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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09cw | $m^4 - m^3 + 1$ тоо нь бүхэл тооны квадрат болж байх охх бүхэл т тоог ол. | [
"$m^4 - m^3 + 1 = n^2$ гэе. Хэрэв $|m| > 2$ бол\n$$\n(m^2 - \\frac{m}{2} - 1)^2 < n^2 < \\left(m^2 - \\frac{m}{2}\\right)^2\n$$\nболохыг төвөггүй шалгаж болно. Иймд $|m| \\le 2$ байх ба\n$m \\in \\{-2, -1, 0, 1, 2\\}$ болж эдгээр утгуудад $m^4 - m^3 + 1 = n^2$-ийн утгуудыг шууд бодож шалгасаар\n$(m, n) = (0, \\pm 1... | Mongolia | ММО-48 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Mongolian | proof and answer | m ∈ {−2, 0, 1, 2} | |
0exl | Problem:
A group of 100 people is formed to patrol the local streets. Every evening 3 people are on duty. Prove that you cannot arrange for every pair to meet just once on duty. | [
"Solution:\n\nEvery time a person is on duty he is paired with two other people, so if the arrangement were possible the number of pairs involving a particular person would have to be even. But it is $99$."
] | Soviet Union | 5th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0b72 | a) Factorize $xy - x - y + 1$.
b) Prove that if integers $a$ and $b$ satisfy $|a+b| > |1+ab|$, then $ab = 0$. | [
"a) $xy - x - y + 1 = (x - 1)(y - 1)$.\n\nb) Both members of the inequality are positive, so we can square and get the equivalent form $a^2 + b^2 + 2ab > 1 + 2ab + a^2b^2$, that is $(a^2 - 1)(b^2 - 1) < 0$.\nThis shows that $a^2 - 1 < 0$ or $b^2 - 1 < 0$. Since $a$ and $b$ are integers, this yields $a = 0$ or $b = ... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) (x - 1)(y - 1)
b) ab = 0 | |
0f59 | Problem:
A group of children form two equal lines side-by-side. Each line contains an equal number of boys and girls. The number of mixed pairs (one boy in one line next to one girl in the other line) equals the number of unmixed pairs (two girls side-by-side or two boys side-by-side). Show that the total number of chi... | [] | Soviet Union | 17th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
06mo | How many pairs of positive integers $(m, n)$ are there such that $m^2n = 20^{22}$? | [
"Answer: 276\nSince $20^{22}$ is a square number, $m^2n$, and hence $n$, must also be a square number. Write $n = k^2$. Then the equation becomes $m^2k^2 = 20^{22}$, or $mk = 20^{11}$. Each positive factor of $20^{11}$ corresponds to a choice of $m$, which in turn corresponds to a solution $(m, n)$ to the original ... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English | proof and answer | 276 | |
0hy8 | Problem:
Given any two positive real numbers $x$ and $y$, then $x \diamond y$ is a positive real number defined in terms of $x$ and $y$ by some fixed rule. Suppose the operation $x \diamond y$ satisfies the equations $(x \cdot y) \diamond y = x(y \diamond y)$ and $(x \diamond 1) \diamond x = x \diamond 1$ for all $x, ... | [
"Solution:\n\nNote first that $x \\diamond 1 = (x \\cdot 1) \\diamond 1 = x \\cdot (1 \\diamond 1) = x \\cdot 1 = x$.\n\nAlso, $x \\diamond x = (x \\diamond 1) \\diamond x = x \\diamond 1 = x$.\n\nNow, we have $(x \\cdot y) \\diamond y = x \\cdot (y \\diamond y) = x \\cdot y$.\n\nSo $19 \\diamond 98 = \\left(\\frac... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 19 | |
03bv | A real nonzero number is assigned to every point in the space. It is known that for any tetrahedron $\tau$ the number written in the incenter equals the product of the four numbers written in the vertices of $\tau$. Prove that all numbers equal 1. | [
"Consider two arbitrary points $X$ and $Y$ and let $x$ and $y$ be the corresponding numbers. Choose points $I$ and $J$ on the line $XY$ such that $XY = YI = IJ$. Let $X'$ on the line $XY$ be such that $XI = JX'$. Consider the plane $\\lambda$ perpendicular to $IJ$ and passing through the midpoint of $IJ$. Let $ABCX... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
0h5t | Inside an equilateral triangle $ABC$ point $M$ is chosen. Let points $M_1$, $M_2$, $M_3$ be symmetric corresponding to sides $BC$, $AC$, $AB$ of the triangle. Prove that $\overrightarrow{MM_1} + \overrightarrow{MM_2} + \overrightarrow{MM_3}$ is equal to $\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC}$.... | [
"Let us draw through point $M$ lines that are parallel to sides $ABC$. Let them intersect $AB$, $BC$, $AC$ at $C_1$, $C_2$; $A_1$, $A_2$; $B_1$, $B_2$ (Fig. 30).\n\n\n\nHence, $C_1A_2 \\parallel AC$, $A_1B_2 \\parallel BA$, $B_1C_2 \\parallel BC$, so $C_1A_2$, $A_1B_2$ and $B_1C_2$ intersec... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0bia | Let $ABC$ be an isosceles triangle, $AB = AC$, and let $M$ and $N$ be points on the sides $BC$ and $CA$, respectively, such that the angles $BAM$ and $CNM$ are equal. The lines $AB$ and $MN$ meet at $P$. Show that the internal angle bisectors of the angles $BAM$ and $BPM$ meet at a point on the line $BC$.
Bogdan Enesc... | [
"Denote $I$ the intersection of the bisector of $\\angle BAM$ with $BC$ and denote $D$ the reflection of $A$ about $BC$. Then $\\angle BMD = \\angle BMA = \\angle CMN$, so $P, M, D$ are collinear. On the other hand, $DI$ is the bisector of $\\angle BDM$ – the reflection of $\\angle BAM$ – and $BI$ is the bisector o... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05s0 | Problem:
Soit $ABC$ un triangle isocèle en $A$ mais pas rectangle. Soit $D$ le point de $(BC)$ tel que $(AD)$ soit perpendiculaire à $(AB)$, et soit $E$ le projeté orthogonal de $D$ sur $(AC)$. Soit enfin $H$ le milieu de $[BC]$.
Montrer que $AHE$ est isocèle en $H$. | [
"Solution:\n\n\n\nCommençons par remarquer que les points $A$, $D$, $E$ et $H$ sont cocycliques sur le cercle de diamètre $[AD]$. On en déduit par chasse aux angles :\n$$\n\\widehat{HEA} = \\widehat{HDA} = 90^\\circ - \\widehat{DBA} = 90^\\circ - \\widehat{BCA} = \\widehat{HAE},\n$$\ndonc l... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0575 | Four boxes with masks and means of disinfection are transported to participants of a carnival. Weighing the boxes pairwise results in six quantities, four largest of which are $125$ kg, $120$ kg, $110$ kg and $101$ kg. Find all possibilities of what can be the weights of the four boxes. | [
"Let the masses of the boxes be $a$, $b$, $c$ and $d$ kilograms, whereby $a \\le b \\le c \\le d$. Clearly the largest result of pairwise weighing occurs in the case of two heaviest boxes, i.e., $c+d = 125$. The second largest result occurs in the case of the heaviest and the third heaviest box, i.e., $b+d = 120$, ... | Estonia | Final Round of National Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Either 38 kg, 48 kg, 53 kg, and 72 kg; or 33.5 kg, 52.5 kg, 57.5 kg, and 67.5 kg. | |
0ctx | Peter chose several consecutive positive integers. He wrote down each of the chosen numbers either in red or in blue (both colors are present). Is it possible that the sum of the l.c.m. of the red numbers and the l.c.m. of the blue numbers is a power of 2? (O. Dmitriev, R. Zhenodarov)
Петя выбрал несколько последовате... | [
"Let $2^k$ be the maximal power of $2$ dividing one of the chosen numbers; since the numbers are consecutive, $2^k$ divides exactly one of them. Then one of the two l.c.m.'s under consideration is divisible by $2^k$, while the other is not.\n\nSuppose the contrary. Consider the powers of $2$ dividing the chosen num... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | No | |
0c3j | Problem:
Arătaţi că ecuaţia $x^{2}+y^{2}+z^{2}=x+y+z+1$ nu are soluţii în mulţimea numerelor raţionale. | [
"Solution:\nEcuaţia se poate scrie echivalent sub forma $(2x-1)^{2}+(2y-1)^{2}+(2z-1)^{2}=7$.\n\nDacă această ecuaţie ar avea o soluţie raţională $(x, y, z)$, notând $2x-1=\\frac{a_{1}}{b_{1}}$, $2y-1=\\frac{a_{2}}{b_{2}}$, $2z-1=\\frac{a_{3}}{b_{3}}$, am obţine că numerele întregi $a_{1}, b_{1}, a_{2}, b_{2}, a_{3... | Romania | Olimpiada Naţională de Matematică Primul baraj pentru Olimpiada Balcanică de Matematică pentru Juniori | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
07k4 | Sahand and Gholam play on a $1403 \times 1403$ grid, initially with all cells white. For each row and each column, there is a button (total $2 \times 1403 = 2806$ buttons). Starting with Sahand, each player, in his turn, presses a button that has not yet been pressed. Then it's the other player's turn, until all button... | [
"In general, for an $n \\times n$ table, we claim the answer is $n$. First, we describe Gholam's strategy to achieve this score. Whenever Sahand presses a row button, Gholam in his next turn presses a column button. By the time Sahand presses a column button, Gholam presses a row button.\n\nSuppose that Sahand pick... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1403 | |
05af | Circle $\omega$ with center $O$ and circle $\alpha$ with center $A$ intersect at two distinct points $C$ and $D$, whereas $\angle OCA = 90^\circ$. A point $E$ is chosen on circle $\omega$ inside circle $\alpha$. Let $F$ be the reflection of point $E$ over the point $O$, and $G$ the intersection of line $CE$ with circle... | [
"The line $OC$ perpendicular to the radius $AC$ of the circle $\\alpha$ is tangent to this circle. Hence we get $\\angle GDC = 180^\\circ - \\angle OCG = 180^\\circ - \\angle OCE$. From the equality of inscribed angles, we get $\\angle CDF = \\angle CEF = \\angle CEO$. Since $OC = OE$, we finally get $\\angle CEO =... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0era | At the start of the Mighty Mathematicians Football Team's first game of the season, their coach noticed that the jersey numbers of the 22 players on the field (11 players per team) were all the numbers from $1$ to $22$. At half-time, the coach substituted her goal-keeper (who had the number $1$ on her jersey) for a res... | [
"The sum from $2$ to $22$ is $2 + 3 + \\cdots + 22 = \\frac{22(23)}{2} - 1 = 11 \\times 23 - 1 = 252$.\n\nTherefore the new number must be even or otherwise the sum can't be exactly divisible by $2$.\n\nThe smallest possible even number we can use is $24$. To see that this is indeed possible take\n$$\n24 + 22 + 21 ... | South Africa | South African Mathematics Olympiad Third Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) 24; b) 122 | |
0fzs | Problem:
Finde die grösste natürliche Zahl $n$, sodass für alle reellen Zahlen $a, b, c, d$ folgendes gilt:
$$
(n+2) \sqrt{a^{2}+b^{2}}+(n+1) \sqrt{a^{2}+c^{2}}+(n+1) \sqrt{a^{2}+d^{2}} \geq n(a+b+c+d)
$$ | [
"Solution:\n\nL'idée est d'appliquer Cauchy-Schwarz sur les trois racines de gauche :\n$$\n\\begin{aligned}\n& (n+2) \\sqrt{a^{2}+b^{2}}+(n+1) \\sqrt{a^{2}+c^{2}}+(n+1) \\sqrt{a^{2}+d^{2}} \\\\\n= & \\sqrt{(4 n+4)+n^{2}} \\sqrt{a^{2}+b^{2}}+\\sqrt{(2 n+1)+n^{2}} \\sqrt{a^{2}+c^{2}}+\\sqrt{(2 n+1)+n^{2}} \\sqrt{a^{2... | Switzerland | SMO - Finalrunde | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 24 | |
09l1 | Bill wants to write a set of pairwise distinct positive integers on the board, ensuring that each number is divisible by at most one other number on the board.
(1)
Prove that Bill can write 14 numbers on the board.
(2)
Prove that Bill cannot write 15 numbers on the board. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 14 | |
0aax | Prove that if $a+\frac{b}{a}-\frac{1}{b}$ is an integer, then it is a perfect square, where $a,b \in \mathbb{N}$. | [
"Let $a,b \\in \\mathbb{N}$ and $a+\\frac{b}{a}-\\frac{1}{b}=k \\in \\mathbb{Z}$. From $a^2b+b^2-a=kab$, we get that $b|a$. Let $a=bq$, $q>0$, $q \\in \\mathbb{Z}$. Then $b^3q^2+b^2-bq=kb^2q$, and after dividing with $b$ ($b>0$) we have $b^2q^2+b-q=kbq$, therefore $q|b$. Let $b=qt$, $t>0$, $t \\in \\mathbb{Z}$. The... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0ics | Problem:
Evaluate the sum
$$
\frac{1}{2\lfloor\sqrt{1}\rfloor+1}+\frac{1}{2\lfloor\sqrt{2}\rfloor+1}+\frac{1}{2\lfloor\sqrt{3}\rfloor+1}+\cdots+\frac{1}{2\lfloor\sqrt{100}\rfloor+1}.
$$ | [
"Solution:\nThe first three terms all equal $1/3$, then the next five all equal $1/5$; more generally, for each $a=1,2,\\ldots,9$, the terms $1/(2\\lfloor\\sqrt{a^{2}}\\rfloor+1)$ to $1/(2\\lfloor\\sqrt{a^{2}+2a}\\rfloor+1)$ all equal $1/(2a+1)$, and there are $2a+1$ such terms. Thus our terms can be arranged into ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 190/21 | |
05i5 | Problem:
Dans le plan on se donne 2011 points deux à deux distincts colorés soit en bleu, soit en rouge.
a. On suppose que pour tout point bleu le disque de centre ce point et de rayon 1 contienne exactement deux points rouges. Quel est le plus grand nombre possible de points bleus?
b. On suppose que pour tout point... | [
"Solution:\n\na. S'il existe un point bleu, il doit exister au moins deux points rouges donc il ne peut y avoir plus de 2009 points bleus.\n\nRéciproquement, si l'on considère 2009 disques de rayon 1 ayant un intérieur commun à tous qui soit non vide, il suffit de marquer deux points rouges dans cette partie commun... | France | Olympiades Françaises de Mathématiques - Épreuve en temps limité de Janvier | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a) 2009; b) 1966 | |
08z0 | Alex and Betty play a game with a row of consecutive $2022$ cells. At the start of the game, the name of Alex is written in the $1$st, $3$rd, ..., $2021$st cells from the left and the name of Betty is written in the $2$nd, $4$th, ..., $2022$nd cells from the left. Starting from Alex, two players do the following operat... | [
"At the start of the game, there are $2021$ pairs of two adjacent cells with different names and the number of such pairs is reduced by two per one operation. If there are three or more pairs of two adjacent cells with different names then a player can do the operation hence the number of such pairs is one at the e... | Japan | Japan 2022 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1011 | |
09ct | $$
\frac{1}{2a_1} + \frac{1}{2a_2} + \dots + \frac{1}{2a_n} = \frac{1}{3a_1} + \frac{1}{3a_2} + \dots + \frac{n}{3a_n} = 1
$$
байх сөрөг биш бүхэл $a_1, \dots, a_n$ тоонууд орших байдаг бүх натурал тоо $n$-ийг ол | [
"Such numbers $a_1, a_2, \\dots, a_n$ exist if and only if $n \\equiv 1 \\pmod 4$ or $n \\equiv 2 \\pmod 4$.\nLet $\\sum_{k=1}^{n} \\frac{k}{3^k} = 1$ with $a_1, a_2, \\dots, a_n$ nonnegative integers. Then $i \\cdot x_1 + 2 \\cdot x_2 + \\dots + n \\cdot x_n = 3^a$ with $x_1, \\dots, x_n$ powers of 3 and $a \\ge 0... | Mongolia | ОУМО-53 | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Mongolian | proof and answer | All n such that n ≡ 1 or 2 mod 4 | |
0k1y | Problem:
Points $A$, $B$, $C$, $D$ are chosen in the plane such that segments $AB$, $BC$, $CD$, $DA$ have lengths $2$, $7$, $5$, $12$, respectively. Let $m$ be the minimum possible value of the length of segment $AC$ and let $M$ be the maximum possible value of the length of segment $AC$. What is the ordered pair $(m,... | [
"Solution:\n\nBy the triangle inequality on triangle $ACD$, $AC + CD \\geq AD$, or $AC \\geq 7$. The minimum of $7$ can be achieved when $A$, $C$, $D$ lie on a line in that order.\n\nBy the triangle inequality on triangle $ABC$, $AB + BC \\geq AC$, or $AC \\leq 9$. The maximum of $9$ can be achieved when $A$, $B$, ... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | (7, 9) | |
063t | Problem:
Ein Rechteck $\mathcal{R}$ mit ungeraden ganzzahligen Seitenlängen ist in Rechtecke unterteilt, die alle ganzzahlige Seitenlängen haben. Man beweise, dass für mindestens eines dieser Rechtecke die Abstände zu jeder der vier Seiten von $\mathcal{R}$ alle gerade oder alle ungerade sind. | [
"Solution:\n\nWir unterteilen $\\mathcal{R}$ in Einheitsquadrate und färben einige dieser Einheitsquadrate in rot oder blau ein, entsprechend der folgenden Illustration.\n\n\n\nDa $\\mathcal{R}$ laut Voraussetzung ungerade Seitenlängen hat, sind alle vier Eckfelder von $\\mathcal{R}$ blau g... | Germany | 1. Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0bve | Let $A = \{x \mid x = n(n+1) \text{ and } n \in \mathbb{N}^*\},\ B = \{y \mid y = 2^{4k+3} \text{ and } k \in \mathbb{N}\}$.
a) Show that $A \cap B = \emptyset$.
b) Determine the positive integer $m$ such that the triple of the sum of the $m$ smallest elements of the set $A$ equals the sum of the three smallest eleme... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 12 | |
0b9r | Let $ABCD$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $AB$ and $CD$ are $F$ and $G$ respectively and $\ell$ is the line through $G$ parallel to $AB$. The feet of the perpendiculars from $E$ onto $\ell$ and $CD$ are $H$ and $K$, respectively. Prove that the lines... | [
"The points $E$, $K$, $H$, $G$ are on the circle of diameter $GE$, so the angles $EHK$ and $EGK$ are equal.\n\n\n\nAlso, from $\\angle DCA = \\angle DBA$ and $CE/CD = BE/BA$ follows\n$$\n\\frac{CE}{CG} = \\frac{2CE}{CD} = \\frac{2BE}{BA} = \\frac{BE}{BF},\n$$\nso the triangles $CGE$ and $BF... | Romania | 28th BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0e3u | Find all prime numbers $p$ and $q$, such that $2^2 + p^2 + q^2$ is also prime. | [
"If the pair $(p, q)$ satisfies the conditions of the problem, then so does the pair $(q, p)$. It is therefore sufficient to only consider the case where $p \\le q$. Obviously, $p = q = 2$ is not a solution. If $p$ and $q$ are both odd primes, then $2^2 + p^2 + q^2$ is an even integer greater than $2$, so it is not... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | (p, q) = (2, 3) or (3, 2) | |
038z | Let $k > 5$ be an integer. Replace given positive integer by the product of the sum of its digits in base $k$ and $(k-1)^2$. Repeat the same with the new number, etc. Prove that the obtained numbers are equal from some point onwards. | [
"Since the sum of digits of an integer divisible by $k-1$, is divisible by $k-1$, too, $(k-1)^3$ divides all the numbers obtained after the second step. $(k-1)^3$. On the other hand, if $a = \\overline{a_n a_{n-1} \\dots a_0(k)}$, $n \\ge 4$ or $a_3 \\ge 2$, $n = 3$, then\n\n$$\n\\begin{align*} a - (k-1)^2 (a_n + a... | Bulgaria | Spring Mathematical Tournament | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0dul | Problem:
Reši enačbo:
$$
\left(2\left(2^{\sqrt{x}+3}\right)^{\frac{1}{2 \sqrt{x}}}\right)^{\frac{2}{\sqrt{x}-1}}=4
$$ | [
"Solution:\n\nPotenciranje potence\n$$\n\\left(2 \\cdot 2^{\\frac{\\sqrt{x}+3}{2 \\sqrt{x}}}\\right)^{\\frac{2}{\\sqrt{x}-1}}=4\n$$\n\nZapis števila $4$ kot potenca\n$$\n4 = 2^2\n$$\n\nMnoženje potenc\n$$\n\\left(2^{\\frac{2 \\sqrt{x}+\\sqrt{x}+3}{2 \\sqrt{x}}}\\right)^{\\frac{2}{\\sqrt{x}-1}}=2^{2}\n$$\n\nPotencir... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 9 | |
06uu | An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the di... | [
"Answer: No, it is not possible.\n\nLet $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and... | IMO | IMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
04zo | In a rectangle $ABCD$ we have $|AB| = a$ and $|BC| = b$, where $a \ge b$. Let $E$ be a point in the interior of side $AB$ such that there is exactly one possibility to choose points $F, G, H$ on the sides $BC, CD, DA$, respectively, in such a way that $EFGH$ is a rectangle, too. Find the ratio of the areas of rectangle... | [
"The rectangles $ABCD$ and $EFGH$ have a common center $O$ (see Fig. 15).\n\nAs rectangles are cyclic quadrangles, the point $F$ lies on the circle with center $O$ and radius $|OE|$. This circle intersects the side $BC$ at two points symmetric with respect to the midpoint of the side. To have exactly one point comm... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 1/2 | |
02cz | Problem:
Rosa tem um papagaio que faz contas de um modo estranho. Cada vez que Rosa diz dois números ele faz a mesma conta. Por exemplo:
- se Rosa diz "4 e 2" o papagaio responde "12";
- se Rosa diz "5 e 3" o papagaio responde "12";
- se Rosa diz "3 e 5" o papagaio responde "14";
- se Rosa diz "9 e 7" o papagaio respo... | [
"Solution:\n\n19."
] | Brazil | Desafios | [
"Math Word Problems"
] | null | final answer only | 19 | |
03av | The positive integers $a_0, a_1, \dots, a_9$ and $b_1, b_2, \dots, b_9$ are such that $a_9 < b_9$, $a_k \neq b_k$ and $1 \le k \le 8$. A cash machine is loaded with $n \ge a_9$ leva. For any $1 \le i \le 9$ it is allowed to withdraw $a_i$ leva (if the machine has at least $a_i$ leva), and after that the bank puts in th... | [
"Set $d_s = |a_s - b_s|$, $0 \\le s \\le 9$, where $b_0 = 0$. Without loss of generality assume that there exists $k$, $0 \\le k \\le 8$ such that $a_s - b_s > 0$ for $0 \\le s \\le k$ and $a_s - b_s < 0$ for $k + 1 \\le s \\le 9$.\n\nIf $n$ is one of the desired values then there exist positive integers $x_0, x_1,... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All n ≥ a9 such that n is divisible by d = gcd(a0, |a1 − b1|, |a2 − b2|, …, |a9 − b9|). | |
0hxo | Problem:
Evaluate
$$
sin \left(1998^{\circ}+237^{\circ}\right) \sin \left(1998^{\circ}-1653^{\circ}\right)
$$ | [
"Solution:\n\n$\\sin \\left(1998^{\\circ}+237^{\\circ}\\right) \\sin \\left(1998^{\\circ}-1653^{\\circ}\\right) = \\sin \\left(2235^{\\circ}\\right) \\sin \\left(345^{\\circ}\\right) = \\sin \\left(75^{\\circ}\\right) \\sin \\left(-15^{\\circ}\\right) = -\\sin \\left(75^{\\circ}\\right) \\sin \\left(15^{\\circ}\\ri... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Trigonometric functions"
] | null | final answer only | -1/4 | |
065b | Determine all integers $n \ge 1$ for which there exists $n$ real numbers $x_1, x_2, \dots, x_n$ in the closed interval $[-4, 2]$ such that the following three conditions are fulfilled:
- the sum of these real numbers is at least $n$;
- the sum of their squares is at most $4n$;
- the sum of their fourth powers is at lea... | [
"Since the data of the problem concern $n$ real numbers $x_1, x_2, \\dots, x_n$ in the closed interval $[-4, 2]$, we consider the polynomial\n$$\nP(x) = (x+4)(x-2)(x-1)^2,\n$$\nwhich in $[-4, 2]$ satisfies the relation\n$$\nP(x) = (x+4)(x-2)(x-1)^2 \\le 0. \\quad (1)\n$$\nAdding by parts the inequalities coming fro... | Greece | Mediterranean Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | All n divisible by 10 | |
0l5w | Problem:
Points $A$ and $B$ lie on circle $\omega$ with center $O$. Let $X$ be a point inside $\omega$. Suppose that $XO = 2\sqrt{2}$, $XA = 1$, $XB = 3$, and $\angle AXB = 90^{\circ}$. Points $Y$ and $Z$ are on $\omega$ such that $Y \neq A$ and triangles $\triangle AXB$ and $\triangle YXZ$ are similar with the same o... | [
"Solution:\n\nConsider a rotation about $X$ by $90^{\\circ}$ followed by a homothety with ratio $\\frac{1}{3}$ that sends $B$ to $A$. This sends $\\omega$ to $\\omega'$ with radius $\\frac{1}{3}$ of the radius of $\\omega$ and center $O'$. Since $A$ is the image of $B$ under this rotation, we know $A$ lies on both ... | United States | HMMT February | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance ... | null | final answer only | 11/5 | |
08oh | Problem:
Each letter of the word OHRID corresponds to a different digit belonging to the set $\{1,2,3,4,5\}$. Decipher the equality $$(O+H+R+I+D)^2 : (O-H-R+I+D) = O^{H^{R^{I_{D}}}}.$$ | [
"Solution:\n\nSince $O$, $H$, $R$, $I$ and $D$ are distinct numbers from $\\{1,2,3,4,5\\}$, we have $O+H+R+I+D=15$ and $O-H-R+I+D=O+H+R+I+D-2(H+R)<15$. From this $O^{H^{R^{I^{D}}}}=\\frac{(O+H+R+I+D)^2}{O-H-R+I+D}=\\frac{225}{15-2(H+R)}$, hence $O^{H^{R^{I^{D}}}}>15$ and divides $225$, which is only possible for $O... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | O=5, H=2, R=1, and either I=3, D=4 or I=4, D=3 | |
00n8 | Let $n \ge 2$ be an integer.
We draw an $n \times n$ grid on a board and label each box with either the number $-1$ or the number $1$. Then we calculate the sum of each of the $n$ rows and the sum of each of the $n$ columns and determine the sum $S$ of these $2n$ sums.
a. Show that there does not exist a labelling of ... | [
"As each number of the grid appears exactly once in the sum of all columns of the grid and the same holds for the sum of all rows, we get that $S$ is twice the sum of all labels of the boxes of the $n \\times n$ grid. Therefore, $S = 0$ holds if and only if the sum of all labels of the boxes vanishes, or equivalent... | Austria | Austria2019 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0gvq | A point $M$ is taken on the midperpendicular of the side $AC$ of an acute triangle $ABC$. The points $M$ and $B$ belong to the same half-plane with respect to the line $AC$, $\angle BAC = \angle MCB$ and $\angle ABC + \angle MBC = 180^\circ$. Find $\angle BAC$ (in degrees). | [
"З умови задачі випливає, що вершина **B** не лежить на серединному перпендикулярі відрізка **AC**. Тому залишилося розглянути такі два випадки.\n\n**Перший випадок.** Припустимо, що $AB < BC$. Нехай точка **M** лежить усерединні трикутника **ABC**. Тоді $\\angle MCB < \\angle ACB < \\angle BAC$, що суперечить умов... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof and answer | 30 | |
073i | The in-circle $\Gamma$ of a triangle $ABC$ touches the side $BC$ at $D$. Let $D'$ be the point which is diametrically opposite to $D$ on the circle $\Gamma$. The tangent through $D'$ to $\Gamma$ meets $AD$ in $X$. The tangent to $\Gamma$ through $X$, other than $XD'$, touches $\Gamma$ at $N$. Prove that the circum-circ... | [
"Observe that $NX$ is the polar of $Y$, $EF$ is the polar of $A$ and $BC$ is the polar of $D$. Since $A$, $Y$, $D$ are collinear, it follows that $NX$, $EF$, $BC$ are concurrent. Let the point of concurrency be $D'$. Let $S$ be the point of intersection of $EF$ and $AD$. Since \\{$E, S, F, D'$\\} form a harmonic ra... | India | Indija TS 2008 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry >... | English | proof only | null | |
00w4 | Problem:
Prove that the modulus of an integer root of a polynomial with integer coefficients cannot exceed the maximum of the moduli of the coefficients. | [
"Solution:\nFor a non-zero polynomial $P(x) = a_{n} x^{n} + \\cdots + a_{1} x + a_{0}$ with integer coefficients, let $k$ be the smallest index such that $a_{k} \\neq 0$. Let $c$ be an integer root of $P(x)$. If $c = 0$, the statement is obvious. If $c \\neq 0$, then using $P(c) = 0$ we get $a_{k} = -c \\left(a_{k+... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0b2p | Problem:
Let $P(x)$ be a polynomial with integer coefficients such that $P(-4)=5$ and $P(5)=-4$. What is the maximum possible remainder when $P(0)$ is divided by 60? | [
"Solution:\n$$\n\\begin{gathered}\n0-(-4) \\mid P(0)-P(-4) \\text{ or } 4 \\mid P(0)-5, \\text{ so } P(0) \\equiv 5 \\pmod{4} \\\\\n5-0 \\mid P(5)-P(0) \\text{ or } 5 \\mid -4-P(0), \\text{ so } P(0) \\equiv -4 \\pmod{5}\n\\end{gathered}\n$$\nBy the Chinese Remainder Theorem, there is a solution $r$ that satisfies ... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 41 | |
0611 | Problem:
Soit $N$ un entier strictement positif. On suppose qu'il existe quatre sous-ensembles $A_{1}$, $A_{2}$, $A_{3}$ et $A_{4}$ de $\{1, \ldots, N\}$, chacun de cardinal $500$ et on suppose que, pour tous $x, y$ dans $\{1, \ldots, N\}$, il existe un indice $i$ tel que $x$ et $y$ sont dans $A_{i}$. Déterminer la pl... | [
"Solution:\n\nNous allons traiter ce problème avec le langage des graphes. La modélisation du problème pousse à considérer le graphe dont les sommets sont les éléments de $S$, qui sont reliés par une arête de couleur $i$ si les deux sommets appartiennent à l'ensemble $A_{i}$.\n\nDans la suite, on dira également qu'... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 833 | |
08r9 | Let $m$, $n$ be integers greater than $1$. A $m \times n$ grid is given. We want to write integers in each square so that
(i) at least one of the entries are nonzero, and
(ii) for each square $S$, $\sigma(S) = 0$, where $\sigma(S)$ denotes the sum of the entry in all the squares which are next to $S$ (i.e., all the squ... | [
"Denote by $X_{i,j}$ the entry that lies in the $i$$-$th row and the $j$$-$th column. We will prove that $(m, n)$ is a good pair if and only if $m+1$ and $n+1$ are not relatively prime.\n\nFirst, we prove that $(k, k)$ is a good pair for any integer $k \\ge 2$. In fact, given a $k \\times k$ grid, write integers as... | Japan | The 4th Japanese Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Functional equations"
] | English | proof and answer | (1) n = 3, 5, 7, 9. (2) 29 | |
077m | Let $r > 0$ be a real number. All the interior points of the disc $D(r)$ of radius $r$ are colored with one of two colors, red or blue.
1. If $r > \frac{\pi}{\sqrt{3}}$, show that we can find two points $A$ and $B$ in the interior of the disc such that the distance $AB = \pi$ and $A$ and $B$ have the same color.
2. Doe... | [
"We will show that the conclusion holds when $r > \\frac{\\pi}{2}$.\n\nWe begin with a circle $C(r)$ with center $C$ and radius $r > \\frac{\\pi}{2}$. Now, a regular polygon $P$ can be inscribed in the circle to have any odd number of sides $2k + 1$. Because the number of sides is odd, each vertex $A$ is opposite a... | India | EGMO TST Day 1 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
0bvd | The point $M$ is on the incircle of the square $ABCD$. Show that $MA \cdot MB \cdot MC \cdot MD \le 5$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
04uy | Given a positive integer $n$, denote by $\tau(n)$ the number of positive divisors of $n$, and by $\sigma(n)$ the sum of all positive divisors of $n$. Find all positive integers $n$ satisfying
$$
\sigma(n) = \tau(n) \cdot \lceil\sqrt{n}\rceil.
$$
(Here, $\lceil x \rceil$ denotes the smallest integer not less than $x$.)
... | [] | Czech Republic | Czech-Polish-Slovak Match | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | English | proof and answer | [1, 3, 5, 6] | |
09ul | Liselotte has a collection of $100$ candies, which are either sweet or bitter.
She wants to choose between the following possibilities.
I) She eats half of the sweet candies. The rest is kept in the bag.
II) She eats half of the bitter candies. The rest is kept in the bag.
The part of the remaining candies in case I ... | [
"$20$"
] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 20 | |
0687 | Vangelis has a box containing $2015$ white and $2015$ black balls. He follows the following procedure: He chooses randomly two balls from the box. If both are black, then paints one of them white and puts it in the box, while drops the other out of the box. If both are white, then he keeps one of them in the box and dr... | [
"We will consider what happens at every step of the procedure for the number of white and black balls. In the case of selection of two black balls, no one is going back to the box and so the number of the black balls **is reducing by 2**. In the second case of selection of two white balls, the number of the black b... | Greece | Selection Examination | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2 white balls and 1 black ball | |
0e2t | Find all functions $f: \mathbb{R} \to \mathbb{R}$, such that
$$ (x - 2)f(y) + f(y + 2f(x)) = f(x + yf(x)) $$
for all real $x$ and $y$. | [
"First, assume that $f(0) = 0$. Inserting $x = 0$ into the functional equation we get $f(y) = 0$ for all $y \\in \\mathbb{R}$. This function satisfies the equation.\n\nNow, let $f(0) \\neq 0$. Inserting $y = 0$ into the equation we get\n$$\n(x - 2)f(0) + f(2f(x)) = f(x)\n$$\nfor all $x \\in \\mathbb{R}$. Obviously,... | Slovenia | Selection Examinations for the IMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = 0 for all x, and f(x) = x - 1 | |
08xg | Circle $C_1$ is internally tangent to circle $C_2$ at point $A$. Let $O$ be the center of circle $C_2$. A tangent line to circle $C_1$ at point $P$ on $C_1$ goes through point $O$. Let $Q$ be the point of intersection of half-line $OP$ and circle $C_2$, and let $R$ be the point of intersection of the line tangent to ci... | [
"3\n\nSince lines $RA$, $RP$ are tangent to circle $C_1$ at $A$, $P$, respectively, we have $AR = PR$. Also, if we let $S$ be the point of intersection, different from $Q$, of line $OQ$ and circle $C_2$, then, by the theorem on the power of a point with respect to a circle, we have $AR^2 = SR \\cdot QR$. From $AR =... | Japan | Japan Mathematical Olympiad Initial Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof and answer | 3 | |
0aqk | Problem:
Let $p$ be a prime number. Let $a$, $b$, and $c$ be integers that are divisible by $p$ such that the equation $x^{3} + a x^{2} + b x + c = 0$ has at least two different integer roots. Prove that $c$ is divisible by $p^{3}$. | [
"Solution:\n\nLet $r$ and $s$ be two different integral roots of $x^{3} + a x^{2} + b x + c = 0$; that is, $r^{3} + a r^{2} + b r + c = 0$ and $s^{3} + a s^{2} + b s + c = 0$. Since $p$ divides $a$, $b$, and $c$, it follows that $p$ divides both $r^{3}$ and $s^{3}$. Being prime, $p$ divides $r$ and $s$.\n\nSubtract... | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0k0m | Problem:
Find all natural numbers $n$ such that when we multiply all divisors of $n$, we will obtain $10^{9}$. Prove that your number(s) $n$ works and that there are no other such numbers.
(Note: A natural number $n$ is a positive integer; i.e., $n$ is among the counting numbers 1, 2, 3, .... A divisor of $n$ is a nat... | [
"Solution:\n\nSolution 1: Since the prime factorizaton of $10^{9}=2^{9} \\cdot 5^{9}$, the prime divisors of our natural number $n$ are exactly 2 and 5; i.e., $n=2^{a} \\cdot 5^{b}$ for some integer exponents $a \\geq 1$ and $b \\geq 1$.\nWhen $a=b=2$, we get the number $n=2^{2} \\cdot 5^{2}=100$, which actually wo... | United States | 19th Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 100 | |
01o0 | Exactly one integer stands in each cell of an $m \times n$ table ($m \ge 4, n \ge 4$). The number in any cell is equal to the arithmetic mean of the numbers in some two neighboring cells (i.e. in the cells having the common side with the given cell).
Find the greatest possible number of all distinct integers in this ta... | [
"**Answer: $mn-6$.**\nLet $A$ be the maximal number in the table and $a$ be the minimal number. Consider any cell $k$ of the table where $A$ stands. Since $A$ is the arithmetic mean of the number in some two neighboring cells $k_1$ and $k_2$. But all numbers in the table are no greater than $A$, the numbers in $k_1... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | mn - 6 | |
0gjt | Let $P$ be a point inside $\triangle ABC$. Let $A_1, B_1, C_1$ be points in the interiors of the segments $PA, PB, PC$, respectively. Let $\overline{BC_1} \cap \overline{CB_1} = \{A_2\}$, $\overline{CA_1} \cap \overline{AC_1} = \{B_2\}$, and $\overline{AB_1} \cap \overline{BA_1} = \{C_2\}$. Let $U$ be the intersection ... | [] | Thailand | Selected Problems from Thailand Training Camp | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"
] | English | proof only | null | |
0a93 | Problem:
Find one solution in positive integers to the equation
$$
x^{2} - 2x - 2007y^{2} = 0
$$ | [
"Solution:\nThe equation can be written in the form\n$$\nx(x - 2) = 223 \\cdot (3y)^{2}\n$$\nHere the prime number $223$ must divide $x$ or $x - 2$. In fact, for $x = 225$ we get $x(x - 2) = 15^{2} \\cdot 223$, which is equivalent to $223 \\cdot (3y)^{2}$ for $y = 5$. Thus, $(x, y) = (225, 5)$ is one solution."
] | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 21 | [
"Number Theory > Diophantine Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | (225, 5) | |
08je | Problem:
The quadrilateral $ABCD$ with perpendicular diagonals is inscribed in the circle with center $O$, the points $M$ and $N$ are the middle points of the sides $[BC]$ and $[CD]$ respectively. Find the value of the ratio of areas of the figures $OMCN$ and $ABCD$. | [] | JBMO | The first selection test for JBMO 2003 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1/4 | |
09sy | Problem:
de Facebookgroep Olympiadetraining heeft minstens vijf leden. Er is een zeker getal $k$ met de eigenschap: voor elk $k$-tal leden geldt dat er minstens één lid van dat $k$-tal bevriend is met de andere $k-1$. (Vriendschap is wederzijds: als $A$ bevriend is met $B$, dan is $B$ ook bevriend met $A$.)
a) Stel $... | [
"Solution:\n\na) Ja, dat kan. Als iedereen bevriend is met iedereen, dan zijn we klaar. Stel dus dat er twee leden zijn, zeg $A$ en $B$, die geen vrienden met elkaar zijn. Als we een groep van vier bekijken met $A, B$ en twee andere leden, dan moet één van die andere twee bevriend zijn met de ander en met $A$ en $B... | Netherlands | Selectietoets | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0dka | A mail carrier delivers mail to the 19 houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible? | [
"Consider 19 consecutive squares and color them yellow (corresponding to houses that receive letters) and white (corresponding to houses that do not receive letters). Then, according to the assumption, there are no 2 consecutive yellow cells and no 3 consecutive white cells. We need to calculate the number of such ... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 351 | |
08il | Problem:
Let $a$, $b$, $c$ be lengths of triangle sides, $p = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ and $q = \frac{a}{c} + \frac{c}{b} + \frac{b}{a}$. Prove that $|p - q| < 1$.
Problem:
Fie $a$, $b$, $c$ lungimile laturilor unui triunghi, $p = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$, $q = \frac{a}{c} + \frac{c}{b}... | [
"Solution:\nOne has\n$$\n\\begin{aligned}\na b c|p-q| & = a b c \\left| \\frac{c-b}{a} + \\frac{a-c}{b} + \\frac{b-a}{c} \\right| \\\\\n& = \\left| b c^{2} - b^{2} c + a^{2} c - a c^{2} + a b^{2} - a^{2} b \\right| \\\\\n& = \\left| a b c - a c^{2} - a^{2} b + a^{2} c - b^{2} c + b c^{2} + a b^{2} - a b c \\right| ... | JBMO | 7th JBMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
02x8 | Problem:
Considere o trapézio $ABCD$ de bases $BC$ e $AD$ de modo que $AB = BC = CD = 5$ e $AD = 10$. Seja $E$ o ponto de interseção das diagonais $AC$ e $BD$. A reta perpendicular a $AC$ traçada por $E$ intersecta o prolongamento de $AB$ em $F$ e a base $AD$ em $H$.
a) Determine o comprimento... | [
"Solution:\na) Inicialmente, verificaremos que $ABCD$ é metade de um hexágono regular. Seja $M$ o ponto médio de $AD$. Como $BC$ e $AM$ são iguais e paralelos, $ABCM$ é um paralelogramo. Além disso, como $AM = AB = BC$, segue que $CM = AB = CD = DM$. Assim, $CDM$ é um triângulo equilátero. De modo semelhante, podem... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | AH = 20/3, AE = 10√3/3, area(AFCH) = 50√3/3 | |
0989 | Problem:
Mulțimea de numere $0,1,2,\ldots, 2022$ este împărțită în două grupe. Prima grupă conține numerele cu suma pară a cifrelor, iar grupa a doua - cu suma impară a cifrelor. Aflați diferența dintre suma numerelor din prima grupă și suma numerelor din grupa a doua. | [
"Solution:\n\n1) Pentru început, vom diviza mulțimea de numere $0,1,2, \\ldots, 999$ în submulțimi a câte 10 numere naturale incluse respectiv în intervalele:\n$$\n[0,9],[10,19],[20,29], \\ldots,[990,999] .\n$$\nÎn fiecare astfel de submulțime numerele au ultima cifră $0,1,2, . ., 9$ în ordine crescătoare. Prin urm... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof and answer | 2021 | |
0e9t | Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that:
* for any positive integers $a$ and $b$ which are not relatively prime we have $f(a)f(b) = f(ab)$,
* for any positive integers $a$ and $b$ there exists a non-degenerate triangle with the sides of lengths $f(a)$, $f(b)$ and $f(a + b - 1)$.
(Note: A triangle is... | [
"The second condition implies that for all $a, b \\in \\mathbb{N}$ the triangle inequalities hold:\n$$\n\\begin{aligned}\nf(a) + f(b) &> f(a+b-1), \\\\\nf(a) + f(a+b-1) &> f(b), \\\\\nf(b) + f(a+b-1) &> f(a).\n\\end{aligned}\n$$\nFirst, consider $a = b = 2$. It then follows from the first condition that $f(4) = f(2... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Algebraic Expressions > Functional Equations",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | Either f(n) = 1 for all positive integers n; or f(n) = n for all n ≥ 2 with f(1) ∈ {1, 2, 3}. | |
0a2i | How many pairs of positive integers $a$ and $b$ are there, with $a > b$, both smaller than $100$, and for which $a + b = (a - b)^3$?
A) 2 B) 3 C) 4 D) 5 E) 6 | [] | Netherlands | Junior Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | C | |
07e5 | $$
S_n = \{x^2 + n y^2 : x, y \in \mathbb{Z}\}.
$$
Find all positive integers $n$ such that there exists an element of $S_n$ which doesn't belong to any of the sets $S_1, S_2, \dots, S_{n-1}$. | [
"**Answer.** Any square-free number.\n\nSince $S_j$ doesn't exist for $j \\le 1$, the statement is held for $i = 1$. On the other hand if for an integer $n > 1$ there exists a prime number $p$, which $p^2 \\mid n$, $n$ is not a valid number; Because for all integer numbers $x, y$\n$$\nx^2 + n y^2 = x^2 + \\left(\\f... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arith... | English | proof and answer | All square-free positive integers. | |
006y | Sea $ABC$ un triángulo y $P$ un punto de la bisectriz del ángulo $\hat{A}$ que está en el interior del triángulo $ABC$. Se sabe que $PC = BC$ y $\hat{A}BP = 30^\circ$. Hallar la medida del ángulo $\hat{APC}$. | [] | Argentina | Argentina 2009 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | Spanish | proof and answer | 150° | |
04qo | Prove that if positive numbers $a, b, c$ satisfy the inequality $5abc > a^3 + b^3 + c^3$, then there is a triangle with sides $a, b, c$. | [
"Let positive numbers $a, b, c$ satisfy the inequality $5abc > a^3 + b^3 + c^3$. Let us show that there exists a triangle with the sides $a, b, c$. On the contrary, suppose that there is no such triangle. Then for $a, b, c$ at least one of the triangle inequalities is not valid. Let, e.g. $c \\ge a+b$, i.e. $c = a+... | Czech Republic | 6-th Czech-Slovak Match | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
09my | Find all strictly increasing arithmetic progressions $a < b < c$ of positive integers such that the sequence $a^2 + c$, $b^2 + b$, $c^2 + a$ is a geometric progression. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (a,b,c) = (2,3,4) and (1,4,7) | |
0942 | Problem:
Let $ABCD$ be a parallelogram with $\angle DAB < 90^{\circ}$. Let $E \neq B$ be the point on the line $BC$ such that $AE = AB$ and let $F \neq D$ be the point on the line $CD$ such that $AF = AD$. The circumcircle of the triangle $CEF$ intersects the line $AE$ again in $P$ and the line $AF$ again in $Q$. Let ... | [
"Solution:\n\nWe have $\\angle ECD = \\angle EBA = \\angle AEC$, so $AECD$ is an isosceles trapezoid. We now deduce $\\angle EPF = 180^{\\circ} - \\angle ECF = \\angle EAD$, so $PF$ is parallel to $AD$. Analogously, $QE$ is parallel to $AB$. Therefore the quadrilaterals $ABEQ$ and $APFD$ are isosceles trapezoids an... | Middle European Mathematical Olympiad (MEMO) | MEMO Individual Competition | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04bk | Depending on the real parameter $a$, solve the equation
$$
(a - 1) (1 + x + x^2)^2 = (a + 1) (1 + x^2 + x^4)
$$ | [
"First, expand both sides:\n\nLeft side:\n$(a - 1)(1 + x + x^2)^2 = (a - 1)(1 + 2x + 3x^2 + 2x^3 + x^4)$\n\nRight side:\n$(a + 1)(1 + x^2 + x^4)$\n\nBring all terms to one side:\n$$(a - 1)(1 + 2x + 3x^2 + 2x^3 + x^4) - (a + 1)(1 + x^2 + x^4) = 0$$\n\nExpand:\n$(a - 1)(1) + (a - 1)(2x) + (a - 1)(3x^2) + (a - 1)(2x^3... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | The equation is equivalent to (x^2 + b x + 1)(x^2 + c x + 1) = 0, where b and c are the roots of t^2 + (a − 1)t − a = 0. Explicitly, b, c = (−(a − 1) ± √(a^2 + 6a + 1)) / 2. Hence all real solutions x satisfy either x^2 + ((−(a − 1) + √(a^2 + 6a + 1)) / 2) x + 1 = 0 or x^2 + ((−(a − 1) − √(a^2 + 6a + 1)) / 2) x + 1 = 0... | |
04es | The upper right quarter of a chess board (with dimensions $8 \times 8$) is covered with paper. How many rooks (at most) can we place on the remaining part of the board so that no two of them attack each other? In how many ways can they be placed?
(Two rooks are mutually attacking each other if they are in the same row ... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 8 rooks; 576 ways | |
0lfa | Let $ABCD$ be a convex quadrilateral with $\angle B < \angle A < 90^\circ$. Let $I$ be the midpoint of $AB$ and $S$ the intersection of $AD$ and $BC$. Let $R$ be a variable point inside the triangle $SAB$ such that $\angle ASR = \angle BSR$. On the lines $AR, BR$, take the points $E, F$, respectively so that $BE, AF$ a... | [
"a) We will prove that $SM$ and $SN$ are isogonal in $\\angle ASB$, since $(SMN)$ touches $(SAB)$ and $J$ belongs to the line connecting $S$ and the center of $(SAB)$. Indeed, according to Steiner's theorem for pairs of isogonals, we need to show that\n$$\n\\frac{MA}{MB} \\cdot \\frac{NA}{NB} = \\frac{SA^2}{SB^2}.\... | Vietnam | Team selection tests | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transforma... | English | proof only | null | |
0kh0 | A regular hexagon of side length $1$ is inscribed in a circle. Each minor arc of the circle determined by a side of the hexagon is reflected over that side. What is the area of the region bounded by these $6$ reflected arcs?
(A) $\frac{5\sqrt{3}}{2} - \pi$ (B) $3\sqrt{3} - \pi$ (C) $4\sqrt{3} - \frac{3\pi}{2}$ (D) $\pi... | [
"Note that the reflected arcs do not overlap except at their endpoints. The area of the region can be found by subtracting from the area of the hexagon the difference between the areas of the circle and the hexagon. This is equivalent to twice the area of the hexagon minus the area of the circle. Therefore the requ... | United States | Fall 2021 AMC 10 B | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | MCQ | B | |
00kl | Let $U$ be the circumcenter of the acute-angled triangle $\triangle ABC$. Furthermore, let $M_A$, $M_B$ and $M_C$ be the circumcenters of the triangles $\triangle UBC$, $\triangle UAC$ and $\triangle UAB$ in this order. For which triangles $\triangle ABC$ is the triangle $\triangle M_A M_B M_C$ similar to the original ... | [
"Since $ABC$ is acute-angled, we first note that $U$ must lie in the interior of $ABC$. Since $AU \\perp M_B M_C$ and $M_C U \\perp AB$, we have $\\angle UAB = \\angle M_B M_C U$, and since analogous results hold all around the perimeter of the figure, we can write\n$$\n\\begin{aligned}\n\\phi &= \\angle M_B M_C U ... | Austria | Austria 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | Equilateral triangles only | |
0cb7 | Let $(G, \cdot)$ be a finite group of order $n \in \mathbb{N}^*$, with $n \ge 2$. We shall call the group $(G, \cdot)$ *arrangeable* if there is an ordering of its elements, such that
$$
G = \{a_1, a_2, \dots, a_k, \dots, a_n\} = \{a_1 \cdot a_2, a_2 \cdot a_3, \dots, a_k \cdot a_{k+1}, \dots, a_n \cdot a_1\}.
$$
a) D... | [
"a. We will show that the group $(\\mathbb{Z}_n, +)$ is arrangeable if and only if $n \\ge 2$ is an odd positive integer.\n\nIf $(G, \\cdot)$ is an abelian arrangeable group, then considering the arrangement $G = \\{a_1, a_2, \\dots, a_k, \\dots, a_n\\} = \\{a_1 \\cdot a_2, a_2 \\cdot a_3, \\dots, a_k \\cdot a_{k+1... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof and answer | a) The group of integers modulo n under addition is arrangeable if and only if n is odd (with n at least three due to n at least two). b) An example of an arrangeable group of even order is Z4 × Z2 (of order eight). | |
0i32 | Problem:
What is the sum of the coefficients of the expansion $(x+2y-1)^6$? | [
"Solution:\nThe sum of the coefficients of a polynomial is that polynomial evaluated at $1$, which for the question at hand is $(1+2 \\cdot 1-1)^6 = 2^6 = 64$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 64 | |
0a4u | Problem:
Let $ABCDEF$ be a convex hexagon containing a point $P$ in its interior such that $PABC$ and $PDEF$ are congruent rectangles with $PA = BC = PD = EF$ (and $AB = PC = DE = PF$). Let $\ell$ be the line through the midpoint of $AF$ and the circumcentre of $PCD$. Prove that $\ell$ passes through $P$. | [
"Solution:\n\nLet $M$ be the midpoint of $AF$ and let $O$ be the circumcentre of triangle $CPD$. Now construct $Q$ to be the point such that $CPDQ$ is a parallelogram, and let $R$ be the centre of this parallelogram (i.e. $R$ is the intersection of $PQ$ with $CD$, and also $R$ is the midpoint of $PQ$).\n\n![](attac... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07v0 | Let $A$, $B$, $C$ be three points on a circle of centre $O$. The perpendicular line from $O$ to $BC$ intersects line $AC$ at $P$, and the perpendicular line from $O$ to $AC$ intersects line $BC$ at $Q$. Let $L$ be the midpoint of $OC$ and $K$ the midpoint of $PQ$.
Prove that $KL$ is perpendicular on $AB$. | [
"The quadrilateral *CNOM* is cyclic because $\\angle CNO = 90^\\circ$ and $\\angle CMO = 90^\\circ$, and $OC$ is a diameter and $L$ the centre of its circumcircle.\n\nThe quadrilateral *MNPQ* is cyclic as well since $\\angle PNQ = \\angle PMQ = 90^\\circ$, and $PQ$ is a diameter and $K$ the... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle"
] | English | proof only | null | |
0aij | Find all infinite sequences $a_1, a_2, a_3, \dots$ of positive integers such that
a) $a_{nm} = a_n a_m$, for all positive integers $n, m$, and
b) there are infinitely many positive integers $n$ such that $\{1, 2, \dots, n\} = \{a_1, a_2, \dots, a_n\}$. | [
"Instead of sequence $a_n$, we'll use notation with the function $f(n)$ with same properties.\n\nThere exists only one such function: $f(n) = n$. We'll solve the problem with many separate facts.\n\n**Fact 1:** $f(1) = 1$\n\n**Proof.** According to a) it holds $f(1) = f(1)f(1) = f(1)^2$. Since $f(1)$ is positive in... | North Macedonia | European Mathematical Cup | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | a_n = n for all n | |
05mm | Problem:
Soit $ABCD$ un quadrilatère convexe dont les diagonales ne sont pas perpendiculaires et tel que les droites $(AB)$ et $(CD)$ ne sont pas parallèles. On note $O$ le point d'intersection de $[AC]$ et $[BD]$. Soit $H_1$ et $H_2$ les orthocentres respectifs des triangles $AOB$ et $COD$. On désigne par $M$ et $N$ ... | [
"Solution:\n\nSoit $A'$, $B'$ les pieds des hauteurs issues respectivement de $A$ et $B$ dans $AOB$. Soit $C'$, $D'$ les pieds des hauteurs issues respectivement de $C$ et $D$ dans $COD$.\n\nLes points $A'$ et $D'$ appartiennent donc au cercle $\\Gamma_1$ de diamètre $[AD]$, et les points $B'$ et $C'$ appartiennent... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0ev3 | Given a positive integer $n \ge 2$, define the set $T$ by
$$
T = \{ (i, j) : 1 \le i < j \le n \text{ and } i \ne j \}.
$$
For nonnegative real numbers $x_1, x_2, \dots, x_n$ satisfying $x_1 + x_2 + \dots + x_n = 1$, find the maximum (as a function of $n$) of
$$
\sum_{(i,j) \in T} x_i x_j.
$$ | [
"Let $M(n)$ be the maximum of $\\sum_{(i,j) \\in T} x_i x_j$. We will show that\n$$\nM(n) = \\frac{\\lfloor \\log_2 n \\rfloor}{2(\\lfloor \\log_2 n \\rfloor + 1)}.\n$$\nFor $k = \\lfloor \\log_2 n \\rfloor$, set $x_{20} = x_{21} = \\dots = x_{2k} = \\frac{1}{k+1}$, and $x_i = 0$ otherwise. Then we have $\\sum_{(i,... | South Korea | The 26th Korean Mathematical Olympiad Final Round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | floor(log_2 n) / (2 (floor(log_2 n) + 1)) | |
0i95 | Problem:
A teacher must divide 221 apples evenly among 403 students. What is the minimal number of pieces into which she must cut the apples? (A whole uncut apple counts as one piece.) | [
"Solution:\nConsider a bipartite graph, with 221 vertices representing the apples and 403 vertices representing the students; each student is connected to each apple that she gets a piece of. The number of pieces then equals the number of edges in the graph. Each student gets a total of $221 / 403 = 17 / 31$ apple,... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 611 | |
0h5g | The number $2013$ is written on the board. Two players are playing the following game. A move consists of replacing the number on the board with the difference of this number and one of its divisors. The player who writes $0$ loses. Who of the two players can guarantee the win?
(Oleksiy Piskun) | [
"It is easy to observe that odd numbers have only odd divisors. So, if the player moves from an odd number, he has to write an even number. This provides a strategy for the second player: subtract $1$ from the current number at any move. Then the first player will always deal with an odd number, so will write an ev... | Ukraine | Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | Second player | |
09wt | Problem:
In een scherphoekige driehoek $A B C$ is $D$ het voetpunt van de hoogtelijn vanuit $A$. Laat $D_{1}$ en $D_{2}$ de spiegelbeelden zijn van $D$ in respectievelijk $A B$ en $A C$. Het snijpunt van $B C$ en de lijn door $D_{1}$ evenwijdig aan $A B$, noemen we $E_{1}$. Het snijpunt van $B C$ en de lijn door $D_{2... | [
"Solution:\n\nHet midden van $D D_{1}$ noemen we $K$ en het midden van $D D_{2}$ noemen we $L$. Dan ligt $K$ op $A B$ en $L$ op $A C$. Wegens $\\angle A K D=90^{\\circ}=\\angle A L D$ is $A K D L$ een koordenvierhoek. Dus $\\angle D L K=\\angle D A K=\\angle D A B=90^{\\circ}-\\angle A B C$. Verder is $K L$ een mid... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h7n | There are four numbers on the board: $1$, $3$, $6$ and $10$. Each time we can erase any two numbers $a$, $b$ written on the board and write numbers $a+b$, $ab$ instead. Can we obtain such four numbers
a) $2015$, $2016$, $2017$, $2018$; after several moves?
b) $2016$, $2017$, $2019$, $2022$ | [
"**Answer:** a), b) that is not possible.\n\na) Let us look at the numbers modulo $3$. Obviously, the amount of numbers divisible by $3$ cannot decrease. Because if both $a$, $b$ are divisible by $3$, then both $a+b$, $ab$ are also divisible by $3$. If one of the numbers is divisible by $3$, then $ab$ is also divis... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) no; b) no | |
066d | (α) Write the expression $A = k^4 + 4$, where $k$ is a positive integer, as a product of two factors each of them being a sum of two squares of integers.
(β) Simplify the expression
$$
K = \frac{\left(2^4 + \frac{1}{4}\right)\left(4^4 + \frac{1}{4}\right)\left(6^4 + \frac{1}{4}\right) \cdots \left((2n)^4 + \frac{1}{4}... | [
"(α) We have\n$$\n\\begin{aligned}\nk^4 + 4 &= (k^2)^2 + 4k^2 + 2^2 - 4k^2 = (k^2 + 2)^2 - (2k)^2 \\\\\n&= (k^2 + 2 - 2k)(k^2 + 2 + 2k) = [(k-1)^2 + 1^2][(k+1)^2 + 1^2].\n\\end{aligned}\n$$\n\n(β) We multiply both terms of the fraction by $(2^4)^n$, to receive:\n$$\n\\begin{aligned}\nK &= \\frac{\\left(2^4 + \\frac... | Greece | Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | A = [(k-1)^2 + 1^2][(k+1)^2 + 1^2]; K = (2n)^2 + (2n+1)^2 | |
01pw | Let $I$ be the center of inscribed circle of the non-isosceles triangle $ABC$. The ray $AI$ meets circumscribed circle of the triangle $ABC$ at point $D$. The circle passing through $C$, $D$, and $I$ meets again the ray $BI$ at point $K$.
Prove that $BK = CK$. | [
"Let $O$ be the circumcenter of the triangle $\\triangle ABC$. Let $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$. We construct the line passing through $D$ and $O$. Let $L$ be the point of intersection of this line and the ray $BI$. Since $AI$ is a bisector of the angle $BAC$, we have $BD... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0cld | Find all the integers $n \ge 2$ with the property: the cells of a $n \times n$ board can be colored with several colors, so that each cell $C$ has exactly two neighbouring cells with the same color as $C$. Here neighbouring cells means cells with a common side. | [
"Indeed, if $n$ is even, then we can divide the board into squares $2 \\times 2$. Now we pick a different color for each such square and use it for the cells of that square. This coloring clearly satisfies the requirement.\n\nTo prove that $n$ must be even, start from any cell $S$ of the board and move from each ce... | Romania | 75th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All even integers n ≥ 2 | |
06em | For a positive integer $k$, let $f_1(k)$ be the square of the sum of the digits of $k$. (For example $f_1(123) = (1+2+3)^2 = 36$.) Let $f_{n+1}(k) = f_1(f_n(k))$. Determine the value of $f_{2007}(2^{2006})$. Justify your claim. | [
"Firstly, since $2^{2006} < 10^{2006}$, we have\n$$\nf(2^{2006}) \\le (9 \\times 2006)^2 < 10^{10}.\n$$\nSecondly, this implies\n$$\nf_2(2^{2006}) \\le (9 \\times 10)^2 = 8100.\n$$\nSimilarly, we find that\n$$\nf_3(2^{2006}) \\le (7 + 9 + 9 + 9)^2 = 1156\n$$\nand\n$$\nf_4(2^{2006}) \\le (9 \\times 3)^2 = 729.\n$$\n... | Hong Kong | CHKMO | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 169 | |
0fba | Problem:
Determina todos los enteros positivos $x$, tales que $2x+1$ sea un cuadrado perfecto, pero entre los números $2x+2, 2x+3, \cdots, 3x+2$, no haya ningún cuadrado perfecto. | [
"Solution:\n\nSea $n$ un número entero tal que $2x+1 = n^{2}$ y $n^{2} \\leq 3x+2 < (n+1)^{2}$. De la primera ecuación se obtiene $x = \\left(n^{2}-1\\right) / 2$ y sustituyendo este valor en la doble desigualdad, resulta\n$$\nn^{2} \\leq \\frac{3n^{2}+1}{2} < n^{2} + 2n + 1 \\Leftrightarrow 2n^{2} \\leq 3n^{2} + 1... | Spain | LIV Olimpiada matemática Española (Concurso Final) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4 | |
0an0 | Problem:
Find the integer $m$ so that
$$
10^{m} < \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \ldots \frac{99}{100} < 10^{m+1}
$$ | [
"Solution:\nLet $a = \\frac{1}{2} \\times \\frac{3}{4} \\times \\frac{5}{6} \\times \\cdots \\times \\frac{99}{100} = \\frac{3}{2} \\times \\frac{5}{4} \\times \\frac{7}{6} \\times \\cdots \\times \\frac{99}{98} \\times \\frac{1}{100}$. Hence, $a > \\frac{1}{100} = 10^{-2}$. Thus, $m \\geq -2$.\n\nNow, let $b = \\f... | Philippines | Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | -2 | |
07pz | An $8 \times 8$ square is subdivided into $64$ unit squares like a chessboard. Some of these $64$ unit squares are black, all others are white. Such a configuration is called *spotty* if it contains at least two black unit squares each of which shares an edge with a white unit square to its left or above it.
How many s... | [
"We will use coordinates $(i, j)$, $0 \\le i, j \\le 8$, to address the vertices of the $64$ small squares so that $(0,0)$ is the bottom left corner. There are $2^{64}$ ways to make each of the $64$ squares either black or white. Each such 'colouring' will be called a configuration.\n\nLet a spot be a black square ... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 2^64 - 421278 | |
0bks | Problem:
Az $ABCD$ paralelogramma átlói az $O$ pontban metszik egymást. A $DAC$ és $DBC$ szögek szögfelezői a $T$ pontban metszik egymást. Tudjuk, hogy $\overrightarrow{TD} + \overrightarrow{TC} = \overrightarrow{TO}$. Határozd meg az $ABT$ háromszög szögeinek mértékét! | [
"Solution:\n\nDin ipoteză rezultă că $DOCT$ este paralelogram.\n\nDin $AO \\parallel DT$ deducem $\\angle DTA \\equiv \\angle OAT \\equiv \\angle DAT$, deci $DA = DT$.\n\nAstfel $DA = DT = OC$; analog $BC = CT = OD$, de unde $BD = AC$. Astfel $ABCD$ este dreptunghi, $AOTD$ este romb, triunghiul $AOD$ este echilater... | Romania | Olimpiada Naţională de Matematică | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 60°, 60°, 60° | |
005l | Se distribuyen los números $1, 2, 3, \ldots, 2008^2$ en un tablero de $2008 \times 2008$, de modo que en cada casilla haya un número distinto. Para cada fila y cada columna del tablero se calcula la diferencia entre el mayor y el menor de sus elementos. Sea $S$ la suma de los $4016$ números obtenidos. Determine el mayo... | [] | Argentina | XXIII Olimpíada Iberoamericana de Matemática | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | Spanish | proof and answer | 16184704896 | |
0805 | Problem:
Quale delle seguenti affermazioni è vera nell'insieme dei numeri razionali?
(A) Per ogni $x$ c'è un $y$ tale che per ogni $z$ si ha $x+y+z=x$
(B) per ogni $x$ c'è un $y$ tale che per ogni $z$ si ha $x+y+z=z$
(C) per ogni $x$ c'è un $y$ tale che per ogni $z$ si ha $x y z=x$
(D) per ogni $x$ c'è un $y$ tale che ... | [] | Italy | Italy Febbraio Contest | [
"Discrete Mathematics > Logic",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
04eu | Exactly three internal angles of a convex polygon are obtuse. How many sides (at most) can this polygon have? (Math Questions 2009) | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 6 | |
0ex2 | Problem:
If $m$, $k$, $n$ are natural numbers and $n > 1$, prove that we cannot have $m(m + 1) = k^n$. | [
"Solution:\n$m$ and $m + 1$ have no common divisors, so each must separately be an $n$th power. But the difference between the two $n$th powers is greater than $1$ (for $n > 1$)."
] | Soviet Union | 4th ASU | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
041w | Let $n$, $k$ be integers greater than $1$ and satisfy $n < 2^k$. Prove that there are $2k$ integers not divisible by $n$, such that if we divide them into two groups, then there must exist a group in which the sum of some integers can be divided by $n$. | [
"At first, we consider the case that $n = 2^r$, $r \\ge 1$. Obviously, at this time $r < k$. We take three $2^{r-1}$'s and $2k-3$ $1$'s — each of them cannot be divided by $n$. If these $2k$ numbers are divided into two groups, then there must exist a group that contains two $2^{r-1}$'s, whose sum is $2^r$ — divisi... | China | China Mathematical Competition (Complementary Test) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
0k25 | Problem:
Let $n$ be an odd positive integer not divisible by $3$. Show that $n^{2}-1$ is divisible by $24$. | [
"Solution:\n\nWe will show it is divisible by $8$ and $3$. Since the least common multiple of $8$ and $3$ is $24$, this implies the result.\n\nWe factor $n^{2}-1 = (n-1)(n+1)$.\n\nTo show divisibility by $8$, note that $n-1$ and $n+1$ are two consecutive even integers. Among any two consecutive even integers, one o... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null |
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