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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0590 | A fly farm contains $100001$ fruit flies. A research group wishes to buy some flies from the farm to perform (once) either experiment A or experiment B (but not both).
For experiment A, the research group needs a set of flies in which no fly is a descendant of any other. For such a set, the research group would pay th... | [
"For $k = 1, 2, \\dots$, we call a fly a $k$-th order parent, if the greatest suitable set for experiment B containing this fly as the oldest fly consists of exactly $k$ flies.\n\nIf there exists a suitable set for experiment B with at least $201$ flies, then the research group would pay at least $1000 + 10 \\cdot ... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
05tl | Problem:
Déterminer tous les triplets $(x, y, z)$ d'entiers naturels satisfaisant l'équation:
$$
2^{x} + 3^{y} = z^{2}
$$ | [
"Solution:\n\nTout d'abord, analysons le problème : on a une équation diophantienne avec une puissance de $2$, une puissance de $3$ et un carré. On s'empresse donc de tester les petites valeurs de $z$ et de trouver la solution. On peut remarquer que $2^{0} + 3^{1} = 2^{2}$, $2^{3} + 3^{0} = 3^{2}$ et $2^{4} + 3^{2}... | France | Envoi 5: Pot Pourri | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | [[0, 1, 2], [3, 0, 3], [4, 2, 5]] | |
00et | Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden mean $\tau=\frac{1+\sqrt{5}}{2}$. Here, an integral power of $\tau$ is of the form $\tau^{i}$, where $i$ is an integer (not necessarily positive). | [
"We will prove this statement by induction using the equality\n$$\n\\tau^{2}=\\tau+1\n$$\nIf $n=1$, then $1=\\tau^{0}$. Suppose that $n-1$ can be written as a finite sum of integral powers of $\\tau$, say\n$$\n\\begin{equation*}\nn-1=\\sum_{i=-k}^{k} a_{i} \\tau^{i} \\tag{1}\n\\end{equation*}\n$$\nwhere $a_{i} \\in... | Asia Pacific Mathematics Olympiad (APMO) | null | [
"Number Theory > Algebraic Number Theory > Algebraic numbers",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0j86 | Problem:
Let $n$ be a positive integer such that $n > 2$. Prove that $\varphi(n)$ is even. | [
"Solution:\n\nLet $A_{n}$ be the set of all positive integers $x \\leq n$ such that $\\operatorname{gcd}(n, x) = 1$. Since $\\operatorname{gcd}(n, x) = \\operatorname{gcd}(n, n - x)$ for all $x$, if $a$ is a positive integer in $A_{n}$, so is $n - a$. Moreover, if $a$ is in $A_{n}$, $a$ and $n - a$ are different si... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
03bp | Let $m$, $n$ and $r$ be positive integers with $n \ge 2$ and $1 \le r \le n-1$. Consider a square table of size $(mn+r) \times (mn+r)$. The table is covered by squares of size $n \times n$ with sides parallel to the sides of the table. Each unit square is covered at least once and some unit squares are covered multiple... | [
"We call a unit square bad if it is covered more than once.\n\n(Bound) Choose an arbitrary row and mark its cells in the columns $r+1$, $r+n+1$, $\\dots$, $r+(m-1)n+1$. Consider all $n \\times n$ squares having nonempty intersection with the chosen row. Since $mn < mn+r < (m+1)n$, we have at least $m+1$ such square... | Bulgaria | Team selection test for the 54th IMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (mn+mr+2r)(n-r) | |
0law | Given a sequence of real numbers ($x_n$):
$$
x_1 = 1 \quad \text{and} \quad x_n = \frac{2n}{(n-1)^2} \sum_{i=1}^{n-1} x_i \quad \text{for all } n \ge 2.
$$
For each positive integer $n$, let $y_n = x_{n+1} - x_n$.
Show that the sequence ($y_n$) has finite limit as $n \to +\infty$. | [
"For all $n \\ge 1$, we have\n$$\nx_{n+1} = \\frac{2(n+1)}{n^2} \\cdot \\sum_{i=1}^{n} x_i = \\frac{2(n+1)}{n^2} \\left( \\frac{(n-1)^2}{2n} + 1 \\right) x_n = \\frac{(n+1)(n^2+1)}{n^3} x_n.\n$$\nConsequently $\\frac{x_{n+1}}{n+1} = \\left(1 + \\frac{1}{n^2}\\right) \\cdot \\frac{x_n}{n} \\quad \\forall n \\ge 1$.\... | Vietnam | Vijetnam 2011 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
029k | Problem:
Jorge ganhou 15 roseiras para seu jardim, com a condição de plantá-las em 6 filas de 5 roseiras cada uma. Isso é possível? Em caso afirmativo faça um desenho indicando para Jorge como plantar as roseiras. | [
"Solution:\n\nÉ possível plantar as roseiras em 6 filas de 5 roseiras cada uma, conforme mostra o desenho a seguir.\n\n"
] | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Yes. Take six straight lines with no two parallel and no three concurrent; plant at the fifteen pairwise intersection points so that each of the six lines forms a row containing five plants. | |
0feu | Problem:
En el sótano del castillo, 7 gnomos guardan su tesoro. El tesoro está detrás de 12 puertas, cada una de ellas con 12 cerraduras. Todas las cerraduras son distintas. Cada gnomo tiene llaves para algunas de las cerraduras. Tres gnomos cualesquiera tienen conjuntamente llaves para todas las cerraduras. Probar que... | [
"Solution:\nDebe haber 4 gnomos de modo que cada uno de ellos tenga al menos 48 llaves (de lo contrario se podrían elegir 3 de ellos de manera que tuvieran conjuntamente menos de $3 \\cdot 48=144$ llaves y por lo tanto no podrían abrir las cerraduras, contra lo supuesto). Los otros 3 gnomos tienen conjuntamente al ... | Spain | TANDA II | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cie | Let $ABC$ be a triangle with $AB = 2AC$. Let $D$ be the midpoint of side $AB$. A point $E$ is chosen on side $BC$ such that $\angle BDE = \angle ACB$.
a) Show that the ray ($EA$ bisects angle $\angle CED$.
b) Let $O$ be the intersection of lines $CD$ and $AE$. Show that $\angle COA = \angle ACB$. | [] | Romania | 75th NMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
071o | Problem:
For any natural number $n$ ($n \geq 3$), let $f(n)$ denote the number of non-congruent integer-sided triangles with perimeter $n$ (e.g., $f(3)=1$, $f(4)=0$, $f(7)=2$). Show that
a. $f(1999) > f(1996)$
b. $f(2000) = f(1997)$. | [
"Solution:\n\na. Let $a, b, c$ be the sides of a triangle with $a + b + c = 1996$, and each being a positive integer. Then $a + 1, b + 1, c + 1$ are also sides of a triangle with perimeter $1999$ because\n$$\na < b + c \\quad \\Longrightarrow \\quad a + 1 < (b + 1) + (c + 1)\n$$\nand so on. Moreover, $(999, 999, 1)... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
084e | Problem:
Su una scacchiera $75 \times 75$ le righe e le colonne sono numerate da 1 a 75. Chiara vuole mettere una pedina in tutte e sole le caselle che abbiano una coordinata pari e l'altra multipla di 3. Quante pedine disporrà in tutto sulla scacchiera? | [
"Solution:\n\nLa risposta è 1706. Infatti, le caselle $(x, y)$ in cui c'è una pedina possono essere divise in due gruppi:\n\na. $x$ multiplo di 2 e $y$ multiplo di 3;\nb. $x$ multiplo di 3 e $y$ multiplo di 2.\n\nPoiché ci sono sulla scacchiera $\\left[\\frac{75}{2}\\right]$ colonne con ascissa multipla di 2 e $\\l... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 1706 | |
01u9 | Point $D$ is marked on the side $AB$ of triangle $ABC$. The bisectors of the angles $ABC$ and $ADC$ meet at point $U$, and the bisectors of the angles $BAC$ and $BDC$ meet at point $V$. Let $S$ be the midpoint of the segment $UV$.
Prove that the lines $SD$ and $AB$ are perpendicular if and only if the inscribed circle... | [
"Note that the triangle $UDV$ is a right-angled triangle regardless of the position of the point $D$ since\n$$\n\\angle UDV = \\angle UDC' + \\angle VDC' = \\frac{1}{2} \\angle ADC + \\frac{1}{2} \\angle BDC = \\frac{1}{2} \\cdot 180^\\circ = 90^\\circ.\n$$\nSo $S$ is the center of the circumcircle of the triangle ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
00kp | Let $x, y$ be positive real numbers with $xy = 4$.
Prove that
$$
\frac{1}{x+3} + \frac{1}{y+3} \le \frac{2}{5}
$$
For which $x$ and $y$ does equality hold? | [
"Clearing denominators, we obtain the equivalent inequality\n$$\n5x + 5y + 30 \\le 2xy + 6x + 6y + 18,\n$$\nwhich simplifies to $x+y \\ge 12-2xy = 4$. This inequality is a direct consequence of the AM-GM inequality\n$$\n\\frac{x+y}{2} \\ge \\sqrt{xy} = 2.\n$$\nEquality holds exactly for $x = y = 2$."
] | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | x = y = 2 | |
03z3 | Suppose positive real numbers $a, b, c, d$ satisfy $abcd = 1$. Prove
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{a+b+c+d} \ge \frac{25}{4}.
$$ | [
"Solution 1.\nFirst, we will prove that, whenever there are two numbers among $a, b, c, d$ that are equal, the inequality holds. We may assume that $a = b$ and let $s = a + b + c + d$. Then we have\n$$\n\\begin{aligned}\n& \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\\\\n&= \\fra... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
03tp | Suppose real number $a$ satisfies $|2x - a| + |3x - 2a| \ge a^2$ for any $x \in \mathbb{R}$. Then $a$ lies exactly in ( ).
(A) $\left[ -\frac{1}{3}, \frac{1}{3} \right]$
(B) $\left[ -\frac{1}{2}, \frac{1}{2} \right]$
(C) $\left[ -\frac{1}{4}, \frac{1}{3} \right]$
(D) $[-3, 3]$ | [
"Let $x = \\frac{2}{3}a$. Then we have $|a| \\le \\frac{1}{3}$. Therefore (B) and (D) are excluded. By symmetry, (C) is also excluded. Then only (A) can be correct.\n\nIn general, for any $k \\in \\mathbb{R}$, let $x = \\frac{1}{2}ka$. Then the original inequality becomes\n$$\n|a| \\cdot |k-1| + \\frac{3}{2} |a| \\... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | MCQ | A | |
0e73 | Lara and Sara will draw $n$ straight lines on a rectangular sheet of paper, one after the other, in turn, and one line each time. Every line will be parallel to one of the edges of the sheet and will go from edge to edge. No line will coincide with an edge or an already drawn line. At the end, the sheet of paper will b... | [
"If $n$ is odd, the winner is Sara, no matter who starts. If $n$ is even, the winner is the one who does not start. Suppose at the end there are $p$ vertical and $r$ horizontal lines on the sheet of paper, where $p + r = n$. Then the sheet of paper is divided into $(p+1)(r+1)$ rectangles. If $n$ is odd, one of the ... | Slovenia | National Math Olympiad 2012 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | If the number of lines is odd, Sara wins regardless of who starts. If the number of lines is even, the player who does not start (the second player) wins. | |
0698 | Find all prime numbers $p$ for which there exist non-negative integers $x$, $y$ and $z$ such that the number
$$
x^p + y^p + z^p - x - y - z
$$
is a product of exactly three distinct prime numbers. | [] | Greece | 23rd Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p = 2 or p = 5 | |
0g0v | Problem:
Sei $n$ eine natürliche Zahl. Zeige, dass $7^{7^{n}}+1$ mindestens $2 n+3$ nicht notwendigerweise verschiedene Primteiler hat. | [
"Solution:\n\nWir lösen das Problem per Induktion. Die Verankerung ist der Fall $n=0$, bei dem $7^{7^{0}}+1=7^{1}+1=2^{3}$ gilt.\n\nFür den Induktionsschritt gilt es zu zeigen, dass wenn $x=7^{2 m-1}$ für eine natürliche Zahl $m$ gilt, dass $\\left(x^{7}+1\\right) /(x+1)$ nicht prim ist. Daraus folgt, dass $x^{7}+1... | Switzerland | IMO-Selektion | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ef5 | Problem:
Marko in France imata vsak svoj sadovnjak z jablanami in hruškami. France ima $20 \%$ manj jablan, a 7 hrušk več kot Marko. Skupaj imata 218 dreves. Naslednje leto namerava France na novo posaditi še 22 jablan in 24 hrušk, da bo imel jablan 1,5-krat toliko kot hrušk. Koliko ima letos vsak od njiju jablan in k... | [
"Solution:\n\nNaj ima Marko $x$ jablan in $y$ hrušk. Potem ima France $0,8 x$ jablan in $y+7$ hrušk. Skupaj imata $1,8 x+2 y+7=218$ dreves, dobimo enačbo $1,8 x+2 y=211$. Naslednje leto bo imel France $0,8 x+22$ jablan in $y+31$ hrušk. Ker je jablan 1,5-krat več kot hrušk, dobimo enačbo $0,8 x+22=1,5(y+31)$. Preobl... | Slovenia | 17. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Marko: 85 apples and 29 pears; France: 68 apples and 36 pears. | |
0gbv | 已知正整數 $n$ 的所有因數除了 1 以外都不是完全平方數。
試證: 不存在任何互質的正整數 $x$ 和 $y$, 使得 $x^n + y^n$ 是 $(x + y)^3$ 的倍數。 | [
"設存在互質的正整數 $x, y$ 使得 $x^n + y^n$ 是 $(x+y)^3$ 的倍數。令 $s = x+y$，則 $s > 2$。\n\n(1) 若 $n$ 為偶數，因 $x^n + y^n = x^n + (s-x)^n = 2x^n \\pmod s$。但同一時間，$x^n + y^n$ 是 $(x+y)^3$ 的倍數，$x^n + y^n = 0 \\pmod s$，故 $2x^n = 0 \\pmod s$。但因 $x$ 和 $y$ 互質，故 $x$ 和 $s$ 互質，因此必有 $2 = 0 \\pmod s$，也就是 $x = y = 1$，矛盾！\n\n(2) 若 $n$ 為奇數，因 $x^n + y... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 獨立研究(三) | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09cc | $$
\sum_{j=0}^{n} \left\{ \binom{3n+2-j}{j} 2^j - \binom{3n+1-j}{j-1} 2^{j-1} \right\} = 2^{3n}
$$
гэж батал.
$$
\text{Энд } \binom{n}{k} = C_n^k \text{ ба } j = 0 \text{ үед } \binom{3n+1-j}{j-1} 2^{j-1} = 0 \text{ гэж үзнэ.}
$$ | [
"$f$ дахь $x^j$-ийн коэффициентийг $ko(f, x^j)$ гэвэл нийлбэр дахь $j$ дугаар илэрхийлэл нь\n$$\nA = ko[(1 + 2x)^{3n+2-j}, x^j] - ko[(1 + 2x)^{3n+1-j}, x^{j-1}]\n$$\nболно.\n$$\n\\begin{aligned}\nA &= ko\\{x^{3n+2-j}[(1 + 2x)^{3n+2-j} - x(1 + 2x)^{3n+1-j}], x^{3n+2}\\} = \\\\\n&= ko\\left[(x + x^2)(x + 2x^2)^{3n+1-... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Mongolian | proof only | null | |
06h1 | Given $\triangle ABC$ with $CA > BC > AB$, let $O$ and $H$ be the circumcentre and orthocentre of $\triangle ABC$ respectively. Denote by $D$ and $E$ the midpoints of arcs $\widehat{AB}$ and $\widehat{AC}$ of the circumcircle of $\triangle ABC$ not containing the opposite vertices. Let $D'$ be the reflection of $D$ in ... | [
"We claim that both statements are equivalent to $\\angle A = 60^\\circ$.\n\nFirstly,\n$A, D', E'$ are collinear\n$$\n\\Leftrightarrow \\angle BAD' + \\angle CAE' = \\angle BAC\n$$\n$$\n\\Leftrightarrow \\angle BAD + \\angle CAE = \\angle BAC\n$$\n$$\n\\Leftrightarrow \\frac{C}{2} + \\frac{B}{2} = A\n$$\n$$\n\\Left... | 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 > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ij0 | Problem:
Let $a \geq b \geq c$ be real numbers such that
$$
\begin{aligned}
a^{2} b c + a b^{2} c + a b c^{2} + 8 &= a + b + c \\
a^{2} b + a^{2} c + b^{2} c + b^{2} a + c^{2} a + c^{2} b + 3 a b c &= -4 \\
a^{2} b^{2} c + a b^{2} c^{2} + a^{2} b c^{2} &= 2 + a b + b c + c a
\end{aligned}
$$
If $a + b + c > 0$, then co... | [
"Solution:\nWe factor the first and third givens, obtaining the system\n$$\n\\begin{aligned}\na^{2} b c + a b^{2} c + a b c^{2} - a - b - c = (a b c - 1)(a + b + c) &= -8 \\\\\na^{2} b + a^{2} c + b^{2} c + b^{2} a + c^{2} a + c^{2} b + 3 a b c = (a b + b c + c a)(a + b + c) &= -4 \\\\\na^{2} b^{2} c + a b^{2} c^{2... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1279 | |
02jz | Problem:
André treina para a maratona dando voltas em torno de uma pista circular de raio $100~\mathrm{m}$. Para percorrer aproximadamente $42~\mathrm{km}$, o número de voltas que André precisa dar está entre:
A) $1$ e $10$
B) $10$ e $50$
C) $50$ e $100$
D) $100$ e $500$
E) $500$ e $1000$ | [
"Solution:\n\nO comprimento de uma circunferência de raio $r$ é $2 \\pi r$. Assim, em cada volta, André percorre $2 \\pi \\times 100~\\mathrm{m} = 200 \\pi~\\mathrm{m}$. Logo, o número de voltas que André precisa dar é $\\frac{42000}{200 \\pi} = \\frac{210}{\\pi}$.\n\nPodemos agora finalizar o problema de duas mane... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles"
] | null | MCQ | C | |
09hc | Do there exist positive integers $a_1, a_2, \dots, a_{2017}$ such that the product
$$
(a_1^{2017} + a_2)(a_2^{2017} + a_3) \dots (a_{2016}^{2017} + a_{2017})(a_{2017}^{2017} + a_1)
$$
is a power of a prime with exponent
a) $2017 \cdot 2018,$
b) $2017 \cdot 2023.$ | [
"Assume that there are positive integers $a_1, a_2, \\dots, a_{2017}$ and a prime $p$ as required. Then, for each $i$, there is a positive integer $k_i$ so that\n$$\na_i^{2017} + a_{i+1} = p^{k_i}. \\qquad (*)\n$$\nHere $a_{2018} = a_1$. The sum of all $p^{k_i}$ equals to the sum of all $a_i^{2017} + a_i$ which is ... | Mongolia | Mongolian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | a) No; b) No | |
002s | Sea $n$ un entero positivo tal que hay $k$ divisores positivos de $n$, $k > 1$, cuya suma es un número primo. Demostrar que el producto de esos $k$ divisores es menor o igual que $n^{k-1}$. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Español | proof only | null | |
0a4r | Problem:
An equilateral triangle is partitioned into smaller equilateral triangular pieces. Prove that two of the pieces are the same size. | [
"Solution:\n\nFor the purpose of this proof, we will consider a vertex to be any point which is a corner of at least one of the triangular pieces. Define an edge to be any line segment between two vertices, which is part of a side of a triangular piece but does not pass through any other vertex. Note that each vert... | New Zealand | New Zealand Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0djq | Each cell of a $4 \times 4$ table contains one of the numbers $1$ or $2$. For every row we calculate the sum of its entries and for every column we calculate the product of its entries. Can the eight obtained results be all different? | [
"But both $1$ and $16$ can appear only as products, which forces a column of $1$'s and a column of $2$'s. And this leaves only three possible row sums, contradiction. $\\square$"
] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | No | |
049x | Bisector of an acute angle in the right triangle divides the opposite side into parts of lengths $4$ and $5$. What is the area of that triangle? | [] | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 54 | |
00cu | Para todo número entero positivo $n$, sea $S(n)$ la suma de los dígitos de $n$. Hallar, si existe, un número entero positivo $n$ de 171 dígitos tal que 7 divide a $S(n)$ y 7 divide a $S(n+1)$. | [
"Sí, existe. Hay muchos ejemplos, damos uno. Observamos que $n$ debe terminar en $9$, pues si no, $S(n+1)=S(n)+1$ y $S(n)$ es coprimo con $S(n+1)$. Consideramos $n=11\\dots199\\dots9$, el número que comienza con $b$ unos y termina en $a$ nueves, donde $a+b=171$. Luego $n+1=11\\dots1200\\dots0$.\n\nTenemos que $S(n)... | Argentina | Nacional OMA | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | Spanish | proof and answer | A valid example is the 171-digit number consisting of 90 ones followed by 81 nines. | |
05zn | Problem:
La suite $(a_{n})$ est définie par $a_{1}=1$ et
$$
a_{n} = \frac{1}{n} + \frac{1}{a_{1} \cdot \ldots \cdot a_{n-1}}
$$
Montrer que pour tout entier $m \geqslant 3$, on a $a_{m} \leqslant 1$. | [
"Solution:\n\nOn pose $P_{n} = \\prod^{n} a_{k}$. On a\n$$\n\\frac{P_{n}}{P_{n-1}} = \\frac{1}{n} + \\frac{1}{P_{n-1}}\n$$\nce qui donne\n$$\nn P_{n} = P_{n-1} + n\n$$\nOn pose alors $M_{n} = n! P_{n}$. On déduit\n$$\nM_{n} = M_{n-1} + n!\n$$\nOn déduit que $M_{n} = \\sum k!$. On déduit alors\n$$\na_{n} = \\frac{P_... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
039t | Let $M$ and $N$ be the midpoints of the sides $AC$ and $BC$ of $\triangle ABC$ ($AC > BC$) and let the bisector of $\angle B$ intersect the segment $MN$ at a point $P$. The incircle of $\triangle ABC$ has center $I$ and is tangent to $BC$ at a point $Q$. Denote by $R$ the intersection point of the perpendiculars from $... | [
"a) Obviously\n$$\n\\vDash ABP = \\vDash BPN = \\vDash PBN = \\beta/2.\n$$\nTherefore $BN = CN = PN$, whence $\\vDash BPC = 90^\\circ$. Since $\\vDash CQI = 90^\\circ$, the quadrilateral *PCQI* is cyclic.\n\n\n\nb) Answer. $BS = \\frac{b+c}{2}$."
] | Bulgaria | Fall Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof and answer | BS = (b+c)/2 | |
07cc | For two points $A = (x_1, y_1)$ and $B = (x_2, y_2)$ in $\mathbb{R}^2$, their distance can be defined as,
$$
d(A, B) = |x_1 - x_2| + |y_1 - y_2|.
$$
a) The perpendicular bisector of two points in the plane is the locus of points equally spaced from those two points. Determine the perpendicular bisector of two arbitrar... | [
"a.\nAssume $A = (x_a, y_a)$ and $B = (x_b, y_b)$ are two arbitrary points in the plane. The new perpendicular bisector of $A$ and $B$ comprises points $(x, y)$ satisfying\n$$\n|x - x_b| + |y - y_b| = |x - x_a| + |y - y_a|.\n$$\nDraw two lines from each of the points $A$ and $B$ parallel to the coordinate axes to c... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory... | null | proof and answer | a) The perpendicular bisector is the locus |x − x_b| + |y − y_b| = |x − x_a| + |y − y_a|; geometrically it is a piecewise linear union of segments whose configuration depends on the slope of AB (when slope is ±1 it includes the mid-segment of the axis-parallel rectangle formed from A and B; otherwise it is a different ... | |
04ra | The number $0$ is written on each of the $n+1$ faces of an $n$-sided pyramid. In a step we choose a vertex and we increase by $1$ each number on the faces, which contain the vertex. Show, that in such way, we cannot get number $1$ written on each face.
(Peter Novotný) | [
"Let $b$ be the sum of numbers on side faces of the pyramid, let $a$ be the number on the base. After a step involving any base vertex, $b$ increases or decreases by $2$ and $a$ increases or decreases by $1$, that means the value $V = b - 2a$ stays the same. If we choose for a step the apex, only $b$ increases or d... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0apd | Problem:
If
$$
\left\{\begin{array}{l}
w x y=10 \\
w y z=5 \\
w x z=45 \\
x y z=12
\end{array}\right.
$$
what is $w + y$? | [
"Solution:\n$\\frac{19}{6}$\nWe multiply the four given equations.\n$$\n\\begin{gathered}\n(w x y)(w y z)(w x z)(x y z) = 10 \\cdot 5 \\cdot 45 \\cdot 12 \\\\\n(w x y z)^3 = 2^3 3^3 5^3 \\\\\nw x y z = 2 \\cdot 3 \\cdot 5 = 30 \\\\\nw = \\frac{w x y z}{x y z} = \\frac{30}{12} = \\frac{5}{2}, \\quad y = \\frac{w x y... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 19/6 | |
056g | Teacher drew a pentagon on the blackboard. The following conditions hold for the pentagon.
a) Two of the pentagon's interior angles are equal.
b) There exist three interior angles such that the first one equals the sum of the other two.
c) There exist four interior angles such that one of them equals the sum of the ... | [
"Let the sizes of the angles of the pentagon be denoted in decreasing order as $\\alpha \\ge \\beta \\ge \\gamma \\ge \\delta \\ge \\varepsilon$. The sum of all the interior angles is $(5-2) \\cdot 180^\\circ$, in other words $\\alpha + \\beta + \\gamma + \\delta + \\varepsilon = 540^\\circ$.\n\nThe angle that equa... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 270°, 135°, 67.5°, 33.75°, 33.75° | |
0alb | Problem:
$ABCD$ is a trapezoid with $AB \parallel CD$, $AB = 6$ and $CD = 15$. If the area of $\triangle AED = 30$, what is the area of $\triangle AEB$?
(a) 20
(b) $40/7$
(c) 12
(d) $8/3$ | [] | Philippines | QUALIFYING STAGE | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | c | |
076c | Problem:
Find all real functions $f$ from $\mathbb{R} \rightarrow \mathbb{R}$ satisfying the relation
$$
f\left(x^{2}+y f(x)\right)=x f(x+y)
$$ | [
"Solution:\nPut $x=0$ and we get $f(y f(0))=0$. If $f(0) \\neq 0$, then $y f(0)$ takes all real values when $y$ varies over real line. We get $f(x) \\equiv 0$. Suppose $f(0)=0$. Taking $y=-x$, we get $f\\left(x^{2}-x f(x)\\right)=0$ for all real $x$.\n\nSuppose there exists $x_{0} \\neq 0$ in $\\mathbb{R}$ such tha... | India | INMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x, or f(x) = x for all real x | |
0joq | Problem:
Let $ABC$ be a triangle that satisfies $AB = 13$, $BC = 14$, $AC = 15$. Given a point $P$ in the plane, let $P_A$, $P_B$, $P_C$ be the reflections of $A$, $B$, $C$ across $P$. Call $P$ good if the circumcircle of $P_A P_B P_C$ intersects the circumcircle of $ABC$ at exactly 1 point. The locus of good points $... | [
"Solution:\n\nBy the properties of reflection, the circumradius of $P_A P_B P_C$ equals the circumradius of $ABC$. Therefore, the circumcircle of $P_A P_B P_C$ must be externally tangent to the circumcircle of $ABC$. Now it's easy to see that the midpoint of the 2 centers of $ABC$ and $P_A P_B P_C$ lies on the circ... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
... | null | proof and answer | 4225/64 * pi | |
081q | Problem:
Quante sono le terne di interi $(a, b, c)$ tutti maggiori di $1$ tali che $a^{b^{c}} < 2002$? | [
"Solution:\n\nDistinguiamo 2 casi:\n\n- se $a=2$ si deve avere $b^{c}<11$ perché $2^{11}=2048>2002$, quindi $b<4$ e se $b=2$ allora $c$ può essere $2$ o $3$, mentre se $b=3$ allora $c=2$: si ottengono così le soluzioni $(2,2,2)$, $(2,2,3)$, $(2,3,2)$;\n\n- se $a>2$ si deve avere $b^{c}<7$ perché $a^{7} \\geq 3^{7}=... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 7 | |
02er | At a party every woman dances with at least one man, and no man dances with every woman. Show that there are men $M$ and $M'$ and women $W$ and $W'$ such that $M$ dances with $W$, $M'$ dances with $W'$, but $M$ does not dance with $W'$, and $M'$ does not dance with $W$. | [
"Let $M$ be one of the men who dance with the maximal number of women, $W'$ one of the women he doesn't dance with, and $M'$ one of the men $W'$ dances with. If $M'$ were to dance with every woman that $M$ dances with, then the maximality of the number of women that $M$ dances with would be contradicted, so there i... | Brazil | XIII OBM | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
06x6 | Let $ABC$ be an acute-angled triangle with $AC > AB$, let $O$ be its circumcentre, and let $D$ be a point on the segment $BC$. The line through $D$ perpendicular to $BC$ intersects the lines $AO$, $AC$ and $AB$ at $W$, $X$ and $Y$, respectively. The circumcircles of triangles $AXY$ and $ABC$ intersect again at $Z \neq ... | [
"Let $AO$ intersect $BC$ at $E$. As $EDW$ is a right-angled triangle and $O$ is on $WE$, the condition $OW = OD$ means $O$ is the circumcentre of this triangle. So $OD = OE$ which establishes that $D, E$ are reflections in the perpendicular bisector of $BC$.\n\nNow observe:\n$$\n180^{\\circ} - \\angle DXZ = \\angle... | IMO | International 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 > Transformations > Spira... | English | proof only | null | |
03f4 | a) Find all values of $a$ for which the inequality
$$
x \log_{\frac{1}{2}} a^4 - x^2 > 3 + 2 \log_2 a^2
$$
has a solution.
b) Calculate the limit
$$
\lim_{a \to -\infty} \left( \sqrt{a^2 - a + 1} + a \right).
$$ | [
"a) Since $\\log_{\\frac{1}{2}}(a^4) = -2 \\cdot \\log_2(a^2)$, then by putting $2\\log_2(a^2) = b$, we get the inequality $x^2 + b \\cdot x + 3 + b < 0$. For this inequality to have at least one solution, it is necessary and sufficient that $D = b^2 - 4b - 12 > 0$ whose solutions are $b < -2$ or $b > 6$, whence $\... | Bulgaria | Bulgarian Spring Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | a ∈ (-∞, -2√2) ∪ (-√2/2, 0) ∪ (0, √2/2) ∪ (2√2, ∞); limit = 1/2 | |
0eao | Problem:
Cena vstopnice za ogled nogometne tekme je bila $9$ evrov. Zaradi nezanimanja za nakup vstopnic so ceno znižali. Število kupcev se je povečalo za $50\%$, izkupiček od prodanih vstopnic pa za $25\%$. Koliko je bila cena vstopnice po znižanju? Za koliko odstotkov je bila nova cena vstopnice nižja od prvotne cen... | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 7.5 euros; decreased by 16 2/3% | |
06ak | Solve in the positive real numbers the system:
$$
\begin{cases} \frac{xyz+1}{x+1} = \frac{yzw+1}{y+1} = \frac{zwx+1}{z+1} = \frac{wxy+1}{w+1} \\ x+y+z+w=48 \end{cases} .$$ | [
"1. (First solution) The given equations can be written as:\n$$\nxy^2z + xyz + y = xyzw + yzw + x\n$$\n$$\nz^2yw + yzw + z = xyzw + y + zwx\n$$\n$$\nw^2xz + zwx + w = xyzw + wxy + z.\n$$\n$$\nx^2wy + wxy + x = xyzw + xyz + w.\n$$\nSumming up the four equations we get:\n$$\nxy^2z + x^2wy + yz^2w + w^2xy = 4xyzw \\Le... | Greece | 40th Hellenic Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (12, 12, 12, 12); (12 + sqrt(143), 12 - sqrt(143), 12 + sqrt(143), 12 - sqrt(143)); (12 - sqrt(143), 12 + sqrt(143), 12 - sqrt(143), 12 + sqrt(143)) | |
0b84 | Prove that there is a similarity between a triangle $ABC$ and the triangle having as sides the medians of the triangle $ABC$ if and only if the squares of the lengths of the sides of triangle $ABC$ form an arithmetical sequence. | [
"Recall that $m_a^2 = \\frac{1}{4}(2(b^2 + c^2) - a^2)$, together with the other similar formulas.\n\nAssume that the squares of the lengths of the sides are in arithmetic progression, for example $2b^2 = a^2 + c^2$. Then $m_a^2 = \\frac{3}{4}c^2$, $m_b^2 = \\frac{3}{4}b^2$, $m_c^2 = \\frac{3}{4}a^2$, implying that... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0g60 | 試求最小的正整數 $n$, 使得存在有理係數多項式 $f_1, f_2, \dots, f_n$, 滿足
$$
x^2 + 7 = f_1(x)^2 + f_2(x)^2 + \dots + f_n(x)^2.
$$ | [
"我們有 $x^2 + 7 = x^2 + 2^2 + 1^2 + 1^2 + 1^2$。下面我們證明 $n = 4$ 是不可能的（從而 $n$ 更小也不可能，因為我們可以取 $f_i = 0$）。\n\n用反證法，假設 $x^2 + 7 = f_1(x)^2 + f_2(x)^2 + f_3(x)^2 + f_4(x)^2$ 都是有理係數的多項式。顯然 $f_i$ 必須都是一次的，記 $f_i = a_i x + b_i$，$i = 1, 2, 3, 4$。我們有\n$$\n\\sum_{i=1}^{4} a_{i}^{2} = 1, \\quad \\sum_{i=1}^{4} a_{i}b_{i} = 0, \\qua... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 5 | |
0juf | Problem:
A rectangular pool table has vertices at $(0,0)$, $(12,0)$, $(0,10)$, and $(12,10)$. There are pockets only in the four corners. A ball is hit from $(0,0)$ along the line $y=x$ and bounces off several walls before eventually entering a pocket. Find the number of walls that the ball bounces off of before enter... | [
"Solution:\n\nConsider the tiling of the plane with the $12 \\times 10$ rectangle to form a grid. Then the reflection of the ball off a wall is equivalent to traveling along the straight line $y=x$ into another $12 \\times 10$ rectangle. Hence we want to find the number of walls of the grid that the line $y=x$ hits... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 9 | |
0bm7 | Two circles $\gamma$ and $\gamma'$ cross one another at points $A$ and $B$. The tangent to $\gamma'$ at $A$ meets $\gamma$ again at $C$, the tangent to $\gamma$ at $A$ meets $\gamma'$ again at $C'$, and the line $CC'$ separates the points $A$ and $B$. Let $\Gamma$ be the circle externally tangent to $\gamma$, externall... | [
"Invert with respect to a circle centred at $A$ and denote by $X^*$ the image of a point $X \\neq A$ under this inversion. The circles $\\gamma$ and $\\gamma'$ invert into straight lines $B^*C^*$ and $B^*C'^*$, and the tangents at $A$ into lines through $A$, parallel to $B^*C^*$ and $B^*C'^*$. The line $CC'$ invert... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0aut | Problem:
The vertices of a triangle are at the points $(0,0)$, $(a, b)$, and $(2016-2a, 0)$, where $a>0$. If $(a, b)$ is on the line $y=4x$, find the value(s) of $a$ that maximizes the triangle's area. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = 504 | |
09vr | Let $a_1, a_2, \dots, a_{2020}$ be real numbers, not necessarily distinct. For all $n \ge 2020$, let $a_{n+1}$ be the minimal real root of the polynomial
$$
P_n(x) = x^{2n} + a_1 x^{2n-2} + a_2 x^{2n-4} + \dots + a_{n-1} x^2 + a_n,
$$
if it exists. Assume that $a_{n+1}$ exists for all $n \ge 2020$. Prove that $a_{n+1} ... | [
"If $x = \\alpha$ is a root of $P_n$, then $x = -\\alpha$ is a root of $P_n$ as well, as all terms of $P_n$ have even degree. The minimal root of $P_n$ therefore cannot be positive. Therefore $a_n \\le 0$ for all $n > 2020$.\n\nWe have $P_{n+1}(x) = x^2 \\cdot P_n(x) + a_{n+1}$. Substitute $x = a_{n+1}$; as that is... | Netherlands | IMO Team Selection Test 2, June 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
0925 | Problem:
Let $K$ be a point inside an acute triangle $A B C$, such that $B C$ is a common tangent of the circumcircles of $A K B$ and $A K C$. Let $D$ be the intersection of the lines $C K$ and $A B$, and let $E$ be the intersection of the lines $B K$ and $A C$. Let $F$ be the intersection of the line $B C$ and the pe... | [
"Solution:\n\nThe line $B C$ is tangent to the circumcircle of $A K C$, so the angles $B C D$ and $C A K$ are equal. Analogously, the angles $C B E$ and $B A K$ are equal.\n\n\n\nTherefore\n$$\n\\pi = |\\angle K B C| + |\\angle K C B| + |\\angle B K C| = |\\angle D A K| + |\\angle E A K| + ... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ibi | Problem:
A given line passes through the center $O$ of a circle. The line intersects the circle at points $A$ and $B$. Point $P$ lies in the exterior of the circle and does not lie on the line $A B$. Using only an unmarked straightedge, construct a line through $P$, perpendicular to the line $A B$. Give complete instr... | [
"Solution:\n\n1. Draw a line from $P$ to $A$, intersecting the circle at $C$.\n2. Draw a line from $P$ to $B$, intersecting the circle at $D$.\n3. Draw lines $A D$ and $B C$, and let $E$ be their point of intersection.\n4. Draw a line from $P$ through $E$; this will be the desired perpendicular line.\n\nThis works ... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0am1 | Problem:
Find the exact value of $\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$. | [
"Solution:\n\nLet us denote the expression as $E$:\n\n$$\nE = \\frac{\\sqrt{3}}{\\sin 20^{\\circ}} - \\frac{1}{\\cos 20^{\\circ}}\n$$\n\nLet us write both terms with a common denominator:\n\n$$\nE = \\frac{\\sqrt{3} \\cos 20^{\\circ} - 1 \\cdot \\sin 20^{\\circ}}{\\sin 20^{\\circ} \\cos 20^{\\circ}}\n$$\n\nRecall t... | Philippines | Area Stage | [
"Precalculus > Trigonometric functions"
] | null | final answer only | 4 | |
0677 | Find all values of the integer $n$ for which the number $A = \frac{8n-25}{n+5}$ is equal with the cube of a rational number.
(A. Fellouris) | [
"Let $p, q \\in \\mathbb{Z}$, $q \\neq 0$, with $(p, q) = 1$ such that\n$$\nA = \\frac{8n-25}{n+5} = \\left(\\frac{p}{q}\\right)^3. \\qquad (1)\n$$\nThen $(p^3, q^3) = 1$, while from relation (1) we get:\n$$\nq^3(8n-25) = p^3(n+5), \\qquad (2)\n$$\nfrom which we conclude that\n$$\np^3 \\mid (8n-25) \\text{ and } q^... | Greece | 31st Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | n = 3 | |
0fev | Problem:
Determinar todos los enteros $n$ tales que
$$
\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}
$$
es entero. | [
"Solution:\nLlamemos\n$$\np=\\sqrt{\\frac{25}{2}+\\sqrt{\\frac{625}{4}-n}}+\\sqrt{\\frac{25}{2}-\\sqrt{\\frac{625}{4}-n}}=\\sqrt{25+2 \\sqrt{n}}\n$$\n(la última igualdad se comprueba elevando al cuadrado la expresión dada a la que se llama $p$ ). Entonces\n$$\nn=\\left(\\frac{p^{2}-25}{2}\\right)^{2}\n$$\ny $p$ es ... | Spain | TANDA II | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 0, 144 | |
0901 | Let $N$ denote the set of positive integers. Find all functions $f : N \to N$ such that the equation
$$
\operatorname{lcm}(m, f(m + f(n))) = \operatorname{lcm}(f(m), f(m) + n)
$$
holds for any positive integers $m$ and $n$. Here, for positive integers $x$ and $y$, $\operatorname{lcm}(x, y)$ denotes their least common m... | [
"We prove that the function $f(n) = n$ is the unique function satisfying the condition in the problem. It is easy to see that this $f$ satisfies the condition in the problem.\n\nSuppose that $f$ is the function that satisfies the condition in the problem. First, for any positive integer $k$, we prove that $f(k)$ is... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | f(n) = n for all positive integers n | |
05jg | Problem:
Le point $D$ appartient au côté $[A C]$ du triangle équilatéral $A B C$. On note $F$ le projeté orthogonal de $D$ sur $(B C)$, puis $K$ le projeté orthogonal de $F$ sur $(A B)$, et enfin $E$ le projeté orthogonal de $K$ sur $(C A)$. Soit $L$ le milieu de $[BC]$, et $P$ l'intersection des droites $(K E)$ et $(... | [
"Solution:\n\n\n\nPremière solution. Comme les côtés de $PKF$ sont perpendiculaires aux côtés de $ABC$, le triangle $PKF$ a les mêmes angles intérieurs que $A B C$ et est donc équilatéral, et donc $P F = K F$.\n\nNotons $M$ l'intersection de $(BP)$ et $(A L)$. Comme les triangles $BLM$ et $... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04ru | We are given a triangle $ABC$. Find the locus of points $X$ in the plane $ABC$ whose reflections through the lines $AB$, $BC$, $CA$ are vertices of an equilateral triangle. | [
"For any point $X$ of the plane $ABC$, let $X_a, X_b$ and $X_c$ denote the reflections of $X$ through the lines $BC$, $CA$ and $AB$, respectively (Fig. 5). First we prove that the distances between any two of the points $X_a, X_b$ and $X_c$ are given in general by formulæ\n$$\n|X_a X_b| = 2|XC| \\sin \\gamma, \\qua... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof and answer | The locus consists of the common intersection points of the three Apollonius circles
k_a: |XB|/|XC| = |AB|/|AC|,
k_b: |XA|/|XC| = |AB|/|BC|,
k_c: |XA|/|XB| = |AC|/|BC|,
which equivalently are the points X satisfying 2|XA|sin α = 2|XB|sin β = 2|XC|sin γ > 0. The number of solutions depends on the triangle: one solution ... | |
0d6i | Let $ABC$ be a triangle whose incircle $(I)$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. The line passing through $A$ and parallel to $BC$ cuts $DE$, $DF$ at $M$, $N$, respectively. The circumcircle of triangle $DMN$ cuts $(I)$ again at $L$.
1. Let $K$ be the intersection of $NE$ and $MF$. Prove that $K$ ... | [
"\n\n1) Because $MN \\parallel BC$ so $\\angle ANF = \\angle FDB = \\angle DFB = \\angle AFN$, this deduces $AN = AF$. Similarly, $AM = AE = AF = AN$.\n\nThus, $M$, $N$, $E$, $F$ lie on the circle center $A$. Let $MF$ cut $NE$ at $K$, because $E$, $F$ lie on circle diameter $MN$ so\n$$\n\\a... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jfs | Problem:
Let $ABCD$ be a quadrilateral inscribed in a unit circle with center $O$. Suppose that $\angle AOB = \angle COD = 135^{\circ}$, $BC = 1$. Let $B'$ and $C'$ be the reflections of $A$ across $BO$ and $CO$ respectively. Let $H_1$ and $H_2$ be the orthocenters of $AB'C'$ and $BCD$, respectively. If $M$ is the mid... | [
"Solution:\n\nLet $B$ have coordinate $b$ and $C$ have coordinate $c$. We obtain easily that $B'$ is $b^2$, $C'$ is $c^2$, and $D$ is $bc$. Therefore, $H_1$ is $1 + b^2 + c^2$ and $H_2$ is $b + c + bc$ (we have used the fact that for triangles on the unit circle, their orthocenter is the sum of the vertices).\n\nFi... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | final answer only | (8 - sqrt(6) - 3*sqrt(2)) / 4 | |
0i9x | At the vertices of a regular hexagon are written six nonnegative integers whose sum is $2003$. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Be... | [
"Define the *sum* and *maximum* of a position to be the sum and maximum of the six numbers at the vertices. We will show that from any position in which the sum is odd, it is possible to reach the all-zero position.\nOur strategy alternates between two steps:\n\na) from a position with odd sum, move to a position w... | United States | USA IMO 2003 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0bhj | Find all real solutions of the system
$$
\begin{cases}
|x + y + z| = 1 \\
|x| + |y| + |z| = 1 \\
|x - y| + |y - z| + |z - x| = 2.
\end{cases}
$$ | [
"Let us denote $a = x$, $b = y$, $c = z$ for simplicity.\n\nFrom $|a| + |b| + |c| = 1$, the numbers $a$, $b$, $c$ are all real and their absolute values sum to $1$.\n\nLet us consider all possible sign combinations for $a$, $b$, $c$.\n\nLet us suppose $a$, $b$, $c$ are all non-negative. Then $a + b + c = 1$ and $|a... | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | All permutations of (1, 0, 0) and (-1, 0, 0). | |
0bmd | Let $n \in \mathbb{N}$, $n \ge 4$. Determine the sets $A = \{a_1, a_2, \dots, a_n\} \subset \mathbb{N}$ that contain $2015$ and for which $|a_i - a_j|$ is a prime, for all distinct $i, j \in \{1, 2, \dots, n\}$. | [
"It is easy to see that $A$ can not contain more than two numbers of any parity. Combined with $n \\ge 4$, this forces $A$ to have exactly $4$ elements, two of each parity. The difference between the two even (odd) numbers must be $2$, therefore we can have two types of sets: $\\{2013, 2015, 2k, 2k + 2\\}$ and $\\{... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | {2008, 2010, 2013, 2015}; {2010, 2012, 2015, 2017}; {2013, 2015, 2018, 2020}; {2015, 2017, 2020, 2022} | |
0gfz | 設 $k$ 為非負整數。有一堆硬幣共有 $2^{2^k}$ 枚，其中恰有一枚是偽幣。我們有無止盡的警犬，其中有一隻生病了，但我們不知道是哪一隻。檢測硬幣的測試包含以下三個步驟：從全部的硬幣中選一些出來；挑一隻警犬；讓這隻警犬同時嗅所選出來的硬幣。當選出的硬幣中有偽幣的話，健康的警犬嗅過之後一定會叫；反之一定不叫。而生病的警犬叫或不叫是隨機的。請設計一個最多進行 $2^k + k + 2$ 個測試的程序以選出該偽幣，並證明該程序可行。 | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 國際競賽實作(一) | [
"Discrete Mathematics > Algorithms",
"Algebra > Linear Algebra > Matrices"
] | Chinese; English | proof only | null | |
06pl | Let $ABC$ be a fixed triangle, and let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $BC, CA, AB$, respectively. Let $P$ be a variable point on the circumcircle. Let lines $PA_{1}, PB_{1}, PC_{1}$ meet the circumcircle again at $A', B', C'$ respectively. Assume that the points $A, B, C, A', B', C'$ are distinct, and ... | [
"Let $A_{0}, B_{0}, C_{0}$ be the points of intersection of the lines $AA', BB'$ and $CC'$ (see Figure). We claim that area $\\left(A_{0}B_{0}C_{0}\\right)=\\frac{1}{2} \\operatorname{area}(ABC)$, hence it is constant.\n\nConsider the inscribed hexagon $ABCC'PA'$. By Pascal's theorem, the points of intersection of ... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
07j1 | Given an odd positive integer $n$. Determine the smallest positive integer $k$ such that it would be possible to fill all the cells of a $3 \times k$ table with non-negative integers such that
i. The sum of the entries of each column be $n$;
ii. Each of numbers $0, 1, \dots, n$ must appear at least once in each row. | [
"We shall claim that the answer would be $\\frac{3(n+1)}{2}$. For this reason, we shall firstly provide the following example\n\n| 0 | 0 | ... | 0 | 0 |\n|---|---|-----|---|---|\n| 0 | 1 | ... | t - 1 | t |\n| 2t - 1 | 2t | ... | t + 2 | t + 1 |\n| 0 | 1 | ... | t - 1 | t |\n| 2t + 1 | 2t | ... | t + 2 | t + 1 |\n|... | Iran | 41th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3(n+1)/2 | |
0azg | Problem:
Let $P(x)$ be the polynomial of minimal degree such that $P(k) = \dfrac{720 k}{k^{2} - 1}$ for $k \in \{2, 3, 4, 5\}$. Find the value of $P(6)$. | [
"Solution:\n\nLet $Q(x) = (x^{2} - 1) P(x) - 720 x$. Then $Q(k) = 0$ for $k \\in \\{2, 3, 4, 5\\}$ so $Q(x) = R(x)(x - 2)(x - 3)(x - 4)(x - 5)$ for some polynomial $R(x)$. Observe that $Q(1) = -720$ and $Q(-1) = 720$, so $R(x)$ cannot be constant. As the degree of $Q$ is as small as possible, we set $R(x) = a x + b... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 48 | |
0d0r | Prove that for every positive real numbers $a$, $b$, $c$,
$$
\frac{1+a^2}{1+b} + \frac{1+b^2}{1+c} + \frac{1+c^2}{1+a} \ge 6(\sqrt{2}-1).
$$ | [
"**Solution 1.** Using AM-GM inequality we have\n$$\n\\frac{1+a^2}{1+b} + \\frac{1+b^2}{1+c} + \\frac{1+c^2}{1+a} \\ge 3\\sqrt[3]{\\frac{1+a^2}{1+a} \\cdot \\frac{1+b^2}{1+b} \\cdot \\frac{1+c^2}{1+c}} \\quad (1)\n$$\nOn the other hand, for every positive real number $x$, the following inequality holds:\n$$\n\\frac... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0j2y | Problem:
What is the remainder when $(1+x)^{2010}$ is divided by $1+x+x^{2}$? | [
"Solution:\nWe use polynomial congruence $\\bmod 1+x+x^{2}$ to find the desired remainder. Since $x^{2}+x+1 \\mid x^{3}-1$, we have that $x^{3} \\equiv 1\\ (\\bmod\\ 1+x+x^{2})$. Now:\n$$\n\\begin{aligned}\n(1+x)^{2010} & \\equiv (-x^{2})^{2010} \\quad (\\bmod\\ 1+x+x^{2}) \\\\\n& \\equiv x^{4020} \\quad (\\bmod\\ ... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | final answer only | 1 | |
0b2x | Problem:
Find the number of functions $f$ from the set $S=\{0,1,2, \ldots, 2020\}$ to itself such that, for all $a, b, c \in S$, all three of the following conditions are satisfied:
(i) If $f(a)=a$, then $a=0$;
(ii) If $f(a)=f(b)$, then $a=b$; and
(iii) If $c \equiv a+b(\bmod 2021)$, then $f(c) \equiv f(a)+f(b)(\bmod 2... | [
"Solution:\nNote that, from (i), our function is completely determined by $f(1)$; i.e., $f(a) \\equiv a f(1)$ $(\\bmod 2021)$. Then, from (i) and (ii), we need that $f(a) \\neq 0$ if $a \\neq 0$; otherwise, if $a \\neq 0$ but $f(a)=0, f(b)=f(a+b)$ for any $b$. Thus, if $a \\neq 0$, we need that $a f(1) \\neq 0(\\bm... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof and answer | 1845 | |
020t | Problem:
Find all positive integers $d$ with the following property: there exists a polynomial $P$ of degree $d$ with integer coefficients such that $|P(m)|=1$ for at least $d+1$ different integers $m$. | [
"Solution:\nNote that $P(x)=c$ for a fixed constant has at most $d$ solutions, since the polynomial $P(x)-c$ of degree $d$ cancels at most $d$ times. This implies that there are integers $m$ satisfying $P(m)=1$, as well as integers $m$ such that $P(m)=-1$.\n\nNext, we prove the following lemma.\n\nLemma. If $a$ and... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [1, 2, 3] | |
0k94 | Problem:
A computer screen shows a $98 \times 98$ chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result, the colors in the selected rectangle switch (black becomes white and white becomes black). Determine the m... | [
"Solution:\n\nThe minimum number of clicks needed is $98$. One way to do this is to click all the even numbered rows one by one ($49$ clicks), then all the even numbered columns one by one (another $49$ clicks).\n\nTo show that this is minimal, note that there are $4 \\times 97$ pairs of adjacent squares along the ... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 98 | |
0f8b | Problem:
$ABCD$ is a convex quadrilateral. The midpoints of the diagonals and the midpoints of $AB$ and $CD$ form another convex quadrilateral $Q$. The midpoints of the diagonals and the midpoints of $BC$ and $CA$ form a third convex quadrilateral $Q'$. The areas of $Q$ and $Q'$ are equal. Show that either $AC$ or $BD... | [
"Solution:\n\nNote that $Q$ is a parallelogram because each side is formed by joining the midpoints of two sides of a triangle, so it is parallel to and half the length of the base of the triangle. But the triangles corresponding to opposite sides have the same base. Hence opposite sides of $Q$ are parallel and equ... | Soviet Union | 22nd ASU | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0eug | Let $a$, $b$ and $c$ be the sides of a triangle, and we set
$$
A = \frac{a^2 + bc}{b+c} + \frac{b^2 + ca}{c+a} + \frac{c^2 + ab}{a+b}
$$
$$
B = \frac{1}{\sqrt{(a+b-c)(b+c-a)}} + \frac{1}{\sqrt{(b+c-a)(c+a-b)}} + \frac{1}{\sqrt{(c+a-b)(a+b-c)}}
$$
Prove that $AB \ge 9$. | [
"Clearly\n$$\nB \\ge \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{a}\n$$\nand\n$$\nA - (a + b + c) = \\frac{a^4 + b^4 + c^4 - a^2b^2 - b^2c^2 - c^2a^2}{(a+b)(b+c)(c+a)} \\ge 0.\n$$\nTherefore we have\n$$\nAB \\ge \\left( \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{a} \\right) (b+c+a) \\ge 9.\n$$\nby Cauchy-Schwarz inequalit... | South Korea | 22nd Korean Mathematical Olympiad Final Round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof only | null | |
0gqz | Alice and Bob play a game on a complete graph $G$ with 2014 vertices. They take moves in turn with Alice beginning. At each move Alice directs one undirected edge of $G$. At each move Bob chooses a positive integer number $m$, $1 \le m \le 1000$ and after that directs $m$ undirected edges of $G$. The game ends when all... | [
"Let us prove by induction that Alice can guarantee the existence of directed path of length $k$ after her $k$-th move. The case $k = 1$ is obvious. Suppose that prior to Alice's $k+1$-th move there is a directed path $\\overrightarrow{v_0, v_1}, \\overrightarrow{v_1, v_2}, \\dots, \\overrightarrow{v_{k-1}, v_k}$ o... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Alice has a winning strategy. | |
02jw | Problem:
Se $x + y = 8$ e $x y = 15$, qual é o valor de $x^{2} + 6 x y + y^{2}$?
A) 64
B) 109
C) 120
D) 124
E) 154 | [
"Solution:\n\nUsando a identidade $(x + y)^{2} = x^{2} + 2 x y + y^{2}$, temos\n\n$x^{2} + 6 x y + y^{2} = \\left(x^{2} + 2 x y + y^{2}\\right) + 4 x y = (x + y)^{2} + 4 x y = 8^{2} + 4 \\times 15 = 124$"
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0fx2 | Problem:
Bestimme alle ungeraden natürlichen Zahlen der Form
$$
\frac{p+q}{p-q}
$$
wobei $p>q$ Primzahlen sind. | [
"Solution:\n\nSei $n$ diese Zahl. Nach Voraussetzung gilt $n(p-q)=p+q=(p-q)+2q$, folglich ist $p-q$ ein Teiler von $2q$. Wegen $\\operatorname{ggT}(p-q, q)=1$ gilt daher $p-q=1$ oder $p-q=2$. Im ersten Fall ist $p$ oder $q$ gerade, also $p=3, q=2$ und $n=5$. Im zweiten Fall sind $p$ und $q$ ungerade und somit ist $... | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 5 | |
06y4 | Let $\mathbb{Q}$ be the set of rational numbers. Let $f: \mathbb{Q} \rightarrow \mathbb{Q}$ be a function such that the following property holds: for all $x, y \in \mathbb{Q}$,
$$
f(x+f(y))=f(x)+y \quad \text{or} \quad f(f(x)+y)=x+f(y).
$$
Determine the maximum possible number of elements of $\{f(x)+f(-x) \mid x \in \... | [
"Solution 1. We begin by providing an example of a function $f$ for which there are two values of $g(x)$. We take the function $f(x)=\\lfloor x\\rfloor-\\{x\\}$, where $\\lfloor x\\rfloor$ denotes the floor of $x$ (that is, the largest integer less than or equal to $x$) and $\\{x\\}=x-\\lfloor x\\rfloor$ denotes th... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | 2 | |
00jk | We are given a sequence $\langle a_1, a_2, a_3, \dots \rangle$ of real numbers. For every positive integer $n$ we define $m_n$ as the arithmetic mean of the numbers from $a_1$ through $a_n$. We assume that a real number $C$ exists, such that
$$
(i-j) \cdot m_k + (j-k) \cdot m_i + (k-i) \cdot m_j = C
$$
holds for all tr... | [
"By exchanging the roles of $i$ and $j$, we see that $(i-j) \\cdot m_k + (j-k) \\cdot m_i + (k-i) \\cdot m_j = C = (j-i) \\cdot m_k + (i-k) \\cdot m_j + (k-j) \\cdot m_i = -C$ must hold, which yields $C = 0$. For $(i, j, k) = (1, 2, 3)$, we obtain\n$$\n(1-2) \\cdot \\frac{a_1+a_2+a_3}{3} + (2-3) \\cdot a_1 + (3-1) ... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0jgr | Problem:
Thaddeus is given a $2013 \times 2013$ array of integers each between $1$ and $2013$, inclusive. He is allowed two operations:
- Choose a row, and subtract $1$ from each entry.
- Choose a column, and add $1$ to each entry.
He would like to get an array where all integers are divisible by $2013$. On how many ... | [
"Solution:\n\nAnswer: $2013^{4025}$\n\nWe claim that the set of grids on which it is possible to obtain an array of all zeroes (mod $2013$) is indexed by ordered $4025$-tuples of residues (mod $2013$), corresponding to the starting entries in the first row and first column of the grid, giving the answer of $2013^{4... | United States | HMMT 2013 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2013^4025 | |
0acz | Let $c$ be a positive integer. The sequence $a_1, a_2, \dots, a_n, \dots$ is defined by $a_1 = c$, and $a_{n+1} = a_n^2 + a_n + c^3$, for every positive integer $n$. Find all values of $c$ for which there exist some integers $k \ge 1$ and $m \ge 2$, such that $a_k^2 + c^3$ is the $m$th power of some positive integer. | [
"First, notice\n$$\na_{n+1}^2 + c^3 = (a_n^2 + a_n + c^3)^2 + c^3 = (a_n^2 + c^3)(a_n^2 + 2a_n + 1 + c^3).\n$$\nWe first prove that $a_n^2 + c^3$ and $a_n^2 + 2a_n + 1 + c^3$ are coprime. We prove by induction that $4c^3 + 1$ is coprime with $2a_n + 1$, for every $n \\ge 1$.\nLet $n = 1$ and $p$ be a prime divisor ... | North Macedonia | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysi... | null | proof and answer | All positive integers c such that c+1 is a perfect square; equivalently, c = s^2 − 1 for some integer s ≥ 2. | |
0fg6 | Problem:
Una circunferencia de centro $O$ pasa por los vértices $A$ y $C$ de un triángulo $A B C$ y corta otra vez los segmentos $A B$ y $B C$ en los puntos distintos $K$ y $N$, respectivamente. Las circunferencias circunscritas a los triángulos $A B C$ y $K B N$ se cortan exactamente en dos puntos distintos $B$ y $M$... | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bml | Prove that the number $1$ can be represented as the sum of a finite number $n$ of real positive numbers, not necessarily distinct, that only use the digits $0$ and/or $7$ in their decimal representation. What is the smallest possible value of $n$? | [
"The problem is equivalent to representing\n$$\n\\frac{1}{7} = 0.142857142857142857\\dots := 0.\\overline{142857}\n$$\nas the sum of a finite number, $n$, of real numbers that only have digits $0$ and $7$ in their decimal representation. In order to get the fourth decimal $8$, it is easy to see that we need $n \\ge... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 8 | |
01nc | Let $A_1A_2A_3A_4$ be a non-cyclic quadrilateral. Let $O_1$ and $r_1$ be the circumcenter and the circumradius of the triangle $A_2A_3A_4$. Define $O_2$, $O_3$, $O_4$ and $r_2$, $r_3$, $r_4$ in a similar way.
Prove that
$$
\frac{1}{O_1A_1^2 - r_1^2} + \frac{1}{O_2A_2^2 - r_2^2} + \frac{1}{O_3A_3^2 - r_3^2} + \frac{1}{O... | [] | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
01l8 | Let $P$ be the intersection point of the diagonals of inscribed quadrilateral $ABCD$. Points $K$ and $L$ are marked on the bisectors of $\angle APD$ and $\angle BPC$, respectively, so that $AP = PK$ and $BP = PL$. Let $M$ be the intersection point of the lines $AK$ and $BL$, and $N$ be the intersection point of the lin... | [
"Let $\\angle APD = \\angle BPC = \\varphi$. Then $\\angle PKA = \\angle PLB = \\frac{\\pi-\\varphi}{2}$ since the triangles $APK$ and $BPL$ are isosceles. Therefore, $MK = ML$.\n\n\n\nFrom the equalities $AP = PK$, $BP = PL$ and the power of a point theorem it follows that $PK \\cdot PC = ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0f8j | Problem:
The numbers $1$ and $2$ are written on an empty blackboard. Whenever the numbers $m$ and $n$ appear on the blackboard the number $m + n + mn$ may be written. Can we obtain
(1) $13121$, (2) $12131$? | [
"Solution:\n\n(1) $13121 = 2 + 4373 + 2$. $4373, 4373 = 2 + 1457 + 2$. $1457, 1457 = 2 + 485 + 2$. $485, 485 = 2 + 161 + 2$. $161, 161 = 2 + 53 + 2$. $53, 53 = 2 + 17 + 2$. $17, 17 = 2 + 5 + 2$. $5, 5 = 2 + 1 + 2$. $1$.\n\nPut $M = m + 1$, $N = n + 1$. Then the number derived from $m$ and $n$ is $MN - 1$. So if $M$... | Soviet Union | 22nd ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (1) Yes, (2) No | |
0k6m | Problem:
2019 points are chosen independently and uniformly at random on the interval $[0,1]$. Tairitsu picks 1000 of them randomly and colors them black, leaving the remaining ones white. Hikari then computes the sum of the positions of the leftmost white point and the rightmost black point. What is the probability t... | [
"Solution:\n\nNote that each point is chosen uniformly and independently from $0$ to $1$, so we can apply symmetry. Given any coloring, suppose that we flip all the positions of the black points: then the problem becomes computing the probability that the leftmost white point is to the left of the leftmost black po... | United States | HMMT February 2019 | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1019/2019 | |
07la | Prove that, for every positive integer $n$ which ends in the digit 5,
$$
20^n + 15^n + 8^n + 6^n
$$
is divisible by 2009. | [
"Observe that\n$$\n20^n + 15^n + 8^n + 6^n = (5^n + 2^n)(4^n + 3^n) \\quad (4)\n$$\nSince $n = 10k + 5 = 5(2k + 1)$ for some integer $k$,\n$$\n5^n + 2^n = 5^{5(2k+1)} + 2^{5(2k+1)}\n$$\nand so $5^5 + 2^5$ divides $5^n + 2^n$. Similarly, $4^5 + 3^5$ divides $4^n + 3^n$. Now calculation shows that $5^5 + 2^5 = 7 \\cd... | Ireland | Irska | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
02ya | Problem:
Na figura a seguir, $ABC$ é um triângulo equilátero de papel com lado $1\ \mathrm{m}$ que foi dobrado ao longo do segmento $EF$ de modo que o vértice $A$ caísse sobre o lado $BC$, onde está o ponto $D$ na figura. Suponha que $DF$ é perpendicular a $BC$.
a) Determine o ângulo $\angle... | [
"Solution:\n\na) Como $\\angle FDC = 90^{\\circ}$, segue que $\\angle DFC = 30^{\\circ}$ e $\\angle AFD = 180^{\\circ} - \\angle DFC = 150^{\\circ}$. A dobradura ao longo de $EF$ nos diz que os triângulos $AEF$ e $DEF$ são congruentes. Daí $\\angle AFE = \\angle EFD = \\frac{150^{\\circ}}{2} = 75^{\\circ}$ e $\\ang... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a) 90 degrees; b) 2 − √3 m; c) (9√3 − 15)/2 | |
0ckb | Consider positive real numbers $m, n, a, b, c$, such that $m > n$ and $|ma - nb| \le c(m - n)$, $|mb - nc| \le a(m - n)$, $|mc - na| \le b(m - n)$. Prove that $a = b = c$.
Lucian Petrescu | [
"If $x \\ge 0$, $y \\ge 0$, and $z \\ge 0$, then $ma - nb = c(m - n)$ and analogously, we have\n$$\nm(a - c) = n(b - c), \\quad m(b - a) = n(c - a), \\quad m(c - b) = n(a - b). \\quad (*)\n$$\nIf all parentheses in $(*)$ are nonzero, then we obtain $m^3(a-c)(b-a)(c-b) = -n^3(b-c)(c-a)(a-b)$, and therefore $m = -n$,... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof only | null | |
02eg | A figure on a computer screen shows $n$ points on a sphere, no four coplanar. Some pairs of points are joined by segments. Each segment is colored red or blue. For each point there is a key that switches the colors of all segments with that point as endpoint. For every three points there is a sequence of key presses th... | [
"Consider three of the $n$ points. The parity of the number of blue segments of the triangle with these points as vertices doesn't change while switching the keys. Since it is possible to make all three segments red, the number of blue segments in each triangle is even.\n\nLet $P$ be one of the $n$ points. Let $A$ ... | Brazil | X OBM | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | floor(n/2) | |
05lp | Problem:
On répartit des poids de respectivement $1~\mathrm{g}, 2~\mathrm{g}, \ldots, 200~\mathrm{g}$ sur les deux plateaux d'une balance de sorte que chaque plateau contienne 100 poids.
Prouver que l'on peut échanger 50 poids d'un plateau avec 50 poids de l'autre plateau pour que la balance devienne équilibrée. | [
"Solution:\n\nDans cette solution, on dira que deux poids sont complémentaires s'ils pèsent 201 grammes à eux deux.\n\nConsidérons alors les 100 poids situés sur le plateau de gauche. On peut les classer en trois catégories:\n\n1. les poids $a_{1}, \\ldots, a_{n}$ de 100 grammes ou moins, dont le complémentaire est... | France | Olympiades Françaises de Mathématiques - Test de Janvier | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
04t3 | Let $n$ be an even positive integer. There are $n$ real numbers written on the blackboard. In every step, we choose two numbers, erase them, and replace each of them by their product. Show that for any initial $n$-tuple it is possible to obtain $n$ equal numbers on the blackboard after a finite number of steps. | [
"We shall prove the claim by induction with respect to $n$. The claim is trivial for $n = 2$ (we can get the desired 2-tuple after a single step $(a, b) \\rightarrow (ab, ab)$) and $n = 4$ (we can follow the scheme $(\\underline{a}, \\underline{b}, c, d) \\rightarrow (ab, ab, \\underline{c}, \\underline{d}) \\right... | Czech Republic | 15th Czech-Polish-Slovak Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0dyh | Let $ABC$ be an isosceles triangle with the apex at $C$ and let $A'$ be the foot of the altitude from $A$. Assuming that $|CA'| = \frac{1}{2}|AB|$, prove that the triangle $ABC$ is equilateral. | [
"Let $C'$ be the foot of the altitude from $C$. Since the triangle $ABC$ is isosceles with the apex at $C$, we have $|AC'| = |C'B| = \\frac{1}{2}|AB| = |CA'|$. Since $ABA'$ and $CBC'$ are right triangles and $\\angle ABA' = \\angle CBC'$, they are similar. This implies\n$$\n\\frac{|AB|}{|BA'|} = \\frac{|CB|}{|BC'|}... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0jgz | A rational number $x$ is given. Prove that there exists a sequence $x_0, x_1, x_2, \dots$ of rational numbers with the following properties:
a. $x_0 = x$;
b. for every $n \ge 1$, either $x_n = 2x_{n-1}$ or $x_n = 2x_{n-1} + \frac{1}{n}$;
c. $x_n$ is an integer for some $n$.
(This problem was suggested by Gabriel Ca... | [
"Let $x$ be written in lowest terms as $p/q$, and write $q = 2^r s$, where $s$ is odd. Let $S$ be the set of residue classes modulo $s$, where arithmetic on elements of $S$ is understood to be done modulo $s$. For any positive integer $N$ and any $t \\in S$, say that $t$ is *attainable* at $N$ if there exists a seq... | United States | TSTST | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0dsa | Consider a polygon with $m+n$ sides where $m, n$ are positive integers. Colour $m$ of its vertices red and the remaining $n$ vertices blue. A side is given the number $2$ if both its end vertices are red, the number $\frac{1}{2}$ if both its end vertices are blue and the number $1$ otherwise. Let the product of these n... | [
"We first show that if two adjacent vertices have different colours, then swapping the colours of these vertices leaves $P$ unchanged. To see this, we only need to consider the four possible cases:\n\n$RRBR$, $BRBB$, $RRBB$, $BRBR$\n\nwhich change to $RBRR$, $BBRB$, $RBRB$, $BBRP$.\n\nThe value of $P$ remains uncha... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^(m-n) | |
0f42 | Problem:
The positive reals $x$, $y$ satisfy $x^{3} + y^{3} = x - y$. Show that $x^{2} + y^{2} < 1$. | [
"Solution:\nSince $x$, $y$ are positive, so is $x^{3} + y^{3}$, and hence $x > y$. So\n$$(x^{2} + y^{2})(x - y) = (x^{3} - y^{3}) - xy(x - y) < x^{3} - y^{3} = x - y.$$\nHence $x^{2} + y^{2} < 1$."
] | Soviet Union | 15th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0er4 | A certain pen can draw a line $0.4$ mm wide and $1$ km long before it runs out of ink. The largest square you can colour in with this pen has an area, in square metres, of
(A) $4\,000$ (B) $400$ (C) $40$ (D) $4$ (E) $0.4$ | [
"Since $0.4 \\text{ mm} = 0.4 \\times 10^{-3} \\text{ m}$ and $1 \\text{ km} = 10^3 \\text{ m}$, it follows that the area of the line is $(0.4 \\times 10^{-3}) \\times 10^3 = 0.4 \\text{ m}^2$, and the area of the square will be the same."
] | South Africa | South African Mathematics Olympiad First Round | [
"Math Word Problems"
] | English | MCQ | E | |
00h0 | Let $n$ be a positive integer. $n$ people take part in a certain party. For any pair of the participants, either the two are acquainted with each other or they are not. What is the maximum possible number of the pairs for which the two are not acquainted but have a common acquaintance among the participants? | [
"When 1 participant, say the person $A$, is mutually acquainted with each of the remaining $n-1$ participants, and if there are no other acquaintance relationships among the participants, then for any pair of participants not involving $A$, the two are not mutual acquaintances, but they have a common acquaintance, ... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | (n^2 - 3n + 2)/2 |
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