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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07dw | Let $a_1, a_2, a_3, \dots$ be an infinite sequence of distinct integers. Prove that there are infinitely many prime numbers like $p$ that distinct positive integers $i, j, k$ can be found such that $p \mid a_i a_j a_k - 1$. | [
"For the sake of contradiction, let $p_1, p_2, \\dots, p_n$ be all the prime divisors of numbers in form of $a_i a_j a_k - 1$. Moreover, let $p$ be the smallest of these primes, and $M = \\max(a_1, \\dots, a_{n+1})$ (mind that $n$ is fixed.). Let $k$ be a positive integer number satisfying $p^k > M$. There are infi... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
07ze | Problem:
Tre paia di calzini, uno rosso, uno blu e uno verde, sono stesi in fila. Sapendo che due calzini dello stesso colore non sono vicini uno all'altro, quante successioni di colori si possono avere?
(A) 15
(B) 24
(C) 30
(D) 36
(E) Nessuna delle precedenti | [] | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | C | |
0gp7 | Alice and Bob play a game on a $1 \times m$ board using $2012$ cards numbered from $1$ through $2012$. At each step, Alice chooses a card and Bob places it on an empty square of the board. Bob wins the game when numbers on the cards on the board are in an increasing order after $k$ steps where $1 \leq k \leq 2012$, oth... | [
"Bob wins for all pairs of $(k, m \\ge 2^k - 1)$ if $k = 1, 2, \\dots, 10$ and for all pairs $(k, m \\ge 2012)$ if $2012 \\ge k \\ge 11$.\n\nLet us show that Bob wins in all cases listed above. If $k = 1$ then $2^k - 1 = 1$ and the result is trivial. Suppose that Bob wins for $k - 1 \\le 9$ when $m \\ge 2^{k-1} - 1... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Bob can guarantee a win for all pairs with k between 1 and 10 and m at least 2^k − 1, and for all pairs with k between 11 and 2012 and m at least 2012. | |
04d4 | In some country there are $c$ cities and $r$ roads, every road connects two different cities and between any two cities there is at most one road. Roads are named by numbers $1, 2, \dots, r$. Tonči travels along some roads in such a way that, when he writes down the names of the roads in the order he passes through the... | [
"We place one of Tonči's friends in every city. In the step $i$ ($i = 1, 2, \\dots, r$) friends which are at that moment in the cities connected by the road $i$ switch their positions. In each step we have exactly two shifts from one city to another, and all together $2r$ shifts. Hence at least one of the $c$ frien... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory"
] | English | proof only | null | |
03gn | Problem:
Let $f$ be a function with the following properties:
1) $f(n)$ is defined for every positive integer $n$;
2) $f(n)$ is an integer;
3) $f(2)=2$;
4) $f(m n)=f(m) f(n)$ for all $m$ and $n$;
5) $f(m)>f(n)$ whenever $m>n$.
Prove that $f(n)=n$. | [
"Solution:\n\nLet us analyze the properties of $f$.\n\nFrom property 4, $f(m n) = f(m) f(n)$ for all positive integers $m, n$. This means $f$ is a multiplicative function.\n\nLet us compute $f(1)$:\n\nTake $m = n = 1$:\n$$\nf(1) = f(1 \\cdot 1) = f(1) f(1) \\implies f(1)^2 = f(1)\n$$\nSo $f(1) = 0$ or $f(1) = 1$.\n... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | null | proof only | null | |
0gha | 令 $n$ 為一正整數。松鼠阿布與阿江準備了 $n$ 顆核桃好過冬。某天，阿江發現阿布把核桃擺成 $n$ 堆，每堆一顆；牠覺得太多堆了，心生不悅。阿江於是決定進行以下操作：每次選兩堆核桃，從中各拿取等量的核桃，並將拿取的核桃合併成新的一堆。阿江的目標是讓非空的核桃堆數 $P(n)$ 越少越好。試對所有正整數 $n$，求阿江能透過有限步操作達到的最小 $P(n)$ 值。
Let $n$ be a positive integer. Two squirrels, Bushy and Jumpy, have collected $n$ walnuts for the winter. One day, Jumpy noticed that... | [
"若 $n$ 為 2 的幂次，則最小 $P = 1$；否則，最小 $P = 2$。\n\n當 $n = 2^k$ 時，我們只要每次取顆數最少的任兩堆，拿取其全部核桃合併，最終便能成為單一一堆，而這顯然是最小可能 $P$ 值。\n\n現在考慮 $2^k < n < 2^{k+1}$。以下用 $t$-堆表示有 $t$ 顆核桃的堆。考慮以下操作：\n\n1. 我們先從起始的 $n$ 堆中選擇 $2^k$ 堆，然後每次取其中顆數最少的任兩堆，拿取其全部核桃合併，最終便能成為一個 $2^k$-堆與 $m = n - 2^k$ 個 1-堆。稱這個 $2^k$-堆為 XL 堆。\n\n2. 接下來，我們從 XL 堆和一個 1-堆中各取一顆... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Chinese (Traditional) | proof and answer | P(n) = 1 if n is a power of two; otherwise P(n) = 2. | |
06kt | The incircle of $\triangle ABC$, with incentre $I$, meets $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. The line $EF$ cuts the lines $BI$, $CI$, $BC$ and $DI$ at points $K$, $L$, $M$ and $Q$ respectively. The line through the midpoint of $CL$ and $M$ meets $CK$ at $P$.
a. Determine $\angle BKC$.
b. Show that ... | [
"a. Since $\\angle FKI = \\angle FKB = \\angle EFA - \\angle KBF = 90^\\circ - \\frac{A}{2} - \\frac{B}{2} = \\frac{C}{2} = \\angle ECI$, the points $C$, $E$, $K$, $I$ are concyclic. Note that $C$, $E$, $I$, $D$ are also concyclic. This shows $C$, $E$, $K$, $I$, $D$ lie on the same circle. It follows that\n$$\n\\an... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Colline... | null | proof and answer | ∠BKC = 90° and PQ ∥ CL | |
04qe | The incircle of the triangle $ABC$ touches the sides $\overline{AB}$ and $\overline{AC}$ in $D$ and $E$, respectively. The excircle of the same triangle opposite to the vertex $A$ touches the rays $AB$ and $AC$ in $F$ and $G$, respectively. Let the bisectors of the internal (external) angles $\angle CBA$ and $\angle AC... | [
"Let $I$ be the incentre, and let $\\alpha$, $\\beta$ and $\\gamma$ be the measures of angles of the triangle $ABC$.\nThe triangle $ADE$ is isosceles, and we have $\\angle BDX = 90^\\circ + \\frac{\\alpha}{2}$. Since $\\angle XBD = \\frac{\\beta}{2}$, it follows that $\\angle EXI = \\angle DXB = \\frac{\\gamma}{2} ... | Croatia | Croatian 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 > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
04fp | The segment $\overline{AB}$ is a diameter of a circle with the centre $O$. On the circle the point $C$ is given such that $OC$ is perpendicular to $AB$. Let $P$ be a point on the shorter arc $\widearc{BC}$. The lines $CP$ and $AB$ intersect at the point $Q$, and the point $R$ is the intersection of the line $AP$ and th... | [
"The triangle $OCB$ is isosceles right triangle because $\\overline{OB}$ and $\\overline{OC}$ are both radii of the circle with the centre $O$. Hence $\\angle CBA = \\angle CBO = 45^\\circ$.\n\n\n\nInscribed angles $\\angle CPA$ and $\\angle CBA$ over the chord $\\overline{CA}$ are equal, s... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01pu | Given triangle $ABC$ with $AC = (AB + BC)/2$. Let $BL$ be the bisector of the angle $ABC$; let $K$ and $M$ be the midpoints of $AB$ and $BC$ respectively.
Find the value of the angle $KLM$ if $\angle ABC = \beta$. | [
"Answer: $90^\\circ - \\beta/2$.\nLet point $N$ be marked on the side $AC$ such that $AN = 0.5 AB$. Then, by condition, $NC = AC - AN = 0.5(AB + BC) - 0.5 AB = 0.5 BC$. Therefore, $AN : NC = AB : BC$. Since $BL$ is a bisector of the angle $ABC$ we have $AL : LC = AB : BC$, so $L$ and $N$ coincide. Therefore the tri... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 90° - β/2 | |
04ec | Determine the non-negative real number $a$ for which the expression
$$
a^3 - a^2 - 2\sqrt{a}
$$
is minimal. | [
"For $a = 1$ the expression is equal to $-2$. We will prove that this is the minimal value, i.e. that for every $a \\ge 0$ we have $a^3 - a^2 - 2\\sqrt{a} \\ge -2$.\nWe have\n$$\na^3 - a^2 - 2\\sqrt{a} + 2 = a^2(a-1) - 2(\\sqrt{a}-1) = (\\sqrt{a}-1)(a^2(\\sqrt{a}+1) - 2).\n$$\nIf $a \\ge 1$, then $a^2(\\sqrt{a}+1) ... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 1 | |
0l8j | The sequence of integers $\{a_n\}$, $n = 0, 1, 2, \ldots$, is defined by:
$a_0 = 1$ and $a_n = a_{n-1} + a_{\lceil n/3 \rceil}$ for every $n = 1, 2, 3, \ldots$.
Prove that for every prime number $p \le 13$, there exists an infinite number of natural numbers $k$ such that $a_k$ is divisible by $p$.
([x] denotes the inte... | [] | Vietnam | VIETNAMESE MATHEMATICAL COMPETITION FOR TEAM SELECTION | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
050s | There are $2^n$ soldiers standing in a line, where $n$ is a positive integer. The soldiers can rearrange themselves into a new line only in the following way: the soldiers standing at odd numbered positions move to the front of the row, keeping their positions with respect to each other, and the soldiers previously sta... | [
"The last soldier does not change its position. The rest of the soldiers regroup just as in the case, when the last soldier was not there, and the number of the soldiers was $2^n - 1$. So, it suffices to prove the claim for $2^n - 1$ soldiers. We show that after $n$ rearrangements the soldiers are in positions, whi... | Estonia | Estonian Math Competitions | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | null | proof only | null | |
0jgk | Problem:
Determine whether there is a polynomial $f(x)$ such that
- Every coefficient of $f$, from the leading coefficient down to the constant term, is either $1$ or $-1$.
- $(x-1)^{2013}$ evenly divides $f(x)$ (this means that their quotient is a polynomial). | [
"Solution:\nThe answer is yes. We replace $2013$ by a variable $n$ and prove that such a polynomial exists for all $n$. Our base case is $n=0$, for which the polynomial $f_{0}(x)=1$ clearly works.\nGiven a polynomial $f_{n}(x)$ of degree $d$ that is divisible by $(x-1)^{n}$, consider the polynomial\n$$\nf_{n+1}(x) ... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | yes | |
0dvf | Problem:
Vemo, da je $A=\frac{a^{-2}-b^{-2}}{a^{-1}-b^{-1}}$
$$
\text{in}\quad B=\left(\frac{a^{-1}}{a^{-1}-b^{-1}}-\frac{b^{-1}}{a^{-1}+b^{-1}}\right) \cdot\left(a^{-1}-b^{-1}\right) \cdot\left(a^{-2}+b^{-2}\right)^{-1}
$$
Dokaži, da je $A=B^{-1}$. | [
"Solution:\n\nNajprej poenostavimo izraz $A$ :\n$$\n\\frac{\\frac{b^{2}-a^{2}}{a^{2} b^{2}}}{\\frac{b-a}{a b}}=\\frac{(b-a)(b+a)}{(b-a) a b}=\\frac{b+a}{a b}\n$$\n\nNato pa še izraz $B$ :\n$$\n\\left(\\frac{\\frac{1}{a}}{\\frac{b-a}{a b}}-\\frac{\\frac{1}{b}}{\\frac{b+a}{a b}}\\right) \\cdot \\frac{b-a}{a b} \\cdot... | Slovenia | 3. matematično tekmovanje dijakov srednjih tehniških in strokovnih sol | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0cmr | A circle with center $I$ touches the sides $AB$, $BC$, $AC$ of non-isosceles triangle $ABC$ at points $C_1$, $A_1$, $B_1$, respectively. Circles $\omega_B$ and $\omega_C$ are inscribed into the quadrilaterals $BA_1IC_1$ and $CA_1IB_1$, respectively. Prove that the common internal tangent to $\omega_B$ and $\omega_C$, d... | [
"Первое решение. Обозначим через $O_B$ и $O_C$ центры окружностей $\\omega_B$ и $\\omega_C$, соответственно. Заметим, что достаточно доказать, что $\\angle O_BAO_C = \\angle BAC/2$: тогда прямые, полученные из $AB$ и $AC$ симметриями относительно $AO_B$ и $AO_C$ соответственно, совпадают; значит, это и есть общая в... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscel... | English; Russian | proof only | null | |
0l3r | Problem:
Let $P(n)=\left(n-1^{3}\right)\left(n-2^{3}\right) \ldots\left(n-40^{3}\right)$ for positive integers $n$. Suppose that $d$ is the largest positive integer that divides $P(n)$ for every integer $n>2023$. If $d$ is a product of $m$ (not necessarily distinct) prime numbers, compute $m$. | [
"Solution:\nWe first investigate what primes divide $d$. Notice that a prime $p$ divides $P(n)$ for all $n \\geq 2024$ if and only if $\\left\\{1^{3}, 2^{3}, \\ldots, 40^{3}\\right\\}$ contains all residues in modulo $p$. Hence, $p \\leq 40$. Moreover, $x^{3} \\equiv 1$ must not have other solution in modulo $p$ th... | United States | HMMT February 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 48 | |
0a7i | Problem:
Prove that among all triangles with inradius $1$, the equilateral one has the smallest perimeter. | [
"Solution:\n(See Figure 4.) The area $T$, perimeter $p$ and inradius $r$ satisfy $2T = r p$. (Divide the triangle into three triangles with a common vertex at the incenter of the triangle.) So for a fixed inradius, the triangle with the smallest perimeter is the one which has the smallest area. To prove that the eq... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 6 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
085m | Problem:
Una pulce si muove saltando avanti e indietro lungo una retta. La tana della pulce è un punto della retta. Le regole di salto sono le seguenti:
- se la pulce si trova ad una distanza minore o uguale a un metro dalla tana, dopo il salto successivo si troverà ad una distanza doppia della precedente allontanando... | [
"Solution:\n\nLa risposta è 6. Osserviamo innanzitutto che, indipendentemente dalla posizione iniziale, dopo il primo salto la pulce si troverà ad una distanza minore o uguale a 2 metri dalla tana e con i salti successivi non potrà raggiungere posizioni a più di due metri dalla tana.\n\nSi osservi poi che in base a... | Italy | Progetto Olimpiadi di Matematica 2007 GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 6 | |
05l3 | Problem:
Trouver tous les couples d'entiers relatifs $(x, y)$ tels que $x^{3}+y^{3}=(x+y)^{2}$. | [
"Solution:\n\nSoit $(x, y)$ un couple solution. Comme $x^{3}+y^{3}=(x+y)(x^{2}-x y+y^{2})$, on a donc soit $x+y=0$, soit $x^{2}-x y+y^{2}=x+y$.\n\nDans le deuxième cas, on remarque qu'alors nécessairement\n$$(x-1)^{2}+(y-1)^{2}+(x-y)^{2}=2.$$\nParmi les trois entiers $x-1$, $y-1$ et $x-y$, deux sont donc égaux à $1... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All integer pairs of the form (t, -t) for any integer t, together with (2,2), (1,2), (2,1), (1,0), and (0,1). | |
0kut | Problem:
Let $n$ be a positive integer, and let $s$ be the sum of the digits of the base-four representation of $2^{n}-1$. If $s=2023$ (in base ten), compute $n$ (in base ten).
Proposed by: Dongyao Jiang | [
"Solution:\n\nEvery power of $2$ is either represented in base $4$ as $100\\ldots 00_{4}$ or $200\\ldots 00_{4}$ with some number of zeros. That means every positive integer in the form $2^{n}-1$ is either represented in base $4$ as $333\\ldots 33_{4}$ or $133\\ldots 33$ for some number of threes. Note that $2023 =... | United States | HMMT February 2023 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 1349 | |
00qf | Let $A_0B_0C_0$ be a triangle with area equal to $\sqrt{2}$. We consider the excenters $A_1$, $B_1$ and $C_1$ then we consider the excenters, say $A_2$, $B_2$ and $C_2$, of the triangle $A_1B_1C_1$. By continuing this procedure, examine if it is possible to arrive to a triangle $A_nB_nC_n$ with all coordinates rational... | [
"The answer is no. Suppose that it is possible. We assert that the previous triangle $A_{n-1}B_{n-1}C_{n-1}$ has rational coordinates. In fact, the points $A_{n-1}$, $B_{n-1}$, $C_{n-1}$ are the feet of the altitudes of the triangle $A_nB_nC_n$. Therefore it is enough to show that, if a line segment has its ends wi... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | No | |
01hf | Let $n$ be a positive integer. Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy the equation
$$ f(x)^n f(x+y) = f(x)^{n+1} + x^n f(y) $$
for all $x, y \in \mathbb{R}$. | [
"The functions we are looking for are $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = 0$ and $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = x$. For $n$ even $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = -x$ is also a solution.\n\nThroughout the solution, $P(x_0, y_0)$ will denote the substitution of $x_0$ and $y_0$ for $x$ ... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | All solutions are: the zero function; the identity function; and, additionally when the parameter is even, the negation function. | |
0jlc | Problem:
Find the number of triples of sets $(A, B, C)$ such that:
a. $A, B, C \subseteq \{1,2,3, \ldots, 8\}$.
b. $|A \cap B|=|B \cap C|=|C \cap A|=2$.
c. $|A|=|B|=|C|=4$.
Here, $|S|$ denotes the number of elements in the set $S$. | [
"Solution:\nWe consider the sets drawn in a Venn diagram.\n\n\n\nNote that each element that is in at least one of the subsets lies in these seven possible spaces. We split by casework, with the cases based on $N=|R_{7}|=|A \\cap B \\cap C|$.\n\nCase 1: $N=2$\nBecause we are given that $|R_... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 45360 | |
0jt7 | Problem:
Zlatan has $2017$ socks of various colours. He wants to proudly display one sock of each of the colours, and he counts that there are $N$ ways to select socks from his collection for display. Given this information, what is the maximum value of $N$? | [
"Solution:\n\nAnswer: $3^{671} \\cdot 4$\n\nSay that there are $k$ sock types labeled $1,2, \\ldots, k$, and $a_{i}$ socks of type $i$. The problem asks to maximize $\\prod_{i=1}^{k} a_{i}$ subject to $\\sum_{i=1}^{k} a_{i}=2017$, over all $k$ and all sequences of positive integers $a_{1}, \\ldots, a_{k}$.\n\nThe o... | United States | HMMT November | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 3^{671} * 4 | |
0bmg | Let $n$ be a positive integer. If $\sigma$ is a permutation of the first $n$ positive integers, let $S(\sigma)$ be the set of all distinct sums of the form $\sum_{i=k}^{\ell} \sigma(i)$, where $1 \le k \le \ell \le n$.
a) Exhibit a permutation $\sigma$ of the first $n$ positive integers such that $|S(\sigma)| \ge \lflo... | [
"a) We show that the permutation $\\sigma$ of the first $n$ positive integers, defined by $\\sigma(i) = (i+1)/2$ for each positive odd $i \\le n$, and $\\sigma(i) = n - i/2 + 1$ for each positive even $i \\le n$, satisfies the required condition.\nMore precisely, we show that the $\\lfloor(n+1)^2/4\\rfloor$ sums of... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
04dq | A rectangle with dimensions $5 \times 6$ is divided into eight rectangles whose sides are parallel to the sides of the original rectangle, and the lengths of their sides are positive integers. Prove that at least two of the eight rectangles are congruent. | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05wz | Problem:
Soit $n \geqslant 3$ et soient $x_{1}, \ldots, x_{n}$ des réels. Montrer que
$$
2\left(x_{1}+\cdots+x_{n}\right)^{2} \leqslant n\left(x_{1}^{2}+\cdots+x_{n}^{2}+x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n-1} x_{n}+x_{n} x_{1}\right)
$$ | [
"Solution:\nLa difficulté de l'exercice est le fait que l'énoncé n'est pas symétrique. Au vu de celui-ci, on veut faire apparaître les produits $x_{i} x_{i+1}$ ; pour cela, on peut penser à utiliser le développement de $\\left(x_{i}+x_{i+1}\\right)^{2}$. On a\n$$\n\\begin{gathered}\n\\left(x_{1}+x_{2}\\right)^{2}+\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
06ki | Let $a_1, a_2, a_3, \dots$ and $b_1, b_2, b_3, \dots$ be infinite sequences of real numbers satisfying $a_{n+1} + b_{n+1} = \frac{a_n + b_n}{2}$ and $a_{n+1}b_{n+1} = \sqrt{a_n b_n}$ for all $n \ge 1$. Suppose $b_{2016} = 1$ and $a_1 > 0$. Find all possible value(s) of $a_1$. | [
"$a_1$ can only be $2^{2015}$.\nLet $s_n = a_n + b_n$ and $p_n = a_n b_n$ for all $n \\ge 1$. The relations become $s_{n+1} = \\frac{s_n}{2}$ and $p_{n+1} = \\sqrt{p_n}$. Inductively, we find that $s_n = \\frac{s_1}{2^{n-1}}$ and $p_n = \\sqrt[2^{n-1}]{p_1}$.\nSince $a_n$ and $b_n$ are real roots of $x^2 - s_n x + ... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2^{2015} | |
0abl | a) Find the smallest positive integer which multiplied by $2520$ gives a square of a positive integer.
b) Prove that the sum of two consecutive odd integers is divisible with $4$. | [
"a) For $2520$ we have $2520 = 2^3 \\cdot 3^2 \\cdot 5 \\cdot 7$. We notice that in order to get a square of a positive integer this number has to be multiplied with at least $2 \\cdot 5 \\cdot 7 = 70$. We obtain the product $2520 \\cdot 70 = 420^2$ and the desired number is $70$.\n\nb) We denote by $2k-1$ and $2k+... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 70; the sum of two consecutive odd integers is divisible by 4 | |
0kqm | Problem:
A sequence that starts with a positive number has the property that each of the following terms is the perimeter of the square with area equal to the preceding term. If the first three terms form an arithmetic sequence, what are the possible values for the first term of the sequence? (Having a common differenc... | [
"Solution:\nLet the first term of the sequence be $a$. If $a$ is the area of a square, then the side length of that square must be $\\sqrt{a}$, so the second term must be $4 \\sqrt{a}$. Similarly, the third term must be $4 \\sqrt{4 \\sqrt{a}} = 8 \\sqrt[4]{a}$. If these terms form an arithmetic sequence, then they ... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 16 and (sqrt(5) - 1)^4 | |
0khm | Problem:
In a $k \times k$ chessboard, a set $S$ of 25 cells that are in a $5 \times 5$ square is chosen uniformly at random. The probability that there are more black squares than white squares in $S$ is $48\%$. Find $k$. | [
"Solution:\n\nWe know that there must be fewer black squares than white squares, and $k$ must be odd. Additionally, we know that there are $k-4$ ways to pick the left column of the $5 \\times 5$ square so that the right column can fit within the $k \\times k$ grid, and $k-4$ ways to pick the top row by similar logi... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9 | |
0h41 | Positive numbers $x$ and $y$ satisfy equation $x^2 + y^2 + \frac{8xy}{x+y} = 16$. Prove that $x + y = 4$. | [
"$$(x^2 + y^2)(x + y) + 8xy = 16(x + y),$$\n$$( (x + y)^2 - 2xy )(x + y) - 16(x + y) + 8xy = 0, $$\n$$(x + y)^3 - 16(x + y) - 2xy(x + y) + 8xy = 0,$$\n$$(x + y)((x + y)^2 - 16) - 2xy(x + y - 4) = 0,$$\n$$(x + y)(x + y - 4)(x + y + 4) - 2xy(x + y - 4) = 0,$$\n$$(x + y - 4)((x + y)(x + y + 4) - 2xy) = 0.$$\n\nFor $x ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 4 | |
0gew | 設三角形 $ABC$ 中有 $AB < AC$，並設 $I_a$ 為位於 $\angle A$ 內的旁心。令 $D$ 為 $I_a$ 到 $BC$ 的投影點。設 $X$ 為 $AI_a$ 與 $BC$ 的交點，並於直線 $AC, AB$ 上分別取點 $Y, Z$ 使得 $X, Y, Z$ 落在一條與 $AI_a$ 垂直的直線上。設三角形 $AYZ$ 的外接圓與 $AI_a$ 再交於點 $U$。已知過 $A$ 並與 $ABC$ 的外接圓相切的直線交 $BC$ 於點 $T$，而線段 $TU$ 與 $ABC$ 的外接圓交於點 $V$。
證明：$\angle BAV = \angle DAC$。
Let $ABC$ be a triang... | [
"Let $\\Gamma$ be the circumcircle of $ABC$, $\\Gamma_I$ be the incircle and $\\Gamma_{I_a}$ be the A-excircle. Let $\\Gamma_1$ and $\\Gamma_2$ be the two circles that are tangent to $AB, AC$ and $\\Gamma$, where $\\Gamma_1$ is inside $\\Gamma$ and $\\Gamma_2$ is outside $\\Gamma$. Let $\\Gamma_1$ be tangent to $\\... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"G... | null | proof only | null | |
0448 | Suppose $f(x) = \cos x + \log_2 x$ ($x > 0$). If positive real number $a$ satisfies $f(a) = f(2a)$, then the value of $f(2a) - f(4a)$ is ______. | [
"By the condition, it follows that $\\cos a + \\log_2 a = \\cos 2a + \\log_2 2a = 2\\cos^2 a - 1 + 1 + \\log_2 a$, so $\\cos a = 2\\cos^2 a$. Thus, we have $\\cos a = 0$ or $\\cos a = \\frac{1}{2}$, and hence correspondingly $\\cos 2a = 2\\cos^2 a - 1 = -1$ or $\\cos 2a = -\\frac{1}{2}$. Therefore,\n$$\n\\begin{ali... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | -3 or -1 | |
003c | Una empresa aérea tiene 9 aviones todos de distintos modelos y 13 pilotos. Entrenar a cada piloto para pilotear en cada avión cuesta $1000. Cada día se sortean 9 de los pilotos para que piloteen los aviones y los otros 4 tienen el día libre.
Hallar la mínima cantidad que se debe invertir en el entrenamiento de los pil... | [] | Argentina | Argentina 2006 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Español | proof and answer | 45000 | |
0flw | Find all positive integers $n$ and $k$ such that $(n+1)^n = 2n^k + 3n + 1$ | [] | Spain | Spanija 2012 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (n, k) = (3, 3) | |
05ne | Problem:
Soit $a$, $b$, $c$, $n$ des entiers, avec $n \geq 2$. Soit $p$ un nombre premier qui divise $a^{2}+a b+b^{2}$ et $a^{n}+b^{n}+c^{n}$, mais qui ne divise pas $a+b+c$.
Prouver que $n$ et $p-1$ ne sont pas premiers entre eux. | [
"Solution:\n\nSi $p$ divise $a$ et $b$, alors $p$ divise $c^{n}$ donc $p$ divise $c$, ce qui contredit la dernière assertion. Donc, quitte à échanger $a$ et $b$, on suppose que $p$ ne divise pas $b$. Quitte à multiplier $a, b, c$ par un inverse de $b$ modulo $p$, on peut supposer que $b=1$ et donc\n$$\np\\left|a^{2... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
096d | Problem:
Comparați numerele $X=2019^{\log _{2018} 2017}$ şi $Y=2017^{\log _{2019} 2020}$. | [
"Solution:\nAvem\n$$\n\\begin{gathered}\n\\ln X=\\ln \\left(2019^{\\log _{2018} 2017}\\right)=\\log _{2018} 2017 \\cdot \\ln 2019= \\\\\n=\\frac{\\ln 2019 \\cdot \\ln 2017}{\\ln 2018}=\\log _{2018} 2019 \\cdot \\ln 2017=\\ln \\left(2017^{\\log _{2018} 2019}\\right)\n\\end{gathered}\n$$\nceea ce implică $X=2017^{\\l... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | X > Y | |
00kt | Let $n \ge 3$ be a fixed integer. The numbers $1, 2, 3, \dots, n$ are written on a board. In every move one chooses two numbers and replaces them by their arithmetic mean. This is done until only a single number remains on the board.
Determine the least integer that can be reached at the end by an appropriate sequence ... | [
"The answer is $2$ for every $n$. Surely we cannot reach an integer less than $2$, since $1$ appears only once and produces an arithmetic mean greater than $1$, as soon as it is used.\n\nOn the other hand, we can prove by induction on $k$ that the number $a+1$ can be reached from the numbers $a, a+1, \\dots, a+k$ b... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2 | |
0am4 | Problem:
Find the least common multiple of $15!$ and $2^{3} 3^{9} 5^{4} 7^{1}$.
(a) $2^{3} 3^{6} 5^{3} 7^{1} 11^{1} 13^{1}$
(b) $2^{3} 3^{6} 5^{3} 7^{1}$
(c) $2^{11} 3^{9} 5^{4} 7^{2} 11^{1} 13^{1}$
(d) $2^{11} 3^{9} 5^{4} 7^{2}$ | [] | Philippines | QUALIFYING STAGE | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | (c) | |
043p | Suppose $m$ is a real number, and complex numbers $z_1 = 1 + 2i$, $z_2 = m + 3i$, where $i$ is the imaginary unit. If $z_1 \cdot \bar{z_2}$ is purely imaginary, then the value of $|z_1 + z_2|$ is ______. | [
"Since $z_1 \\cdot \\bar{z_2} = (1 + 2i)(m - 3i) = m + 6 + (2m - 3)i$ is purely imaginary, we get $m = -6$. Therefore, $|z_1 + z_2| = |-5 + 5i| = 5\\sqrt{2}$."
] | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 5*sqrt(2) | |
0jkj | Problem:
Compute the greatest common divisor of $4^{8}-1$ and $8^{12}-1$. | [
"Solution:\n\nAnswer: 15 Let $d=\\operatorname{gcd}(a, b)$ for some $a, b \\in \\mathbb{Z}^{+}$.\nThen, we can write $d=a x-b y$, where $x, y \\in \\mathbb{Z}^{+}$, and\n$$\n\\begin{aligned}\n& 2^{a}-1 \\mid 2^{a x}-1 \\\\\n& 2^{b}-1 \\mid 2^{b y}-1\n\\end{aligned}\n$$\nMultiplying the right-hand side of (2) by $2^... | United States | HMMT November 2014 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 15 | |
02wd | Problem:
Um ladrilho, em forma de polígono regular, foi retirado do lugar que ocupava em um painel. Observou-se, então, que se esse ladrilho sofresse uma rotação de $40^{\circ}$ ou de $60^{\circ}$ em torno do seu centro, poderia ser encaixado perfeitamente no lugar que ficou vago no painel. Qual o menor número de lado... | [
"Solution:\n\nPara que seja possível efetuar tais rotações, com o encaixe do polígono, é necessário e suficiente que o ângulo central seja um divisor de $40^{\\circ}$ e $60^{\\circ}$. Se $n$ é o número de lados do ladrilho, o ângulo central é dado por $\\frac{360^{\\circ}}{n}$. Assim, as razões\n$$\n\\frac{40^{\\ci... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 18 | |
08cm | Problem:
Angela ha a disposizione i polinomi $x-1$, $(x-1)(x-2)$, $(x-1)(x-2)(x-3)$, $\ldots$ fino a $(x-1)(x-2) \cdots (x-2017)(x-2018)$, e li divide in due gruppi. Detto $p(x)$ il prodotto dei polinomi del primo gruppo e $q(x)$ quello dei polinomi del secondo gruppo, Angela si accorge che il polinomio $p(x)$ divide ... | [
"Solution:\n\nLa risposta è 1009. Osserviamo innanzitutto che Angela può assegnare i polinomi $(x-1)(x-2)$, $(x-1)(x-2)(x-3)(x-4)$, $\\cdots$, $(x-1)(x-2)\\cdots(x-2017)(x-2018)$ (ovvero quelli di grado pari) al gruppo corrispondente a $p(x)$, e gli altri al gruppo corrispondente a $q(x)$. Con questa scelta, il rap... | Italy | GARA di FEBBRAIO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1009 | |
03hx | Problem:
Let $P(x, y)$ be a polynomial in two variables $x, y$ such that $P(x, y) = P(y, x)$ for every $x, y$ (for example, the polynomial $x^{2} - 2 x y + y^{2}$ satisfies this condition). Given that $(x - y)$ is a factor of $P(x, y)$, show that $(x - y)^{2}$ is a factor of $P(x, y)$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
08bt | Problem:
Siano $a$ e $b$ due numeri reali positivi. Consideriamo un esagono regolare di lato $a$, e costruiamo sui suoi lati sei rettangoli di lati $a$ e $b$, disposti esternamente all'esagono. I dodici nuovi vertici giacciono su una circonferenza. Ripetiamo l'operazione precedente, ma scambiando fra loro i valori di ... | [
"Solution:\n\nPrima Soluzione: Consideriamo il primo esagono, e chiamiamo $O$ il suo centro, $MN$ un suo lato, e $MNPQ$ il rettangolo costruito su $MN$. Il raggio della circonferenza che passa per i vertici esterni è $OQ$. Siccome l'esagono è regolare, $OMN$ è un triangolo equilatero, e quindi $MN = OM = a$, e l'an... | Italy | XXXIII Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0dip | Find the maximum integer $k$ such that
$$
\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + k \frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} \ge 3
$$
for all positive real numbers $a, b, c$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 15 | |
0j82 | A polynomial $P(x)$ is called *nice* if $P(0) = 1$ and the nonzero coefficients of $P(x)$ alternate between $1$ and $-1$ when written in order. Suppose $P(x)$ is nice, and let $m$ and $n$ be two relatively prime positive integers. Show that
$$
Q(x) = P(x^n) \cdot \frac{(x^{mn} - 1)(x - 1)}{(x^m - 1)(x^n - 1)}
$$
is nic... | [
"$Q(x)$ is a polynomial, so $Q(x)$ is as well.\n\nWe now establish a lemma giving an alternate characterization of nice polynomials.\n\n**Lemma 2.** If $P(x)$ is a polynomial with constant term 1, then $P(x)$ is nice if and only if each nonzero term in the power series expansion of $P(x)/(1-x)$ has coefficient 1.\n... | United States | Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Generating functions",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
02ws | Problem:
Maria e Pedro jogam em um tabuleiro $9 \times 9$. Maria começa pintando de vermelho 46 quadradinhos do tabuleiro. Em seguida, Pedro deve escolher um quadrado $2 \times 2$. Se o quadrado escolhido por Pedro tem 3 ou mais casinhas pintadas de vermelho, ele vence o jogo. Caso contrário, vence Maria. Qual dos doi... | [
"Solution:\n\nPedro vence o jogo. Divida o tabuleiro em 20 tabuleiros $2 \\times 2$ como indicado na figura abaixo. Existem 5 quadradinhos do tabuleiro que não fazem parte desses tabuleiros e que estão indicados com a letra $X$. Com a pintura dos 46 quadradinhos, pelo menos $46-5=41=2 \\cdot 20+1$ serão colocados n... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Pedro | |
0grz | Find all pairs of real numbers $(x, y)$ satisfying the following conditions:
$$x^2 + y^2 + x + y = xy(x + y) - \frac{10}{27}$$
$$|xy| \le \frac{25}{9}.$$ | [
"Answer: $(x, y) = (-1/3, -1/3)$, $(5/3, 5/3)$.\nFirstly, we obtain that\n$$\nx^2 + y^2 + x + y - xy(x + y) + 2 = -(1-x)(1-y)(x+y+2)\n$$\nand hence we get\n$$\n(1-x)(1-y)(x+y+2) = \\frac{64}{27}.\n$$\nLet $k^3 = 1-x$, $\\ell^3 = 1-y$, $m^3 = x+y+2$. Then we have\n$$\nk^3 + \\ell^3 + m^3 = 4 \\quad \\text{and} \\qua... | Turkey | 26th Turkish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (x, y) = (-1/3, -1/3) and (x, y) = (5/3, 5/3) | |
025j | Problem:
João, Jorge, José e Jânio são bons amigos. Certa vez, João estava sem dinheiro, mas seus amigos tinham algum. Então Jorge deu a João um quinto de seu dinheiro, José deu um quarto de seu dinheiro e Jânio deu um terço de seu dinheiro. Se todos eles deram a mesma quantidade de dinheiro para João, que fração do d... | [
"Solution:\n\nSe $A$ é a quantidade de dinheiro que João recebeu de cada um de seus amigos, então ele recebeu um total de $3A$. Como ele recebeu, de Jorge, um quinto do seu dinheiro, então Jorge tinha $5A$. Da mesma maneira, José tinha $4A$ e Jânio tinha $3A$. Assim, os três amigos tinham, juntos, $5A + 4A + 3A = 1... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1/4 | |
077z | The numbers $1$, $2$, $3$, $4$, $\ldots$, $39$ are written on a blackboard. In one step we are allowed to choose two numbers $a$ and $b$ on the blackboard such that $a$ divides $b$, and replace $a$ and $b$ by the single number $\frac{b}{a}$. This process is continued till no number on the board divides any other number... | [] | India | IMOTC Practice Test 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 4 | |
06yr | Problem:
The reals $x, y, z$ satisfy $x \neq 1$, $y \neq 1$, $x \neq y$, and $\dfrac{y z - x^{2}}{1 - x} = \dfrac{x z - y^{2}}{1 - y}$. Show that $\dfrac{y z - x^{2}}{1 - x} = x + y + z$. | [
"Solution:\n\nWe have $y z - x^{2} - y^{2} z + y x^{2} = x z - y^{2} - x^{2} z + x y^{2}$. Hence $z (y - x - y^{2} + x^{2}) = -y^{2} + x y^{2} - x^{2} y + x^{2}$. Hence $z = \\dfrac{x + y - x y}{x + y - 1}$.\n\nSo $y z = x + y + z - x y - x z$, so $y z - x^{2} = x + y + z - x^{2} - x y - x z = (x + y + z)(1 - x)$, ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ebg | Let $a, b > 1$ be positive integers such that the number $a + b$ divides the number $D(a, b) + v(a, b)$. Here $D(a, b)$ and $v(a, b)$ denote the greatest common divisor and the least common multiple of the numbers $a$ and $b$ respectively. Prove that
$$
\frac{D(a,b) + v(a,b)}{a+b} \le \frac{a+b}{4}.
$$ | [
"Let $d = D(a, b)$. We can write $a = a_1 d$ and $b = b_1 d$, where $a_1$ and $b_1$ are co-prime. We have $v(a, b) = a_1 b_1 d$. We insert this in the condition of the problem to get\n$$\na_1 d + b_1 d \\mid d + a_1 b_1 d \\Rightarrow a_1 + b_1 \\mid 1 + a_1 b_1,\n$$\nand we can rearrange the inequality to\n$$\n\\f... | Slovenia | Selection Examinations for the IMO 2015 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
0e7q | Problem:
Naj bo $n$ naravno število. Pokaži, da je vrednost izraza
$$
\frac{\left(9^{n-1}+3^{2 n-1}\right)^{3}}{\left(3^{3 n-1}+27^{n}\right)^{2}}
$$
neodvisna od $n$. | [] | Slovenia | Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0far | Problem:
A plane intersects a sphere in a circle $C$. The points $A$ and $B$ lie on the sphere on opposite sides of the plane. The line joining $A$ to the center of the sphere is normal to the plane. Another plane $p$ intersects the segment $AB$ and meets $C$ at $P$ and $Q$. Show that $BP \cdot BQ$ is independent of t... | [
"Solution:\n\nAll points of the circle $C$ are equidistant from $A$. The plane $p$ also meets the sphere in a circle $C'$. Let $C''$ be the circle center $A$ radius $AP$. Provided that $AB$ is not a diameter of $C'$, one of the lines $BP$, $BQ$ will meet $C''$ again at some point $R$ (see diagram).\n\n![](attached_... | Soviet Union | 1st CIS | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08j0 | Problem:
The points $M$ and $N$ are the tangent points of the sides $[AB]$ and $[AC]$ of the triangle $ABC$ to the incircle with the center $I$. The internal bissectrices, drawn from the vertices $B$ and $C$, intersect the straight line $MN$ at points $P$ and $Q$ respectively. If $F$ is the intersection point of the s... | [] | JBMO | The second selection test for IMO 2003 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | null | proof only | null | |
093i | Problem:
Let $\alpha$ be a real number. Determine all polynomials $P$ with real coefficients such that
$$
P(2x+\alpha) \leq (x^{20} + x^{19}) P(x)
$$
holds for all real numbers $x$. | [
"Solution:\n\nZero polynomial obviously satisfies the problem. Further, let us suppose that polynomial $P$ is non-zero. Let $n$ be its degree and $a_n \\neq 0$ be its coefficient at $x^n$. Polynomial $(x^{20} + x^{19}) P(x) - P(2x+\\alpha)$ has degree $n+20$, coefficient $a_n$ at $x^{n+20}$ and it is non-negative f... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | P(x) ≡ 0 | |
0fr9 | Los vértices, $A$, $B$ y $C$, de un triángulo equilátero de lado $1$ están en la superficie de una esfera de radio $1$ y centro $O$. Sea $D$ la proyección ortogonal de $A$ sobre el plano, $\alpha$, determinado por $B$, $C$ y $O$. Llamamos $N$ a uno de los cortes con la esfera de la recta perpendicular a $\alpha$ por $O... | [
"Es obvio que $A$, $B$, $C$ y $O$ son vértices de un tetraedro regular de arista igual a $1$, puesto que la distancia entre dos cualesquiera de ellos es $1$. Como $D$ es la proyección ortogonal de $A$ sobre la cara opuesta del tetraedro, $D$ es el centro de la cara $BCO$. Así pues, la distancia de $D$ a $O$ (distan... | Spain | LVII Olimpiada Matemática Española Concurso Final Nacional | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordi... | null | proof and answer | 30 degrees | |
0l7g | Problem:
Let $P$ and $Q$ be points selected uniformly and independently at random inside a regular hexagon $ABCDEF$. Compute the probability that segment $\overline{PQ}$ is entirely contained in at least one of the quadrilaterals $ABCD$, $BCDE$, $CDEF$, $DEFA$, $EFAB$, or $FABC$. | [
"Solution:\n\nLet $O$ be the center of the hexagon. Without loss of generality, assume $P$ is in $\\triangle ABO$. Then, segment $PQ$ is entirely contained in one of the given quadrilaterals if and only if $Q$ is not in $\\triangle DEO$. The probability that $Q$ is in $\\triangle DEO$ is $\... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 5/6 | |
05br | Call a natural number *twistable* if it does not contain digits $3$, $4$, $7$ and its last digit is not zero. The *twisting* of a twistable number is the number obtained after the following two steps:
* Reverse the order of digits of the given number;
* Twist each digit: $0$, $1$ and $8$ remain unchanged, $2$ and $5$ a... | [
"Since the numbers $n$ and $k$ are positive integers of the same length, the quotient $\\frac{n}{k}$ must be a single-digit positive number, because multiplying by a multi-digit number increases the number of digits. The quotient $1$ is obviously possible (e.g. $\\frac{69}{69} = 1$). We show that no other quotient ... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 1 | |
0dr2 | Let $N$ be a positive integer. Prove that there exists a positive integer $n$ such that $n^{2013} - n^{20} + n^{13} - 2013$ has at least $N$ distinct prime factors. | [
"The result is true for any nonconstant polynomial $f(n) = a_m n^m + a_{m-1} n^{m-1} + \\dots + a_0$ with integer coefficients. We may assume that $a_m > 0$. Thus there exists a positive integer $n_0$ such that $f(n)$ is positive and increasing on $(n_0, \\infty)$.\n\nIt suffices to show that if for some $n_1 > n_0... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
07dq | A simple graph is called **divisibility** if it's possible to label its vertices with positive integers such that there is an edge between two vertices if and only if the label of one vertex is divisible by the other one.
A simple graph is called a **permutation** graph, if it's possible to label its vertices by $1, 2,... | [
"Let's call a simple graph **good**, if we can assign a direction to the edges such that for any two vertices $v, u$ with $v \\to u$, if there's a vertex $w$ with $u \\to w$, then there is necessarily a directed edge from $v$ to $w$.\n**Lemma.** A graph is divisibility if and only if it's a good graph.\n*Proof.*\n*... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0j0c | Problem:
Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. How many (potentially empty) subsets $T$ of $S$ are there such that, for all $x$, if $x$ is in $T$ and $2x$ is in $S$ then $2x$ is also in $T$? | [
"Solution:\n\nWe partition the elements of $S$ into the following subsets: $\\{1,2,4,8\\}$, $\\{3,6\\}$, $\\{5,10\\}$, $\\{7\\}$, $\\{9\\}$.\n\nConsider the first subset, $\\{1,2,4,8\\}$. Say $2$ is an element of $T$. Because $2 \\cdot 2 = 4$ is in $S$, $4$ must also be in $T$. Furthermore, since $4 \\cdot 2 = 8$ i... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 180 | |
02wf | Problem:
A tabela abaixo mostra alguns dos resultados do último Festival de Pesca de Pirajuba, exibindo quantos competidores $q$ pescaram $n$ peixes para alguns valores de $n$.
| $n$ | 0 | 1 | 2 | 3 | $\ldots$ | 13 | 14 | 15 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $q$ | 9 | 5 | 7 ... | [
"Solution:\nSejam $N$ e $P$, os números de competidores e peixes pescados no evento, respectivamente. Analisando os competidores que pescaram 3 ou mais peixes, temos:\n$$\nN-9-5-7=N-21 \\text{ pescadores e } P-0 \\cdot 9-1 \\cdot 5-2 \\cdot 7=P-19 \\text{ peixes, }\n$$\nsendo a média escrita como:\n$$\n\\begin{alig... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) 943; b) 123 | |
0jk0 | Problem:
Evaluate the infinite sum
$$
\sum_{n=2}^{\infty} \log_{2}\left(\frac{1-\frac{1}{n}}{1-\frac{1}{n+1}}\right)
$$ | [
"Solution:\nAnswer: $\\quad -1$\n\nUsing the identity $\\log_{2}\\left(\\frac{a}{b}\\right) = \\log_{2} a - \\log_{2} b$, the sum becomes\n$$\n\\sum_{n=2}^{\\infty} \\log_{2}\\left(\\frac{n-1}{n}\\right) - \\sum_{n=2}^{\\infty} \\log_{2}\\left(\\frac{n}{n+1}\\right)\n$$\nMost of the terms cancel out, except the $\\... | United States | HMMT November 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | -1 | |
0dey | Let $A, B, C, D$ be points on the line $d$ in that order and $AB = CD$. Denote $(P)$ as some circle that passes through $A, B$ with its tangent lines at $A, B$ are $a, b$. Denote $(Q)$ as some circle that passes through $C, D$ with its tangent lines at $C, D$ are $c, d$. Suppose that $a$ cuts $c, d$ at $K, L$ respectiv... | [
"Consider the points that arranged as following figure, the other cases will be proved similarly.\nDenote $R, S$ as intersection of the pairs of lines $PB, QC$ and $PA, QD$. Note that $BMCR$ and $ALDS$ are cyclic quadrilateral. Thus\n$$\n\\begin{aligned}\n\\angle KMN &= \\angle BRC = 180^\\circ - (\\angle BCR + \\a... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cz6 | Find all pairs $(a, b)$ of nonnegative integers $(a, b)$ such that
$$
a + 2b - b^{2} = \sqrt{2a + a^{2} + |2a + 1 - 2b|}
$$ | [
"It is clear that $2a + 1 - 2b$ is an odd integer, hence\n$$\n|2a + 1 - 2b| \\geq 1\n$$\nIf $(a, b)$ is a pair of nonnegative integers satisfying\n$$\na + 2b - b^{2} = \\sqrt{2a + a^{2} + |2a + 1 - 2b|}\n$$\nthen we get\n$$\na + 2b - b^{2} \\geq \\sqrt{2a + a^{2} + 1} = \\sqrt{(a + 1)^{2}} = a + 1\n$$\nTherefore $0... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | [(0,1), (1,1)] | |
02jv | Problem:
A maior raiz da equação $(x-37)^2-169=0$ é:
A) 39
B) 43
C) 47
D) 50
E) 53 | [
"Solution:\nSolução 1: Usando a fatoração $a^2-b^2=(a-b)(a+b)$ : $(x-37)^2-13^2=0 \\Leftrightarrow (x-37-13)(x-37+13)=0 \\Leftrightarrow (x-50)(x-24)=0$.\nLogo, as raízes são 24 e 50.\n\n\nSolução 2: Extraindo a raiz quadrada em ambos os lados:\n$(x-37)^2=13^2 \\Leftrightarrow x-37=13$ ou $x-37=-13$. Assim, $x=50$ ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
04ax | Let $a$, $b > 1$ be relatively prime positive integers. Define a sequence
$$
x_1 = a, \quad x_2 = b, \quad x_n = \frac{x_{n-1}^2 + x_{n-2}^2}{x_{n-1} + x_{n-2}} \quad \text{for } n \ge 3.
$$
Prove that $x_n$ is not an integer for $n \ge 3$. (Tonći Kokan) | [
"Notice that $x_n > 1$, for all $n \\in \\mathbb{N}$. We also notice that all $x_n$ are rational so we can write $x_n = \\frac{p_n}{q_n}$, where $p_n$ and $q_n$ are positive integers and $M(p_n, q_n) = 1$.\n\nFirst let us prove that $p_n$ and $p_{n+1}$ are relatively prime for every $n \\in \\mathbb{N}$. We will pr... | Croatia | Hrvatska 2011 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
09da | $\forall x, y \in \mathbb{R}$-ийн хувьд
$$
f(x + f(y)) = f(x) + \frac{1}{8} x f(4y) + f(f(y))
$$
байх бүх $f : \mathbb{R} \to \mathbb{R}$ функцийг ол. | [
"$\\forall x \\in \\mathbb{R}: f(x) = 0$ илэрхий хариу. $\\exists t \\in \\mathbb{R}: f(t) \\neq 0$.\n$(x, y) \\to (0, 0)$ гэе. $f(0) = 0$ гэж гарна. $(x, y) \\to (f(x), f(t))$ гэе.\n$$\nf(f(x) + f(t)) = f(f(x)) + \\frac{1}{8}f(x)f(4t) + f(f(t))\n$$\n$(x, y) \\to (f(t), f(x))$ гэе.\n$$\nf(f(x) + f(t)) = f(f(x)) + \... | Mongolia | ММО-48 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | Mongolian | proof and answer | f(x) = 0 or f(x) = x^2 | |
0k4c | Problem:
A bug is on a corner of a cube. A healthy path for the bug is a path along the edges of the cube that starts and ends where the bug is located, uses no edge multiple times, and uses at most two of the edges adjacent to any particular face. Find the number of healthy paths. | [
"Solution:\nThere are $6$ symmetric ways to choose the first two edges on the path. After these are chosen, all subsequent edges are determined, until the starting corner is reached once again."
] | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 6 | |
08ud | Find the average value of all those integers $n$ satisfying $0 \le n \le 10000$ for which the digit $1$ does not appear in their decimal expansions. | [
"Call a non-negative integer less than or equal to $10000$ for which the digit $1$ does not appear in its decimal expansion a good integer.\nLet us denote by $A$ the average value of all good integers. Since $10000$ is not a good integer, it suffices to consider only good integers of $4$ or less digits. Write a goo... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 48884/9 | |
07ct | $ABC$ is a triangle with circumcircle $\omega_1$ and $\widehat{C} = 2\widehat{B}$. A tangent line to $\omega_1$ at $A$ intersects $BC$ at $E$. Let $\omega_2$ be a circle passing through $B$ and tangent to $AC$ at $C$. This circle intersects $AB$ for the second time at $F$. A line through $E$ is tangent to $\omega_2$ at... | [
"Considering the power of point $E$ with respect to circles $\\omega_1$, $\\omega_2$ we have\n$$\n\\begin{cases}\nEC \\cdot EB = EA^2 \\\\\nEC \\cdot EB = EK^2\n\\end{cases} \\implies EA = EK.\n$$\n\n\n\nNow we claim that the angle bisectors of angles $\\widehat{BAC}$ and $\\widehat{BKC}$ i... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bkw | Find all positive primes $p$ and $q$ such that $p^3 + p = q^2 + q$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (p, q) = (3, 5) | |
0j9h | Problem:
In a game of rock-paper-scissors with $n$ people, the following rules are used to determine a champion:
a. In a round, each person who has not been eliminated randomly chooses one of rock, paper, or scissors to play.
b. If at least one person plays rock, at least one person plays paper, and at least one per... | [
"Solution:\n\nAnswer: $\\frac{45}{14}$\n\nFor each positive integer $n$, let $E_{n}$ denote the expected number of rounds required to determine a winner among $n$ people. Clearly, $E_{1}=0$. When $n=2$, on the first move, there is a $\\frac{1}{3}$ probability that there is a tie, and a $\\frac{2}{3}$ probability th... | United States | HMMT November 2012 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 45/14 | |
0jq3 | Problem:
For a convex quadrilateral $P$, let $D$ denote the sum of the lengths of its diagonals and let $S$ denote its perimeter. Determine, with proof, all possible values of $\frac{S}{D}$. | [
"Solution:\n\nAnswer: $1 < \\frac{S}{D} < 2$\n\nSuppose we have a convex quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at $E$ (convexity is equivalent to having $E$ on the interiors of segments $AC$ and $BD$).\n\nTo prove the lower bound, note that by the triangle inequality, $AB + BC > AC$ and $AD... | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | (1, 2) | |
0azz | Problem:
Let $\triangle ABC$ be a right triangle with right angle at $B$. Let the points $D$, $E$, and $F$ be on $AB$, $BC$, and $CA$, respectively, such that $\triangle DEF$ is an equilateral triangle and $EC = FC$. If $DB = 5\sqrt{3}$, $BE = 3$, and $\sin \angle ACB = 4\sqrt{3}/7$, find the perimeter of $\triangle A... | [
"Solution:\n\nBy Pythagorean Theorem, $DE = \\sqrt{(5\\sqrt{3})^{2} + 3^{2}} = \\sqrt{84} = EF$.\n\nFrom $\\sin \\angle ACB = 4\\sqrt{3}/7$, we have $\\cos \\angle ACB = 1/7$.\n\nLet $EC = FC = x$, then by Cosine Law on side $EF$ of $\\triangle ECF$, we have\n$$\nEF^{2} = x^{2} + x^{2} - 2x^{2} \\cos \\angle ACB.\n... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 35√3 + 63 + 2√21 | |
004d | Se tiene en el plano una circunferencia $\Gamma$ de radio $1$. En un punto a distancia $2006$ del centro de $\Gamma$ se encuentra un grillo. Este grillo quiere entrar en $\Gamma$ mediante saltos que satisfacen la siguiente condición: Si $G$ y $G'$ son las posiciones del grillo antes y después de un salto, entonces la m... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | Español | proof and answer | 1003 | |
0k4q | Problem:
Rachel has the number $1000$ in her hands. When she puts the number $x$ in her left pocket, the number changes to $x+1$. When she puts the number $x$ in her right pocket, the number changes to $x^{-1}$. Each minute, she flips a fair coin. If it lands heads, she puts the number into her left pocket, and if it ... | [
"Solution:\n\nCall a real number very large if $x \\in [1000, 1008]$, very small if $x \\in \\left[0, \\frac{1}{1000}\\right]$, and medium-sized if $x \\in \\left[\\frac{1}{8}, 8\\right]$. Every number Rachel is ever holding after at most $8$ steps will fall under one of these categories. Therefore the main contrib... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 13 | |
0g8q | 在 $\triangle ABC$ 中, 設點 $D$ 在 $BC$ 邊上且 $AD$ 平分 $\angle BAC$, 並設 $AD$ 的中點為 $M$。設以 $AC$ 為直徑的圓 $\omega_1$ 與 $BM$ 交於點 $E$, 以 $AB$ 為直徑的圓 $\omega_2$ 與 $CM$ 交於點 $F$。證明 $B, E, F, C$ 四點共圓。
Let $M$ be the midpoint of the internal bisector $AD$ of $\triangle ABC$. Circle $\omega_1$ with diameter $AC$ intersects $BM$ at $E$ and c... | [
"If $AB = AC$, then the statement is obvious. Without loss of generality, we assume that $AB < AC$. Let $AH$ be the common chord of the given circles, as shown in the figure. We draw the line which passes through $A$ and is perpendicular to $AD$ and denote by $K$ and $L$ the points of intersection of the given line... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02bh | Problem:
Para comemorar seu aniversário, Ana vai preparar tortas de pera e tortas de maçã. No mercado, uma maçã pesa $300~\mathrm{g}$ e uma pera $200~\mathrm{g}$. A sacola de Ana aguenta um peso máximo de $7~\mathrm{kg}$. Qual é o número máximo de frutas que ela pode comprar para poder fazer tortas das duas frutas? | [
"Solution:\n\nDenotemos por $m$ o número de maçãs e $p$ o número de peras que Ana comprou, assim o peso que ela leva na sacola é $300 m + 200 p$ gramas. Como a sacola aguenta no máximo $7000$ gramas, temos que\n$$\n300 m + 200 p \\leq 7000, \\text{ que é equivalente a } 3 m + 2 p \\leq 70\n$$\nComo as peras pesam m... | Brazil | Nível 2 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 34 | |
02dr | Show that it is possible to color each point of a circle red or blue so that no right-angled triangle inscribed in the circle has its vertices all the same color. | [
"Use any coloring with all pairs at opposite ends of a diameter having opposite colors. For example, take $AB$ to be a diameter. Color $A$ red, $B$ blue, all points in the interior of one arc $AB$ red and all points in the interior of the other blue."
] | Brazil | V OBM | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
01pv | Find the smallest real number $x$ such that the inequality $x + c \le (x + a)(x + b)$ holds for any triangle, where $a \le b \le c$ are the sides of the triangle. | [
"**Answer:** $x = 1$.\n\nFirst, we prove that if $a$, $b$, $c$ are the sides of a triangle, then the inequality\n$$\nx + c \\le (x + a)(x + b) \\quad (*)\n$$\nholds for $x = 1$. Indeed, we can rewrite (*) as $x + c \\le x^2 + (a + b)x + ab$. It is easy to see that this inequality holds for $x = 1$ since $x^2 = x = ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1 | |
001y | Dada una circunferencia $C$ y un punto $P$ exterior a ella, se trazan por $P$ las dos tangentes a la circunferencia, siendo $A$ y $B$ los puntos de tangencia.
Se toma un punto $Q$ sobre el arco menor $AB$ de $C$. Sea $M$ la intersección de la recta $AQ$ con la perpendicular a $AQ$ trazada por $P$ y sea $N$ la intersecc... | [] | Argentina | 15ª Olimpiada Matemática del Cono Sur | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscell... | español | proof only | null | |
0jik | Let $\triangle ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $\widehat{BC}$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\... | [
"Let $H$ denote the orthocenter of $\\triangle ABC$. We claim that $\\ell$ always passes through $H$.\n\nLemma. Line $PQ$ bisects segment $XH$.\n\nProof. Let $X_A, X_B$ be the reflections of $X$ across $BC$ and $AC$ respectively, and let $H_A$ be the reflection of $H$ across $BC$. It is easy to see that since $\\an... | United States | IMO Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bqo | Problem:
Fie $(A,+, \cdot)$ un inel (unitar) comutativ și $U(A)$ mulțimea elementelor inversabile ale inelului.
a) Dacă $x \in A$ și $x^{2}=0$ să se arate că $1+x \in U(A)$.
b) Fie $x \in A$ pentru care există $n \in \mathbb{N}^*$ astfel încât $x^{n}=0$ și fie $u \in U(A)$. Să se arate că $u+x \in U(A)$. | [
"Solution:\n\na.\nDacă $x \\in A$ și $x^2 = 0$, atunci $(1 + x) \\cdot (1 - x) = 1 - x^2 = 1$, deci $1 + x$ este inversabil, cu inversul $1 - x$.\n\nb.\nFie $x \\in A$ cu $x^n = 0$ pentru un $n \\in \\mathbb{N}^*$ și $u \\in U(A)$. Atunci $u$ este inversabil, deci există $u^{-1} \\in A$ cu $u u^{-1} = 1$.\n\nObserv... | Romania | Olimpiada Natională de Matematică | [
"Algebra > Abstract Algebra > Other"
] | null | proof only | null | |
07wd | Prove that any finite sum of terms
$$
\frac{1}{abc(a + b + c + 1)}
$$
where $a, b, c$ are positive integers, is smaller than 6. | [
"We will use the following notation:\n$$\nH_i = \\sum_{k=1}^{i} \\frac{1}{k} \\qquad P_i(m) = \\sum_{k=m}^{N} \\frac{H_{k+i}}{k(k+1)}.\n$$\n**Lemma 1.** For all $i \\ge 1$ and all $1 \\le M \\le N$ we have\n$$\n\\sum_{k=M}^{N} \\frac{1}{k(k+i)} < \\frac{1}{i} \\sum_{k=M}^{M+i-1} \\frac{1}{k} .\n$$\n*Proof.* Using $... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation"
] | English | proof only | null | |
0igv | Problem:
If $a, b, c > 0$, what is the smallest possible value of $\left\lfloor\frac{a+b}{c}\right\rfloor + \left\lfloor\frac{b+c}{a}\right\rfloor + \left\lfloor\frac{c+a}{b}\right\rfloor$? (Note that $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.) | [
"Solution:\nSince $\\lfloor x \\rfloor > x - 1$ for all $x$, we have that\n$$\n\\begin{aligned}\n\\left\\lfloor\\frac{a+b}{c}\\right\\rfloor + \\left\\lfloor\\frac{b+c}{a}\\right\\rfloor + \\left\\lfloor\\frac{c+a}{b}\\right\\rfloor &> \\frac{a+b}{c} + \\frac{b+c}{a} + \\frac{c+a}{b} - 3 \\\\\n&= \\left(\\frac{a}{b... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4 | |
0b74 | Given an integer number $n \ge 3$, consider $n$ distinct points on a circle, labeled 1 through $n$. Determine the maximum number of closed chords $[ij]$, $i \neq j$, having pairwise non-empty intersections. | [
"We shall prove that any such configuration contains at most $n$ chords and the upper bound is achieved, so the required maximum is $n$.\n\nTo this end, fix an orientation of the circle and relabel the points $1$ through $n$ in the corresponding circular order. Consider a configuration of chords $[ij]$, $i \\neq j$... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n | |
06lz | Does there exist a nonzero polynomial $P(x)$ with integer coefficients satisfying both of the following conditions?
* $P(x)$ has no rational root;
* For every positive integer $n$, there exists an integer $m$ such that $n$ divides $P(m)$. | [
"Yes. We shall prove that $P(x) = (x^2+1)(x^2-2)(x^2+2)(x^2+7)$ satisfies all properties. It is not hard to see that $P$ has no rational root. It remains to show it satisfies the second condition.\n\nConsider any odd prime $p$. Using facts about quadratic residues, we have the following.\n* If $p \\equiv 1 \\pmod 4... | Hong Kong | Year 2021 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Algebra > Algebraic Expressions > Polynomi... | null | proof and answer | P(x) = (x^2+1)(x^2-2)(x^2+2)(x^2+7) | |
0k4z | Problem:
Let $ABC$ be an equilateral triangle with side length $8$. Let $X$ be on side $AB$ so that $AX = 5$ and $Y$ be on side $AC$ so that $AY = 3$. Let $Z$ be on side $BC$ so that $AZ$, $BY$, $CX$ are concurrent. Let $ZX$, $ZY$ intersect the circumcircle of $AXY$ again at $P$, $Q$ respectively. Let $XQ$ and $YP$ in... | [
"Solution:\n\nLet $BY$ and $CX$ meet at $O$. $O$ is on the circumcircle of $AXY$, since $\\triangle AXC \\cong \\triangle CYB$.\n\nWe claim that $KA$ and $KO$ are tangent to the circumcircle of $AXY$. Let $XY$ and $BC$ meet at $L$. Then, $LBZC$ is harmonic. A perspectivity at $X$ gives $AYOP$ is harmonic. Similarly... | United States | HMMT February | [
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 304 | |
0429 | Given positive integers $m$ and $n$. Find the smallest integer $N$ ($\ge m$) with the following property: if an $N$-element set of integers contains a complete residue system modulo $m$, then it has a nonempty subset such that the sum of its elements is divisible by $n$. | [
"$$\nN = \\max\\{m, m+n - \\frac{1}{2}m[(m, n) + 1]\\}.\n$$\n\nFirst we show that $N \\ge \\max\\{m, m+n - \\frac{1}{2}m[(m,n)+1]\\}$.\nLet $d = (m,n)$, and write $m = d m_1$, $n = d n_1$. If $n > \\frac{1}{2}m(d+1)$, there exists a complete residue system modulo $m$, $x_1, x_2, \\dots, x_m$, such that their residu... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | N = \max\{\, m,\; m + n - \tfrac{1}{2} m\big(\gcd(m,n) + 1\big) \,\}. | |
09ug | Ahmed, Babeth, Casper, Daan, Emine, and Freek are sitting in a row, in this order. Ahmed and Babeth both write a positive integer on a piece of paper. Then Casper adds the numbers on the papers of Ahmed and Babeth and writes the result on his piece of paper. Afterwards, Daan adds the numbers on the papers of Babeth and... | [
"B) 76"
] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | MCQ | B | |
0dxr | Problem:
Krožnici $\mathcal{K}_1$ in $\mathcal{K}_2$ s središčema $O_1$ in $O_2$ se sekata v točkah $A$ in $B$. Razdalja med središčema je večja od polmerov obeh krožnic. Naj bosta $C_1$ in $C_2$ tisti presečišči premice $O_1 O_2$ s krožnicama $\mathcal{K}_1$ in $\mathcal{K}_2$, ki ne ležita na daljici $O_1 O_2$. Ozna... | [
"Solution:\n\nKer je štirikotnik $A O_1 B O_2$ tetiven in leži središče njemu očrtane krožnice na simetrali tetive $A B$, torej na daljici $O_1 O_2$, po Talesovem izreku sledi $\\angle O_1 A O_2 = \\angle O_2 B O_1 = \\frac{\\pi}{2}$.\n\n\n\nNaj bo $\\angle A O_2 O_1 = \\alpha$. Potem je $\... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jzi | Problem:
A positive integer $n$ is magical if
$$
\lfloor\sqrt{\lceil\sqrt{n}\rceil}\rfloor=\lceil\sqrt{\lfloor\sqrt{n}\rfloor}\rceil,
$$
where $\lfloor\cdot\rfloor$ and $\lceil\cdot\rceil$ represent the floor and ceiling function respectively. Find the number of magical integers between 1 and 10,000, inclusive. | [
"Solution:\nFirst of all, we have $\\lfloor\\sqrt{n}\\rfloor=\\lceil\\sqrt{n}\\rceil$ when $n$ is a perfect square and $\\lfloor\\sqrt{n}\\rfloor=\\lceil\\sqrt{n}\\rceil-1$ otherwise. Therefore, in the first case, the original equation holds if and only if $\\sqrt{n}$ is a perfect square itself, i.e., $n$ is a four... | United States | HMMT November 2017 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1330 | |
0f87 | Problem:
$n^2$ real numbers are written in a square $n \times n$ table so that the sum of the numbers in each row and column equals zero. A move is to add a row to one column and subtract it from another (so if the entries are $a_{ij}$ and we select row $i$, column $h$ and column $k$, then column $h$ becomes $a_{1h} +... | [] | Soviet Union | 22nd ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0glt | Find all positive integers $n$ such that $4n! - 4n + 1$ is a perfect square. | [
"For $n \\le 6$, we can easily check that $n = 1, 2, 4$ are the only positive integers satisfying the condition.\n\nWe will prove that there are no other positive integers which satisfy the condition. Assume to the contrary that there exists a positive integer $n > 6$ that satisfies the condition.\nLet $4n! - 4n + ... | Thailand | T3MO 2017 | [
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"... | English | proof and answer | 1, 2, 4 | |
0f81 | Problem:
$A$, $B$, $C$ are the angles of a triangle. Show that
$$
2\frac{\sin A}{A} + 2\frac{\sin B}{B} + 2\frac{\sin C}{C} \leq \left(\frac{1}{B} + \frac{1}{C}\right) \sin A + \left(\frac{1}{C} + \frac{1}{A}\right) \sin B + \left(\frac{1}{A} + \frac{1}{B}\right) \sin C.
$$ | [
"Solution:\n\nAssume $A \\leq B \\leq C$. Then $\\sin A \\leq \\sin B$. Also $A \\leq C < 180^{\\circ} - A$, so $\\sin A \\leq \\sin C$. Similarly $\\sin B \\leq \\sin C$. Hence $(1/A - 1/B)(\\sin B - \\sin A)$, $(1/B - 1/C)(\\sin C - \\sin B)$ and $(1/C - 1/A)(\\sin A - \\sin C)$ are all non-negative. Hence their ... | Soviet Union | 22nd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.