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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01lf | Let $BK$ be the bisector of the angle $B$ of a triangle $ABC$. Find the angles of the triangle $ABC$ if $AK = 1$, $BK = KC = 2$. | [
"We mark the point $L$ on the extension of $BA$ over $A$ so that $AL = AB$. Let $M$ be the point of intersection of the medians of $LBC$. Since $AK : KC = 1 : 2$, point $K$ of the lines $BK$ and $CL$. Then $CA$ is the point of intersection of the medians of $LBC$. Therefore, $BM$ is also the median of $LBC$. Since ... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ∠A = 90°, ∠B = 60°, ∠C = 30° | |
0awy | Problem:
Let $S$ be a subset of $\{1,2, \ldots, 2017\}$ such that no two elements of $S$ have a sum divisible by $37$. Find the maximum number of elements that $S$ can have. | [
"Solution:\nLet us consider the set $\\{1,2,\\ldots,2017\\}$ modulo $37$. Since $2017 = 37 \\times 54 + 19$, the residues modulo $37$ that appear are $1,2,\\ldots,37$ repeated $54$ times each, and $1,2,\\ldots,19$ appearing one more time each.\n\nLet us group the numbers by their residues modulo $37$.\n\nFor each p... | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1046 | |
0ley | Let $ABCD$ be a parallelogram and $AC$ intersects $BD$ at $I$. Let $G$ be the point inside triangle $IAB$ that satisfies
$$
\angle IAG = \angle IBG \neq 45^\circ - \frac{\angle AIB}{4}.
$$
Let $E, F$ be projections of $C$ on $AG$ and $D$ on $BG$. The median respect to vertex $E$ of triangle $BEF$ and the median respect... | [
"a) Let $E'$, $F'$ be the midpoints of $AE$, $BF$. Note that triangles $DFB$ and $CEA$ are right at $F$, $E$ and $I$ is the midpoint of $BD$ and $AC$ then triangles $IBF$, $ICE$ are isosceles at $I$. On the other hand, because $\\angle GAI = \\angle GBI$ then $\\triangle IFB \\sim \\triangle IAE$, which implies\n$$... | Vietnam | IMO Team Selection Test | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fxe | Problem:
Betrachte $n$ kollineare Punkte $P_{1}, \ldots, P_{n}$ und alle Kreise mit Durchmesser $P_{i} P_{j}$ für $1 \leq i<j \leq n$. Jeder dieser Kreise wird mit einer von $k$ Farben gefärbt. Eine solche Menge von gefärbten Kreisen heisst ein $(n, k)$-Gewusel. Eine einfarbige Acht sind zwei Kreise derselben Farbe, d... | [
"Solution:\n\nWir nennen ein Paar $(n, k)$ gut, falls jedes $(n, k)$-Gewusel eine einfarbige Acht enthält. Ist $(n, k)$ gut, dann offenbar auch alle $(n', k)$ mit $n' \\geq n$. Für die Lösung verwenden wir folgende graphentheoretische Uminterpretation des Problems: Für jede endliche Menge $D$ natürlicher Zahlen sei... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | every such configuration contains a monochromatic externally tangent pair if and only if n > 2^k | |
05eo | Problem:
Soient $x, y$ des réels strictement positifs. Montrer que:
$$
x + \frac{y^{2}}{x} \geqslant 2y
$$
et trouver les cas d'égalité. | [
"Solution:\nL'inégalité de la moyenne donne\n$$\nx + \\frac{y^{2}}{x} \\geqslant 2 \\sqrt{x \\cdot \\frac{y^{2}}{x}} = 2y\n$$\nSupposons qu'on a égalité, alors $x = \\frac{y^{2}}{x}$ donc $x^{2} = y^{2}$ donc $x = y$. Réciproquement si $x = y$, $x + \\frac{y^{2}}{x} = y + \\frac{y^{2}}{y} = 2y$ on a bien égalité."
... | France | ENVOI 2 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds if and only if x equals y. | |
019s | Prove the inequality
$$
\frac{a}{a + \sqrt{(a+b)(a+c)}} + \frac{b}{b + \sqrt{(b+c)(b+a)}} + \frac{c}{c + \sqrt{(c+a)(c+b)}} \le 1,
$$
in which $a$, $b$ and $c$ are assumed to be positive numbers. | [
"From the Cauchy–Schwarz Inequality follows\n$$\n\\sqrt{(x + y)(z + x)} \\geq \\sqrt{xz} + \\sqrt{yx},\n$$\nfrom which\n$$\n\\begin{align*}\n& \\frac{a}{a + \\sqrt{(a+b)(a+c)}} + \\frac{b}{b + \\sqrt{(b+c)(b+a)}} + \\frac{c}{c + \\sqrt{(c+a)(c+b)}} \n\\le \\\\\n& \\le \\frac{a}{a + \\sqrt{ab} + \\sqrt{ac}} + \\frac... | Baltic Way | Baltic Way 2013 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
009l | The quadrilateral $ABCD$ in the figure has three angles equal to $45^\circ$, at vertices $A$, $B$ and $C$. ($ABCD$ is not convex). It is allowed to measure the length of exactly one line segment in the figure. Find the area of the quadrilateral.
 | [
"It is enough to measure $BD$ because $(ABCD) = \\frac{1}{2} BD^2$.\nIndeed, extend $AD$ to meet $BC$ at $P$. Since $\\angle ABP = \\angle BAP = 45^\\circ$, we have $AP = BP$ and $\\angle APB = 90^\\circ$. Hence triangle $ABP$ is right and isosceles, so that $(ABP) = \\frac{1}{2} BP^2$.\n\nNext, triangle $CDP$ has ... | Argentina | NATIONAL XXX OMA | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | BD^2/2 | |
0cng | Given an isosceles right-angled triangle $ABC$. Point $K$ is a midpoint of hypotenuse $AB$. Points $L$ and $M$ are chosen on sides $BC$ and $AC$ respectively so that $BL = CM$. Prove that the triangle $LMK$ is also isosceles and right-angled. (R. Zhenodarov) | [
"Медиана $CK$ треугольника $ABC$ является также высотой и биссектрисой, так как треугольник равнобедренный. Поэтому $\\angle KBC = \\angle KCB = \\angle KCA = 45^\\circ$. Отсюда $KC = KB$, и, значит, треугольники $KBL$ и $KCM$ равны по двум сторонам ($KC = KB$, $BL = CM$) и углу между ними. Поэтому $KL = KM$, и из ... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
04k8 | Find all three-digit positive integers $n$ for which the numbers $n$ and $n^2$ coincide in the last three digits. (AIME 2014) | [
"Let $n$ be a three-digit positive integer such that the last three digits of $n^2$ are the same as those of $n$. This means $n^2 \\equiv n \\pmod{1000}$.\n\nSo, $n^2 - n \\equiv 0 \\pmod{1000}$, or $n(n-1) \\equiv 0 \\pmod{1000}$.\n\nSince $n$ and $n-1$ are consecutive integers, they are coprime. Therefore, $1000 ... | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 376, 625 | |
0kv1 | Problem:
A real number $x$ satisfies $9^{x} + 3^{x} = 6$. Compute the value of $16^{1 / x} + 4^{1 / x}$. | [
"Solution:\nSetting $y = 3^{x}$ in the given equation yields\n\n$$\ny^{2} + y = 6 \\Longrightarrow y^{2} + y - 6 = 0 \\Longrightarrow y = -3, 2\n$$\n\nSince $y > 0$ we must have\n\n$$\n3^{x} = 2 \\Longrightarrow x = \\log_{3}(2) \\Longrightarrow 1 / x = \\log_{2}(3)\n$$\n\nThis means that\n\n$$\n16^{1 / x} + 4^{1 /... | United States | HMMT November 2023 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 90 | |
04me | Let $ABC$ be a right-angled triangle with the right angle at $C$, having sides of length $|AB| = 26$ and $|BC| = 24$. A semicircle with diameter on $\overline{BC}$ containing point $C$ is inscribed in the triangle. The semicircle touches the side $\overline{AB}$. What is the radius of the semicircle? (Nikola Adžaga) | [] | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 20/3 | |
007s | Let $n$ be a positive integer. We call *smooth* a sequence of integers $a_1, a_2, \dots, a_k$, with $1 \le a_i \le n$, if there exists an integer $m$, with $1 \le m < k$, such that $a_1 = a_{k-m+1}, a_2 = a_{k-m+2}, \dots, a_m = a_k$. Furthermore a sequence is *universal* if each of the sequences obtained through repla... | [
"The minimum length in question is $2^n$.\n\nDelete the last term of a given universal sequence and consider the shortened sequence $\\alpha$. For each $i = 1, \\dots, n$ the hypothesis implies that $\\alpha$ starts with a block $B_i$, followed by a term $i$ which we called distinguished, and ends with a block $B'_... | Argentina | Mathematical Olympiad Rioplatense | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2^n | |
0gkh | Denote by $\nu(n)$ the exponent of $2$ in the prime factorization of $n!$. Show that for arbitrary positive integers $a$ and $m$ there exists an integer $n > 1$ for which $\nu(n) \equiv a \mod m$. | [
"We will use the well-known fact that $\\nu(n) = \\sum_{k=1}^{\\infty} \\lfloor \\frac{n}{2^k} \\rfloor$. If the base-2 form of the number $n$ is $n = \\overline{d_l d_{l-1} \\dots d_1 d_0} = \\sum_{i=0}^{\\infty} d_i \\cdot 2^i$, where the digits $d_0, \\dots, d_l$ are all $0$ or $1$, then\n$$\n\\begin{aligned}\n\... | Thailand | Thai Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
035m | Problem:
Let $\triangle A'B'C$ be the image of $\triangle ABC$ under a rotation with center $C$. Denote by $M$, $E$ and $F$ the midpoints of the segments $BA'$, $AC$ and $B'C$, respectively. If $AC \neq BC$ and $EM = FM$, find $\Varangle EMF$. | [
"Solution:\nSet $BC = B'C = a$, $AC = A'C = b$, $AB = A'B' = c$ and $\\Varangle C = \\gamma$. Since the triangles $AA'C$ and $BB'C$ are similar we have\n$$\n\\frac{AA'}{AC} = \\frac{BB'}{BC} = k\n$$\nTherefore $AA' = k b$ and $BB' = k a$.\n\n\n\nUsing the median formula several times we get... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 60° | |
01gz | Let $\omega$ and $I$ be the incircle and the incentre of a triangle $ABC$, respectively. Let $E$ and $F$ be the tangency points of $\omega$ with the sides $AC$ and $AB$, respectively. The perpendicular bisector of $AI$ intersects $AC$ at a point $P$. Point $Q$ lies on $AB$ and satisfies $QI \perp FP$. Prove that $EQ \p... | [
"Let $PF$ intersect $QI$ at $X$. Let $Y$ be the point symmetric to $E$ with respect to $PI$.\nLet $\\omega_1$ be the circle with diameter $PI$. Let $\\omega_2$ be the circle with diameter $FI$.\nNote that $E, X$, and $Y$ all lie on $\\omega_1$, and that $X$ lies on $\\omega_2$. The line $XI$ is the radical axis of ... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid,... | null | proof only | null | |
07l6 | Find all pairs $(x, y)$ of non-negative integers such that $x^3 + 7x^2 + 35x + 27 = y^3$. | [
"Observe that for non-negative integers $x$ and $y$ that satisfy the equation $x^3 + 7x^2 + 35x + 27 = y^3$, we obtain\n$$\n\\begin{aligned}\ny^3 - (x+2)^3 &= x^3 + 7x^2 + 35x + 27 - x^3 - 6x^2 - 12x - 8 \\\\\n&= x^2 + 23x + 19 > 0\n\\end{aligned}\n$$\nand similarly\n$$\n(x+4)^3 - y^3 = 5x^2 + 13x + 37 > 0\n$$\nThu... | Ireland | Irska | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (0, 3) and (4, 7) | |
0k5z | Problem:
A polynomial $P$ with integer coefficients is called tricky if it has $4$ as a root.
A polynomial is called teeny if it has degree at most $1$ and integer coefficients between $-7$ and $7$, inclusive.
How many nonzero tricky teeny polynomials are there? | [
"Solution:\nIf a degree $0$ polynomial has $4$ as a root, then it must be the constant zero polynomial. Thus, we will only consider polynomials of degree $1$.\n\nIf $P$ has degree $1$, integer coefficients, and $4$ as a root, then it must be of the form $P(x) = a(x-4) = a x - 4 a$ for some nonzero integer $a$. Sinc... | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2 | |
04x8 | Given positive integers $a$ and $k$, the sequence $(a_n)_{n=1}^{\infty}$ is defined by
$a_1 = a$ and $a_{n+1} = a_n + k \cdot \varphi(a_n)$ for $n = 1, 2, \dots,$
where $\varphi(m)$ stands for the product of digits of $m$ in its decimal representation (e.g., $\varphi(413) = 12$, $\varphi(308) = 0$). Prove that there ex... | [
"Obviously, the sequence $(a_n)_{n=1}^{\\infty}$ is increasing until the first term with the zero digit occurs and is constant following this term. Our aim is to find such values $a$ and $k$ that the zero digit first occurs in $a_{2009}$. We will solve the problem in general – given an integer $m > 4$, we present $... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0l6m | Suppose $\triangle ABC$ has angles $\angle BAC = 84^\circ$, $\angle ABC = 60^\circ$, and $\angle ACB = 36^\circ$. Let $D$, $E$, and $F$ be the midpoints of sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$, respectively. The circumcircle of $\triangle DEF$ intersects $\overline{BD}$, $\overline{AE}$, and $\ov... | [] | United States | AIME II | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 336 | |
03tj | Let $\{a_n\}$ be an arithmetic progression with common difference $d$ ($d \neq 0$) and $\{b_n\}$ be a geometric progression with common ratio $q$, where $q$ is a positive rational number less than $1$. If $a_1 = d$, $b_1 = d^2$ and $\frac{a_1^2 + a_2^2 + a_3^2}{b_1 + b_2 + b_3}$ is a positive integer, then $q$ equals _... | [
"As\n$$\n\\begin{aligned}\n\\frac{a_1^2 + a_2^2 + a_3^2}{b_1 + b_2 + b_3} &= \\frac{a_1^2 + (a_1+d)^2 + (a_1+2d)^2}{b_1 + b_1q + b_1q^2} \\\\\n&= \\frac{14}{1+q+q^2} = m\n\\end{aligned}\n$$\nis a positive integer, we get $1 + q + q^2 = \\frac{14}{m}$. Then\n$$\nq = -\\frac{1}{2} + \\sqrt{\\frac{1}{4} + \\frac{14}{m... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | final answer only | 1/2 | |
088e | Problem:
Un fabbro sta costruendo una cancellata orizzontale in ferro formata da tante barre verticali, parallele tra loro, ciascuna delle quali è posizionata a $18~\mathrm{cm}$ di distanza dalle 2 vicine. Il fabbro collega le estremità di ciascuna coppia di sbarre contigue con una barra incurvata ad arco di circonfer... | [
"Solution:\n\nLa risposta è (D). Con riferimento alla figura, siano $A$ e $B$ estremità di sbarre contigue, $V$ il vertice dell'arco $\\overparen{A B}$, $M$ il punto medio di $A B$ e $O$ il centro della circonferenza cui appartiene l'arco $\\widehat{A B}$. Deve valere $O B=O V$, posto dunque $O M=x$, per il Teorema... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
083v | Problem:
a. Determinare se $2005^{2004}$ è somma di due quadrati perfetti positivi.
b. Determinare se $2004^{2005}$ è somma di due quadrati perfetti positivi. | [
"Solution:\n\na. $2005^{2004}$ è somma di due quadrati perfetti positivi.\nSi osservi che $5^{2} = 3^{2} + 4^{2}$ e che $2005^{2004} = 5^{2} m^{2}$, dove $m = 5^{1001} \\cdot 401^{1002}$. Moltiplicando la prima relazione per $m^{2}$ si ottiene\n$$\n2005^{2004} = (5m)^{2} = (3m)^{2} + (4m)^{2}.\n$$\n\nb. $2004^{2005... | Italy | Olimpiade Italiana di Matematica | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a: yes; b: no | |
020d | Problem:
Determine the smallest positive integer $q$ with the following property:
for every integer $m$ with $1 \leqslant m \leqslant 1006$, there exists an integer $n$ such that
$$
\frac{m}{1007} q < n < \frac{m+1}{1008} q .
$$ | [
"Solution:\nFor $m=1006$, we have\n$$\nq - q / 1007 < n < q - q / 1008\n$$\nfor some integer $n$. If $q \\leqslant 1007$, then $q - q / 1007$ and $q - q / 1008$ are both numbers that are at least $q-1$ and smaller than $q$, so there can be no integer $n$ in between. Hence $q > 1007$ and $q - q / 1008 \\leqslant q-1... | Benelux Mathematical Olympiad | BxMO | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2015 | |
067v | In a mathematical competition some students are friends, and friendship is always mutual. Prove that there exists a subset $M$ of the students, such that every member of $M$ has at most three friends in $M$ and such that every non-member of $M$ has at least four friends in $M$. | [
"We start with the set $M_0 = \\emptyset$, and we modify/improve it step by step until it satisfies the two desired conditions. There are two possibilities for an improving step:\n\n1. If there is a student $x \\in M$ who has $\\ge 4$ friends in $M$, then $x$ is removed from $M$.\n\n2. If there is a student $y \\no... | Greece | Mediterranean Mathematical Competition PETER O' HALLORAN MEMORIAL | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0eg6 | Problem:
Poišči vse realne rešitve enačbe
$$
\sqrt{x+4}+\sqrt{2x+1}+\sqrt{4x+2}=\sqrt{3-x}
$$ | [
"Solution:\n\nEnačbo preoblikujemo v\n$$\n\\sqrt{2x+1}+\\sqrt{x+4}=\\sqrt{3-x}-\\sqrt{4x+2}\n$$\njo kvadriramo\n$$\n(2x+1)+2\\sqrt{2x+1}\\sqrt{x+4}+(x+4)=(3-x)-2\\sqrt{3-x}\\sqrt{4x+2}+(4x+2)\n$$\nin poenostavimo, da dobimo\n$$\n\\sqrt{2x+1}\\sqrt{x+4}=-\\sqrt{3-x}\\sqrt{4x+2}\n$$\nKo enačbo še enkrat kvadriramo in... | Slovenia | 61. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -1/2 | |
0e7y | Let $ABC$ be a triangle with $D$ a point on the side $AB$ and $E$ a point on the side $AC$ such that $|AE| = |ED| = |DB|$ and $|AD| = |DC| = |CB|$. Determine the sizes of the angles of the triangle $ABC$. | [
"Write $\\angle EBD = \\alpha$. Then $\\angle DEB = \\alpha$, so $\\angle EDA = 2\\alpha$ and $\\angle DAE = 2\\alpha$. This implies $\\angle DEC = 4\\alpha$, or $\\angle BEC = 3\\alpha$. At the same time we have $\\angle ACD = \\angle DAC = 2\\alpha$, so $\\angle BDC = 4\\alpha$ and $\\angle CBD = 4\\alpha$, or $\... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠A = 36°, ∠B = 72°, ∠C = 72° | |
0ai7 | Give all integer solutions of the equation:
$$
3^{2a+1} b^2 + 1 = 2^c.
$$
Во множеството на цели броеви да се реши равенка
$$
3^{2a+1}b^2 + 1 = 2^c.
$$ | [
"Case 1. $a \\ge 0$.\nClearly $c \\ge 0$ where $c=0$ implies $b=0$. We get that $(a,0,0)$ is a solution for an arbitrary non-negative integer $a$. From the equality $3^{2a+1}b^2+1=2^c$ it follows that $b$ is an odd integer. We can write the left-hand side in the following form\n$$\n3^{2a+1} b^2 + 1 = (3^{2a+1} + 1)... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | All integer solutions (a,b,c) are:
- (a, 0, 0) for any integer a;
- (a, ±3^(−a), 2) for any integer a ≤ 0. (In particular, a=0 gives b=±1, c=2.) | |
040k | As shown in the figure below, the in-circle $I$ of $\triangle ABC$ is tangent to sides $AB$ and $AC$ at points $D$ and $E$ respectively, and $O$ is the circum-center of $\triangle BCI$. Prove that $\angle ODB = \angle OEC$.
 | [] | China | China Girls' Mathematical Olympiad | [
"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"
] | English | proof only | null | |
06bm | In triangle $ABC$ with incentre $I$, let $M_A, M_B$ and $M_C$ be the midpoints of $BC, CA$ and $AB$ respectively, and $H_A, H_B$ and $H_C$ be the feet of altitudes from $A, B$ and $C$ to the respective sides. Denote by $\ell_b$ the line being tangent to the circumcircle of triangle $ABC$ and passing through $B$, and de... | [
"Firstly, since $M_C M_A \\parallel AC$, we have $\\angle M_C BP_B = \\angle ACB = \\angle M_C M_A B$ so that $\\ell'_b$ is tangent to $(BM_A M_C)$. This implies $P_B B^2 = P_B M_C \\times P_B M_A$, and hence $P_B$ lies on the radical axis of $B$ and the nine-point circle $\\Gamma$ of $\\triangle ABC$.\n\nSecondly,... | Hong Kong | 1997-2023 IMO HK TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | null | proof only | null | |
0f33 | Problem:
Find a sequence $a_{1}$, $a_{2}$, ..., $a_{25}$ of 0s and 1s such that the following sums are all odd:
\[
\begin{array}{rl}
& a_{1}a_{1} + a_{2}a_{2} + \ldots + a_{25}a_{25} \\
& a_{1}a_{2} + a_{2}a_{3} + \ldots + a_{24}a_{25} \\
& a_{1}a_{3} + a_{2}a_{4} + \ldots + a_{23}a_{25} \\
& \ldots \\
& a_{1}a_{2... | [] | Soviet Union | ASU | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
07un | Show that the only choice of integers $x, y, z$ that satisfies the equation
$$
x^{2022} + 20y^{2022} = 22z^{2022},
$$
is $x = y = z = 0$. | [
"It suffices to consider non-negative $x, y, z$. Assume that there is a non-trivial solution, i.e. at least one of the numbers $x, y, z$ is positive. We pick such a solution with minimal sum $x + y + z$.\nNow, $x^{2022}$ must be even since the other two terms are even, so we can write\n$x = 2u$ and divide the new e... | Ireland | IRL_ABooklet | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
04m7 | Find all positive integers $n$ for which there exist positive integers $a$ and $b$ such that
$$
(n^2 + 2)^a = (2n - 1)^b.
$$ | [
"Notice that $n^2 + 2 \\ge 3$, $n^2 + 2 > 2n - 1$ and also that numbers $n^2 + 2$ and $2n - 1$ have the same prime factors.\nLet $p$ be a prime factor of $n^2 + 2$ and $2n - 1$.\n$$\np \\mid n^2 + 2 \\quad \\text{and} \\quad p \\mid 2n - 1 \\implies p \\mid (n^2 + 2) + (2n - 1) = (n + 1)^2 \\implies p \\mid n + 1.\... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 5 | |
0889 | Problem:
Su una lavagna sono scritti dei numeri interi, compresi fra 1 e 7. È possibile che non tutti i numeri da 1 a 7 siano presenti, ed è anche possibile che uno, alcuni o tutti i numeri siano ripetuti, una o più volte.
Una mossa consiste nello scegliere uno o più numeri presenti sulla lavagna, purché tutti divers... | [
"Solution:\n\nChiamiamo $n_{1}$ il numero di cifre 1 presenti in un certo momento sulla lavagna, $n_{2}$ il numero di cifre 2, e così via fino ad $n_{7}$.\n\nOgni volta che si fa una mossa, ognuna di queste molteplicità cambia di 1 (e quindi inverte la sua parità), perché ogni numero tra 1 e 7 viene scritto o cance... | Italy | Cesenatico | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
034c | Problem:
Solve the equation
$$
(x+6) 5^{1-|x-1|}-x=(x+1)\left|5^{x}-1\right|+5^{x+1}+1
$$ | [
"Solution:\nFor $x \\geq 1$ the equation becomes $(x+6)\\left(5^{x}-5^{2-x}\\right)=0$, whence $x=2-x$, i.e. $x=1$.\n\nFor $0 \\leq x<1$ the equation becomes an identity, i.e. every $x \\in [0,1)$ is a solution.\n\nFor $x<0$ we obtain $2\\left(5^{x}-1\\right)(x+1)=0$, whence $x=-1$.\n\nThe solutions of the problem ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | [0,1] ∪ {-1} | |
0krl | Problem:
Let $ABCD$ be a square of side length $2$. Let points $X$, $Y$, and $Z$ be constructed inside $ABCD$ such that $ABX$, $BCY$, and $CDZ$ are equilateral triangles. Let point $W$ be outside $ABCD$ such that triangle $DAW$ is equilateral. Let the area of quadrilateral $WXYZ$ be $a + \sqrt{b}$, where $a$ and $b$ a... | [
"Solution:\n\n$WXYZ$ is a kite with diagonals $XZ$ and $WY$, which have lengths $2\\sqrt{3} - 2$ and $2$, so the area is $2\\sqrt{3} - 2 = \\sqrt{12} - 2$."
] | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 10 | |
0he4 | After $x$ matches in a championship a team had exactly $n\%$ of wins, where $x, n$ are positive integers. What is the least $x$ for which it is possible that after $x+1$-st match the team had exactly $(n+1)\%$ of wins? | [
"It follows from the conditions of the problem that after $x$ matches the team had $y$ wins, such that $\\frac{y}{x} = \\frac{n}{100}$. Then, the following has to hold: $\\frac{y+1}{x+1} = \\frac{n+1}{100}$.\n\nThe first equation implies $100y = nx$, and so $100$ has to divide $nx$.\nThe second equation implies $10... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 24 | |
07fx | Let $a$, $m$ be positive integers such that $\text{ord}_m(a)$ is odd and for any integers $x$, $y$ such that:
a) $xy \equiv a \pmod{m}$.
b) $\text{ord}_m(x) \le \text{ord}_m(a)$.
c) $\text{ord}_m(y) \le \text{ord}_m(a)$.
We have either $\text{ord}_m(x) \mid \text{ord}_m(a)$ or $\text{ord}_m(y) \mid \text{ord}_m(a)$. Pr... | [
"Assume for a contradiction that $\\text{ord}_m(a)$ has at least two prime factors. Then we can write $\\text{ord}_m(a) = rs$ where $\\gcd(r, s) = 1$ and $r$, $s$ are both greater than $1$. Now take a natural solution pair $(k, \\ell)$ of the equation $ks + r\\ell \\equiv 1 \\pmod{\\varphi(m)}$, that definitely exi... | Iran | 37th Iranian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
043i | In a plane rectangular coordinate system $xOy$, the graph of function $y = \frac{1}{|x|}$ is $\Gamma$. Let points $P, Q$ on $\Gamma$ satisfy: $P$ is in the first quadrant, $Q$ is in the second quadrant, and line $PQ$ is tangent to the part of $\Gamma$ in the second quadrant at point $Q$. Find the minimum of $|PQ|$. | [
"When $x > 0$, $y = \\frac{1}{x}$. When $x < 0$, $y = -\\frac{1}{x}$, and its corresponding derivative is $y' = -\\frac{1}{x^2}$.\n\nSuppose $Q(-a, \\frac{1}{a})$, where $a > 0$. By the condition, the slope of $PQ$ is $y'|_{x=-a} = \\frac{1}{a^2}$.\n\nThe equation of line $PQ$ is $y = \\frac{1}{a^2}(x + a) + \\frac... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2 | |
0e48 | The first term of an arithmetic progression is equal to $\frac{1}{3}$ and the third term is equal to $\frac{1}{5}$. What is the second term?
(A) $\frac{1}{4}$
(B) $\frac{4}{15}$
(C) $\frac{5}{24}$
(D) $\frac{7}{30}$
(E) $\frac{2}{9}$ | [
"The second term of the arithmetic progression is equal to $a_2 = \\frac{1}{2}(a_1 + a_3) = \\frac{1}{2}\\left(\\frac{1}{3} + \\frac{1}{5}\\right) = \\frac{4}{15}$."
] | Slovenia | National Math Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
00le | Let $ABCDE$ be a convex pentagon with five equal sides and right angles at $C$ and $D$. Let $P$ denote the intersection point of the diagonals $AC$ and $BD$.
Prove that the segments $PA$ and $PD$ have the same length. | [
"$BCDE$ is a square since $\\overline{BC} = \\overline{CD} = \\overline{DE}$ and $BC \\perp CD$, $CD \\perp DE$. Hence also the length of the segment $BE$ coincides with the side length of the pentagon $ABCDE$ and we have $BC \\perp BE$ and $BE \\perp DE$. Furthermore $\\overline{AB} = \\overline{AE} = \\overline{B... | Austria | Beginners' Competition | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0gbw | 令 $a, b, c, d$ 為非負實數使得 $a + b + c + d = 4$. 試證:
$$
a\sqrt{3a+b+c}+b\sqrt{3b+c+d}+c\sqrt{3c+d+a}+d\sqrt{3d+a+b} \ge 4\sqrt{5}.
$$ | [
"Let\n$$\n\\sqrt{3a+b+c} = x, \\sqrt{3b+c+d} = y, \\sqrt{3c+d+a} = z, \\sqrt{3d+a+b} = w,\n$$\nthen we have\n$$\nx^2 = 3a + b + c, \\quad y^2 = 3b + c + d,\\\\\nz^2 = 3c + d + a, \\quad w^2 = 3d + a + b.\n$$\nTherefore, we have\n$$\n\\begin{aligned}\n15a &= 5(3a + b + c) - 2(3b + c + d) - (3c + d + a) + (3d + a + b... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
08he | Problem:
The rational numbers $p$, $q$, $r$ satisfy the relation $p q + p r + q r = 1$. Prove that the number $(1 + p^{2})(1 + q^{2})(1 + r^{2})$ is a square of a rational number. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
02s5 | Problem:
Juarez utilizou os algarismos $1$, $2$, $3$, $4$ e $5$ para escrever o número $\overline{abcde}$ de cinco algarismos distintos. Sem revelar qual é esse número, ele disse a Luciana que:
- o número $\overline{abc}$ é divisível por $4$;
- o número $\overline{bcd}$ é divisível por $5$;
- o número $\overline{cde}$ ... | [
"Solution:\nTodo número divisível por $5$ tem o algarismo das unidades igual a $5$ ou igual a $0$. Como o algarismo $0$ não foi usado por Juarez, então para que $\\overline{bcd}$ seja divisível por $5$, devemos ter, necessariamente, $d=5$.\n\nTodo número divisível por $3$ é tal que a soma dos seus algarismos também... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 12453 | |
08nt | Problem:
Find the largest possible value of the expression $\left|\sqrt{x^{2}+4x+8}-\sqrt{x^{2}+8x+17}\right|$ where $x$ is a real number. | [
"Solution:\nWe observe that\n$$\n\\left|\\sqrt{x^{2}+4x+8}-\\sqrt{x^{2}+8x+17}\\right|=\\left|\\sqrt{(x-(-2))^{2}+(0-2)^{2}}-\\sqrt{(x-(-4))^{2}+(0-1)^{2}}\\right|\n$$\nis the absolute difference of the distances from the point $P(x, 0)$ in the $xy$-plane to the points $A(-2,2)$ and $B(-4,1)$.\nBy the Triangle Ineq... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | sqrt(5) | |
07su | Find all triples $(p, q, r)$ of prime numbers that satisfy $p \le q \le r$ and
$$
pqr \mid (7^p - 3^p)(7^q - 3^q)(7^r - 3^r).
$$ | [
"We start with establishing the following statement, which is valid for not necessarily different prime numbers $m$ and $n$:\n$$\nm \\mid 7^n - 3^n \\quad \\text{implies} \\quad m = 2 \\text{ or } m > n. \\qquad (5)\n$$\nFirst note that $m$ cannot be equal to $3$ or $7$, hence there exists an integer $a$ such that ... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (2, 2, 2), (2, 2, 5), (2, 5, 41), (2, 5, 101) | |
008v | David is a very curious child. He has a container in the shape of a rectangular cuboid such that its three dimensions (width, height and depth) are different positive integers. The container is filled with two different unmixable liquids, $A$ and $B$, and the volume of $A$ is $36\%$ of the volume of $B$.
David observe... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Geometry > Solid Geometry > Volume",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 235824 | |
0ets | A cube of side length $n$ is made up of $n^3$ smaller unit cubes. Some of the six faces of the large cube are fully painted. When the large cube is taken apart, 245 smaller cubes do not have any paint on them.
Determine the value(s) of $n$ and how many faces of the large cube were painted. | [
"Let us first determine how the number of painted faces influences the formula for the total number of unpainted unit cubes.\nIf zero faces are painted, the number of cubes is $n^3$, and if all six are painted, the number of unpainted cubes is $(n - 2)^3$. Since 245 is not a perfect cube, both of these are ruled ou... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Solid Geometry > 3D Shapes",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 7; two faces painted, and they are opposite each other | |
0b4p | Problem:
Let $x$, $y$, $z$ be three real numbers such that
- $y$, $x$, $z$ form a harmonic sequence; and
- $3 x y$, $5 y z$, $7 z x$ form a geometric sequence.
The numerical value of $\frac{y}{z} + \frac{z}{y}$ can be expressed in the form $p/q$, where $p$ and $q$ are relatively prime positive integers. What is $p+q$? | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 59 | |
0gj6 | 令 $x$ 與 $y$ 為滿足 $x + y = 2$ 的正實數。證明
$$
2 + xy(1 + xy) \geq 2x^y + 2y^x.
$$
Let $x$ and $y$ be positive real numbers such that $x + y = 2$. Prove that
$$
2 + xy(1 + xy) \geq 2x^y + 2y^x.
$$ | [
"由廣義算幾不等式有\n$$\n\\begin{aligned}\n\\frac{2x \\times y + 2 \\times (2-y)}{y + (2-y)} &\\ge \\sqrt{(2x)^y \\times 2^{2-y}} \\\\\n\\Rightarrow xy + (2-y) &\\ge \\sqrt{4x^y} \\\\\n\\Rightarrow (xy + 2 - y)^2 &\\ge 4x^y.\n\\end{aligned}\n$$\n同理有\n$$\n(xy + 2 - x)^2 \\ge 4y^x.\n$$\n兩式相加得\n$$\n\\begin{aligned}\n4x^y + 4y^... | Taiwan | IMO 1J, Independent Study 1 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Chinese; English | proof only | null | |
0ftr | Problem:
Finde alle injektiven Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle reellen Zahlen $x \neq y$ gilt
$$
f\left(\frac{x+y}{x-y}\right)=\frac{f(x)+f(y)}{f(x)-f(y)}
$$ | [
"Solution:\nEinsetzen von $y=0$ liefert für $x \\neq 0$ die Gleichung $f(1)=(f(x)+f(0))/(f(x)-f(0))$, also\n$$\nf(x)(f(1)-1)=f(0)(f(1)+1)\n$$\nDaraus folgt $f(1)=1$ und damit auch $f(0)=0$, denn sonst wäre $f$ konstant auf $\\mathbb{R} \\backslash \\{0\\}$, also nicht injektiv. Setze $y=-x \\neq 0$, dann folgt\n$$\... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
04y3 | Prove that for every non-negative integer $n$ there exist integers $x, y, z$ with $\gcd(x, y, z) = 1$, such that
$$
x^2 + y^2 + z^2 = 3^{2n}.
$$ | [
"We will use the following algebraic formula:\n$$\n(x^2 + y^2 + z^2)^2 = (x^2 + y^2 - z^2)^2 + (2xz)^2 + (2yz)^2.\n$$\nThis means that if a positive integer can be represented as a sum of 3 squares, then so can its square. Consequently, if we put\n$$\n\\begin{aligned}\n(x_0, y_0, z_0) &= (1, 1, 1) \\\\\n(x_{n+1}, y... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
07hx | Points $B$ and $C$ are fixed on the circle $\omega$. Denote by $M$ the midpoint of $BC$. Let $A$ be a variable point on $\omega$, and $H$ be the orthocenter of triangle $ABC$. The perpendicular from $H$ to $MH$, intersects lines $AB$ and $AC$ at points $X$ and $Y$, respectively. Prove that as point $A$ moving on $\omeg... | [
"Let $O$ be the center of $\\omega$, and $H'$ be the orthocenter of triangle $AXY$. Denote the intersection of $AH'$ and $OM$ as $D$. It is evident that quadrilateral $AHMD$ is a parallelogram, so $2OM = AH = MD$. This result shows that $D$ is a fixed point. If we prove $\\frac{AH'}{AD}$ is constant, it follows tha... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Tr... | null | proof only | null | |
0h6q | Son, his Dad and his Grandfather has run from their home to a shop and back. Son's velocity was constant. Grandfather's velocity was two times greater than Son's while he was running to the shop and three times less when he was running back. Dad's velocity was two times less than Son's on the way to shop and 3 times gr... | [
"Denote the velocity of Son by $x$ and the distance by $S$. Then we can compute the time which was spent on running:\n$$\nt_1 = \\frac{S}{x} + \\frac{S}{x}, \\quad t_2 = \\frac{S}{\\frac{1}{2}x} + \\frac{S}{3x}, \\quad t_3 = \\frac{S}{2x} + \\frac{S}{\\frac{1}{3}x}\n$$\nHence, we need to compare the following numbe... | Ukraine | UkraineMO | [
"Math Word Problems"
] | null | proof and answer | Son first, Grandfather last | |
01kh | Points $M$ and $N$ are the midpoints of the sides $AC$ and $BC$ of the triangle $ABC$, respectively.
Prove that a circle passing through $C, M, N$ touches the side $AB$ if and only if $AB = \frac{AC + BC}{\sqrt{2}}$.
(I. Voronovich) | [
"Let $I$ be the center of the circle $\\Gamma$ passing through $C, M, N$; let $r$ be the radius of $\\Gamma$. Let $P$ and $Q$ be the intersection points of segments $AI, BI$ and $\\Gamma$; let $H$ be the foot of the perpendicular from $I$ on $AB$. Set $a = BC, b = AC, c = AB, x = AP, y = BQ, l = IH$.\n\n![](attache... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0d01 | Divide a rectangle of dimension $m \times n$ by lines parallel to the sides into $mn$ congruent squares. Find the number of squares in this configuration. | [
"There are $mn$ squares of length side $1$, $(m-1)(n-1)$ squares of length side $2$, $(m-2)(n-2)$ squares of side length $3$, etc.\nAssume that $m \\ge n$. The total number of squares in this configuration is\n$$\n\\sum_{k=0}^{n-1} (m-k)(n-k). \\quad (1)\n$$\nWe have\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n-1} (m-k)(n... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Discrete Mathematics > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | n(n+1)(3m - n + 1)/6 | |
0dqp | Let $n$ be an integer $\ge 2$. Let $S$ be a set of triples of nonnegative integers $(a, b, c)$ such that $a + b + c = n$ and if $(a, b, c)$ and $(a', b', c')$ are distinct triples in $S$, then $a \ne a'$, $b \ne b'$ and $c \ne c'$. Find the maximum value of $|S|$. | [
"Suppose $|S| = s$. Then $\\sum_{(a, b, c) \\in S} a \\ge \\sum_{i=0}^{s-1} i = \\frac{s(s-1)}{2}$. Likewise $\\sum_S b \\ge \\frac{s(s-1)}{2}$ and $\\sum_S c \\ge \\frac{s(s-1)}{2}$. Therefore\n$$\n\\frac{3s(s-1)}{2} \\le \\sum_{S} a + b + c = ns \\Rightarrow s \\le \\left\\lfloor \\frac{2n}{3} \\right\\rfloor + 1... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | ⌊2n/3⌋ + 1 | |
0cgu | Three friends color the positive integers from $1$ to $2025$ as follows: Alexia colors in red the numbers $1$ and $2$, then Bianca colors in yellow the numbers $3$, $4$ and $5$, and Cristina colors in blue the numbers $6$, $7$, $8$ and $9$. Afterwards, the operation is repeated: Alexia colors in red the next two number... | [
"For simplicity we will call the numbers red, yellow, respectively blue, and a sequence of $9$ consecutive numbers in which the first two are red, the next three are yellow, and the last four are blue will be called a *complete coloring*.\n\na) Since $2025 = M_9$, the last four numbers are blue, hence $2024$ is blu... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) Blue
b) 111 | |
03kz | Problem:
Find all real solutions to the equation $4x^{2} - 40[x] + 51 = 0$.
Here, if $x$ is a real number, then $[x]$ denotes the greatest integer that is less than or equal to $x$. | [
"Let $n = [x]$, where $n$ is an integer and $n \\leq x < n+1$.\n\nThen the equation becomes:\n$$\n4x^2 - 40n + 51 = 0\n$$\nwhich gives\n$$\n4x^2 = 40n - 51\n$$\n$$\nx^2 = 10n - 12.75\n$$\n$$\nx = \\pm \\sqrt{10n - 12.75}\n$$\nBut $n \\leq x < n+1$, so $x$ must be real, i.e., $10n - 12.75 \\geq 0$, or $n \\geq 1.275... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = sqrt(29)/2, sqrt(189)/2, sqrt(229)/2, sqrt(269)/2 | |
0art | Problem:
Find the square root of $25 \cdot 26 \cdot 27 \cdot 28 + 1$. | [
"Solution:\n$a(a+1)(a+2)(a+3) + 1 = a(a+3)(a+1)(a+2) + 1 = [a^2 + 3a][a^2 + 3a + 2] + 1 = [(a^2 + 3a + 1) - 1][(a^2 + 3a + 1) + 1] + 1 = (a^2 + 3a + 1)^2$.\n\nIf $a = 25$, the square root is $25^2 + 3 \\cdot 25 + 1 = 701$."
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 701 | |
09hq | Let $ABCD$ be a quadrilateral with $AB = BC = CD$ and $\angle ABC + \angle BCD = 240^\circ$. Find the angle $\angle CAD$. | [] | Mongolia | Round 2 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 30° | |
0c50 | Let $n \in \mathbb{N}^*$ and $E(z) = |z+1| + |z^{2n} + 1|$, $z \in \mathbb{C}$. Find
$$ \min\{E(z) \mid z \in \mathbb{C}\}. $$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 2\sin\left(\frac{\pi}{4n}\right) | |
0ab3 | Is the sum $1^{2008} + 2^{2008} + 3^{2008} + 4^{2008} + 5^{2008} + 6^{2008}$ divisible with $5$? Explain your answer. | [
"Notice that $1^{2008} = 1$. $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 16$, $2^5 = 32$, ... From this we conclude that all the powers of $2$ are ending on $2$, $4$, $8$, $6$ and they are repeating in that order, depending on the residuum of the power $1$, $2$, $3$ or $0$ when divided with $4$. Because $2008$ is divis... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | Yes, it is divisible by 5. | |
09f2 | Draw median $AM$ and bisector $AL$ of a scalene triangle $ABC$. Tangent lines of circumcircle of the triangle $ABC$ at points $B$, $C$ intersect at point $T$ and line $AT$ intersects the circumcircle at point $K$ which is different from $A$. The line $AM$ intersects circumcircle of the triangle $AKL$ at point $P$. Prov... | [
"\n\nNote that $AF$, $FK$ are sim medians of triangles $\\Delta ABF$, $\\Delta BKF$ respectively.\nTherefore we get $\\angle CMK = \\angle CMP$. Since $LK$ angle bisector and quadrilateral\n$APLK$ is inscribed in a circle, $\\angle LPM = \\angle LKM$ and from this follows\n$PM = MK$. Since ... | Mongolia | Mongolian Mathematical Olympiad | [
"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 > Advanced Configurations... | English | proof only | null | |
09w5 | Annemiek and Bart each have a note on which they have written three different positive integers. It appears that there is exactly one number that is on both their notes. Moreover, if you add any two different numbers from Annemiek's note, you get one of the numbers on Bart's note. One of the numbers on Annemiek's note ... | [
"3."
] | Netherlands | First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 5 | |
0aiy | Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 3$. Find the minimum value of the expression
$$
A = \frac{2 - a^3}{a} + \frac{2 - b^3}{b} + \frac{2 - c^3}{c}
$$ | [
"\\begin{align*}\nA &= \\frac{2 - a^3}{a} + \\frac{2 - b^3}{b} + \\frac{2 - c^3}{c} = 2\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) - a^2 - b^2 - c^2 \\\\\n&= 2\\frac{ab + bc + ca}{abc} - (a^2 + b^2 + c^2) \\\\\n&= 2\\frac{ab + bc + ca}{abc} - \\left((a + b + c)^2 - 2(ab + bc + ca)\\right) \\\\\n&= 2\\... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 3 | |
0jkz | Problem:
Consider a number line, with a lily pad placed at each integer point. A frog is standing at the lily pad at the point $0$ on the number line, and wants to reach the lily pad at the point $2014$ on the number line. If the frog stands at the point $n$ on the number line, it can jump directly to either point $n+... | [
"Solution:\n\nAnswer: 0.9102805441016536\n\nFirst, we establish a rough upper bound for the probability $p$. Let $q$ be the probability that the frog can reach the lily pad at the point $2014$ on the number line if it is allowed to jump from a point $n$ on the number line to the point $n+1$, in addition to the poin... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | final answer only | 0.9102805441016536 | |
0hqe | Problem:
Let $ABC$ be a triangle with $\angle ABC = 90^{\circ}$. Points $D$ and $E$ on $AC$ and $BC$ respectively satisfy $BD \perp AC$ and $DE \perp BC$. The circumcircle of $\triangle CDE$ intersects $AE$ at two points, $E$ and $F$. Prove that $BF \perp AE$. | [
"Solution:\n\nBy Power of a Point,\n$$\nAF \\cdot AE = AD \\cdot AC\n$$\nbecause $\\triangle ABC$ is right,\n$$\nAD \\cdot AC = AB^2\n$$\nCombining,\n$$\nAF \\cdot AE = AB^2.\n$$\nOn the other hand, if $F'$ is the foot of the altitude from $B$ to $AE$, then\n$$\nAF' \\cdot AE = AB^2\n$$\nConsequently $AF' = AF$ and... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0gcv | 令 $R$ 表示所有實數所成的集合。試求所有的函數 $f: R \to R$ 使得對所有的實數 $x, y$,
$$
f(xf(y) - f(x) - y) = yf(x) - f(y) - x \text{ 皆成立。}
$$ | [
"令 $P(x, y)$ 表示題目的函數等式。\n\n$$\nP(x, 0) \\rightarrow f(xf(0) - f(x)) = -x - f(0).\n$$\n\n由此可知 $f$ 為滿射，且 $f(f(x)) = x$。注意到\n$$\nP(1, 1) \\rightarrow f(-1) = -1 \\text{ 且 } P(0, -1) \\rightarrow f(1 - f(0)) = 1 - f(0).\n$$\n\n考慮 $x = 0, y = 1 - f(0)$：\n$$\n-1 = f(-1) = (1-f(0))f(0) - f(1-f(0)) = -(1-f(0))^2 \\Leftrigh... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x for all real x | |
006w | Sea $ABC$ un triángulo tal que $\hat{A} = 3\hat{B}$. Si $BC = 5$ y $CA = 3$, calcular la medida del lado $AB$. | [] | Argentina | Argentina 2009 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Spanish | proof and answer | 4*sqrt(6)/3 | |
00lq | In the isosceles triangle $ABC$ with $\overline{AC} = \overline{BC}$ we denote by $D$ the foot of the altitude through $C$. The midpoint of $CD$ is denoted by $M$. The line $BM$ intersects $AC$ in $E$. Prove that the length of $AC$ is three times that of $CE$.
Figure 4: Problem 16 | [
"We consider the centroid $S$ of triangle $DBC$, which lies on the axis $BM$; see Figure 4. The centroidal axis $DS$ bisects the segment $BC$, thus $DS$ is parallel to $AC$ by the intercept theorem. Since the centroid divides the centroidal axis in the ratio $2 : 1$, we have the same division ratio on the parallel ... | Austria | 48th Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0cgx | The quadrilateral regular pyramid $VABCD$ has basis $ABCD$. The points $M$, $N$ and $P$ are the midpoints of the edges $AD$, $BC$, respectively $VA$. Prove that the angle of the line $CP$ with the plane $(BAD)$ has $45^\circ$ if and only if the angle of the line $CP$ with the plane $(VMN)$ has $30^\circ$. | [
"\nThe projection of the line $CP$ on the plane $(BAD)$ is the line $OC$, hence the angle of the line $CP$ with the plane $(BAD)$ is $\\angle OCQ$.\nFrom $CN \\perp MN$, $CN \\perp VO$ and $MN \\cap VO = \\{O\\}$ follows $CN \\perp (VMN)$. So, the projection of the line $CP$ on the plane $(... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
024q | Problem:
Determine o número inteiro mais próximo de
a. $\frac{19}{15}+\frac{19}{3}$
b. $\frac{85}{42}+\frac{43}{21}+\frac{29}{14}+\frac{15}{7}$
c. $-\frac{11}{10}-\frac{1}{2}-\frac{7}{5}+\frac{2}{3}$ | [
"Solution:\n\na.\nTemos:\n$$\n\\frac{19}{15}+\\frac{19}{3}=1+\\frac{4}{15}+6+\\frac{1}{3}=7+\\frac{9}{15}=7+\\frac{3}{5}\n$$\nLogo, a soma dada está entre $7$ e $8$. Como $\\frac{3}{5}>\\frac{1}{2}$, o número inteiro mais próximo é $8$.\n\n\n\nb.\nTemos:\n$$\n\\begin{aligned}\n\\frac{85}{42... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | a: 8; b: 8; c: -2 | |
0aj9 | A $5 \times 5$ board, consisting of 25 unit squares, a positive integer $k \le 25$ and an unlimited supply of $L$-shapes are given. Two players, $A$ and $B$, play the following game: starting with $A$ they alternatively mark a previously unmarked unit square until they mark a total of $k$ unit squares.
We say that a p... | [
"If $k=1$, player $A$ marks the upper left corner of the square and then fills it as follows.\n\nIf $k=2$, player $A$ marks the upper left corner of the square. Whatever square player $B$ marks, then player $A$ can fill in the square in exactly the same pattern as above except that he doesn't put the trimino which ... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 4 | |
0iyf | Problem:
A single burger is not enough to satisfy a guy's hunger. The five guys go to Five Guys' Restaurant, which has 20 different meals on the menu. Each meal costs a different integer dollar amount between $1$ and $20$. The five guys have $20$ to split between them, and they want to use all the money to order five ... | [
"Solution:\n\nAnswer: $7$\n\nSuppose the meals, sorted in descending order, cost $5 + x_{1}, 4 + x_{2}, \\ldots, 1 + x_{5}$. To satisfy the conditions in the problem, the $x_{i}$ must be a non-increasing sequence of non-negative integers which sums to $5$. Therefore, there is exactly one order for each partition of... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 7 | |
04jj | In a box, there is one red and five white balls marked by numbers $1$, $2$, $3$, $4$ and $5$. Without looking, Domagoj draws one ball at a time until he draws the red ball and then he stops. Drawn balls are not returned into the box. What is the probability that the sum of the numbers on the drawn balls is at least $10... | [] | Croatia | Croatia Mathematical Competitions | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 2/5 | |
0jqq | Problem:
Consider a $6 \times 6$ grid of squares. Edmond chooses four of these squares uniformly at random. What is the probability that the centers of these four squares form a square? | [
"Solution:\n\nFirstly, there are $\\binom{36}{4}$ possible combinations of points. Call a square proper if its sides are parallel to the coordinate axes and improper otherwise. Note that every improper square can be inscribed in a unique proper square. Hence, an $n \\times n$ proper square represents a total of $n$... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1/561 | |
00mf | Anna and Berta play a game in which they take turns in removing marbles from a table. Anna takes the first turn. When at the beginning of a turn there are $n \ge 1$ marbles on the table, then the player whose turn it is removes $k$ marbles, where $k \ge 1$ either is an even number with $k \le \frac{n}{2}$ or an odd num... | [
"We claim that the losing situations are those with exactly $n = 2^a - 2$ marbles left on the table for all integers $a \\ge 2$. All other situations are winning situations.\n\n*Proof:* By induction for $n \\ge 1$. For $n = 1$ the player wins by taking the single remaining marble. For $n = 2$ the only possible move... | Austria | 48th Austrian Mathematical Olympiad National Competition (Final Round, part 1) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 131070 | |
0h0s | Find minimal number $n$ such that $n^3 + n^2 + 330n + 330$ is divisible by $2011$? | [
"$$\nn^3 + n^2 + 330n + 330 = (n+1)(n^2 + 330),\n$$\nSo this expression is divisible by $2011$ if at least one bracket is divisible by $2011$. If first bracket is divisible by $2011$ then minimal $n=2010$, for $n^2 + 330$, since $n^2$ is increasing, we find that for $n=41$, $n^2 + 330 = 2011$."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 41 | |
0eie | Problem:
Največ koliko je lahko največji skupni delitelj števil $a-2b+3$, $2a-3b-1$ in $3a+b-2$, če sta $a$ in $b$ naravni števili? | [
"Solution:\n\nNaj bo $d$ največji skupni delitelj danih števil. Tedaj $d$ deli tudi števila\n$$\n\\begin{aligned}\n& (2a-3b-1)-2(a-2b+3) = b-7 \\\\\n& (a-2b+3)+2(b-7) = a-11 \\\\\n& (3a+b-2)-(b-7) = 3a+5 \\\\\n& (3a+5)-3(a-11) = 38\n\\end{aligned}\n$$\nTorej je $d$ lahko največ $38$. Če izberemo $a=11$ in $b=7$, po... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 38 | |
0gnv | Alice, who works for the Graph County Electric Works, is commissioned to wire the newly erected utility poles in $k$ days. Each day she either chooses a pole and runs wires from it to as many poles as she wishes, or chooses at most $17$ pairs of poles and runs wires between each pair. Bob, who works for the Graph Count... | [
"Let $G = \\bigcup_{i=1}^{k} G_i$ be a graph that can be vertex-colored using not more than $2009$ colors where each $G_i$ is either a star or $e(G_i) \\le 17$ for $1 \\le i \\le k$. We want to find the maximum possible value of $k$.\n\nSuppose that $G_i$ is a star for $1 \\le i \\le j$, and $G_i$ satisfies $e(G_i)... | Turkey | 17th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2000 | |
0a3h | Problem:
Zij $P(x)$ een polynoom met gehele coëfficienten van graad $n>1$ waarvoor geldt dat $Q(x)=P(P(P(x)))-P(x)$ precies $n^{3}$ verschillende reële nulpunten heeft. Bewijs dat de nulpunten van $Q(x)$ in twee groepen verdeeld kunnen worden waarvan het rekenkundig gemiddelde hetzelfde is. | [
"Solution:\n\nWe schrijven $\\gamma_{i}$ met $1 \\leq i \\leq n^{3}$ voor de verschillende wortels zijn van $Q(x)$. Dat betekent dat $P\\left(P\\left(P\\left(\\gamma_{i}\\right)\\right)\\right)-P\\left(\\gamma_{i}\\right)=0$, dus $P\\left(\\gamma_{i}\\right)$ is een nulpunt van het polynoom $P(P(x))-x$. De graad va... | Netherlands | IMO-selectietoets III | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
02ur | Problem:
Sabemos que
$$
\frac{8^{x}}{2^{x+y}}=64 \mathrm{e} \frac{9^{x+y}}{3^{4 y}}=243
$$
Determine o valor de $2 x y$. | [
"Solution:\nComo $8=2^{3}$ e $9=3^{2}$, temos\n$$\n\\begin{aligned}\n64 & =\\frac{8^{x}}{2^{x+y}} \\\\\n2^{6} & =2^{3 x-(x+y)} \\\\\n& =2^{2 x-y}\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n243 & =\\frac{9^{x+y}}{3^{4 y}} \\\\\n3^{5} & =3^{(2 x+2 y)-4 y} \\\\\n& =3^{2 x-2 y}\n\\end{aligned}\n$$\nAssim, temos o segui... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 7 | |
09v9 | If you try to divide the number $19$ by $5$, you will get a remainder. The number $5$ fits $3$ times in $19$ and you will be left with $4$ as remainder. There are two positive integers $n$ having the following property: if you divide $n^2$ by $2n + 1$, you will get a remainder of $1000$. What are these two integers? | [
"$666$ and $1999$"
] | Netherlands | Second Round, March 2019 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 666 and 1999 | |
0318 | Problem:
Let $A$ be the set of all 4-tuples of $0$ and $1$. Two such 4-tuples are called neighbors if they coincide exactly at three positions. Let $M$ be a subset of $A$ with the following property: any two elements of $M$ are not neighbors and there exists an element of $M$ which is neighbor of exactly one of them. F... | [
"Solution:\nConsider a table with rows corresponding to the elements of $A$ and columns corresponding to the elements of $M$. We write $\\times$ in a cell if the element of $A$ in the respective row is adjacent to the element of $M$ in the respective column. Let $|M| = k$. It follows from the given condition that t... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
0hnd | Problem:
Prove that
$$
A=\sqrt{4-2 \sqrt{3}}-\frac{\sqrt{3}+1}{\sqrt{3}-1}
$$
is an integer. | [
"Solution:\n$\\sqrt{4-2 \\sqrt{3}} = \\sqrt{3^{2} - 2 \\sqrt{3} + 1^{2}} = \\sqrt{(\\sqrt{3}-1)^{2}} = |\\sqrt{3}-1| = \\sqrt{3}-1$.\n\nIf we multiply both numerator and denominator of $\\frac{\\sqrt{3}+1}{\\sqrt{3}-1}$ by $\\sqrt{3}+1$ we get:\n$$\n\\frac{\\sqrt{3}+1}{\\sqrt{3}-1} = \\frac{(\\sqrt{3}+1)^{2}}{\\sqr... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof only | null | |
0ld2 | Given a circumcircle $O$ and two fixed points $B, C$ on $O$ ($BC$ is not the diameter of $O$). A point $A$ is moving on $O$ such that $ABC$ is an acute triangle. Let $E, F$ be the feet of the altitudes from $B, C$ of triangle $ABC$. Let $I$ be an arbitrary circle passing through $E$ and $F$.
a) Let $I$ touch $BC$ at $... | [
"In case $AB = AC$, it is obvious that $DB = DC$ and the bisector of $\\angle MTN$ passes through the midpoint of the minor arc $BC$ of $O$ (which is the fixed point that is revealed in part b), hence we only need to consider the case where $AB \\neq AC$.\n\na) Let $R, S$ be the second intersections of $I$ with $AB... | Vietnam | VMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, i... | English | proof and answer | a) DB/DC = sqrt(cot B / cot C). b) The interior angle bisector of angle MTN always passes through the midpoint of the minor arc BC of the circumcircle O. | |
04oh | Prove that, for every positive integer $n \ge 2$, there exist positive integers $a_1, a_2, \dots, a_n$ such that for all $1 \le i < j \le n$ the expression
$$
\frac{a_j + a_i}{a_j - a_i}
$$
is a positive integer. | [
"We will prove the claim using mathematical induction on $n$. We will call the set $\\{a_1, a_2, \\dots, a_n\\}$ good if it satisfies the given condition.\n\nFor $n=2$ the set $\\{1,2\\}$ is clearly good. Let us assume that for some positive integer $n \\ge 2$ there exists a good set $\\{a_1, a_2, \\dots, a_n\\}$. ... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
06bt | $ABCDEF$ is a hexagon inscribed in a circle. Show that the diagonals $AD$, $BE$, $CF$ are concurrent if and only if $AB \cdot CD \cdot EF = BC \cdot DE \cdot FA$. | [
"By Ceva's theorem, $AD$, $BE$, $CF$ are concurrent if and only if\n$$\n\\frac{\\sin \\angle CAD}{\\sin \\angle EAD} \\times \\frac{\\sin \\angle AEB}{\\sin \\angle CEB} \\times \\frac{\\sin \\angle ECF}{\\sin \\angle ACF} = 1.\n$$\n\nBy the extended sine law, we have\n$$\n\\frac{\\sin \\an... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0h11 | In a triangle $ABC$ with $AC > BC > AB$. Points $D$ and $K$ are chosen on the sides $BC$ and $AC$ respectively so that $CD = AB$, $AK = BC$. $F$ and $L$ are the midpoints of the segments $BD$ and $KC$ respectively. $R$ and $S$ are the midpoints of the sides $AC$ and $AB$ respectively. The line segments $SL$ and $FR$ in... | [
"\nFig. 41\n\n$\\angle FOS = \\angle BJT = 55^\\circ \\Rightarrow \\angle BJC = 125^\\circ$, and because $J$ is the incenter $\\angle BJC = 90^\\circ + \\frac{1}{2}\\angle BAC \\Rightarrow \\angle BAC = 70^\\circ$."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 70° | |
0gbq | 設 $A$, $B$, $C$ 分別為 $\triangle A'B'C'$ 的邊 $B'C'$, $C'A'$, $A'B'$ 的中點。設點 $P$ 位於 $\triangle ABC$ 內部, $AP$, $BP$, $CP$ 分別與 $BC$, $CA$, $AB$ 交於點 $P_a$, $P_b$, $P_c$。
直線 $P_aP_b$, $P_aP_c$ 分別與 $B'C'$ 交於 $R_b$, $R_c$; 直線 $P_bP_c$, $P_bP_a$ 分別與 $C'A'$ 交於 $S_c$, $S_a$; 直線 $P_cP_a$, $P_cP_b$ 分別與 $A'B'$ 交於 $T_a$, $T_b$。已知 $S_c$,... | [
"先證明 $A$ 為 $R_bR_c$ 的中點：因 $BC \\parallel B'C'$，知 $\\frac{AR_c}{BP_a} = \\frac{AP_c}{P_cB}$ 及 $\\frac{AR_b}{CP_a} = \\frac{AP_b}{P_bC}$。由這兩式及西瓦定理知 $\\frac{AR_c}{AR_b} = \\frac{AP_c}{P_cB} \\cdot \\frac{BP_a}{P_aC} \\cdot \\frac{CP_b}{P_bA} = 1$。顯然 $R_b$, $R_c$ 在 $A$ 的異側，所以 $A$ 為 $R_bR_c$ 的中點。\n\n同理可證 $B$ 為 $S_cS_a$ ... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 獨立研究(二) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | null | proof only | null | |
0di5 | Several boxes of weight no more than $1$ kg weigh $36$ kg in total. Your friend can carry at most $4$ kg at a time. Prove that he can carry all the weights in no more than $11$ times. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
06rd | Determine all sequences $\left(x_{1}, x_{2}, \ldots, x_{2011}\right)$ of positive integers such that for every positive integer $n$ there is an integer $a$ with
$$
x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}=a^{n+1}+1 .
$$ | [
"Throughout this solution, the set of positive integers will be denoted by $\\mathbb{Z}_{+}$.\nPut $k=2+3+\\cdots+2011=2023065$. We have\n$$\n1^{n}+2 k^{n}+\\cdots 2011 k^{n}=1+k \\cdot k^{n}=k^{n+1}+1\n$$\nfor all $n$, so $(1, k, \\ldots, k)$ is a valid sequence. We shall prove that it is the only one.\nLet a vali... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | The unique sequence is x1 = 1 and x2 = x3 = … = x2011 = 2023065. | |
0b7p | Each point of the plane is coloured in one of two colours. Given an odd integer number $n \ge 3$, prove that there exist (at least) two similar triangles whose similitude ratio is $n$, each of which has a monochromatic vertex-set. | [
"We first show that there exists a rectangle with monochromatic vertices, then subdivide it into $n^2$ rectangles by subdividing each side into $n$ congruent segments, and show that (at least) one of these smaller rectangles has (at least) three monochromatic vertices – here is where the assumption that $n$ be odd ... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
02ss | Problem:
Na cidade de Cordisburgo, foi realizada uma corrida de bicicleta num circuito circular, da qual participaram três ciclistas, Guimarães, Rosa e João. Na primeira hora da corrida, Guimarães fez exatamente 230 voltas completas, João fez exatamente 111 voltas completas, porém não se sabe quantas voltas Rosa reali... | [
"Solution:\n\nSeja $x$ o número de voltas que Rosa realizou nessa primeira hora de corrida. Como Guimarães deu mais voltas que Rosa, e ambos andaram com velocidade constante, o número de vezes que Guimarães ultrapassou Rosa foi\n$$\n230 - x\n$$\nComo Rosa deu mais voltas que João, e ambos andaram com velocidade con... | Brazil | Brazilian Mathematical Olympiad | [
"Math Word Problems"
] | null | final answer only | 238 | |
01ma | In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of other singers such that he wishes to perform later than all the singers from the set.
Can it happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied?
(IMO-2010 Shortlist, Problem C1) | [
"3. See IMO-2010 Shortlist, Problem C1."
] | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Yes | |
0lej | Let $ABC$ be a triangle and $N$ be a point that differs from $A$, $B$ and $C$. Let $A_b$ be the reflection of $A$ through $NB$, and $B_a$ be the reflection of $B$ through $NA$. We define $B_c$, $C_b$, $A_c$ and $C_a$ similarly. Let $m_a$ be the line passing through $N$ and perpendicular to $B_cC_b$. Define $m_b$, $m_c$... | [
"First, we present two following lemmas. Let $ABC$ be a triangle inscribed in $(O)$ and $N$ be the nine-point center of $\\triangle ABC$.\n\n**Lemma 1.** Let $K$ be the center of $(BOC)$, then $AN$, $AK$ are isogonal with respect to $\\angle BAC$.\n\n*Proof.* Let $X$ be the reflection of $O$ through $BC$. Let $H$ b... | Vietnam | TST | [
"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 > Circles > Radical axis theorem",
... | English | proof only | null | |
01mu | Several chess players took part in a chess tournament. Each participant played exactly one game with any other participant. A participant received $1$ point for a win, $0.5$ point for a draw, and $0$ point for a lose. Any two players received different numbers of points and the participant taking the last place receive... | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2k | |
01bm | Let $ABC$ be a triangle with circumcircle $\omega$. Let $D$, $E$ and $F$ be points on the sides $BC$, $CA$ and $AB$ such that the circumcircle of the triangle $DEF$ touches $\omega$ at $A$. Let $G$ and $H$ be the intersection points of the circumcircles of the triangles $BDE$ and $CDF$ with $\omega$ (different from $B$... | [
"The dilation which maps the circumcircle of the triangle $DEF$ onto $\\omega$ maps $E$ and $F$ to $C$ and $B$, respectively. Hence $EF$ is parallel to $BC$.\n\nLet $X$ be the intersection point of $AD$ with $\\omega$ (different from $A$), let $Y$ be the intersection point of $AD$ with $GE$ and let $Z$ be the inter... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04f5 | Let $n \ge 2$ be a positive integer and $x_1, x_2, \dots, x_n$ real numbers ($x_i \ne 0$, for $i=1, 2, \dots, n$) such that $x_1 + x_2 + \dots + x_n = 0$. Prove that there exist distinct positive integers $i$ and $j$ ($i, j \le n$) such that
$$
\frac{1}{2} \le \left| \frac{x_i}{x_j} \right| \le 2.
$$ | [
"If there are two equal numbers between the given numbers, we can choose those two. So we assume that all the numbers are distinct and without loss of generality we assume that $x_1 > x_2 > \\dots > x_n$. Since the total sum of the numbers is zero there exists a positive integer $k$ ($1 < k < n$) such that $x_k > 0... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0cml | The two circles $\omega_1$ and $\omega_2$ touch each other externally at point $O$. Let $A$ and $B$ be points on $\omega_1$, and $C$ and $D$ be points on $\omega_2$ such that lines $AC$ and $BD$ are common external tangents to the circles. Line $AO$ intersects segment $CD$ at point $M$, while line $CO$ intersects $\ome... | [
"Из симметрии относительно линии центров $\\ell$ имеем: $AB \\parallel CD \\perp \\ell$, $OA = OB$, $\\angle OAB = \\angle OBA = \\alpha$, $\\angle OCD = \\angle ODC = \\beta$ (см. рис. 9).\n\nПо теореме об угле между касательной и хордой $\\angle OAC = \\angle OBA = \\alpha$, $\\angle OCA = \\angle ODC = \\beta$. ... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null |
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