paese l'uno per cento della popolazione è affetto da una certa malattia. Il test per sapere se si è contagiati sbaglia nell'uno per cento dei casi. Lorenzo si sottopone al test e risulta malato. Qual è la probabilità che egli sia sano?
(A) $\frac{99}{10000}$
(B) $\frac{1}{100}$
(C) $\frac{99}{5000}$
(D... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | D | |
0j2i | Let $T$ be a finite set of positive integers greater than $1$. A subset $S$ of $T$ is called *good*, if for every $t \in T$ there exists some $s \in S$ with $\gcd(s, t) > 1$. Prove that the number of good subsets of $T$ is odd. | [
"Consider the set $\\mathcal{A}$ of all (ordered) pairs $(X, Y)$ with $X, Y \\subseteq T$ and $\\gcd(x, y) = 1$ for all $x \\in X$ and $y \\in Y$. Clearly $X$ and $Y$ are disjoint for any $(X, Y) \\in \\mathcal{A}$. We have the following claims.\n\ni. If $X'$ is good, then the number of pairs $(X', Y) \\in \\mathca... | United States | Team Selection Test 2010 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
06ho | Let $\lfloor x \rfloor$ denote the greatest integer not exceeding $x$. Find the last three digits of
$$
\left\lfloor \left( \sqrt[3]{\sqrt{5}+2} + \sqrt[3]{\sqrt{5}-2} \right)^{2014} \right\rfloor
$$ | [
"Let $x = \\sqrt[3]{\\sqrt{5}+2} + \\sqrt[3]{\\sqrt{5}-2}$. Using the formula $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$, we have\n$$\nx^3 = (\\sqrt{5}+2) + (\\sqrt{5}-2) + 3\\sqrt[3]{(\\sqrt{5}+2)(\\sqrt{5}-2)}x \\\\ = 2\\sqrt{5} + 3x.\n$$\nRewrite this as $(x - \\sqrt{5})(x^2 + \\sqrt{5}x + 2) = 0$. Since the quadratic equ... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | English | proof and answer | 125 | |
07jt | In triangle $ABC$, altitudes $AD$, $BE$, and $CF$ meet at $H$. Line $EF$ and the perpendicular bisector of $HD$ intersect at point $P$. Point $N$ is the center of the nine-point circle of triangle $ABC$. Point $L$ is on the circumcircle of the triangle such that $\angle PLN = 90^\circ$ and points $A$, $L$ are on opposi... | [
"Let $B'$, $C'$ be the intersections of the perpendicular bisector of $HD$ with $BH$, $CH$ respectively.\n\nDenote by $M$, $M'$, $M_0$ the midpoints of $BC$, $B'C'$, $EF$, respectively. Since $N$ is the center of the circle $FEC'B'$, we have\n$$\n\\angle NM_0P = \\angle NM'P = 90^\\circ\n$$... | Iran | Iranian Mathematical Olympiad | [
"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... | null | proof only | null | |
09cb | **ДБ-В1.** (Ц.Дашдорж) $n \ge 2$ байг. $n \mid 3^n + 4^n$ бол $7 \mid n$ гэж батал. | [
"$p$ нь $n$-ийн хамгийн бага анхны тоон хуваагч гэвэл\n$3 \\neq p \\neq 2$ ба $(n, p-1) = 1$ тул $\\exists a \\in \\mathbb{N}, an \\equiv 1 \\pmod{p-1}$.\n$p-1$ тэгш тоо тул $a$ сондгой тоо.\nИймд $p \\mid 3^n + 4^n \\mid 3^{an} + 4^{an}$. Фермагийн теоремоор $3^{p-1} \\equiv 1 \\pmod{p}$.\n$p-1 \\mid an-1$ гэдгээс... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Mongolian | proof only | null | |
06qa | Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $\operatorname{gcd}\left(a_{i}, a_{i+1}\right)>a_{i-1}$. Prove that $a_{n} \geq 2^{n}$ for all $n \geq 0$. | [
"Since $a_{i} \\geq \\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$, the sequence is strictly increasing. In particular $a_{0} \\geq 1, a_{1} \\geq 2$. For each $i \\geq 1$ we also have $a_{i+1}-a_{i} \\geq \\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$, and consequently $a_{i+1} \\geq a_{i}+a_... | IMO | 49th International Mathematical Olympiad Spain | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0dmt | Problem:
Таблица димензија $n \times n$, на чијим пољима су бројеви $1,2, \ldots, n^{2}$ (на сваком пољу тачно један број и сваки број на тачно једном пољу) назива се нишка ако сви производи од по $n$ бројева који се налазе на $n$ „разбацаних" поља дају исти остатак при дељењу са $n^{2}+1$. Да ли постоји нишка таблица... | [
"Solution:\n\na.\nПретпоставимо да постоји нишка таблица $8 \\times 8$ и да производ ма којих 8 разбацаних бројева даје остатак $r$ по модулу $8^{2}+1=65=5 \\cdot 13$. Сви бројеви у таблици се могу поделити на 8 дисјунктних осморки разбацаних бројева. Међу овим осморкама постоји једна која садржи умножак броја 13 и... | Serbia | Serbian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
04ea | Collinear points $A$, $B$ and $M$ are given so that $M$ is between $A$ and $B$. Let $k$ be the circle with diameter $\overline{AB}$ and let $N$ be any point on $k$ different from $A$ and $B$. Prove that the expression
$$
\frac{\operatorname{tg}(\angle ANM)}{\operatorname{tg}(\angle MAN)}
$$
is a constant, i.e. that it ... | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
03gl | Problem:
Let
$$
p(x) = a_{0} x^{n} + a_{1} x^{n-1} + \cdots + a_{n-1} x + a_{n}
$$
where the coefficients $a_{i}$ are integers. If $p(0)$ and $p(1)$ are both odd, show that $p(x)$ has no integral roots. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0dum | Problem:
Za katere vrednosti realnega parametra $a$ ima sistem enačb
$$
x + y = a^{3} - a \quad \text{in} \quad x y = a^{2}
$$
realni rešitvi $x$ in $y$? | [
"Solution:\n\nIzrazimo $x = a^{3} - a - y$, vstavimo $x$ v drugo enačbo in dobimo $(a^{3} - a - y) y = a^{2}$ oziroma $y^{2} + y(a - a^{3}) + a^{2} = 0$.\n\nKvadratna enačba ima realni rešitvi, če je njena diskriminanta nenegativna, torej $(a - a^{3})^{2} - 4 a^{2} \\geq 0$.\n\nNeenačbo poenostavimo in dobimo $a^{2... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | a = 0 or a^2 >= 3 | |
0f5m | Problem:
The projection of a tetrahedron onto the plane $P$ is $ABCD$. Can we find a distinct plane $P'$ such that the projection of the tetrahedron onto $P'$ is $A'B'C'D'$ and $AA'$, $BB'$, $CC'$ and $DD'$ are all parallel? | [] | Soviet Union | 17th ASU | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
02wj | Problem:
No triângulo $ABC$, retângulo em $A$, tem-se $AB = 8~\mathrm{cm}$ e $AC = 6~\mathrm{cm}$.
a. O ponto $P$, interior ao triângulo, dista $1~\mathrm{cm}$ do lado $AB$ e $2~\mathrm{cm}$ do lado $AC$. Qual é a distância de $P$ ao lado $BC$?
b. Calcule o raio da circunferência que é tangente ao lado $AC$ e aos pr... | [
"Solution:\n\na.\nSe $x$ é a distância de $P$ ao lado $BC$ e usando a notação $(XYZ)$ para indicar a área do polígono $XYZ$, temos\n$$\n\\begin{aligned}\n(APB) + (BPC) + (CPA) &= (ABC) \\\\\n\\frac{8 \\cdot 1}{2} + \\frac{10 \\cdot x}{2} + \\frac{6 \\cdot 2}{2} &= \\frac{6 \\cdot 8}{2} \\\\\n8 + 10x + 12 &= 48\n\\e... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a) 2.8 cm; b) 4 cm | |
0fzp | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck mit $AB \neq BC$ und Umkreis $k$. Seien $P$ und $Q$ die Schnittpunkte von $k$ mit der Winkelhalbierenden beziehungsweise der Außenwinkelhalbierenden von $\angle CBA$. Sei $D$ der Schnittpunkt von $AC$ und $PQ$. Bestimme das Verhältnis $AD : DC$. | [
"Solution:\n\nPar le théorème de l'angle inscrit et le fait que $BP$ est la bissectrice de $\\angle ABC$, $\\angle PCA = \\angle PBA = \\angle PBC = \\angle PAC$, donc $\\triangle APC$ est isocèle. De même, prenons $R$ un point sur la droite $AB$ tel que $B$ est entre $A$ et $R$. Alors $\\angle QAC = \\angle QBC = ... | Switzerland | SMO - Finalrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1 | |
0cxy | Consider a circle of center $O$ and a chord $AB$ of it (not a diameter). Take a point $T$ on the ray $OB$. The perpendicular at $T$ onto $OB$ meets the chord $AB$ at $C$ and the circle at $D$ and $E$. Denote by $S$ the orthogonal projection of $T$ onto the chord $AB$. Prove that $AS \cdot BC = TE \cdot TD$. | [
"\n\nDenote $BB'$ the diameter corresponding to point $B$ and consider $\\alpha$ the angle $\\widehat{ABB'}$. Writing the power of $T$ with respect to the circle, we get\n$$\n\\begin{gathered}\nTE \\cdot TD = TB \\cdot TB' = TB \\cdot (2R - TB) = 2R \\cdot \\frac{BS}{\\cos \\alpha} - TB^2 \... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0132 | Problem:
Prove that any real solution of
$$
x^{3}+p x+q=0
$$
satisfies the inequality $4 q x \leq p^{2}$. | [
"Solution:\nLet $x_{0}$ be a root of the cubic, then $x^{3}+p x+q=(x-x_{0})(x^{2}+a x+b)=x^{3}+(a-x_{0}) x^{2}+(b-a x_{0}) x-b x_{0}$. So $a=x_{0}$, $p=b-a x_{0}=b-x_{0}^{2}$, $-q=b x_{0}$. Hence $p^{2}=b^{2}-2 b x_{0}^{2}+x_{0}^{4}$. Also $4 x_{0} q=-4 x_{0}^{2} b$. So $p^{2}-4 x_{0} q=b^{2}+2 b x_{0}^{2}+x_{0}^{4... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
072j | Problem:
1. In a non-equilateral triangle $A B C$, the sides $a, b, c$ form an arithmetic progression. Let $I$ and $O$ denote the incentre and circumcentre of the triangle respectively.
a. Prove that $I O$ is perpendicular to $B I$.
b. Suppose $B I$ extended meets $A C$ in $K$, and $D, E$ are the midpoints of $B C, ... | [
"Solution:\n\na.\nExtend $B I$ to meet the circumcircle in $F$. Then we know that $F A = F I = F C$. (See Figure)\n\n\n\nLet $B I : I F = \\lambda : \\mu$. Applying Stewart's theorem to triangle $B A F$, we get\n$$\n\\lambda A F^{2} + \\mu A B^{2} = (\\lambda + \\mu)\\left(A I^{2} + B I \\c... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellane... | null | proof only | null | |
00x6 | Problem:
There is a finite number of towns in a country. They are connected by one direction roads. It is known that, for any two towns, one of them can be reached from the other one. Prove that there is a town such that all the remaining towns can be reached from it. | [
"Solution:\n\nConsider a town $A$ from which a maximal number of towns can be reached. Suppose there is a town $B$ which cannot be reached from $A$. Then $A$ can be reached from $B$ and so one can reach more towns from $B$ than from $A$, a contradiction."
] | Baltic Way | Baltic Way 1992 | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0fgo | Problem:
Un triángulo dado $T$ se descompone en triángulos $T_{1}, T_{2}, \ldots, T_{n}$ de manera que:
a) Ningún par de triángulos $T_{i}$ tiene puntos interiores comunes.
b) La unión de todos los triángulos $T_{i}$ es $T$.
c) Cada segmento que es lado de algún triángulo $T_{i}$, o bien es lado de otro triángulo $T_{... | [
"Solution:\n\nEn el triángulo $ABC$ tenemos $n=1$, $s=3$, $v=3$. Tomando un punto $P$ en el interior del triángulo unido con los vértices resultan $3$ triángulos; se tiene ahora $n=3$, $s=6$, $v=4$.\n\n\n\nDe una forma general, tomemos un nuevo punto $Q$ en el interior de los triángulos for... | Spain | OME 23 | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | s = 3(n+1)/2, v = (n+5)/2, and n must be odd (no such decomposition exists for even n). | |
0icf | Problem:
Find the number of 20-tuples of integers $x_{1}, \ldots, x_{10}, y_{1}, \ldots, y_{10}$ with the following properties:
- $1 \leq x_{i} \leq 10$ and $1 \leq y_{i} \leq 10$ for each $i$;
- $x_{i} \leq x_{i+1}$ for $i=1, \ldots, 9$;
- if $x_{i}=x_{i+1}$, then $y_{i} \leq y_{i+1}$. | [
"Solution:\nBy setting $z_{i}=10 x_{i}+y_{i}$, we see that the problem is equivalent to choosing a nondecreasing sequence of numbers $z_{1}, z_{2}, \\ldots, z_{10}$ from the values $11,12, \\ldots, 110$. Making a further substitution by setting $w_{i}=z_{i}-11+i$, we see that the problem is equivalent to choosing a... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | C(109,10) | |
0aom | Problem:
Determine the number of factors of $5^{x} + 2 \cdot 5^{x+1}$.
(a) $x$
(b) $x+1$
(c) $2x$
(d) $2x+2$ | [] | Philippines | QUALIFYING STAGE | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | d | |
0ffv | Problem:
Sean $n$ y $k$ dos números naturales primos entre sí, con $0 < k < n$. Cada número del conjunto $\mathcal{M} = \{1, 2, \ldots, n-1\}$ se colorea o bien en azul, o bien en blanco. Se sabe que
1) Para cada $i \in \mathcal{M}$, los elementos $i$ y $n-i$ tienen el mismo color.
2) Para cada $i \in \mathcal{M}$, ... | [] | Spain | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
0g3o | Problem:
Find all polynomials $P$ with real coefficients having no repeated roots, such that for any complex number $z$, the equation $z P(z)=1$ holds if and only if $P(z-1) P(z+1)=0$. | [
"Solution:\nAssume that we have a polynomial $P$ that satisfies the desired conditions. We begin by examining some easy cases:\nIf $P$ is a constant polynomial, that is, $P(z)=c$ for some $c \\in \\mathbb{R}$, the condition becomes $c z=1$ if and only if $c^{2}=0$. If $c=0$, then $c^{2}=0$ is always satisfied while... | Switzerland | IMO Selection | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | P(z) = z | |
01oi | Given a right-angled triangle $ABC$ with $AC = BC$, and $\angle C = 90^\circ$. Let points $M$ and $N$ belong to the sides $AC$ and $BC$ respectively; $MN = BC$. For each pair of such points $M$ and $N$ a circle passing through $M$, $N$ and touching the hypotenuse $AB$ is constructed.
Find the locus of the centers of th... | [
"Answer: The segment $XY$ such that $AXYB$ is a rectangle and the midpoint of $XY$ coincides with $C$.\n\n(Solution of E. Dauhiala, B. Gilevich, K. Kostevich.) Let $O$ be the point in the same half-plane as $C$ with respect to the line $MN$ such that $\\angle MON = 90^\\circ$ and $OM = ON$. Then $OM = ON = \\dfrac{... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0jsl | Problem:
Create a cube $C_{1}$ with edge length $1$. Take the centers of the faces and connect them to form an octahedron $O_{1}$. Take the centers of the octahedron's faces and connect them to form a new cube $C_{2}$. Continue this process infinitely. Find the sum of all the surface areas of the cubes and octahedrons... | [
"Solution:\n\nThe lengths of the second cube are one-third of the lengths of the first cube, so the surface area decreases by a factor of one-ninth. Since the first cube has surface area $6$ and the first octahedron has surface area $\\sqrt{3}$, the total area is $$(6+\\sqrt{3}) \\cdot\\left(1+\\frac{1}{9}+\\frac{1... | United States | HMMT November | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (54+9 sqrt(3))/8 | |
0b5v | Determine the least real number $c$, such that for any integer $n \ge 1$ and any positive real numbers $a_1, a_2, \dots, a_n$, the following holds
$$
\sum_{k=1}^{n} \frac{k}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_k}} < c \sum_{k=1}^{n} a_k.
$$ | [
"We claim $c_{\\min} = 2$.\nTaking $a_j = \\frac{1}{j}$ for $j = 1, 2, \\dots, n$, we have\n$$\n\\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} = \\sum_{k=1}^{n} \\frac{k}{1 + 2 + \\dots + k} = 2 \\sum_{k=1}^{n} \\frac{1}{k+1}, \\text{ while } c \\sum_{k=1}^{n} a_k =\n$$\n\nWe w... | Romania | Local Mathematical Competitions | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 2 | |
0hdd | Determine all positive integers $n$ which are less than 1% of the number 2020, and such that $n+1$ is more than 1% of the number 2019. | [
"The conditions of the problem can be written the following way: $\\frac{n}{2020} < \\frac{1}{100} < \\frac{n+1}{2019}$.\n\nThe left inequality implies that $\\frac{n}{2020} < \\frac{1}{100}$, so $n < \\frac{2020}{100} = 20.2$, hence $n \\le 20$.\n\nIf $n=20$ is substituted into the right inequality, one can see th... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 20 | |
0fqp | Problem:
¿Existen $m, n$ números naturales de forma que
$$
n^{2}+2018 m n+2019 m+n-2019 m^{2}
$$
es un número primo? | [
"Solution:\n\nTratamos de factorizar la expresión del enunciado. Igualando esta expresión a $0$, tendremos\n$$\nn^{2}+(2018 m+1) n+2019 m-2019 m^{2}=0\n$$\nque podemos tratar como una ecuación en la variable $n$, obteniendo que\n$$\nn=\\frac{-(2018 m+1) \\pm \\sqrt{(2018 m+1)^{2}-4\\left(2019 m-2019 m^{2}\\right)}}... | Spain | OME fase local | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
0gr2 | In the chess tournament organized in a school consisting of $2017$ students every two students played at most one match among themselves. At the end of the tournament it turned out that if two students played a match then at least one of them played at most $22$ matches in total. What is the maximal possible number of ... | [
"The answer is $43890 = 1995 \\cdot 22$. Let us divide the students into two groups consisting of $1995$ and $22$ students. If each student played a match with each student of other group and no matches are played between students from the same group then the conditions are satisfied and there are $1995 \\cdot 22$ ... | Turkey | 22nd Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 43890 | |
0hye | Problem:
For any finite set $S$, let $f(S)$ be the sum of the elements of $S$ (if $S$ is empty then $f(S)=0$). Find the sum over all subsets $E$ of $S$ of $\frac{f(E)}{f(S)}$ for $S=\{1,2, \ldots, 1999\}$. | [
"Solution:\n\nAn $n$ element set has $2^{n}$ subsets, so each element of $S$ appears in $2^{1998}$ subsets $E$, so our sum is $2^{1998} \\cdot \\frac{1+2+\\ldots+1999}{1+2+\\ldots+1999} = 2^{1998}$."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2^{1998} | |
0fff | Problem:
Un disco microsurco gira a velocidad de $33 \frac{1}{3}$ revoluciones por minuto y su audición dura 24 min $30 \mathrm{~s}$. La parte grabada tiene $29 \mathrm{~cm}$ de diámetro exterior y 11.5 $\mathrm{cm}$ de diámetro interior. Con estos datos, calcular la longitud del surco grabado. | [
"Solution:\n\nAdoptemos las siguientes notaciones: $R_{0}=29 / 2$, radio exterior del surco; $R_{1}=11.5 / 2$, radio interior del mismo; $\\Delta R=R_{0}-R_{1}=17.5 / 2$; $T=1470 \\mathrm{~s}$, tiempo de audición; $\\omega=33 \\frac{1}{3} \\mathrm{rpm}=10 \\pi / 9 \\mathrm{rad} / \\mathrm{s}$, velocidad angular del... | Spain | Olimpiadas Matemáticas Españolas | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 16537.5π cm (approximately 519.541 m) | |
0bvk | Triangle $ABC$ has $AB = 10$, $AC = 17$ and $BC = 21$. Take on the side $BC$ the points $X, Y, Z$ – the feet of the altitude, the bisector and, respectively, the median from $A$. Find the smallest positive integer $n$ with the property: if the side $BC$ is divided into $n$ equal parts by $n - 1$ points, then $X, Y, Z$ ... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | 378 | |
056v | Let $a > 2$ be an integer. Let
$$x = (a-1) \cdot a^{a-2} + (a-2) \cdot a^{a-3} + \dots + 2 \cdot a^1 + 1 \cdot a^0,$$
$$y = 1 \cdot a^{a-2} + 2 \cdot a^{a-3} + \dots + (a-2) \cdot a^1 + (a-1) \cdot a^0.$$
Prove that $x - 1$ is divisible by $y + 1$. | [
"Note that $x + y = a \\cdot a^{a-2} + a \\cdot a^{a-3} + \\dots + a \\cdot a^1 + a \\cdot a^0 = a^{a-1} + a^{a-2} + \\dots + a^1$.\nHence\n$$\nx + 2y = a^{a-1} + 2 \\cdot a^{a-2} + 3 \\cdot a^{a-3} + \\dots + (a-1) \\cdot a^1 + (a-1) \\cdot a^0 \\\\\n= a(1 \\cdot a^{a-2} + 2 \\cdot a^{a-3} + \\dots + (a-1) \\cdot ... | Estonia | Final Round of National Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0l7i | Problem:
Two points are selected independently and uniformly at random inside a regular hexagon. Compute the probability that a line passing through both of the points intersects a pair of opposite edges of the hexagon. | [
"Solution:\n\nFirst, we compute the probability that the line through two random points in a triangle $ABC$ passes through segments $\\overline{AB}$ and $\\overline{AC}$. We can take an affine transform of the two random points and the triangle such that $ABC$ becomes equilateral. Since the... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 4/9 | |
02c7 | Problem:
Soma das raízes de uma equação - Determine a soma das raízes distintas da equação $x^{2} + 3x + 2 = |x + 1|$. | [
"Solution:\n\nTemos que considerar dois casos.\n\nCaso 1: $x \\geq -1$.\n\nNesse caso, $x^{2} + 3x + 2 = x + 1$, e $x^{2} + 2x + 1 = 0$ que só possui a solução $x = -1$.\n\nCaso 2: $x < -1$.\n\nNesse caso, $x^{2} + 3x + 2 = -x - 1$, logo $x^{2} + 4x + 3 = 0$ que tem, no intervalo, apenas a solução $x = -3$.\n\nAssi... | Brazil | Lista 6 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | -4 | |
0i0a | Problem:
Consider the equation $F O R T Y + T E N + T E N = S I X T Y$, where each of the ten letters represents a distinct digit from $0$ to $9$. Find all possible values of $SIXTY$. | [
"Solution:\n\nSince $Y + N + N$ ends in $Y$, $N$ must be $0$ or $5$. But if $N = 5$ then $T + E + E + 1$ ends in $T$, which is impossible, so $N = 0$ and $E = 5$. Since $F \\neq S$ we must have $O = 9$, $R + T + T + 1 > 10$, and $S = F + 1$. Now $I \\neq 0$, so it must be that $I = 1$ and $R + T + T + 1 > 20$. Thus... | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 31486 | |
017b | The entries of an $8 \times 8$ chessboard are numbered by the numbers $1, 2, \ldots, 64$ in such a way that the sum of the four numbers in each of its parts of one of the forms
is divisible by the same integer $N$. For which of the integers $3$, $4$, $5$ is this possible? | [
"Numbers in cells \"A\" and \"B\" must have the same remainder modulo $N$, because shaded cells are common for two forms (shaded cells + \"A\" and shaded cells + \"B\"). Investigating all possible form placements, we will get that numbers in cells marked by the same lowercase letter must have the same remainder mod... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | 4 | |
0ggh | 一個獵人和一隻隱形的兔子在整點座標平面 $\mathbb{Z}^2 := \{(x, y) : x, y \in \mathbb{Z}\}$ 上玩遊戲 ($\mathbb{Z}$ 為所有整數所成的集合)。遊戲開始前, 獵人先用有限多種顏色, 將 $\mathbb{Z}^2$ 中的每個點各塗上恰一個顏色, 然後兔子在看到獵人的塗色方式後, 秘密地選擇一個點作為起點。在接下來的每分鐘, 兔子都會先告訴獵人牠所在點的顏色, 接著從牠上下左右的相鄰四點中, 秘密地選擇一個牠從未去過的點, 並移動到該點上。
若在有限時間內, 兔子無法再移動, 或是獵人可以確知兔子在遊戲開始時所選的起點, 則獵人獲勝。試問: 是否存在在有限時間內讓獵人獲勝的必... | [
"首先注意到,如果有兩個塗色方法,第一個方法將點 $(x,y)$ 塗 $c_1(x,y)$ 色,第二個方法將點 $(x,y)$ 塗 $c_2(x,y)$ 色,則我們可以令 $c(x,y) = 2^{c_1(x,y)}3^{c_2(x,y)}$,並將 $(x,y)$ 塗 $c(x,y)$ 色,藉此得到同時達到兩個塗色方法功能的塗色法。因此,我們以下給出五種塗色方法,綜合起來可以確知兔子的位置,便達到題目所求。\n\n1. 首先令 $c_1: \\mathbb{Z}^2 \\rightarrow \\{1,2,3\\}$ 為 $c_1(x,y) \\equiv x(\\text{mod } 3)$。此塗色告訴我們兔子是往左、往右... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 獨立研究(一) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Chinese; English | proof and answer | Yes | |
047y | Given nonzero real numbers $\lambda_1, \lambda_2, \dots, \lambda_{2025}$ and a real number $d$. Let $X$ be a finite set of real numbers. Define the sets:
$$
A = \{(x_1, \dots, x_{2025}) \in X^{2025} \mid \lambda_1 x_1 + \dots + \lambda_{2025} x_{2025} = d\};
$$
$$
B = \{(x_1, \dots, x_{2024}) \in X^{2024} \mid x_1 + \d... | [
"**Proof 1:** Let $\\Lambda$ be the set consisting of $\\pm\\lambda_i$ for $1 \\le i \\le 2025$. For positive integer $n$, define functions $S, K: \\Lambda^{2n} \\to \\mathbb{Z}_{\\ge 0}$ as:\n$$\nS(c_1, \\dots, c_{2n}) = \\#\\{(x_1, \\dots, x_{2n}) \\in X^{2n} \\mid c_1 x_1 + \\dots + c_{2n} x_{2n} = 0\\},\n$$\n$$... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
029r | Problem:
A superfície do globo terrestre consiste de água (70\%) e de terra (30\%). Dois quintos da terra são desertos ou cobertos por gelo e, um terço é pastagem, floresta ou montanha; o resto é cultivado. Que percentual da superfície total do globo terrestre é cultivada? | [
"Solution:\nA fração da terra que é cultivada é\n$$\n1-\\frac{2}{5}-\\frac{1}{3}=\\frac{15-6-5}{15}=\\frac{4}{15}\n$$\nComo a terra é $\\frac{3}{10}$ do globo, temos que área cultivada é $\\frac{4}{15} \\times \\frac{3}{10}=\\frac{2}{25}$ do globo, isto é o $\\frac{2}{25} \\times 100 \\%=8 \\%$ do globo terrestre."... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 8% | |
0gna | Show that
$$
\frac{1}{ab + 2c^2 + 2c} + \frac{1}{bc + 2a^2 + 2a} + \frac{1}{ca + 2b^2 + 2b} \geq \frac{1}{ab + bc + ca}
$$
for all positive real numbers $a$, $b$, $c$ satisfying the equality $a + b + c = 1$. | [
"Using the condition $a + b + c = 1$ and the inequality $a^2 + b^2 \\ge 2ab$, we get\n$$\n\\begin{aligned}\n(ab + bc + ca)^2 &= a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc + 2b^2ca + 2c^2ab \\\\\n&= a^2b^2 + (a^2 + b^2)c^2 + 2abc(a + b + c) \\\\\n&\\ge a^2b^2 + 2abc^2 + 2abc.\n\\end{aligned}\n$$\nWriting this inequality in t... | Turkey | Team Selection Test for IMO 2007 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
00cb | Decimos que tres enteros positivos $a$, $b$, $c$ forman una *familia* si se cumplen las siguientes dos condiciones
$$
\bullet\ a+b+c=900;
$$
• existe un entero $n$, $n \ge 2$, tal que $\frac{a}{n-1} = \frac{b}{n} = \frac{c}{n+1}$.
Hallar la cantidad de familias que hay. | [] | Argentina | Nacional OMA 2019 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof and answer | 17 | |
0et9 | Marjorie is the drum major of the world's largest marching band, with more than one million members. She would like the band members to stand in a square formation. To this end, she determines the smallest integer $n$ such that the band would fit in an $n \times n$ square and lets the members form rows of $n$ people. H... | [
"The answer is $1000977$. Let $M$ be the number of members of the marching band. We prove by induction that Marjorie's approach always yields a perfect square at some point, unless $M$ is of the form $M = (2^a + b)^2 + 2b + 1$ ($a, b$ nonnegative integers, $0 \\le b < 2^a$) or $(2^a + b)^2 + 2^a + 3b + 2$ ($a, b$ n... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 1000977 | |
008l | Determine all the natural numbers $n$ for which there are $2n$ different positive integers $x_1, \dots, x_n, y_1, \dots, y_n$ such that the product
$$
(11x_1^2 + 12y_1^2)(11x_2^2 + 12y_2^2) \dots (11x_n^2 + 12y_n^2)
$$
is a perfect square. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | all even natural numbers n | |
0g0n | Problem:
Sei $ABC$ ein Dreieck mit $\angle BCA = 90^\circ$ und $H$ der Höhenfusspunkt von $C$. Sei $D$ ein Punkt innerhalb des Dreiecks $BCH$, sodass $CH$ die Strecke $AD$ halbiert. Sei $P$ der Schnittpunkt der Geraden $BD$ und $CH$. Sei $\omega$ der Halbkreis mit Durchmesser $BD$, der die Strecke $CB$ schneidet. Die ... | [
"Solution:\n\nSei $K$ die Projektion von $D$ auf $AB$. Dann gilt $AH = HK$. Da $PH \\parallel DK$ haben wir\n$$\n\\frac{PD}{PB} = \\frac{HK}{HB} = \\frac{AH}{HB}\n$$\nSei $L$ die Projektion von $Q$ auf $DB$. Da $PQ$ die Tangente an $\\omega$ ist, und $\\angle DQB = \\angle BLQ = 90^\\circ$, folgt $\\angle PQD = \\a... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h40 | On the sides $AB$ and $AC$ of an acute triangle $ABC$ triangles $ABC_1$ and $AB_1C$ are constructed outside with $\angle BAC_1 = \angle B_1AC = 30^\circ$, $\angle AC_1B = \angle CB_1A = 60^\circ$. A point $A_1$ inside $ABC$ is such that $\angle CBA_1 = \angle BCA_1 = 30^\circ$. Prove that the points $A_1$, $B_1$ and $C... | [
"Побудуємо на стороні $BC$ даного трикутника зовні нього рівносторонній трикутник $BCA_2$. Унаслідок умови задачі, $\\angle ABC_1 = \\angle ACB_1 = 90^\\circ$ і $\\angle A_1BA_2 = \\angle A_1CA_2 = 90^\\circ$, а внаслідок рівності трикутників $A_1BA_2$ і $A_1CA_2$ (за двома катетами) маємо: $\\angle A_1A_2B = \\ang... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points"
] | English | proof only | null | |
0854 | Problem:
Quale fra le seguenti espressioni è equivalente a $(x+y+z)^{3}-x^{3}-y^{3}-z^{3}$?
(A) $3 x^{2}(y+z)+3 y^{2}(x+z)+3 z^{2}(x+y)$
(B) $3 x(y+z)^{2}+3 y(x+z)^{2}+3 z(x+y)^{2}$
(C) $3(x+y)(x+z)(y+z)$
(D) $3 x\left(y^{2}+z^{2}\right)+3 y\left(x^{2}+z^{2}\right)+3 z\left(x^{2}+y^{2}\right)$
(E) $3 x y(1-z)+3 x ... | [
"Solution:\nLa risposta è $(\\mathbf{C})$. Svolgendo il cubo, si ha\n$$(x+y+z)^{3}-x^{3}-y^{3}-z^{3}=6 x y z+3 x^{2} y+3 x y^{2}+3 x^{2} z+3 x z^{2}+3 y^{2} z+3 y z^{2}$$\n$$=3 x y(x+y+z)+3 x z(x+y+z)+3 y z(y+z)$$\n$$=3 x(x+y+z)(y+z)+3 y z(y+z)$$\n$$=3(y+z)(x(x+y+z)+y z)$$\n$$=3(y+z)(x(x+y)+z x+z y)$$\n$$=3(y+z)(x(... | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | MCQ | C | |
01cb | For $n \ge 2$, an equilateral triangle is divided into $n^2$ congruent smaller equilateral triangles. Determine all ways in which real numbers can be assigned to the $\frac{n(n+1)}{2}$ vertices so that three such numbers sum to zero whenever the three vertices form an equilateral triangle with edges parallel to the sid... | [
"We label the vertices (and the corresponding real numbers) as follows.\n\n\nFor $n = 2$, the only requirement is obviously $a_1 = -a_2 - a_3$.\n\nFor $n = 3$, we see that\n$$\na_2 + a_4 + a_5 = 0 = a_2 + a_3 + a_5,\n$$\nwhich shows that $a_3 = a_4$ and similarly $a_1 = a_5$ and $a_2 = a_6$... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | Let n ≥ 2, with values assigned to the n(n+1)/2 lattice points.
- n = 2: The only condition is that the three corner values sum to zero; two can be chosen arbitrarily and the third is their negative sum.
- n = 3: The six points split into three pairs of equal values; if the three pair-values are p, q, r, then p + q + r... | |
03th | Let $T = \{1, 2, 3, 4, 5, 6, 7, 8\}$. Find the number of all nonempty subsets $A$ of $T$ such that $3|S(A)$ and $5 \nmid S(A)$, where $S(A)$ is the sum of all elements of $A$. | [
"Define $S(\\emptyset) = 0$. Let $T_0 = \\{3, 6\\}$, $T_1 = \\{1, 4, 7\\}$, $T_2 = \\{2, 5, 8\\}$. For $A \\subseteq T$, let $A_0 = A \\cap T_0$, $A_1 = A \\cap T_1$, $A_2 = A \\cap T_2$, then\n\n$$\nS(A) = S(A_0) + S(A_1) + S(A_2) \\equiv |A_1| - |A_2| \\pmod{3},\n$$\n\nSo $3|S(A)$ if and only if $|A_1| \\equiv |A... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Other"
] | English | proof and answer | 70 | |
00za | Problem:
The graph of the function $f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ (where $n>1$), intersects the line $y=b$ at the points $B_{1}, B_{2}, \ldots, B_{n}$ (from left to right), and the line $y=c$ ($c \neq b$) at the points $C_{1}, C_{2}, \ldots, C_{n}$ (from left to right). Let $P$ be a point on the lin... | [
"Solution:\n\nLet the points $B_{i}$ and $C_{i}$ have the coordinates $(b_{i}, b)$ and $(c_{i}, c)$, respectively, for $i=1,2, \\ldots, n$. Then we have\n$$\ncot \\angle B_{1} C_{1} P+\\cdots+\\cot \\angle B_{n} C_{n} P=\\frac{1}{b-c} \\sum_{i=1}^{n}\\left(b_{i}-c_{i}\\right)\n$$\nThe numbers $b_{i}$ and $c_{i}$ ar... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 0 | |
0csg | Учитель записал Пете в тетрадь четыре различных натуральных числа. Для каждой пары этих чисел Петя нашёл их наибольший общий делитель. У него получились шесть чисел: 1, 2, 3, 4, 5 и $N$, где $N > 5$. Какое наименьшее значение может иметь число $N$? | [
"Число $N$ может равняться 14, как показывает, например, четвёрка чисел 4, 15, 70, 84. Осталось показать, что $N \\ge 14$.\n\n**Лемма.** Среди попарных НОД четырёх чисел не может быть ровно двух чисел, делящихся на некоторое натуральное $k$.\n**Доказательство.** Если среди исходных четырёх чисел есть не больше двух... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 14 | |
07hj | Is it possible to write $1400$ natural numbers (not necessarily distinct) around a circle such that $2021$ is used at least once and each number is the sum of the greatest common divisor of the two previous numbers and the greatest common divisor of the two next numbers? For example, if $a, b, c, d, e$ are five consecu... | [
"We shall show that this is impossible. Assume that there are such $1400$ numbers. First of all, note that if we divide all the numbers by $k$ then the new numbers satisfy the second condition and one of them is a divisor of $2021$, so we can assume that the greatest common divisor of the numbers is $1$.\n\n**Lemma... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | No, it is impossible. | |
034l | Problem:
Let $H$ be the orthocenter of $\triangle ABC$, $M$ be the midpoint of $AB$ and $H_1$ and $H_2$ be the feet of the perpendiculars from $H$ to the inner and the outer bisector of $\Varangle ACB$, respectively. Prove that the points $H_1$, $H_2$ and $M$ are colinear. | [
"Solution:\nDenote by $D$ and $E$ the feet of the altitudes of $\\triangle ABC$ from the vertices $A$ and $B$, respectively. The quadrilateral $HDCE$ is inscribed in the circle with diameter $CH$. The points $H_1$ and $H_2$ are the midpoints of the two arcs $\\overparen{DE}$ since $CH_1$ and $CH_2$ are the internal... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0foq | Problem:
En la pizarra está escrito un número entero. Dos jugadores $A$ y $B$ juegan alternadamente, empezando por $A$. Cada jugador en su turno reemplaza el número existente por el que resulte de realizar una de estas dos operaciones: restar 1 o dividir entre 2, siempre que se obtenga un entero positivo. El jugador q... | [
"Solution:\n\nSi el valor de $N$ inicial es par, veamos que gana $A$: ya sea restando 1 o bien dividiendo entre 2 (en el caso $N=4k+2$, $\\frac{N}{2}=2k+1$ impar), $A$ siempre le dejará a $B$ un impar, obligándolo a restar 1 por no poder dividir entre 2, con lo cual al jugar $A$ volverá a encontrarse con un par, me... | Spain | LI Olimpiada matemática Española (Concurso Final) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 6*2^{1007} - 4 | |
0f9m | Problem:
For which positive integers $n$ is $3^{2n + 1} - 2^{2n + 1} - 6^n$ composite? | [
"Solution:\n\nAnswer all $n \\neq 1$\n\n$3^{2n + 1} - 2^{2n + 1} - 6^n = (3^n - 2^n)(3^{n + 1} + 2^{n + 1})$, so it is certainly composite for $n > 1$. For $n = 1$, it is $27 - 8 - 6 = 13$, which is prime."
] | Soviet Union | 24th ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | all n > 1 | |
07wp | Alice and Brian play a game. Before they start, Alice chooses a positive integer $n$ and Brian then chooses the initial value of another positive integer $m$. They then place $n$ counters on a board and take turns removing them; Alice always moves first.
Each move, for some player $P$, always has two parts:
(a) $P$ eit... | [
"We will first define a function $f: \\mathbb{N} \\times \\mathbb{Z}_{\\ge 0} \\to \\{0, 1\\}$, where $\\mathbb{Z}_{\\ge 0}$ indicates the non-negative integers, and then show that if the current player is faced with a particular pair of values $(m, n)$ then that player has a guaranteed win if $f(m, n) = 1$ but a g... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2026 | |
08al | Problem:
Dimostrare che esiste un intero positivo che può essere scritto come somma di 2015 potenze 2014-esime distinte di interi positivi $x_{1}<x_{2}<\cdots<x_{2015}$ in almeno due modi. | [] | Italy | Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0a2o | What is the smallest integer $a > 10$ for which:
* $a$ is divisible by $10$ and
* $a + 1$ is divisible by $11$ and
* $a + 2$ is divisible by $12$? | [] | Netherlands | Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | final answer only | 670 | |
08pe | Problem:
Determine the largest positive integer $n$ that divides $p^{6}-1$ for all primes $p>7$. | [
"Solution:\nNote that\n$$\np^{6}-1=(p-1)(p+1)\\left(p^{2}-p+1\\right)\\left(p^{2}+p+1\\right)\n$$\nFor $p=11$ we have\n$$\np^{6}-1=1771560=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 37\n$$\nFor $p=13$ we have\n$$\np^{6}-1=2^{3} \\cdot 3^{2} \\cdot 7 \\cdot 61 \\cdot 157\n$$\nFrom the last two calculation... | JBMO | Junior Balkan Mathematics Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 504 | |
09xl | On a rectangular board consisting of $m \times n$ squares ($m, n \ge 3$), dominos have been placed (2 × 1- or 1 × 2-tiles), not overlapping each other. Each domino covers exactly two squares of the board. Suppose that the placement of the dominos has the property that no extra domino can be placed on the board, and the... | [
"Assign each empty square to the domino directly right of this square (unless the square is on the right edge of the board). Now suppose that two empty squares are assigned to the same domino, then this domino must be placed vertically and both squares left of this domino are empty. However, that would mean that an... | Netherlands | IMO Team Selection Test 1 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof only | null | |
00p9 | Given an integer number $n \ge 3$, determine the maximum value the product of $n$ non-negative real numbers $x_1, x_2, \dots, x_n$ may achieve, subject to
$$
\frac{x_1}{1+x_1} + \frac{x_2}{1+x_2} + \dots + \frac{x_n}{1+x_n} = 1.
$$ | [
"The required maximum is $1/(n-1)^n$ and is achieved if and only if the $x_i$ are all equal to $1/(n-1)$.\nThe constraint on the $x_i$ is equivalent to\n$$\n\\sum_{k=1}^{n} (k-1)\\sigma_k = 1,\n$$\nwhere\n$$\n\\sigma_k = \\sum_{1 \\le i_1 < \\dots < i_k \\le n} x_{i_1} \\cdots x_{i_k}, \\quad k = 1, 2, \\dots, n.\n... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 1/(n-1)^n | |
0g7y | 設三角形 $ABC$ 的外心為 $O$, 內心為 $I$。$D, E, F$ 三點分別位於 $BC, CA, AB$ 邊上, 並滿足 $BD + BF = CA$ 且 $CD + CE = AB$。令三角形 $BFD$ 的外接圓與三角形 $CDE$ 的外接圓的兩交點為 $D, P$。試證: $OP = OI$。 | [
"以下用 $(XYZ)$ 代表 $\\triangle XYZ$ 的外接圓。設 $D, E, F$ 三點分別在 $BC, CA, AB$ 邊上。由 Miquel 定理，$(AEF) = \\omega_A$，$(BFD) = \\omega_B$，$(CDE) = \\omega_C$ 三個外接圓有共同點 $P \\neq D$。\n設 $\\omega_A, \\omega_B, \\omega_C$ 分別交 $AI, BI, CI$ 於點 $A \\neq A', B \\neq B', C \\neq C'$。此處的要點是 $A', B', C'$ 這三點並不依賴於點 $D, E, F$，只要它們滿足 $BD + BF... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 模擬競賽(二) | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Mis... | null | proof only | null | |
070o | Problem:
$P_{1}, P_{2}, \ldots, P_{n}$ are points in the plane and $r_{1}, r_{2}, \ldots, r_{n}$ are real numbers such that the distance between $P_{i}$ and $P_{j}$ is $r_{i} + r_{j}$ (for $i$ not equal to $j$). Find the largest $n$ for which this is possible. | [
"Solution:\n\nDraw a circle radius $r_{i}$ at $P_{i}$. Then each pair of circles must touch. But that is possible iff $n \\leq 4$.\n\n"
] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 4 | |
0h36 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$
$$
f(xy) + f(xz) \geq f(x)f(yz) + 1.
$$ | [
"Покладемо у вихідній нерівності $x = y = z = 0$. Тоді $(f(0) - 1)^2 \\le 0$, тобто $f(0) = 1$.\n\nВізьмемо $y = z = 0$ і одержимо, що $f(0) + f(0) \\ge f(x)f(0) + 1$. Звідси $f(x) \\le 1$ для всіх $x \\in \\mathbb{R}$.\n\nЯкщо $x = y = z = 1$, то $(f(1) - 1)^2 \\le 0$, $f(1) = 1$.\n\nДля $y = z = 1$ маємо: $f(x) +... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1 for all real x | |
03vl | Find the largest positive number $\lambda$ such that
$$
| \lambda xy + yz | \le \frac{\sqrt{5}}{2}, \text{ where } x^2 + y^2 + z^2 = 1.
$$ | [
"Note that\n$$\n\\begin{aligned}\n1 &= x^2 + y^2 + z^2 \\\\\n &= x^2 + \\frac{\\lambda^2}{1+\\lambda^2}y^2 + \\frac{1}{1+\\lambda^2}y^2 + z^2 \\\\\n &\\ge \\frac{2}{\\sqrt{1+\\lambda^2}}(|x| + |y| + |z|) \\\\\n &\\ge \\frac{2}{\\sqrt{1+\\lambda^2}}(|\\lambda xy + yz|),\n\\end{aligned}\n$$\nand the two equalities... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 2 | |
0be5 | Problem:
Az $A$ és $B$ másodrendű négyzetes mátrixok teljesítik az
$$
AB = A^2 B^2 - (AB)^2 \text{ és } \quad \operatorname{det}(B) = 2
$$
összefüggéseket.
a) Igazold, hogy az $A$ mátrix nem invertálható!
b) Számítsd ki a $\operatorname{det}(A+2B) - \operatorname{det}(B+2A)$ különbséget! | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | 6 | |
0ap3 | Problem:
Let $ABC$ be an acute-angled triangle. Let $D$ and $E$ be points on $\overline{BC}$ and $\overline{AC}$, respectively, such that $\overline{AD} \perp \overline{BC}$ and $\overline{BE} \perp \overline{AC}$. Let $P$ be the point where $\overrightarrow{AD}$ meets the semicircle constructed outwardly on $\overlin... | [
"Solution:\n\n(This problem is taken from the British Mathematical Olympiad 2005.) Refer to Figure 3. By the Pythagorean Theorem, we have $QC^{2} = EQ^{2} + EC^{2}$. On the other hand, since $\\triangle AQC$ is right-angled at $Q$ and $\\overline{QE} \\perp \\overline{AC}$, we have $EQ^{2} = AE \\cdot EC$. It follo... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01s0 | Let $AM$ be the median of the triangle $ABC$, $B_1$ be the foot of the perpendicular $BB_1$ from $B$ onto the bisector of the angle $BMA$, $C_1$ be the foot of the perpendicular $CC_1$ from $C$ onto the bisector of the angle $AMC$. The ray $MA$ intersects the segment $B_1C_1$ at a point $A_1$.
Find the value of the rat... | [
"Answer: 1.\nLet $\\angle BMA = 2x$, then $\\angle AMC = 180^\\circ - 2x$. Since $MB_1$ is the bisector of the angle $BMA$, we have $\\angle BMB_1 = \\angle B_1MA = x$.\nSimilarly, $\\angle AMC_1 = \\angle C_1MC = 90^\\circ - x$. The triangle $BMB_1$ is a right-angled triangle, so $\\angle B_1BM = 90^\\circ - \\ang... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 1 | |
0isq | Problem:
Let $S$ be the smallest subset of the integers with the property that $0 \in S$ and for any $x \in S$, we have $3x \in S$ and $3x+1 \in S$. Determine the number of non-negative integers in $S$ less than $2008$. | [
"Solution:\nWrite the elements of $S$ in their ternary expansion (i.e. base $3$). Then the second condition translates into: if $\\overline{d_{1} d_{2} \\cdots d_{k}} \\in S$, then $\\overline{d_{1} d_{2} \\cdots d_{k} 0}$ and $\\overline{d_{1} d_{2} \\cdots d_{k} 1}$ are also in $S$. It follows that $S$ is the set... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 128 | |
06cq | $k$ cards numbered $1$ to $k$ are arranged at random in a row. In a move, we may change any block of consecutive cards whose numbers are in ascending or descending order and switch the block around. For example, in the case $k = 4$, $4132$ may be changed to $1432$ and $3214$ may be changed to $1234$. Suppose that in at... | [
"a. Consider the same problem but we require the numbers at the end to be in ascending order (i.e. descending order is not allowed). Let $m_k$ be the least number of moves needed to guarantee that this can be done when there are $k$ numbers. By symmetry, this number is the same if we require the numbers at the end ... | Hong Kong | Test 1 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n4 = 3; n5 = 4 | |
0l60 | Four unit squares form a $2 \times 2$ grid. Each of the $12$ unit line segments forming the sides of the squares is colored either red or blue in such a way that each unit square has $2$ red sides and $2$ blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings.
![](attac... | [
"Call the $8$ unit segments along the outside of the figure *outer edges*, and call the remaining $4$ unit segments *inner edges*. The count can be organized by the number of red inner edges.\n* If all $4$ of the inner edges are red, then in order for each square to be bounded by $2$ red segments and $2$ blue segme... | United States | AIME II | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 82 | |
0j0i | Problem:
You are the general of an army. You and the opposing general both have an equal number of troops to distribute among three battlefields. Whoever has more troops on a battlefield always wins (you win ties). An order is an ordered triple of non-negative real numbers $(x, y, z)$ such that $x+y+z=1$, and correspo... | [
"Solution:\n\n$\\boxed{\\frac{5}{8}}$\n\nLet $x$ be the portion of soldiers the opposing general sends to the first battlefield, and $y$ the portion he sends to the second. Then $1-x-y$ is the portion he sends to the third. Then $x \\geq 0$, $y \\geq 0$, and $x+y \\leq 1$. Furthermore, you win if one of the three c... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 5/8 | |
0eiy | Problem:
Vsi liki na slikah so omejeni s polkrožnimi loki, pri čemer je največji polkrožni lok pri vseh likih enak. Kateri izmed likov z najmanjšim obsegom ima največjo ploščino?
(A)
(B)
(C)
(D)
(E)
![](attached_ima... | [
"Solution:\n\nSkupna dolžina polkrožnih lokov nad daljico dolžine $d$ je enaka $\\frac{\\pi d}{2}$ neglede na to, koliko je teh polkrožnih lokov in kako veliki so (glej primer na sliki).\n\nTorej je obseg vseh 5 likov enak, saj je enak $2 \\cdot \\frac{\\pi d}{2}$, kjer je $d$ premer največ... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | MCQ | C | |
06j7 | Find the remainder when $19^{17^{15}}$ is divided by 100. | [
"For convenience we define $f(1) = 1$ and $f(n + 2) = (n + 2)^{f(n)}$ for odd positive integers $n$. Then the question asks for the last two digits of $f(19)$.\n\nSince $f(13) = 13^{f(11)}$ is odd, we have $f(15) = 15^{f(13)} \\equiv (-1)^{f(13)} = -1 \\equiv 3 \\pmod 4$.\n\nAs $f(17) = 17^{f(15)}$ and the units di... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | final answer only | 59 | |
099o | Let $f(x)$ be a polynomial with integer coefficients, which has no multiple roots and $\deg f \ge 1$. Prove that for every positive integer $k$,
$$
P_k(f) = |\{p \in \mathbb{P} \mid \mathrm{ord}_p(f(x)) = k, \exists x \in \mathbb{Z}\}| = \infty.
$$
(proposed by G. Batzaya) | [
"**Lemma.** Let $f(x)$ be a polynomial of degree greater than or equal to $1$.\n$$\n\\mathbb{P}'(f(x)) = |\\{p \\in \\mathbb{P} \\mid p|f(x), \\exists x \\in \\mathbb{Z}\\}| = \\infty\n$$\n**Proof of lemma:** Assume $f(x) = b_kx^k + \\dots + b_0$, $k \\ge 1$. Suppose the contrary, $\\mathbb{P}'(f(x)) < \\infty$; we... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0031 | Sea $k$ un entero positivo. Demostrar que para todo $n > k$ se verifica lo siguiente:
Existen figuras convexas $F_1, \dots, F_n$ y $F$ tales que ningún subconjunto de $k$ figuras elegidas entre $F_1, \dots, F_n$ cubre por completo a $F$, pero todo subconjunto de $k+1$ figuras elegidas entre $F_1, \dots, F_n$ cubre por ... | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | Español | proof only | null | |
09by | $a_1 = 1,\ a_n = \frac{n+1}{n-1} (a_1 + a_2 + \dots + a_{n-1})\ (n > 1)$ байх $a_n$ дарааллын $a_{2010}$-г ол. | [
"$a_{n-1} = \\frac{n}{n-2}(a_1 + \\dots + a_{n-2})$ нөхцөлөөс\n$$\na_1 + \\dots + a_{n-2} = \\frac{n-2}{n} a_{n-1}\n$$\nболох ба үүнийг анхны рекурент томъёонд орлуулбал\n$$\na_n = \\frac{n+1}{n-1} \\left( \\frac{n-2}{n} a_{n-1} + a_{n-1} \\right) =\n$$\n$$\n= \\frac{n+1}{n-1} \\cdot \\frac{2n-2}{n} \\cdot a_{n-1} ... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Mongolian | proof and answer | 2^{2008} \cdot 2011 | |
0gi1 | Find all positive integers $n \ge 3$ such that
$$
n! \prod_{\substack{p<q \le n, \\ p,q \text{ 為質數}}} (p+q).
$$
That is, $n!$ divides the product $\prod_{p<q \le n,\ p,q\ \text{為質數}} (p+q)$.
Find all positive integers $n \ge 3$ such that
$$
n! \prod_{\substack{p<q \le n, \\ p,q \text{ primes}}} (p+q).
$$
That is, $n!... | [
"This only holds for $n = 7$.\n\nLet $2 = p_1 < p_2 < \\cdots < p_m \\le n$ be the primes $\\le n$. Each such prime divides $n!$.\n\nIn particular, $p_m \\mid p_i + p_j$ for some $p_i < p_j \\le n$. But\n$$\n0 < \\frac{p_i + p_j}{p_m} < 2,\n$$\nso $p_m = p_i + p_j$ which implies $m \\ge 3$, $p_i = 2$, and $p_m = 2 ... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Chinese (Traditional) | proof and answer | n = 7 | |
0b8i | Let $G$ be a finite group of order $n$. Define the set
$$
H = \{x : x \in G \text{ and } x^2 = e\},
$$
where $e$ is the neutral element of $G$. Let $p = |H|$ be the cardinal of $H$. Prove that
a) $|H \cap xH| \geq 2p - n$, for any $x \in G$, where $xH = \{xh : h \in H\}$.
b) If $p > \frac{3n}{4}$, then $G$ is commuta... | [
"a) Since $G$ is a group, it follows that $|xH| = p$. Therefore\n$$\nn = |G| \\ge |H \\cup xH| = |H| + |xH| - |H \\cap xH| = 2p - |H \\cap xH|,\n$$\nwhence $|H \\cap xH| \\ge 2p - n$.\n\nb) Let have $x \\in H$ and $y \\in H \\cap xH$. Then $y = y^{-1}$ and $y = xh$, $h \\in H$. Since $xy = x^2h = h \\in H$, it foll... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof only | null | |
0ivi | Problem:
Five cards labeled $A$, $B$, $C$, $D$, and $E$ are placed consecutively in a row. How many ways can they be re-arranged so that no card is moved more than one position away from where it started? (Not moving the cards at all counts as a valid re-arrangement.) | [
"Solution:\n\nThe only things we can do is leave cards where they are or switch them with adjacent cards. There is 1 way to leave them all where they are, 4 ways to switch just one adjacent pair, and 3 ways to switch two different adjacent pairs, for 8 possibilities total."
] | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 8 | |
04z8 | Four musketeers together bought a plot of rectangular shape and paid for it equally. They divided the plot by two cuts into four pieces of rectangular shape, from which every musketeer got one. It turned out that one musketeer obtained as much land as the other three in total. Prove that the price per acre of one muske... | [
"Let $a$ and $b$ be the side lengths of the plot. Assume that the cuts divided the side of length $a$ to parts of length $x$ and $a-x$ where $x$ being the greater part, and the side of length $b$ to parts of length $y$ and $b-y$ where $y$ being the greater part. Then the area of the largest piece was $xy$. The cond... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
077n | Let $N \ge 3$ be an integer. In the country of Sibyl, there are $N^2$ towns arranged as the vertices of an $N \times N$ grid, with each pair of towns corresponding to an adjacent pair of vertices on the grid connected by a road. Several automated drones are each given the instruction to traverse a rectangular path star... | [
"**Answer:** The desired minimum is $N$ if $N$ is odd, and $(N-1)$ if $N$ is even.\nWe say a drone covers a vertex if it takes a $90^\\circ$ turn at that vertex.\n\n**Construction:** Let $v_1, \\dots, v_N$ denote the $N$ vertical lines and $h_1, \\dots, h_N$ denote the $N$ horizontal lines of the grid.\nFor $N = 2n... | India | EGMO TST Day 1 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Minimum number of drones is N if N is odd, and N−1 if N is even. | |
078v | Let $p$ be an odd prime number and $a, b, c$ be integers so that the integers
$$
a^{2023} + b^{2023}, \quad b^{2024} + c^{2024}, \quad c^{2025} + a^{2025}
$$
are all divisible by $p$. Prove that $p$ divides each of $a, b$, and $c$. | [
"Set $k = 2023$. If one of $a, b, c$ is divisible by $p$, then all of them are. Indeed, for example, if $p \\mid a$, then $p \\mid a^k + b^k$ implies $p \\mid b$, and then $p \\mid b^{k+1} + c^{k+1}$ implies $p \\mid c$. The other cases follow similarly.\nSo for the sake of contradiction assume none of $a, b, c$ is... | India | INMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0laq | For each positive integer $n$, consider the set
$$
T_n = \{ 11(h + k) + 10(n^h + n^k) \mid h, k \in \mathbb{N} \text{ và } 1 \le h \le k \le 10 \}.
$$
Find all positive integers $n$, such that the set $T_n$ has the following property: For all $a, b$ in $T_n$ with $a \ne b$, we always have $a - b$ is not divisible by $1... | [] | Vietnam | Vietnamese Team Selection for IMO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | All positive integers n with n ≡ 2, 6, 7, or 8 (mod 11). | |
09ue | Inside an equilateral triangle, a circle is drawn that touches all three sides. The radius of the circle is $10$. A second, smaller, circle touches the first circle and two sides of the triangle. A third, even smaller, circle touches the second circle and two sides of the triangle (see the figure). What is the radius o... | [
"$\\frac{10}{9}$"
] | Netherlands | First Round, January 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | final answer only | 10/9 | |
0i5b | Problem:
A conical flask contains some water. When the flask is oriented so that its base is horizontal and lies at the bottom (so that the vertex is at the top), the water is 1 inch deep. When the flask is turned upside-down, so that the vertex is at the bottom, the water is 2 inches deep. What is the height of the co... | [
"Solution: \nLet $h$ be the height, and let $V$ be such that $V h^{3}$ equals the volume of the flask. When the base is at the bottom, the portion of the flask not occupied by water forms a cone similar to the entire flask, with a height of $h-1$; thus its volume is $V(h-1)^{3}$. When the base is at the top, the w... | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Volume",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1/2 + sqrt(93)/6 | |
03i2 | Problem:
A right circular cone of base radius $1$ cm and slant height $3$ cm is given. $P$ is a point on the circumference of the base and the shortest path from $P$ around the cone and back to $P$ is drawn (see diagram). What is the minimum distance from the vertex $V$ to this path?
 | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1.5 cm | |
0hbc | Andrew and Olesya in turn cut some squares $2 \times 2$ or $1 \times 1$ from the rectangle $2 \times 2n$, following the lines in such a way that after every turn the remaining figure stays connected. The one who cannot make a move loses. Who is going to win if both children play the best they can, and Olesya is the fir... | [
"We proceed by induction on $n$. One can verify that if the field is $2 \\times 2$ Olesya wins (by cutting entire $2 \\times 2$ square), and if the field is $2 \\times 4$ then Andrew wins (casework). If $n = 2m+1$, then Olesya can cut $2 \\times 2$ square from the end of a strip and place herself in the position of... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Olesya wins when n is odd; Andrew wins when n is even. | |
0bsd | Let $f: \mathbb{R} \to (0, \infty)$ be a continuous periodic function. Under the assumption that $2$ is a period of $f$, prove that:
a.
$$
\int_0^2 \frac{f(x+1)}{f(x)} \, dx \ge 2;
$$
b.
$$
\int_0^2 \frac{f(x+1)}{f(x)} \, dx = 2 \text{ if and only if } 1 \text{ is a period of } f.
$$ | [
"(a) To establish the required inequality write:\n$$\n\\begin{align*}\n\\int_0^2 \\frac{f(x+1)}{f(x)} \\, dx &= \\int_0^1 \\frac{f(x+1)}{f(x)} \\, dx + \\int_1^2 \\frac{f(x+1)}{f(x)} \\, dx \\\\\n&= \\int_0^1 \\frac{f(x+1)}{f(x)} \\, dx + \\int_1^2 \\frac{f(x-1+2)}{f(x-1+1)} \\, dx \\\\\n&= \\int_0^1 \\frac{f(x+1)}... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0caj | Problem:
Fie $\alpha$ un număr din intervalul $(0,1)$. Arătaţi că există un şir de numere $\left(\varepsilon_{n}\right)_{n>1}$, cu valori 0 sau 1, astfel încât şirul $\left(s_{n}\right)_{n \geq 1}$ definit prin
$$
s_{n}=\frac{\varepsilon_{1}}{n(n+1)}+\frac{\varepsilon_{2}}{(n+1)(n+2)}+\ldots+\frac{\varepsilon_{n}}{(2 ... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ Barajul 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
03ex | We will call a natural number Yambolian if it can be represented in the form $a^2 + 6ab + b^2$, where $a$ and $b$ are (not necessarily different) natural numbers. The number $36^{2024}$ is written as the sum of $k$ number of (not necessarily distinct) Yambol numbers. What is the smallest possible value of $k$?
(Mirosla... | [
"We will first show that $k = 1$ is not possible, i.e. $a^2 + 6ab + b^2 = 2^{4048} \\cdot 3^{4048}$ has no solution in natural numbers. If such $a, b$ exist, then $a^2 + b^2$ is divisible by $3$ and therefore $a$ and $b$ are divisible by $3$. Writing $a = 3a_1$, $b = 3b_1$ and dividing by $3^2$, we get $a_1^2 + 6a_... | Bulgaria | Bulgarian Winter Tournament | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 2 | |
0k2v | Problem:
How many noncongruent triangles are there with one side of length $20$, one side of length $17$, and one $60^{\circ}$ angle? | [
"Solution:\nThere are $3$ possible vertices that can have an angle of $60^{\\circ}$, we will name them. Call the vertex where the sides of length $20$ and $17$ meet $\\alpha$, denote the vertex where $17$ doesn't meet $20$ by $\\beta$, and the final vertex, which meets $20$ but not $17$, we denote by $\\gamma$.\n\n... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2 | |
0a88 | Problem:
Consider $7$-gons inscribed in a circle such that all sides of the $7$-gon are of different length. Determine the maximal number of $120^{\circ}$ angles in this kind of a $7$-gon. | [
"Solution:\n\nIt is easy to give examples of heptagons $A B C D E F G$ inscribed in a circle with all sides unequal and two angles equal to $120^{\\circ}$. These angles cannot lie on adjacent vertices of the heptagon. In fact, if $\\angle A B C = \\angle B C D = 120^{\\circ}$, and arc $B C$ equals $b^{\\circ}$, the... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 13 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2 |
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