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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09rj | Problem:
Zij $n \geq 3$ een geheel getal en bekijk een $n \times n$-bord, opgedeeld in $n^{2}$ eenheidsvierkantjes. We hebben voor elke $m \geq 1$ willekeurig veel $1 \times m$-rechthoeken (type I) en willekeurig veel $m \times 1$-rechthoeken (type II) beschikbaar. We bedekken het bord met $N$ van deze rechthoeken, di... | [
"Solution:\n\nWe bewijzen dat de minimale waarde $N=2n-1$ is. We construeren eerst een voorbeeld door middel van inductie.\n\nInductiebasis. Voor $n=3$ is $N=5$ mogelijk, door in het middelste veld een $1 \\times 1$-rechthoek te leggen en de overige velden te bedekken met vier $2 \\times 1$- en $1 \\times 2$-rechth... | Netherlands | MO-selectietoets | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2n-1 | |
0jvp | Problem:
Let $X$ be the collection of all functions $f:\{0,1, \ldots, 2016\} \rightarrow \{0,1, \ldots, 2016\}$. Compute the number of functions $f \in X$ such that
$$
\max _{g \in X}\left(\min _{0 \leq i \leq 2016}(\max (f(i), g(i)))-\max _{0 \leq i \leq 2016}(\min (f(i), g(i)))\right)=2015 .
$$
Proposed by: | [
"Solution:\nFor each $f, g \\in X$, we define\n$$\nd(f, g):=\\min _{0 \\leq i \\leq 2016}(\\max (f(i), g(i)))-\\max _{0 \\leq i \\leq 2016}(\\min (f(i), g(i)))\n$$\nThus we desire $\\max _{g \\in X} d(f, g)=2015$.\n\nFirst, we count the number of functions $f \\in X$ such that\n$$\n\\exists g: \\min _{i} \\max \\{f... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2*(3^2017 - 2^2017) | |
0d00 | Prove that there exists a finite sequence of distinct positive integers $n_1, n_2, \dots, n_k$, such that
$$
2012 < \sum_{i=1}^{k} \frac{1}{n_i} < 2012 + \left(\frac{1}{2012}\right)^{2012}.
$$ | [
"We use the fact that the harmonic series $\\sum_{i=1}^{\\infty} \\frac{1}{i}$ diverges. This means that for any real $M$, we can find an $n$ such that $\\sum_{i=1}^{n} \\frac{1}{i} > M$. But $\\sum_{i=1}^{k} \\frac{1}{i}$ is finite for any fixed $k$, so therefore for any real $M$ and integer $k$ we can find an $n$... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0fwv | Problem:
Sei $ABC$ ein Dreieck mit $\angle ABC \neq \angle BCA$. Der Inkreis $k$ des Dreiecks $ABC$ berühre die Seiten $BC$, $CA$ bzw. $AB$ in den Punkten $D$, $E$ bzw. $F$. Die Strecke $AD$ schneide $k$ ein weiteres Mal in $P$. Sei $Q$ der Schnittpunkt von $EF$ mit der Rechtwinkligen zu $AD$ durch $P$. Sei $X$ bzw. $... | [
"Solution:\n\nSei $I$ der Inkreismittelpunkt des Dreiecks $ABC$ und $M$ der Mittelpunkt der Strecke $EF$. Wir zeigen zuerst, dass $IMPD$ ein Sehnenviereck ist. Wegen $AF = AE$ liegen $A$, $M$ und $I$ auf einer Geraden und $\\angle FMI = 90^\\circ$. Da auch $\\angle AFI = 90^\\circ$ gilt, liegt der Umkreis von $FIM$... | Switzerland | IMO Selektion 2008 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
01nz | Given $0 < a < b < c$ prove that
$$
a^{20}b^{12} + b^{20}c^{12} + c^{20}a^{12} < b^{20}a^{12} + a^{20}c^{12} + c^{20}b^{12}.
$$ | [
"Lemma. Let $0 < x < y < z$ and $\\lambda > 1$. Then $\\sum_{cyc} x^{\\lambda}y < \\sum_{cyc} x y^{\\lambda}$.\n\nProof. Consider the function $f(y) = \\sum_{cyc} (x^{\\lambda}y - x y^{\\lambda})$. It takes the value $0$ for $y = x$ and $y = z$. We have also $f''(y) = \\lambda(\\lambda - 1)y^{\\lambda}(z-x) > 0$, h... | Belarus | Belorusija 2012 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0i3w | Problem:
Compute $\sum_{i=1}^{\infty} \frac{a i}{a^{i}}$ for $a>1$. | [
"Solution:\n\nThe sum $S = a + a x + a x^{2} + a x^{3} + \\cdots$ for $x < 1$ can be determined by realizing that $x S = a x + a x^{2} + a x^{3} + \\cdots$ and $(1-x) S = a$, so $S = \\frac{a}{1-x}$.\n\nUsing this, we have\n$$\n\\sum_{i=1}^{\\infty} \\frac{a i}{a^{i}} = a \\sum_{i=1}^{\\infty} \\frac{i}{a^{i}} = a\... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a^2/(a-1)^2 | |
0bo2 | Problem:
Adottak az $m$ és $n$ természetes számok $(n \geq 2)$. Az $A$ gyűrűnek pontosan $n$ eleme van és $a$ egy olyan eleme $A$-nak, amelyre $1-a^{k}$ invertálható, bármely $k \in\{m+1, m+2, \ldots, m+n-1\}$ esetén. Igazold, hogy $a$ nilpotens (azaz létezik olyan nem nulla $p$ természetes szám, amelyre $a^{p}=0$).
... | [
"Solution:\n\nFie $k \\in \\{1,2, \\ldots, n-1\\}$. Există $p \\in \\{1,2, \\ldots, n-1\\}$, astfel încât $m+p$ este divizibil cu $k$, deci $m+p=k \\ell$. Cum $1-a^{m+p}=(1-a^{k})(1+a^{k}+\\cdots+a^{k(\\ell-1)})$ şi $1-a^{m+p} \\in U(A)$, rezultă că $1-a^{k} \\in U(A)$.\n\nDin $1-a^{k}=(1-a)(1+a+\\cdots+a^{k-1})$, ... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
0cyp | A Geostationary Earth Orbit is situated directly above the equator and has a period equal to the Earth's rotational period. It is at the precise distance of $22,236$ miles above the Earth that a satellite can maintain an orbit with a period of rotation around the Earth exactly equal to $24$ hours. Because the satellite... | [
"Let $A$ and $B$ be two neighboring stations. We have $\\widehat{A O B}=\\frac{\\pi}{5}$, hence $A B=2 R \\sin \\frac{\\pi}{10}$, where $R=22236+3960=26196$. We will prove that $\\sin \\frac{\\pi}{10}=\\frac{\\sqrt{5}-1}{4}$.\n\n\n\nSince $\\sin \\pi=0$, then $\\sin \\left(\\frac{2 \\pi}{5}... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | -13098 + 13098 sqrt 5 | |
031v | Problem:
Let $a$, $b$ and $c$ be rational numbers such that $a+b+c$ and $a^{2}+b^{2}+c^{2}$ are equal integers. Prove that the number $a b c$ can be written as a ratio of a perfect cube and a perfect square that are coprime. | [
"Solution:\nLet $a+b+c=a^{2}+b^{2}+c^{2}=t$. Then $t \\geq 0$. On the other hand, the Root mean square - Arithmetic mean inequality implies that\n$$\n\\frac{a^{2}+b^{2}+c^{2}}{3} \\geq \\frac{(a+b+c)^{2}}{9} \\Longleftrightarrow 3 t \\geq t^{2}\n$$\nHence $t \\in\\{0,1,2,3\\}$. If $t=0$ or $t=3$, then $a=b=c=0$ or ... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0k08 | Problem:
Does there exist an irrational number $\alpha>1$ such that
$$
\left\lfloor\alpha^{n}\right\rfloor \equiv 0 \quad(\bmod 2017)
$$
for all integers $n \geq 1$? | [
"Solution:\nAnswer: Yes\nLet $\\alpha>1$ and $0<\\beta<1$ be the roots of $x^{2}-4035 x+2017$. Then note that $\\left\\lfloor\\alpha^{n}\\right\\rfloor=\\alpha^{n}+\\beta^{n}-1$. Let $x_{n}=\\alpha^{n}+\\beta^{n}$ for all nonnegative integers $n$. It's easy to verify that $x_{n}=4035 x_{n-1}-2017 x_{n-2} \\equiv x_... | United States | February 2017 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Yes | |
0gbg | 對於所有 $n \ge 3$, 試求最大的正整數 $m$, 滿足: 我們可以從正 $n$ 邊形的對角線中選出 $m$ 條, 使得所選對角線中的任兩條, 若它們在正 $n$ 邊形內部有交點, 則它們必互相垂直。 | [
"解：若 $n$ 為奇數，則 $m = n - 3$；若 $n$ 為偶數，則 $m = n - 2$。\n\n1. 首先證明 $n$ 為奇數的狀況。\n\na. **Claim.** 正 $n$ 邊形內的任兩條對角線不垂直。\n假設對角線 $\\overline{AB}$ 和 $\\overline{CD}$ 垂直。做 $AB$ 中垂線, 交此正 $n$ 邊形於頂點 $E$ (注意到 $n$ 為奇數, 故 $E$ 存在。) 令 $E'$ 為正 $n$ 邊形上 $E$ 的對稱點:\n- 注意到 $n$ 為奇數, 故 $E'$ 必不是正 $n$ 邊形的頂點。\n- 然而由對稱性, 易知 $\\overline{EC} = \\o... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | If n is odd, m = n − 3; if n is even, m = n − 2. | |
09hj | A $10 \times 10$ square is divided into $1 \times 1$ squares. A *point light* at a vertex of $1 \times 1$ square lights all the $1 \times 1$ squares that the vertex belongs. (A point light can be positioned on a vertex on the edge of the big square). Find the minimum number of point lights required such that all the sq... | [
"The minimum number of light is $55$.\nEach $1 \\times 1$ black square needs at least $2$ lights and each figure that consist of three $1 \\times 1$ square needs at least $3$ lights. (See picture 1.) So we need at least $2 \\times 20 + 3 \\times 5 = 55$ lights.\nPicture 2 shows that $55$ lights could be placed as r... | Mongolia | Mongolian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 55 | |
0k05 | Problem:
Marisa has a collection of $2^{8}-1=255$ distinct nonempty subsets of $\{1,2,3,4,5,6,7,8\}$. For each step she takes two subsets chosen uniformly at random from the collection, and replaces them with either their union or their intersection, chosen randomly with equal probability. (The collection is allowed t... | [
"Solution:\n\nIt suffices to compute the probability of each number appearing in the final subset. For any given integer $n \\in [1,8]$, there are $2^{7}=128$ subsets with $n$ and $2^{7}-1=127$ without. When we focus on only this element, each operation is equivalent to taking two random sets and discarding one of ... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1024/255 | |
0htf | Problem:
a. Let $a$, $b$, and $n$ be positive integers such that $a b = n^{2} + 1$. Prove that
$$
|a-b| \geq \sqrt{4 n - 3}
$$
b. Prove that there are infinitely many such triples $(a, b, n)$ such that equality occurs. | [
"Solution:\n\na.\nWe may assume that $a \\geq b$ (thus eliminating the absolute value sign). Write $t = a - b$, so $a = b + t$. Then the equality $n^{2} + 1 = a b$ can be rewritten as $n^{2} + 1 = (b + t) b$ or $b^{2} + t b - (n^{2} + 1) = 0$.\n\nReplacing $b$ by $x$, we consider the quadratic equation $x^{2} + t x... | United States | Berkeley Math Circle Monthly Contest | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0av0 | Problem:
How many solutions does the equation $x+y+z=2016$ have, where $x$, $y$ and $z$ are integers with $x>1000$, $y>600$, and $z>400$? | [
"Solution:\n\nFirst, note that the equation is equivalent to $(x-1001)+(y-601)+(z-401)=13$. Letting $x' = x-1001$, $y' = y-601$, and $z' = z-401$, we can instead count the number of solutions of $x' + y' + z' = 13$, where $x'$, $y'$, $z'$ are nonnegative integers. Now note that each solution $(x', y', z')$ to this ... | Philippines | 18th PMO National Stage Oral Phase | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 105 | |
0dnf | Problem:
За природан број $n$ кажемо да је шашав ако и само ако постоје природни бројеви $a>1$ и $b>1$ такви да је $n=a^{b}+b$. Да ли постоји 2014 узастопних природних бројева међу којима је тачно 2012 шашавих бројева?
(Милош Милосављевић) | [
"Solution:\n\nПрво ћемо дати пример 2012 узастопних шашавих бројева. Довољно је узети бројеве $N+2, N+3, \\ldots, N+2013$, где је $N=2^{2013!}$.\nЗа природан број $n$, означимо са $f(n)$ број шашавих бројева међу $n, n+1, \\ldots, n+2013$. Како је $f(1)<2012$ (бројеви $1,2,3,4,5$ нису шашави), $f(N) \\geqslant 2012... | Serbia | 8. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes | |
0dan | Find the smallest integer $m$ for which there are positive integers $n > k > 1$ satisfying the equation
$$
\underbrace{11 \ldots 1}_{n} = \underbrace{11 \ldots 1}_{k} \cdot m.
$$ | [
"Obviously $m > 9$. If $m = \\overline{ab}$, where $a \\geq 1$ then we must have $b = 1$ to ensure last digit of $1 \\cdot m$ is equal $1$. In this case regardless of value of $a$ the second last digit of $\\underbrace{11 \\ldots 1}_{k} \\cdot m$ is equal to the last digit of $a + 1$ and can't be equal $1$. So $m \... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 101 | |
0h5n | Circles $w_1$ and $w_2$ centered at points $O_1$ and $O_2$ respectively intersect at points $A$ and $B$. Let $w$ be the circumscribed circle of $O_1O_2B$ centered at $O$, which intersects $w_1$ and $w_2$ again at points $K$ and $L$ respectively. The straight line $OA$ intersects $w_1$ and $w_2$ at points $M$ and $N$ re... | [
"We use the following lemma by Archimedes:\n\n**Lemma (Archimedes).** Circles $w_1$ and $w_2$ intersect at points $A$ and $B$, with the center of $w_2$ lying on $w_1$. A chord $AC$ of $w_2$ intersects $w_1$ again at a point $M$. Then $CM = CB$.\n\nPut $\\alpha = \\angle KBA = \\angle KMA$, since they intercept the ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
0faf | Problem:
Find unequal integers $m$, $n$ such that $mn + n$ and $mn + m$ are both squares. Can you find such integers between $988$ and $1991$? | [
"Solution:\n\nAnswer no\n\nFor example, $49 = 7^2$, $50 = 2 \\cdot 5^2$, $8 = 2 \\cdot 2^2$, $9 = 3^2$, so $49 \\cdot 8 + 8 = 20^2$, $49 \\cdot 8 + 49 = 21^2$.\n\nWLOG $m < n$. Then $mn + m = (m + h)^2$, $mn + n = (m + k)^2$, with $k > h$. So $n - m = (m + k)^2 - (m + h)^2 = (k - h)(2m + k + h) > 2m$, so $n > 3m$. ... | Soviet Union | 25th ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | One example is m = 49 and n = 8. No such pair exists with both integers between 988 and 1991. | |
0gdy | 設 $a, b$ 為正整數。若對於所有 $an \ge b$ 的正整數 $n$, $\binom{an}{b} - 1$ 都可被 $an + 1$ 整除, 我們就稱 $b$ 為 $a$-好正整數。設 $b$ 是 $a$-好正整數但 $b + 2$ 不是。證明 $b + 1$ 是質數。
Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\binom{an}{b} - 1$ is divisible by $an+1$ for all positive integers $n$ with $an \ge b$. Suppo... | [
"We first show that $b$ is $a$-good iff $b$ is even and $p \\mid a$ for all primes $p \\le b$. Suppose there is a prime $p \\le b$ with $p \\nmid a$, take $t = v_p(b!)$. Then there exist positive integers $c$ such that $ac \\equiv 1 \\pmod{p^{t+1}}$. Take large enough $c$ (so that $an \\ge b$) and let $n = (p-1)c$,... | Taiwan | 2020 Taiwan IMO 3J | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0ipr | Problem:
In a game show, Bob is faced with 7 doors, 2 of which hide prizes. After he chooses a door, the host opens three other doors, of which one is hiding a prize. Bob chooses to switch to another door. What is the probability that his new door is hiding a prize? | [
"Solution:\n\nIf Bob initially chooses a door with a prize, then he will not find a prize by switching.\n\nWith probability $\\frac{5}{7}$ his original door does not hide the prize. After the host opens the three doors, the remaining three doors have equal probability of hiding the prize. Therefore, the probability... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 5/21 | |
0jzl | Problem:
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of positive real numbers which satisfies
$$
a_{n+1} \geq a_{n}^{2}+\frac{1}{5}
$$
for every positive integer $n$. Prove that $\sqrt{a_{n+5}} \geq a_{n-5}$ for each positive integer $n$. | [
"Solution:\nFrom the given we can deduce that\n$$\na_{n+1} \\geq a_{n}^{2}+\\frac{1}{4}-\\frac{1}{20} \\geq a_{n}-\\frac{1}{20}.\n$$\nThus for any $n$ we have\n$$\na_{n+5} \\geq a_{n+1}-4 \\cdot \\frac{1}{20}=a_{n+1}-\\frac{1}{5}=a_{n}^{2}\n$$\nThus $\\sqrt{a_{n+5}} \\geq a_{n} \\geq a_{n-5}$ follows."
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0bqz | Let $a$, $b$, $c \in (0, \infty)$ be such that $\sqrt{a} + \sqrt{b} + \sqrt{c} \le 2$. Prove that
$$
\frac{\sqrt{a^2 - ab + b^2}}{a^2 + b^2} + \frac{\sqrt{b^2 - bc + c^2}}{b^2 + c^2} + \frac{\sqrt{c^2 - ca + a^2}}{c^2 + a^2} \le \frac{1}{\sqrt{abc}}
$$ | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
06at | $$
\frac{a^2 + b^2}{2ab} + \frac{b^2 + c^2}{2bc} + \frac{c^2 + a^2}{2ca} + \frac{2(ab + bc + ca)}{3} \geq 5.
$$
When does equality hold? | [
"Since $a^2 + b^2 + c^2 = 3$, the inequality is equivalent to\n\n$$\n\\left( \\frac{a^2 + b^2}{2ab} + 1 \\right) + \\left( \\frac{b^2 + c^2}{2bc} + 1 \\right) + \\left( \\frac{c^2 + a^2}{2ca} + 1 \\right) + \\left( \\frac{2(ab + bc + ca)}{a^2 + b^2 + c^2} + 1 \\right) \\geq 9.\n$$\nwhich can be written as\n$$\n\\fr... | Greece | Selection Examination | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | Equality holds when a = b = c = 1. | |
0ean | Problem:
V vrsti stoji 8 škatel, oštevilčenih s števili od 1 do 8, in prazna vreča. V vsaki škatli je 1 žeton. Miha, ki ima veliko žetonov, se igra igro, v kateri sta dovoljeni naslednji dve potezi:
- odstrani 1 žeton iz škatle, oštevilčene z $i$ ($i < 8$), in doda 2 žetona v škatlo, oštevilčeno z $(i+1)$,
- odstrani ... | [
"Solution:\n\nNaj $A_{i}$ oziroma $B_{i}$ označuje prvo oziroma drugo potezo, izvedeno na škatlah $i$ in $i+1$. Naj bo $i \\leq 6$. Če Miha izvede potezo $B_{i}$, potem se število žetonov v škatlah $i$ in $i+1$ zmanjša za 1, število žetonov v škatli $i+2$ ostane enako, število žetonov v vreči pa se poveča za 1. Če ... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 85 | |
01zs | Given $n \in \mathbb{N}$. Consider polynomials of degree $n$ of the form $P(x) = x^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ with non-negative integer coefficients not exceeding $100$. We call such a polynomial *expandable* if it can be represented as a product of two non-constant polynomials with non-negative integer c... | [
"If the polynomial $P(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0$ is expandable, then it can be represented as\n$$\nP(x) = (x^m + b_{m-1}x^{m-1} + \\cdots + b_1x + b_0) \\cdot (x^k + c_{k-1}x^{k-1} + \\cdots + c_1x + c_0),\n$$\nwhere $m \\ge k \\ge 1$, $m + k = n$ and numbers $b_i, c_i$ are all non-negative i... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0gkq | Determine all positive integers $x, y$ such that $\frac{xy^3}{x+y}$ is a cube of a prime. | [
"Let $p$ be such a prime and let $d = \\gcd(x, y)$. Thus, $x = da, y = db$. The condition becomes $p^3d(a+b) = d^4ab^3$, i.e.,\n$$\np^3(a+b) = d^3ab^3. \\qquad (1)\n$$\nSince $1 = \\gcd(a, b) = \\gcd(a, a+b) = \\gcd(ab^3, a+b)$, we get $ab^3 \\mid p^3$, and so $b \\mid p$.\n\n**Case 1:** $b=1$. From (1), we have $p... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (x, y) = (2, 14) | |
078u | Find all functions $f : \mathbb{N} \to \mathbb{N}$ so that for any positive integer $n$ and finite sequence of positive integers $a_0, \dots, a_n$, whenever the polynomial $a_0 + a_1x + \dots + a_nx^n$ has at least one integer root, so does $f(a_0) + f(a_1)x + \dots + f(a_n)x^n$. | [
"The answer is all functions of the form $f(x) = kx$ for some $k \\in \\mathbb{N}$. These clearly work: now let us prove they are the only possibilities.\n\nSince $x + a$ has an integer root (namely, $-a$) for any $a \\in \\mathbb{N}$, so does $f(1)x + f(a)$, implying $f(1)|f(a)$ for all $a \\in \\mathbb{N}$. Now t... | India | EGMO TST | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | All functions of the form f(n) = k n for some positive integer k. | |
0iuv | Problem:
You are trapped in ancient Japan, and a giant enemy crab is approaching! You must defeat it by cutting off its two claws and six legs and attacking its weak point for massive damage. You cannot cut off any of its claws until you cut off at least three of its legs, and you cannot attack its weak point until yo... | [
"Solution:\n\nAnswer: $14400$\n\nThe answer is given by $6!\\,2!\\binom{5}{2}$, because we can cut off the claws and legs in any order and there are $\\binom{5}{2}$ ways to decide when to cut off the two claws (since we can do it at any time among the last 5 cuts)."
] | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 14400 | |
00rq | Find all monic polynomials $f$ with integer coefficients satisfying the following condition:
There exists a positive integer $N$ such that for every prime $p > N$, $p$ divides $2(f(p))! + 1$. | [
"From the divisibility relation $p|2(f(p))! + 1$ we conclude that:\n$$\nf(p) < p, \\text{ for all primes } p > N \\quad (*)\n$$\nIn fact, if for some prime number $p$ we have $f(p) \\ge p$, then $p|(f(p))!$ and then $p|1$, which is absurd.\nNow suppose that $\\deg f = m > 1$. Then $f(x) = x^m + Q(x)$, $\\deg Q(x) \... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | f(x) = x - 3 | |
0c72 | Let $ABC$ be a triangle. Let $M$ be a variable point interior to the segment $AB$, and let $\gamma_B$ be the circle through $M$ and tangent at $B$ to $BC$. Let $P$ and $Q$ be the touch points of $\gamma_B$ and its tangents from $A$, and let $X$ be the midpoint of the segment $PQ$. Similarly, let $N$ be a variable point... | [
"We show that the line through the centres of the circles $AMN$ and $AXY$ passes through the centre of the circle $ABC$. Alternatively, but equivalently, we prove that the three circles share a point different from $A$.\n\nInvert the whole configuration from $A$ and let $Z'$ denote the image of $Z$ under inversion:... | Romania | Stars of Mathematics Competition | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
04tk | Let $D$ be an arbitrary point on the base $AB$ of an isosceles triangle $ABC$. Let $E$ be such that $ADEC$ is a parallelogram. Point $F$ on the ray opposite to $ED$ satisfies $EF = EB$. Prove that the length of a chord cut by line $BE$ in the circumcircle of triangle $ABF$ is twice the length of $AC$.
(Jan Kuchařík, Pa... | [
"Denote by $S$ the circumcenter of triangle $ABF$ and by $K$, $L$, $M$ the feet of perpendiculars from $S$ to lines $AB$, $BF$, $BE$. It's easy to check that angle $ABF$ is obtuse, point $S$ lies on the perpendicular bisector $CK$ of side $AB$ in the half-plane opposite to $CEB$, and that point $E$ lies inside segm... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07t0 | Are there infinitely many pairs $(a, b)$ of positive integers such that
$$
a \mid b^2 + 1 \quad \text{and} \quad b \mid a^2 + 1?
$$ | [
"**Solution 1.** If $(a, b)$ is a solution with $a < b$, then so is $(b, c)$ where\n$$\nc = \\frac{b^2 + 1}{a}.\n$$\nIt is immediate that $c \\mid b^2 + 1$ so we have to show only that $b \\mid c^2 + 1$. We can calculate:\n$$\na^2(c^2 + 1) = (b^2 + 1)^2 + a^2 = b^2 + 2b + (a^2 + 1) \\equiv 0 \\pmod{b}\n$$\nAs $b \\... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
05zh | Problem:
Soient $a$, $b$ deux entiers relatifs. Montrer que, si ni $a$, ni $b$ n'est multiple de $3$, alors $a^{4}-b^{2}$ est multiple de $3$. | [
"Solution:\n\nRappelons que si $k$ n'est pas un multiple de $3$, alors $k$ vaut $1$ ou $2$ modulo $3$. Ainsi $k^{2}$ vaut $1^{2}=1$ ou $2^{2} \\equiv 4 \\equiv 1 \\pmod{3}$ : tout carré d'un nombre non divisible par $3$ vaut $1$ modulo $3$.\n\nAinsi $a^{4}-b^{2} \\equiv (a^{2})^{2}-b^{2} \\equiv 1^{2}-1 \\equiv 0 \... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
08gz | Problem:
Let $ABC$ be an acute triangle such that $AB$ is the shortest side of the triangle. Let $D$ be the midpoint of the side $AB$ and $P$ be an interior point of the triangle such that
$$
\Varangle CAP=\Varangle CBP=\Varangle ACB
$$
Denote by $M$ and $N$ the feet of the perpendiculars from $P$ to $BC$ and $AC$, re... | [
"Solution:\n\nIf $\\gamma=\\Varangle ACB$ then $\\Varangle CAP=\\Varangle CBP=\\Varangle ACB=\\gamma$. Let $E=KN \\cap AP$ and $F=KM \\cap BP$. We show that points $E$ and $F$ are midpoints of $AP$ and $BP$, respectively.\n\n\n\nIndeed, consider the triangle $AEN$. Since $KN \\parallel BC$,... | JBMO | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fss | Problem:
Für $x>0$ sei $f(x)=4^{x} /(4^{x}+2)$. Bestimme den Wert der Summe
$$
\sum_{k=1}^{1290} f\left(\frac{k}{1291}\right)
$$ | [
"Solution:\n\nEs gilt für $0 < x < 1/2$\n$$\nf\\left(\\frac{1}{2} + x\\right) + f\\left(\\frac{1}{2} - x\\right) = \\frac{2 \\cdot 4^{x}}{2 \\cdot 4^{x} + 2} + \\frac{2 \\cdot 4^{-x}}{2 \\cdot 4^{-x} + 2} = \\frac{4^{x}}{4^{x} + 1} + \\frac{1}{1 + 4^{x}} = 1\n$$\nwobei wir beide Brüche mit $2$ gekürzt und den zweit... | Switzerland | IMO - Selektion | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 645 | |
007f | a) Several distinct positive integers have the property that the sum of every three of them is a prime number. At most how many of them are there?
b) Several distinct integers (not necessarily positive) have the property that the sum of every three of them is positive and also a prime number. At most how many of them ... | [
"Among 5 arbitrary integers either there are three with the same remainder modulo 3 or three with different remainders modulo 3. In both cases the sum of these three is divisible by 3.\n\nIf in addition the integers are positive and distinct like in a) then the sum in question is at least $1 + 2 + 3 > 3$, hence it ... | Argentina | National Olympiad of Argentina | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | a) 4; b) 5 | |
00n9 | Determine all non-negative integers $n$ smaller than $128^{97}$ which have exactly 2019 positive divisors. | [
"Numbers with exactly 2019 positive divisors are either of the form $p^{2018}$ or $p^{672} \\cdot q^2$ for distinct prime numbers $p$ and $q$. The number $128^{97}$ can be written as\n$$\n128^{97} = (2^7)^{97} = 2^{679}.\n$$\nAs $p$ is at least $2$, the number $p^{2018}$ is greater than $2^{679}$ and therefore, the... | Austria | AUT_ABooklet_2020 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 2^672 * 3^2, 2^672 * 5^2, 2^672 * 7^2, 2^672 * 11^2 | |
073q | Problem:
All the points with integer coordinates in the $x y$-plane are coloured using three colours, red, blue and green, each colour being used at least once. It is known that the point $(0,0)$ is coloured red and the point $(0,1)$ is coloured blue. Prove that there exist three points with integer coordinates of dis... | [
"Solution:\n\nConsider the lattice points (points with integer coordinates) on the lines $y=0$ and $y=1$, other than $(0,0)$ and $(0,1)$. If one of them, say $A=(p, 1)$, is coloured green, then we have a right-angled triangle with $(0,0)$, $(0,1)$ and $A$ as vertices, all having different colours. (See Figures 1 an... | India | INMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0au4 | Problem:
Find all integer values of $n$ that will make $\frac{6 n^{3}-n^{2}+2 n+32}{3 n+1}$ an integer. | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 0, 10 | |
04aj | Let $a$ and $b$ be positive integers. What are the possible units digits of the number $(a+b)^5 - (a^5+b^5)$? | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 0 | |
00k6 | Determine all real numbers $x$ and $y$ such that
$$
x^2 + x = y^3 - y.
$$
$$
y^2 + y = x^3 - x.
$$ | [
"We factorise both equations and obtain\n$$\nx(x + 1) = (y - 1)y(y + 1), \\quad (1)\n$$\n$$\ny(y + 1) = (x - 1)x(x + 1). \\quad (2)\n$$\nWe insert (2) into (1) and obtain\n$$\nx(x + 1) = (y - 1)(x - 1)x(x + 1). \\quad (3)\n$$\nWe first consider the cases $x = 0$ and $x = -1$, where (2) results in $y \\in \\{0, -1\\... | Austria | Austria 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (0, 0), (0, -1), (-1, 0), (-1, -1), (2, 2) | |
0kus | Let $n \ge m \ge 1$ be integers. Prove that
$$
\sum_{k=m}^{n} \left( \frac{1}{k^2} + \frac{1}{k^3} \right) \ge m \cdot \left( \sum_{k=m}^{n} \frac{1}{k^2} \right)^2 .
$$ | [
"**First solution (authors)** By Cauchy-Schwarz, we have\n$$\n\\begin{align*}\n\\sum_{k=m}^{n} \\frac{k+1}{k^3} &= \\sum_{k=m}^{n} \\frac{\\left(\\frac{1}{k^2}\\right)^2}{\\frac{1}{k(k+1)}} \\\\\n&\\ge \\frac{\\left(\\frac{1}{m^2} + \\frac{1}{(m+1)^2} + \\dots + \\frac{1}{n^2}\\right)^2}{\\frac{1}{m(m+1)} + \\frac{... | United States | USA TSTST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
04ak | Among all the points $z$ in the complex plane satisfying $|z+3| + |z-3| = 10$ determine the point closest to the line through $(-3+4i)$ and $(-8+i)$. | [] | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance... | English | proof and answer | -3 + (16/5)i | |
06s4 | An integer $a$ is called friendly if the equation $(m^{2}+n)(n^{2}+m)=a(m-n)^{3}$ has a solution over the positive integers.
a. Prove that there are at least 500 friendly integers in the set $\{1,2, \ldots, 2012\}$.
b. Decide whether $a=2$ is friendly. | [
"a. Every $a$ of the form $a=4k-3$ with $k \\geq 2$ is friendly. Indeed the numbers $m=2k-1>0$ and $n=k-1>0$ satisfy the given equation with $a=4k-3$:\n$$\n((2k-1)^{2}+(k-1))((k-1)^{2}+(2k-1)) = (4k-3)k^{3} = a(m-n)^{3}.\n$$\nHence $5, 9, \\ldots, 2009$ are friendly and so $\\{1,2, \\ldots, 2012\\}$ contains at lea... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | There are at least 502 friendly integers in {1,2,…,2012} (for example, all numbers 5, 9, …, 2009). The integer 2 is not friendly. | |
0g3m | Problem:
Let $m \geq n$ be positive integers. Frieder is given $mn$ posters of Linus with different integer dimensions $k \times l$ with $1 \leq k \leq m$ and $1 \leq l \leq n$. He must put them all up one by one on his bedroom wall without rotating them. Every time he puts up a poster, he can either put it on an empt... | [
"Solution:\n\nWe introduce the following definitions:\n- A chain is a sequence of posters, each of which covers the previous one completely.\n- An antichain is a group of posters, none of which can be placed over any of the others because it would not cover any of the others completely.\n- The minimal and maximal e... | Switzerland | Final round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | m n(n+1)/2 | |
072i | Let $n$ be a positive integer divisible by $4$. Find the number of permutations $\sigma$ of $(1, 2, 3, \ldots, n)$ which satisfy the condition $\sigma(j) + \sigma^{-1}(j) = n + 1$, for all $j \in \{1, 2, 3, \ldots, n\}$. | [
"Let us take $n = 4k$. First we observe that such $\\sigma$ cannot have a fixed point; for if $\\sigma(j) = j$ for some $j$, then $\\sigma^{-1}(j) = j$ and hence $n + 1 = \\sigma(j) + \\sigma^{-1}(j) = 2j$, which is impossible because $n$ is divisible by $4$.\n\nSuppose $\\sigma(a) = b$, where $a \\neq b$. Then $\\... | India | Indija TS 2006 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | (n/2)! / (n/4)! | |
0itl | Problem:
Find the product of all real $x$ for which
$$
2^{3x+1} - 17 \cdot 2^{2x} + 2^{x+3} = 0
$$ | [
"Solution:\nWe can re-write the equation as $2^{x}\\left(2 \\cdot (2^{x})^{2} - 17 \\cdot 2^{x} + 8\\right) = 0$, or $2 \\cdot (2^{x})^{2} - 17 \\cdot 2^{x} + 8 = 0$. Make the substitution $y = 2^{x}$. Then we have $2y^{2} - 17y + 8 = 0$, which has solutions (by the quadratic formula) $y = \\frac{17 \\pm \\sqrt{289... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | -3 | |
0ckx | Let $ABC$ be a right isosceles triangle with right angle at $A$, $O$ the midpoint of the hypotenuse $BC$, $E$ the midpoint of the segment $CO$, $M$ the midpoint of $AC$, and $D$ the midpoint of $AM$. Let $F$ be the intersection of lines $OD$ and $AE$.
a) Prove that $MF \perp OD$.
b) Prove t... | [
"*First solution for a).* Since $ME \\perp OC$, it suffices to show that the quadrilateral $MEOF$ is cyclic.\n\n(1) Since $ME \\parallel AO$, we have $\\angle MEF = \\angle OAE$, therefore it remains to prove that $\\angle MOD = \\angle OAE$. This follows from similarity: $\\triangle MOD \\sim \\triangle OAE$, beca... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
04fy | Determine all positive integers $n$ such that
$$
3^{2n+1} - 4^{n+1} + 6^n
$$
is a prime number. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 1 | |
0kjp | A cube is constructed from 4 white unit cubes and 4 blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
(A) 7 (B) 8 (C) 9 (D) 10 (E) 11 | [
"Rotate the cube so that the number of small blue cubes showing in the front face is maximized.\n\n* If the front face contains four blue cubes, this gives 1 possible construction.\n\n* If the front face contains three blue cubes, then the back face must contain one blue cube, which can be in any of 4 positions. No... | United States | Fall 2021 AMC 10 B | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | MCQ | A | |
0f9s | Problem:
The point $P$ lies inside the triangle $ABC$. A line is drawn through $P$ parallel to each side of the triangle. The lines divide $AB$ into three parts length $c$, $c'$, $c''$ (in that order), and $BC$ into three parts length $a$, $a'$, $a''$ (in that order), and $CA$ into three parts length $b$, $b'$, $b''$ ... | [
"Solution:\n\n\n\nThe three small triangles are similar, so $a/a'' = c'/c = b''/b'$ and $a/a' = c'/c'' = b''/b$. Hence $(a/a'')(b/b'') = (c'/c)(c''/c') = c''/c$, so $abc = a''b''c''$. Similarly, $(a/a')(c/c') = (b''/b)(b''/b'') = b/b$, so $abc = a'b'c'.$"
] | Soviet Union | 24th ASU | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02un | Problem:
Encontre todas as soluções, no conjunto dos números reais positivos, do sistema de equações:
$$
\left\{\begin{array}{l}
x(x+y+z)=26 \\
y(x+y+z)=27 \\
z(x+y+z)=28
\end{array}\right.
$$ | [
"Solution:\nSomando as três equações, obtemos $(x+y+z)^2=81$, ou seja, $x+y+z=9$, pois queremos soluções positivas.\n\nSubstituindo tal valor em cada equação, temos:\n\n$x = 26/9$, $y = 27/9 = 3$ e $z = 28/9$.\n\nAssim, a única solução do sistema é $(x, y, z) = (26/9, 3, 28/9)$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (26/9, 3, 28/9) | |
09tb | Problem:
Bepaal alle mogelijke waarden van $\frac{1}{x}+\frac{1}{y}$ als $x$ en $y$ reële getallen (ongelijk aan $0$) zijn die voldoen aan $x^{3}+y^{3}+3 x^{2} y^{2}=x^{3} y^{3}$. | [
"Solution:\n\nWe herschrijven het gegeven naar $x^{3}+y^{3}-x^{3} y^{3}=-3 x^{2} y^{2}$, waarmee we vinden\n$$\n\\begin{aligned}\n(x+y)^{3}-x^{3} y^{3} & =x^{3}+3 x^{2} y+3 x y^{2}+y^{3}-x^{3} y^{3} \\\\\n& =-3 x^{2} y^{2}+3 x^{2} y+3 x y^{2}=3 x y(-x y+x+y)\n\\end{aligned}\n$$\nDoor te gebruiken dat $a^{3}-b^{3}=(... | Netherlands | MO-selectietoets | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | -2 and 1 | |
06se | In the plane, 2013 red points and 2014 blue points are marked so that no three of the marked points are collinear. One needs to draw $k$ lines not passing through the marked points and dividing the plane into several regions. The goal is to do it in such a way that no region contains points of both colors.
Find the min... | [
"Answer. $k=2013$.\n\nFirstly, let us present an example showing that $k \\geqslant 2013$. Mark 2013 red and 2013 blue points on some circle alternately, and mark one more blue point somewhere in the plane. The circle is thus split into 4026 arcs, each arc having endpoints of different colors. Thus, if the goal is ... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2013 | |
0hs1 | Problem:
Let $A$ be the number of 4-tuples $(x, y, z, t)$ of positive integers smaller than $2006^{2006}$ such that
$$
x^{3}+y^{2}=z^{3}+t^{2}+1,
$$
and let $B$ the number of 4-tuples $(x, y, z, t)$ of positive integers smaller than $2006^{2006}$ such that
$$
x^{3}+y^{2}=z^{3}+t^{2}.
$$
Prove that $B>A$. | [
"Solution:\nFor each natural number $k$, denote by $l_{k}$ the number of pairs $(x, y)$ such that $x^{3}+y^{2}=k$. Then $B=l_{2}^{2}+l_{3}^{2}+\\cdots+l_{r}^{2}$ where $r$ is the maximal $k$ for which $l_{k} \\neq 0$. Similarly, $A=l_{2} l_{3}+l_{3} l_{4}+\\cdots+l_{r-1} l_{r}$. Now\n$$\nB=\\frac{l_{2}^{2}}{2}+\\fr... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
09ie | Compute the sum:
$$
\frac{1}{\left(\frac{1}{2022}\right)^2 + 1} + \frac{1}{\left(\frac{2}{2021}\right)^2 + 1} + \frac{1}{\left(\frac{3}{2020}\right)^2 + 1} + \dots + \frac{1}{\left(\frac{2021}{2}\right)^2 + 1} + \frac{1}{\left(\frac{2022}{1}\right)^2 + 1}
$$ | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1011 | |
021p | Problem:
Um heptágono regular está imprensado entre dois círculos, como mostra a figura abaixo. Seus vértices estão inscritos na circunferência maior e seus lados são tangentes à circunferência menor. Os lados do polígono medem $2~\mathrm{cm}$ e são tangentes ao círculo menor e, além disso, seus vértices pertencem à c... | [
"Solution:\n\nSe $R_{1}$ e $R_{2}$ denotam os raios do círculo externo e interno ao heptágono, respectivamente, podemos concluir que a área procurada é $\\pi \\cdot R_{1}^{2} - \\pi \\cdot R_{2}^{2}$. O apótema do heptágono é igual a $R_{2}$ e forma com metade do lado e o raio $R_{1}$ um triângulo retângulo. Daí, p... | Brazil | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | π | |
0kqe | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD = BC = 333$. The angle bisectors of $\angle A$ and $\angle D$ meet at $P$, and the angle bisectors of $\angle B$ and $\angle C$ meet at $Q$. Find $PQ$. | [] | United States | 2022 AIME I | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 242 | |
0brd | Let $ABC$ be a triangle and let $D$ be a point on the segment $BC$, $D \neq B$ and $D \neq C$. The circle $ABD$ meets the segment $AC$ again at an interior point $E$. The circle $ACD$ meets the segment $AB$ again at an interior point $F$. Let $A'$ be the reflection of $A$ in the line $BC$. The lines $A'C$ and $DE$ meet... | [
"Let $\\sigma$ denote reflection in the line $BC$. Since $\\angle BDF = \\angle BAC = \\angle CDE$, by concyclicity, the lines $DE$ and $DF$ are images of one another under $\\sigma$, so the lines $AC$ and $DF$ meet at $P' = \\sigma(P)$, and the lines $AB$ and $DE$ meet at $Q' = \\sigma(Q)$.\nConsequently, the line... | Romania | 2016 Eighth Romanian Master of Mathematics | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08st | Determine all the real-valued functions $f$ defined on the real line, which satisfies for all real numbers $x$ and $y$
$$
f(x+y)f(f(x)-y) = xf(x) - yf(y).
$$ | [
"Let $f(0) = a$ and in the given functional equation\n$$\nf(x+y)f(f(x)-y) = xf(x) - yf(y) \\quad (1)\n$$\nsubstitute $x = y = 0$, then we get\n$$\nf(0)f(a) = 0,\n$$\nfrom which we obtain $f(0) = 0$ or $f(a) = 0$. Since $f(0) = 0$ implies that $a = 0$ by definition, we see that $f(a) = 0$ always holds.\nLet $t$ be a... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | Either f(x) = x for all real x, or f(x) = 0 for all nonzero x and f(0) is an arbitrary real constant. | |
00k2 | Let $ABC$ be an acute triangle and $D$ be a point on the altitude through $C$.
Prove that the mid-points of the line segments $AD$, $BD$, $BC$ and $AC$ form a rectangle. | [
"We denote with $M_{XY}$ the mid-point of the line segment $XY$.\nUsing the intercept theorem, we deduce that\n* $M_{AD}M_{BD}$ is parallel to $AB$.\n* $M_{AC}M_{BC}$ is parallel to $AB$.\n* $M_{AC}M_{AD}$ is parallel to $CD$.\n* $M_{BC}M_{BD}$ is parallel to $CD$.\nTherefore, $M_{AD}M_{BD}$ is parallel to $M_{AC}M... | Austria | AustriaMO2013 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0d56 | The total number of languages used in KAUST is $n$. For each positive integer $k \leq n$, let $A_{k}$ be the set of all those people in KAUST who can speak at least $k$ languages; and let $B_{k}$ be the set of all people $P$ in KAUST with the property that, for any $k$ pairwise different languages (used in KAUST), $P$ ... | [
"Let $\\mathcal{L}$ be the set of all the languages used in KAUST.\n\na.\nAssume $2k \\geq n+1$. Consider $P \\in A_{k}$, and let $\\mathcal{L}_P \\subseteq \\mathcal{L}$ be the set of all languages spoken by $P$. The set $\\mathcal{L}_P$ contains at least $k$ languages.\n\nLet $\\mathcal{L}' \\subseteq \\mathcal{L... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English, Arabic | proof only | null | |
0fuq | Problem:
Sei $ABC$ ein Dreieck mit $\Varangle BAC = 60^\circ$. Finde alle Punkte $P$ im Innern dieses Dreiecks mit folgender Eigenschaft:
Ist $D$ die Projektion von $P$ auf die Gerade $BC$, $E$ die Projektion von $P$ auf $CA$ und $F$ die Projektion von $P$ auf $AB$, dann gilt $\Varangle EDF = 30^\circ$. | [
"Solution:\n\nWir setzen $\\alpha = \\Varangle CAB$, $\\beta = \\Varangle ABC$, $\\gamma = \\Varangle BCA$, sowie $\\varphi = \\Varangle EDP$ und $\\psi = \\Varangle FDP$. Wegen $\\Varangle PEC = \\Varangle PDC = 90^\\circ$ ist $EPDC$ ein Sehnenviereck, daher gilt $\\Varangle ECP = \\varphi$. Analog ist $FPDB$ ein ... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | All interior points P such that ∠BPC = 90°, i.e., the intersection of the triangle’s interior with the Thales circle having diameter BC. | |
0fpz | Sean $k$ un entero positivo y $a_1, a_2, \dots, a_n$ dígitos. Probar que existe un entero positivo $n$ tal que los últimos $2k$ dígitos de $2^n$ son, es este orden, $a_1, a_2, \dots, a_k, b_1, b_2, \dots, b_k$, para ciertos dígitos $b_1, b_2, \dots, b_k$. | [] | Spain | XXXI Olimpiada Iberoamericana de Matemáticas | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | Spanish | proof only | null | |
00wn | Problem:
The vertices of a convex 1991-gon are enumerated with integers from $1$ to $1991$. Each side and diagonal of the $1991$-gon is coloured either red or blue. Prove that, for an arbitrary renumeration of vertices, one can find integers $k$ and $l$ such that the line connecting vertices with numbers $k$ and $l$ b... | [
"Solution:\n\nAssume there exists a renumeration such that for any numbers $1 \\leq k < l \\leq n$ the segment connecting vertices numbered $k$ and $l$ before the renumeration has a different colour than the segment connecting vertices with the same numbers after the renumeration. Then there has to be an equal numb... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0i37 | Problem:
What is the last digit of $1^{1} + 2^{2} + 3^{3} + \cdots + 100^{100}$? | [
"Solution:\nLet $L(d, n)$ be the last digit of a number ending in $d$ to the $n$th power. For $n \\geq 1$, we know that $L(0, n) = 0$, $L(1, n) = 1$, $L(5, n) = 5$, $L(6, n) = 6$. All numbers ending in odd digits in this series are raised to odd powers; for odd $n$, $L(3, n) = 3$ or $7$, $L(7, n) = 3$ or $7$, $L(9,... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 0 | |
00we | Problem:
Let $a_{0}>0$, $c>0$ and
$$
a_{n+1} = \frac{a_{n} + c}{1 - a_{n} c}, \quad n = 0, 1, \ldots
$$
Is it possible that the first 1990 terms $a_{0}, a_{1}, \ldots, a_{1989}$ are all positive but $a_{1990} < 0$? | [
"Solution:\nObviously we can find angles $0 < \\alpha, \\beta < 90^{\\circ}$ such that $\\tan \\alpha > 0$, $\\tan (\\alpha + \\beta) > 0$, $\\ldots$, $\\tan (\\alpha + 1989 \\beta) > 0$ but $\\tan (\\alpha + 1990 \\beta) < 0$. Now it suffices to note that if we take $a_{0} = \\tan \\alpha$ and $c = \\tan \\beta$ t... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | Yes | |
0jyw | Problem:
Find all pairs $(a, b)$ of positive integers such that $a^{2017}+b$ is a multiple of $a b$. | [
"Solution:\n\nWe want $a b \\mid a^{2017}+b$. This gives that $a \\mid b$. Therefore, we can set $b = b_{2017} a$. Substituting this gives $b_{2017} a^{2} \\mid a^{2017} + b_{2017} a$, so $b_{2017} a \\mid a^{2016} + b_{2017}$. Once again, we get $a \\mid b_{2017}$, so we can set $b_{2017} = b_{2016} a$. Continuing... | United States | February 2017 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(1, 1), (2, 2^{2017})] | |
06mu | Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$. Diagonals $AC$ and $BD$ meet at $E$. $F$ and $G$ are points on segments $AB$ and $CD$ respectively. Suppose $AF = 15$, $FB = 13$, $BE = 30$, $ED = 13$, $DG = 7.5$ and $GC = 6.5$. Let $P$ be a point such that $PF \perp AB$ and $PG \perp CD$. Find $\frac{PE}{PO}... | [
"The answer is $\\frac{PE}{PO} = 5$.\nNote that $\\frac{AF}{FB} = \\frac{DG}{GC}$. Now if we let $M$ and $N$ be the perpendicular feet from $O$ to $AB$ and $CD$ respectively, we have $\\frac{AM}{MB} = \\frac{DN}{NC}$. Also if we let $H$ and $K$ be the perpendicular feet from $E$ to $AB$ and $CD$ respectively, then ... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 5 | |
02zr | Problem:
Seja $f: \mathbb{R} \rightarrow \mathbb{R}$ uma função ímpar, isto é, uma função que satisfaz $-f(x)=f(-x)$ para todo $x \in \mathbb{R}$. Suponha que $f(x+5)=f(x)$ para todo $x \in \mathbb{R}$ e que $f(1 / 3)=1$. Determine o valor da soma:
$$
f(16 / 3)+f(29 / 3)+f(12)+f(-7)
$$ | [
"Solution:\n\nTemos\n$$\n\\begin{aligned}\nf\\left(\\frac{1}{3}\\right) & =1 \\\\\nf\\left(\\frac{1}{3}+5\\right) & =1 \\\\\nf\\left(\\frac{16}{3}\\right) & =1\n\\end{aligned}\n$$\nComo $f$ é uma função ímpar, i.e., $f(-x)=-f(x)$, segue que\n$$\n\\begin{aligned}\nf\\left(-\\frac{1}{3}\\right) & =-1 \\\\\nf\\left(-\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 0 | |
0c90 | Let $R$ be a ring with only finitely many invertible elements. Prove that the two statements below are equivalent:
(a) For every non-invertible element $x$ of $R$, there exists a non-invertible element $y$ of $R$ such that $xy = x + y$.
(b) Every non-invertible element of $R$ is nilpotent.
(An element $z$ of $R$ is cal... | [
"We show that (a) implies (b). Let $x \\in R \\setminus U(R)$, where $U(R)$ is the set of invertible elements of $R$, and let $y \\in R \\setminus U(R)$ be such that $xy = x + y$, i.e., $(x - 1)(y - 1) = 1$. Similarly, $(y - 1)(z - 1) = 1$ for some $z \\in R \\setminus U(R)$. Write\n$$\nx - 1 = (x - 1) \\cdot 1 = (... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
0evp | Let $a$, $b$, $c$, $d$ be pairwise coprime positive odd numbers. For positive integers $n$, we define
$$
f(n) = \lfloor \frac{n}{a} \rfloor + \lfloor \frac{n}{b} \rfloor + \lfloor \frac{n}{c} \rfloor + \lfloor \frac{n}{d} \rfloor.
$$
Prove that
$$
\sum_{n=1}^{abcd} (-1)^{f(n)} = 1.
$$ | [
"As $a$, $b$, $c$, $d$ are all odd, we have\n$$\n\\begin{aligned}\nf(n) &\\equiv 4n - \\left(a\\lfloor\\frac{n}{a}\\rfloor + b\\lfloor\\frac{n}{b}\\rfloor + c\\lfloor\\frac{n}{c}\\rfloor + d\\lfloor\\frac{n}{d}\\rfloor\\right) \\\\\n&\\equiv \\left(n - a\\lfloor\\frac{n}{a}\\rfloor\\right) + \\left(n - b\\lfloor\\f... | South Korea | The 37th Korean Mathematical Olympiad Final Round | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
005a | Hugo hace la lista, en orden ascendente, de los primeros 2007 números naturales cuya suma de dígitos es igual a 5.
¿Cuál es el último número de la lista de Hugo? | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | Spanish | proof and answer | 10000000040 | |
01rr | Given two hyperbolae $H_1$ and $H_2$ with the equations $y = 1/x$ and $y = -1/x$, respectively. Let $M$ be an arbitrary point on $H_1$. Let $M_1$ and $M_2$ be the tangency points of the lines $MM_1$ and $MM_2$ with the hyperbola $H_2$.
Prove that the line $M_1M_2$ is tangent line of $H_1$. | [
"Without loss of generality we may assume that the point $M(a; 1/a)$ lies on the upper-right half-hyperbola $H_1$, $y = 1/x$. $M_1(b_1; -1/b_1)$ and $M_2(b_2; -1/b_2)$ be the points of tangency with $H_2$, mentioned in the problem condition (see the Fig.).\n\n\n\nSince the derivative of the... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0e50 | Problem:
Tabela $7 \times 7$ je razdeljena na 49 kvadratkov. Na to tabelo postavljamo ploščice oblike
ki pokrijejo vsaka po dve polji. Najmanj koliko ploščic moramo postaviti na tabelo, da bo imelo vsako nepokrito polje vsaj eno sosednje polje pokrito? (Polji sta sosednji, če imata skupno str... | [
"Solution:\n\nNa tabelo lahko položimo devet ploščic, kot prikazuje prva slika. Vsako nepokrito polje ima vsaj eno sosednje polje pokrito. Utemeljimo, da z manj kot devetimi ploščicami tega ne moremo narediti. Opazujmo ploščico, ki leži na tabeli.\n\nOznačimo še polja, ki so sosednja pokrit... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 9 | |
08w8 | Suppose each digit of a positive integer is greater than or equal to $1$ and less than or equal to $7$. Suppose further that $n$ is a multiple of $7$ and every integer obtained by permuting digits of $n$ is also a multiple of $7$. Determine all such integers $n$. | [
"Let us show first that any positive integer $n$ satisfying the condition of the problem must have the same number in all of its digits. In order to show this it is enough to show that if the integers $a$ and $b$ appear among the digits of $n$, where $1 \\le a, b \\le 7$ and if $n$ satisfies the condition of the pr... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | All positive integers whose digits are all equal. If the common digit is seven, any length is allowed. If the common digit is between one and six, the number of digits must be a multiple of six. | |
06oj | We have $n \geq 2$ lamps $L_{1}, \ldots, L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows:
- if the lamp $L_{i}$ and its neighbours (only one neighbour for $i=1$ or $i=n$, two neighbours for other $i$) are in the same state, then $L_{i}$ is sw... | [
"a.\nExperiments with small $n$ lead to the guess that every $n$ of the form $2^{k}$ should be good. This is indeed the case, and more precisely: let $A_{k}$ be the $2^{k} \\times 2^{k}$ matrix whose rows represent the evolution of the system, with entries $0,1$ (for off and on respectively). The top row shows the ... | IMO | IMO 2006 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
09m8 | A semicircle $\Omega$ with diameter $AD$ is given. Let $O$ be the midpoint of $AD$. Let $B$ be a point on $\Omega$ and $M$ be a point on the segment $OA$ such that $2AM = BM$. Let the midpoint of segment $BM$ be $N$ and the midpoint of segment $BN$ be $K$. On the semicircle $\Omega$, the point $C$ is taken so that $OC$... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance ... | English | proof only | null | |
0crk | Does there exist a positive real number $a$ such that for every real $x$ the inequality
$$
|\cos x| + |\cos a x| > \sin x + \sin a x
$$
holds? | [
"**Первое решение.** Предположим, что $0 < a \\le 1$. Тогда при $x = \\pi/2$ левая часть примет значение $|\\cos(a\\pi/2)|$, то есть будет не больше 1, в то время как правая часть будет равна $1 + \\sin(a\\pi/2)$, то есть она больше 1. Итак, неравенство не выполнено.\n\nЕсли же $a > 1$, то, обозначив $a x = t$ и $b... | Russia | XL Russian mathematical olympiad | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | No | |
0ily | Problem:
Find an explicit, closed form formula for
$$
\sum_{k=1}^{n} \frac{k \cdot(-1)^{k} \cdot\binom{n}{k}}{n+k+1}
$$ | [
"Solution:\nAnswer: $\\frac{-1}{\\binom{2 n+1}{n}}$ or $-\\frac{n!(n+1)!}{(2 n+1)!}$ or obvious equivalent.\n\nConsider the interpolation of the polynomial $P(x)=x \\cdot n!$ at $x=0,1, \\ldots, n$. We obtain the identity\n$$\n\\begin{aligned}\nP(x) & =x \\cdot n!=\\sum_{k=0}^{n} k \\cdot n!\\prod_{j \\neq k} \\fra... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | -n!(n+1)!/(2n+1)! | |
02zd | Problem:
Dizemos que um número é racional se ele pode ser escrito da forma $\frac{p}{q}$, com $p$ e $q$ números inteiros. Se um número real não é racional, dizemos que ele é irracional. Por exemplo, $\frac{1}{4}$ é um número racional e $\sqrt{3}$ é irracional. Encontre todos os pares de números racionais $(a, b)$ tais... | [
"Solution:\n\nElevando ao quadrado, obtemos as equações equivalentes:\n$$\n\\begin{aligned}\na+b+2 \\sqrt{a b} & =2+\\sqrt{3} \\\\\n2 \\sqrt{a b} & =2-a-b+\\sqrt{3}\n\\end{aligned}\n$$\nComo $a$ e $b$ são números racionais, $r=2-a-b$ também é um número racional. Eleve a última equação ao quadrado:\n$$\n\\begin{alig... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | (1/2, 3/2) and (3/2, 1/2) | |
08ot | Problem:
Let $n \geq 1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^{2}$ squares of side length $1$. Find the number of parallelograms which have vertices among the vertices of the $n^{2}$ squares of side length $1$, with both sides smaller or equal to $2$, and w... | [
"Solution:\nWe can divide all these parallelograms into 7 classes (types I-VII), according to Figure.\n\n\n\nType 1: There are $n$ ways to choose the strip for the horizontal (shorter) side of the parallelogram, and $(n-1)$ ways to choose the strip (of the width $2$) for the vertical (longe... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 7n^2 - 12n + 1 | |
01yi | Point $D$ is marked on the half-circle with the diameter $AB$ and the center $O$. The points $E$ and $F$ are the midpoints of lesser arcs $AD$ and $BD$ respectively. It is known that $O$ lies on the line passing through the orthocenters of the triangles $ADF$ and $BDE$.
Find all possible values of the angle $AOD$. | [
"Let $H_1$ and $H_2$ be the intersection points of the altitudes of the triangles $ADF$ and $BDE$. Note that the point $O$ is the circumcenter of both triangles $ADF$ and $BDE$. Therefore the coinciding lines $H_1O$ and $H_2O$ are the Euler lines of these triangles and they contain the intersection points $M_1$ and... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 30° or 150° | |
0hew | a) For non-zero integers $a, b, c$ it holds that: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} = \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a}$.
Does it follow then that: $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{a}{c} + \frac{c}{b} + \frac{b}{a}$?
6) Consider all triples of non-zero integers $(a, b, c)$, whi... | [
"Consider the following transformation:\n$$\n\\left( \\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{a} \\right) - \\left( \\frac{a^2}{c} + \\frac{c^2}{b} + \\frac{b^2}{a} \\right) = \\frac{a^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c}{abc}\n$$\nNow we transform only numerator:\n$$\n\\begin{aligned} a^3 c - a^3 b + ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) No. 6) 65890313340 | |
0bjr | Let $ABCDEF$ be a regular hexagon with side length $a$. At point $A$, the perpendicular $AS$, with length $2a\sqrt{3}$, is erected on the hexagon's plane. The points $M, N, P, Q$, and $R$ are the projections of point $A$ onto the lines $SB, SC, SD, SE$, and $SF$, respectively.
a) Prove that the points $M, N, P, Q, R$ ... | [
"a) Using the Three Perpendiculars Theorem, from $SA \\perp (ABC)$ and $AB \\perp BD$, it results $SB \\perp BD$. Since $BD \\perp AB$ and $BD \\perp SB$, it follows that $BD \\perp (SAB)$, hence $BD \\perp AM$.\n\nSince $AM \\perp SB$, it results $AM \\perp (SBD)$, hence $AM \\perp SD$. We also have $SD \\perp AP$... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 30° | |
03bl | Find the least positive integer $n$, such that there exist positive integers $a$, $b$ and $c$, none of which is a perfect square and
$$
a^3 + b^3 + c^3 - 3abc = 2013^n.
$$ | [
"It is easy to be seen that the left hand side is divisible by $9$ implying that the equation has no solution for $n = 1$.\n\nLet $n=2$. We have $2013^2 = (3 \\cdot 11 \\cdot 61)^2$, $3^2 = 2^3 + 1^3$ and $11 \\cdot 61 = 8^3 + 4^3 + 1^3 + 3 \\cdot 8 \\cdot 4 \\cdot 1$. We use the identity\n$$\n(a^3 + b^3 + c^3 - 3a... | Bulgaria | Team selection test for the 54th IMO | [
"Number Theory > Other",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 2 | |
01en | Quadrilateral $ABCD$ is circumscribed about a circle $\omega$. $E$ is the intersection point of $\omega$ and the diagonal $AC$, which is nearest to $A$. Point $F$ is diametrically opposite to point $E$ in the circle $\omega$. The line which is tangent to $\omega$ in the point $F$ intersects lines $AB$ and $BC$ in point... | [
"Denote by $X$ the intersection point of the lines $A_1A_2$ and $AC$.\nProve that $X$ is a contact point of escribed circle of $\\triangle AA_1A_2$ with side $A_1A_2$. Indeed, consider a homothety with center $A$ which maps incircle $\\omega$ of $\\triangle AA_1A_2$ to its escribed circle. This homothety maps the l... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler li... | English | proof only | null | |
04i3 | Circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Line $l$ intersects circle $k_1$ in points $C$ and $E$, and circle $k_2$ in points $D$ and $F$ in such a way that $D$ is between $C$ and $E$, and $E$ is between $D$ and $F$. Lines $CA$ and $BF$ intersect in point $G$, and lines $DA$ and $BE$ in point $H$.
Prove t... | [
"It suffices to show that $\\angle ECA = \\angle HGA$.\nSince the quadrilateral $ACBE$ is cyclic, we have $\\angle ECA = \\angle EBA$, so it suffices to show that $ABGH$ is a cyclic quadrilateral.\nFrom triangle $DEH$ we have $\\angle DHE = 180^\\circ - \\angle EDH - \\angle HED$, i.e.\n$$\n\\angle AHB = \\angle DH... | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0c7c | In the regular hexagonal prism $ABCDEFA_1B_1C_1D_1E_1F_1$ one constructs the points $P, Q$, the projections of the point $A$ on the lines $A_1B$, respectively $A_1C$, and $R, S$, the projections of the point $D_1$ on the lines $A_1D$, respectively $C_1D$.
a) Determine the measure of the angle of the planes $(AQP)$ and... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 0 degrees | |
04ah | A regular tetrahedron $ABXY$ is given inside the unit cube $ABCD'A'B'C'D'$ so that the point $X$ lies in the plane $ABCD$. Determine the distance between the points $Y$ and $A'$. | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | sqrt(2 - 2*sqrt(2/3)) | |
0ao0 | Problem:
If $x^{6}=64$ and $\left(\frac{2}{x}-\frac{x}{2}\right)^{2}=b$, then a function $f$ that satisfies $f(b+1)=0$ is
(a) $f(x)=1-2^{x-1}$
(c) $f(x)=x^{2}+x$
(b) $f(x)=2^{x-1}$
(d) $f(x)=2 x-1$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | (a) f(x)=1-2^{x-1} | |
0ib3 | Problem:
Let $P(x) = x^{3} - \frac{3}{2} x^{2} + x + \frac{1}{4}$. Let $P^{[1]}(x) = P(x)$, and for $n \geq 1$, let $P^{[n+1]}(x) = P^{[n]}(P(x))$. Evaluate $\int_{0}^{1} P^{[2004]}(x) \, d x$. | [
"Solution:\n$1 / 2$\n\nNote that $P(1-x) = 1 - P(x)$. It follows easily by induction that $P^{[k]}(1-x) = 1 - P^{[k]}(x)$ for all positive integers $k$. Hence\n$$\n\\begin{aligned}\n\\int_{0}^{1} P^{[2004]}(x) \\, d x & = \\int_{0}^{1} 1 - P^{[2004]}(1-x) \\, d x \\\\\n& = 1 - \\int_{0}^{1} P^{[2004]}(1-x) \\, d x ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/2 | |
0045 | Dada una circunferencia $\tau$, considere un cuadrilátero $ABCD$ con sus cuatro lados tangentes a $\tau$, con $AD$ tangente a $\tau$ en $P$ y $CD$ tangente a $\tau$ en $Q$. Sean $X$ e $Y$ los puntos donde $BD$ corta a $\tau$, y $M$ el punto medio de $XY$. Demuestre que $\angle AMP = \angle CMQ$. | [] | Argentina | 21º OLIMPIADA IBEROAMERICANA DE MATEMÁTICA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Inversion"
] | Español | proof only | null | |
07ay | Let $C$ be a circle and $P$ a point outside of it. $PA$ and $PB$ are the two tangent lines to this circle and point $K$ is chosen arbitrarily on the segment $AB$. The circumcircle of triangle $PBK$ intersects circle $C$ for the second time at $T$. Let $P'$ be the reflection of $P$ with respect to $A$. Show that $\angle... | [
"In this solution, all of the arcs considered are from circle $C$.\n\n\nSince quadrilateral $KTPB$ is cyclic, $\\angle AKT = \\angle BPT$.\n$$\n\\left.\n\\begin{array}{l}\n\\angle TAK = \\frac{\\widehat{TB}}{2} = \\angle TBP \\\\\n\\angle AKT = \\angle BPT\n\\end{array}\n\\right\\} \\Righta... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0c7z | Ana and Bogdan play the following turn based game: Ana starts with a pile of $n$ ($n \ge 3$) stones. At his turn each player has to split one pile. The winner is the player who can make at his turn all the piles to have at most two stones. Depending on $n$, determine which player has a winning strategy. | [
"If $n = 3$ or $n = 4$ Ana wins at her first move. If $n$ is odd, greater than 3, Bogdan will win. In this case, Ana has to start by making a pile with an even number of stones. Bogdan will split this pile into a pile with one stone and the rest into an odd pile. Ana has to make an even pile again and Bogdan contin... | Romania | 70th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Ana wins for n = 3 and for all even n; Bogdan wins for all odd n ≥ 5. | |
0d84 | Let $ABC$ be a non isosceles triangle with circumcircle $(O)$ and incircle $(I)$. Denote $(O_{1})$ as the circle that is externally tangent to $(O)$ at $A'$, and also tangent to the lines $AB$, $AC$ at $A_b$, $A_c$ respectively. Define the circles $(O_{2})$, $(O_{3})$ and the points $B'$, $C'$, $B_c$, $B_a$, $C_a$, $C_... | [
"1) Consider the inversion with center $J$ and ratio equal to the power of $J$ to the three circles $(O_{1})$, $(O_{2})$, $(O_{3})$ as a function $f$.\nIt is easy to see that\n$$\nf\\left((O_{1})\\right) = (O_{1}), \\quad f\\left((O_{2})\\right) = (O_{2}), \\quad f\\left((O_{3})\\right) = (O_{3}) .\n$$\nOn the othe... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"... | English | proof only | null | |
0kxe | Problem:
Let $ABC$ be a triangle. Point $D$ lies on segment $BC$ such that $\angle BAD = \angle DAC$. Point $X$ lies on the opposite side of line $BC$ as $A$ and satisfies $XB = XD$ and $\angle BXD = \angle ACB$. Analogously, point $Y$ lies on the opposite side of line $BC$ as $A$ and satisfies $YC = YD$ and $\angle CY... | [
"Solution:\n\nLet $I$ and $I_{A}$ be the incenter and the $A$-excenter of $\\triangle ABC$. The key observation is that $X$ is the circumcenter of $\\triangle BDI_{A}$. To see why this is true, note that\n$$\n\\angle BXD = \\angle C = 2\\angle ICB = 2\\angle II_{A}B = 2\\angle DI_{A}B.\n$$\... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k7k | Problem:
A palindrome is a string that does not change when its characters are written in reverse order. Let $S$ be a 40-digit string consisting only of 0's and 1's, chosen uniformly at random out of all such strings. Let $E$ be the expected number of nonempty contiguous substrings of $S$ which are palindromes. Compute... | [
"Solution:\nNote that $S$ has $41-n$ contiguous substrings of length $n$, so we see that the expected number of palindromic substrings of length $n$ is just $(41-n) \\cdot 2^{-\\lfloor n / 2\\rfloor}$. By linearity of expectation, $E$ is just the sum of this over all $n$ from 1 to 40. However, it is much easier to ... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 113 | |
078m | Given that $a_1, a_2, \dots, a_{10}$ are positive real numbers, determine the smallest possible value of
$$
\sum_{i=1}^{10} \left\lfloor \frac{7a_i}{a_i + a_{i+1}} \right\rfloor
$$
where we define $a_{11} = a_1$. | [
"Let $m = 7$ and $n = 10$. We now claim that the minimum possible value is $m-1$ and this is achieved by letting $a_i = m^i$. Now, to show that this is indeed the minimum, we assume without loss of generality that $a_n$ is the largest amongst all $a_i$. Now,\n$$\n\\sum_{i=1}^{n} \\left\\lfloor \\frac{ma_i}{a_i + a_... | India | EGMO TST | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 6 |
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