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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09ec | Some natural numbers can be written as a sum of 2 or more consecutive natural numbers. For instance $24 = 7+8+9$, $51 = 25+26$ etc. Find all such numbers which do not exceed 2014. | [
"First we shall prove that a number which can be represented as sum of consecutive natural numbers can not be represented in the form $n = 2^k$.\n$$\nn = m + (m + 1) + (m + 2) + \\dots + (m + k) = \\frac{(k + 1)(2m + k)}{2}\n$$\nNote that the numbers $k+1$ and $2m+k$ are different by (mod 2). Hence one of these num... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof and answer | All positive integers at most 2014 that are not powers of two. | |
025n | Problem:
Paulinho estava estudando o Máximo Divisor Comum (MDC) na escola e decidiu praticar em casa. Ele chamou de $a, b$ e $c$ as idades de três pessoas que moram com ele. Em seguida, fez algumas operações com os fatores primos deles e obteve os máximos divisores comuns dos 3 pares de números. Alguns dias depois, el... | [
"Solution:\n\nAnalisando os máximos divisores comuns listados, podemos garantir que $a$ é múltiplo de $15$ e de $5$, $b$ é múltiplo de $15$ e de $20$ e $c$ é múltiplo de $5$ e de $20$. Usando os fatores primos destes números, temos que $a$ é múltiplo de $15$, $b$ é múltiplo de $60$ e $c$ é múltiplo de $20$. Deste m... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a = 15, b = 60, c = 20 | |
0kwb | Problem:
After the Guts round ends, the HMMT organizers will calculate $A$, the total number of points earned over all participating teams on questions 33, 34, and 35 of this round (that is, the other estimation questions). Estimate $A$.
Submit a positive integer $E$. You will receive $\max (0, 25 - 3 \cdot |E - A|)$... | [
"Solution:\n\nOnly 8 teams scored a positive number of combined points on questions 33, 34, and 35. A total of 3 points were scored on question 33, 6 points on question 34, and 4 points on question 35. Extended results can be found in our archive."
] | United States | HMMT February 2023 | [
"Math Word Problems"
] | null | final answer only | 13 | |
05z9 | Problem:
Soient $ABC$ un triangle de cercle circonscrit $\Gamma$, $D$ un point sur $(AB)$ et $E$ un point sur $(AC)$ tel que $(DE)$ et $(BC)$ sont parallèles. Le cercle circonscrit à $ABC$ rencontre le cercle circonscrit à $BDE$ une seconde fois en $K$ et le cercle circonscrit à $CDE$ une seconde fois en $L$. Soit $T$... | [
"Solution:\n\n\n\nOn reconnait ici une situation classique :\n- $KBDE$ cyclique,\n- $KBLC$ cyclique,\n- $CLDE$ cyclique.\n\nOn sait que dans cette situation, les droites $(KB)$, $(CL)$, $(ED)$ sont concourantes (il s'agit ici du fait que les axes radicaux de 3 cercles sont concourants). Or ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05y7 | Problem:
Trouver les triplets d'entiers $(x, y, n)$ tels que $n^{2}=17 x^{4}-32 x^{2} y^{2}+41 y^{4}$. | [
"Solution:\n\nOn va montrer par descente infinie que $(0,0,0)$ est la seule solution.\n\nComme un carré ne vaut que $0$ ou $1$ modulo $3$ (une façon de le voir est de faire une disjonction de cas sur les valeurs modulo $3$), il est pertinent de tenter une étude modulo $3$ pour essayer de voir où ça nous mène. Pour ... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | (0, 0, 0) | |
0kfr | Problem:
Farmer James wishes to cover a circle with circumference $10\pi$ with six different types of colored arcs. Each type of arc has radius $5$, has length either $\pi$ or $2\pi$, and is colored either red, green, or blue. He has an unlimited number of each of the six arc types. He wishes to completely cover his c... | [
"Solution:\n\nFix an orientation of the circle, and observe that the problem is equivalent to finding the number of ways to color ten equal arcs of the circle such that each arc is one of three different colors, and any two arcs which are separated by exactly one arc are of different colors. We can consider every o... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 93 | |
0eea | Problem:
V katerega izmed navedenih izrazov lahko preoblikujemo izraz $\left(x+y+\frac{1}{4}\right)^{2}-\left(x+y-\frac{1}{4}\right)^{2}$?
(A) $4 x y$
(B) $\frac{1}{16}$
(C) $\frac{1}{8}$
(D) 0
(E) $x+y$ | [
"Solution:\nDani izraz razstavimo po pravilu razlike kvadratov in dobimo\n$\\left(x+y+\\frac{1}{4}-\\left(x+y-\\frac{1}{4}\\right)\\right)\\left(x+y+\\frac{1}{4}+\\left(x+y-\\frac{1}{4}\\right)\\right)=\\frac{1}{2}(2 x+2 y)=x+y$. Dani izraz lahko preoblikujemo v izraz $x+y$."
] | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | E | |
0l6a | Problem:
Estimate the total number of pages that teams submitted to the Team Round this year. (All pages associated to at least one problem number count as submitted pages, even blank cover sheets for a problem.)
Submit a positive integer $E$. If the correct answer is $A$, you will receive $\max \left(0, \left[20 \l... | [
"Solution:\n\nIncluding individual teams, 106 teams registered this year, of which 101 teams submitted a nonzero number of pages to the Team Round. A surprisingly accurate estimate of 1000, which scores 19 points, can be obtained by simply assuming 100 teams competed and each team submitted an average of one page p... | United States | HMMT February | [
"Math Word Problems"
] | null | final answer only | 1000 | |
01ed | Points $A$, $B$, $C$, $D$ lie, in this order, on a circle $\omega$, where $AD$ is a diameter of $\omega$. Furthermore, $AB = BC = a$ and $CD = c$ for some relatively prime positive integers $a$ and $c$. Show that if the diameter $d$ of $\omega$ is also an integer, then $d$ is a perfect square or $2d$ is a perfect squar... | [
"By Pythagoras, the lengths of the diagonals of quadrangle $ABCD$ are $\\sqrt{d^2 - a^2}$ and $\\sqrt{d^2 - c^2}$. Applying Ptolemaios' Theorem to the quadrilateral $ABCD$ gives\n$$\n\\sqrt{d^2 - a^2} \\cdot \\sqrt{d^2 - c^2} = ab + ac,\n$$\nwhich after squaring and simplifying becomes\n$$\nd^3 - (2a^2 + c^2)d - 2a... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0e83 | Let $K_1$ be a circle with the center $S_1$ and the radius $r$. Let $K_2$ be a circle with the center $S_2$, lying on the circle $K_1$, and the radius $\frac{2}{3}r$. Let $A$ be the point of intersection of the line $S_1S_2$ and the circle $K_2$ that lies in the exterior of the circle $K_1$. Let $C$ denote one of the i... | [
"The quadrilateral $ES_2CD$ is cyclic, so $\\angle S_2ED = \\angle S_2CA = \\angle CAS_2$ and $EAD$ is an isosceles triangle with the apex at $D$. This implies $|AH| = |EH|$, or $|AH| = \\frac{1}{2}|EA| = \\frac{1}{2}(2r + \\frac{2}{3}r) = \\frac{4}{3}r$. Since $\\frac{4}{3}r$ is precisely the radius of the circle ... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0eih | Problem:
Za notranja kota $\alpha$ in $\beta$ trikotnika $ABC$ z obsegom 24 velja $\cos \alpha=\frac{2}{3}$ in $\cos \beta=\frac{2}{7}$. Izračunaj ploščino trikotnika $ABC$. | [
"Solution:\n\nOpazimo, da sta kota $\\alpha$ in $\\beta$ ostra, saj sta $\\cos \\alpha$ in $\\cos \\beta$ pozitivna. Označimo stranice trikotnika $ABC$ kot običajno z $a, b$ in $c$. Naj bo $C'$ nožišče višine $v$ skozi oglišče $C$. Dolžini daljic $AC'$ in $BC'$ označimo zaporedoma z $b_1$ in $a_1$. Tedaj velja\n$$\... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 12 sqrt(5) | |
0084 | Players $A$ and $B$ play a game as follows. Initially $A$ arranges the numbers $1, 2, \dots, n$ in a row as he wishes; $n$ is a given positive integer. Next, $B$ chooses one number and puts a stone on it. Then $A$ moves the stone to an adjacent number, $B$ does the same and so on. The stone can be placed on number $k$ ... | [
"Player $A$ has a winning strategy if $n$ is $0$ or $-1$ modulo $4$, otherwise $B$ has one.\n\nPutting the stone on a number can be viewed as subtracting $1$ from it. We may assume that $B$ chooses a number in $A$'s arrangement and subtracts $1$ from it; then $A$ must subtract $1$ from an adjacent number etc. Opera... | Argentina | Mathematical Olympiad Rioplatense | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | A wins if and only if n ≡ 0 or 3 (mod 4); otherwise B wins. | |
010d | Problem:
Determine all positive integers $n$ for which there exists a set $S$ with the following properties:
(i) $S$ consists of $n$ positive integers, all smaller than $2^{n-1}$;
(ii) for any two distinct subsets $A$ and $B$ of $S$, the sum of the elements of $A$ is different from the sum of the elements of $B$. | [
"Solution:\n\nDirect search shows that there is no such set $S$ for $n=1,2,3$. For $n=4$ we can take $S=\\{3,5,6,7\\}$. If, for a certain $n \\geqslant 4$ we have a set $S=\\left\\{a_{1}, a_{2}, \\ldots, a_{n}\\right\\}$ as needed, then the set $S^{*}=\\left\\{1,2 a_{1}, 2 a_{2}, \\ldots, 2 a_{n}\\right\\}$ satisfi... | Baltic Way | Baltic Way 1998 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All integers n greater than or equal to four | |
0g44 | Problem:
Let $n > 6$ be a perfect number. Let $p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdot \ldots \cdot p_{k}^{a_{k}}$ be the prime factorisation of $n$ where we assume that $p_{1} < p_{2} < \ldots < p_{k}$ and $a_{i} > 0$ for all $i = 1, \ldots, k$. Prove that $a_{1}$ is even. | [
"Solution:\n\nSince $n$ is perfect, we can write\n$$\n2 n = \\sum_{1 \\leq d \\mid n} d = \\sum_{0 \\leq b_{i} \\leq a_{i}} p_{1}^{b_{1}} p_{2}^{b_{2}} \\cdots p_{k}^{b_{k}} = \\prod_{i=1}^{k}\\left(1 + p_{i} + \\cdots + p_{i}^{a_{i}}\\right)\n$$\nNow assuming $a_{1}$ is odd, we find that\n$$\n\\left(1 + p_{1} + \\... | Switzerland | Final round | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0cis | What are the finite unitary rings for which we can find 3 nonzero elements (not necessarily distinct), whose sum equals their product? | [] | Romania | 75th NMO | [
"Algebra > Abstract Algebra > Ring Theory"
] | English | proof and answer | All finite unitary rings except the field with three elements. | |
03i0 | Problem:
If $f(x) = x^{2} + x$, prove that the equation $4 f(a) = f(b)$ has no solutions in positive integers $a$ and $b$. | [
"Solution:\nLet $f(x) = x^2 + x$. The equation is $4 f(a) = f(b)$, i.e.,\n\n$$\n4(a^2 + a) = b^2 + b.\n$$\n\nExpanding:\n$$\n4a^2 + 4a = b^2 + b\n$$\nBring all terms to one side:\n$$\n4a^2 + 4a - b^2 - b = 0\n$$\n$$\n4a^2 + 4a = b^2 + b\n$$\n\nLet us try to solve for $b$ in terms of $a$:\n\n$$\nb^2 + b - 4a^2 - 4a ... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0jk1 | Problem:
Let $S = \{-100, -99, -98, \ldots, 99, 100\}$. Choose a 50-element subset $T$ of $S$ at random. Find the expected number of elements of the set $\{|x| : x \in T\}$. | [
"Solution:\nLet us solve a more generalized version of the problem: Let $S$ be a set with $2n+1$ elements, and partition $S$ into sets $A_{0}, A_{1}, \\ldots, A_{n}$ such that $|A_{0}| = 1$ and $|A_{1}| = |A_{2}| = \\cdots = |A_{n}| = 2$. (In this problem, we have $A_{0} = \\{0\\}$ and $A_{k} = \\{k, -k\\}$ for $k ... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 8825/201 | |
0ktz | Problem:
Let $ABC$ be an equilateral triangle with side length $2$ that is inscribed in a circle $\omega$. A chord of $\omega$ passes through the midpoints of sides $AB$ and $AC$. Compute the length of this chord.
 | [
"Solution:\nLet $O$ and $r$ be the center and the circumradius of $\\triangle ABC$. Let $T$ be the midpoint of the chord in question.\n\nNote that $AO = \\frac{AB}{\\sqrt{3}} = \\frac{2\\sqrt{3}}{3}$. Additionally, we have that $AT$ is half the distance from $A$ to $BC$, i.e. $AT = \\frac{\\sqrt{3}}{2}$. This means... | United States | HMMT November 2023 | [
"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 > Distance chasing"
] | null | proof and answer | sqrt(5) | |
08mh | Problem:
There are two piles of coins, each containing $2010$ pieces. Two players $A$ and $B$ play a game taking turns ($A$ plays first). At each turn, the player on play has to take one or more coins from one pile or exactly one coin from each pile. Whoever takes the last coin is the winner. Which player will win if ... | [
"Solution:\n\n$B$ wins.\n\nIn fact, we will show that $A$ will lose if the total number of coins is a multiple of $3$ and the two piles differ by not more than one coin (call this a balanced position). To this end, firstly notice that it is not possible to move from one balanced position to another. The winning str... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | B | |
04rq | Determine the number of all coverings of a chessboard $3 \times 10$ by (nonoverlapping) pieces $2 \times 1$ which can be placed both horizontally and vertically. | [
"Let us solve a more general problem of determining the number $a_n$ of all coverings of a chessboard $3 \\times 2n$ by pieces $2 \\times 1$, for a given natural $n$. We will attack the problem by a recursive method, starting with $n = 1$.\n\nThe value $a_1 = 3$ (for the chessboard $3 \\times 2$) is evident (see Fi... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 571 | |
0jlg | Problem:
Suppose that there are initially eight townspeople and one goon. One of the eight townspeople is named Jester. If Jester is sent to jail during some morning, then the game ends immediately in his sole victory. (However, the Jester does not win if he is sent to jail during some night.)
Find the probability tha... | [
"Solution:\n\nAnswer: $\\frac{1}{3}$\n\nLet $a_{n}$ denote the answer when there are $2n-1$ regular townies, one Jester, and one goon. It is not hard to see that $a_{1} = \\frac{1}{3}$.\n\nMoreover, we have a recursion\n$$\na_{n} = \\frac{1}{2n+1} \\cdot 1 + \\frac{1}{2n+1} \\cdot 0 + \\frac{2n-1}{2n+1}\\left(\\fra... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1/3 | |
0iyx | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX$, $BX$, $AZ$, $BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where $O$ is the midpoint of segment $AB... | [
"Let $T$ be the foot of the perpendicular from $Y$ to line $AB$. We note that $P$, $Q$, $T$ are the feet of the perpendiculars from $Y$ to the sides of triangle $ABX$. Because $Y$ lies on the circumcircle of triangle $ABX$, points $P$, $Q$, and $T$ are collinear by Simson's theorem. Likewise, points $S$, $R$, and $... | United States | USAMO 2010 | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kpe | Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probabili... | [
"There are two cases, depending on whether Azar and Carl meet in the semifinals. If they do, which occurs with probability $\\frac{1}{3}$, Carl will win the tournament if and only if he beats Azar and goes on to beat the winner of the other semifinal match, which occurs with probability $\\frac{1}{3} \\cdot \\frac{... | United States | 2022 AIME II | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 125 | |
0b4j | Problem:
Let $ABC$ be an acute scalene triangle with orthocenter $H$. Let $M$ be the midpoint of $BC$, and suppose that the line through $H$ perpendicular to $AM$ intersects $AB$ and $AC$ at points $E$ and $F$ respectively. Denote by $O$ the circumcenter of triangle $AEF$, and $D$ the foot of the perpendicular from $H... | [
"Solution:\n\nWLOG assume $AB < AC$. Let $AM$ intersect the circumcircle of $ABC$ again at $Y \\neq A$. We first need to prove a claim.\n\n\n\nClaim: Quadrilateral $B H D C$ is cyclic.\n\nProof of Claim: Consider the reflection with respect to $M$. This maps $B$ and $C$ to each other. It is... | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | null | proof only | null | |
0ej7 | Problem:
Za neko celo število $x$ je končno zaporedje $\sqrt{x}+2, 3 \sqrt{x+1}, 2 \sqrt{x}+4$ geometrijsko. Kolikšen je količnik tega zaporedja?
(A) 2
(B) $\sqrt{3}$
(C) $\sqrt{2}$
(D) 3
(E) $3 \sqrt{2}$ | [
"Solution:\n\nUpoštevamo zvezo med zaporednimi členi geometrijskega zaporedja. Rešimo iracionalno enačbo in dobimo rešitvi $x_{1}=1$ in $x_{2}=\\frac{1}{49}$. Edina celoštevilska rešitev je $1$. Nato izračunamo člene zaporedja $a_{1}=3$, $a_{2}=3 \\sqrt{2}$, $a_{3}=6$. Izračunamo količnik $q=\\sqrt{2}$. Pravilen je... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0crq | Все клетки квадратной таблицы $n \times n$ пронумерованы в некотором порядке числами от 1 до $n^2$. Петя делает ходы по следующим правилам. Первым ходом он ставит фишку в любую клетку. Каждым последующим ходом Петя может либо поставить новую фишку на какую-то клетку, либо переставить фишку из клетки с номером $a$ ходом... | [
"$n$.\n\nПокажем, что $n$ фишек достаточно. Для этого заметим, что на каждую строку хватит одной фишки: можно поставить её в клетку строки с минимальным номером, а затем обойти все клетки строки в порядке возрастания номеров.\n\nС другой стороны, покажем, что меньше, чем $n$ фишек, может и не хватить. Для этого про... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n | |
0giz | 喬老大有一條 $1 \times 46^2$ 的棟木板，其上有 $46^2$ 個 $1 \times 1$ 大小的正方格子，依序編號為 1 至 $46^2$ 號。喬老大將這條木板鋸成 $N$ 段，每一段皆為連續編號的若干個格子，並在**不旋轉或翻面**的情況下，用這 $N$ 段木板排出滿足以下條件的 $46 \times 46$ 方陣：若位於第 $i$ 列第 $j$ 行格子的編號為 $a_{ij}$，則 $a_{ij} - (i + j - 1)$ 被 46 整除。試求 $N$ 的最小可能值。
Joe has a $1 \times 46^2$ rectangular dogwood strip consisting of $46... | [
"答案為 91；一般性地，對於 $1 \\times n^2$ 的木板，$N$ 的最小可能值為 $2n-1$。\n\n構造：將 $1 \\times n^2$ 的長條切成長度為 $n, 1, n, \\dots, 1, 1$ 的 $2n-1$ 段。用第一段 $n$ 木條構成第一行，依此類推，構成下方的 $(n-1) \\times n$ 方陣，再用所有 $1$ 木條構成最後一行即可。(備註：這並非唯一的構造方法。)\n\n估計：由於題目要求僅與編號對 $n$ 的餘數有關，以下討論都在 mod $n$ 的同餘下進行。\n\n考慮點集 $V = \\{0, 1, \\dots, n-1\\}$，並依以下規則連邊：對於鋸出的每一段... | Taiwan | IMO 2J, Mock Exam 2 | [
"Discrete Mathematics > Graph Theory",
"Number Theory > Modular Arithmetic"
] | Chinese; English | proof and answer | 91 | |
01yt | The numbers $-1011, -1010, \dots, -1, 1, 2, \dots, 1010, 1011$ are arranged as $a_1, a_2, \dots, a_{2022}$ in some order.
Find the maximal possible value of
$$|a_1| + |a_1 + a_2| + |a_1 + a_2 + a_3| + \dots + |a_1 + a_2 + \dots + a_{2022}|.$$ (Yahor Dubovik) | [
"Note that the sum of all numbers is $0$, so the required sum can be presented as the sum of the following two sums:\n$$\nA_1 = |a_1| + |a_1 + a_2| + \\dots + |a_1 + a_2 + \\dots + a_{1011}|\n$$\nand\n$$\nA_2 = |a_{2022}| + |a_{2022} + a_{2021}| + \\dots + |a_{2022} + \\dots + a_{1013}|.\n$$\nLet's bound each term ... | Belarus | Belarus2022 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1011*1012*4043/6 | |
0dh1 | Let $BB_1$ and $CC_1$ be the altitudes of acute-angled triangle $ABC$, and $A_0$ is the midpoint of $BC$. Lines $A_0B_1$ and $A_0C_1$ meet the line passing through $A$ and parallel to $BC$ in points $P$ and $Q$. Prove that the incenter of triangle $PA_0Q$ lies on the altitude of triangle $ABC$. | [
"Since triangles $BCB_1$ and $BCC_1$ are right-angled, their medians $B_1A_0$, $B_1C_0$ are equal to the half of hypotenuse $B_1A_0 = A_0C = A_0B = C_1A_0$.\n\n\n\nNow\n$$\n\\angle PB_1A = \\angle CB_1A_0 = \\angle B_1CA_0 = \\angle PAC,\n$$\nthus $PA = PB_1$. Similarly, $QA = QC_1$. Then t... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02ek | $A$ and $B$ play a game. Each has $10$ tokens numbered from $1$ to $10$. The board is two rows of squares. The first row is numbered $1$ to $1492$ and the second row is numbered $1$ to $1989$. On the $n$th turn, $A$ places his token number $n$ on any empty square in either row and $B$ places his token on any empty squa... | [
"Note that $B$ will lose if he does not space his tokens widely enough. Call the rows $R$ and $S$, so that $S^n$ denotes the number $n$ in row $S$. Suppose $A$ plays $1$ on $R5$, then $2$ on $R6$. If $B$ plays $1$ on $S5$ and $2$ on $S7$, then he loses, because $A$ swaps rows and plays $3$ on $S6$. However, $B$ doe... | Brazil | XI OBM | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | With ten tokens on rows of lengths one thousand four hundred ninety two and one thousand nine hundred eighty nine, player B has a winning strategy. For k tokens on this board, B wins if k is at most ten and A wins if k is greater than ten. If both rows are all integers, B has a winning strategy. If both rows are all ra... | |
0df5 | Determine if there exist functions $f, g: \mathbb{R} \to \mathbb{R}$ satisfying for every $x \in \mathbb{R}$ the following equations
$$
f(g(x)) = x^3 \quad \text{and} \quad g(f(x)) = x^2.
$$ | [
"*Solution.* Denote $b_n = \\frac{a_n}{\\text{rad}(a_n)}$. Since rad($a_n$) divides rad($a_{n+1}$) we have $b_{n+1} | b_n + 1$. If there are indices $i < j$ with $b_i < 2022 < b_{i+1}$, we will be done by “continuity”. If, to the contrary, this does not happen, there are two possible cases.\n* $b_n < 2022$ for all ... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | Yes, such functions exist. | |
01fp | $AC$ is hypotenuse of right triangle $ABC$, $BH$ is its altitude. Points $M$ and $N$ are the midpoints of segments $AH$ and $CH$ correspondingly. Lines $BM$ and $BN$ intersect for second time the circumscribed circle of triangle $ABC$ in points $P$ and $Q$ correspondingly. Segments $AQ$ and $CP$ intersect in point $R$.... | [
"Let $K$ be the midpoint of segment $BH$, $S$ be the intersection point of $AK$ and $BP$, $T$ be the intersection point of $CK$ and $BQ$. Then $SK$ and $KT$ is one third of the corresponding medians and $ST$ is parallel to $AC$.\n\nThe triangles $ABH$ and $BCH$ are similar. From this similarity and properties of in... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
074i | Is there a positive integer $n$, which is a multiple of $103$, such that $2^{2n+1} \equiv 2 \pmod n$? | [
"We show that there is no such positive integer $n$. Suppose the contrary; assume that a positive integer $n$ exists such that $2^{2n+1} \\equiv 2 \\pmod n$ and $103 \\nmid n$. Then $2^{2n+1} \\equiv 2 \\pmod{103}$ as well; and as such $2^{2n} \\equiv 1 \\pmod{103}$. Since $103$ is prime, Fermat's little theorem gi... | India | Indija TS 2010 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | No | |
0iqy | Problem:
Let $P$ be a polyhedron where every face is a regular polygon, and every edge has length $1$. Each vertex of $P$ is incident to two regular hexagons and one square. Choose a vertex $V$ of the polyhedron. Find the volume of the set of all points contained in $P$ that are closer to $V$ than to any other vertex. | [
"Solution:\nAnswer: $\\frac{\\sqrt{2}}{3}$\n\nObserve that $P$ is a truncated octahedron, formed by cutting off the corners from a regular octahedron with edge length $3$. So, to compute the value of $P$, we can find the volume of the octahedron, and then subtract off the volume of truncated corners.\n\nGiven a squ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | sqrt(2)/3 | |
0hu2 | Problem:
Determine all pairs $(n, k)$ of integers such that $0 < k < n$ and
$$
\binom{n}{k-1} + \binom{n}{k+1} = 2 \binom{n}{k}
$$ | [
"Solution:\nIn the factorial form,\n$$\n\\frac{n!}{(k-1)!(n-k+1)!} + \\frac{n!}{(k+1)!(n-k-1)!} = \\frac{2 \\cdot n!}{k!(n-k)!}\n$$\nwe multiply through by $(k+1)!(n-k+1)!$ to clear the fractions and then divide through by $n!$:\n$$\nk(k+1) + (n-k)(n-k+1) = 2(k+1)(n-k+1)\n$$\nTo decrease the number of terms, we let... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | All integer pairs given by n = c^2 − 2 and k = (c^2 + c − 2)/2 for any integer c with c ≤ −3 or c ≥ 3. | |
02kw | Problem:
Esportistas de uma escola - Em um grupo de $40$ estudantes, $20$ jogam futebol, $19$ jogam vôlei e $15$ jogam exatamente uns destes dois esportes. Quantos estudantes não praticam futebol e vôlei?
(a) $7$
(b) $5$
(c) $13$
(d) $9$
(e) $10$ | [
"Solution:\n\nDenotemos por $x$ o número de estudantes que praticam simultaneamente os dois esportes. Logo, temos que o número de estudantes que pratica somente futebol é $20-x$ e o que pratica somente vôlei é $19-x$. Portanto os estudantes que praticam exatamente um esporte são\n$$\n(20-x)+(19-x)=15\n$$\nSegue-se ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | c | |
0deb | Let $ABC$ be a triangle with incircle $(I)$, tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. On the line $DF$, take points $M$, $P$ such that $CM \parallel AB$, $AP \parallel BC$. On the line $DE$, take points $N$, $Q$ such that $BN \parallel AC$, $AQ \parallel BC$. Denote $X$ as intersection of $PE$, $QF$ a... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geomet... | null | proof only | null | |
08cy | Problem:
Veronica osserva che $81 \cdot 3=243$ e $81 \cdot 4=324$ e si chiede quanti siano i numeri $m$ con $10 \leq m \leq 99$ e tali che $3 m=A B C$ e $4 m=C A B$, con $A, B$ e $C$ cifre decimali (si considerano validi anche i casi in cui una o più delle cifre $A, B, C$ siano uguali a zero).
(A) 1
(B) 2
(C) 3
(D) 4... | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. Poiché $3 m=A B C$ è ovviamente un multiplo di 3, tale è la somma delle sue cifre $A+B+C$. Inoltre, sappiamo che $3 m=100 A+10 B+C$ e che $4 m=100 C+10 A+B$; per differenza, dunque, $m=99 C-9 B-90 A=9(11 C-B-10 A)$. D'altro canto $11 C-B-10 A=12 C-9 A-(A+B+C)$ risulta ... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
09e9 | In how many manner can the number $\frac{3}{2014}$ be represented in the form
$$
\frac{1}{p} + \frac{1}{q}, \ p, q \in \mathbb{N}
$$ | [
"$\\frac{3}{2014} = \\frac{1}{p} + \\frac{1}{q} \\Rightarrow 3pq = 2014(p+q) = 2 \\cdot 19 \\cdot 53(p+q).$\n\ni) Consider the case $2 \\cdot 19 \\cdot 53 \\mid p$. Setting $p = 2 \\cdot 19 \\cdot 53r$ we get $3rq = 2 \\cdot 19 \\cdot 53r + q \\Rightarrow q = \\frac{2 \\cdot 19 \\cdot 53r}{3r-1}$. Since $q \\in \\m... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 6 | |
05a4 | Determine all functions $f : \mathbb{R} \to \mathbb{R}$ which satisfy the inequality $f(x) + f(x+y) \le f(xy) + f(y)$ for all real numbers $x, y$. | [
"**Answer:** All constant functions $f(x) = c$ where $c$ is arbitrary real number.\n\nDenote the given inequality by $V(x,y)$. Then $V(x,0)$ together with simplification gives\n$$\nf(x) \\le f(0) \\qquad (3)\n$$\nfor every real number $x$. On the other hand, adding $V(x,y)$ and $V(y,x)$ gives $f(x+y) \\le f(xy)$, w... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All constant functions: f(x) = c for all real x, where c is any real constant. | |
03gr | Problem:
Show that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by $3$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
00v1 | Let $n \ge 3$ be a natural number. Anna and Bob play the following game on the vertices of a regular $n$-gon: Anna places her token on a vertex of the $n$-gon. Afterwards Bob places his token on another vertex of the $n$-gon. Then, with Anna playing first, they move their tokens alternately as follows for $2n$ rounds: ... | [
"**Solution.** We will show that Bob wins if and only if $4|n$ and $n \\ne 4$. We will often say that Anna and Bob are at a distance $d$ if we can move one token $d$ positions clockwise or anticlockwise to reach the other token. Note that the value of this distance is not unique.\nWe first treat the case $4 \\nmid ... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Bob wins if and only if the number of sides is divisible by four and not equal to four; otherwise Anna wins. | |
01jq | Let $p > 7$ be a prime number and let $A \subseteq \{0, 1, \dots, p-1\}$ consist of at least $\frac{p-1}{2}$ elements. Show that for each integer $r$, there are elements $a, b, c, d \in A$ such that
$$
ab - cd \equiv r \pmod{p}.
$$ | [
"Let $P$ be the set of possible products $ab$, for $a, b \\in A$. Clearly, $|P| \\ge |aA| \\ge \\frac{p-1}{2}$, for any $a \\in A$. If $|P| \\ge \\frac{p+1}{2}$, then $|r + P| \\ge \\frac{p+1}{2}$, too. Hence, $|P| + |r + P| \\ge p + 1 > p$, so, by the Pigeonhole Principle, $P$ and $r + P$ must have an element in c... | Baltic Way | Baltic Way 2023 Shortlist | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Abstract Algebra > Group Theory"... | English | proof only | null | |
069p | Let $v > 1$, a positive integer. Each cell of a $v \times v$ table contains one integer. Suppose that the following conditions are satisfied:
a. Each number in the table is congruent to $1$ modulo $v$.
b. The sum of numbers in any row, as well the sum of numbers in any column is congruent to $v$ modulo $v^2$.
Let $\... | [] | Greece | SELECTION EXAMINATION | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0ill | Problem:
A cube of edge length $s > 0$ has the property that its surface area is equal to the sum of its volume and five times its edge length. Compute all possible values of $s$. | [
"Solution:\nThe volume of the cube is $s^{3}$ and its surface area is $6s^{2}$, so we have\n$$\n6s^{2} = s^{3} + 5s\n$$\nor\n$$\n0 = s^{3} - 6s^{2} + 5s = s(s-1)(s-5)\n$$\nThus, the possible values of $s$ are $1$ and $5$ (since $s > 0$)."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1 and 5 | |
00ev | Let $A$, $B$ be two distinct points on a given circle $O$ and let $P$ be the midpoint of the line segment $AB$. Let $O_{1}$ be the circle tangent to the line $AB$ at $P$ and tangent to the circle $O$. Let $\ell$ be the tangent line, different from the line $AB$, to $O_{1}$ passing through $A$. Let $C$ be the intersecti... | [
"Let $S$ be the tangent point of the circles $O$ and $O_{1}$ and let $T$ be the intersection point, different from $S$, of the circle $O$ and the line $SP$. Let $X$ be the tangent point of $\\ell$ to $O_{1}$ and let $M$ be the midpoint of the line segment $XP$. Since $\\angle TBP = \\angle ASP$, the triangle $TBP$ ... | Asia Pacific Mathematics Olympiad (APMO) | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
07xq | Let $A$ be a set of five distinct integers and $S$ the set that contains all sums $x + y$ with $x, y \in A$ and $x \neq y$. The two smallest elements of $S$ are $25$ and $31$, while the two largest elements of $S$ are $57$ and $71$. Determine all possible sets $A$. | [
"Let $a < b < c < d < e$ be the elements of $A$. The two smallest sums are $25 = a + b < 31 = a + c$. The two largest are $57 = c + e < 71 = d + e$. We then obtain\n$$\na + d = (a + c) - (c + e) + (d + e) = 31 - 57 + 71 = 45\n$$\n$$\ne - a = (c + e) - (a + c) = 57 - 31 = 26.\n$$\n\nThis allows us to express the ele... | Ireland | IRL_ABooklet_2025 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | {10, 15, 21, 35, 36}; {11, 14, 20, 34, 37}; {12, 13, 19, 33, 38} | |
07wv | A group of $100$ people, no two of them are of the same height, are placed in random order in a line. They play a game as follows. At each step, a group of at most $4$ people swap places so that they are now arranged in increasing order of height. Everybody else stays put. Prove that it is always possible to arrange th... | [
"At each step, choose the tallest and second tallest person not yet in their places, as well as the two people in the last two places which are not in the right order. By rearranging these, we get the tallest and second tallest persons in place. After $48$ steps we will be left with at most $4$ people not in place ... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0kkz | Problem:
A set of 6 distinct lattice points is chosen uniformly at random from the set $\{1,2,3,4,5,6\}^2$. Let $A$ be the expected area of the convex hull of these 6 points. Estimate $N=\left\lfloor 10^4 A\right\rfloor$.
An estimate of $E$ will receive $\max \left(0,\left\lfloor 20-20\left(\frac{|E-N|}{10^4}\right)^... | [
"Solution:\n\nThe main tools we will use are linearity of expectation and Pick's theorem. Note that the resulting polygon is a lattice polygon, and thus the expected area $A$ satisfies\n$$\nA = I + \\frac{B}{2} - 1\n$$\nwhere $I$ is the expected number of interior points and $B$ is the expected number of boundary p... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 104552 | |
0a9q | Problem:
In a football tournament there are $n$ teams, with $n \geq 4$, and each pair of teams meets exactly once. Suppose that, at the end of the tournament, the final scores form an arithmetic sequence where each team scores 1 more point than the following team on the scoreboard. Determine the maximum possible score... | [
"Solution:\n\nNote that the total number of games equals the number of different pairings, that is, $n(n-1)/2$. Suppose the lowest scoring team ends with $k$ points. Then the total score for all teams is\n$$\nk + (k+1) + \\cdots + (k+n-1) = n k + \\frac{(n-1)n}{2}\n$$\nSome games must end in a tie, for otherwise, a... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n - 2 | |
0jem | Problem:
You are standing at a pole and a snail is moving directly away from the pole at $1~\mathrm{cm}/\mathrm{s}$. When the snail is $1$ meter away, you start "Round 1". In Round $n$ ($n \geq 1$), you move directly toward the snail at $(n+1)~\mathrm{cm}/\mathrm{s}$. When you reach the snail, you immediately turn aro... | [
"Solution:\n\nSuppose the snail is $x_n$ meters away at the start of round $n$, so $x_1 = 1$, and the runner takes $\\frac{100 x_n}{(n+1)-1} = \\frac{100 x_n}{n}$ seconds to catch up to the snail. But the runner takes the same amount of time to run back to the start, so during round $n$, the snail moves a distance ... | United States | HMMT November 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 5050 | |
0eyg | Problem:
9 judges each award 20 competitors a rank from 1 to 20. The competitor's score is the sum of the ranks from the 9 judges, and the winner is the competitor with the lowest score. For each competitor the difference between the highest and lowest ranking (from different judges) is at most 3. What is the highest ... | [
"Solution:\n\nAt most 4 competitors can receive a rank 1. For a competitor with a rank 1 can only receive ranks 1, 2, 3 or 4. There are only 36 such ranks available and each competitor with a rank 1 needs 9 of them.\n\nIf only one competitor receives a rank 1, then his score is 9. If only 2 competitors receive a ra... | Soviet Union | 2nd ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 24 | |
0a4l | Problem:
Find all real solutions to the equation
$$(x^{2} + 3x + 1)^{x^{2} - x - 6} = 1.$$ | [
"Solution:\nLet $a = x^{2} + 3x + 1$ and let $b = x^{2} - x - 6$. The only way to have $a^{b} = 1$ is if $a = \\pm 1$ or $b = 0$.\n\n- If $b = 0$, then we solve the quadratic $x^{2} - x - 6 = 0$ which has solutions $x = -2, 3$ (we would also have to check that $a \\neq 0$ in this case)\n\n- If $a = 1$, then we solv... | New Zealand | NZMO Round One | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | x = -3, -2, -1, 0, 3 | |
0ee0 | Problem:
Za neničelni realni števili $a$ in $b$, $a \neq -1$ in $b \neq -1$, velja
$$
\frac{a}{b+1} + \frac{b}{a+1} = 1.
$$
Katera trditev o izrazu
$$
\frac{a}{b} + \frac{b}{a} - \frac{1}{ab}
$$
je pravilna?
(A) Izraz lahko zavzame poljubno vrednost z intervala $(0,1]$.
(B) Izraz lahko zavzame poljubno vrednost z int... | [
"Solution:\n\nV dani enakosti odpravimo ulomke in jo poenostavimo do $a^{2} + b^{2} = ab + 1$. Dani izraz postavimo na skupni imenovalec, da dobimo\n$$\n\\frac{a^{2} + b^{2} - 1}{ab}.\n$$\nIz enakosti sledi, da je vrednost izraza enaka\n$$\n\\frac{ab}{ab} = 1.\n$$\nPravilen odgovor je (C)."
] | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0ifp | Problem:
Show that no rectangle of the form $1 \times k$ or $2 \times n$, where $4 \nmid n$, is $(1,2)$-tileable. | [
"Solution:\n\nThe claim is obvious for $1 \\times k$ rectangles. For the others, color the first two columns black, the next two white, the next two black, etc. Each $(1,2)$ domino will contain one square of each color, so in order to be tileable, the rectangle must contain the same number of black and white square... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
003b | En un triángulo $ABC$ sea $H$ el punto de corte de sus alturas. Se sabe que la medida del ángulo $\angle BAC$ es de $60^\circ$. Si se toma $J$ perteneciente al lado $AC$ tal que $AJ$ es el doble de $JC$, se cumple que $JH = JC$.
Dada la ubicación de $A$ y de $H$, construya con regla y compás el triángulo $ABC$. | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Español | proof only | null | |
0bxz | Consider a finite collection of 3-element sets $A_i$ no two of which share more than one element, whose union has cardinality $2017$. Show that the elements of this union can be coloured one of two colours, blue and red, so that at least $64$ elements are blue, and each $A_i$ contains at least one red element. | [
"Let $U$ be the union of the $A_i$. It is sufficient to show that a maximal (relative to set-theoretic inclusion) subset $T$ of $U$ containing no $A_i$ satisfies the required cardinality condition.\n\nBy maximality, for each element $x$ of $U \\setminus T$, there exists a 2-element subset $S$ of $T$ such that $S \\... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0e1j | Every school in the region has sent 3 students to a contest. Andrej, Blaž and Žan represented the same school. When all the contestants lined up to receive their start numbers, Andrej realized that there were exactly as many contestants in the line before him as there were behind. Both his friends were behind him: Blaž... | [
"Let $x$ denote the number of contestants in line before Andrej. Then there were also $x$ contestants behind Andrej and there were $2x+1$ contestants altogether. Hence, the total number of contestants was odd. Since Andrej was standing in line before Blaž, who was 19th, there were at most 17 contestants in line bef... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 11 | |
07uu | A convex non-regular octagon $A B C D E F G H$ is inscribed in a circle. Prove that $\angle A + \angle C + \angle E + \angle G = \angle B + \angle D + \angle F + \angle H$. | [
"Connect $C$ to $F$ and $B$ to $G$. This produces three cyclic quadrilaterals $ABGH$, $BCFG$, and $CDEF$.\n\nBecause opposite angles in a cyclic quadrilateral add to $180^\\circ$, we see that $\\angle A + \\angle C + \\angle E + \\angle G = 3 \\cdot 180^\\circ$ which is half of the interior... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cuh | A convex polygon is dissected into isosceles triangles by several non-intersecting diagonals. Prove that this polygon has two sides of equal lengths. | [
"10.7. See problem 9.7.",
"10.7. См. задачу 9.7."
] | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English; Russian | proof only | null | |
01f9 | Find the smallest positive integer that cannot be written in the form $\binom{a}{2} + \binom{b}{2} + c$ with nonnegative integers $a, b, c$ satisfying $a \ge b \ge c$ and $a + b \le 2019$. | [
"The number is $m = \\binom{1957}{2} + \\binom{63}{2} + 1 = 1,915,900$.\nAssume that $m$ has a representation as above. Then $a \\le 1957$ as $\\binom{1958}{2} > m$. On the other hand, by $\\binom{a}{2} + \\binom{b}{2} + c \\le \\binom{a+1}{2} + \\binom{b-1}{2} + (b-1)$ it follows that the largest number that can b... | Baltic Way | Baltic Way 2019 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | English | proof and answer | 1915900 | |
0kes | Problem:
Let $ABC$ be a triangle with $AB = 5$, $BC = 6$, $CA = 7$. Let $D$ be a point on ray $AB$ beyond $B$ such that $BD = 7$, $E$ be a point on ray $BC$ beyond $C$ such that $CE = 5$, and $F$ be a point on ray $CA$ beyond $A$ such that $AF = 6$. Compute the area of the circumcircle of $DEF$. | [
"Solution:\n\nLet $I$ be the incenter of $ABC$. We claim that $I$ is the circumcenter of $DEF$.\n\n\n\nTo prove this, let the incircle touch $AB$, $BC$, and $AC$ at $X$, $Y$, and $Z$, respectively. Noting that $XB = BY = 2$, $YC = CZ = 4$, and $ZA = AX = 3$, we see that $XD = YE = ZF = 9$. ... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"G... | null | proof and answer | 251/3 * pi | |
0gs7 | For every positive integer $n$, let $\sigma(n)$ denote the number of positive divisors of $n$ and let $s(n)$ denote the number of positive divisors $d$ of $n$ such that $d+1|n+1$. Find the maximum value of
$$
2s(n) - \sigma(n).
$$ | [
"Answer: 2.\nIt is easy to verify that for any odd prime number $p$ we have $s(p) = \\sigma(p) = 2$ and hence $2s(n) - \\sigma(n) = 2$. We will show that $2s(n) - \\sigma(n) \\le 2$ for every positive integer $n$. Let $1 = d_1 < d_2 < \\dots < d_k = n$ be positive divisors of $n$. It is well known that $d_i d_{k+1-... | Turkey | Team Selection Test for EGMO 2019 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 2 | |
0jfy | Problem:
Find the number of positive integers less than $1000000$ that are divisible by some perfect cube greater than $1$. Your score will be $\max \left\{0,\left\lfloor 20-200\left|1-\frac{k}{S}\right|\right\rfloor\right\}$, where $k$ is your answer and $S$ is the actual answer. | [
"Solution:\n\nAnswer: $168089$\n\nUsing the following code, we get the answer (denoted by the variable $ans$):\n\nans $=0$\nfor $n$ in xrange $(1,1000000)$ :\n```\ndivisible_by_cube = True\nfor i in xrange(2,101):\n if n%(i*i*i)==0:\n divisible_by_cube = False\n break\nif divisible_by_cube: ans = a... | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 168089 | |
0ffb | Problem:
Se considera un fichero con 1000 fichas numeradas, ordenadas en su orden natural. A ese fichero se le aplica la siguiente operación:
La primera ficha del fichero se coloca intercalada entre la penúltima y la última del mismo, y la segunda, al final de todas, quedando, por tanto, en primer lugar la que antes o... | [
"Solution:\n\nLa operación efectuada sobre $n$ fichas es una permutación del conjunto\n$$\nN=\\{1,2,3, \\ldots, n-2, n-1, n\\}\n$$\nes decir una función biyectiva definida así:\n$$\n\\begin{aligned}\nf(k) & =k+2 \\quad \\text{ si } \\quad k \\leq n-3 \\\\\nf(n-2) & =1 \\\\\nf(n-1) & =n \\\\\nf(n) & =2\n\\end{aligne... | Spain | Olimpiadas Matemáticas Españolas | [
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
0day | Determine whether there exists a positive integer $n$ such that $n+2$ divides the following sum
$$
S = 1^{2019} + 2^{2019} + \ldots + n^{2019}.
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | No such positive integer n exists. | |
03ab | The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect at point $E$, $M$ is the midpoint of $AE$ and $N$ is the midpoint of $CD$. It is known that the diagonal $BD$ bisects $\angle ABC$. Prove that the quadrilateral $ABCD$ is cyclic if and only if the quadrilateral $MBCN$ is cyclic. | [
"Let $ABCD$ be a cyclic quadrilateral. Since $\\angle ABD = \\angle CBD$ it follows that $AD = CD$. Denote by $S$ the midpoint of $DE$. Then $SM = \\frac{AD}{2} = \\frac{CD}{2} = CN$ and $SN \\parallel AC$. Hence $MCNS$ is an isosceles trapezoid and therefore it is cyclic. On the other hand, we have $\\angle MSB = ... | Bulgaria | Fall Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
022c | Problem:
1. Carro flex - Um carro é denominado flex se ele pode ser abastecido com gasolina ou com álcool. Considere que os preços do álcool e da gasolina sejam, respectivamente, $\mathrm{R}\$ 1,59$ e $\mathrm{R}\$ 2,49$ por litro.
a. Suponha que um carro flex rode $12,3~\mathrm{km}$ por litro de gasolina, que indica... | [
"Solution:\n\na. Com gasolina o carro faz $\\frac{12,3}{2,49}=4,94~\\mathrm{km}$ por $\\mathrm{R}\\$ 1,00$. Para que o álcool seja mais vantajoso precisamos que o carro rode, com álcool, mais que $4,94~\\mathrm{km}$ com $\\mathrm{R}\\$ 1,00$. Logo, se o desempenho com álcool é $y~\\mathrm{km}/\\mathrm{l}$, precisam... | Brazil | Nível 3 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) Alcohol must achieve y > 7.85 km/l.
b) g(x) = 249/x and a(x) = 318/(x+2).
c) Equal cost occurs at gasoline x ≈ 7.22 km/l and alcohol (x/2 + 1) ≈ 3.61 km/l.
d) Condition: y > (1.59/2.49) x ≈ 0.64 x. Example: if x = 10 km/l, then y must exceed 6.4 km/l. | |
0jc5 | Determine all positive integers $n$, $n \ge 2$, such that the following statement is true:
If $(a_1, a_2, \ldots, a_n)$ is a sequence of positive integers with $a_1 + a_2 + \cdots + a_n = 2n - 1$, then there is a block of (at least two) consecutive terms in the sequence with their (arithmetic) mean being an integer. | [
"The statement is true for all $n \\ge 4$ but not for $n = 2$ or $n = 3$. In those two cases, the sequences $(1, 2)$ and $(2, 1, 2)$ provide counterexamples.\n\nNow, let $(a_1, \\dots, a_n)$ be any sequence of positive integers, and let $s_k = a_1 + \\dots + a_k - 2k$ for $k = 1, 2, \\dots, n$, and define $s_0 = 0$... | United States | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All integers n ≥ 4 | |
04pp | Find all pairs $(a, b)$ of integers such that $b \ge 0$ and
$$
a^2 + 2ab + b! = 131.
$$
(Olimpiada Matemática del Istmo Centroamericano 2017) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (1, 5), (-11, 5) | |
096i | Problem:
Un plan, care conține o muchie, divizează un tetraedru regulat în două corpuri, volumele cărora se raportă ca $3:5$. Determinați măsurile unghiurilor în care planul secant divizează unghiul diedru al tetraedrului. | [
"Solution:\n\nFie $a$ lungimea muchiei tetraedrului, iar $VKC$ - planul secant.\nConsiderăm că $\\frac{v_{VKCB}}{v_{VKCA}}=\\frac{3}{5}$. Atunci $\\frac{BK}{AK}=\\frac{3}{5}$ și $AK=\\frac{5}{8} a$, $BK=\\frac{3}{8} a$.\nConsiderăm unghiul liniar $AMB$ al unghiului diedru. Atunci $AM=BM=\\frac{a \\sqrt{3}}{2}$. Apl... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | arccos(7√11/33) and arccos(3√11/11) | |
0fxp | Problem:
Für welche natürlichen Zahlen $m, n$ lässt sich ein $m \times n$-Rechteck mit lauter Quadraten der Seitenlänge 2 oder 3 bedecken? | [
"Solution:\n\nGenau dann, wenn $m$ und $n$ beide gerade oder beide durch 3 teilbar sind, oder wenn eine der Zahlen durch 6 teilbar und die andere grösser als 1 ist.\n\nWir zeigen zuerst, dass diese Bedingungen hinreichend sind. Für $2 \\mid m, n$ bzw. $3 \\mid m, n$ lässt sich das Rechteck mit lauter $2 \\times 2$-... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Exactly those pairs where both dimensions are even, or both are divisible by three, or one dimension is divisible by six and the other is greater than one. | |
0l7h | Problem:
Let $ABCD$ be a rectangle with $BC = 24$. Point $X$ lies inside the rectangle such that $\angle AXB = 90^{\circ}$. Given that triangles $\triangle AXD$ and $\triangle BXC$ are both acute and have circumradii $13$ and $15$, respectively, compute $AB$.
Proposed by: Pitchayut Saengrungkongka | [
"Solution:\nLet $M$ be the midpoint of $AB$. Let $O_{1}$ and $O_{2}$ be the circumcenters of $\\triangle AXD$ and $\\triangle BXC$, respectively. Since $O_{1}M$ is the perpendicular bisector of $AX$ and $O_{2}M$ is the perpendicular bisector of $BX$, we get that $\\angle O_{1}MO_{2} = 90^{\\circ}$.\nLet $P_{1}$ and... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 14 + 4√37 | |
0fpo | Las tres raíces del polinomio $x^3 - 14x^2 + Bx - 84$ son los lados de un triángulo rectángulo. Hallar $B$. | [
"Sean $u$, $v$ y $w$ las tres raíces y supongamos que $w^2 = u^2 + v^2$. Por las relaciones de Cardano, $u+v+w = 14$, $uv+uw+vw = B$ y $uvw = 84$. Si $s = u+v$ y $p = uv$, se tiene entonces que $s+w = 14$, $pw = 84$ y $s^2 = w^2 + 2p$. Sustituyendo en esta última ecuación los valores de $s$ y $p$ en función de $w$ ... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | Spanish | proof and answer | 62 | |
0evh | Show that there are no rational numbers $x$ and $y$ such that
$$
x - \frac{1}{x} + y - \frac{1}{y} = 4.
$$ | [
"Suppose that there are rational numbers $x$ and $y$ such that $x - \\frac{1}{x} + y - \\frac{1}{y} = 4$.\nSince $\\left(-\\frac{1}{x}, y\\right)$, $\\left(x, -\\frac{1}{y}\\right)$ and $\\left(-\\frac{1}{x}, -\\frac{1}{y}\\right)$ are solutions of the equation, we can assume that $x > 0$ and $y > 0$.\n\nLetting $u... | South Korea | Korean Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
017f | Assume that all angles of a triangle $ABC$ are acute. Let $D$ and $E$ be points on the sides $AC$ and $BC$ of the triangle such that $A$, $B$, $D$, and $E$ lie on the same circle. Further suppose the circle through $D$, $E$, and $C$ intersects the side $AB$ in two points $X$ and $Y$. Show that the midpoint of $XY$ is t... | [
"We write the power of the point $A$ with respect to the circle $\\gamma$ through $D$, $E$, and $C$:\n$$\n|AX||AY| = |AD||AC| = |AC|^2 - |AC||CD|.\n$$\nSimilarly, if we calculate the power of $B$ with respect to $\\gamma$ we get\n$$\n|BX||BY| = |BC|^2 - |BC||CE|.\n$$\nWe have also that $|AC||CD| = |BC||CE|$, the po... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0eqc | A rectangular sheet of paper can be used to form a cylinder by joining two opposite sides together:
Should the short edges or the long edges be joined together to obtain the largest volume of the cylinder?
NB: Show all your working! | [
"Suppose that the rectangular sheet of paper has dimensions $a$ and $b$, with $b$ being the longer side. We calculate the volume of the two cylinders formed by joining the long sides and short sides, respectively.\n\n* Suppose the short sides are glued together. Then the height of the cylinder is $a$ and the circum... | South Africa | South African Mathematics Olympiad | [
"Geometry > Solid Geometry > Volume"
] | English | proof and answer | Join the short edges. | |
07re | Suppose $u, v$ are real numbers and $w = u + iv$ is a complex number. Show that the quadratic $x^2 - 2ix + w$ has precisely one real root iff $v^2 + 4u = 0$. | [
"Suppose $v^2 + 4u = 0$, and let $\\tau = v/2$. Then, $\\tau$ is real and\n$$\nr^2 - 2ir = \\frac{v^2}{4} - iv = -u - iv = -w.\n$$\nThus, the quadratic has a real root.\n\nConversely, if $\\tau$ is a real root of $x^2 - 2ix + w$, then it is also a real root of $x^2 + 2ix + \\bar{w}$. In other words, $\\tau$ satisfi... | Ireland | Ireland_2017 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
0591 | The radius of the circumcircle of an acute triangle $ABC$ is $R$ and its orthocenter is $H$. Show that $AH^2 + BC^2 = 4R^2$. | [
"Let $O$ and $G$ be the circumcenter and centroid of $ABC$ respectively and let $K$ be the midpoint of $BC$ (Fig. 33).\n\nWe know that $AG = 2GK$. Also we know that $H$, $G$ and $O$ are collinear with $HG = 2GO$ (Euler line). So triangles $AHG$ and $KOG$ are similar with scale factor $2$ (by $2$ proportional sides ... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0itu | Let $n$ be a positive integer and let $a_1, \dots, a_k$ ($k \ge 2$) be distinct integers in the set $\{1, \dots, n\}$ such that $n$ divides $a_i(a_{i+1} - 1)$ for $i = 1, \dots, k-1$. Prove that $n$ does not divide $a_k(a_1 - 1)$. | [
"Assume on the contrary that $n$ divides $a_k(a_1 - 1)$. Then $n$ divides $a_i(a_{i+1} - 1)$ for $i \\ge 1$ (where $a_{k+j} = a_j$); that is, $a_i \\equiv a_i a_{i+1} \\pmod{n}$ for all $i \\ge 1$. It follows that\n$$\na_i \\equiv a_i a_{i+1} \\equiv a_i a_{i+1} a_{i+2} \\equiv \\dots \\equiv a_i a_{i+1} \\dots a_{... | United States | IMO 2009 | [
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0fka | Problem:
Sea $P$ una familia de puntos en el plano tales que por cada cuatro puntos de $P$ pasa una circunferencia. ¿Se puede afirmar que necesariamente todos los puntos de $P$ están en la misma circunferencia? Justifica la respuesta. | [
"Solution:\n\nSea $T = \\{x_{1}, x_{2}, x_{3}, x_{4}\\}$ un subconjunto de $P$ con cuatro elementos. Por hipótesis existe una circunferencia $\\alpha$ que pasa por estos cuatro puntos. Supongamos que exista un punto $x \\in P$, tal que $x \\notin \\alpha$. Por la condición del enunciado existe una circunferencia $\... | Spain | FASE LOCAL DE LA XLIV OME | [
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | Yes, all points lie on the same circle. | |
0c4m | Let $a$ and $b$ be real numbers, $a < b$, and let $f$ be a polynomial of degree 3, with real coefficients, so that the polynomials $f - a$ and $f - b$ have only real roots.
Prove that the set of the real numbers $x$ such that $a < f(x) < b$ is the disjoint union of three open intervals, and the length of one of these i... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
08v6 | Let $n$ be a positive integer greater than or equal to $2$. Determine the maximum possible value the quantity
$$
\left( \sum_{i=1}^{n} i a_i \right) \left( \sum_{i=1}^{n} \frac{a_i}{i} \right)^2
$$
can take where $a_1, a_2, \dots, a_n$ are non-negative real numbers satisfying $a_1 + a_2 + \dots + a_n = 1$. | [
"Let $X = \\sum_{i=1}^{n} i a_{i}$, $Y = \\sum_{i=1}^{n} \\frac{a_{i}}{i}$. We have to find the maximum possible value of the quantity $X Y^{2}$.\n\nFirst, we note that for each $i \\in \\{1, 2, \\dots, n\\}$, $i + \\frac{n}{i} \\le n + 1$ holds. This follows since $(n+1) - (i + \\frac{n}{i}) = \\frac{1}{i}(i-1)(n-... | Japan | Japan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4(n+1)^3/(27n^2) | |
06yl | Determine all positive integers $a$ and $b$ such that there exists a positive integer $g$ such that $\operatorname{gcd}\left(a^{n}+b, b^{n}+a\right)=g$ for all sufficiently large $n$.
(Indonesia) | [
"It is clear that we may take $g=2$ for $(a, b)=(1,1)$. Supposing that $(a, b)$ satisfies the conditions in the problem, let $N$ be a positive integer such that $\\operatorname{gcd}\\left(a^{n}+b, b^{n}+a\\right)=g$ for all $n \\geqslant N$.\n\nLemma. We have that $g=\\operatorname{gcd}(a, b)$ or $g=2 \\operatornam... | IMO | IMO2024 Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | (1,1) | |
0j9b | Problem:
Find the number of integers between $1$ and $200$ inclusive whose distinct prime divisors sum to $16$. (For example, the sum of the distinct prime divisors of $12$ is $2+3=5$.) | [
"Solution:\nThe primes less than $16$ are $2, 3, 5, 7, 11$, and $13$. We can write $16$ as the sum of such primes in three different ways and find the integers less than $200$ with those prime factors:\n\n- $13+3$: $3 \\cdot 13 = 39$ and $3^{2} \\cdot 13 = 117$.\n- $11+5$: $5 \\cdot 11 = 55$ and $5^{2} \\cdot 11 = ... | United States | HMMT November | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 6 | |
0ii4 | Problem:
A $3 \times 3 \times 3$ cube is built from 27 unit cubes. Suddenly five of those cubes mysteriously teleport away. What is the minimum possible surface area of the remaining solid? Prove your answer. | [
"Solution:\nOrient the cube so that its edges are parallel to the $x$-, $y$-, and $z$-axes. A set of three unit cubes whose centers differ only in their $x$-coordinate will be termed an \"$x$-row\"; there are thus nine $x$-rows. Define \"$y$-row\" and \"$z$-row\" similarly.\nTo achieve 50, simply take away one $x$-... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 50 | |
04cv | We say that two cells of the $10 \times 10$ table are *friendly* if they have at least one common vertex. Into each cell of the table a positive integer less than or equal to $10$ is written, so that the numbers in friendly cells are relatively prime. Prove that some number appears in the table at least $17$ times.
(S... | [
"Let's divide the given table into $25$ smaller squares $2 \\times 2$. In each of these squares there is at most one even number and at most one number divisible by $3$. Hence, at most $50$ numbers in the table are divisible by $2$ or $3$. At least $50$ numbers remain, and each of them is equal to $1$, $5$ or $7$. ... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0jtc | Problem:
Consider an infinite grid of equilateral triangles. Each edge (that is, each side of a small triangle) is colored one of $N$ colors. The coloring is done in such a way that any path between any two nonadjacent vertices consists of edges with at least two different colors. What is the smallest possible value o... | [
"Solution:\n\nAnswer: 6\n\nNote that the condition is equivalent to having no edges of the same color sharing a vertex by just considering paths of length two. Consider a hexagon made out of six triangles. Six edges meet at the center, so $N \\geq 6$. To prove $N=6$, simply use two colors for each of the three poss... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 6 | |
038m | The functions $f(x) = 2x^2 + 2x - 4$ and $g(x) = x^2 - x + 2$ are given. Find all real values of $x$ such that:
a) $\frac{f(x)}{g(x)}$ is a positive integer;
b) the inequality $\sqrt{f(x)} + \sqrt{g(x)} \ge \sqrt{2}$ holds. | [
"a) *Hint.* Set $\\frac{f(x)}{g(x)} = k$, where $k$ is a positive integer. Then $(2-k)x^2 + (2+k)x - 2(2+k) = 0$ and use the fact that the discriminant of this quadratic equation is nonnegative.\n\n*Answer.* $x = \\frac{-3+\\sqrt{33}}{2}, \\frac{-3-\\sqrt{33}}{2}, 2$.\n\nb) *Answer.* $x \\in (-\\infty, -2] \\cup [1... | Bulgaria | Winter Mathematical Competition | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a) x = (-3 + sqrt(33)) / 2, (-3 - sqrt(33)) / 2, 2. b) x ∈ (-∞, -2] ∪ [1, ∞). | |
0ayr | Problem:
Compute the number of ordered 6-tuples $(a, b, c, d, e, f)$ of positive integers such that
$$
a+b+c+2(d+e+f)=15
$$ | [
"Solution:\nLet $x = a + b + c$ and $y = d + e + f$. Then the equation becomes\n$$\nx + 2y = 15\n$$\nwhere $a, b, c, d, e, f$ are positive integers, so $x \\geq 3$ and $y \\geq 3$.\n\nLet us solve for all possible integer values of $y$ such that $y \\geq 3$ and $x = 15 - 2y \\geq 3$.\n\nWe have:\n$$\n15 - 2y \\geq ... | Philippines | 21st PMO Area Stage | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 119 | |
0ink | Problem:
Compute the largest positive integer such that $\frac{2007!}{2007^{n}}$ is an integer. | [
"Solution:\n\nAnswer: 9. Note that $2007 = 3^{2} \\cdot 223$. Using the fact that the number of times a prime $p$ divides $n!$ is given by\n$$\n\\left\\lfloor\\frac{n}{p}\\right\\rfloor + \\left\\lfloor\\frac{n}{p^{2}}\\right\\rfloor + \\left\\lfloor\\frac{n}{p^{3}}\\right\\rfloor + \\cdots\n$$\nit follows that the... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 9 | |
00ee | The river city of Platense consists of several platforms and bridges between them. Each bridge connects two platforms and no two bridges are connecting the same two platforms. The mayor wants to change some bridges through a series of moves as follows: if there are three platforms $A$, $B$ and $C$, and bridges $AB$ and... | [
"Let us interpret the problem in terms of graphs. We can think of the initial configuration as a graph $G$ whose vertices are the platforms and whose edges are the bridges. This graph is connected. The claim is, then, that $G$ can be converted into another graph $G'$ by rotating edges as in the statement if $G'$ is... | Argentina | Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
09vf | Problem:
Zij $n$ een positief geheel getal. Bewijs dat $n^{2}+n+1$ niet te schrijven is als het product van twee positieve gehele getallen die minder dan $2 \sqrt{n}$ van elkaar verschillen. | [
"Solution:\n\nOplossing I. Stel dat $a$ en $b$ positieve gehele getallen zijn met $a b=n^{2}+n+1$. We gaan bewijzen dat $|a-b| \\geq 2 \\sqrt{n}$. Merk op dat $(a-b)^{2} \\geq 0$. Aan beide kanten $4 a b$ optellen geeft\n$$\n(a+b)^{2}=(a-b)^{2}+4 a b \\geq 4 a b=4 n^{2}+4 n+4>4 n^{2}+4 n+1=(2 n+1)^{2} .\n$$\nOmdat ... | Netherlands | IMO-selectietoets III | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
07xv | Let $n \ge 3$ be an integer. Determine, as a function of $n$, the number of circular arrangements $x_1, x_2, \dots, x_n$ of the numbers $1, 2, \dots, n$ such that
$$
\sum_{i=1}^{n} |x_i - x_{i+2}| = 2n - 4,
$$
where the indices $i$ and $i+2$ are to be interpreted modulo $n$. Note that any rotation of a circular arrange... | [
"**Solution 1.**\nFirst consider the case when $n$ is odd. It is easy to see that\n$$\nS = \\sum_{i=1}^{n} |x_i - x_{i+2}| = \\sum_{i=1}^{n} |y_i - y_{i+1}|\n$$\nfor another circular arrangement $y_1, y_2, \\dots, y_n$. So we can consider\n$$\nS = \\sum_{i=1}^{n} |x_i - x_{i+1}|\n$$\ninstead. Now, the numbers $1$ a... | Ireland | IRL_ABooklet_2025 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 0 if n is odd; n*2^{n-5} if n is even | |
05l7 | Problem:
Trouver tous les couples d'entiers positifs $(m, n)$ tels que $1+(m+n) m$ divise $(m+n)(n+1)-1$. | [
"Solution:\n\nSoit $(m, n)$ un couple solution. Alors $1+(m+n) m$ divise $(m+n)(n+1)-1+1+(m+n) m = (m+n)(m+n+1)$. Or, $1+(m+n) m$ est premier avec $m+n$. Ainsi, $1+(m+n) m$ divise $m+n+1$. Donc $m^{2}+m n+1 \\leqslant m+n+1$. Donc $m=0$ ou $m=1$.\n\nRéciproquement, on vérifie que les couples $(0, n)$ et $(1, n)$, o... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (m, n) = (1, n) for any positive integer n | |
05du | Problem:
Find all lists $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ of non-negative real numbers such that the following three conditions are all satisfied:
(i) $x_{1} \leq x_{2} \leq \ldots \leq x_{2020}$;
(ii) $x_{2020} \leq x_{1}+1$;
(iii) there is a permutation $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ of $\l... | [
"Solution:\nWe first prove the inequality\n$$\n((x+1)(y+1))^{2} \\geq 4\\left(x^{3}+y^{3}\\right)\n$$\nfor real numbers $x, y \\geq 0$ satisfying $|x-y| \\leq 1$, with equality if and only if $\\{x, y\\}=\\{0,1\\}$ or $\\{x, y\\}=\\{1,2\\}$.\nIndeed,\n$$\n\\begin{aligned}\n4\\left(x^{3}+y^{3}\\right) & =4(x+y)\\lef... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2020 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Either the list with the first 1010 entries equal to 0 and the last 1010 entries equal to 1, or the list with the first 1010 entries equal to 1 and the last 1010 entries equal to 2. | |
08tk | Let $n$ be an integer greater than or equal to $2$. Assign to each vertex of a regular $2n$-gon a distinct number chosen from $\{1, 2, \dots, 2n\}$.
(1) Show that there exists a method of assigning these numbers in such a way that the differences of the numbers assigned to every neighboring pair of vertices are all gr... | [
"(1): Pick a vertex $P$ of the given $2n$-gon and label the vertices consecutively as $1, 2, \\dots, 2n$ starting with $P$ as $1$ and going around clockwise. Then for $1 \\le k \\le n$ reassign the number $k$ to the vertex labeled $2k-1$, respectively, and the number $n+k$ to the vertex labeled $2k$, respectively. ... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
096a | Problem:
Determinați valorile de extrem a funcției $f:\left(0, \frac{\pi}{2}\right) \rightarrow \mathbb{R}$, $f(x)=\sin x+\cos x+\operatorname{tg} x+\operatorname{ctg} x$. | [
"Solution:\n$$\n\\begin{aligned}\n& f'(x)=\\cos x-\\sin x+\\frac{1}{\\cos^2 x}-\\frac{1}{\\sin^2 x}=\\cos x-\\sin x+\\frac{\\sin^2 x-\\cos^2 x}{\\sin^2 x \\cdot \\cos^2 x}= \\\\\n& =\\cos x-\\sin x-\\frac{(\\cos x-\\sin x)(\\cos x+\\sin x)}{\\sin^2 x \\cos^2 x}=(\\cos x-\\sin x)\\left[1-\\frac{\\cos x+\\sin x}{\\si... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Differential Calculus > Applications",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 2+sqrt(2) | |
07dm | $k \in \mathbb{Z}^+$ is a fixed number. Find all functions $f : \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ such that for infinitely many prime numbers like $q$, $q^k$ is in the range of $f$ and also for all $m, n \in \mathbb{Z}^+$
$$
f(m) + f(n) \mid f(m + n)
$$ | [
"We prove by induction that $f(n) = n f(1)$.\n\nLet $c_1, c_2, \\dots$, be the sequence of positive integers such that $f(c_i) = p_i^k$. Then one has\n$$\nf(c_i - (d+1)) + f(d+1) \\mid f(c_i) = p_i^k\n$$\nand hence $f(c_i - (d+1)) = p_i^j - f(d+1)$ for some positive integer $j \\le k$. By pigeonhole principle, ther... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | f(n) = n | |
0jgb | Let $P$ and $P'$ be two convex quadrilateral regions in the plane (regions contain their boundary). Let them intersect, with $O$ a point in the intersection. Suppose that for every line $\ell$ through $O$ the segment $\ell \cap P$ is strictly longer than the segment $\ell \cap P'$. Is it possible that the ratio of the ... | [
"Let $\\mathcal{P}$ denote the square $ABCD$ in both diagrams below, and let $\\mathcal{Q}$ be the isosceles triangle $XYZ$ (with $XY = XZ$) in the diagram shown on the left below. Suppose that $OY = OZ = 2OA = 2OC$ so that $YZ$ and $AC$ share the common midpoint $O$. It is not difficult to see that $\\mathcal{P}$ ... | United States | RMM | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Yes |
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