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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0iup | Problem:
A spider is making a web between $n > 1$ distinct leaves which are equally spaced around a circle. He chooses a leaf to start at, and to make the base layer he travels to each leaf one at a time, making a straight line of silk between each consecutive pair of leaves, such that no two of the lines of silk cross... | [
"Solution:\nThere are $n$ ways to choose a starting vertex, and at each vertex he has only two choices for where to go next: the nearest untouched leaf in the clockwise direction, and the nearest untouched leaf in the counterclockwise direction. For, if the spider visited a leaf which is not nearest in some directi... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | n · 2^(n−2) | |
0beq | Problem:
Adott az $A$-edrendű ($n>1$) nem invertálható négyzetes mátrix. Az $A$ elemei mind 1 moduluszú komplex számok.
a) Igazold, hogy ha $n=3$, akkor az $A$ mátrix két sorának vagy két oszlopának elemei egyenesen arányosak.
b) Igaz-e az előző alpontbeli tulajdonság $n=4$ esetén? | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
05i1 | Problem:
Trouver tous les entiers strictement positifs $a$ tels que l'entier
$$
1-8 \cdot 3^{a}+2^{a+2}\left(2^{a}-1\right)
$$
soit un carré parfait. | [
"Solution:\n\nLa quantité (4) est égale à $\\left(2^{a+1}-1\\right)^{2}-2^{3} \\cdot 3^{a}$. On remarque que cette quantité est un carré pour $a=3$ (elle vaut $9=3^{2}$) et $a=5$ (elle vaut $2025=45^{2}$) mais pas pour $a \\in\\{1,2,4,6,7\\}$. On suppose dorénavant $a \\geqslant 8$.\n\nSupposons que la quantité (4)... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES, ENVOI No. 3 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorizati... | null | proof and answer | 3 and 5 | |
03tg | Let $AB$ be a chord of circle $O$, $M$ the midpoint of arc $AB$, and $C$ a point outside of the circle $O$. From $C$ draw two tangents to the circle at points $S$, $T$. $MS \cap AB = E$, $MT \cap AB = F$. From $E$, $F$ draw a line perpendicular to $AB$, and intersecting $OS$, $OT$ at $X$, $Y$ respectively. Now draw a l... | [
"**Proof** Refer to the figure, join points $O$ and $M$. Then $OM$ is the perpendicular bisector of $AB$. So $\\triangle XES \\sim \\triangle OMS$, and thus $SX = XE$.\nNow draw a circle with center $X$ whose radius is $XE$. Then the circle $X$ is tangent to chord $AB$ and line $CS$. Draw the circumcircle of $\\tri... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0e88 | Problem:
Kvadratu $ABCD$ s stranico $a$ je včrtan enakokrak trikotnik $ABE$ tako, da je vrh $E$ razpolovišče stranice $CD$. Točka $F$ je nožišče višine na krak $AE$ trikotnika $ABE$ iz oglišča $B$. Dokaži, da je $|EF| : |FB| : |BE| = 3 : 4 : 5$. Nariši ustrezno skico. | [] | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0awv | Problem:
An equilateral triangle is divided into three congruent trapezoids, as shown in the figure. If the perimeter of each trapezoid is $10+5 \sqrt{3}$, what is the side length of the triangle?
 | [
"Solution:\n\nEach of the trapezoids will have the following interior angles.\n\n\n\nWe therefore have an isosceles trapezoid. By the manner in which the trapezoids are tiled, the shorter base and the two legs will all have the same length; let it be $a$. The longer base is then equal to $a... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 6+3\sqrt{3} | |
04vw | We say that a positive integer is *flat* if all its digits are the same (even single-digit integers are considered flat). Determine whether every positive integer that is not flat can be expressed as a sum of two or more mutually different flat numbers. (Jozef Rajník) | [] | Czech Republic | School Round | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Yes | |
07zm | Problem:
Dato un foglio rettangolare di lati $a$ e $b$, con $a > b$, determinare l'area del triangolo che risulta dalla sovrapposizione dei due lembi che si ottengono piegando il foglio lungo una diagonale (il triangolo colorato in grigio nella figura).
 | [
"Solution:\n\nIl triangolo $ABC$ è isoscele (gli angoli $\\widehat{BAC}$ e $\\widehat{ABC}$ corrispondono a due angoli alterni interni nel rettangolo diviso da una diagonale) e la sua area è $S = CB \\cdot AD / 2$ ($AD$ è l'altezza relativa a $CB$).\n\nPosto $DB = a$, $AD = b$, $CB = AC = x$ e $DC = a - x$, il teor... | Italy | XV Gara Nazionale di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | S = \frac{b\left(a^2 + b^2\right)}{4a} | |
00er | Let $x_1, x_2, \dots, x_n$ be positive real numbers. For each positive integer $k$, define
$$
S_k = x_1^k + x_2^k + \dots + x_n^k.
$$
(a) Prove that if $S_1 < S_2$ the sequence $S_1, S_2, S_3, \dots$ is strictly increasing.
(b) Prove that it is possible that $S_1 > S_2$ and yet the sequence $S_1, S_2, S_3, \dots$ is n... | [
"We will actually prove a stronger statement: if for some $k$ it is true that $S_k < S_{k+1}$, then $S_j < S_{j+1}$ for all $j \\geq k$ (i.e., the sequence is strictly increasing starting from that point).\n\nFor our proof, we will use the following key observation. For any positive real number $x$ and positive int... | Argentina | Cono Sur Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0kvf | Problem:
A right triangle and a circle are drawn such that the circle is tangent to the legs of the right triangle. The circle cuts the hypotenuse into three segments of lengths $1$, $24$, and $3$, and the segment of length $24$ is a chord of the circle. Compute the area of the triangle. | [
"Solution:\n\n\nLet the triangle be $\\triangle ABC$, with $AC$ as the hypotenuse, and let $D, E, F, G$ be on sides $AB, BC, AC, AC$, respectively, such that they all lie on the circle. We have $AG = 1$, $GF = 24$, and $FC = 3$.\nBy power of a point, we have\n$$\n\\begin{aligned}\n& AD = \\... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 192 | |
0a0k | How many three-digit numbers, with each digit unequal to zero, are there such that the three digits add up to $7$?
A) $4$ B) $7$ C) $10$ D) $15$ E) $21$ | [
"D) $15$"
] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | MCQ | D | |
000h | Sean $\Gamma$ la circunferencia circunscrita y $O$ el circuncentro de un triángulo $ABC$ con $AC \neq BC$. La recta tangente a $\Gamma$ trazada por $C$ corta a la recta $AB$ en $M$. La recta perpendicular a $OM$ trazada por $M$ corta a las rectas $BC$ y $AC$ en $P$ y $Q$, respectivamente. Demostrar que los segmentos $P... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellane... | español | proof only | null | |
0fhk | Problem:
Sea la sucesión (progresión aritmética)
$$
3, 7, 11, 15, \ldots
$$
Demostrar que en dicha sucesión hay infinitos números primos. | [
"Solution:\nDemostrar que en la sucesión\n$$\n3, 7, 11, 15, \\ldots, 4n+3, \\ldots\n$$\nhay infinitos primos es lo mismo que probar que hay infinitos primos de la forma $4n+3$. Para demostrarlo, utilizamos el hecho de que el producto de dos números de la forma $4n+1$ es a su vez de la misma forma pues $\\left(4n_1+... | Spain | OME 28 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0601 | Problem:
Soit $n \geqslant 2$ un entier et $C>0$ une constante réelle. On suppose qu'il existe une suite $x_{1}, x_{2}, \ldots, x_{n}$ de réels non tous nuls telle que :
$-x_{1}+x_{2}+\ldots+x_{n}=0$.
- Pour tout indice $1 \leqslant i \leqslant n$, on a soit $x_{i} \leqslant x_{i+1}$, soit $x_{i} \leqslant x_{i+1}+C... | [
"Solution:\n\nDans toute la solution, on considèrera tous les indices modulo $n$. Remarquons que la condition de l'énoncé signifie que pour tout $i$, si $x_{i+2} \\leqslant 0$, alors $x_{i} \\leqslant x_{i+1}$ car la condition $x_{i} \\leqslant x_{i+1}+C x_{i+2}$ implique aussi $x_{i} \\leqslant x_{i+1}$. De la mêm... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
01vs | The positive integers $a$, $b$ and $c$ satisfy the equality
$$
a^3 b^3 + b^3 c^3 + c^3 a^3 = abc(a^3 + b^3 + c^3).
$$
Prove that the product of some two of these numbers is the square of a positive integer. | [
"Note that\n$$\n\\begin{align*}\n(ab - c^2)(ca - b^2)(bc - a^2) &= (a^2bc - ab^3 - ac^3 + b^2c^2)(bc - a^2) = \\\\\n&= a^2b^2c^2 - ab^4c - abc^4 + b^3c^3 - a^4bc + a^3b^3 + a^3c^3 - a^2b^2c^2 = \\\\\n&= a^3b^3 + b^3c^3 + a^3c^3 - a^4bc - a^4bc - abc^4 = a^3b^3 + b^3c^3 + a^3c^3 - abc(a^3 + b^3 + c^3).\n\\end{align*... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0j0g | Problem:
Let $\triangle ABC$ be a scalene triangle. Let $h_{a}$ be the locus of points $P$ such that $|PB - PC| = |AB - AC|$. Let $h_{b}$ be the locus of points $P$ such that $|PC - PA| = |BC - BA|$. Let $h_{c}$ be the locus of points $P$ such that $|PA - PB| = |CA - CB|$. In how many points do all of $h_{a}$, $h_{b}$... | [
"Solution:\n\nAnswer: 2 The idea is similar to the proof that the angle bisectors concur or that the perpendicular bisectors concur. Assume WLOG that $BC > AB > CA$. Note that $h_{a}$ and $h_{b}$ are both hyperbolas. Therefore, $h_{a}$ and $h_{b}$ intersect in four points (each branch of $h_{a}$ intersects exactly ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2 | |
00qy | Let $ABC$ be a triangle and $C$ its circumcircle. The point $D$ lies on the arc $BC$ of $C$ and is different from $B$, $C$ and the midpoint of $BC$. The tangent line to $C$ at $D$ intersects the lines $BC$, $CA$, $AB$ at $A'$, $B'$, $C'$, respectively. The lines $BB'$ and $CC'$ intersect at $E$. The line $AA'$ intersec... | [
"The problem is equivalent to proving that $\\angle CDE = \\angle CDF$. Let us denote $\\alpha_1 = \\angle BAD$, $\\alpha_2 = \\angle DAC$, $\\beta = \\angle CBA$, $\\gamma = \\angle ACB$. Since $A'B'C'$ is the tangent line at $D$, we have $\\angle BDC' = \\angle BCD = \\alpha_1$ and $\\angle B'DC = \\angle DBC = \... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscell... | null | proof only | null | |
04vg | Consider a sequence $(a_n)_{n=1}^{\infty}$ of positive integers satisfying for each $n \ge 3$ the condition
$$
a_n = a_1a_2 + a_2a_3 + \dots + a_{n-2}a_{n-1} - 1.
$$
a) Prove that some prime number is a divisor of infinitely many terms of this sequence.
(Tomáš Bárta)
b) Prove that there are infinitely many such prime n... | [
"Since all terms $a_i$ are positive integers, we have\n$$\na_5 = a_1a_2 + a_2a_3 + a_3a_4 - 1 \\ge 1 + 1 + 1 - 1 = 2, \\text{ and therefore } a_5 \\ne 1.\n$$\nThe number $a_5$ is thus divisible by at least one prime number.\nFor every $n \\ge 4$, the following holds\n$$\n\\begin{align*}\na_n &= (a_1a_2 + a_2a_3 + \... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0det | Find all positive integers $k$ such that the product of the first $k$ primes increased by 1 is a power of an integer (with an exponent greater than 1). | [
"Denote the first $n$ primes as\n$$\np_1 = 2 < p_2 = 3 < \\dots < p_n\n$$\nSuppose that $p_1p_2\\cdots p_n + 1 = x^k$ for some integers $x, k \\ge 2$. We can assume WLOG that $k$ is prime since $x^{kt} = (x^t)^k$. Obviously, $x$ has no prime factors not exceeding $p_n$, so $x > p_n$ and consequently $k < n < p_n$ i... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | no such k | |
03qv | Let $O$ be an interior point of $\triangle ABC$ such that $\overrightarrow{OA} + 2 \overrightarrow{OB} + 3 \overrightarrow{OC} = 0$. Then the ratio of the area of $\triangle ABC$ to the area of $\triangle AOC$ is ( ).
(A) 2
(B) $\frac{3}{2}$
(C) 3
(D) $\frac{5}{3}$ | [
"In the diagram, let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$, respectively. Then we have\n$$\n\\overrightarrow{OA} + \\overrightarrow{OC} = 2 \\overrightarrow{OD}, \\qquad (1)\n$$\nand\n$$\n2(\\overrightarrow{OB} + \\overrightarrow{OC}) = 4 \\overrightarrow{OE}. \\qquad (2)\n$$\nBy equations (1) and... | China | China Mathematical Competition (Hainan) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles"
] | English | MCQ | C | |
05bb | Let $n$ be a positive integer and $d$ one of its positive divisors. Integers $1$ to $n$ are written in columns of length $d$ (the first column contains the numbers $1$ to $d$, starting from the top, the second column contains the numbers $d+1$ to $2d$ etc.). Then, one finds the greatest common divisor of each row and f... | [
"The first two numbers of the $i$-th row are $i$ and $i+d$. Denote the greatest common divisor of this row by $a$; since $a$ divides both $i$ and $i+d$, it must also divide their difference $d$. Thus all the greatest common divisors divide $d$, meaning their least common multiple is at most $d$. But as the final ro... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
0kqr | Problem:
Let $ABCD$ and $AEFG$ be unit squares such that the area of their intersection is $\frac{20}{21}$. Given that $\angle BAE < 45^{\circ}$, $\tan \angle BAE$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100a + b$. | [
"Solution:\n\nSuppose the two squares intersect at a point $X \\neq A$. If $\\mathcal{S}$ is the region formed by the intersection of the squares, note that line $AX$ splits $\\mathcal{S}$ into two congruent pieces of area $\\frac{10}{21}$. Each of these pieces is a right triangle with one leg of length $1$, so the... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 4940 | |
0ae4 | Определи ги сите природни броеви $n$ за кои бројот $z = \left(\frac{3+i}{2-i}\right)^n$ е реален. | [
"Комплексниот број $\\frac{3+i}{2-i}$ можеме да го запишеме во облик\n$$\n\\frac{3+i}{2-i} = \\frac{3+i2+i}{2-i2+i} = \\frac{6+5i+i^2}{4-i^2} = \\frac{5+5i}{5} = 1+i.\n$$\n\nЗначи, $z = (1+i)^n$. Ако $n$ е парен број, односно $n=2k$ за некој $k \\in \\mathbb{N}$, тогаш\n$$\nz = (1+i)^{2k} = [(1+i)^2]^k = (2i)^k.\n$... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Intermediate Algebra > Complex numbers"
] | Macedonian, English | proof and answer | n is divisible by 4 | |
0ei9 | Problem:
Na sliki je graf polinoma $p$, ki seka os $y$ v $-4$, se dotika abscisne osi v $1$ in seka os $x$ v $4$. Kolikšna je vrednost izraza $p'(1) - 2 \cdot p(0) - 4 \cdot p(4)$?
(A) 2
(B) 4
(C) 8
(D) 0
(E) -1
 | [
"Solution:\n\nIz slike preberemo, da je $p'(1) = 0$, $p(0) = -4$ in $p(4) = 0$, torej je\n$$\np'(1) - 2 \\cdot p(0) - 4 \\cdot p(4) = 0 - 2 \\cdot (-4) - 4 \\cdot 0 = 0 + 8 - 0 = 8.\n$$"
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Functions"
] | null | MCQ | C | |
01in | Let $n$ be a positive integer, $x_1, \dots, x_n$ and $y_1, \dots, y_n$ be positive real numbers satisfying $x_n = n$ and $y_i y_{i+1} \ge 1$ for $1 \le i < n$ and $y_n y_1 \ge 1$. Moreover, let $x_0 = 0$.
Determine the minimal possible value of the expression
$$
\sum_{i=1}^{n} \sqrt{(x_i - x_{i-1})^2 + y_i}.
$$ | [
"We claim that the minimal value of the desired expression is $n\\sqrt{2}$ achieved by $x_i = i$ and $y_i = 1$ for all $i$. In what follows below, $Y$ denotes the sum $Y = \\sum_{i=1}^{n} \\sqrt{y_i}$.\nThe solution consists of proving two separate inequalities.\n\nBy taking the global product of all the inequaliti... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | n*sqrt(2) | |
04x5 | Given is a regular pentagon $ABCDE$. Determine the least value of the expression
$$
\frac{PA + PB}{PC + PD + PE}
$$
where $P$ is an arbitrary point lying in the plane of the pentagon $ABCDE$. | [
"Without loss of generality assume that the given pentagon $ABCDE$ has the side equal to $1$. Then the length of its diagonal is equal to\n$$\n\\lambda = \\frac{1 + \\sqrt{5}}{2}.\n$$\nSet $a = PA$, $b = PB$, $c = PC$, $d = PD$, $e = PE$ (Fig. 2).\n\n\n\nFig. 2\n\nApplying the Ptolemy inequ... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(5) - 2 | |
0189 | A sequence $a_1, a_2, a_3, \ldots$ of positive integers is such that $a_{n+1}$ is the last digit of $a_n + a_{n-1}$ for all $n > 2$. Is it always true that for some $n_0$ the sequence $a_{n_0}, a_{n_0+1}, a_{n_0+2}, \ldots$ is periodic? | [
"Since for $n > 2$, we actually consider the sequence mod $10$, and $\\varphi(10) = 4$, we have that the recursive formula itself has a period of $4$. Furthermore, the subsequent terms of the sequence are uniquely determined by two consecutive terms. Therefore if there exist integers $n_0 > 2$ and $k > 0$ such that... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | Yes | |
0dvt | Problem:
Maček iz sosednje vasi hodi v Butale dražit vaške pse. Vsak večer, ko že vsi spijo, se prikrade v Butale, na ves glas zamijavka, nato pa jo ucvre nazaj domov. Ko maček zamijavka, zalajajo vsi psi, ki so od njega oddaljeni do $90~\mathrm{m}$. Ker so Butale majhna vas, sta vsaka 2 psa v vasi med seboj oddaljena... | [
"Solution:\n\nMaček se lahko postavi tako, da nanj zalajajo vsi psi hkrati.\n\nNaj bosta $A$ in $B$ psa, ki sta med seboj najbolj oddaljena. Področje na sliki je presek krogov s središčema v $A$ in $B$ in polmerom $|AB|$. Izven tega območja ni nobenega psa. Če se maček postavi v točko $M$, ki je razpolovišče daljic... | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | Yes | |
08pf | Problem:
Find the maximum number of natural numbers $x_{1}, x_{2}, \ldots, x_{m}$ satisfying the conditions:
a) No $x_{i}-x_{j}, 1 \leq i<j \leq m$ is divisible by 11 ; and
b) The sum $x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1}$ is divisible by 11 . | [
"Solution:\nAccording to a), the numbers $x_{i}, 1 \\leq i \\leq m$, are all different $(\\bmod 11)$\nHence, the number of natural numbers satisfying the conditions is at most 11.\nIf $x_{j} \\equiv 0(\\bmod 11)$ for some $j$, then\n$$\nx_{2} x_{3} \\ldots x_{m}+x_{1} x_{3} \\ldots x_{m}+\\cdots+x_{1} x_{2} \\ldots... | JBMO | Junior Balkan Mathematics Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 10 | |
0kim | Problem:
Suppose there exists a convex $n$-gon such that each of its angle measures, in degrees, is an odd prime number. Compute the difference between the largest and smallest possible values of $n$. | [
"Solution:\nWe can't have $n=3$ since the sum of the angles must be $180^{\\circ}$ but the sum of three odd numbers is odd. On the other hand, for $n=4$ we can take a quadrilateral with angle measures $83^{\\circ}, 83^{\\circ}, 97^{\\circ}, 97^{\\circ}$.\n\nThe largest possible value of $n$ is $360$. For larger $n$... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 356 | |
0fe5 | Problem:
Demostrar que la ecuación
$$
x^{2}+y^{2}-z^{2}-x-3 y-z-4=0
$$
posee infinitas soluciones en números enteros. | [
"Solution:\n$$\n\\begin{aligned}\nx^{2}+y^{2}-z^{2}-x-3 y-z-4=0 & \\Leftrightarrow \\left(x-\\frac{1}{2}\\right)^{2}+\\left(y-\\frac{3}{2}\\right)^{2}-\\left(z+\\frac{1}{2}\\right)^{2}-\\frac{25}{4}=0 \\\\\n& \\Leftrightarrow \\left(x-\\frac{1}{2}\\right)^{2}-\\left(z+\\frac{1}{2}\\right)^{2}=\\frac{25}{4}-\\left(y... | Spain | null | [
"Number Theory > Diophantine Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
00sp | Let $ABC$ be an acute triangle, and $AX$, $AY$ two isogonal lines. Also, suppose that $K$, $S$ are the feet of perpendiculars from $B$ to $AX$, $AY$, and $T$, $L$ are the feet of perpendiculars from $C$ to $AX$, $AY$ respectively. Prove that $KL$ and $ST$ intersect on $BC$. | [
"Denote $\\phi = \\widehat{XAB} = \\widehat{YAC}$, $\\alpha = \\widehat{CAX} = \\widehat{BAY}$. Then, because the quadrilaterals $ABSK$ and $ACTL$ are cyclic, we have\n$$\n\\widehat{BSK} + \\widehat{BAK} = 180^\\circ = \\widehat{BSK} + \\phi = \\widehat{LAC} + \\widehat{LTC} = \\widehat{LTC} + \\phi,\n$$\nso, due t... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogo... | English | proof only | null | |
07kd | Let $\triangle ABC$ be a triangle the lengths of whose sides $BC, CA, AB$, respectively, are denoted by $a, b, c$, respectively. Let the internal bisectors of the angles $\angle BAC, \angle ABC, \angle BCA$, respectively, meet the sides $BC, CA, AB$, respectively, at $D, E, F$, respectively. Denote the lengths of the l... | [
"There are various formulae for the lengths of the bisectors. The one appropriate for the solution is given by\n$$\nd = \\frac{2\\sqrt{bc}\\sqrt{s(s-a)}}{b+c},\n$$\nwhere $2s = a + b + c$. One way to derive this formula is as follows. Since\n$$\n\\Delta(ABD) + \\Delta(DAC) = \\Delta(ABC) = \\Delta = \\frac{1}{2}bc ... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0lbv | Let $a, b, c$ be positive real numbers. It is known that the system of equations
$$
\begin{cases} a^2x + b^2y + c^2z = 1 \\ xy + yz + zx = 1 \end{cases} \quad \text{(where $a, b, c$ are parameters)}
$$
has only one solution $(x, y, z)$. Prove that $a, b, c$ are the sides of a certain triangle. | [
"Putting $m = a^2$, $n = b^2$, $p = c^2$ then $m, n, p > 0$. We have the system\n$$\n\\begin{cases} mx + ny + pz = 1 \\\\ xy + yz + zx = 1 \\end{cases}.\n$$\nFrom this system, we get $mx = 1 - ny - pz$ and $m(xy + yz + zx) = m$, therefore\n$$\nny^2 + y(nz + pz - mz - 1) + m - z + pz^2 = 0 \\quad (*)\n$$\nEquation $... | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
08ya | Suppose $D$, $E$ are points on the sides $AB$, $AC$, respectively, of a triangle $ABC$, and the following are known: $AB = 6$, $AC = 9$, $AD = 4$, $AE = 6$. Suppose, furthermore, the circum-circle of the triangle $ADE$ intersects the side $BC$ at two points $F$, $G$ and the points $B$, $F$, $G$, $C$ are lined up in thi... | [
"$$\n\\frac{-3 + \\sqrt{33}}{6}\n$$\nLet $H$ be the point of intersection of the lines $DF$ and $EG$, and let $I'$, $I$ be the point of intersection of the lines $AH$ and $DE$, $BC$, respectively. Let $H'$ be the point of intersection, different from $A$, of the circum-circle of the triangle $ADE$ and the line $AH$... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | (-3 + sqrt(33))/6 | |
00gf | Let $a$, $b$ and $c$ be positive real numbers such that $a b c = 8$. Prove that
$$
\frac{a^{2}}{\sqrt{\left(1+a^{3}\right)\left(1+b^{3}\right)}} + \frac{b^{2}}{\sqrt{\left(1+b^{3}\right)\left(1+c^{3}\right)}} + \frac{c^{2}}{\sqrt{\left(1+c^{3}\right)\left(1+a^{3}\right)}} \geq \frac{4}{3}
$$ | [
"Observe that\n$$\n\\frac{1}{\\sqrt{1+x^{3}}} \\geq \\frac{2}{2+x^{2}} \\tag{1}\n$$\nIn fact, this is equivalent to $\\left(2+x^{2}\\right)^{2} \\geq 4\\left(1+x^{3}\\right)$, or $x^{2}(x-2)^{2} \\geq 0$. Notice that equality holds in (1) if and only if $x=2$.\n\nWe substitute $x$ by $a$, $b$, $c$ in (1), respectiv... | Asia Pacific Mathematics Olympiad (APMO) | XVII APMO - March, 2005 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
07zz | Problem:
Alberto e Barbara fanno il seguente gioco. Su di un tavolo ci sono 1999 cerini: a turno ogni giocatore deve togliere dal tavolo un numero di cerini a sua scelta, purché maggiore o uguale ad uno, e minore o uguale alla metà del numero dei cerini che in quel momento sono sul tavolo. Il giocatore che lascia sul ... | [
"Solution:\n\nLa strategia vincente si può determinare procedendo a ritroso. Chi lascia un solo cerino sul tavolo perde. Chi ne lascia due vince, perché costringe l'altro giocatore a lasciarne uno solo. Chi ne lascia 3 o 4 perde, perché alla mossa successiva l'altro potrà lasciarne due. Chi ne lascia 5 invece vince... | Italy | XV Gara Nazionale di Matematica | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Barbara has a winning strategy. The winning heap sizes are 2, 5, 11, 23, 47, 95, 191, 383, 767, 1535. Starting from 1999, Barbara should first leave 1535 matches and thereafter always reduce the pile to the next smaller winning size (767, 383, 191, 95, 47, 23, 11, 5, 2), after which Alberto is forced to leave one and l... | |
0gsq | Rows and columns of a $29 \times 29$ grid are numerated from bottom to top and from left to right, respectively by numbers $1, 2, \ldots, 29$. Some unit squares of the grid are marked. For each marked unit square there is at most one other marked unit square whose row number is not less than the row number and column n... | [
"Answer: $43$.\nLet us give an example for $43$ marked unit squares. A unit square lying in the intersection row number $a$ and column number $b$ will be denoted by $(a, b)$. If $(29, 1)$ and for each $k = 1, 2, \\ldots, 14$ unit squares $(2k-1, 31-2k)$, $(2k, 31-2k)$, $(2k, 30-2k)$ are marked then there are $1+3 \... | Turkey | Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 43 | |
0317 | Problem:
Three nonintersecting circles $k_{i}(O_{i}, r_{i})$, $i=1,2,3$, where $r_{1}<r_{2}<r_{3}$, are tangent to the arms of an angle. One of the arms is tangent to $k_{1}$ and $k_{3}$ at points $A$ and $B$ and the other one is tangent to $k_{2}$ at point $C$. Let $K=AC \cap k_{1}$, $L=AC \cap k_{2}$, $M=BC \cap k_{... | [
"Solution:\n\nIf $E$ and $F$ are the second tangent points of $k_{1}$ and $k_{2}$ with the arms of the angle, then the equalities $AF^{2} = AL \\cdot AC$, $CE^{2} = CK \\cdot CA$ and $AF = CE$ imply that $AL = CK$. Hence $AK = CL$ and analogously $CM = BN$.\n\n\n\nOn the other hand, Ceva's ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | null | proof only | null | |
089l | Problem:
Matteo deve fare un test a crocette con 11 domande. Ciascuna domanda ha una sola risposta giusta. La prima domanda ha 2 possibili risposte (A e B), la seconda domanda ha 3 possibili risposte ($\{A\}, \{B\}, \{C\}$), e così via, fino all'undicesima domanda che ha 12 possibili risposte. Qual è la probabilità ch... | [
"Solution:\n\nLa risposta è (D). Calcoliamo infatti la probabilità $p$ che Matteo sbagli tutte le risposte (la probabilità richiesta è allora $1-p$): per la prima domanda c'è una risposta sbagliata su due risposte totali, quindi Matteo ha probabilità $1/2$ di sbagliare. Per la seconda domanda le opzioni non corrett... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | MCQ | D | |
0jwj | Problem:
Compute the number of possible words $w = w_{1} w_{2} \ldots w_{100}$ satisfying:
- $w$ has exactly $50$ $A$'s and $50$ $B$'s (and no other letters).
- For $i = 1, 2, \ldots, 100$, the number of $A$'s among $w_{1}, w_{2}, \ldots, w_{i}$ is at most the number of $B$'s among $w_{1}, w_{2}, \ldots, w_{i}$.
- For ... | [
"Solution:\nCall the last property in the problem statement $P(i, j)$ where in the statement $i = 44$, $j = 57$. We show that the number of words satisfying the first two conditions and $P(m, m + k)$ is the same independent of $m$ (assuming $k$ is fixed). It suffices to show that the number of words satisfying $P(m... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof and answer | (85 choose 35) - (85 choose 34) | |
0fcm | Problem:
Prueba que si los números $\log_{a} x$, $\log_{b} x$ y $\log_{c} x$ con $(x \neq 1)$ están en progresión aritmética, entonces
$$
c^{2}=(a . c)^{\log_{c} b}
$$ | [
"Solution:\nPor ser tres números en progresión aritmética\n$2 \\log_{b} x = \\log_{a} x + \\log_{c} x$\n\n$2 \\frac{\\log x}{\\log b} = \\frac{\\log x}{\\log a} + \\frac{\\log x}{\\log c}$ y como $\\log x \\neq 0 \\rightarrow 2 = \\frac{\\log b}{\\log a} + \\frac{\\log b}{\\log c} = \\log_{a} b + \\log_{c} b$\n\nUn... | Spain | Fase Local | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
0iqo | Problem:
If $x$ and $y$ are real numbers such that $\frac{(x-4)^{2}}{4}+\frac{y^{2}}{9}=1$, find the largest possible value of $\frac{x^{2}}{4}+\frac{y^{2}}{9}$. | [
"Solution:\n\nThe first equation is an ellipse with major axis parallel to the $y$-axis. If the second expression is set equal to a certain value $c$, then it is also the equation of an ellipse with major axis parallel to the $y$-axis; further, it is similar to the first ellipse. So the largest value of $c$ occurs ... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 9 | |
0e3c | Problem:
Dano je naravno število $n$. V ravnini leži $2n+2$ točk, izmed katerih nobene tri ne ležijo na isti premici. Premica v ravnini je ločnica, če na njej ležita dve izmed danih točk, na vsakem bregu te premice pa je natanko $n$ točk. Določi največje število $m$, za katerega velja, da je v ravnini vedno vsaj $m$ l... | [
"Solution:\n\nPokazali bomo, da je najmanj $n+1$ premic, ki potekajo skozi dve izmed danih točk, ločnic. Najprej podajmo primer, v katerem je ločnic točno $n+1$. Naj bodo točke oglišča pravilnega $(2n+2)$-kotnika. Označimo jih z $A_{1}, A_{2}, \\ldots, A_{2n+2}$. Za vsak $1 \\leq i \\leq n+1$ je očitno premica $A_{... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | n+1 | |
0cl2 | Let $n \ge 2$ and consider a table with $n+1$ rows and $n$ columns, in which, on each of the first $n$ rows, Nicuşor writes, in some order, the numbers $1, 2, \dots, n$. Then, he chooses a permutation $a_1, a_2, \dots, a_n$ of the numbers $1, 2, \dots, n$ and completes the last row as follows: for each $j \in \{1, 2, \... | [
"For $n=2$, no matter how Nicuşor chooses $a_1$ and $a_2$, on the third row we will have two equal values. For $n=3$, suppose without loss of generality that $a_3=3$. Then, in the third column, we have only values equal to $3$. Since the set $(1, 2)$ admits only two permutations, among the first three rows there wi... | Romania | 75th NMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | all integers n ≥ 4 | |
0ere | Exactly two years ago the Benson family had 4 members, and their average age was $19$. The Bensons then adopted another child. If the average age of the family today is still $19$, what is the present age of the adopted child? | [
"2 years ago the sum of all the family's ages was $4 \\times 19 = 76$. That should have increased by $2 \\times 4 = 8$, but has actually become $5 \\times 19 = 95$, i.e. increased by $19$. So the new person is now $19 - 8 = 11$ years old."
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 11 | |
03dv | Determine all prime numbers $p$, for which there exist positive integers $x$ and $y$, such that
$$
\left| \begin{array}{l} p + 49 = 2x^2 \\ p^2 + 49 = 2y^2 \end{array} \right. .
$$ | [
"We will prove that $p = 23$ is the unique solution. Subtracting the second equality from the first one, we get $p(p-1) = 2(y-x)(y+x)$. From the first equality $p$ is odd, thus $p \\ne 2$ and $p \\mid (y-x)(y+x)$. Clearly $y > x$, so assuming $p \\mid (y-x)$, we get $p \\le y-x < y+x$. Therefore,\n$$\n2(y-x)(y+x) >... | Bulgaria | Bulgaria 2022 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 23 | |
0cyb | Let $AMNB$ be a quadrilateral inscribed in a semicircle of diameter $AB = x$. Denote $AM = a$, $MN = b$, $NB = c$. Prove that
$$
x^{3} - (a^{2} + b^{2} + c^{2})x - 2abc = 0.
$$ | [] | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
02oq | Emerald and Jade play the following game: Emerald writes a list with 2011 positive integers, but does not show it to Jade. Jade's goal is finding the product of the 2011 numbers in Emerald's list. In order to do so, she is allowed to ask Emerald the gcd or the lcm of any subset with at least two of the 2011 numbers (as... | [
"She can obtain the product of any two numbers $a$ and $b$ by asking $\\gcd(a, b)$ and $\\text{lcm}(a, b)$, since $\\text{lcm}(a, b) \\cdot \\gcd(a, b) = ab$. The identity\n$$\nabc = \\frac{\\text{lcm}(a, b) \\cdot \\text{lcm}(a, c) \\cdot \\text{lcm}(b, c) \\cdot \\gcd(a, b, c)}{\\text{lcm}(a, b, c)}\n$$\nessentia... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
0e92 | Find all integer solutions of the equation $m^4 + 2n^2 = 9mn$. | [
"If the pair $(m, n)$ is a solution of the equation, then so is the pair $(-m, -n)$. So, we may assume that $m$ is non-negative.\n\nRearrange the equation into $2m^2 - 9mn + m^4 = 0$ and treat it as a quadratic equation in $n$. Its discriminant is $81m^2 - 8m^4$. If the equation is to have integer solutions, the di... | Slovenia | National Math Olympiad 2013 - First Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-3, -9), (-2, -8), (-2, -1), (0, 0), (2, 1), (2, 8), (3, 9) | |
0cfw | Prove that there exist infinitely many positive integers $n$ so that $n + s(n)$ is a perfect square, where $s(x)$ stands for the sum of the decimal digits of $x$. | [] | Romania | 74th NMO Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Other"
] | English | proof only | null | |
0ha6 | In triangle $ABC$, perpendicular bisector of the side $AC$ intersects the angle bisector $AK$ in point $P$, $M$ is such point that $\angle MAC = \angle PCB$, $\angle MPA = \angle CPK$, and points $M$ and $K$ lie on different sides from the segment $AC$. Prove that the line $AK$ divides the segment $BM$ in two equal seg... | [
"Let $T$ be the point, symmetrical to $M$ with respect to $AK$ (Fig. 4). Obviously, to prove the statement, it suffices to prove that $BT \\parallel AK$. Notice that points $C, P, T$ lie on the same line. Also,\n$$\n\\angle TAB = \\angle TAK - \\angle BAK = \\angle MAK - \\angle KAC = \\angle CAM = \\angle TCB,\n$$... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06uv | Let $k$ be a positive integer. The organising committee of a tennis tournament is to schedule the matches for $2k$ players so that every two players play once, each day exactly one match is played, and each player arrives to the tournament site the day of his first match, and departs the day of his last match. For ever... | [
"Enumerate the days of the tournament $1,2, \\ldots, \\binom{2k}{2}$. Let $b_{1} \\leqslant b_{2} \\leqslant \\cdots \\leqslant b_{2k}$ be the days the players arrive to the tournament, arranged in nondecreasing order; similarly, let $e_{1} \\geqslant \\cdots \\geqslant e_{2k}$ be the days they depart arranged in n... | IMO | IMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | k(4k^2 + k - 1) / 2 | |
0fqo | Problem:
Sea $p \geq 3$ un número primo y consideramos el triángulo rectángulo de cateto mayor $p^{2}-1$ y cateto menor $2p$. Inscribimos en el triángulo un semicírculo cuyo diámetro se apoya en el cateto mayor del triángulo y que es tangente a la hipotenusa y al cateto menor del triángulo. Encuentra los valores de $p... | [
"Solution:\n\nEn el triángulo rectángulo $ABC$, consideramos $\\overline{AB}=p^{2}-1$, $\\overline{AC}=2p$.\n\n\n\nPor el Teorema de Pitágoras, tendremos que $\\overline{BC}^{2}=\\overline{AC}^{2}+\\overline{AB}^{2}$, así que\n$$\n\\overline{BC}^{2}=(2p)^{2}+\\left(p^{2}-1\\right)^{2}=4p^{2... | Spain | OME fase local | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | p = 3 | |
0jol | Problem:
Compute the smallest positive integer $n$ for which
$$
0 < \sqrt[4]{n} - \lfloor \sqrt[4]{n} \rfloor < \frac{1}{2015}
$$ | [
"Solution:\nAnswer: 4097\nLet $n = a^{4} + b$ where $a, b$ are integers and $0 < b < 4a^{3} + 6a^{2} + 4a + 1$. Then\n$$\n\\begin{aligned}\n\\sqrt[4]{n} - \\lfloor \\sqrt[4]{n} \\rfloor &< \\frac{1}{2015} \\\\\n\\sqrt[4]{a^{4} + b} - a &< \\frac{1}{2015} \\\\\n\\sqrt[4]{a^{4} + b} &< a + \\frac{1}{2015} \\\\\na^{4}... | United States | HMMT November 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4097 | |
0887 | Problem:
Nel bosco dell'albero viola ci sono tre tipi di animali in grado di parlare: volpi, serpenti e tartarughe. Le prime mentono solo i giorni di pioggia, i secondi mentono sempre, le terze dicono sempre la verità. Un giorno l'esploratore Berny parla con quattro animali. Le loro affermazioni, riportate nell'ordine... | [
"Solution:\n\nLa risposta è (B). Consideriamo le due possibilità: piove oppure non piove.\n\nSe piove le volpi e i serpenti mentono, mentre le tartarughe dicono il vero. Allora $A$, che dice la verità, è necessariamente una tartaruga. $B$ mente, quindi è una volpe o un serpente. Allo stesso modo mente $C$, che è qu... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | B | |
09g0 | Let $X$ be a set of $n$ different positive integers. Denote by $S(I)$ the sum of all elements in a subset $I$ of $X$ and by $X_m$ the set $\{S(I) : I \subset X \text{ and } |I| = m\}$ for a positive integer $m \leq n$. If $|X_m| = m(n-m)+1$ for an integer $m$ such that $n-1 > m > 1$ then show that the elements of $X$ f... | [
"Let $a_0 < a_1 < \\dots < a_{n-1}$ be elements of $X$. For each positive integers $i < n-m$ and $j \\leq m$, construct a subset $A_{ij}$ of $X$ as $A_{ij} = \\{a_i, \\dots, a_{i+m}\\} \\setminus \\{a_{i+j}\\}$. Then it is clear that $S(A_{ij}) \\in X_m$ and\n$$\nS(A_{i0}) > S(A_{i1}) > \\dots > S(A_{im-1}) > S(A_{... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05vh | Problem:
Soit $a$ un réel strictement positif et $n \geqslant 1$ un entier. Montrer que
$$
\frac{a^{n}}{1+a+\ldots+a^{2 n}}<\frac{1}{2 n}
$$ | [
"Solution:\nPuisque la difficulté réside dans le dénominateur du membre de droite, et pour plus de confort, on peut chercher à montrer la relation inverse, à savoir :\n$$\n\\frac{1+a+\\ldots+a^{2 n}}{a^{n}}>2 n\n$$\nL'idée derrière la solution qui suit est \"d'homogénéiser\" le numérateur du membre de gauche, c'est... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kuy | Problem:
A regular $n$-gon $P_{1} P_{2} \ldots P_{n}$ satisfies $\angle P_{1} P_{7} P_{8} = 178^{\circ}$. Compute $n$. | [
"Solution:\n\nLet $O$ be the center of the $n$-gon. Then\n$$\n\\angle P_{1} O P_{8} = 2\\left(180^{\\circ} - \\angle P_{1} P_{7} P_{8}\\right) = 4^{\\circ} = \\frac{360^{\\circ}}{90}\n$$\nwhich means the arc $\\widehat{P_{1} P_{8}}$ that spans 7 sides of the $n$-gon also spans $1 / 90$ of its circumcircle. Thus $n ... | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 630 | |
0blc | Let $n \in \mathbb{N}^*$. Find all real numbers $x_1, x_2, \dots, x_n$ such that $x_k \in [0, 2k)$, for every $k \in \{1, 2, \dots, n\}$, and
$$ \frac{1}{2-x_1} + \frac{2}{4-x_2} + \dots + \frac{n}{2n-x_n} = \frac{x_1}{1} + \frac{x_2}{2} + \dots + \frac{x_n}{n}. $$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x_k = k for all k = 1, 2, ..., n | |
0egf | Problem:
Poenostavi izraz:
$$
2\left(x y^{-1}-1\right)^{-p}\left(x^{2} y^{-2}-1\right)^{p}-\left(\frac{x+y}{y}\right)^{p}
$$
Za $x=-2$, $y=-\frac{1}{2}$ in $p=-3$ izračunaj vrednost izraza. | [
"Solution:\n\nUredimo prvi oklepaj $\\left(x y^{-1}-1\\right)^{-p}=\\left(\\frac{x}{y}-1\\right)^{-p}=\\left(\\frac{x-y}{y}\\right)^{-p}=\\left(\\frac{y}{x-y}\\right)^{p}$ in drugi oklepaj $\\left(x^{2} y^{-2}-1\\right)^{p}=\\left(\\frac{x^{2}-y^{2}}{y^{2}}\\right)^{p}$. Oklepaja pomnožimo med seboj, števec drugega... | Slovenia | 18. tekmovanje v znanju matematike za dijake srednjih tehniških i strokovnih šol, Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/125 | |
06az | We consider the functions $f: \mathbb{R} \to \mathbb{R}$ for which:
$f(x+y)f(x-y) \geq (f(x))^2 - (f(y))^2$, for all $x, y \in \mathbb{R}$.
We suppose that there exists $x_0, y_0 \in \mathbb{R}$ such that
$$
f(x_0 + y_0)f(x_0 - y_0) > (f(x_0))^2 - (f(y_0))^2.
$$
Prove that $f(x) \geq 0$, for every $x \in \mathbb{R}$ or... | [] | Greece | Selection Examination | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
08cl | Problem:
Per ogni intero positivo $n$ chiamiamo $f(n)$ il prodotto di tutti i numeri naturali dispari minori o uguali a $2n+1$ (per esempio $f(4)=1 \cdot 3 \cdot 5 \cdot 7 \cdot 9$). Poniamo poi $g(n)=\frac{n}{f(n)}$. Cosa si può dire della somma $S=g(1)+g(2)+\cdots+g(30)$?
(A) $S \leq 0.35$
(B) $0.35<S \leq 0.49$
(C... | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Sia $S(m)=g(1)+g(2)+\\cdots+g(m)$. Notiamo che\n$$\n\\begin{aligned}\ng(m)=\\frac{m}{1 \\cdot 3 \\cdots (2m+1)} &=\\frac{1}{2} \\frac{2m}{1 \\cdot 3 \\cdots (2m+1)}=\\frac{1}{2} \\frac{2m+1-1}{1 \\cdot 3 \\cdots (2m+1)} \\\\\n&=\\frac{1}{2}\\left(\\frac{1}{1 \\cdot 3 \\c... | Italy | GARA di FEBBRAIO | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | C | |
0egi | Problem:
Kolikšna je vrednost funkcije $f(x) = \log_{3}\left(\frac{\sin\left(\frac{x \pi}{2}\right)}{\log_{\frac{1}{2}} 2^{x}}\right)$ za $x = -3$?
(A) 0
(B) -3
(C) ne obstaja
(D) 1
(E) -1 | [
"Solution:\n\n$f(-3) = \\log_{3}\\left(\\frac{\\sin\\left(\\frac{-3 \\pi}{2}\\right)}{\\log_{\\frac{1}{2}} 2^{-3}}\\right) = \\log_{3}\\left(\\frac{1}{3}\\right) = -1$."
] | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | E | |
09n9 | Let $\mathbb{R}_{>0} = \{x \in \mathbb{R} \mid x > 0\}$ denote the set of positive real numbers.
Find all pairs $f, g: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ of functions satisfying
$$
f(g(x)) = f(x)g(x), \quad f(x) = x(1 + g(x))
$$
and such that the sequence $g(x), g(g(x)), g(g(g(x))), \dots$ takes finitely many differe... | [
"Answer: $f(x) = x + 1$ and $g(x) = 1/x$.\nThe pair above is a solution. In order to prove that there is no other solution, fix $x \\in \\mathbb{R}_{>0}$ and denote $g^0 = x$ and $g^n = g(g^{n-1})$ for $n \\ge 1$.\nWe have $g(x)(1 + g(g(x))) = f(g(x)) = f(x)g(x) = x(1 + g(x))g(x)$. Since $g(x) \\ne 0$, we have $1 +... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | f(x) = x + 1, g(x) = 1/x | |
04hv | Let $a$ and $b$ be positive real numbers such that $a^3 + b^3 = 2ab(a + b)$.
Determine $\frac{a^2}{b^2} + \frac{b^2}{a^2}$. (Ratko Višak) | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 7 | |
0ds0 | For each integer $n > 1$, find a set of $n$ integers $\{a_1, a_2, \dots, a_n\}$ such that the set of numbers $\{a_i + a_j \mid 1 \le i \le j \le n\}$ leave distinct remainders when divided by $n(n+1)/2$. If such a set of integers does not exist, give a proof. | [
"For $n = 2$, $n(n+1)/2 = 3$. Thus $a_1 = 1$, $a_2 = 2$ work.\n\nNow suppose that $n \\ge 3$ and that the set $\\{a_1, \\dots, a_n\\}$ has the desired properties. We may assume without loss of generality, that $1 \\le a_1 < a_2 < \\dots < a_n \\le \\frac{n(n+1)}{2}$. Also let $a_{n+1} = a_1 + \\frac{n(n+1)}{2}$.\n\... | Singapore | Singapur | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Such a set exists only when the set size is two (for example, {1, 2}); for any larger size, no such set exists. | |
0afr | Дадена е функцијата $f: \mathbb{R} \rightarrow \mathbb{R}$ таква што
$$
f(x+1)+f(x-1)=\sqrt{2}f(x).
$$
Докажи дека $f$ е периодична функција. | [
"Од даденото равенство што го исполнува функцијата $f$ имаме:\n$$\nf(x+2)+f(x)=\\sqrt{2}f(x+1)=\\sqrt{2}(\\sqrt{2}f(x)-f(x-1))=2f(x)-\\sqrt{2}f(x-1),\n$$\nодносно\n$$\nf(x+2)=f(x)-\\sqrt{2}f(x-1).\n$$\nПонатаму,\n$$\n\\begin{aligned}\nf(x+4) &= f(x+2) - \\sqrt{2}f(x+1) = f(x) - \\sqrt{2}(f(x-1)+f(x+1)) = f(x) - \\s... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Macedonian, English | proof only | null | |
09ca | $M = \{1, \dots, 2010\}$-г хөх, шар, улаан өнгө бүр орсон ба тоо бүр зөвхөн нэг өнгөөр будагдсан байв.
$$
S_1 = \{(x, y, z) \in M^3 \mid x, y, z \text{ нь адил өнгөтэй ба } x + y + z \equiv 0 \pmod{2010}\}
$$
$$
S_2 = \{(x, y, z) \in M^3 \mid x, y, z \text{ нь өөр өнгөтэй ба } x + y + z \equiv 0 \pmod{2010}\}
$$
гэвэл ... | [
"Бид ерөнхий тохиолдолд бодъё.\n$M = \\{1, \\dots, n\\}$ олонлогийн хөх, шар, улаан элементүүдийн олонлогуудыг харгалзана. $B, Y, R$ ба\n$$\n|B|=b,\\ |Y|=y,\\ |R|=r\\ (b+y+r=n) \\text{ гээ.}\n$$\nТэгвэл\n$$\nf(x) = \\sum_{i=1}^{n} x^{i}, \\quad f_{b}(x) = \\sum_{i \\in B} x^{i}, \\quad f_{y}(x) = \\sum_{i \\in Y} x... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Mongolian | proof only | null | |
0bi5 | Find the irrational numbers $x$ with the property that $x^2 + x$ and $x^3 + 2x^2$ are integer numbers. | [
"Denote $x^2 + x = a$ and $x^3 + 2x^2 = b$. Then $b - a x = x^2 = a - x$, hence $x(a - 1) = b - a$. Since $x$ is an irrational number and $a, b$ are integers, we deduce that $a = b = 1$, and, finally, $x = \\frac{-1 \\pm \\sqrt{5}}{2}$."
] | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-1 + sqrt(5))/2 and (-1 - sqrt(5))/2 | |
0hgy | Given a positive integer $k$. The product of some $k$ consecutive positive integers ends with the number $k$. What value can the number $k$ attain? | [
"**Answer:** $k \\in 1, 2, 4$\n\nSuppose that $k \\ge 5$. It is clear that among any $k$ consecutive numbers, there is one that is divisible by $5$, and one that is divisible by $2$, so their product ends in $0$, hence $k$ is divisible by $10$. It is clear that then $k \\ge 10$, so in the product of $k$ consecutive... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 1, 2, 4 | |
07we | A triomino is formed by joining three equal squares edge to edge. Up to rotation there are only two types, namely straight and corner triominoes:
The diagram shown below is known as the *Aztec Diamond* of order $7/2$.
Which number is larger, ... | [
"There are many ways to fit $8$ corner triominoes into the Aztec Diamond. Here is one possibility.\n\n\nTo show that at most $7$ straight triomino fit into the Aztec Diamond, we colour the squares as shown below with three colours in a diagonal fashion.\n\n\nThere a... | Ireland | IRL_ABooklet_2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | corner triominoes | |
0gz0 | Find all natural numbers $n$, such that $n^2 - 10n + 23$, $n^2 - 9n + 31$ and $n^2 - 12n + 46$ are primes. | [
"Let $n$ be a natural number that satisfies the condition of the problem. We first calculate the sum of all given numbers:\n$$\n3n^2 - 31n + 100 = 2n^2 - 30n + n(n-1) + 100\n$$\nwhich is even. This implies that at least one of them is even, in other words, it equals to $2$. Solving the following three equations\n$$... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | n = 3 or n = 7 | |
0cum | Let $O$ be the circumcenter of an acute-angled isosceles triangle $ABC$ with $AB = AC$. The rays $BO$ and $CO$ meet the sides $AC$ and $AB$ at $B'$ and $C'$, respectively. Let $l$ be the line through $C'$ parallel to $AC$. Prove that $l$ is tangent to the circumcircle of the triangle $B'OC$. | [
"11.2. See problem 10.2.",
"11.2. См. задачу 10.2."
] | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0kpi | Problem:
What is the smallest $r$ such that three disks of radius $r$ can completely cover up a unit disk? | [
"Solution:\n\nLook at the circumference of the unit disk. Each of the disks must be capable of covering up at least $\\frac{1}{3}$ of the circumference, which means it must be able to cover a chord of length $\\sqrt{3}$. Thus, $\\frac{\\sqrt{3}}{2}$ is a lower bound for $r$. This bound is achievable: place the thre... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | sqrt(3)/2 | |
0i7z | Problem:
What fraction of the area of a regular hexagon of side length $1$ is within distance $\frac{1}{2}$ of at least one of the vertices? | [
"Solution:\n$\\text{The hexagon has area } 6\\left(\\frac{\\sqrt{3}}{4}\\right)(1)^2 = \\frac{3\\sqrt{3}}{2}$. The region we want consists of six $120^\\circ$ arcs of circles of radius $\\frac{1}{2}$, which can be reassembled into two circles of radius $\\frac{1}{2}$. So its area is $\\frac{\\pi}{2}$, and the ratio... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | π√3/9 | |
0e0v | Problem:
Poišči vsa praštevila $p$, za katera je $p^{2}+7^{3}$ popoln kub. | [
"Solution:\n\nNaj bo $p^{2}+7^{3}=n^{3}$. Tedaj velja $p^{2}=n^{3}-7^{3}=(n-7)\\left(n^{2}+7 n+49\\right)$. Očitno je $n>7$ in $n-7<n^{2}+7 n+49$, zato je možno le, da je $n-7=1$ in $n^{2}+7 n+49=p^{2}$. Tedaj je $n=8$ in $p^{2}=169$, zato je $p=13$. Edino tako praštevilo je $p=13$."
] | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 13 | |
054y | Kati and Peeter play the following game. First, Kati writes a positive integer $a > 2016$ on the blackboard. Then Peeter starts to write more numbers on the blackboard, adding at each step the number $2016b + 1$ where $b$ is the biggest number on the blackboard. Peeter wins if at some point he writes a number divisible... | [
"The number $2016b + 1$ gives the same remainder upon division by $2017$ as $-b + 1$. Hence the remainders upon division by $2017$ are as in the sequence $b, -b+1, -(b+1)+1, \\dots$. Since $-(b+1)+1 = b$, this sequence has period $2$, whence there are at most $2$ different remainders. Hence the number $a$ gives the... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Other"
] | English | proof and answer | Yes; 2019 | |
09um | Farida makes a list of the integers between $1$ and $10,\!000$ that are divisible by $7$. For every number on the list she adds the digits of the number. What is the smallest number that occurs as an outcome?
A) $1$ B) $2$ C) $3$ D) $4$ E) $5$ | [
"B) $2$"
] | Netherlands | Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | MCQ | B | |
0fov | Problem:
Sean $M$ y $N$ puntos del lado $BC$ del triángulo $ABC$ tales que $BM = CN$, estando $M$ en el interior del segmento $BN$. Sean $P, Q$ puntos que están respectivamente en los segmentos $AN, AM$ tales que $\angle PMC = \angle MAB$ y $\angle QNB = \angle NAC$. ¿Es cierto que $\angle QBC = \angle PCB$? | [
"Solution:\n\nLa idea clave de la solución es considerar las circunferencias circunscritas de los triángulos $BNQ$ (en verde en la figura) y $PMC$ (en rojo). Si $AM$ corta a la circunferencia $(BNQ)$ en $X$, y $AN$ corta a la circunferencia $(PMC)$ en $Y$, es evidente que los cuadriláteros $BQN X$ y $MPCY$ son cíli... | Spain | LI Olimpiada matemática Española (Concurso Final) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Yes | |
0abh | Let $x + y + z = a$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{a}$, for $x, y, z, a \in \mathbb{R}$. Prove that at least one of $x, y, z$ is equal to $a$. | [
"From $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = \\frac{1}{a}$ we have that $x + y + z \\neq 0$ and from $x + y + z = a$ we have $\\frac{1}{x + y + z} = \\frac{1}{a}$.\n\nNow from $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = \\frac{1}{a}$ and $\\frac{1}{x + y + z} = \\frac{1}{a}$ we have $\\frac{1}{x} + \\frac{... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
08w3 | For a triplet $(a, b, c)$ of positive integers, an $a \times b \times c$ rectangular parallelepiped is constructed by assembling $abc$ unit cubes. Then, all the faces of the parallelepiped are colored. How many possible triplets $(a, b, c)$ are there if the number of colored unit cubes equals the number of uncolored on... | [
"If one of $a, b, c$ is less than or equal to $2$, then we see all of the cubes used will be colored and therefore, the condition of the problem will not be satisfied. So, we assume that each of $a, b, c$ is greater than or equal to $3$. We see that the uncolored cubes form an $(a-2) \\times (b-2) \\times (c-2)$ re... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | 20 | |
00zw | Problem:
a) Prove the existence of two infinite sets $A$ and $B$, not necessarily disjoint, of non-negative integers such that each non-negative integer $n$ is uniquely representable in the form $n=a+b$ with $a \in A, b \in B$.
b) Prove that for each such pair $(A, B)$, either $A$ or $B$ contains only multiples of som... | [
"Solution:\na) Let $A$ be the set of non-negative integers whose only non-zero decimal digits are in even positions counted from the right, and $B$ the set of non-negative integers whose only non-zero decimal digits are in odd positions counted from the right. It is obvious that $A$ and $B$ have the required proper... | Baltic Way | Baltic Way 1997 | [
"Number Theory > Other",
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0kom | Let $ABCD$ be a parallelogram with $\angle BAD < 90^\circ$. A circle tangent to sides $DA$, $AB$, and $BC$ intersects diagonal $AC$ at points $P$ and $Q$ with $AP < AQ$, as shown. Suppose that $AP = 3$, $PQ = 9$, and $QC = 16$. Then the area of $ABCD$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are posi... | [] | United States | 2022 AIME I | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 150 | |
05jm | Problem:
Soit $ABC$ un triangle dont tous les angles sont aigus, et dont l'angle $\widehat{B}$ est de 60 degrés ($=\pi/3$ radian). Les hauteurs $[AD]$ et $[CE]$ se coupent au point $H$. Prouver que le centre du cercle circonscrit au triangle $ABC$ est situé sur la bissectrice commune des angles $\widehat{AHE}$ et $\wi... | [
"Solution:\n\nPuisque $B, E, H, D$ sont sur le cercle de diamètre $[BH]$, on a $\\widehat{DHE} = 180^{\\circ} - \\widehat{EBD} = 120^{\\circ}$, donc $\\widehat{AHC} = 120^{\\circ}$. Comme $\\widehat{AOC} = 2\\widehat{ABC} = 120^{\\circ}$, on en déduit que $A, C, O, H$ sont cocycliques.\n\nPar conséquent, $\\widehat... | France | OFM | [
"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"
] | null | proof only | null | |
0cgi | For any real number $x$, let $A(x) = x^2 + 4\lfloor x \rfloor$.
a) Find the real numbers $x$ for which $A(x) = \{x\}^2$.
b) Find the real numbers $y > 0$ for which $A(y)$ is the square of a natural number. | [
"a) $(\\lfloor x \\rfloor + \\{x\\})^2 + 4\\lfloor x \\rfloor = \\{x\\}^2$ yields $\\lfloor x \\rfloor^2 + 2\\lfloor x \\rfloor\\{x\\} + 4\\lfloor x \\rfloor = 0$, (*).\nIf $x \\ge 0$, then $\\lfloor x \\rfloor = 0$, thus any $x \\in [0, 1)$ is a solution.\nIf $x < 0$, equality (*) leads to $\\{x\\} = -\\frac{\\lfl... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a) x ∈ [0, 1) ∪ {−4, −9/2}.
b) All y of the form y = sqrt(k^2 + 4) with integers k ≥ 2. | |
0hyt | Problem:
Let $\triangle ABC$ be inscribed in a circle $k$, and let $M$ be an arbitrary point on $k$ different from $A$, $B$, $C$. Prove that the feet of the three perpendiculars from $M$ to the sides of $\triangle ABC$ are collinear. (Note: You may have to extend some sides to find these feet: see Figure.) | [
"Solution:\n\nIt will suffice to show that $\\angle AQR = \\angle CQP$ (compare with Problem 3.) Note that both quadrilaterals $MQAR$ and $MQPC$ are cyclic: $\\angle MRA = 90^{\\circ} = \\angle MQA$, and $\\angle MQC = 90^{\\circ} = \\angle MPC$. Using inscribed angles in these quadrilaterals, we obtain that the tw... | United States | BAMO | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a6o | Problem:
Find the largest integer $k$ such that any string of $2025$ letters consisting only of $A$'s and $B$'s contains a palindromic substring of length $k$ or longer. A palindromic substring is a string of consecutive letters which reads the same backwards as forwards. | [
"Solution:\nWe claim that the largest integer is $4$. We first prove that all strings $S$ of $2025$ letters contain a palindromic substring of length $4$ or longer, which implies that $k \\geq 4$. Then we shall provide a construction to show that $k$ cannot be $5$ or more.\n\nWe first begin by proving $k \\geq 4$. ... | New Zealand | NZMO Round One | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
03ia | Problem:
Given: (i) $a, b > 0$; (ii) $a, A_{1}, A_{2}, b$ is an arithmetic progression; (iii) $a, G_{1}, G_{2}, b$ is a geometric progression. Show that
$$
A_{1} A_{2} \geq G_{1} G_{2}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0fyf | Problem:
Finde alle Paare $(u, v)$ natürlicher Zahlen, sodass
$$
\frac{u v^{3}}{u^{2}+v^{2}}
$$
eine Primpotenz ist. | [
"Solution:\nSei $p^{n}$ diese Primpotenz, Umformen liefert die äquivalente Gleichung $u v^{3}=p^{n}\\left(u^{2}+v^{2}\\right)$. Setze $d=\\operatorname{ggT}(u, v)$ und schreibe $u=d x, v=d y$, dann sind $x$ und $y$ teilerfremd. Die Gleichung wird zu\n$$\nd^{2} x y^{3}=p^{n}\\left(x^{2}+y^{2}\\right)\n$$\nNun ist $x... | Switzerland | Vorrundenprüfung | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (u, v) = (2^k, 2^k) for integers k ≥ 1 | |
0cv9 | Initially, we put $100$ cards onto a table, each card contains a positive integer. Exactly $43$ of these cards contain odd numbers. Then, on each minute the following operation has been performed. We compute the product of numbers on every set of three cards on the table, add up all these products, write this number on... | [
"Let the table contain $k$ odd numbers; then the parity of the next number coincides with that of $\\binom{k}{3}$. So, on the first minute $k$ increases by $1$, and then it remains stable. Let $D_n$ ($T_n$) be the sum of products of all pairs (triples) of the numbers on the table after the $n$th minute; $D_n \\equi... | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English; Russian | proof only | null | |
0g26 | Problem:
Für eine natürliche Zahl $n$ sei ein $n \times n$ Brett gegeben. Wir färben nun $k$ der Felder schwarz ein, sodass es für jeweils drei Spalten maximal eine Reihe gibt, in der alle Kreuzungsfelder mit den drei Spalten schwarz gefärbt sind. Zeige, dass gilt:
$$
\frac{2 k}{n} \leq \sqrt{8 n-7}+1
$$ | [
"Solution:\n\nWir zählen für jedes Paar von Reihen die Anzahl Spalten, sodass beide Reihen in dieser Spalte schwarz sind. Offensichtlich sind das höchstens 2 für jede der $\\binom{n}{2}$ Kombinationen, also insgesamt maximal $n(n-1)$. Wenn es in der $i$-ten Spalte also $a_{i}$ schwarze Quadrate sind, zählen wir jed... | Switzerland | SMO-Selektion | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0fmg | Show that in any set of three distinct integers there are two of them say $a$ and $b$ such that the number $a^5b^3 - a^3b^5$ is a multiple of 10. | [
"First we observe that the statement holds if the set includes $a = 0$ or $b = 0$. Let us denote by $N(a, b) = a^5b^3 - a^3b^5$. Since $N(-a, -b) = N(a, b)$ and $N(-a, b) = N(a, -b) = -N(a, b)$, then WLOG we may assume that the 3 distinct integers are all positive. Now, it is easy to check that $a^5b^3 - a^3b^5$ is... | Spain | International Mathematical Arhimede Contest | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof only | null | |
0k5h | Let $S = \{1, \dots, 100\}$, and for every positive integer $n$ define
$$
T_n = \{(a_1, \dots, a_n) \in S^n \mid a_1 + \dots + a_n \equiv 0 \pmod{100}\}.
$$
Determine which $n$ have the following property: if we color any 75 elements of $S$ red, then at least half of the $n$-tuples in $T_n$ have an even number of coord... | [
"We claim this holds exactly for $n$ even.\n**First solution by generating functions** Define\n$$\nR(x) = \\sum_{s \\text{ red}} x^s, \\quad B(x) = \\sum_{s \\text{ blue}} x^s.\n$$\n(Here “blue” means “not-red”, as always.) Then, the number of tuples in $T_n$ with exactly $k$ red coordinates is exactly equal to\n$$... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Inva... | null | proof and answer | all even positive integers n | |
0544 | Let $d$ be a positive number. On the parabola, whose equation has the coefficient $1$ at the quadratic term, points $A$, $B$ and $C$ are chosen in such a way that the difference of the $x$-coordinates of points $A$ and $B$ is $d$ and the difference of the $x$-coordinates of points $B$ and $C$ is also $d$. Find the area... | [
"Without loss of generality assume that equation of the parabola is $y = x^2$ (Fig. 2). Let the abscissas of the points $A$, $B$, and $C$ be $a$, $b$, and $c$. Let $A'$, $B'$, and $C'$ be the projections of $A$, $B$ and $C$ onto the $x$-axis. Denoting the area of a region $K$ by $S_K$ we have\n$$S_{ABC} = S_{ACC'A'... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof and answer | d^3 | |
09eh | Let $ABCD$ be quadrilateral inscribed in the circle $\omega$. Simedian of the angle $B$ of the triangle $\triangle ABD$ intersects $\omega$ at point $P$ and simedian of the angle $B$ of the triangle $\triangle CBD$ intersects $\omega$ at point $Q$. Prove that if $CP \cap AB = X$, $AQ \cap BC = Y$ then points $X$, $D$, ... | [
"Let $A', D', C', P', Q', X', Y'$ be images of points $A, D, C, P, Q, X, Y$ transformed by inversion $I_{\\omega(B,1)}$ respectively.\n\n---\n\n\n\n\n\nThen points $A', P', D', Q', C'$ are colinear and $A'P' = P'D'$, $D'Q' = Q'C'$. This implies from $\\triangle BA'D... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07dy | In isosceles trapezoid $ABCD$ where $BC = AD$ and $AB \parallel CD$, point $P$ is the intersection of diagonals $AC$ and $BD$. Let $X$ be the second intersection point of $BC$ and circumcircle of triangle $APB$ and let $Y$ be a point on $AX$ such that $DY \parallel BC$ ($C$ and $Y$ are on different sides of line $AD$).... | [
"Since $YD \\parallel BC$, it's concluded that $\\angle YDP = \\angle CBP = \\angle XAP$. So $YAPD$ and $ABCD$ are cyclic quadrilaterals. Therefore\n$$\n\\angle DYP = \\angle DAP = \\angle DAC = \\angle DBC = \\angle PDY.\n$$\nThus $PY = PD$. On the other hand it's clear that $PD = PC$. From these two it's obtained... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j8b | Problem:
Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. A player wins if, after he rolls, the number on the running tally is a multiple of $7$. Play continues until either player wins, or else indefinitely. I... | [
"Solution:\n\nAnswer: $\\frac{5}{11}$\n\nFor $1 \\leq k \\leq 6$, let $x_{k}$ be the probability that the current player, say $A$, will win when the number on the tally at the beginning of his turn is $k$ modulo $7$. The probability that the total is $l$ modulo $7$ after his roll is $\\frac{1}{6}$ for each $l \\not... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 5/11 | |
0kq0 | Problem:
The function $f(x)$ is of the form $a x^{2}+b x+c$ for some integers $a, b$, and $c$. Given that
$$
\begin{aligned}
\{f(177883), f(348710), & f(796921), f(858522)\} \\
= & \{1324754875645,1782225466694,1984194627862,4388794883485\}
\end{aligned}
$$
compute $a$. | [
"Solution:\nWe first match the outputs to the inputs. To start, we observe that since $a \\geq 0$ (since the answer to the problem is nonnegative), we must either have $f(858522) \\approx 4.39 \\cdot 10^{12}$ or $f(177883) \\approx 4.39 \\cdot 10^{12}$. However, since $858522$ is relatively close to $796921$, the f... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 23 | |
0hhy | Petryk solved 33 problems at the exam. For the lesser part of them, including the first problem, he got $a$ points, while for the rest of them he got $b$ points. It is known that the natural numbers $a$ and $b$ satisfy the condition: $1 \le b < a \le 10$. After the exam, Petrik calculated the average score for all prob... | [
"Let Petryk solve $n$ problems for $a$ points and $(33-n)$ problems for $b$ points. Then the average result of Petrik is:\n$$\nS = \\frac{na + (33 - n)b}{33} = \\frac{n(a - b) + 33b}{33} = \\frac{n(a - b)}{33} + b.\n$$\nFor this number to be an integer, $n(a-b) \\equiv 0 \\pmod{33}$ is required. Since $1 \\le a-b \... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 11 |
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