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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
038w | In $\triangle ABC$ with $\angle ACB = 60^\circ$ let $AA_1$ and $BB_1$ ($A_1 \in BC$, $B_1 \in AC$) be the bisectors of $\angle BAC$ and $\angle ABC$. The line $A_1B_1$ meets the circumcircle of $\triangle ABC$ at points $A_2$ and $B_2$.
a) If $O$ and $I$ are the circumcenter and the incenter of $\triangle ABC$ prove t... | [
"a) Since $\\angle AOB = 2\\gamma = 120^\\circ$ and points $A$, $O$, $I$ and $B$ lie on a circle\n$\\alpha + \\beta$\n\n\nb) Since $OQ \\perp A_2B_2$, $IP \\perp A_2B_2$ ($\\triangle A_1IB_1$ is isosceles) and $OI \\parallel A_2B_2$, we have that $OIPQ$ is a rectangle. The perpendicular bis... | Bulgaria | Winter Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0gbn | 令 $p$ 為一質數。在以下遊戲中, 艾德與阿飛輪流進行各自的回合。輪到某人的回合時, 他先從 $\{0, 1, \dots, p-1\}$ 裡還沒被任何一方選過的數字中選一個當作 $i$, 接著再從 $\{0, 1, 2, \dots, 9\}$ 中選一個元素當作 $a_i$。艾德先進行回合。等 $\{0, 1, \dots, p-1\}$ 全部都被挑過後, 遊戲結束並計算以下數字
$$
M = a_0 + 10 \times a_1 + \cdots + 10^{p-1} \times a_{p-1} = \sum_{j=0}^{p-1} a_j \times 10^j.
$$
若 $M$ 被 $p$ 整除, 艾德勝; 否則, ... | [
"We say that a player makes the move $(i, a_i)$ if he chooses the index $i$ and then the element $a_i$ of the $\\{0, 1, \\dots, 9\\}$ in this move.\n\n(1) If $p=2$ or $p=5$, then the first player simply chooses $(0, 0)$, which forces $10|M$, which guarantees he wins.\n\n(2) So assume $p \\notin \\{2, 5\\}$. Let the... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
08n3 | Problem:
Let $AD$, $BF$ and $CE$ be the altitudes of $\triangle ABC$. A line passing through $D$ and parallel to $AB$ intersects the line $EF$ at the point $G$. If $H$ is the orthocenter of $\triangle ABC$, find the angle $\widehat{CGH}$. | [
"Solution:\nWe can see easily that points $C$, $D$, $H$, $F$ lie on a circle of diameter $[CH]$.\nTake $\\{F, G'\\} = \\odot(CHF) \\cap EF$. We have $\\widehat{EFH} = \\widehat{BAD} = \\widehat{BCE} = \\widehat{DFH}$ since the quadrilaterals $AEDC$, $AEHF$, $CDHF$ are cyclic. Hence $FB$ is the bisector of $\\wideha... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 90° | |
067m | Let $P(x) = a x^3 + (b - a) x^2 - (c + b) x + c$, $Q(x) = x^4 + (b - 1) x^3 + (a - b) x^2 - (c + a) x + c$ be polynomials of the indeterminate $x$, where $a, b, c$ are nonzero real numbers and $b > 0$. If the polynomial $P(x)$ has three different real roots $x_0, x_1, x_2$, which are also roots of the polynomial $Q(x)$... | [
"(a) The sum of the coefficients of the polynomial $P(x)$ is equal to $0$. It means that $1$ is one of its roots and so\n$$\nP(x) = a x^3 + (b - a) x^2 - (c + b) x + c = (x - 1)(a x^2 + b x - c)\n$$\nIf $x_0 = 1$, from Vieta's formulas we have:\n$$\nx_1 + x_2 = -\\frac{b}{a} \\quad \\text{and} \\quad x_1 x_2 = -\\f... | Greece | Hellenic Mathematical Olympiad ARCHIMEDES | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | abc > 28; and for integers with the given conditions the only possibility is a = 2, b = 4, c = 4. | |
026h | Problem:
Soma constante - Preencha as 5 casas em branco da tabela $3 \times 3$ com os números de $3$ a $8$, sem repeti-los, de modo que as somas dos $4$ números escritos nas subtabelas formadas por quadrados $2 \times 2$ sejam a mesma nas $4$ subtabelas.
 | [
"Solution:\n\nSejam $a, b, c, d, e$ e $f$ os números que colocaremos na tabela.\n\nDe acordo com a regra para as $4$ subtabelas $2 \\times 2$:\n\n| 1 | $a$ | 2 |\n| :---: | :---: | :---: |\n| $b$ | 9 | $c$ |\n| $d$ | $e$ | $f$ |\n\ntemos:\n$$\n\\begin{aligned}\n& 1+a+b+9 = a+2+9+c \\Rightarrow b = c+1 \\\\\n& 1+a+b... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Two valid completions are: 1 6 2 / 8 9 7 / 4 3 5 and 1 8 2 / 5 9 4 / 6 3 7. | |
012a | Problem:
Suppose there exists a point $A$ such that $A$ is connected to twelve points. Then there exist three points $B$, $C$ and $D$ such that $\angle BAC \leqslant 60^{\circ}$, $\angle BAD \leqslant 60^{\circ}$ and $\angle CAD \leqslant 60^{\circ}$. | [
"Solution:\nWe can assume that $|AD| > |AB|$ and $|AD| > |AC|$. By the cosine law we have\n$$\n\\begin{aligned}\n|BD|^2 & = |AD|^2 + |AB|^2 - 2|AD||AB| \\cos \\angle BAD \\\\\n& < |AD|^2 + |AB|^2 - 2|AB|^2 \\cos \\angle BAD \\\\\n& = |AD|^2 + |AB|^2 (1 - 2 \\cos \\angle BAD) \\\\\n& \\leqslant |AD|^2\n\\end{aligned... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0hbq | Find the average of all 5-digit numbers that satisfy the following:
* The number is of the form $\overline{ab0cd}$, that is, its third digit is zero;
* Digits are pairwise distinct;
* Both numbers $\overline{ab0cd}$, and $\overline{dc0ba}$ are divisible by 7. | [
"Let us rewrite the problem the following way: since $1001$ is divisible by $7$, then\n$$\n\\begin{align*}\n\\overline{ab0cd} &= 1000 \\cdot \\overline{ab} + \\overline{cd} = 1001 \\cdot \\overline{ab} + (\\overline{cd} - \\overline{ab}) \\\\\n&\\Rightarrow (\\overline{cd} - \\overline{ab}) \\text{ is divisible by ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 55055 | |
0833 | Problem:
In a square $ABCD$ with side $2$, a segment $MN$ of length $1$ is constrained to have endpoint $M$ on side $AB$ and endpoint $N$ on side $BC$. This segment divides the square into a triangle $T$ and a pentagon $P$. What is the maximum value that the ratio of the area of $T$ to that of $P$ can take?
(A) $\fra... | [
"Solution:\n\nThe answer is $\\mathbf{(E)}$. The ratio between the two areas is maximized when the area of $T$ is maximized: in this case, the area of $P$ assumes its minimum value. $T$ is a right triangle whose hypotenuse is $1$, and it has maximum area when it is isosceles. Indeed, such a triangle can be inscribe... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | E | |
04vi | In the triangle $ABC$, let us denote $M, N, P$ the midpoints of the sides $BC, CA, AB$ respectively and let $G$ be the centroid of $ABC$. Let the circumcircle of $BGP$ intersects the line $MP$ at a point $K$ different from $P$, and let the circumcircle of $CGN$ intersects the line $MN$ at a point $L$ different from $N$... | [
"Obviously, $MP$ intersects the median $BN$ between points $B$ and $G$, so the point $K$ lies on the ray $PM$ and $BKGP$ is cyclic. Similarly, the point $L$ lies on the ray $NM$ and $CLGN$ is cyclic. Due to $MP \\parallel CA$ and $MN \\parallel BA$ we have\n$$\n|\\angle BPK| = |\\angle BPM| = |\\angle BAC| = |\\ang... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06ck | a. Let $a_1, a_2, \dots, a_n > 0$. Prove that $a_1^{a_1} a_2^{a_2} \cdots a_n^{a_n} \ge (a_1 a_2 \cdots a_n)^{\frac{a_1+a_2+\cdots+a_n}{n}}$.
b. Let $x, y, z \ge 0$ and $x + y + z = 1$. Prove that $0 \le xy + yz + zx - 3xyz \le \frac{1}{4}$. | [
"a. Taking logarithm, it is the same as proving\n$$\na_1 \\ln a_1 + a_2 \\ln a_2 + \\cdots + a_n \\ln a_n \\ge \\frac{a_1 + a_2 + \\cdots + a_n}{n} \\ln (a_1 a_2 \\cdots a_n).\n$$\nThis means\n$$\n\\sum_{k=1}^{n} a_k \\ln a_k \\ge \\frac{1}{n} \\left( \\sum_{k=1}^{n} a_k \\right) \\left( \\sum_{k=1}^{n} \\ln a_k \\... | Hong Kong | Test 1 | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
01zk | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for any numbers $x \neq y$ the following equality is true:
$$
(f(x + y))^2 = f(x + y) + f(x) + f(y).
$$ | [
"Only two functions $f(x) \\equiv 0$ or $f(x) \\equiv 3$ are solutions to the equation. Indeed, let's put $c = f(0)$. By substituting $y = 0$, we obtain a quadratic equation\n$$\n(f(x))^2 - 2f(x) - c = 0,\n$$\nwhence $f(x) = 1 - \\sqrt{1+c}$ or $1 + \\sqrt{1+c}$ for any $x \\neq 0$.\nSuppose that there are two non-... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all real x, or f(x) = 3 for all real x | |
0duh | Problem:
Določi najmanjše število polj na tabli razsežnosti $8 \times 8$, ki jih moramo pobarvati, da bo na vsakem kosu tabele v obliki črke L (glej sliko) vsaj eno polje pobarvano.
| [
"Solution:\n\nNajmanjše število polj, ki jih moramo pobarvati, je 32. Tablo najprej razdelimo na 16 kvadratov s po 4 polji. Če bi pobarvali samo 31 polj table, potem je vsaj v enem od teh kvadratov največ eno polje pobarvano. V tem kvadratu je potem lik oblike črke L brez pobarvanih polj. Če pobarvamo tablo kot šah... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 32 | |
0aif | Let $k_1$, $k_2$ and $k_3$ be three circles with centers $O_1$, $O_2$ and $O_3$ respectively, such that none of the centers lies inside any of the two other circles. The circles $k_1$ and $k_2$ intersect in $A$ and $P$, $k_1$ and $k_3$ intersect in $C$ and $P$ and $k_2$ and $k_3$ intersect in $B$ and $P$. Let $X$ be a ... | [
"We will first show that the points $Y$, $B$ and $Z$ are collinear. Since the quadrilateral $BYAP$ is inscribed we have $\\angle PBY = \\angle PAX$. Since the quadrilateral $AXCP$ is inscribed we have $\\angle PAX = \\angle PCZ$. Since the quadrilateral $CPBZ$ is inscribed we obtain $\\angle PBZ + \\angle PCZ = 180... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | Triangles XYZ and O1O2O3 are similar. The area of triangle XYZ is at most four times the area of triangle O1O2O3, and this maximum is attainable. | |
06q7 | There is given a convex quadrilateral $A B C D$. Prove that there exists a point $P$ inside the quadrilateral such that
$$
\angle P A B+\angle P D C=\angle P B C+\angle P A D=\angle P C D+\angle P B A=\angle P D A+\angle P C B=90^{\circ} \tag{1}
$$ | [
"if and only if the diagonals $A C$ and $B D$ are perpendicular.\n\nFor a point $P$ in $A B C D$ which satisfies (1), let $K, L, M, N$ be the feet of perpendiculars from $P$ to lines $A B, B C, C D, D A$, respectively. Note that $K, L, M, N$ are interior to the sides as all angles in (1) are acute. The cyclic quadr... | IMO | 49th International Mathematical Olympiad Spain | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry ... | English | proof only | null | |
0jgt | Problem:
Rahul has ten cards face-down, which consist of five distinct pairs of matching cards. During each move of his game, Rahul chooses one card to turn face-up, looks at it, and then chooses another to turn face-up and looks at it. If the two face-up cards match, the game ends. If not, Rahul flips both cards face... | [
"Solution:\n\nAnswer: 4 Label the 10 cards $a_{1}, a_{2}, \\ldots, a_{5}, b_{1}, b_{2}, \\ldots, b_{5}$ such that $a_{i}$ and $b_{i}$ match for $1 \\leq i \\leq 5$.\n\nFirst, we'll show that Rahul cannot always end the game in less than 4 moves, in particular, when he turns up his fifth card (during the third move)... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 4 | |
0j5n | Problem:
In how many ways can each square of a $4 \times 2011$ grid be colored red, blue, or yellow such that no two squares that are diagonally adjacent are the same color? | [
"Solution:\n\nAnswer: $64 \\cdot 3^{4020}$\n\nIf we first color the board in a checkerboard pattern, it is clear that the white squares are independent of the black squares in diagonal coloring, so we calculate the number of ways to color the white squares of a $4 \\times n$ board and then square it.\n\nLet $a_{n}$... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 64 * 3^4020 | |
0fs0 | Problem:
1. $67$ Schüler schreiben eine Prüfung. Die Prüfung besteht aus $6$ multiple-choice Fragen, die alle mit ja oder nein beantwortet werden müssen. Jeder Schüler beantwortet dabei alle $6$ Fragen. Eine richtige Antwort auf die $k$-te Frage gibt $k$ Punkte, eine falsche Antwort $-k$ Punkte.
a) Zeige, dass mindes... | [
"Solution:\n\na) Es sind sechs Fragen zu beantworten. Zu jeder Frage gibt es genau $2$ mögliche Antworten, insgesamt kann man das Prüfungsblatt also auf $2^{6} = 64$ Arten ausfüllen. Da in der Klasse $67 > 64$ Schüler sind, müssen nach dem Schubfachprinzip zwei Schüler das Prüfungsblatt gleich ausgefüllt haben.\n\n... | Switzerland | Vorselektionsprüfung | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0it0 | Problem:
Determine the value of $\lim_{n \rightarrow \infty} \sum_{k=0}^{n} \binom{n}{k}^{-1}$. | [
"Solution:\nLet $S_{n}$ denote the sum in the limit. For $n \\geq 1$, we have $S_{n} \\geq \\binom{n}{0}^{-1} + \\binom{n}{n}^{-1} = 2$.\n\nOn the other hand, for $n \\geq 3$, we have\n$$\nS_{n} = \\binom{n}{0}^{-1} + \\binom{n}{1}^{-1} + \\binom{n}{n-1}^{-1} + \\binom{n}{n}^{-1} + \\sum_{k=2}^{n-2} \\binom{n}{k}^{... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2 | |
0hiy | Problem:
Define
$$
A = 1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{4 + \ddots_{+} \frac{1}{2006 + \frac{1}{2007}}}}} \text{ and } B = 1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{4 + \ddots + \frac{1}{2005 + \frac{1}{2006}}}}}
$$
Which of the two numbers is greater, $A$ or $B$? Explain your answer! | [
"Solution:\n\nWe will determine the sign of $A - B$. If that number happens to be positive then $A > B$, otherwise $A < B$. For each $n$ such that $1 \\leq n \\leq 2005$, let us define\n$$\nA_{n} = n + \\frac{1}{n+1 + \\frac{1}{n+2 + \\frac{1}{n+3 + \\ddots' + \\frac{1}{2006 + \\frac{1}{2007}}}}} \\quad B_{n} = n +... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | B > A | |
0bg6 | Show that there exists a proper non-empty subset $S$ of the set of real numbers such that, for every real number $x$, the set $\{nx+S: n \in \mathbb{N}\}$ is finite, where $nx + S = \{nx + s: s \in S\}$. | [
"Let $H$ be a Hamel basis; that is, $H$ is a set of real numbers such that every real number $x$ can uniquely be written in the form\n$$\nx = \\sum_{h \\in H} q(x, h) \\cdot h,\n$$\nwhere the $q(x, h)$ are all rational and vanish for all but a finite number (depending on $x$) of $h$'s. The existence of Hamel bases ... | Romania | The Danube Mathematical Competition | [
"Algebra > Linear Algebra > Vectors",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
09gc | Find all triplets $(m, n, p)$ such that
$$
3^m p^2 + 1 = n^{2017},
$$
where $m$, $n$ are positive integers and $p$ is a prime number. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | no solutions | |
0b21 | Problem:
A right triangle has legs of lengths $3$ and $4$. Find the volume of the solid formed by revolving the triangle about its hypotenuse. | [
"Solution:\n\nThe solid consists of two conical solids with a common circular base of radius $\\frac{3 \\cdot 4}{5} = \\frac{12}{5}$. If $h_{1}$ and $h_{2}$ are the heights of the two cones, then $h_{1} + h_{2} = 5$. Hence, the volume of the solid is\n$$\n\\frac{\\pi}{3} \\cdot \\left(\\frac{12}{5}\\right)^{2} \\cd... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume"
] | null | final answer only | 48*pi/5 | |
0guf | Any two pupils in a school are either friends or not, the friendship is mutual. For each integer $1 \le l \le 99$ there is a school pupil having exactly $l$ friends in the school. Given that there is no triple of school pupils such that any two of them are friends, find the minimal possible number of pupils in this sch... | [
"**6.** First $XY \\parallel l_0$ implies that $XYBC$ is concyclic. Now let $S$ be the second intersection point of the circle ($XYZ$) with the circumcircle; then the lines $ZS, XY$ and $BC$ are concurrent since they are the pairwise radical axes of the circles ($XYBC$) and ($XYZS$) and the circumcircle (let $P$ de... | Turkey | Team Selection Test for EGMO 2023 | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 149 | |
038t | Find all values of the real parameter $a$ such that the equation
$$
x^3 - a x^2 + (a^2 - 1)x - a^2 + a = 0
$$
has three distinct real roots which (in some order) form an arithmetic progression. | [
"Writing the equation in the form\n$$\n(x-1)(x^2 + (1-a)x - a + a^2) = 0,\n$$\nwe obtain $x_1 = 1$. Let $x_2$ and $x_3$ be the roots of the quadratic equation. If $1$ is the second term of the progression then $x_2 + x_3 = 2$, giving $a - 1 = 2$, i.e. $a = 3$. When $a = 3$ the roots of the quadratic equation are no... | Bulgaria | Winter Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | a = 0 or a = 6/7 | |
0agi | A strip of width $w$ is the set of all points which lie on, or between, two parallel lines distance $w$ apart. Let $S$ be a set of $n$ ($n \ge 3$) points on the plane such that any three different points of $S$ can be covered by a strip of width $1$.
Prove that $S$ can be covered by a strip of width $2$. | [
"Firstly we shall prove the following statement.\n**Lemma.** If a triangle can be covered by a strip of breadth $b$, then at least one altitude of the triangle is at most $b$ long.\n**Proof.** At least one of the perpendicular lines through the vertices of the triangle to the border lines of the strip meets the opp... | North Macedonia | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | English | proof only | null | |
07kz | For any positive integer $n$ define
$$
E(n) = n(n + 1)(2n + 1)(3n + 1) \cdots (10n + 1)
$$
Find the greatest common divisor of $E(1)$, $E(2)$, $E(3)$, \ldots $E(2009)$. | [
"Let $m$ be the g.c.d. of $E(1), E(2), E(3), \\dots, E(2009)$. Since $m|E(1) = 2 \\cdot 3 \\dots 11$, it follows that any prime divisor of $m$ is less than or equal to $11$. Let $p$ be a prime number such that $p \\nmid m$. Since $p \\le 11 < 2009$, it follows that $m|E(p) = p(p+1)(2p+1)(3p+1)\\cdots(10p+1)$. Obser... | Ireland | Irska | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | 2310 | |
00u4 | Let $ABC$ be a triangle with circumcircle $\omega$, circumcenter $O$, and orthocenter $H$. Let $K$ be the midpoint of $AH$. The perpendicular to $OK$ at $K$ intersects $AB$ and $AC$ at $P$ and $Q$, respectively. The lines $BK$ and $CK$ intersect $\omega$ again at $X$ and $Y$, respectively. Prove that the second interse... | [
"**Claim 1.** $PK = KQ$.\n**Proof of Claim 1.** Let $L$ and $N$ be the midpoints of $AB$ and $AC$, respectively. Since $L$ and $K$ are midpoints of $AB$ and $AH$, then $LK \\parallel BH$ and so $LK \\perp AC$. Since also $LO \\perp AB$, then $\\angle KLO = \\angle BAC = \\alpha$. Also, $\\angle OLP = 90^\\circ = \\... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"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 > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
03tc | Suppose points $F_1, F_2$ are the left and right foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$ respectively, and point $P$ is on line $l$: $x - \sqrt{3}y + 8 + 2\sqrt{3} = 0$. When $\angle F_1PF_2$ reaches the maximum, the value of ratio $\frac{|PF_1|}{|PF_2|}$ is ______. | [
"Euclidean geometry tells us that, $\\angle F_1PF_2$ reaches the maximum only if the circle through points $F_1, F_2, P$ is tangent to the line $l$ at $P$. Now suppose $l$ intercepts the $x$-axis at point $A(-8-2\\sqrt{3}, 0)$. Then $\\angle APF_1 = \\angle AF_2P$, and that means $\\triangle APF_1 \\sim \\triangle ... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | sqrt(3) - 1 | |
0hn9 | Problem:
Let $P$ be a polynomial with integer coefficients. Let $S$ be the set of integers $n$ for which $P(n) / n$ is an integer. Show that $S$ contains either finitely many integers, or all but finitely many integers. | [
"Solution:\nLet $c$ be the constant coefficient of $P$, so that $P(x)$ is of the form $Q(x) x + c$. If $n \\mid P(n)$, we then have $n \\mid (Q(n) n + c)$, or $n \\mid c$. $S$ is thus the set of divisors of $c$. If it is not finite, $c$ must then be $0$, and $S$ is the set of all nonzero integers, as desired."
] | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
06hd | Points $A$ and $C$ lie on the circumference of a circle with radius $\sqrt{50}$. $B$ is a point inside the circle such that $\angle ABC = 90^\circ$. If $AB = 6$ and $BC = 2$, find the distance from $B$ to the centre of the circle.
$A$ 和 $C$ 是一個半徑為 $\sqrt{50}$ 的圓的圓周上的兩點。$B$ 是該圓內的一點，使得 $\angle ABC = 90^\circ$。若 $AB = 6$... | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Chinese | proof and answer | sqrt(26) | |
079b | There is a $m \times n$ board divided into $mn$ unit squares, and we have drawn one of the two diagonals of each unit square. Prove that there is a path using only these diagonals that either connects the upper side of the board to the lower side or connects the left side of it to the right side. | [
"A connected component of diagonals is some of the diagonals that are connected to each other by a path of diagonals, and if we add another diagonal to it, it will no longer be connected. Easily, it can be seen that the diagonals that are not in this component but are on the border of this component, are connected.... | Iran | 27th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
01z2 | Call the set $\{a_1, a_2, \dots, a_n\}$ of positive integers good if
$$
\gcd(a_k, \sum_{i \neq k} a_i) > 1
$$
for each $k = \overline{1, n}$.
Prove that for each $c \in (0, 1)$ there exists a positive integer $M$ such that for any positive integer $K \ge M$ one can find a good set with at least $cK$ elements all elemen... | [
"Recall (with proof) two well-known facts:\n**Lemma 1.** The series $\\sum_{i=1}^{+\\infty} \\frac{1}{i}$ is divergent. In other words for each $c \\in \\mathbb{R}$ there exists $m$ such that $\\sum_{i=1}^{m} \\frac{1}{i} \\ge c$.\n**Proof.** Note that $\\sum_{i=2}^{2^{a+1}} \\frac{1}{i} \\ge \\sum_{i=2}^{2^{a+1}} ... | Belarus | SELECTION and TRAINING SESSION | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0ht8 | Problem:
Let $n \geq 1$ be an integer. How many ways can the rectangle having vertices $(0,0)$, $(n, 0)$, $(n, 1)$, $(0,1)$ be dissected into $2n$ triangles, all vertices of which have integer coordinates?
The triangles are considered as positioned on the coordinate plane; in particular, tilings related by rotation an... | [
"Solution:\nSince the vertices of the tiles must lie within the rectangle and cannot be collinear, all tiles have the form\n$$\n(a, 0)(b, 0)(c, 1) \\text{ or } \\quad (a, 1)(b, 1)(c, 0) \\quad (a < b).\n$$\nSuch a triangle has area $(b-a)/2$; for $2n$ such triangles to tile a rectangle of area $n$, it is necessary ... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | binomial(2n, n) | |
0bvs | Show that there is no positive integer with $2017$ divisors and the sum of the digits $2017$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
08wd | In the exterior of a triangle $ABC$, 3 squares $PQBA$, $RSCB$ and $TUAC$, each having a side of the triangle as one of its sides, were drawn. If $AB = 3$, $BC = 4$ and $CA = 3$, determine the area of the hexagon $PQRSTU$. Here we denote for a line segment $XY$ its length also by $XY$.
 | [
"First, let us show that the areas of the triangles $AUP$, $BQR$ and $CST$ are all equal to the area of the triangle $ABC$. Since we have $\\angle QBR + \\angle ABC = 180^\\circ$, $QB = AB$ and $BR = BC$, if we rotate the triangle $BQR$ clockwise for $90^\\circ$ around the axis through $B$, it lands onto a triangle... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 34 + 8√5 | |
0fxu | Problem:
Die Punkte $A$, $M_{1}$, $M_{2}$ und $C$ liegen in dieser Reihenfolge auf einer Geraden. Sei $k_{1}$ der Kreis mit Mittelpunkt $M_{1}$ durch $A$ und $k_{2}$ der Kreis mit Mittelpunkt $M_{2}$ durch $C$. Die beiden Kreise schneiden sich in den Punkten $E$ und $F$. Eine gemeinsame Tangente an $k_{1}$ und $k_{2}$... | [
"Solution:\n\nSei $S$ der Schnittpunkt von $A B$ und $C D$. Wir zeigen, dass $S$ auf der Potenzlinie $E F$ von $k_{1}$ und $k_{2}$ liegt. Wir müssen also zeigen, dass die Potenz zu beiden Kreisen gleich groß ist, d.h. $S B \\cdot S A = S C \\cdot S D$. Dies ist gleichbedeutend damit, dass $A B C D$ ein Sehnenvierec... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b3w | Problem:
Let $\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_{A}$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_{A}$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_{B}, P_{C}, Q_{B}$, and $Q_{C}$ similarly. Show th... | [
"Solution:\nRefer to the figure shown below:\n\nLet $M_{A}, M_{B}$, and $M_{C}$ be the midpoints of $BC, CA$, and $AB$ respectively.\n\nFirst, it may be shown that $P_{B}$ and $Q_{C}$ lie on $M_{B}M_{C}$.\n\nNote that $\\angle AM_{C}Q_{C} = 2\\angle ABQ_{C} = 180^{\\circ} - \\angle ABC = \\... | Philippines | Philippine 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 > Angle chasing"
] | null | proof only | null | |
07j0 | For a function $f : \mathbb{N} \to \mathbb{N}$ define $P(n) = f(1) \cdots f(n)$. Find all functions $f$ such that for all positive integers $a$ and $b$, we have
$$
P(a) + P(b) \mid a! + b!
$$ | [
"Plugging $(a, b) = (1, 1)$ to obtain $f(1) = 1$. We can then inductively prove that $f(n) = n$. Assume that it has already been held for all positive integers less than $n$. Plugging $(a, b) = (n, 1)$, then after a simple algebra it follows that $f(n) \\le n$ and $(n-1)!f(n) + 1 \\mid (n-1)!(n - f(n))$. Since $\\g... | Iran | 41th Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(n) = n for all positive integers n | |
0ba3 | Let $m$ be a positive integer. Determine the smallest positive integer $n$ for which there exist real numbers $x_1, x_2, \dots, x_n \in (-1, 1)$ such that $|x_1| + |x_2| + \dots + |x_n| = m + |x_1 + x_2 + \dots + x_n|$. | [
"Let us consider $x_1, x_2, \\dots, x_n$ a solution of the equation above. We may suppose, without loss of generality, that $x_1 + x_2 + \\dots + x_n \\ge 0$ (otherwise we change the signs of all the numbers) and that $x_1 \\le \\dots \\le x_p \\le 0 < x_{p+1} \\le \\dots \\le x_n$. Then the equation becomes $-2(x_... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n = m + 1 if m is odd; n = m + 2 if m is even | |
0h0t | Find all values of parameter $b$, such that for all $x$ at least one function $f_1(x) = x^2 + 2011x + b$ or $f_2(x) = x^2 - 2011x + b$ is positive. | [
"For $x = 0$ we have $f_1(0) = f_2(0) = b$, thus all $b \\le 0$ does not satisfy the condition of the problem.\n\nLet $b > 0$. We add two values up and get $f_1(x) + f_2(x) = 2x^2 + 2b > 0$, thus at least one function is positive."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | b > 0 | |
0juo | Problem:
Steph Curry is playing the following game and he wins if he has exactly 5 points at some time. Flip a fair coin. If heads, shoot a 3-point shot which is worth 3 points. If tails, shoot a free throw which is worth 1 point. He makes $\frac{1}{2}$ of his 3-point shots and all of his free throws. Find the probabi... | [
"Solution:\n\nAnswer: $\\frac{140}{243}$\n\nIf he misses the shot, then the state of the game is the same as before he flipped the coin. Since the probability of making a free throw is $\\frac{1}{2}$ and the probability of making a 3-point shot is $\\frac{1}{4}$. Therefore, given that he earns some point, the proba... | United States | HMMT November 2016 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 140/243 | |
0isf | Two sequences of integers, $a_1, a_2, a_3, \dots$ and $b_1, b_2, b_3, \dots$, satisfy the equation
$$
(a_n - a_{n-1})(a_n - a_{n-2}) + (b_n - b_{n-1})(b_n - b_{n-2}) = 0
$$
for each integer $n$ greater than $2$. Prove that there is a positive integer $k$ such that $a_k = a_{k+2008}$. | [
"Define $d(i, j) = (a_i - a_j)^2 + (b_i - b_j)^2$. Notice that\n$$\n(a_n - a_{n-1})^2 + (a_n - a_{n-2})^2 - (a_{n-1} - a_{n-2})^2 = 2(a_n - a_{n-1})(a_n - a_{n-2})\n$$\nand\n$$\n(b_n - b_{n-1})^2 + (b_n - b_{n-2})^2 - (b_{n-1} - b_{n-2})^2 = 2(b_n - b_{n-1})(b_n - b_{n-2}).\n$$\nAdding these two equations and using... | United States | Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
01yw | A convex quadrilateral $ABCD$ has the incircle $\omega$. The diagonal $AC$ intersects $\omega$ at the points $P$ and $Q$. Let $M$ and $N$ be the midpoints of the arcs $PQ$ of $\omega$ such that the points $B$ and $M$ lie in one halfplane with respect to the line $AC$ while the points $D$ and $N$ — in another.
Prove tha... | [
"Without loss of generality assume that $P$ lies between $A$ and $Q$. Let the line passing through $M$ parallel to $AC$ intersect the sides $AB$ and $BC$ at the points $K$ and $L$ respectively. Since $M$ is the midpoint of the arc $PQ$, the line $KL$ is parallel to $AC$ whence $\\omega$ is the tangency point of $B$... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0jsg | Problem:
Let $ABC$ be a triangle such that $AB = 13$, $BC = 14$, $CA = 15$ and let $E, F$ be the feet of the altitudes from $B$ and $C$, respectively. Let the circumcircle of triangle $AEF$ be $\omega$. We draw three lines, tangent to the circumcircle of triangle $AEF$ at $A$, $E$, and $F$. Compute the area of the tri... | [
"Solution:\n\nNote that $AEF \\sim ABC$. Let the vertices of the triangle whose area we wish to compute be $P, Q, R$, opposite $A, E, F$ respectively. Since $H, O$ are isogonal conjugates, line $AH$ passes through the circumcenter of $AEF$, so $QR \\parallel BC$.\n\nLet $M$ be the midpoint of $BC$. We claim that $M... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circu... | null | proof and answer | 462/5 | |
0jkd | Problem:
Natalie has a copy of the unit interval $[0,1]$ that is colored white. She also has a black marker, and she colors the interval in the following manner: at each step, she selects a value $x \in [0,1]$ uniformly at random, and
a. If $x \leq \frac{1}{2}$ she colors the interval $\left[x, x+\frac{1}{2}\right]$ ... | [
"Solution:\n\nAnswer: 5\n\nThe first choice always wipes out half the interval. So we calculate the expected value of the amount of time needed to wipe out the other half.\n\nSolution 1 (non-calculus):\nWe assume the interval has $2n$ points and we start with the last $n$ colored black. We let $f(k)$ be the expecte... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 5 | |
0i8h | Problem:
How many lattice points are enclosed by the triangle with vertices $(0,99)$, $(5,100)$, and $(2003, 500)$? Don't count boundary points. | [
"Solution:\nUsing the determinant formula, we get that the area of the triangle is\n$$\n\\left|\\begin{array}{cc}\n5 & 1 \\\\\n2003 & 401\n\\end{array}\\right| / 2 = 1\n$$\nThere are 4 lattice points on the boundary of the triangle (the three vertices and $(1004, 300)$), so it follows from Pick's Theorem that there... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | final answer only | 0 | |
0clb | Let $f : [0, 1] \to \mathbb{R}$ be a continuous function. We define the function $\tilde{f} : [0, 1] \to \mathbb{R}$ by
$$
\tilde{f}(x) = \begin{cases} \frac{1}{x} \cdot \int_{0}^{x} f(t) \, dt, & \text{if } x > 0, \\ f(0), & \text{if } x = 0. \end{cases}
$$
Show that:
a) the function $\tilde{f}$ is continuous in $0$ ... | [
"a) The function $f$ being continuous on $[0, 1]$, it follows that the function $F: [0, 1] \\to \\mathbb{R}$ defined by $F(x) = \\int_{0}^{x} f(t) dt$ is differentiable on $[0, 1]$, with $F' = f$. It follows that the function $\\tilde{f}$ is differentiable on $(0, 1]$, as a product of differentiable functions.\nIt ... | Romania | 75th Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Applications",
"Calculus > Differential Calculus > Derivatives"
] | English | proof only | null | |
0dih | Does there exist the infinite sequence of real numbers $(x_n)$ satisfying $x_1 = 2$ and
$$
\frac{2x_n^2 + 2}{x_n + 3} < x_{n+1} \le \frac{2x_n + 2}{x_n + 3} + 2023
$$
for all positive integers $n = 1, 2, 3, \dots$? | [
"The answer is No. Suppose by contradiction that there is such a sequence. First, we will prove by induction that $x_n > 2$ for every $n \\ge 2$. One can check with $n = 2$, then assume that the assertion is true for $n = k \\ge 2$, i.e. $x_k > 2$ then\n$$\n\\frac{2x_k^2 + 2}{x_k + 3} - 2 = \\frac{2x_k^2 - 2x_k - 4... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | No | |
0ect | A polynomial $p(x) = 2015x^{2013} - 2$ and a real number $h$ are given so that $p(h) = -2015$. What is the value of $p(-h)$?
(A) 2011 (B) 2012 (C) 2013 (D) 2014 (E) 2015 | [
"From data it follows that $2015h^{2013} - 2 = -2015$ which means $2015h^{2013} = -2013$. Thus $p(-h) = 2015(-h)^{2013} - 2 = -2015h^{2013} - 2 = -(-2013) - 2 = 2011$."
] | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | A | |
0cwn | A right prism $ABC A_1 B_1 C_1$ is given. It is known that triangles $A_1 BC$, $AB_1 C$, $ABC_1$, and $ABC$ are acute-angled. Prove that the orthocenters of these triangles, and the centroid of $ABC$ lie on a sphere. | [
"Let $M$ and $H$ denote the centroid and orthocenter of triangle $ABC$ respectively, and let $T$ be a point such that $3\\overrightarrow{MT} = \\overrightarrow{AA_1} = \\overrightarrow{BB_1} = \\overrightarrow{CC_1}$. Let $\\omega$ be the sphere with diameter $HT$. Since the line $MT$ is perpendicular to the plane ... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Me... | Russian | proof only | null | |
0coe | In a square grid $n \times n$, a set consisting of all cells lying on or under its main diagonal is called an $n$-staircase (4-staircase is shown in the figure). Find the number of ways to partition an $n$-staircase into several grid rectangles with pairwise distinct areas. (D. Khramtsov)
На... | [
"Ответ. $2^{n-1}$.\n\nОтметим в каждом столбце лестницы по одной верхней клетке; назовём их объединение **верхним слоем**. Никакие две из $n$ клеток этого слоя не могут лежать в одном прямоугольнике разбиения, поэтому в любом разбиении лестницы не менее $n$ прямоугольников. С другой стороны, минимальная суммарная п... | Russia | Regional round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Russian | proof and answer | 2^{n-1} | |
0f7s | Problem:
Show that $$(2n + 1)^n \geq (2n)^n + (2n - 1)^n$$ for every positive integer $n$. | [] | Soviet Union | 21st ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0gw5 | Prove that
$$
\left( \frac{a+b}{c^2} + \frac{b+c}{a^2} + \frac{c+a}{b^2} \right) \cdot \left( \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \right) \ge 3 + \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b}
$$
for any positive real numbers $a$, $b$ and $c$. | [
"It is easy to prove the inequalities\n$$\n\\frac{a^2}{b+c} \\geq a - \\frac{b+c}{4}, \\quad \\frac{b^2}{c+a} \\geq b - \\frac{c+a}{4}, \\quad \\frac{c^2}{a+b} \\geq c - \\frac{a+b}{4},\n$$\nadding which we obtain\n$$\n\\frac{a^2}{b+c} + \\frac{b^2}{c+a} + \\frac{c^2}{a+b} \\ge \\frac{a+b+c}{2}. \\quad (1)\n$$\n\nS... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0ivm | Problem:
Four points, $A$, $B$, $C$, and $D$, are chosen randomly on the circumference of a circle with independent uniform probability. What is the expected number of sides of triangle $A B C$ for which the projection of $D$ onto the line containing the side lies between the two vertices? | [
"Solution:\n\nBy linearity of expectations, the answer is exactly 3 times the probability that the orthogonal projection of $D$ onto $A B$ lies interior to the segment. This happens exactly when either $\\angle D A B$ or $\\angle D B A$ is obtuse, which is equivalent to saying that $A$ and $B$ lie on the same side ... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3/2 | |
0a14 | We play a game of musical chairs with $n$ chairs numbered $1$ to $n$. You attach $n$ leaves, numbered $1$ to $n$, to the chairs in such a way that the number on a leaf does not match the number on the chair it is attached to. One player sits on each chair. Every time you clap, each player looks at the number on the lea... | [
"If $m = n$, then attach to chair $i$ the leaf with number $i+1$. Everyone then moves up a chair every clap and for everyone, the first time they return to the chair they started on is after $n$ claps. So after $n$ claps for the first time, everyone has returned to the chair they started on. Now suppose $m < n$.\n\... | Netherlands | BxMO Team Selection Test | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
0hme | Problem:
The integers from $1$ to $16$ are arranged in a $4 \times 4$ array so that each row, column and diagonal adds up to the same number.
a. Prove that this number is $34$.
b. Prove that the four corners also add up to $34$. | [
"Solution:\n\na. Add up all the numbers in the square in two ways. On the one hand, it consists of four rows, each adding to the common sum $S$, so the entire square adds to $4S$. But the numbers in the square are also the integers from $1$ to $16$, whose sum is\n$$\n\\frac{16 \\cdot 17}{2} = 136\n$$\nHowever, the ... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
04yh | Let $\{a_n\}_{n=1}^{\infty}$ be a sequence of positive integers such that for every positive integer $n$
$$
a_{n+1} = (n + 1)(a_n - n + 1).
$$
In terms of $a_1$, determine the greatest positive integer $k$ such that $\gcd(a_i, a_{i+1}) = k$
for some positive integer $i \ge 2$. (Note that $\gcd(x, y)$ denotes the greate... | [
"First, we will prove by induction that $a_n = (a_1 - 1)n! + n$ for all $n \\ge 1$. The base case $a_1$ is trivial. Now suppose that the closed form holds for some $a_n$. Then\n$$\n\\begin{align*}\na_{n+1} &= (n + 1) (((a_1 - 1)n! + n) - n + 1) \\\\\na_{n+1} &= (n + 1) ((a_1 - 1)n! + 1) \\\\\na_{n+1} &= (a_1 - 1)(n... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2025 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete... | null | proof and answer | The greatest k is the largest odd prime divisor of a1; if a1 is a power of two, then k = 1. | |
00sr | Given semicircle $(c)$ with diameter $AB$ and center $O$. On the $(c)$ we take point $C$ such that the tangent at the $C$ intersects the line $AB$ at the point $E$. The perpendicular line from $C$ to $AB$ intersects the diameter $AB$ at the point $D$. On the $(c)$ we get the points $H$, $Z$ such that $CD = CH = CZ$. Th... | [
"Since $CH = CZ$ we have $OC \\perp HZ$. So from the cyclic quadrilateral $SODI$ we get\n$$\nCS \\cdot CO = CI \\cdot CD. \\qquad (1)\n$$\n\nFigure 9: G9\nWe draw the perpendicular line $(v)$ to $HC$ at the point $H$. Let $J$ be the intersection point of lines $(v)$ and $CO$. Then $CJ$ is d... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"... | English | proof only | null | |
0grs | In the round robin chess tournament organized in a school every two students played one match among themselves. Find the minimal possible number of students in the school if each girl student has at least 21 wins in matches against boy students and each boy student has at least 12 wins in matches against girl students. | [
"The answer is $65$. Suppose that there are $x$ girl and $y$ boy students in the school. Obviously $xy \\ge 21x + 12y$ is a necessary condition for the existence of the tournament with given conditions. Let us show that this inequality is also a sufficient condition for the existence of the tournament. Let us consi... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 65 | |
0kxw | Let $p$ be a fixed prime and let $a \ge 2$ and $e \ge 1$ be fixed integers. Given a function $f: \mathbb{Z}/a\mathbb{Z} \to \mathbb{Z}/p^e\mathbb{Z}$ and an integer $k \ge 0$, the $k$th finite difference, denoted $\Delta^k f$, is the function from $\mathbb{Z}/a\mathbb{Z}$ to $\mathbb{Z}/p^e\mathbb{Z}$ defined recursive... | [
"Let $d = \\nu_p(a)$, so $a = p^d \\cdot b$ with $p \\nmid b$. We call a function $f: \\mathbb{Z}/a\\mathbb{Z} \\to \\mathbb{Z}/p^e\\mathbb{Z}$ *essential* if $\\Delta^k f = f$ for some $k \\ge 1$.\n\n**Claim (Characterization of essential functions):**\nA function $f$ is essential if and only if\n$$\nf(x) + f(x + ... | United States | USA TSTST | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Ex... | null | proof and answer | p^{e(a - p^{v_p(a)})} | |
0klk | Problem:
Teresa the bunny has a fair 8-sided die. Seven of its sides have fixed labels $1, 2, \ldots, 7$, and the label on the eighth side can be changed and begins as $1$. She rolls it several times, until each of $1, 2, \ldots, 7$ appears at least once. After each roll, if $k$ is the smallest positive integer that s... | [
"Solution:\n\nLet $n = 7$ and $p = \\frac{1}{4}$.\nLet $q_{k}$ be the probability that $n$ is the last number rolled, if $k$ numbers less than $n$ have already been rolled. We want $q_{0}$ and we know $q_{n-1} = 1$.\nWe have the relation\n$$\nq_{k} = (1-p) \\frac{k}{n-1} q_{k} + \\left[1 - (1-p) \\frac{k+1}{n-1}\\r... | United States | HMMT Spring 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 104 | |
00jf | We call an isosceles trapezoid interesting if it is inscribed in the unit square $ABCD$ such that one vertex of the trapezoid lies on each side of the square, and if the lines joining the mid-points of adjacent sides of the trapezoid are parallel to the sides of the square. Determine all interesting trapezoids and thei... | [
"Let $E$, $F$, $G$ and $H$ be the mid-points of $PQ$, $QR$, $RS$ and $SP$ respectively. Since the sides of $EFGH$ are parallel to the sides of $ABCD$, $EFGH$ is certainly a rectangle. Since $PQRS$ is isosceles, the line $FH$ joining the parallel sides must be an axis of symmetry of the trapezoid, and therefore also... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | All such trapezoids are precisely those whose axis of symmetry lies along a diagonal of the square and whose diagonals are parallel to the sides of the square; their area is 1/2. | |
0a5c | Problem:
In a sequence of numbers, a term is called golden if it is divisible by the term immediately before it. What is the maximum possible number of golden terms in a permutation of $1,2,3,\ldots ,2021$? | [
"Solution:\n\nLet $k$ be the number of golden terms. We claim that $k \\leq 1010$.\n\nProof: Define the term immediately before a golden term to be a silver term. The number of silver terms is also $k$. If $a$ is any silver term and $b$ is the corresponding golden term then we must have\n$$a \\leq \\frac{b}{2} \\le... | New Zealand | NZMO Round One | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1010 | |
0h1g | Five years ago the total age of all sons in the family was two years more than the total age of all daughters. Since that time, one more child was born and now the total age of all daughters is two years more than the total age of all sons. What was the difference between the total age of sons and daughters two years a... | [
"Suppose that five years ago family had $n$ sons and $m$ daughters, total age of sons was $N$, total age of daughters was $M$.\nIf $k$ years ago one daughter was born, then the total age of sons now is $N+5n$, for daughters it is $M+5m+k$. Since $N=M+2$, we get that\n$$\n(N+5n)+2 = M+5m+k \\Rightarrow M+5n+4 = M+5m... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 0 or 1 | |
06pq | Let $X$ be a set of $10000$ integers, none of them is divisible by $47$. Prove that there exists a $2007$-element subset $Y$ of $X$ such that $a-b+c-d+e$ is not divisible by $47$ for any $a, b, c, d, e \in Y$. | [
"Call a set $M$ of integers good if $47 \\nmid a-b+c-d+e$ for any $a, b, c, d, e \\in M$.\n\nConsider the set $J=\\{-9,-7,-5,-3,-1,1,3,5,7,9\\}$. We claim that $J$ is good. Actually, for any $a, b, c, d, e \\in J$ the number $a-b+c-d+e$ is odd and\n$$\n-45 = (-9) - 9 + (-9) - 9 + (-9) \\leq a-b+c-d+e \\leq 9 - (-9)... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0e82 | Let $E$ be a point on the side $CD$ of a rectangle $ABCD$ such that $\angle AEB$ is a right angle and $3|EA| = 2|EC|$. Determine the ratio of the lengths of the sides of the rectangle $ABCD$. | [
"Denote $|AB| = |CD| = a$, $|BC| = |DA| = b$ and $|EC| = c$. Then $|EA| = \\frac{2}{3}c$ and $|ED| = a - c$. By Pythagoras' theorem for the triangle $AED$ we have $b^2 + (a-c)^2 = \\frac{4}{9}c^2$, or $b^2 = -a^2 + 2ac - \\frac{5}{9}c^2$.\n\nBy Pythagoras' theorem for the triangles $BCE$ and $ABE$ we have $c^2 + b^... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4*sqrt(3)/3 | |
0dob | Problem:
Наћи све моничне полиноме $P(x)$ такве да је полином $P(x)^2-1$ дељив полиномом $P(x+1)$. | [
"Solution:\n\nЈедина решења су полиноми $P(x)=1$ и $P(x)=x-c$, где је $c$ константа.\n\nПретпоставимо да је $P(x)=(x-c)\\left(x-x_{2}\\right) \\cdots\\left(x-x_{n}\\right)$ неконстантан полином, при чему је $c$ његова комплексна нула чији је реални део најмањи.\n\nПо услову задатка, $x+1-c$ дели $P(x)^2-1$, одакле ... | Serbia | 14. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | P(x) = 1 or P(x) = x - c (where c is a constant) | |
0cne | A sequence of integers is written on an infinite tape. The first number is $1$; each number except the first one is obtained by adding to the previous number its minimal nonzero digit (in decimal representation). Find the number of digits in the decimal representation of the number at $9 \cdot 1000^{1000}$th place in t... | [
"Answer: $3001$.\n\nSince each number in the sequence, starting from the second, is greater than the previous one by at least $1$, the $9 \\cdot 1000^{1000}$-th number is at least $9 \\cdot 1000^{1000}$, so it has at least $3001$ digits. Denote the $n$-th number of the sequence by $a_n$, and let $k$ be the smallest... | Russia | Euler olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English; Russian | proof and answer | 3001 | |
04br | For which $x \in \mathbb{R}$ is the number $\sqrt[3]{4+4x}$ greater than the number $1+\sqrt[3]{x}$? | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof and answer | (-1, 1) ∪ (1, ∞) | |
0f8a | Problem:
A polygonal line connects two opposite vertices of a cube with side $2$. Each segment of the line has length $3$ and each vertex lies on the faces (or edges) of the cube. What is the smallest number of segments the line can have? | [
"Solution:\n\nAnswer $6$\n\n\n\nSuppose one endpoint of a segment length $3$ is at $A$. Evidently the other end could be at the edge midpoints $B$, $C$, $D$. It could also be on the circular arc connecting $B$ and $C$ (with center $O$ and radius $\\sqrt{5}$). Similarly, it could be on arcs ... | Soviet Union | 22nd ASU | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 6 | |
0fm7 | Problem:
Dado un entero positivo $n$, hallar la suma de todos los enteros positivos inferiores a $10 n$ que no son múltiplos de 2 ni de 5. | [
"Solution:\nSean los conjuntos\n$$\n\\begin{aligned}\nA & =\\{1,2, \\ldots, 10 n\\} \\\\\nB & =\\{2,4, \\ldots, 2(5 n)\\} \\\\\nC & =\\{5,10, \\ldots, 5(2 n)\\} \\\\\nB \\cap C & =\\{10,20, \\ldots, 10 n\\}\n\\end{aligned}\n$$\nNos piden la suma de los elementos de $A$ que no son de $B$ ni de $C$. Las sumas de los ... | Spain | Spain | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 20n^2 | |
0ekf | Problem:
Želimo splesti $20~\mathrm{m}$ dolg navijaški šal. V koliko dneh ga bomo dokončali, če prvi dan spletemo $18~\mathrm{cm}$, nato pa vsak naslednji dan za $4~\mathrm{cm}$ več kot predhodni dan?
(A) v 27 dneh
(B) v 18 dneh
(C) v 36 dneh
(D) v 28 dneh
(E) v 497 dneh | [
"Solution:\n\nUgotovimo, da gre za aritmetično zaporedje s prvim členom $18$ in diferenco $4$ ter vsoto prvih $n$ členov $2000$. Zapišemo njegov splošni člen $a_{n} = 18 + (n-1) 4$ in formulo za vsoto prvih $n$ členov aritmetičnega zaporedja $S_{n} = \\frac{n}{2} (a_{1} + a_{n}) = 2000$. Dobimo enačbo oblike $n^{2}... | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | D | |
0ii3 | Problem:
The lottery cards of a certain lottery contain all nine-digit numbers that can be formed with the digits $1$, $2$ and $3$. There is exactly one number on each lottery card. There are only red, yellow and blue lottery cards. Two lottery numbers that differ from each other in all nine digits always appear on ca... | [
"Solution:\n\nFirst, it can in fact be red, if, say, cards are colored based on the first digit only ($1=$ red, $2=$ yellow, $3=$ blue$)$. We now endeavor to show it must be red.\n\nConsider the cards $333133133$ and $331331331$: they each differ in all their digits from $122222222$ and from $222222222$, so they mu... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Logic"
] | null | proof and answer | red | |
0by6 | Consider an $m \times n$ board where $m, n \ge 3$ are positive integers, divided into unit squares. Initially all the squares are white. What is the minimum number of squares that need to be painted red such that each $3 \times 3$ square contains at least two red squares? | [
"We label the rows from 1 to $m$ and the columns from 1 to $n$.\nIf $m$ and $n$ are not congruent to 2 modulo 3, i.e. $m = 3a + r_1$ and $n = 3b + r_2$ with $r_1, r_2 \\in \\{0, 1\\}$, we can tile the rectangle formed by the first $3a$ lines and $3b$ columns with $a \\cdot b$ disjoint $3 \\times 3$ squares. Each of... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | If m ≡ 2 mod 3 and n ≡ 2 mod 3, the minimum is 2·floor(m/3)·floor(n/3) + min(floor(m/3), floor(n/3)); otherwise, the minimum is 2·floor(m/3)·floor(n/3). | |
0ckz | Determine all polynomials $P$ with integer coefficients, satisfying $0 \le P(n) \le n!$, for all non-negative integers $n$. | [
"The required polynomials are $P = 0$, $P = 1$, $P = (X - 1)^2$, $P = X(X - 1)\\cdots(X - k)$ and $P = X(X - 1)\\cdots(X - k)(X - k - 2)^2$ for some non-negative integer $k$. The verification is routine and is hence omitted.\n\nWe first deal with the case $P(0) = 1$. The polynomials $P_1 = 1$ and $P_2 = (X-1)^2$ bo... | Romania | 75th NMO Selection Tests | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All such polynomials are exactly the following:
- P(X) = 0;
- P(X) = 1;
- P(X) = (X − 1)^2;
- P(X) = X(X − 1)⋯(X − k) for some nonnegative integer k;
- P(X) = X(X − 1)⋯(X − k) (X − k − 2)^2 for some nonnegative integer k. | |
02v4 | Problem:
Queremos cobrir um tabuleiro quadriculado com certas pecinhas sem sobreposição e de modo que nenhuma parte delas fique fora do tabuleiro. Usaremos pecinhas, formadas por quadradinhos, chamadas L-triminós e I-triminós e que podem ser rotacionadas nas posições descritas na figura a seguir.
![](attached_image_1... | [
"Solution:\n\na) A figura a seguir mostra uma maneira de cobrir o tabuleiro $3 \\times 4$ usando apenas L-triminós.\n\n\n\nb) Considere o tabuleiro $3 \\times 5$ a seguir e os 6 quadradinhos pintados.\nComo um L-triminó não pode cobrir duas dessas casinhas pintadas, se fosse possível cobrir... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) A covering of the three by four board using only L-triminos exists (one explicit arrangement is shown). b) It is impossible to cover the three by five board using only L-triminos. c) The straight trimino must cover two marked squares; it can occupy seven positions: four horizontal positions in the top and bottom row... | |
0dc3 | Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\frac{a^{3}}{a^{2}+b c}+\frac{b^{3}}{b^{2}+c a}+\frac{c^{3}}{c^{2}+a b} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a)}{a^{3}+b^{3}+c^{3}+3 a b c}
$$
When will the equality hold? | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | Equality holds when a = b = c. | |
0irg | Problem:
There are $5$ dogs, $4$ cats, and $7$ bowls of milk at an animal gathering. Dogs and cats are distinguishable, but all bowls of milk are the same. In how many ways can every dog and cat be paired with either a member of the other species or a bowl of milk such that all the bowls of milk are taken? | [
"Solution:\n\nAnswer: $20$\n\nSince there are $9$ dogs and cats combined and $7$ bowls of milk, there can only be one dog-cat pair, and all the other pairs must contain a bowl of milk. There are $4 \\times 5$ ways of selecting the dog-cat pair, and only one way of picking the other pairs, since the bowls of milk ar... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 20 | |
0iza | Problem:
In a $16 \times 16$ table of integers, each row and column contains at most 4 distinct integers. What is the maximum number of distinct integers that there can be in the whole table? | [
"Solution:\nAnswer: 49\n\nFirst, we show that 50 is too big. Assume for sake of contradiction that a labeling with at least 50 distinct integers exists. By the Pigeonhole Principle, there must be at least one row, say the first row, with at least 4 distinct integers in it; in this case, that is exactly 4, since tha... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 49 | |
0fg2 | Problem:
Dada la ecuación $x^{5}-p x-1=0$, estudiar el valor de $p$ de forma que existan dos soluciones de la ecuación, $x_{1}, x_{2}$, que a la vez sean soluciones de $x^{2}-a x+b=0$, con $a, b$ enteros. | [
"Solution:\n\nDe acuerdo con las condiciones del problema, deberá ser\n$$\n\\begin{aligned}\nx^{5}-p x-1= & \\left(x^{2}-a x+b\\right)\\left(x^{3}+c x^{2}+d x+e\\right)= \\\\\n& =x^{5}+(c-a) x^{4}+(d-a c+b) x^{3}+(e-a d+b c) x^{2}+(b d-a e) x+b e\n\\end{aligned}\n$$\nasí que, identificando coeficientes, se obtiene ... | Spain | OME 21 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | p = -1 | |
098u | Problem:
Baza piramidei $V A B C$ este triunghiul isoscel $A B C$, în care $A B = A C = 6 \sqrt{2}\ \mathrm{~cm}$ şi $B C = 4 \sqrt{6}\ \mathrm{~cm}$. Muchiile laterale ale piramidei sunt de $\sqrt{51}\ \mathrm{~cm}$. Determinați distanța dintre dreptele $A B$ și $V C$. | [
"Solution:\n\nFie $O$ - proiecția vârfului $V$ pe planul $(A B C)$. Deoarece muchiile laterale sunt congruente, $O$ este centrul cercului circumscris triunghiului $A B C$.\n\n\n\nFie $K \\in B C$, astfel încât $A K \\perp B C$. Atunci $A K = 4 \\sqrt{3}\\ \\mathrm{~cm}$.\n\nFie $\\alpha = m... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geome... | null | proof and answer | 16 sqrt 6 / 7 cm | |
0bun | Problem:
Fie $A \in \mathcal{M}_{n}(\mathbb{R})$ cu proprietatea că suma elementelor de pe fiecare linie este pozitivă și suma elementelor de pe fiecare coloană este negativă.
Calculați $\operatorname{det}(A)$. | [] | Romania | Olimpiada Nationala de Matematica | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
05vz | Problem:
Soit $p$ un nombre premier. Montrer qu'il existe une permutation $\left(a_{1}, \ldots, a_{p}\right)$ de $(1, \ldots, p)$ telle que les entiers $a_{1}, a_{1} \cdot a_{2}, \ldots, a_{1} \cdot a_{2} \cdot \ldots \cdot a_{p}$ donnent $p$ restes deux à deux distincts lorsque qu'on réalise leur division euclidienne... | [
"Solution:\n\nOn peut être très optimiste sur cet exercice et vouloir que $a_{1} a_{2} \\cdots a_{k} \\equiv k \\bmod p$ pour tout $k$ compris entre $1$ et $p$. Il faut donc que $a_{1}=1$ et pour tout $k \\neq 1$ :\n$$\na_{k} \\equiv \\frac{a_{1} \\cdots a_{k}}{a_{1} \\cdots a_{k-1}} \\equiv \\frac{k}{k-1} \\quad \... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
043s | Given positive integers $n$ and $k$, $n > k^2 > 4$. In an $n \times n$ grid, any $k$ squares in distinct rows and distinct columns are called a $k$-set. Find the largest positive integer $N$ satisfying that: one can choose $N$ squares of the $n \times n$ grid and colour them in a certain way, such that for any coloured... | [
"$N = (k - 1)^2 n$.\n\nChoose $(k-1)^2$ rows of the grid: colour the first $k-1$ rows in colour $c_1$; the second $k-1$ rows in colour $c_2$; ...; the last $k-1$ rows in colour $c_{k-1}$. Altogether, $(k-1)^2 n$ squares are coloured. For any coloured $k$-set, as there are only $k-1$ colours, some two squares must h... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (k - 1)^2 n | |
0doh | In a triangle $ABC$ with incenter $I$, let $P$ be the intersection point of the bisector of the angle $A$ with the circumcircle other than $A$, $D$ the point of tangency of the incircle to the side $BC$, and $Q$ the intersection point of $PD$ with the circumcircle other than $P$. Show that $PI = QI$ if $PD$ is equal to... | [
"\nIn the triangle $BIA$, $\\angle BIP = \\angle IBA + \\angle IAB = \\angle B/2 + \\angle A/2$. On the other hand, $\\angle PBI = \\angle PBC + \\angle CBI = \\angle A/2 + \\angle B/2$, as $\\angle PBC = \\angle PAC = \\angle A/2$. Hence, in the triangle $BPI$, $BP = IP$.\n\nSince $\\angle... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"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 | |
0cem | Prove that a convex polygon $A_1A_2...A_n$ has three vertices $A_i, A_j, A_k$ such that
$$
[A_iA_jA_k] > \frac{1}{4}[A_1A_2...A_n],
$$
where $[X_1X_2...X_m]$ denotes the area of the polygon $X_1X_2...X_m$. | [
"Let $A_iA_jA_k$ be a triangle of maximal area. Let $A'_i$ be the reflection of $A_i$ across the midpoint of the side $A_jA_k$; the points $A'_j$ and $A'_k$ are defined similarly. Note that at most one of these three reflections can be a vertex of the polygon — otherwise, the polygon would have three collinear vert... | Romania | Eighteenth STARS OF MATHEMATICS Competition | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
06lh | Two circles $\Gamma$ and $\Omega$ intersect at two distinct points $A$ and $B$. Let $P$ be a point on $\Gamma$. The tangent at $P$ to $\Gamma$ meets $\Omega$ at the points $C$ and $D$, where $D$ lies between $P$ and $C$, and $ABCD$ is a convex quadrilateral. The lines $CA$ and $CB$ meet $\Gamma$ again at $E$ and $F$ re... | [
"Since\n$$\n\\angle TEC = \\angle TEA = \\angle TBA = \\angle DBA = \\angle DCA = \\angle DCE,\n$$\nwe get $PC//ET$. Similarly, since\n$$\n\\angle FSD = \\angle FSA = \\angle CBA = \\angle PDA = \\angle PDS,\n$$\nwe get $PC//SF$."
] | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l30 | Problem:
Let $f$ be a function on nonnegative integers such that $f(0)=0$ and
$$
f(3 n+2)=f(3 n+1)=f(3 n)+1=3 f(n)+1
$$
for all integers $n \geq 0$. Compute the sum of all nonnegative integers $m$ such that $f(m)=13$. | [
"Solution:\nLet $\\underline{x}_{k}$ denote the number $x$ in base $k$. Observe that if $f\\left(\\underline{x}_{3}\\right)=\\underline{y}_{3}$, then\n$$\nf\\left(\\underline{x 0}_{3}\\right)=f(3 x)=3 f(x)=\\underline{y 0}_{3}\n$$\nand\n$$\nf\\left(\\underline{x 1}_{3}\\right)=f\\left(\\underline{x 2}_{3}\\right)=3... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 156 | |
03ya | Given integer $n \ge 2$ and real numbers $x_1, x_2, \dots, x_n$ in the interval $[0, 1]$, prove that there exist real numbers $a_0, a_1, \dots, a_n$ satisfying simultaneously the following conditions:
$$
(1)
\begin{aligned}
a_0 &+ a_n = 0; \\
a_i &\le 1, \text{ for every } i = 0, 1, \dots, n;
\end{aligned}
$$
$$
(2) \... | [
"For any $a \\in [0, 1)$, define a sequence $\\{a_i\\}_{i=0}^n$ generated by $a$ as follows: $a_0 = a$; for $1 \\le i \\le n$, $a_i = a_{i-1} - x_i$ if $a_{i-1} \\ge 0$, and $a_i = a_{i-1} + x_i$ if $a_{i-1} < 0$.\n\nSet $f(a) = a_n$. It is easy to show by induction that $|a_i| \\le 1$ for every $0 \\le i \\le n$.\... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0iny | Problem:
The elliptic curve $y^{2}=x^{3}+1$ is tangent to a circle centered at $(4,0)$ at the point $(x_{0}, y_{0})$. Determine the sum of all possible values of $x_{0}$. | [
"Solution:\n\nAnswer: $\\frac{1}{3}$.\n\nNote that $y^{2} \\geq 0$, so $x^{3} \\geq -1$ and $x \\geq -1$.\n\nLet the circle be defined by $(x-4)^{2}+y^{2}=c$ for some $c \\geq 0$.\n\nNow differentiate the equations with respect to $x$, obtaining $2y \\frac{\\mathrm{d}y}{\\mathrm{d}x} = 3x^{2}$ from the given and $2... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 1/3 | |
01uw | Given the isosceles triangle $ABC$ ($CA = CB$). The bisector of the angle $\angle B$ intersects the side $AC$ at point $L$ and the circumcircle of the triangle $ABC$ at point $D$. It is known that $\angle C > 60^\circ$.
Prove that $DC + DL \le BC$. | [
"We prove the statement for $\\angle C \\ge 36^\\circ$. Let $\\angle ABC = \\angle BAC = 2x$. We have $DC + DL \\le BC \\Leftrightarrow \\frac{BC}{DC} \\ge 1 + \\frac{DL}{DC}$. Now, by the law of sines,\n$$\n\\frac{BC}{DC} = \\frac{\\sin \\angle BDC}{\\sin \\angle DBC} = \\frac{\\sin 2x}{\\sin x} = 2 \\cos x, \\\\\... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0c4c | Problem:
Fie $n \in \mathbb{N}^*, n \geq 2$. Pentru numerele reale $a_{1}, a_{2}, \ldots, a_{n}$, notăm $S_{0}=1$ şi
$$
S_{k}=\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}
$$
suma tuturor produselor de câte $k$ numere alese dintre $a_{1}, a_{2}, \ldots, a_{n}$, $k \in \{1,2, \ldots... | [
"Solution:\n\nAre loc identitatea\n$$\n\\prod_{k=1}^{n}\\left(a_{k}+i\\right)=\\left(S_{n}-S_{n-2}+S_{n-4}-\\cdots\\right)+i\\left(S_{n-1}-S_{n-3}+S_{n-5}-\\cdots\\right)\n$$\n\nRezultă\n$$\n\\begin{aligned}\n& \\left(S_{n}-S_{n-2}+S_{n-4}-\\cdots\\right)^{2}+\\left(S_{n-1}-S_{n-3}+S_{n-5}-\\cdots\\right)^{2} \\\\\... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2^{n-1} | |
0kvz | Problem:
Richard starts with the string $HHMMMMTT$. A move consists of replacing an instance of $HM$ with $MH$, replacing an instance of $MT$ with $TM$, or replacing an instance of $TH$ with $HT$. Compute the number of possible strings he can end up with after performing zero or more moves. | [
"Solution:\n\nThe key claim is that the positions of the $M$s fully determines the end configuration. Indeed, since all $H$s are initially left of all $T$s, the only successful swaps that can occur will involve $M$s. So, picking $\\binom{8}{4} = 70$ spots for $M$s and then filling in the remaining 4 spots with $H$s... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 70 | |
005v | Diremos que un número entero positivo es *lindo* si es divisible por cada uno de sus dígitos no nulos. Demostrar que no puede haber más de 13 números lindos consecutivos y hallar 13 números enteros consecutivos lindos. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof and answer | The maximum possible length is 13; an explicit sequence of 13 consecutive such integers exists. | |
08d3 | Problem:
Siano $a$ e $b$ due numeri reali distinti. Si sa che le due equazioni
$$
\begin{aligned}
& x^{2}+a x+3 b=0 \\
& x^{2}+b x+3 a=0
\end{aligned}
$$
hanno una soluzione in comune: quali sono i possibili valori per la somma $a+b$ ?
(A) 0 o -3
(B) 0 o 3
(C) Soltanto 0
(D) Soltanto -3
(E) Esistono infiniti valori p... | [
"Solution:\n\nLa risposta è (D). Ogni soluzione di entrambe le equazioni dev'essere soluzione anche della loro differenza, che è\n$$\n\\left(x^{2}+a x+3 b\\right)-\\left(x^{2}+b x+3 a\\right)=(a-b) x+3(b-a)=(a-b)(x-3)=0 .\n$$\nSiccome $a$ e $b$ sono distinti, l'unica possibilità è $x=3$, che è quindi l'unica possib... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | D | |
023m | Problem:
Escolhi quatro frações dentre $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{6}$, $\frac{1}{8}$, $\frac{1}{10}$ e $\frac{1}{12}$, cuja soma é $1$. Quais foram as frações que eu não escolhi? | [
"Solution:\n\n"
] | Brazil | Desafios | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 1/8 and 1/10 | |
0h8y | $N$ positive integer numbers are given, such that greatest common divisors of all nonempty sets of these numbers are pairwise distinct. Determine the smallest possible number of distinct prime divisors of the product of these $N$ numbers.
(OleksandrGolovanov) | [
"Firstly we provide an example of such $N$ numbers. Consider numbers $a_k = p_k p_{k+1} p_{k+2} \\dots p_{k+N-1}$, where $p_1, p_2, \\dots, p_N$ are $N$ distinct prime numbers and $p_{N+i} = p_i$ ($1 \\le i \\le N-1$). Indeed, GCD of any set will include $p_k$ exactly in a power 1 if and only if $a_k$ belongs to th... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | N | |
03sq | Find the smallest positive real number $k$ such that for any four given distinct real numbers $a$, $b$, $c$ and $d$, each greater than or equal to $k$, there exists a permutation $p$, q, r$ and $s$ of $a$, $b$, $c$ and $d$ such that the equation
$$
(x^2 + px + q)(x^2 + rx + s) = 0
$$
has four distinct real roots. | [
"Suppose $k < 4$. Take $a$, $b$, $c$, $d \\in [k, \\sqrt{4k}]$. Then for any permutation $p$, $q$, $r$, $s$ of $a$, $b$, $c$, $d$, consider the equation $x^2 + px + q = 0$, its discriminant\n$$\n\\Delta = p^2 - 4q < 4k - 4q \\le 4k - 4k = 0.\n$$\nTherefore it has no real roots. So $k \\ge 4$.\n\nSuppose $4 \\le a <... | China | China Western Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 4 | |
00em | A list of $n$ positive integers $a_1, a_2, a_3, \dots, a_n$ is called *good* if both of the following conditions are satisfied:
* $a_1 < a_2 < a_3 < \dots < a_n$,
* $a_1 + a_2^2 + a_3^3 + \dots + a_n^n \le 2023$.
For each $n \ge 1$, find how many good lists of $n$ numbers are there. | [
"Since $a_1 < a_2 < a_3 < \\dots < a_n$ and they are positive integers, we have that $a_i \\ge i$ for all $i = 1, 2, \\dots, n$. In particular, if $n \\ge 5$, then\n$$\n2023 < 5^5 \\le a_n^6 < a_1 + a_2^2 + a_3^3 + \\dots + a_n^5 \\le 2023,\n$$\nwhich is impossible. Therefore, there are no good lists with $n \\ge 5... | Argentina | Cono Sur Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | For n=1: 2023; n=2: 946; n=3: 220; n=4: 15; for n>=5: 0 | |
07n8 | Suppose $x$, $y$ and $z$ are positive numbers such that
$$
1 = 2xyz + xy + yz + zx. \tag{1}
$$
Prove that
$$
(i) \quad \frac{3}{4} \le xy + yz + zx < 1;
$$
$$
(ii) \quad xyz \le \frac{1}{8}.
$$
Using (i) or otherwise, deduce that
$$
x + y + z \ge \frac{3}{2}, \tag{2}
$$
and derive the case of equality in (2). | [
"**First approach.** Suppose that the LHS of (i) is false for some triple of positive numbers $a$, $b$, $c$ that satisfy (1), so that $1 = 2abc + ab + bc + ca$, but $ab + bc + ca < 3/4$. Then, by the AM-GM inequality,\n$$\n\\begin{aligned}\nabc &= \\sqrt{(ab)(bc)(ca)} = (\\sqrt[3]{(ab)(bc)(ca)})^{3/2} \\\\\n&\\leq ... | Ireland | Ireland | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | The minimum of x + y + z is 3/2, achieved when x = y = z = 1/2. Additionally, 3/4 ≤ xy + yz + zx < 1 and xyz ≤ 1/8. |
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