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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
03u3 | In triangle $ABC$ the bisector of angle $BCA$ intersects the circumcircle at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area. | [
"If $AC = BC$, $\\triangle ABC$ is an isosceles triangle, and $CR$ is the symmetry axis of $\\triangle RQL$ and $\\triangle RPK$. The conclusion is obviously true.\n\nIf $AC \\ne BC$, without loss of generality, let $AC < BC$. Denote the center of circumcircle of $\\triangle ABC$ by $O$.\n\nSince the right triangle... | China | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordin... | English | proof only | null | |
0cn2 | Let $ABC$ be an acute non-isosceles triangle. Let $AA'$, $H$, $O$ be its altitude, orthocenter, and circumcenter respectively. Prove that the point symmetrical to the circumcenter of triangle $HOA'$ about line $HO$ lies on a midline of triangle $ABC$. (D. Prokopenko) | [
"Пусть $A_0, B_0, C_0$ — середины сторон $BC, CA, AB$ соответственно. Так как $OA_0 \\perp BC$ и $B_0C_0 \\parallel BC$, то $A_0O$ — высота треугольника $A_0B_0C_0$. Аналогично, $B_0O$ и $C_0O$ — высоты треугольника $A_0B_0C_0$. Поэтому $O$ является точкой пересечения высот треугольника $A_0B_0C_0$. Треугольник $AB... | Russia | Russian 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 > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null | |
054d | Let $O$ be the circumcentre of the acute triangle $ABC$. Let $c_1$ and $c_2$ be the circumcircles of triangles $ABO$ and $ACO$. Let $P$ and $Q$ be points on $c_1$ and $c_2$ respectively, such that $OP$ is a diameter of $c_1$ and $OQ$ is a diameter of $c_2$. Let $T$ be the intersection of the tangent to $c_1$ at $P$ and... | [
"Since $\\angle OAP = \\angle OAQ = 90^\\circ$, the points $P$, $A$ and $Q$ are collinear. Since $\\angle OPT = \\angle OQT = 90^\\circ$, $OPTQ$ is cyclic. Since $OA = OB$, the diameter $OP$ of $c_1$ is perpendicular to the chord $AB$. Therefore $PT$ and $AB$ are parallel.\n\nNow $\\angle TOQ = \\angle TPQ = \\angl... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0hi4 | Find all pairs of non-negative integers $x \ge y$ for which $x + 3^y$ and $y + 3^x$ are two consecutive integers. | [
"Obviously, $x > y$ (since $x + 3^y \\neq y + 3^x$). First, we prove the following lemma.\n\n**Lemma.** For any natural number $n > 1$, the inequality $3^n > n + 2$ holds.\n\n*Proof.* We prove the statement by induction. For $n=2$, we have that $3^2 = 9 > 2 + 2 = 4$. Suppose that the statement holds for some $n \\g... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | (x, y) = (1, 0) | |
019r | A set $a_1 \ge a_2 \ge \dots \ge a_{100n}$ of positive numbers has the following property. For any subset of $2n + 1$ numbers, the sum of $n$ maximal numbers in this subset is greater than the sum of the remaining $n + 1$ numbers. Prove that
$$
(n + 1)(a_1 + a_2 + \dots + a_n) > a_{n+1} + a_{n+2} + \dots + a_{100n}.
$$ | [
"Let us sum up all the inequalities for each subset of consecutive $(2n + 1)$ numbers in our set:\n$$\n\\begin{align*}\na_1 + a_2 + \\dots + a_n &\\ge a_{n+1} + a_{n+2} + \\dots + a_{2n+1}, \\\\\na_2 + a_3 + \\dots + a_{n+1} &\\ge a_{n+2} + a_{n+3} + \\dots + a_{2n+2}, \\\\\n\\dots\\dots\\dots\\dots \\\\\na_{n+1} +... | Baltic Way | Baltic Way 2013 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
037p | Problem:
Prove that if $x$, $y$ and $a$ are real numbers from the interval $(0,1)$, then
$$
\frac{|x-y|}{1-x y} \leq \frac{\left|x^{a}-y^{a}\right|}{1-x^{a} y^{a}}
$$ | [
"Solution:\nWe have to prove that\n$$\n\\frac{u^{q}-v^{q}}{1-(u v)^{q}} \\leq \\frac{u^{p}-v^{p}}{1-(u v)^{p}}\n$$\nfor $v \\leq u$ and $p<q$. To do this we shall show that\n$$\n\\frac{u^{p+1}-v^{p+1}}{1-(u v)^{p+1}} \\leq \\frac{u^{p}-v^{p}}{1-(u v)^{p}}\n$$\nwhich implies by induction the above inequality. It is ... | Bulgaria | Team selection test for 23. BMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0cm9 | Problem:
Let $T_{1}$, $T_{2}$, $T_{3}$, $T_{4}$ be pairwise distinct collinear points such that $T_{2}$ lies between $T_{1}$ and $T_{3}$, and $T_{3}$ lies between $T_{2}$ and $T_{4}$. Let $\omega_{1}$ be a circle through $T_{1}$ and $T_{4}$; let $\omega_{2}$ be the circle through $T_{2}$ and internally tangent to $\om... | [
"Solution:\n\nLet $O_{i}$ be the centre of $\\omega_{i}$, $i=1,2,3,4$. Notice that the isosceles triangles $O_{i} T_{i} T_{i-1}$ are similar (indices are reduced modulo $4$), to infer that $\\omega_{4}$ is internally tangent to $\\omega_{1}$ at $T_{4}$, and $O_{1} O_{2} O_{3} O_{4}$ is a (possibly degenerate) paral... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
03ty | Let $n$ be an integer greater than $3$, and let $a_1, a_2, \dots, a_n$ be nonnegative real numbers with $a_1 + a_2 + \dots + a_n = 2$.
Determine the minimum value of
$$
\frac{a_1}{a_2^2+1} + \frac{a_2}{a_3^2+1} + \cdots + \frac{a_n}{a_1^2+1}.
$$ | [
"The answer is $\\frac{3}{2}$.\n\nThe given problem is equivalent to finding the minimum value of\n$$\n\\begin{aligned}\nm &= 2 - \\left( \\frac{a_1}{a_2^2+1} + \\frac{a_2}{a_3^2+1} + \\cdots + \\frac{a_n}{a_1^2+1} \\right) \\\\\n &= \\left( a_1 - \\frac{a_1}{a_2^2+1} \\right) + \\left( a_2 - \\frac{a_2}{a_3^2+1} ... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 3/2 | |
0j6x | Problem:
Evaluate $2011 \times 20122012 \times 201320132013 - 2013 \times 20112011 \times 201220122012$. | [
"Solution:\nBoth terms are equal to $2011 \\times 2012 \\times 2013 \\times 1 \\times 10001 \\times 100010001$.\n\nTherefore, the answer is $0$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 0 | |
09dc | $\{a_n\}_{n \ge 1}$ натурал тоон дарааллын хувьд $\forall n \ge 1: 0 < a_{n+1} - a_n \le m$ (энд $m \in \mathbb{N}$ тогтмол тоо) бол $a_r \mid a_s$ ба $1 \le r < s \in \mathbb{N}$ байх гишүүд олдоно гэж батад. | [
"Ямар ч дэс дараалсан $m$-ш $a_1$-ээс их буюу тэнцүү тооны хувьд ядаж 1-нь $\\{a_n\\}$ дарааллын гишүүн байх нь илэрхий.\n$$\n(m+1) \\times m \\text{ хүснэгтийн хувьд}\n$$\n$$\nx_{0,1} = a_1, \\quad x_{0,j} = x_{0,j-1} + 1, \\quad j = 2, m\n$$\n$$\n\\begin{array}{l}\nx_{0,1}, x_{0,2}, \\dots, x_{0,m} \\\\\nx_{1,1},... | Mongolia | ММО-48 | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Mongolian | proof only | null | |
0c5c | Let $n$ be a positive integer. Find all functions $f: [0, \infty) \to \mathbb{R}$, which are $n$ times differentiable, their $n^{th}$ derivative is bounded below and
$$
x \int_{x}^{x+1} f(t) \, dt = \int_{0}^{x} f(t) \, dt,
$$
for every real $x \ge 0$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Calculus > Integral Calculus > Applications",
"Calculus > Differential Calculus > Derivatives"
] | English | proof and answer | All constant functions: f(x) = c for some real constant c. | |
01zw | Problem:
On a line $l$ there are three different points $A$, $B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to... | [
"Solution:\n\na.\nSince $P$, $R$ and $Q$ are collinear, we have $\\triangle P A Q \\sim \\triangle P B R$, hence\n$$\n\\frac{|A Q|}{|B R|} = \\frac{|A P|}{|B P|}\n$$\nConversely, $P$, $T$ and $S$ are collinear if it holds that\n$$\n\\frac{|A S|}{|B T|} = \\frac{|A P|}{|B P|}\n$$\nSo it suffices to prove\n$$\n\\frac... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bzm | Vlad, Luca and Adina bought school supplies worth of $118$ lei. Vlad bought $5$ pens, $4$ notebooks and $3$ pencil boxes, Luca bought $7$ pens, $3$ notebooks and $4$ pencil boxes, and Adina bought $8$ pens, $7$ notebooks and $7$ pencil boxes. Given that Luca payed $5$ lei more than Vlad, and Adina $4$ lei less than Vla... | [] | Romania | 69th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | pen 1 lei, notebook 2 lei, pencil box 5 lei | |
0eo1 | An altitude from $A$ to the triangle $ABC$ intersects the side $BC$ in $D$. A circle touches $BC$ in $D$, intersects $AB$ in $M$ and $N$ and intersects $AC$ in $P$ and $Q$. Prove that
$$
\frac{AM + AN}{AC} = \frac{AP + AQ}{AB}.
$$ | [
"Let $P(X)$ denote the power of point $X$ with respect to the circle $M, N, D$. Then\n$$\n\\begin{align*}\nAB(AM + AN) &= AB \\cdot AM + AB \\cdot AN \\\\\n&= (AN + NB)AM + AB(AB - BN) \\\\\n&= AN \\cdot AM + BN \\cdot AM + AB^2 - AB \\cdot BN \\\\\n&= P(A) + AD^2 + BD^2 - BN(AB - AM) \\\\\n&= P(A) + AD^2 + P(B) - ... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0jjr | Problem:
Let $x_{1}, \ldots, x_{100}$ be defined so that for each $i$, $x_{i}$ is a (uniformly) random integer between $1$ and $6$ inclusive. Find the expected number of integers in the set $\{x_{1}, x_{1}+x_{2}, \ldots, x_{1}+x_{2}+\ldots+x_{100}\}$ that are multiples of $6$. | [
"Solution:\n\nAnswer: $\\boxed{\\dfrac{50}{3}}$\n\nNote that for any $i$, the probability that $x_1 + x_2 + \\ldots + x_i$ is a multiple of $6$ is $\\dfrac{1}{6}$ because exactly $1$ value out of $6$ possible values of $x_{i}$ works. Because these $100$ events are independent, the expected value is $100 \\cdot \\df... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 50/3 | |
092s | Problem:
Let $ABC$ be an acute-angled triangle with $AB \neq AC$, and let $O$ be its circumcentre. The line $AO$ intersects the circumcircle $\omega$ of $ABC$ a second time in point $D$, and the line $BC$ in point $E$. The circumcircle of $CDE$ intersects the line $CA$ a second time in point $P$. The line $PE$ interse... | [
"Solution:\n\nLet us denote $\\Varangle ABC$ by $\\beta$ and $\\Varangle BCA$ by $\\gamma$. Without loss of generality, $AB > AC$, or equivalently $\\beta < \\gamma$, as in the figure.\nSegment $AD$ is a diameter of $\\omega$, so by Thales' theorem we have $\\Varangle DCA = 90^{\\circ}$. Since the quadrilateral $CE... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle cen... | null | proof only | null | |
02k4 | Problem:
Larissa e Jorge estão jogando com cartões numerados de $1$ a $6$ que devem ser colocados nas casas do tabuleiro abaixo de modo a formar um número de seis algarismos.
Jorge coloca o primeiro cartão e a seguir as jogadas são alternadas entre os dois. O objetivo de Larissa é obter o m... | [
"Solution:\n\nA formação de um número de $6$ algarismos é ilustrada a seguir.\n\n| centena <br> de milhar | dezena de <br> milhar | unidade <br> de <br> milhar | centena | dezena | unidade |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n\nPara se obter o menor número possível, os menores algarismos devem esta... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | MCQ | B | |
0e9h | One day $n$ students participated at a sporting competition with several different disciplines. Exactly $k$ students competed in each of the disciplines where $3 \le k \le n$. Each pair of students had exactly three disciplines in common and each triple has competed in exactly two common disciplines. Determine all poss... | [
"The number of pairs of students over all disciplines can be counted in two ways. On the one hand the number of pairs is equal to the product of the number of all pairs and the number of times every pair of students competes together, which is $\\binom{n}{2} \\cdot 3$. On the other hand it must be equal to the prod... | Slovenia | National Math Olympiad in Slovenia | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | n = 5, k = 4 | |
01sx | The sequence $a_1, a_2, \dots$ of positive real numbers satisfies
$$
a_{k+1} \ge \frac{ka_k}{a_k^2 + k - 1}
$$
for every positive integer $k$.
Prove that $a_1 + a_2 + \dots + a_n \ge n$ for every $n \ge 2$. | [
"2. See IMO-2015 Shortlist, Problem A1."
] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
06i0 | If $x$ is a real number, find the minimum value of $|x+1| + 2|x-5| + |2x-7| + \frac{|x-11|}{2}$.
若 $x$ 為實數, 求 $|x+1| + 2|x-5| + |2x-7| + \frac{|x-11|}{2}$ 的最小值。 | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Chinese | proof and answer | 45/4 | |
0kvh | Problem:
Compute the smallest multiple of $63$ with an odd number of ones in its base two representation. | [
"Solution:\n\nNotice that $63 = 2^{6} - 1$, so for any $a$ we know\n$$\n63a = 64a - a = 2^{6}(a-1) + (64 - a)\n$$\nAs long as $a \\leq 64$, we know $a-1$ and $64-a$ are both integers between $0$ and $63$, so the binary representation of $63a$ is just $a-1$ followed by $64-a$ in binary (where we append leading $0$s ... | United States | HMMT November 2023 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 4221 | |
0aja | The circles $k_1$, $k_2$ intersect at points $A$ and $B$. A line through $B$ intersects the circles $k_1$ and $k_2$ for the second time at points $C$ and $D$, respectively, in such a way that $C$ lies outside of $k_2$, and $D$ lies outside of $k_1$. Let $M$ be the point of intersection of the tangents to $k_1$ and $k_2... | [
"**Due to symmetry reasons, it is enough to show that $KP \\parallel MC$ holds.**\n\nFirst we show that the quadrilateral $ACMD$ is inscribed. **(1 point)** Namely, let us notice that $B$ lies on the line segment $\\overline{CD}$, and $A$ and $M$ are on different sides of the line $CD$. From $\\angle BDM = \\angle ... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j8a | Problem:
Ten $C$s are written in a row. Some $C$s are upper-case and some are lower-case, and each is written in one of two colors, green and yellow. It is given that there is at least one lower-case $C$, at least one green $C$, and at least one $C$ that is both upper-case and yellow. Furthermore, no lower-case $C$ ca... | [
"Solution:\n\nAnswer: $36$\n\nBy the conditions of the problem, we must pick some point in the line where the green $C$s transition to yellow, and some point where the upper-case $C$s transition to lower-case. We see that the first transition must occur before the second, and that they cannot occur on the same $C$.... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 36 | |
0bxa | Let $n$ be an odd integer greater than $1$. Each vertex of a regular $n$-gon is coloured one of three colours so that the total number of vertices of each colour is odd. Show that there exists an isosceles triangle whose vertices have pairwise distinct colours. | [
"Suppose, if possible, no such triangles exist, and consider the parity of the total number of pairs $(T, e)$, where $T$ is an isosceles triangle, and $e$ is an edge of $T$ whose endpoints have different colors.\n\nOn the one hand, under the above assumption, the vertices of a triangle in a pair form a bichromatic ... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
05cj | A positive integer $m$ is called *usual* if the square of every prime divisor of $m$ is less than $m$.
a. Prove that there are infinitely many positive integers $n$ such that both $n$ and $n+1$ are usual.
b. Is there a positive integer $n$ such that $n$, $n+1$ and $n+2$ are all usual? | [
"Let $a$ be a composite number such that $a+1$ is composite, too. Then $a^2$ is usual since each of its prime divisors is a prime divisor of $a$ and therefore less than $a$. We show that $a^2-1$ is usual, too. To this end, observe that $a^2-1 = (a-1)(a+1)$. As $a+1$ is composite, all prime divisors of $a+1$ are les... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a) Infinitely many such n exist. For example, if a and a+1 are both composite, then n = a^2 − 1 works. b) Yes. For instance, n = 440 gives 440, 441, 442 all usual (also n = 9800 gives 9800, 9801, 9802). | |
0f6p | Problem:
Given a line $L$ and a point $O$ not on the line, can we move an arbitrary point $X$ to $O$ using only rotations about $O$ and reflections in $L$? | [] | Soviet Union | 19th ASU | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof only | null | |
0cdi | Let $A, B \in \mathcal{M}_3(\mathbb{C})$ be two matrices with the property $A^2 = B^2 = O_3$. Prove that $AB = BA$ implies $AB = O_3$. Show that the reciprocal implication is false. | [
"Assume $AB = BA$. We obtain $(A+B)^2 = 2AB$ and $(A+B)^3 = 2AB(A+B) = 2A^2B + 2AB^2 = O_3$. Then, $(A+B)(A-B) = A^2 - B^2 = O_3$. From Sylvester rank inequality, we get $\\text{rank}(A+B) + \\text{rank}(A-B) \\le 3$. Thus, $\\text{rank}(A+B) \\le 1$ or $\\text{rank}(A-B) \\le 1$.\n\nSuppose $\\text{rank}(A+B) \\le... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0f82 | Problem:
Form 10A has 29 students who are listed in order on its duty roster. Form 10B has 32 students who are listed in order on its duty roster. Every day two students are on duty, one from form 10A and one from form 10B. Each day just one of the students on duty changes and is replaced by the following student on t... | [
"Solution:\n\nAnswer: no.\n\nSuppose such an arrangement is possible. Suppose that it includes $m$ cycles through the form 10A roster and $n$ cycles through the 10B roster. Then the total number of changes is $29m + 32n = 29 \\times 32$ (since each pair occurs once). But that means $29$ divides $n$ and $32$ divides... | Soviet Union | 22nd ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | no | |
06vz | Suppose that $a, b, c, d$ are positive real numbers satisfying $(a+c)(b+d)=a c+b d$. Find the smallest possible value of
$$
S=\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}
$$
(Israel) | [
"To show that $S \\geqslant 8$, apply the AM-GM inequality twice as follows:\n$$\n\\left(\\frac{a}{b}+\\frac{c}{d}\\right)+\\left(\\frac{b}{c}+\\frac{d}{a}\\right) \\geqslant 2 \\sqrt{\\frac{a c}{b d}}+2 \\sqrt{\\frac{b d}{a c}}=\\frac{2(a c+b d)}{\\sqrt{a b c d}}=\\frac{2(a+c)(b+d)}{\\sqrt{a b c d}} \\geqslant 2 \... | IMO | IMO 2020 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 8 | |
0cba | Find all four-digit numbers $\overline{abcd}$, so that $\overline{ab}$, $\overline{cb}$ and $d$ are prime numbers, and $\overline{ab}^2 + \overline{cb}^2 + d^2 = 2022$. | [
"If $c = 0$, then $b^2 + d^2 \\le 2 \\cdot 7^2$, which means that $\\overline{ab}$ must be a prime number such that $1924 \\le \\overline{ab}^2 < 2022$, false. Therefore, $c \\ne 0$. As $\\overline{ab}$ and $\\overline{cb}$ are primes, from the given equality we deduce that $d = 2$. We obtain $\\overline{ab}^2 + \\... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1342, 4312 | |
05nr | Problem:
Soit $n \geq 5$ un entier, et $E_{1}, E_{2}, \ldots, E_{2 n-1}$ des parties distinctes à deux éléments de $\{1,2, \ldots, n\}$.
Prouver que, parmi les $2 n-1$ parties $E_{i}$, on peut en choisir $n$ de sorte que la réunion de ces $n$ parties ne contienne pas plus de $\frac{2}{3} n+1$ éléments. | [
"Solution:\n\nOn va prouver par récurrence sur $k \\leq \\frac{2 n-1}{3}$ que l'on peut toujours éliminer $3 k$ des $2 n-1$ parties de sorte que la réunion des $2 n-1-3 k$ restantes ne contienne pas plus de $n-k$ éléments.\n\nLe cas $k=0$ est immédiat.\n\nSoit maintenant $1 \\leq k \\leq \\frac{2 n-1}{3}$. On suppo... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
05wu | Problem:
Soient $A, B, C, D$ et $E$ cinq points dans cet ordre sur un cercle tels que $A E = D E$. Soit $P$ le point d'intersection de $(A C)$ et $(B D)$. Soit $Q$ le point de la demi-droite $[B A)$ tel que $A Q = D P$. Soit $R$ le point de la demi-droite $[C D)$ tel que $D R = A P$. Montrer que les droites $(P E)$ et ... | [
"Solution:\nOn se place dans le cas de la figure suivante :\n\nDans cet énoncé on a de nombreuses égalités de longueur ($A Q = D P$, $D R = A P$, $A E = D E$). Comme sur la figure, il ne faut pas hésiter à colorier les longueurs égales pour voir des potentiels triangles isométriques, et ava... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0d4k | Two circles $\omega_{1}$ and $\omega_{2}$, with center $O_{1}$ and $O_{2}$ respectively, meet at points $A$ and $B$. Let $X$ and $Y$ be points on $\omega_{1}$. Lines $X A$ and $Y A$ meet $\omega_{2}$ at $Z$ and $W$, respectively, such that $A$ lies between $X$ and $Z$ and between $Y$ and $W$. Let $M$ be the midpoint of... | [
"Solution suggested by the Student Alzubair Habibullah. Because $M$ is the midpoint of $O_{1} O_{2}$, we have from Apollonius' theorem\n$$\n4 M S^{2} = 2 S O_{1}^{2} + 2 S O_{2}^{2} - O_{1} O_{2}^{2}\n$$\nBecause $S$ is the midpoint of segment $A X$, segment $O_{1} S$ is perpendicular to segment $A X$ and therefore... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English, Arabic | proof only | null | |
0709 | Problem:
$\mathrm{M}$ is the midpoint of the median $\mathrm{AD}$ of the triangle $\mathrm{ABC}$. The ray $\mathrm{BM}$ meets $\mathrm{AC}$ at $N$. Show that $\mathrm{AB}$ is tangent to the circumcircle of $\mathrm{NBC}$ iff $\mathrm{BM} / \mathrm{BN}=(\mathrm{BC} / \mathrm{BN})^{2}$. | [
"Solution:\n\n\n\nApplying Menelaus to the triangle $\\mathrm{ADC}$, we have ($\\mathrm{AM} / \\mathrm{MD})(\\mathrm{BD} / \\mathrm{DC})(\\mathrm{CN} / \\mathrm{NA})=1$, so $(\\mathrm{CN} / \\mathrm{NA})=2$. Hence $\\mathrm{AN} / \\mathrm{AC}=1 / 3$. Applying Menelaus to the triangle $\\mat... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
054u | Find the number of solutions of the equation $|a - b| = |b - c|$ in integers from $0$ to $36$. | [
"The equation $|a - b| = |b - c|$ is satisfied if and only if either $a - b = b - c$ or $a - b = c - b$. The first equality is equivalent to $a + c = 2b$, the second is equivalent to $a = c$.\n\nTo fulfill the condition $a + c = 2b$, the numbers $a$ and $c$ must have the same parity. Then their sum is even and $b$ ... | Estonia | National Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2017 | |
09dn | $0 \ne f(x),\ g(x) \in \mathbb{R}[x]$, $f(x^3) + g(x) = f(x) + x^5 g(x)$ байх $f(x)$ нь хамгийн бага зэрэгтэй байх олон гишүүнтүүдийн жишээ гарга. | [
"$f(x^3) - f(x) = g(x) \\cdot (x^5 - 1)$.\n\n$f(x^3) - f(x) = x^3 - x = x(x-1)(x+1)$ байх шаардлагатай бөгөөд $f(x^3) - f(x) : x^3 - x = x(x-1)(x+1)$ ба $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$ гэдгийг тооцвол $f(x^3) - f(x) = 0 \\pmod{x^4 + x^3 + x^2 + x + 1}$ болно.\n\n$f_1^k = 1 \\pmod{x^4 + x^3 + x^2 + x + ... | Mongolia | ММО-48 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | Mongolian | proof and answer | Minimal degree of f is 4. One family of solutions is f(x) = a(x^4 + x^3 + x^2 + x) + e and g(x) = a(x^7 + x^4 + x^2 + x), where a ≠ 0 and e is any real constant. | |
0kwq | Problem:
A complex quartic polynomial $Q$ is quirky if it has four distinct roots, one of which is the sum of the other three. There are four complex values of $k$ for which the polynomial $Q(x) = x^{4} - k x^{3} - x^{2} - x - 45$ is quirky. Compute the product of these four values of $k$. | [
"Solution:\n\nLet the roots be $a, b, c, d$ with $a + b + c = d$. Since $a + b + c = k - d$ by Vieta's formulas, we have $d = k / 2$. Hence\n\n$$\n0 = Q\\left(\\frac{k}{2}\\right) = \\left(\\frac{k}{2}\\right)^{4} - k\\left(\\frac{k}{2}\\right)^{3} - \\left(\\frac{k}{2}\\right)^{2} - \\left(\\frac{k}{2}\\right) - 4... | United States | HMMT November 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 720 | |
09k4 | Let $n$ be a positive integer and suppose that $n = a_1 + a_2 + \cdots + a_m$ for $m \ge 2$ and $a_1 > a_2 > \cdots > a_m$ distinct positive integers satisfying
$$
\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_m} = 1.
$$
Prove that the inequality $n + 2m + 1 \le a_1 a_2$ holds and determine the conditions under w... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | Equality holds if and only if m = 3 and (a1, a2, a3) = (6, 3, 2); in all other cases the inequality is strict. | |
077o | In the fictional country of Mahishmati, there are $50$ cities, including a capital city. Some pairs of cities are connected by two-way flights. Given a city $A$, an ordered list of cities $C_1, \dots, C_{50}$ is called an *antitour* from $A$ if
* every city (including $A$) appears in the list exactly once, and
* for ea... | [
"Rephrase in terms of graph theory:\nLet $G$ be a connected graph on $50$ vertices. Given a vertex $A$, a permutation $(V_1, \\dots, V_{50})$ of the vertices of $G$ is called an antitour from $A$ if the following condition holds for every integer $k$ with $1 \\le k \\le 50$: there is no walk of length $k$ from $A$ ... | India | IMO TST Day 1 | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | (25!)^2 | |
0cgb | Is it possible to arrange the first 49 positive integers in a $7 \times 7$ array, so that each cell contains exactly one number and no two primes are neighbours?
Two neighbours are numbers from cells with a common side or a common vertex. | [
"It is possible. The positive primes less than $49$ are $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, that is $15$ numbers.\n\nNumber the lines and the columns from $1$ to $7$, set the primes in $15$ of the $16$ squares with both odd indexes and the o... | Romania | 74th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | Yes, it is possible. | |
0ala | Problem:
Let $ABCD$ be a rectangle with $AB = 3$ and $BC = 1$. Let $E$ be the midpoint of $AB$ and let $F$ be a point on the extension of $DE$ such that $|CF| = |CD|$. Find the area of $\triangle CDF$. | [] | Philippines | Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 54/13 | |
09i6 | Let $\alpha$ and $\beta$ be distinct real numbers. Consider the sequence $A_1, A_2, \dots$ of real numbers given by
$$
A_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}
$$
for $n \ge 1$. If the members $A_p, A_{p+1}, A_{p+2}$ are integers for some prime number $p$, prove that all members of the sequence are integers.
(Pr... | [] | Mongolia | Round 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
07u8 | For $n$ a positive integer, $n! = 1 \cdot 2 \cdot 3 \cdots (n-1) \cdot n$ is the product of the positive integers from 1 to $n$. Determine, with proof, all positive integers $n$ for which $n! + 3$ is a power of 3. | [
"The only solutions are $3! + 3 = 3^2$ and $4! + 3 = 3^3$. We can calculate directly that $5! + 3 = 123$ is not a power of 3. For $n \\ge 6$, we have $9 \\mid n!$ as 3 and 6 are two of the factors defining $n!$. But also 9 is a factor of all powers of 3 beyond $3^1$, so we would have the contradiction that 9 divide... | Ireland | IRL_ABooklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | n = 3 and n = 4 | |
0l00 | Problem:
Let $ABCD$ be a rectangle whose vertices are labeled in counterclockwise order with $AB=32$ and $AD=60$. Rectangle $AB^{\prime}C^{\prime}D^{\prime}$ is constructed by rotating $ABCD$ counterclockwise about $A$ by $60^{\circ}$. Given that lines $BB^{\prime}$ and $DD^{\prime}$ intersect at point $X$, compute $C... | [
"Solution:\n\n\n\nThe key claim is the following.\nClaim 1. $\\angle BXD = 90^{\\circ}$.\n\nProof. We see that $\\angle ABB^{\\prime} = \\angle AD^{\\prime}D = 60^{\\circ}$ and $\\angle BAD^{\\prime} = 90^{\\circ} + \\angle DAD^{\\prime} = 150^{\\circ}$, from which we get $\\angle BX D^{\\p... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 34 | |
0dku | Let $P(x)$ be a polynomial with integer coefficients, the leading one being positive. Prove that there are finitely many positive integers $n$ such that $n! + 1$ is a power of $P(n)$. | [
"For each positive integer $n$ such that $n! + 1$ is a power of $P(n)$, we write $n! + 1 = P(n)^{k_n}$ and in this case call $n$ good. Due to the left-hand side and $P(x) \\ge x$ for $x$ big enough, we have $k_n < n$ and $P(n)$ must be odd for $n \\ge 2$. Hence, using the LTE, we obtain\n$$\n\\nu_2 (P(n)^{k_n} - 1)... | Saudi Arabia | Saudi Booklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
03mr | Fix positive integers $n$ and $k \ge 2$. A list of $n$ integers is written in a row on a blackboard. You can choose a contiguous block of integers, and I will either add 1 to all of them or subtract 1 from all of them. You can repeat this step as often as you like, possibly adapting your selections based on what I do. ... | [
"We will think of all numbers as being residues mod $k$. Consider the following strategy:\n\n* If there are less than $k-1$ non-zero numbers, then stop.\n* If the first number is 0, then recursively solve on the remaining numbers.\n* If the first number is $j$ with $0 < j < k$, then choose the interval stretching f... | Canada | Kanada 2014 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
00x2 | Problem:
Find all fourth degree polynomials $p(x)$ such that the following four conditions are satisfied:
(i) $p(x) = p(-x)$ for all $x$.
(ii) $p(x) \geq 0$ for all $x$.
(iii) $p(0) = 1$.
(iv) $p(x)$ has exactly two local minimum points $x_1$ and $x_2$ such that $|x_1 - x_2| = 2$. | [
"Solution:\nLet $p(x) = a x^4 + b x^3 + c x^2 + d x + e$ with $a \\neq 0$.\nFrom (i)-(iii) we get $b = d = 0$, $a > 0$ and $e = 1$.\nFrom (iv) it follows that $p'(x) = 4 a x^3 + 2 c x$ has at least two different real roots.\nSince $a > 0$, we have $c < 0$ and $p'(x)$ has three roots $x = 0$, $x = \\pm \\sqrt{ -c / ... | Baltic Way | Baltic Way 1992 | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | All polynomials of the form p(x) = a(x^2 − 1)^2 + 1 − a with 0 < a ≤ 1. | |
08i1 | Problem:
Let $a$ and $b$ be positive integers such that $a + b \leq 10$. Find all pairs $(a, b)$ such that the fraction $(2n + a)/(5n + b)$ is irreducible for every natural number $n$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | [(1,2), (1,3), (3,7)] | |
05zt | Problem:
Soit $ABC$ un triangle et soit $J$ le centre de son cercle $A$-exinscrit. Soit $K$ le symétrique du point $J$ par rapport au segment $[BC]$. Soient $E$ et $F$ les points sur les droites $(BJ)$ et $(CJ)$ tels que $\widehat{EAB} = \widehat{CAF} = 90^{\circ}$. Montrer que $\widehat{FKE} + \widehat{FJE} = 180^{\c... | [
"Solution:\n\n\n\nPuisque la droite $(BJ)$ est la bissectrice extérieure de l'angle $\\widehat{ABC}$, on a $\\widehat{ABJ} = \\widehat{CBJ} = 90^{\\circ} - \\widehat{ABC}/2$. Puisque la symétrie axiale préserve les angles, on a ainsi $\\widehat{KBC} = \\widehat{ABE}$.\n\nNotons $A_1$ le pro... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
011l | Problem:
Fredek runs a private hotel. He claims that whenever $n \geqslant 3$ guests visit the hotel, it is possible to select two guests who have equally many acquaintances among the other guests, and who also have a common acquaintance or a common unknown among the guests. For which values of $n$ is Fredek right?
(... | [
"Solution:\n\nFredek is right for all $n \\neq 4$.\n\nSuppose that any two guests of Fredek having the same number of acquaintances have neither a common acquaintance nor a common unknown. From the set $\\mathcal{K}$ of Fredek's guests choose any two guests $A$ and $B$ having the same number of acquaintances (the e... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | All integers n>=3 except n=4 | |
0d77 | Find all pairs of polynomials $P(x), Q(x)$ with integer coefficients such that
$$
P(Q(x)) = (x-1)(x-2) \ldots (x-9)
$$
for all real numbers $x$. | [
"We consider 3 cases.\n\n1) Suppose that $\\deg Q = 1$. By looking at the leading coefficient, we have $Q(x) = \\pm x + a$. If $Q(x) = x + a$, we have\n$$\nP(x) = (x - (a+1))(x - (a+2)) \\ldots (x - (a+9)).\n$$\nAnd if $Q(x) = -x + a$ we have\n$$\nP(x) = (a-1-x)(a-2-x) \\ldots (a-9-x).\n$$\n\n2) Suppose that $\\deg... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | All solutions are when one polynomial is linear. Specifically:
- If Q(x) = x + a with a an integer, then P(x) = ∏_{k=1}^{9} (x − (a + k)). If Q(x) = −x + a, then P(x) = ∏_{k=1}^{9} (a − k − x).
- If P(x) = x + a with a an integer, then Q(x) = (x − 1)(x − 2)⋯(x − 9) − a. If P(x) = −x + a, then Q(x) = −(x − 1)(x − 2)⋯(x ... | |
0b61 | Prove that if $a$, $b$, $c$ are positive real numbers such that $abc = 1$, then
$$
(1 + a^3)(1 + b^3)(1 + c^3) \ge 2 \left(1 + \frac{b}{a} + \frac{c}{b} + \frac{a}{c}\right).
$$ | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
03bm | Prove that if $\alpha, \beta, \gamma \in [0, \frac{\pi}{2}]$ and $\tan\alpha + \tan\beta + \tan\gamma \le 3$, then
$$
\cos 2\alpha + \cos 2\beta + \cos 2\gamma \ge 0.
$$ | [
"*First solution.* The inequality follows by the fact that if $\\theta \\in [0, \\frac{\\pi}{2}]$, then\n$$\n\\cos 2\\theta = 1 - \\sin 2\\theta \\cdot \\tan \\theta \\ge 1 - \\tan \\theta.\n$$\n\n\n*Second solution.* Set $x = \\tan\\alpha$, $y = \\tan\\beta$ and $z = \\tan\\gamma$. Then $x, y, z \\ge 0$, $x+y+z \\... | Bulgaria | Bulgarian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0gwz | (Lishunov Vitalii) $\frac{1}{x-y} + \frac{1}{y-z} + \frac{1}{z-x} = \frac{3}{2}$ is true for some real numbers $x$, $y$, $z$. What value may expression $\frac{1}{(x-y)^2} + \frac{1}{(y-z)^2} + \frac{1}{(z-x)^2}$ have?
**Answer:** $\frac{9}{4}$. | [
"Let $B = \\frac{1}{x-y} + \\frac{1}{y-z} + \\frac{1}{z-x} = \\frac{3}{2}$ and $A = \\frac{1}{(x-y)^2} + \\frac{1}{(y-z)^2} + \\frac{1}{(z-x)^2}$ then\n$$B^2 = \\frac{9}{4} = A + 2\\left(\\frac{1}{(x-y)(y-z)} + \\frac{1}{(x-y)(z-x)} + \\frac{1}{(y-z)(z-x)}\\right).$$\nLet's rearrange the second summand of the last ... | Ukraine | Ukrajina 2008 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 9/4 | |
0i29 | Problem:
How many times does $24$ divide into $100!$ (factorial)? | [
"Solution:\n\nWe first determine the number of times $2$ and $3$ divide into $100! = 1 \\cdot 2 \\cdot 3 \\cdots 100$. Let $\\langle N \\rangle_{n}$ be the number of times $n$ divides into $N$ (i.e., we want to find $\\langle 100! \\rangle_{24}$).\n\nSince $2$ only divides into even integers, $\\langle 100! \\rangl... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 32 | |
0fif | Problem:
Hallar las tangentes de los ángulos de un triángulo sabiendo que son números enteros positivos. | [
"Solution:\n\nSean $\\alpha, \\beta, \\gamma$ los tres ángulos y supongamos $\\alpha \\leq \\beta \\leq \\gamma$.\nSi fuera $\\gamma \\geq \\frac{\\pi}{2}$, tendría que ser $\\alpha<\\frac{\\pi}{4}$ y entonces $\\tan \\alpha$ no sería entero.\nSi $\\tan \\alpha>1$, entonces $\\alpha \\geq \\arctan 2>\\arctan \\sqrt... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | tan(alpha) = 1, tan(beta) = 2, tan(gamma) = 3 | |
05ps | Problem:
Soit $ABC$ un triangle.
Pour un point $P$ de $(BC)$ donné, on note $E(P)$ et $F(P)$ les deuxièmes points d'intersection des droites $(AB)$ et $(AC)$ avec le cercle de diamètre $[AP]$. Soit $T(P)$ l'intersection des tangentes à ce cercle en $E(P)$ et $F(P)$.
Montrer que quand $P$ varie sur $(BC)$, le lieu gé... | [
"Solution:\n\n\n\n**Étape 1 : une première remarque importante.**\n\nPosons tout d'abord $H$ le pied de la hauteur issue de $A$ dans $ABC$. Soient deux points $P$ et $P'$ de la droite $(BC)$. On note $E = E(P)$, $F = F(P)$, $T = T(P)$, $E' = E(P')$, $F' = F(P')$, $T' = T(P')$. On introduit ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle ... | null | proof only | null | |
0ji4 | Problem:
Let points $A$ and $B$ be on circle $\omega$ centered at $O$. Suppose that $\omega_{A}$ and $\omega_{B}$ are circles not containing $O$ which are internally tangent to $\omega$ at $A$ and $B$, respectively. Let $\omega_{A}$ and $\omega_{B}$ intersect at $C$ and $D$ such that $D$ is inside triangle $A B C$. Su... | [
"Solution:\n\nLet $H = C A \\cap \\omega$, and $G = B H \\cap \\omega_{B}$. There are homotheties centered at $A$ and $B$ taking $\\omega_{A} \\rightarrow \\omega$ and $\\omega_{B} \\rightarrow \\omega$ that take $A: F \\mapsto E$, $A: C \\mapsto H$, $B: C \\mapsto E$ and $B: G \\mapsto H$. In particular $C F \\par... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
058p | Find all quadruples $(p, q, r, s)$ of primes that satisfy the following system of equations:
$$
\begin{cases} 6p + 5q + 5r + 3s = 130 \\ 3p + 3q + 5r + 6s = 130 \end{cases}
$$ | [
"Subtracting the second equation from the first one gives $3p + 2q - 3s = 0$ which implies $2q = 3(s - p)$. Thus $2q$ is divisible by $3$. As both $2$ and $q$ are primes, this implies $q = 3$. Substituting $q = 3$ into the initial system of equations and simplifying gives\n$$\n\\begin{cases} 6p + 5r + 3s = 115, \\\... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (11, 3, 2, 13) | |
03dg | An isosceles triangle $ABC$ with sides $AB = 4$, $BC = AC = 6$ is given.
The points $X_1, X_2, X_3, \dots$ in that order are taken on side $AB$, such that the lengths of segments $AX_1, X_1X_2, X_2X_3, \dots$ form an infinite geometric series with first term $3$ and common ratio $1/4$.
The points $Y_1, Y_2, Y_3, \dots$... | [
"Firstly we will prove the following:\n**Lemma.** Let $n > 1$ be a natural number. On the line segment $PQ$ with length $\\frac{a_1}{1 - \\frac{1}{n}}$ are taken the points $T_1, T_2, \\dots$, such that $PT_1, T_1T_2, T_2T_3, \\dots$ form an infinite geometric series with first term $a_1$ and common ratio $\\frac{1... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (1, 2k, k) and (2k, 1, k) for k in natural numbers | |
0gmm | Is there an arithmetic sequence with
a. $2003$
b. infinitely many
terms every term of which is a power of a positive integer greater than $1$? | [] | Turkey | IMO TEAM SELECTION EXAM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a) No; b) No | |
004r | Sea $\Gamma$ la circunferencia circunscrita al triángulo acutángulo $ABC$ y $P$ un punto sobre el arco $BC$ que no contiene a $A$. Sean $K$, $L$ y $S$ los pies de las perpendiculares desde $P$ a las rectas $AB$, $AC$ y $BC$ respectivamente. Sean $M \neq P$ y $N \neq P$ las intersecciones de $PK$ y $PL$ con $\Gamma$ res... | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inversion"
] | Spanish | proof only | null | |
0cj8 | Let $ABC$ be an isosceles triangle with $\angle BAC = 100^\circ$. The angle bisector of $\angle ABC$ intersects $AC$ at $D$. The line perpendicular to $AC$ at $A$ intersects $BD$ at $M$. Point $E$ is on side $BC$ such that $AE = AM$ and $BE < CE$.
a) Show that $CA = CE$.
b) Find the measure of angle $\angle AEM$. | [] | Romania | 75th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | CA = CE and angle AEM = 80° | |
0aha | $$
\sum_{\text{cyc}} (x+y)\sqrt{(z+x)(z+y)} \ge 4(xy+yz+zx)
$$
for all positive real numbers $x$, $y$, $z$. | [
"We will obtain the inequality by adding the inequalities\n$$\n(x+y)\\sqrt{(z+x)(z+y)} \\geq 2xy+yz+zx\n$$\nfor cyclic permutation of $x$, $y$, $z$.\nSquaring both sides of this inequality we obtain\n$$\n(x+y)^2(z+x)(z+y) \\geq 4x^2y^2 + y^2z^2 + z^2x^2 + 4xyz^2 + 4x^2yz + 2xyz^2\n$$\nwhich is equivalent to\n$$\nx^... | North Macedonia | 29-th Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nin... | English | proof only | null | |
0k33 | Problem:
An $n \times m$ maze is an $n \times m$ grid in which each cell is one of two things: a wall, or a blank. A maze is solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom right cell going through no walls. (In particular, the top left and bottom right cells must both ... | [
"Solution:\n\nAnswer: 3\n\nWe must have both top-left and bottom-right cells blank, and we cannot have both top-right and bottom-left cells with walls. As long as those conditions are satisfied, the maze is solvable, so the answer is 3."
] | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 3 | |
084d | Problem:
$ABC$ è un triangolo con $AC = BC$ e $\angle ACB < 60^{\circ}$. Siano $A'$ e $B'$ due punti sui lati $BC$ e $AC$ rispettivamente tali che $AA' = BB' = AB$. Sia $C'$ l'intersezione di $AA'$ con $BB'$. Sapendo che $AC' = AB'$ e $BC' = BA'$, quanto vale l'ampiezza in gradi dell'angolo $\angle ACB$? | [
"Solution:\n\nLa risposta è $36$. Infatti i triangoli $CAB$ e $BAB'$ sono isosceli con in comune un angolo adiacente alla base, per cui sono simili. Analogamente $BAB'$ e $AB'C'$ sono simili. Riapplicando lo stesso ragionamento ai triangoli $ABA'$ e $BA'C'$, si trova che tutti e cinque tali triangoli sono simili. A... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 36 | |
08xx | Suppose that three points $A$, $B$, $C$ lie on the circumference of a circle $\Gamma$. Let $P$ be the point of intersection of the lines tangent to $\Gamma$ at $B$ and $C$. Suppose that the lines $AB$ and $CP$ are parallel, and that $AB = 3$ and $BP = 4$. Find the length of the line segment $BC$. Here we represent the ... | [
"$2\\sqrt{3}$\n\nSince $AB$ and $CP$ are parallel, $\\angle ABC = \\angle BCP$. By a well-known theorem we also see that $\\angle CAB = \\angle PBC$ must hold. Therefore the triangles $ABC$ and $BCP$ are similar, which implies that $AB : BC = BC : CP$. From this it follows that $BC = \\sqrt{AB \\cdot CP}$. Since $P... | Japan | Japan 2015 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2√3 | |
03tl | Let $a_n = \sum_{k=1}^{n} \frac{1}{k(n+1-k)}$. Prove that $a_{n+1} < a_n$ for $n \ge 2$. | [
"As\n$$\n\\frac{1}{k(n+1-k)} = \\frac{1}{n+1}\\left(\\frac{1}{k} + \\frac{1}{n+1-k}\\right),\n$$\nwe get $a_n = \\frac{2}{n+1} \\sum_{k=1}^{n} \\frac{1}{k}$. Then for $n \\ge 2$ we have\n$$\n\\begin{aligned}\n\\frac{1}{2}(a_n - a_{n+1}) &= \\frac{1}{n+1} \\sum_{k=1}^{n} \\frac{1}{k} - \\frac{1}{n+2} \\sum_{k=1}^{n+... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
075i | Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $f(x + y + xy) = f(x) + f(y) + f(xy)$ for all $x, y \in \mathbb{R}$. Prove that $f$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$. | [
"Put $x = y = 0$; we get $f(0) = 3f(0)$ so that $f(0) = 0$. Taking $y = -1$, we get $f(-1) = f(x) + f(-1) + f(-x)$ so that $f(-x) = -f(x)$ for all $x \\in \\mathbb{R}$. Taking $y = 1$, we get $f(2x + 1) = 2f(x) + 1$, for all $x \\in \\mathbb{R}$. Replace $x$ by $u + v + uv$ in this, we get\n$$\nf(2(u + v + uv) + 1)... | India | Indija TS 2012 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof only | null | |
09bp | 
Зураг дээр 14 хотыг холбосон авто замын сүлжээний зургийг дүрслэв. Аль нэг хотоос гараад хот бүрийг яг нэг удаа дайран гарах маршрут олдох уу? | [
"Зураг дээрх 14 хотыг холбосон авто замын сүлжээг ажвал 3 зам гарсан хот бүрээс зөвхөн 4 зам гарсан хотод очиж чадна. 4 зам гарсан хот бүрээс 3 зам гарсан хотод очиж чадах нь харагдана (нэг ирмэг дамжин).\n\nХэрэв аль нэг хотоос гараад хот бүрийг яг нэг удаа дайран гардаг маршрут олддог гэвэл 3 зам гарсан хотын тоо... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Graph Theory"
] | Mongolian | proof and answer | No | |
0h19 | There is a number of toy kittens in a box. Each kitten has its head and tail painted with one of 2011 colors, not necessarily the same. One can take some kittens out of the box to form a collection. A collection is called "right", if it consists of exactly 2011 kittens so that their heads are painted with different col... | [
"Enumerate the colors with the natural numbers from $1$ to $2011$. Construct a graph with the vertices $A_1, A_2, \\dots, A_{2011}, B_1, B_2, \\dots, B_{2011}$. For each kitten we draw an edge in the graph in the following way. If its head is painted with the color $i$, and the tail – with the color $j$, then the e... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof only | null | |
00gv | Let three circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, which are non-overlapping and mutually external, be given in the plane. For each point $P$ in the plane, outside the three circles, construct six points $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$ as follows: For each $i=1,2,3$, $A_{i}, B_{i}$ are distinct points o... | [
"Let $O_{i}$ be the center and $r_{i}$ the radius of circle $\\Gamma_{i}$ for each $i=1,2,3$. Let $P$ be an exceptional point, and let the three corresponding lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ concur at $Q$. Construct the circle with diameter $P Q$. Call the circle $\\Gamma$, its center $O$ and its radi... | Asia Pacific Mathematics Olympiad (APMO) | XXI Asian Pacific Mathematics Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gcu | 平面上給定一個半徑為 $1$ 的圓 $\omega$。對於一組三角形所成的集合 $T$, 如果滿足下列兩條件:
(i) $T$ 中的每一個三角形都內接於 $\omega$,
(ii) $T$ 中任兩個三角形都沒有共同內部點,
就說 $T$ 是好棒棒。
試決定所有的正實數 $t$, 使得對於每一個正整數 $n$, 都可以找到一組由 $n$ 個三角形所組成的好棒棒集合 $T$, 其中的每一個三角形的周長都大於 $t$。 | [
"$t \\in (0, 4]$.\n\nFirst, we show how to construct a good collection of $n$ triangles, each of perimeter greater than $4$. This will show that all $t \\le 4$ satisfy the required conditions.\n\nConstruct inductively an $(n+2)$-gon $BA_1A_2\\dots A_nC$ inscribed in $\\omega$ such that $BC$ is a diameter, and $BA_1... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營, 獨立研究 (三) | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | (0, 4] | |
0h73 | An acute-angled triangle $ABC$ is given, where $AB < BC$. Its incircle with the centre in point $I$ is touching the side $BC$ in point $K$. The line $AK$ intersects a second time the circumscribed circle of triangle $ABC$ in point $T$. Let $M$ be the midpoint of $BC$, and $N$ be the midpoint of arc $BAC$ of the circums... | [
"Let's assume $W$ is the midpoint of arc $BC$. As is known, points $W$, $M$, and $N$ are collinear, and $W$ is the midpoint of the circumscribed circle of triangle $BIC$ (see fig. 47). Also, it is known that the center of the excircle of triangle $ABC$ that touches $BC$ belongs to this circle. Let $IK$ cross $TN$ a... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis ... | null | proof only | null | |
09vm | Problem:
Vind alle functies $f: \mathbb{Z} \rightarrow \mathbb{Z}$ die voldoen aan
- $f(p)>0$ voor alle priemgetallen $p$,
- $p \mid (f(x)+f(p))^{f(p)}-x$ voor alle $x \in \mathbb{Z}$ en alle priemgetallen $p$. | [
"Solution:\nVul $x=p$ in: $p \\mid (2 f(p))^{f(p)}-p$ en $p$ is priem, dus $p \\mid 2 f(p)$. We zien dat $p=2$ of $p \\mid f(p)$. Vul nu $x=0$ in: $p \\mid (f(0)+f(p))^{f(p)}-0$ en $p$ is priem, dus $p \\mid f(0)+f(p)$. Aangezien $p \\mid f(p)$ voor $p \\neq 2$, zien we nu ook $p \\mid f(0)$ voor $p \\neq 2$. Dus $... | Netherlands | IMO-selectietoets II | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x | |
08lm | Problem:
Let $ABC$ be an isosceles triangle with $AC = BC$. The point $D$ lies on the side $AB$ such that the semicircle with diameter $BD$ and center $O$ is tangent to the side $AC$ at the point $P$ and intersects the side $BC$ at the point $Q$. The radius $OP$ intersects the chord $DQ$ at the point $E$ such that $5 \... | [
"Solution:\nWe denote $OP = OD = OB = R$, $AC = BC = b$ and $AB = 2a$. Because $OP \\perp AC$ and $DQ \\perp BC$, then the right triangles $APO$ and $BQD$ are similar and $\\angle BDQ = \\angle AOP$. So, the triangle $DEO$ is isosceles with $DE = OE$. It follows that\n$$\n\\frac{PE}{DE} = \\frac{PE}{OE} = \\frac{3}... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 6/5 | |
0f11 | Problem:
Find the smallest positive integer which can be represented as $36^{m} - 5^{n}$. | [
"Solution:\nObviously $11 = 36^{1} - 5^{2}$, and we guess that this is the best possible. We cannot have $36^{m} - 5^{n} = k$, where $2$, $3$ or $5$ divides $k$ (because then $2$ would divide $5^{n}$ and similarly in the other cases). So the only possible values of $k < 11$ are $1$, $7$.\nWe have $36^{m} - 5^{n} = ... | Soviet Union | ASU | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 11 | |
071m | Problem:
In a convex quadrilateral $P Q R S$, $P Q = R S$, $(\sqrt{3}+1) Q R = S P$ and $\angle R S P - \angle S P Q = 30^{\circ}$. Prove that
$$
\angle P Q R - \angle Q R S = 90^{\circ}
$$ | [
"Solution:\n\nLet $[$Fig$]$ denote the area of Fig. We have\n$$\n[P Q R S] = [P Q R] + [R S P] = [Q R S] + [S P Q]\n$$\nLet us write $P Q = p$, $Q R = q$, $R S = r$, $S P = s$. The above relations reduce to\n$$\np q \\sin \\angle P Q R + r s \\sin \\angle R S P = q r \\sin \\angle Q R S + s p \\sin \\angle S P Q\n$... | India | INMO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l5h | Problem:
A polyomino is a connected figure constructed by joining one or more unit squares edge-to-edge. Determine, with proof, the number of non-congruent polyominoes with no holes, perimeter $180$, and area $2024$. | [
"Solution:\nDefine the bounding box of a polyomino to be the smallest axis-aligned rectangle that contains the entire polyomino. Suppose a polyomino satisfying the given conditions has a bounding box with dimensions $w \\times h$.\n\nClaim 1. $w + h \\leq 90$.\n\nProof. The polyomino has at least $2w$ horizontal ed... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2 | |
0l35 | The product of three integers is $60$. What is the least possible positive sum of the three integers?
(A) 2 (B) 3 (C) 5 (D) 6 (E) 13 | [
"**Answer (B):** Note that $60 = 10 \\cdot (-1) \\cdot (-6)$, and the sum of these factors is $3$. It remains to show that no positive sum can be less than $3$. Such a sum would have to consist of one positive integer and two negative integers with smaller absolute value. If the positive integer is greater than or ... | United States | AMC 12 A | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | B | |
08um | Suppose a regular 2010-gon is given. How many combinations of 3 distinct vertices $A$, $B$, $C$ of this polygon are there for which all the inner angles of the triangle $ABC$ have integer-valued degrees (i.e., integral multiples of $1^\circ$)? Regard 2 triples as representing the same combination if one is a permutatio... | [
"$$\n\\boxed{272020}\n$$\nLet $a$ be the number of the sides of the 2010-gon lying in between the vertices $B$ and $C$ and on the opposite side from the vertex $A$, and $b$ be the number of the sides lying in between the vertices $C$ and $A$ and on the opposite side from the vertex $B$. Let $\\Gamma$ be the circum-... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 272020 | |
0bad | Consider a convex pentagon $A_0A_1A_2A_3A_4$ so that the rays ($A_iA_{i+1}$ and ($A_{i+3}A_{i+2}$ meet at $B_{i+4}$, for each $i = 0, 1, 2, 3, 4$ – indices are considered modulo 5. Show that
$$
\prod_{i=0}^{4} A_i B_{i+3} = \prod_{i=0}^{4} A_i B_{i+2}.
$$ | [
"Let $C_i$ be the projection of $B_i$ on the line $A_{i+2}A_{i+3}$, $i = 0, 1, 2, 3, 4$. Notice that the right-angled triangles (possibly degenerate) $A_iB_{i+3}C_{i+3}$ and $A_iB_{i+2}C_{i+2}$ are similar, for angles $A_i$ are vertical. Hence $\\frac{A_iB_{i+3}}{A_iB_{i+2}} = \\frac{B_{i+3}C_{i+3}}{B_{i+2}C_{i+2}}... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ev0 | Let $M$ be the set of positive integers which is not divisible by any prime number greater than $3$. For arbitrary chosen subsets $A_1, A_2, A_3, \dots$ of $M$, prove that there exist two distinct positive integers $i$ and $j$ such that:
For each $x$ in $A_i$, $A_j$ has a divisor of $x$. | [
"Regard $A_n$ as a set of lattice points by\n$$\n2^x 3^y \\mapsto (x, y) \\in \\mathbb{N}_0^2,\n$$\nwhere $\\mathbb{N}_0$ is the set of nonnegative integers. For two elements $(a, b), (c, d) \\in \\mathbb{N}_0^2$, we denote $(a, b) \\le (c, d)$ if $a \\le c$ and $b \\le d$. Also we denote $(a, b) < (c, d)$ if $(a, ... | South Korea | 25th Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0ka2 | Problem:
A regular hexagon $P R O F I T$ has area $1$. Every minute, greedy George places the largest possible equilateral triangle that does not overlap with other already-placed triangles in the hexagon, with ties broken arbitrarily. How many triangles would George need to cover at least $90\%$ of the hexagon's area? | [
"Solution:\nIt's not difficult to see that the first triangle must connect three non-adjacent vertices (e.g. $P O I$), which covers area $\\frac{1}{2}$, and leaves three $30$-$30$-$120$ triangles of area $\\frac{1}{6}$ each. Then, the next three triangles cover $\\frac{1}{3}$ of the respective small triangle they a... | United States | HMMT February 2019 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 46 | |
01so | Find all functions $f: \mathbb{R} \to \mathbb{R}$, $g: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x - 2f(y)) = xf(y) - yf(x) + g(x)
$$
for all real $x, y$. | [
"Answer: $f(x) = a(x + 2a)$, $g(x) = a(1 - 2a)(x + 2a)$, $a \\in \\mathbb{R}$.\n\n(Solution by A. Goloubitskaya, L. Manzhulina.) First, we claim that $f$ takes the value $0$. If $f(0) = 0$, then there is nothing to prove. If $f(0) = b \\neq 0$, then setting $x = 0$ in the given equality\n$$\nf(x - 2f(y)) = xf(y) - ... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All solutions are given by f(x) = a(x + 2a) and g(x) = a(1 − 2a)(x + 2a) for real a. | |
01wq | Let $AA_1$ be the bisector of the triangle $ABC$. The points $D$ and $F$ are chosen on the line $BC$ such that $A_1$ is the midpoint of the segment $DF$. A line $l$, different from $BC$, passes through $A_1$ and intersects the lines $AB$ and $AC$ at points $B_1$ and $C_1$, respectively.
Find the locus of the points of ... | [
"Answer: The line passing through $A$ parallel to the line $KF$ where $K$ is the point on $AC$ such that $\\angle AA_1K = 90^\\circ$ (excluding $A$, and the intersection points of this line with the line $BC$ and the lines passing through $F$ and $D$ parallel to $AC$ and $AB$, respectively).\nChoose the points $L$ ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | The locus is the line through A parallel to KF, where K is the point on AC such that the angle between AA1 and AK is a right angle. Exclude the point A, and the intersection points of this line with BC and with the lines through F parallel to AC and through D parallel to AB. | |
0cau | Let $f : [-\frac{\pi}{2}, \frac{\pi}{2}] \to \mathbb{R}$ be a twice differentiable function such that
$$
(f''(x) - f(x)) \cdot \operatorname{tg}(x) + 2 \cdot f'(x) \ge 1, \quad \text{for any } x \in (-\frac{\pi}{2}, \frac{\pi}{2}).
$$
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \cdot \sin x \, dx \geq \pi - 2.
$$ | [
"Since $\\cos x > 0$ for any $x \\in (-\\frac{\\pi}{2}, \\frac{\\pi}{2})$, the inequality in the statement can be written equivalently\n$$\n(f''(x) - f(x)) \\cdot \\sin x + 2 \\cdot f'(x) \\cdot \\cos x \\geq \\cos x, \\quad \\text{for any } x \\in (-\\frac{\\pi}{2}, \\frac{\\pi}{2}),\n$$\nor $g''(x) \\ge 0, \\fora... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0cfs | a) Give an example of a function with the above property, which is primitivable and is non-identically vanishing on any interval included in $[0, 1]$.
b) If $f$ is continuous on $(0, 1]$, show that $f$ is primitivable on $[0, 1]$.
where $f : [0, 1] \to \mathbb{R}$ is an integrable function with $f(0) = 0$, such that
... | [] | Romania | 74th NMO Shortlisted Problems | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | English | proof only | null | |
02og | For each real number $t$, let $P_t(x) = x^3 - 12x + t$ and let
$$
\Delta(t) = \max\{c \in \mathbb{R} \mid P_t(c) = 0\} - \min\{c \in \mathbb{R} \mid P_t(c) = 0\}
$$
the difference between the largest and the smallest real roots of $P_t(x)$. Determine the range of values that $\Delta(t)$ can assume as $t$ varies. | [
"Let $Q(x) = x^3 - 12x$. Then $Q'(x) = 3x^2 - 12$ has roots $-2$ and $2$, and $Q$ has a local minimum at $(2, -16)$ and a local maximum at $(-2, 16)$. So $Q(x) = -t$ has three (not necessarily distinct) real roots for $-16 \\le t \\le 16$ and one real root for $t < -16$ and $t > 16$, which means that $\\Delta(t) = ... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | {0} ∪ [6, 4√3] | |
07lg | A triangle *ABC* is inscribed in a circle, centre *O*, and $D$, $E$, $F$ are the mid-points of $BC$, $CA$, $AB$ respectively. The point $A$ is free to move on the circumference of the circle. Find the locus of $L$, the midpoint of $FE$. | [
"Let $M$ and $N$ be the midpoints of $BD$ and $DC$ respectively. Let $AD$ meet $EF$ at $L'$. Then $FL' \\parallel BD$ and $|FL'| = \\frac{1}{2}|BD|$. Similarly $|L'E| = \\frac{1}{2}|DC|$, hence $|FL'| = |L'E|$, i.e. $L' = L$ is the midpoint of $FE$.\n\n\n\nLet $P$ be the midpoint of $OD$. T... | Ireland | Irska | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | A fixed circle: the circle centered at the midpoint of the segment joining the circle’s center to the midpoint of the fixed side, with radius equal to half of the circumradius. | |
07u6 | A tangent is drawn at $A$ to a circle centre $O$. Point $D$ is on this tangent and $B$, $C$ are on the circle such that $B$, $C$ and $D$ are collinear. Points $P$ and $Q$ are the circumcentres of triangles $ACD$ and $ABD$, respectively.
Prove $\angle AOB = \angle APD = \angle BQD$. | [
"Because $P$ is the circumcentre of triangle $ACD$, we have\n$$\n\\frac{1}{2} \\angle APD = \\angle ACD = 180^\\circ - \\angle ACB.\n$$\nBecause $O$ is the circumcentre of triangle $ABC$ we see that\n$$\n\\frac{1}{2} \\angle AOB = 180^\\circ - \\angle ACB\n$$\nand we obtain $\\angle APD = \\angle AOB$.\n\nBecause $... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dj4 | Given an acute triangle $ABC$ with its altitudes $AD$, $BE$, $CF$ concurrent at $H$. The point $K$ changes on the segment $AH$. Let $M$, $N$ be the projection of $H$ on the lines $KE$, $KF$. Prove that the line joining the circumcenter of triangles $HEN$ and $HFM$ always passes through a fixed point. | [
"Let $S$, $T$ be the intersection of the lines $KF$, $KE$ with $CA$, $AB$ and $P$, $Q$ the midpoint of $HT$, $HS$. Since $\\angle HFT = \\angle HMT = 90^\\circ$ so $HT$ is the diameter of $(HMF)$ and $P$ is the center of $(HMF)$. Similarly, $Q$ is the center of $(HNE)$.\n\nLet $R$ be the intersection of $ST$, $EF$.... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometr... | English | proof only | null | |
0fwj | Problem:
Gegeben sind fünf Teiler von $10^{2008}$. Zeige, dass es zwei dieser Teiler gibt, deren Produkt eine Quadratzahl ist. | [
"Solution:\n\nJeder Teiler einer Zehnerpotenz ist von der Form $2^{a} 5^{b}$ mit ganzen Zahlen $a, b \\geq 0$. Die fünf Teiler können wir auf vier Schubfächer verteilen, je nachdem welche der beiden Zahlen $a$ und $b$ gerade sind. Zwei dieser Teiler, sagen wir $2^{a_{1}} 5^{b_{1}}$ und $2^{a_{2}} 5^{b_{2}}$, liegen... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0by7 | Determine the smallest real number $a$ satisfying
$$
a \ge \sum_{k=1}^{n} a_k \cos(a_1 + \dots + a_k),
$$
for all positive integers $n$ and all positive real numbers $a_1, \dots, a_n$ that add up to at most $\pi$. | [
"The required minimum is $1$. To show that $1$ is an upper bound, let $n$ be a positive integer, and let $a_1, \\dots, a_n$ be positive real numbers such that $a_1+\\dots+a_n \\le \\pi$. If $a_1 \\ge \\pi/2$, then the sum in question is non-positive, so let $a_1 < \\pi/2$, and let $m$ be the largest positive intege... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1 | |
00ns | Let $x$, $y$, $z$ be nonzero real numbers with
$$
\frac{ x + y }{ z } = \frac{ y + z }{ x } = \frac{ z + x }{ y }.
$$
Determine all possible values of
$$
\frac{ (x + y)(y + z)(z + x) }{ xyz }.
$$ | [
"*Answer.* The only possible values are $8$ and $-1$.\n\nWe add $1$ to the equations and obtain\n$$\n\\frac{ x + y + z }{ z } = \\frac{ x + y + z }{ x } = \\frac{ x + y + z }{ y }.\n$$\nFor $x + y + z = 0$, this is clearly true, and the expression in the problem statement becomes\n$$\n\\frac{ (-z)(-x)(-y) }{ xyz } ... | Austria | AUT_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 8 and -1 | |
0cpy | Let $a, b, c$ be three positive numbers such that $a + b^2 + c^2 = a^2 + b + c^2 = a^2 + b^2 + c$. Determine whether all three numbers $a, b, c$ should necessarily be equal. (L. Emelyanov)
Про три положительных числа известно, что если выбрать одно из них и прибавить к нему сумму квадратов двух других, то получится од... | [
"**Ответ.** Нет, неверно.\nПодойдёт, например, тройка $\\frac{1}{3}, \\frac{1}{3}, \\frac{2}{3}$.\n\n**Замечание.** Все такие тройки можно получить, решив соответствующую систему: $a + b^2 + c^2 = a^2 + b + c^2 = a^2 + b^2 + c$. Из первых двух равенств имеем $a^2 - a = b^2 - b$; перенося всё в левую часть, получаем... | Russia | Russian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English, Russian | proof and answer | No; for example, a = 1/3, b = 1/3, c = 2/3. | |
0937 | Problem:
For an integer $n \geqslant 3$ we define the sequence $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k}$ as the sequence of exponents in the prime factor decomposition of $n! = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$, where $p_{1} < p_{2} < \cdots < p_{k}$ are primes.
Determine all integers... | [
"Solution:\n\nLet $p_{i}$ be the $i$\\textsuperscript{th} prime number and let $\\alpha_{i}$ be the exponent of $p_{i}$ in $n!$. It is well known that\n$$\n\\alpha_{i} = \\left\\lfloor \\frac{n}{p_{i}} \\right\\rfloor + \\left\\lfloor \\frac{n}{p_{i}^{2}} \\right\\rfloor + \\left\\lfloor \\frac{n}{p_{i}^{3}} \\righ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3, 4, 6, 10 | |
0cn7 | 11 positive integers are arranged on a circle. It appears that the difference between each two neighboring numbers is not less than 20, while the sum of each two neighboring numbers is not less than 100. Find the least possible value of the sum of all 11 integers. | [
"Consider an arrangement that satisfies the conditions, and connect each pair of neighboring numbers with an arrow from the smaller to the larger. Since the total number of arrows is odd, there must be two consecutive arrows pointing in the same direction: $a \\to b \\to c$. Then $b \\leq c - 20$, $a \\leq b - 20 \... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 580 | |
0kyp | Problem:
Suppose point $P$ is inside quadrilateral $A B C D$ such that
$$
\begin{aligned}
& \angle P A B = \angle P D A \\
& \angle P A D = \angle P D C, \\
& \angle P B A = \angle P C B, \text{ and } \\
& \angle P B C = \angle P C D
\end{aligned}
$$
If $P A = 4$, $P B = 5$, and $P C = 10$, compute the perimeter of $... | [
"Solution:\n\n\n\nFirst of all, note that the angle conditions imply that $\\angle B A D + \\angle A B C = 180^{\\circ}$, so the quadrilateral is a trapezoid with $A D \\parallel B C$. Moreover, they imply $A B$ and $C D$ are both tangent to $(P A D)$ and $(P B C)$; in particular $A B = C D... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Reflection",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Ge... | null | proof and answer | 9*sqrt(410)/5 | |
013m | Problem:
Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers with arithmetic mean $X$. Prove that there is a positive integer $K$ such that the arithmetic mean of each of the lists $\{x_{1}, x_{2}, \ldots, x_{K}\}$, $\{x_{2}, x_{3}, \ldots, x_{K}\}$, $\ldots$, $\{x_{K-1}, x_{K}\}$, $\{x_{K}\}$ is not greater than $X$. | [
"Solution:\nSuppose the conclusion is false. This means that for every $K \\in \\{1,2, \\ldots, n\\}$, there exists a $k \\leq K$ such that the arithmetic mean of $x_{k}, x_{k+1}, \\ldots, x_{K}$ exceeds $X$. We now define a decreasing sequence $b_{1} \\geq a_{1} > a_{1}-1 = b_{2} \\geq a_{2} > \\cdots$ as follows:... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null |
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