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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ifw | Problem:
One hundred people are in line to see a movie. Each person wants to sit in the front row, which contains one hundred seats, and each has a favorite seat, chosen randomly and independently. They enter the row one at a time from the far right. As they walk, if they reach their favorite seat, they sit, but to av... | [
"Solution:\n\nLet $S(i)$ be the favorite seat of the $i$th person, counting from the right. Let $P(n)$ be the probability that at least $n$ people get to sit. At least $n$ people sit if and only if $S(1) \\geq n, S(2) \\geq n-1, \\ldots, S(n) \\geq 1$. This has probability:\n$$\nP(n) = \\frac{100-(n-1)}{100} \\cdot... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 10 | |
0k2o | Problem:
An isosceles right triangle $ABC$ has area $1$. Points $D$, $E$, $F$ are chosen on $BC$, $CA$, $AB$ respectively such that $DEF$ is also an isosceles right triangle. Find the smallest possible area of $DEF$.
Proposed by: Yuan Yao | [
"Solution:\n\nWithout loss of generality, suppose that $AB$ is the hypotenuse.\n\nIf $F$ is the right angle, then $F$ must be the midpoint of $AB$. To prove this, let $X$ and $Y$ be the feet from $F$ to $BC$ and $AC$. Since $\\angle XFY = \\angle DFE = 90^{\\circ}$, we have $\\angle XFD = \\angle YFE$ so\n$$\nXF = ... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane... | null | proof and answer | 1/5 | |
05dx | Problem:
Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$. The circumcircle $\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$.
Pro... | [
"Solution:\n\nA. The angle bisector of $\\angle ECD$ intersects the circumcircle of $CDE$ (which is $\\Gamma$) at the midpoint $M$ of arc $DBE$. It is well-known that the incentre is the intersection of the angle bisector segment $CM$ and the circle with centre at $M$ and passing through $D, E$. We will verify this... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance ... | null | proof only | null | |
0ktg | Ellina has twelve blocks, two each of red (R), blue (B), yellow (Y), green (G), orange (O), and purple (P). Call an arrangement of blocks *even* if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement RBBYGGYROPPO is even. Ellina arranges her blocks in a row in r... | [] | United States | 2022 AIME I | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 247 | |
038q | Prove that if $x, y, z > 0$, then
$$
\frac{(x+1)(y+1)^2}{3\sqrt[3]{z^2x^2}+1} + \frac{(y+1)(z+1)^2}{3\sqrt[3]{x^2y^2}+1} + \frac{(z+1)(x+1)^2}{3\sqrt[3]{y^2z^2}+1} \geq x+y+z+3.
$$ | [
"Denote by $S$ the left-hand side of the given inequality. The arithmetic mean–geometric mean inequality implies that $xy + x + y \\geq 3\\sqrt[3]{x^2y^2}$ and hence\n$$\nS \\geq \\frac{(x+1)(y+1)^2}{(z+1)(x+1)} + \\frac{(y+1)(z+1)^2}{(x+1)(y+1)} + \\frac{(z+1)(x+1)^2}{(y+1)(z+1)}.\n$$\nSetting $a = x + 1$, $b = y ... | Bulgaria | Spring Mathematical Tournament | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0eo0 | Prove that the positive integer $n$ is the product of exactly two primes that differ by 2 if and only if
$$
\phi(n)\sigma(n) = (n-3)(n+1).
$$
(Recall that $\phi(n)$ equals the number of positive integers less than or equal to $n$, relatively prime to $n$, and $\sigma(n)$ equals the sum of the positive divisors of $n$, ... | [
"Let $\\phi(n)\\sigma(n) = (n-3)(n+1)$, and let prime $p$ divide $n$ such that $n = p^a m$ for some $m$ not divisible by $p$. Then we have\n$$\n\\begin{align*}\n\\phi(n) &= \\phi(p^a)\\phi(m) = p^{a-1}(p-1)\\phi(m) \\\\\n\\sigma(n) &= \\sigma(p^a)\\sigma(m) = \\left(\\frac{p^{a+1}-1}{p-1}\\right)\\sigma(m),\n\\end{... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0cix | Determine the natural numbers $n$ that are perfect squares, $2 \le n \le 1000$, such that $n = a^2 + b^2 + c^2$, where $a$ is the smallest proper divisor of $n$, and $b$ and $c$ are divisors of $n$, not necessarily distinct, greater than 1. | [] | Romania | 75th NMO | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 36 | |
095a | Problem:
Să se determine toate numere naturale nenule $n$, pentru care numărul
$$
S_{n}=1 \cdot C_{2 n}^{1}+2 \cdot C_{2 n}^{2}+3 \cdot C_{2 n}^{3}+\ldots+n \cdot C_{2 n}^{n}
$$
est un pătrat perfect mai mic decât $1000000$. | [
"Solution:\n\nTransformăm termenul general al sumei\n$$\na_{k}=k \\cdot C_{2 n}^{k}=k \\cdot \\frac{(2 n)!}{k!(2 n-k)!}=\\frac{(2 n)!}{(k-1)!(2 n-k)!}=2 n \\cdot \\frac{(2 n-1)!}{(k-1)!(2 n-k)!}=2 n \\cdot C_{2 n-1}^{k-1}\n$$\nSuma $S_{n}$ ia forma:\n$$\nS_{n}=2 n\\left(C_{2 n-1}^{0}+C_{2 n-1}^{1}+C_{2 n-1}^{2}+\\l... | Moldova | Olimpiada de Matematică a Republicii Moldova | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Other"
] | null | proof and answer | n = 2 and 8 | |
0d65 | Let $p$, $q$ be given primes and the sequence $(p_{n})_{n \geq 1}$ defined recursively as follows: $p_{1} = p$; $p_{2} = q$ and $p_{n+2}$ is the largest prime divisor of the number $(p_{n} + p_{n+1} + 2016)$ for all $n \geq 1$. Prove that this sequence is bounded, that is there exists a positive real number $M$ such th... | [
"Put $k = 2016$ and $a_{n} = \\max \\{p_{n}, p_{n+1}\\}$ for each $n \\geqslant 1$.\nWe shall show that $a_{n+1} \\leqslant a_{n} + k + 2$. In fact, since $p_{n+1} \\leqslant a_{n}$, it remains to show that $p_{n+2} \\leqslant a_{n} + k + 2$.\nIf $p_{n+2} = 2$ then the inequality is trivially true. It suffices to c... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
06vh | For any two different real numbers $x$ and $y$, we define $D(x, y)$ to be the unique integer $d$ satisfying $2^{d} \leqslant |x-y| < 2^{d+1}$. Given a set of reals $\mathcal{F}$, and an element $x \in \mathcal{F}$, we say that the scales of $x$ in $\mathcal{F}$ are the values of $D(x, y)$ for $y \in \mathcal{F}$ with $... | [
"We first construct a set $\\mathcal{F}$ with $2^{k}$ members, each member having at most $k$ different scales in $\\mathcal{F}$. Take $\\mathcal{F} = \\{0, 1, 2, \\ldots, 2^{k} - 1\\}$. The scale between any two members of $\\mathcal{F}$ is in the set $\\{0, 1, \\ldots, k-1\\}$.\n\nWe now show that $2^{k}$ is an u... | IMO | IMO 2019 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2^k | |
073j | Suppose $n$ is a composite number such that $\varphi(n)$ divides $n-1$, where $\varphi$ denotes Euler's totient function. Show that $n$ has at least four distinct prime factors. | [
"Suppose $n$ has only one prime factor. Then $n = p^\\alpha$ for some $\\alpha > 1$ and prime $p$. Then $\\varphi(n) = p^{\\alpha-1}(p-1)$ divides $p^\\alpha - 1$, which is impossible since $\\alpha \\ge 2$. Thus $n$ has at least two prime factors.\n\nSuppose $n = p^\\alpha q^\\beta$, where $p \\ne q$ are primes, $... | India | Indija TS 2008 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
00hm | Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways. | [
"We claim that $k=2^{a}$ for all $a \\geq 0$.\nLet $A=\\{1,2,4,8, \\ldots\\}$ and $B=\\mathbb{N} \\backslash A$. For any set $T$, let $s(T)$ denote the sum of the elements of $T$. (If $T$ is empty, we let $s(T)=0$.)\nWe first show that any positive integer $k=2^{a}$ satisfies the desired property. Let $B'$ be a sub... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | English | proof and answer | k = 2^a for all integers a ≥ 0 | |
054v | a) The general form $\overline{ABC}$ of a three-digit number is initially written on a blackboard. Ann and Enn replace by turns letters with digits, exactly one at a time, with Ann starting. Can Ann write digits in such a way that, irrespectively of Enn's move, the resulting three-digit number would be divisible by 11?... | [
"a) Let Ann replace on her first move one of letters $B$ or $C$ with a digit $k$. If Enn now replaces the other one of $B$ and $C$ with digit $k$, too, the resulting number $\\overline{Akk}$ is divisible by 11 if and only if $\\overline{A00}$ is divisible by 11, which in turn is the case if $A$ is divisible by 11. ... | Estonia | National Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | a) No. b) Yes. | |
06g3 | A rectangle with sides of integral lengths can be divided into two or more squares with sides of integral lengths and such that there is exactly one square with smallest side length. Determine the smallest possible area of the rectangle. | [
"The smallest possible area is $35$.\n\nLet $a$ be the length of the smallest square $A$. If $A$ touches a side but not a corner, then it is surrounded by two larger squares $B$ and $C$. Then there is a gap between $B$ and $C$ with length $a$. It is not possible to place some larger squares in this gap, contradicti... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 35 | |
0964 | Problem:
Arătaţi că partea întreagă a numărului $(7+4 \sqrt{3})^{n}$ este un număr impar, oricare ar fi numărul natural $n$. | [
"Solution:\n\n1. Pentru $n=0$, avem $\\left[(7+4 \\sqrt{3})^{0}\\right]=1$ - număr impar.\n\n2. Fie $n \\in \\mathbb{N}^{*}$. Vom arăta că putem scrie $(7+4 \\sqrt{3})^{n}=A_{n}+B_{n} \\sqrt{3}$ cu $A_{n}, B_{n} \\in \\mathbb{N}^{*}$.\n\n$(7+4 \\sqrt{3})^{n}=C_{n}^{0} \\cdot 7^{n}+C_{n}^{1} \\cdot 7^{n-1} \\cdot 4 ... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0gxk | Angular bisectors $BD$ and $CE$ are drawn in the triangle $ABC$ with angle $\angle BAC \ge 60^\circ$. Prove that $AD + AE \le BC$.
Fig.30 | [
"Denote the sides of triangle by $a$, $b$, $c$. By the property of angular bisectors,\n\n$AD = \\frac{ch}{a+c}$, $AE = \\frac{ch}{a+b}$ (pic.30), thus it is enough to prove: $\\frac{ch}{a+c} + \\frac{ch}{a+b} \\le a$.\n\nAfter simple transformations, we have: $(a^2 - bc)(a+b+c) \\ge 0$, and so it is enough to prove... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0dz2 | Let $m$ and $n$ be positive integers of different parity. Show that
$$
(m + 3n)(5m + 7n)(7m + 5n)(3m + n)
$$
can never be a perfect square. | [
"Let us assume there exist numbers $m$ and $n$, one even and one odd, such that $A = (m + 3n)(5m + 7n)(7m + 5n)(3m + n)$ is a perfect square. Let $d$ be the greatest common divisor of $m$ and $n$, and write $m = d m_1$, $n = d n_1$. Then $m_1$ and $n_1$ are coprime and $A$ can be written as\n$$\nA = d^4 (m_1 + 3n_1... | Slovenia | Slovenija 2008 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0bed | Determine all injective functions $f: \mathbb{Z} \to \mathbb{Z}$ that satisfy
$$
|f(x) - f(y)| \le |x - y|,
$$
for all $x, y \in \mathbb{Z}$. | [
"The given relation yields $|f(x+1) - f(x)| \\le 1$, so $f(x+1)-f(x) \\in \\{-1, 0, 1\\}$. Because $f$ is one-to-one, we get $f(x+1)-f(x) \\in \\{-1, 1\\}$, for all $x$ in $\\mathbb{Z}$. We suppose WLOG that $f(1)-f(0) = 1$ (if $f$ verifies the hypothesis then $-f$ verifies it also). We have $f(2)-f(1) = \\pm 1$. I... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All functions of the form f(x) = x + k or f(x) = -x + k, where k is an integer. | |
0dq1 | A set $S$ of real numbers is such that $1 + \frac{1}{x} \in S$ for each $x \in S$. Is it possible that $S$ contains exactly 2025 elements? | [
"**Answer.** The answer is no.\n\nLet $f(x) = 1 + \\frac{1}{x}$ and $f^{(n)}(x) = f(f(\\dots f(x)))$ (where $f$ is applied $n$ times). For each $x \\in S$ the set $S$ contains all numbers of the form $f^{(n)}(x)$. If $S$ is finite, then $f^{(k)}(x) = f^{(\\ell)}(x)$ for some $k > \\ell$. Since $f(a) = f(b)$ implies... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICAL COMPETITION | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | No | |
07qs | Given a positive integer $m$, a sequence of real numbers $a = (a_1, a_2, a_3, ...)$ is called $m$-powerful if it satisfies
$$
\left( \sum_{k=1}^{n} a_k \right)^m = \sum_{k=1}^{n} a_k^m \quad \text{for all positive integers } n.
$$
a.
Show that a sequence is 30-powerful if and only if at most one of its terms is non-ze... | [
"If at most one term in a sequence $a$ is non-zero, it is immediate that it is $m$-powerful for all $m \\in \\mathbb{N}$. We will see that there are no other 30-powerful sequences (or indeed, $m$-powerful for any even $m$), but that there are many other sequences that are $m$-powerful for all odd $m$.\n\nWe begin o... | Ireland | Ireland_2017 | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | a) Exactly the sequences with at most one nonzero term. b) For example, a_n = (−1)^{n−1} (more generally, a_n = c·(−1)^{n−1} with c ≠ 0). | |
06g1 | Let $n$ be a positive integer and let $a$ be an integer with $\text{gcd}(a, n) = 1$. Prove that
$$
\frac{a\varphi(n) - 1}{n} \equiv \sum_{\substack{j=1 \\ \text{gcd}(j,n)=1}}^{n} \frac{1}{aj} \left[ \frac{aj}{n} \right] \pmod{n},
$$
where $\varphi(n)$ is the Euler's totient function and $[x]$ is the greatest integer no... | [
"For each integer $j$ relatively prime to $n$, let $r_j$ be the remainder when $aj$ is divided by $n$. Then we have\n$$\n\\begin{aligned}\na^{\\varphi(n)} &= \\prod_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\frac{aj}{j} = \\prod_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\frac{1}{j} \\left( \\left\\lfloor \\frac{aj}{n} \\right\... | Hong Kong | CHKMO | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0ewm | Problem:
Given an isosceles triangle, find the locus of the point $P$ inside the triangle such that the distance from $P$ to the base equals the geometric mean of the distances to the sides. | [
"Solution:\nLet the triangle be $ABC$, with $AB = AC$. Take the circle through $B$ and $C$ which has $AB$ and $AC$ as tangents. The required locus is the arc $BC$.\n\nSuppose $P$ lies on the arc. Let the perpendiculars from $P$ meet $BC$ in $L$, $AB$ in $N$ and $AC$ in $M$. Join $PB$ and $PC$. The triangles $PNB$ a... | Soviet Union | 3rd ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The arc BC of the circle through B and C that is tangent to AB and AC. | |
0540 | The circumcentre of an acute triangle $ABC$ is $O$. Line $AC$ intersects the circumcircle of $AOB$ at a point $X$, in addition to the vertex $A$. Prove that the line $XO$ is perpendicular to the line $BC$. | [
"By the properties of inscribed angles $\\angle CXO = \\angle ABO$ (Fig. 8) independent of whether the point $X$ lies on the side $CA$ or on the extension of $CA$ over $A$. Since $O$ is the circumcentre of $ABC$, we have $\\angle ABO = \\angle BAO$. Let the intersection of lines $XO$ and $BC$ be $Y$ and let the poi... | Estonia | Estonian Math 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"
] | null | proof only | null | |
0j95 | In acute triangle $ABC$, $\angle A < \angle B$ and $\angle A < \angle C$. Let $P$ be a variable point on side $BC$. Points $D$ and $E$ lie on sides $AB$ and $AC$, respectively, such that $BP = PD$ and $CP = PE$. Prove that as $P$ moves along side $BC$, the circumcircle of triangle $ADE$ passes through a fixed point oth... | [
"We will prove that the fixed point is the orthocenter $H$ of triangle $ABC$. Let $X$ be the foot of the perpendicular from $C$ to $AB$, and let $Y$ be the foot of the perpendicular from $B$ to $AC$. Note that if $M$ is the midpoint of $BC$, then $MB = MX = MY = MC$. Suppose without loss of generality that $P$ is b... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry... | null | proof only | null | |
055a | There is a finite number of lamps in an electrical scheme. Some pairs of lamps are directly connected by a wire. Every lamp is lit either red or blue. With one switch all lamps that have a direct connection with a lamp of the other colour change their colour (from red to blue or vice versa). Prove that after some numbe... | [
"If some connected lamps are lit in different colours they both change colour upon switching, hence they are also lit differently after the switch. The same holds on each following switch. Hence no pair of connected lamps lit in different colours can disappear but more of such pairs can appear. As there are only fi... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
07h1 | We call an infinite set $S \subseteq \mathbb{N}$ *good* if for all pairwise distinct integers $a$, $b$, $c \in S$, all positive divisors of $\frac{a^c - b^c}{a - b}$ are in $S$. For all positive integers $n > 1$, prove that there exists a good set $S$ such that $n \notin S$. | [
"Let $L(z)$ be the smallest prime divisor of $z$.\n\n**Lemma.** If $a$, $b$, $c \\in \\mathbb{N}$ such that $a \\neq b$ we have\n$$\nL\\left(\\frac{a^c - b^c}{a - b}\\right) \\geq L(abc)\n$$\n*Proof.* Assume the contrary, then there should be an integer $p$ such that $\\gcd(p, a) = \\gcd(p, b) = 1$ and $p \\mid \\f... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equation... | English | proof only | null | |
0c1y | Ann and Bob play a game on an infinite checkered plane making moves in turn; Ann makes the first move. A move consists in orienting any unit grid-segment that has not been oriented before. If at some stage some oriented segments form an oriented cycle, Bob wins. Does Bob have a strategy that guarantees him to win? | [
"The answer is in the negative: Ann has a strategy allowing her to prevent Bob's victory.\nWe say that two unit grid-segments form a *low-left corner* (or *LL-corner*) if they share an endpoint which is the lowest point of one and the leftmost point of the other. An *up-right corner* (or *UR-corner*) is defined sim... | Romania | Eleventh Romanian Master of Mathematics | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No; Bob does not have a winning strategy. | |
0488 | A finite point set $P = \{(x_i, y_i) \mid i = 1, 2, \dots, n\}$ in the plane is called **sparse** if for any $1 \le i < j \le n$,
$$
|x_i - x_j| \cdot |y_i - y_j| \ge 1.
$$
(1) Let $\{Q_1, Q_2, Q_3\}$ be a sparse set where no $Q_i$ lies inside the axis-aligned rectangle with the other two points as diagonal vertices. P... | [
"(1) Let the coordinates of $Q_i$ be $(x_i, y_i)$. Without loss of generality, assume $x_1 < x_2 < x_3$. Since both the conditions and conclusion are symmetric under reflection about the $x$-axis and $y$-axis, we can further assume $y_1 < y_3 < y_2$. After appropriate translation, we may set $x_1 = y_1 = 0$, so $0 ... | China | China-TST-2025A | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0if9 | Problem:
Find
$$
\{\ln (1+e)\}+\left\{\ln \left(1+e^{2}\right)\right\}+\left\{\ln \left(1+e^{4}\right)\right\}+\left\{\ln \left(1+e^{8}\right)\right\}+\cdots,
$$
where $\{x\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$. | [
"Solution:\n$1-\\ln (e-1)$\n\nSince $\\ln \\left(1+e^{2^{k}}\\right)$ is just larger than $2^{k}$, its fractional part is $\\ln \\left(1+e^{2^{k}}\\right)-\\ln e^{2^{k}}=\\ln \\left(1+e^{-2^{k}}\\right)$. But now notice that\n$$\n\\prod_{k=0}^{n}\\left(1+x^{2^{k}}\\right)=1+x+x^{2}+\\cdots+x^{2^{n+1}-1}\n$$\n(This ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1 - ln(e - 1) | |
0e69 | Problem:
Naj leži točka $E$ na stranici $C D$ kvadrata $A B C D$. Točka $F$ leži na premici $A B$, a ne na daljici $A B$, in zadošča $|B F|=|D E|$. Dokaži, da sta premici $A C$ in $E F$ pravokotni. | [
"Solution:\n\nOznačimo s $T$ presečišče premic $A C$ in $E F$. Ker je $|B F|=|D E|$, $|B C|=|D A|$ in $\\angle C B F=\\angle A D E=\\frac{\\pi}{2}$, sta trikotnika $A D E$ in $C B F$ skladna. Zato je $|A E|=|C F|$ in štirikotnik $A F C E$ je enakokrak trapez. Sledi $\\angle E F A=\\angle C A F$, ta kot pa je enak $... | Slovenia | 56. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0kdy | Problem:
You are blindfolded and have a spinning table with four switches on it in front of you. The switches are always either up or down, and you don't know what configuration they start in. On each move, you can spin the table some unknown amount, then reach out and choose two switches (either next to each other or... | [
"Solution:\n\nIt is indeed possible. One strategy is as follows:\n\na) Choose two switches that are next to each other. Turn them so they are both up.\n\nb) Choose two switches that are diagonally opposite each other (so exactly one is a switch you picked last time). Turn them so they are both up. Now at least thre... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | It is always possible to win. | |
0j0b | Problem:
In the following figure, a regular hexagon of side length $1$ is attached to a semicircle of diameter $1$. What is the longest distance between any two points in the figure?
 | [
"Solution:\n\nAnswer: $\\frac{1+\\sqrt{13}}{2}$\n\nInspection shows that one point must be on the semicircle and the other must be on the side of the hexagon directly opposite the edge with the semicircle, the bottom edge of the hexagon in the above diagram. Let $O$ be the center of the semicircle and let $M$ be th... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | (1+sqrt(13))/2 | |
00dz | Prove that for every positive integer $n$, there exists a positive integer $k$ such that each of the numbers $k, k^2, \dots, k^n$ has at least one 2022 block in its decimal representation.
(For example, the numbers $4202213$ and $544202212022$ have at least one 2022 block in their decimal representation.) | [
"We begin by proving that for every positive integer $n$, there exists a positive integer $k$ such that the number $k^n$ has at least one 2022 block in its decimal representation. Even more, we prove that the block is formed by the first four digits of the number. In that case we would have\n$$\n2022 \\cdot 10^d \\... | Argentina | XXXIII Cono Sur Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
08bs | Problem:
La sequenza dei numeri di Fibonacci $F_{1}, F_{2}, \ldots, F_{n}, \ldots$ è tale che $F_{1}=F_{2}=1$ e che l'n-esimo termine ( $n \geq 3$ ) sia la somma dei due precedenti (i primi termini della successione sono quindi $F_{1}=1, F_{2}=1, F_{3}=2=1+1, F_{4}=3=2+1, F_{5}=5=3+2$ ). Quanti sono i numeri di Fibona... | [
"Solution:\n\nLa risposta è $\\mathbf{( B )}$. Osserviamo innanzitutto che (per $n \\geq 2$ ) la relazione $F_{n+1}=F_{n}+F_{n-1}$, insieme al fatto che $F_{n-1} \\leq F_{n}$, implica che $F_{n}$ è almeno metà di $F_{n+1}$. Sia ora $k \\geq 1$ e $F_{n}$ il più piccolo numero di Fibonacci con $k+1$ cifre decimali (q... | Italy | Gara di Febbraio | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
01g0 | Let us consider a polynomial $P(x)$ with integer coefficients satisfying
$$
P(-1) = -4, \quad P(-3) = -40, \quad \text{and} \quad P(-5) = -156.
$$
What is the largest possible number of integers $x$ satisfying
$$
P(P(x)) = x^2?
$$ | [
"No such numbers $x$ can ever exist. To see this, let us recall that when $x$ is an integer, the right-hand side of the equation is always $\\equiv 0 \\pmod{3}$ or $1 \\pmod{3}$. We recall also that when $x$ and $y$ are integers, we have\n$$\nP(x) \\equiv P(y) \\pmod{3}, \\quad \\text{whenever} \\quad x \\equiv y \... | Baltic Way | Baltic Way 2019 | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | 0 | |
04dm | The lines drawn parallel to the sides of an isosceles triangle divide the triangle into three congruent equilateral quadrilaterals, three congruent trapeziums and one isosceles triangle in the middle. If the area of the triangle in the middle is twice the area of each equilateral quadrilateral, what is the share of the... | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1/5 | |
049h | Let $\triangle ABC$ be a right triangle with the right angle at vertex $C$ and let $M$ be the midpoint of the leg $\overline{BC}$. Prove that $\sin(\angle MAB) \le \frac{1}{3}$. When is the equality achieved? | [
"Since $\\angle MAB = \\angle CAB - \\angle CAM$, according to the sine addition theorem we have\n$$\n\\sin(\\angle MAB) = \\sin(\\angle CAB) \\cos(\\angle CAM) - \\cos(\\angle CAB) \\sin(\\angle CAM).\n$$\n\nLet $|BC| = a$, $|CA| = b$, $|AB| = c$ and $|AM| = t$. From right triangles $ABC$ ... | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sin(angle MAB) ≤ 1/3, with equality when BC = √2 · CA (equivalently, tan(angle A) = √2). | |
0dit | Consider integers $b_1, b_2, \dots, b_n$ have the sum is $2022$ and take $a$ as a divisor of $\left\lfloor \dfrac{2022!}{3^{1006}} \right\rfloor$. Prove that the following polynomial does not have rational root
$$
P(x) = a x^{2n+2} + b_1 x^{2n} + b_2 x^{2n-2} + \dots + b_n x^2 + a.
$$ | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0hq2 | Problem:
Count the number of sequences $1 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{5}$ of integers with $a_{i} \leq i$ for all $i$. | [
"Solution:\n\n$C$ (number of terms) $= C(5) = 42$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 42 | |
0180 | Does there exist a non-equilateral triangle, such that the angle between any two of its medians equals $120^\circ$? | [
"No. Let $ABC$ be a triangle satisfying the condition. Assume that $AA'$ is its shortest median, and denote by $M$ the centroid. The Law of Sines gives\n$$\n\\frac{\\sin \\angle MBC}{\\sin \\angle MA'B} = \\frac{MA'}{MB} = \\frac{MA}{2MB} \\le \\frac{1}{2}.\n$$\nHence $\\angle MBC \\le 30^\\circ$, with equality iff... | Baltic Way | BALTIC WAY | [
"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 and answer | No; only the equilateral triangle has medians with pairwise angles of one hundred twenty degrees. | |
0ipq | Problem:
You have a $2 \times 3$ grid filled with integers between $1$ and $9$. The numbers in each row and column are distinct, the first row sums to $23$, and the columns sum to $14$, $16$, and $17$ respectively.
| | $14$ | $16$ | $17$ |
| :---: | :---: | :---: | :---: |
| $23$ | $a$ | $b$ | $c$ |
| | $x$ | $y$... | [
"Solution:\n\nThe sum of all $6$ numbers is $14+16+17=47$, so $x+y+z=47-23=24$. If three distinct digits sum to $24$, they must be $7$, $8$, and $9$, because any other triple of digits would have a smaller sum. So, we try placing these digits in for $x$, $y$, and $z$, and the only arrangement that does not force eq... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 49 | |
0k6u | Problem:
Your math friend Steven rolls five fair icosahedral dice (each of which is labelled $1,2, \ldots, 20$ on its sides). He conceals the results but tells you that at least half of the rolls are $20$. Suspicious, you examine the first two dice and find that they show $20$ and $19$ in that order. Assuming that Ste... | [
"Solution:\n\nThe given information is equivalent to the first two dice being $20$ and $19$ and there being at least two $20$'s among the last three dice. Thus, we need to find the probability that given at least two of the last three dice are $20$'s, all three are.\n\nSince there is only one way to get all three $... | United States | HMMT February 2019 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 1/58 | |
0ecy | How many quadruples of positive integers $(a, b, c, d)$ satisfy the inequalities
$$
a > b > c > d \quad \text{and} \quad \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} > 2?
$$ | [
"If $d \\ge 2$ we get $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\le 4 \\cdot \\frac{1}{2} = 2$ which is a contradiction. Thus $d = 1$.\n\nIf $c \\ge 3$ we get $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\le 3 \\cdot \\frac{1}{3} + 1 = 2$ which is again a contradiction. Thus $c = 2$... | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2 | |
0ik6 | Problem:
Triangle $A B C$ has side lengths $A B=2 \sqrt{5}$, $B C=1$, and $C A=5$. Point $D$ is on side $A C$ such that $C D=1$, and $F$ is a point such that $B F=2$ and $C F=3$. Let $E$ be the intersection of lines $A B$ and $D F$. Find the area of $C D E B$. | [
"Solution:\n\nDraw segment $A F$. Then notice $A F=4$, and we have a right triangle. Now draw line $C E$, let it intersect $A F$ at $G$. By Ceva, $F G=\\frac{4}{3}$ and $A G=\\frac{8}{3}$. Using mass points we find that $\\frac{A E}{E B}=6$ so $\\frac{[A E F]}{[B E F]}=6$, and since $[A B F]=4$, $[B E F]=\\frac{4}{... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 22/35 | |
0i68 | Problem:
Find the greatest common divisor of the numbers $2002+2, 2002^{2}+2, 2002^{3}+2, \ldots$. | [
"Solution:\nNotice that $2002+2$ divides $2002^{2}-2^{2}$, so any common divisor of $2002+2$ and $2002^{2}+2$ must divide $(2002^{2}+2)-(2002^{2}-2^{2})=6$. On the other hand, every number in the sequence is even, and the $n$th number is always congruent to $1^{n}+2 \\equiv 0$ modulo $3$. Thus, $6$ divides every nu... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 6 | |
0geq | 考慮一正整數數列 $a_1, a_2, a_3, \dots$，滿足 $a_1 = 2021$ 且
$$
\sqrt{a_{n+1} - a_n} = \lfloor \sqrt{a_n} \rfloor.
$$
證明此數列含有無窮多個奇數和無窮多個偶數。
Let $a_1, a_2, a_3, \dots$ be a sequence of positive integers such that $a_1 = 2021$ and
$$
\sqrt{a_{n+1} - a_n} = \lfloor \sqrt{a_n} \rfloor.
$$
Show that there are infinitely many odd numb... | [
"假設不存在無窮多個奇數或無窮多個偶數，則存在一個 $N$ 使得數列 $a_N, a_{N+1}, a_{N+2}, \\dots$ 擁有同樣的奇偶性。所以當 $n \\ge N$ 時，\n$$\nb_n := \\lfloor \\sqrt{a_n} \\rfloor = \\sqrt{a_{n+1} - a_n}\n$$\n為偶數。令 $k_n = a_n - b_n^2$，則 $0 \\le k_n \\le 2b_n$。我們有\n$$\na_{n+1} = a_n + b_n^2 = 2b_n^2 + k_n \\implies b_{n+1} \\le \\sqrt{2b_n^2 + k_n} < b_{n+1} ... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
00cm | Bruno elige un número entero positivo $X$. A continuación, Flor elige cuatro números enteros $a, b, c, d$ y calcula $N = (a-b)(b-c)(c-d)(d-a)(a-c)(b-d)$, la multiplicación de las seis diferencias entre esos cuatro números. Determinar el mayor valor de $X$ con el que Bruno tiene la certeza de que $N$ será múltiplo de $X... | [] | Argentina | Nacional OMA 2019 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof and answer | 12 | |
0c2b | Let $n$ be a natural number with $n \ge 2$ and let $\mathbf{A}$ be a matrix in $M_n(\mathbb{C})$, such that $\mathbf{A}$ and $\mathbf{A}^2$ have different ranks. Prove that there exists a non-zero matrix $\mathbf{B}$ in $M_n(\mathbb{C})$, such that $\mathbf{A}\mathbf{B} = \mathbf{B}\mathbf{A} = \mathbf{B}^2 = \mathbf{O... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof only | null | |
021u | Problem:
Partindo do número $265863$ e utilizando uma única vez cada uma das operações $+$, $-$, $\times$, $:$, e também uma única vez os números $51$, $221$, $6817$, $13259$, podemos obter vários números, por exemplo $54911$:
$265863 \xrightarrow{+221} 1203 \xrightarrow{\times 51} 61353 \xrightarrow{-13259} 48094 \x... | [
"Solution:\n\n$265863 \\xrightarrow{\\div 6817} 39 \\xrightarrow{+221} 260 \\xrightarrow{\\times 51} 13260 \\xrightarrow{-13259} 1$"
] | Brazil | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 265863 / 6817 = 39, then +221 → 260, then ×51 → 13260, then −13259 → 1 | |
05f9 | Problem:
Soit $BCDE$ un carré et soit $O$ son centre. Soit $A$ un point situé à l'extérieur du carré $BCDE$ tel que le triangle $ABC$ est rectangle en $A$. Montrer que le point $O$ appartient à la bissectrice de l'angle $\widehat{BAC}$.
 | [
"Solution:\n\nLes diagonales du carré $BCDE$ se coupent perpendiculairement en $O$ donc $\\widehat{BOC} = 90^{\\circ} = \\widehat{BAC}$ donc les points $A$, $B$, $O$ et $C$ sont cocycliques. Comme $O$ est sur la médiatrice du segment $[BC]$, $O$ est le pôle Sud de $A$ dans le triangle $ABC$ donc $O$ est sur la biss... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07ds | Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy the following conditions:
(a) $x + f(y + f(x)) = y + f(x + f(y))$, $\forall x, y \in \mathbb{R}$
(b) The set $I = \left\{\frac{f(x)-f(y)}{x-y} \mid x, y \in \mathbb{R}, x \neq y\right\}$ is an interval. (means for any two numbers $u, v \in I$ with $u \le v... | [
"By changing $x, y$, it is obvious that the given set is the same as $\\left\\{\\frac{f(x)-f(y)}{x-y} \\mid x, y \\in \\mathbb{R}, x > y\\right\\}$. Let's define $F(x, y) = \\frac{f(x)-f(y)}{x-y}$. First, we prove two lemmas:\n\n**Lemma.** $0, 1 \\notin I$. So $f(x)$ is an injective function.\n*Proof.* Assume the c... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | No such function exists. | |
0gwi | A point $M$ lies inside a cube $ABCD-A_1B_1C_1D_1$. Points $A', B', C', D', A'_1, B'_1, C'_1$ and $D'_1$ belong to the half-lines $MA, MB, MC, MD, MA_1, MB_1, MC_1$ and $MD_1$ respectively. Prove that if the polyhedron $A'B'C'D'A'_1B'_1C'_1D'_1$ is a parallelepiped (i.e., all its faces are parallelograms), then in fact... | [
"Доведемо спочатку, що якщо основою чотирикутної піраміди $SABCD$ є квадрат $ABCD$, і точки $A_0, B_0, C_0, D_0$ належать променям $SA, SB, SC, SD$ відповідно, причому $A_0B_0C_0D_0$ є паралелограмом, то $A_0B_0C_0D_0$ — квадрат. Нехай $l = (SCD) \\cap (SAB)$ і $m = (SAD) \\cap (SBC)$. За відомою лемою про три пара... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
015t | Let $n > 2$ be a natural number. In the country there are $n$ cities and every two of them are connected by a direct road. We enumerate roads with numbers $1, 2, \ldots, m$ (we can assign the same number to several roads, even if they are going from the same city), where $m$ is a natural number. The *ID* of a city is a... | [
"We will show that $m = 3$ for every $n > 2$.\n\nFirstly we show that $m \\ge 3$. It is clear that $m \\ge 2$, so suppose that $m = 2$. We have $n$ cities, and the possible numbers for the sum of numbers assigned to roads going out from the given city are $n-1, n, n+1, \\dots, 2n-2$ (the smallest possible sum is $1... | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 3 | |
0bsb | Given positive integers $m$ and $n \ge m$, determine the largest number of dominoes ($1 \times 2$ or $2 \times 1$ rectangles) that can be placed on a rectangular board with $m$ rows and $2n$ columns consisting of unit cells ($1 \times 1$ squares) so that:
1) each domino covers exactly two adjacent cells of the board;
2... | [
"The required maximum is $mn - \\lfloor m/2 \\rfloor$ and is achieved by the brick-like vertically symmetric arrangement of blocks of $n$ and $n-1$ horizontal dominoes placed on alternate rows, so that the bottom row of the board is completely covered by $n$ dominoes.\nTo show that the number of dominoes in an arra... | Romania | 2016 Eighth Romanian Master of Mathematics | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | mn - floor(m/2) | |
027k | Problem:
A seguinte figura mostra um triângulo $ABC$ que foi dividido em 3 triângulos isósceles.
a) Mostre que todo triângulo retângulo pode ser dividido em 2 triângulos isósceles.
b) Mostre que qualquer triângulo pode ser dividido em 4 triângulos isósceles.
c) Mostre que qualquer triângul... | [
"Solution:\n\na) Consideremos o triângulo retângulo com ângulos agudos $\\alpha^\\circ$ e $\\beta^\\circ$, como mostra a seguinte figura:\n\n\nTraçamos um segmento $BM$, com $M$ sobre o lado $AC$, de modo que $\\measuredangle MBC = \\beta^\\circ$.\n\n\nTemos então q... | Brazil | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06vu | Let $H=\{\lfloor i \sqrt{2}\rfloor: i \in \mathbb{Z}_{>0}\}=\{1,2,4,5,7, \ldots\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \subset\{1,2, \ldots, n\}$ satisfies $|A| \geqslant C \sqrt{n}$, then there exist $a, b \in A$ such that $a-b \in H$. (Here $\mathbb{Z}_{>0}$ is ... | [
"First, observe that if $n$ is a positive integer, then $n \\in H$ exactly when\n$$\n\\left\\{\\frac{n}{\\sqrt{2}}\\right\\}>1-\\frac{1}{\\sqrt{2}} \\tag{1}\n$$\nTo see why, observe that $n \\in H$ if and only if $0<i \\sqrt{2}-n<1$ for some $i \\in \\mathbb{Z}_{>0}$. In other words, $0<i-n / \\sqrt{2}<1 / \\sqrt{2... | IMO | IMO 2019 Shortlisted Problems | [
"Number Theory > Diophantine Equations > Pell's equations",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Floors... | English | proof only | null | |
044p | An even cube dice with six faces is marked with six numbers $1$, $2$, $3$, $4$, $5$ and $6$, respectively. The dice are randomly tossed three times, with each toss being independent from each other, and the resulting dots are $a_1, a_2, a_3$ in order. Then the probability of event $|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_... | [
"Note that $|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_1| = 2 \\max_{1 \\le i \\le 3} a_i - 2 \\min_{1 \\le i \\le 3} a_i$.\nTherefore, the three $a_1, a_2, a_3$ tossed satisfy the condition if and only if the difference between the largest number and the smallest number is $3$.\nThis is to say that $a_1, a_2, a_3$ is a ... | China | China Mathematical Competition | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 1/4 | |
01qo | Determine all pairs $(x, y)$ of positive integers such that
$$
\sqrt[3]{7x^2 - 13xy + 7y^2} = |x - y| + 1.
$$ | [
"2. See IMO-2014 Shortlist, Problem N2."
] | Belarus | SELECTION and TRAINING SESSION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | {(x, y) = (1, 1)} ∪ { (x, y) = (n^3 + 2n^2 − n − 1, n^3 + n^2 − 2n − 1), (n^3 + n^2 − 2n − 1, n^3 + 2n^2 − n − 1) for integers n ≥ 2 } | |
00lx | Determine all integers $n \ge 2$ that have a representation
$$
n = a^2 + b^2,
$$
where $a$ is the smallest divisor of $n$ different from $1$ and $b$ is an arbitrary divisor of $n$. | [
"If $n$ is odd, then both $a$ and $b$ are odd and therefore $n = a^2 + b^2$ is even, contradiction. Therefore, $n$ is even and $a = 2$. This also shows that $b$ is even. Furthermore, $b \\mid (n - b^2) = a^2 = 4$. Thus $b \\in \\{2, 4\\}$, which results in $n = 8$ and $n = 20$, respectively. $\\square$"
] | Austria | 48th Austrian Mathematical Olympiad Regional Competition (Qualifying Round) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 8, 20 | |
06hc | Let $ABCDEF$ be a hexagon whose diagonals $AD$, $BE$ and $CF$ intersect at one point, which is the midpoint of each diagonal. Prove that the area of the given hexagon is twice the area of the triangle $ACE$. | [
"Suppose $AD$, $BE$, $CF$ meet at $P$. Since $P$ is the midpoint of $AD$, $BE$, $CF$, all of $ABDE$, $BCEF$ and $CDFA$ are parallelograms. Let $A'$ be the point such that $ACA'E$ is a parallelogram. Since $EA' = AC = FD$ and $EA' \\parallel AC \\parallel FD$, $FDA'E$ is also a parallelogram. Thus, we have $DA' = FE... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0cmz | Prove that $x \cos x \le \frac{\pi^2}{16}$ for every $0 \le x \le \frac{\pi}{2}$. | [
"Так как $\\sin t \\le t$ при $0 \\le t \\le \\frac{\\pi}{2}$, то имеем\n\n$x \\cos x = x \\sin (\\frac{\\pi}{2} - x) \\le x (\\frac{\\pi}{2} - x) = \\frac{\\pi^2}{16} - (x - \\frac{\\pi}{4})^2 \\le \\frac{\\pi^2}{16}$,\nчто и требовалось.\n\n\nВторое решение. Перепишем доказываемое неравенство в виде $\\sqrt{x \\c... | Russia | Russian mathematical olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English; Russian | proof only | null | |
08or | Problem:
2015 points are given in a plane such that from any five points we can choose two points with distance less than 1 unit. Prove that 504 of the given points lie on a unit disc. | [
"Solution:\n\nStart from an arbitrary point $A$ and draw a unit disc with center $A$. If all other points belong to this disc then we are done. Otherwise, take any point $B$ outside of the disc. Draw a unit disc with center $B$. If two drawn discs cover all 2015 points, by the pigeonhole principle (PHP), one of the... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0eh8 | Problem:
V zavetišču imajo 15 psov in 20 mačk. Izmed njih naključno izberemo 8 živali. Izračunaj verjetnosti dogodkov na pet decimalnih mest natančno:
a) Izbrali smo 8 psov,
b) Izbrali smo 5 psov in 3 mačke,
c) Izbrali smo več psov kot mačk,
d) Izbrali smo vsaj 6 mačk. | [
"Solution:\n\nIzračun števila vseh izidov:\n\nSkupno število živali je $15 + 20 = 35$. Izberemo $8$ živali:\n\n$$\nN = \\binom{35}{8}\n$$\n\na) Izbrali smo 8 psov:\n\nŠtevilo ugodnih izidov:\n$$\nN_a = \\binom{15}{8}\n$$\nVerjetnost:\n$$\nP_a = \\frac{\\binom{15}{8}}{\\binom{35}{8}} \\approx 0.00113\n$$\n\nb) Izbra... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | a) 0.00113; b) 0.09113; c) 0.11038; d) 0.23713 | |
03pc | 8 balls of radius $1$ are placed in a cylinder in two layers, with each layer containing $4$ balls. Each ball is in contact with $2$ balls in the same layer, $2$ balls in the other layer, one base and the lateral surface of the cylinder. Then the height of the cylinder is ________. | [
"As in the diagram, let $A$, $B$, $C$, $D$ be the centers of the $4$ balls in the bottom layer, and $A'$, $B'$, $C'$, $D'$ the centers of the $4$ balls in the upper layer. Then $A$, $B$, $C$, $D$ and $A'$, $B'$, $C'$, $D'$ are the $4$ vertices of squares of length $2$, respectively. Now, the circumscribed circles w... | China | China Mathematical Competition (Shaanxi) | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | sqrt(8) + 2 | |
01jr | Given real non-zero numbers $a$, $b$, $c$ ($b > 0$) so that two distinct roots of the equation $ax^2 + bx + c = 0$ are also the roots of the equation $x^3 + bx^2 + ax + c = 0$.
Prove that
a) $abc > 16$;
b) $abc \ge \frac{3125}{108}$.
(V. Karamzin) | [
"Let $x_1$, $x_2$ be distinct roots of the equation $ax^2 + bx + c = 0$, i.e. be zeroes of the functions $f(x) = ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + ax + c$. By condition, $f(0) = g(0) = c \\neq 0$. Let $F(x) = g(x) - f(x)$. Then $0$, $x_1$, $x_2$ are the distinct zeroes of the polynomial $F(x)$. So $F(x) = x(x... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
05g4 | Problem:
Soit $ABCD$ un trapèze tel que $(AB)$ et $(CD)$ sont parallèles. On suppose que $AD < CD$ et que $ABCD$ est inscrit dans un cercle $\Gamma$. Soit $P$ sur $\Gamma$ tel que $(DP)$ est parallèle à $(AC)$. La tangente à $\Gamma$ en $D$ recoupe $(AB)$ en $E$, et les cordes $[BP]$ et $[CD]$ s'intersectent en $Q$. M... | [
"Solution:\n\nOn sait que $(CQ)$ est parallèle à $(AE)$, donc l'énoncé revient à montrer que $ACQE$ est un parallélogramme, donc il suffit de montrer que $(EQ)$ et $(AC)$ sont parallèles. Le quadrilatère $BQDE$ est un trapèze, avec $(BE)$ et $(DQ)$ parallèles. On va d'abord montrer que c'est un trapèze isocèle. En ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d1x | An acute triangle $ABC$ is inscribed in circle $\omega$ centered at $O$. Line $BO$ and side $AC$ meet at $B_1$. Line $CO$ and side $AB$ meet at $C_1$. Line $B_1C_1$ meets circle $\omega$ at $P$ and $Q$. If $AP = AQ$, prove that $AB = AC$. | [
"Assume $AP = AQ$. Because $OP = OQ$, the line $AO$ is perpendicular to $PQ$. But since $\\angle C_1AO = 90^{\\circ} - \\angle ACB$ then $\\angle B_1C_1A = \\angle ACB$, and therefore, quadrilateral $BCB_1C_1$ is cyclic.\n\nUsing the power of the point $O$ with respect to the circumcircle of $BCB_1C_1$ we get\n$$\n... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
0dr6 | Let $\mathbb{R}$ be the set of real numbers. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,
$$
f(xf(y) + x) = xy + f(x).
$$ | [
"Let $x = 1$, we have $f(f(y) + 1) = y + f(1)$. From this we conclude that $f$ is a bijection.\nLet $y = 0$, we have $f(xf(0) + x) = f(x)$. Since $f$ is injective, we have $xf(0) + x = x$, i.e., $xf(0) = 0$ for all $x$. Hence $f(0) = 0$.\nFor $x \\neq 0$, take $y = -f(x)/x$, then $f(xf(y) + x) = 0$. Thus $xf(y) + x... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x and f(x) = -x | |
0dwb | Problem:
Na krožnici $k$ s središčem v točki $O$ izberemo točki $A$ in $B$ tako, da izbrani točki nista krajišči premera. Na trikotniku $OAB$ očrtani krožnici izberemo točko $C$, ki ne sovpada niti z $A$ niti z $B$. Premica $AC$ seka krožnico $k$ v točkah $A$ in $D$. Dokaži, da je trikotnik $DCB$ enakokrak. | [
"Solution:\n\nČe točka $C$ leži na istem loku nad $AB$ kot točka $O$, je $\\angle AOB = \\angle ACB = \\angle CDB + \\angle DBC$. Ker je $\\angle CDB = \\frac{1}{2} \\angle AOB$, je tudi $\\angle DBC = \\frac{1}{2} \\angle AOB$. Torej je $DCB$ enakokrak trikotnik z vrhom $C$.\n\n\n\nČe pa t... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03jt | Problem:
A competition involving $n \geq 2$ players was held over $k$ days. On each day, the players received scores of $1,2,3, \ldots, n$ points with no two players receiving the same score. At the end of the $k$ days, it was found that each player had exactly $26$ points in total. Determine all pairs $(n, k)$ for whi... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (3, 13), (12, 4), (25, 2) | |
03yv | Let $a$, $b$ and $c$ be coprime positive integers so that $a^2 \mid (b^3 + c^3)$, $b^2 \mid (a^3 + c^3)$ and $c^2 \mid (a^3 + b^3)$. Find the values of $a$, $b$ and $c$. (posed by Yang Xiaoming) | [
"By the condition of the problem, we have $a^2 \\mid (a^3 + b^3 + c^3)$, $b^2 \\mid (a^3 + b^3 + c^3)$ and $c^2 \\mid (a^3 + b^3 + c^3)$. Since $a$, $b$ and $c$ are coprime, we see that $a^2 b^2 c^2 \\mid (a^3 + b^3 + c^3)$.\n\nWithout loss of generality, suppose that $a \\ge b \\ge c$, so\n$$\n3a^3 \\ge a^3 + b^3 ... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | The only solutions are (1,1,1) and all permutations of (1,2,3). | |
0678 | Find all nonzero polynomials with real coefficients satisfying the equality: $$(P(x))^3 + 3(P(x))^2 = P(x^3) - 3P(-x),$$ for all $x \in \mathbb{R}$. | [
"Let $\\deg P(x) = 0$. Then $P(x) = a \\neq 0$ and from relation (1) we have: $a^3 + 3a^2 = -2a \\Leftrightarrow a = -1$ or $a = -2$.\nHence the constant (nonzero) polynomials $P(x) = -1$ and $P(x) = -2$, are solutions of the problem.\n\nLet $\\deg P(x) = n > 0$. Then the polynomial $P(x)$ can be written as\n$$\nP(... | Greece | SELECTION EXAMINATION | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All solutions are P(x) = -1, P(x) = -2, P(x) = x^{2m} - 1, and P(x) = -x^{2m} - 1, where m is any positive integer. | |
0i0u | Problem:
Let $ABC$ be a triangle and $D, E, F, G, H, I, J$ points on sides $BC, CA, AB, BC, CA, AB, BC$ respectively, such that $EF$ and $HI$ are parallel to $BC$; $IJ$ and $FG$ are parallel to $CA$; and $DE, GH$ are parallel to $AB$. Prove that $D = J$. | [
"Solution:\n\nFirst we prove the following lemma: If $PQR$ is a triangle with points $S, T$ on sides $PQ, QR$ such that $ST$ is parallel to $RP$, then (a) $SQ / PQ = TQ / RQ$ and (b) $PS / PQ = RT / RQ$.\n\nProof: First, observe that triangles $PQR, SQT$ are similar (because parallel lines imply equal angles). Equa... | United States | Berkeley Math Circle Monthly Contest #1 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
05g7 | Problem:
Soit $ABC$ un triangle, et $M$ le milieu de $[BC]$. On note $I_{b}$ et $I_{c}$ les centres des cercles inscrits à $AMB$ et $AMC$. Montrer que le second point d'intersection des cercles circonscrits aux triangles $ABI_{b}$ et $ACI_{c}$ se situe sur la droite $(AM)$. | [
"Solution:\n\nSoit $T$ l'intersection entre le cercle de diamètre $[BC]$ et la demi-droite $[MA)$. Il suffit de montrer que $T$ appartient au cercle $ABI_{b}$ (par symétrie des rôles de $B$ et $C$, ceci montrera qu'il appartient également au cercle $ACI_{c}$).\n\nComme $\\widehat{BTC}$ est droit, on a $\\widehat{AT... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"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 | |
088d | Problem:
Ad una fiera c'è un gioco molto invitante, perché si può partecipare gratis; chi vince guadagna un premio. Il premio pattuito per le prime quattro partite è una moneta, per la quinta è di due monete. Nicola ad ogni partita ha probabilità $\frac{2}{3}$ di vincere il premio e decide di giocare 5 partite. Qual è... | [
"Solution:\n\nLa risposta è (A). Consideriamo separatamente i casi in cui Nicola abbia rispettivamente perso e vinto l'ultima partita. Nel primo caso (che avviene con probabilità $1 / 3$ ) Nicola deve aver vinto le prime quattro partite, evento che ha probabilità pari a $\\left(\\frac{2}{3}\\right)^{4}$. Nel second... | Italy | Progetto Olimpiadi di Matematica | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | A | |
0kb9 | Problem:
Sir Alex is coaching a soccer team of $n$ players of distinct heights. He wants to line them up so that for each player $P$, the total number of players that are either to the left of $P$ and taller than $P$ or to the right of $P$ and shorter than $P$ is even. In terms of $n$, how many possible orders are the... | [
"Solution:\n\nWe want to determine the number of permutations $\\pi$ of $1, \\ldots, n$ such that every number is involved in an even number of inversions. The key observation is that the number of inversions $k$ is involved in has the same parity as $\\pi(k)-k$. This is since $\\pi(k)-k$ can be interpreted as the ... | United States | HMIC | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | floor(n/2)! * ceil(n/2)! | |
07gc | Let $x$, $y$, $z > 0$ be real numbers such that $x + y + z = 1399$. Determine the maximum value of
$$
y\lfloor x \rfloor + z\lfloor y \rfloor + x\lfloor z \rfloor,
$$
where $\lfloor x \rfloor$ is the largest integer less than or equal to $x$. | [
"The answer is $652400$. Notice that $y\\lfloor x \\rfloor + z\\lfloor y \\rfloor + x\\lfloor z \\rfloor \\leq xy + yz + zx$. On the other hand, $xy + yz + zx \\leq \\frac{1}{3}(x + y + z)^2$. It follows that\n$$\ny\\lfloor x \\rfloor + z\\lfloor y \\rfloor + x\\lfloor z \\rfloor \\leq \\frac{1}{3}(x + y + z)^2 = 6... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 652400 | |
0iab | Problem:
Let $R S T U V$ be a regular pentagon. Construct an equilateral triangle $P R S$ with point $P$ inside the pentagon. Find the measure (in degrees) of angle $P T V$. | [
"Solution:\n\nWe have $\\angle P R V=\\angle S R V-\\angle S R P=108^{\\circ}-60^{\\circ}=48^{\\circ}$. Since $P R=R S=R V$, triangle $P R V$ is isosceles, so that $\\angle V P R=\\angle R V P=\\left(180^{\\circ}-\\angle P R V\\right) / 2=66^{\\circ}$. Likewise, we have $\\angle T P S=66^{\\circ}$, so that\n$$\n\\a... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 6° | |
0bmk | Let $n$ be an integer greater than $1$, and let $p$ be a prime divisor of $n$. A confederation consists of $p$ states, each of which has exactly $n$ airports. There are $p$ air companies operating interstate flights only such that every two airports in different states are joined by a direct (two-way) flight operated b... | [
"The required maximum is $n$. The following example shows that $N$ cannot exceed $n$. Split the $n$ airports in the $i$-th state, $i = 1, \\dots, p$, into $p$ disjoint groups of $n/p$ airports each, $A_{i,j}$, $j = 1, \\dots, p$. Let the $k$-th air company, $k = 1, \\dots, p$, operate direct flights between every a... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | n | |
04f1 | Determine all positive integers $n$ such that
$$
2^n \cdot (n!)^2 < (2n)!
$$
*(Ilko Brnetić)* | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | all positive integers n ≥ 2 | |
0i2c | Problem:
Let $ABCD$ be a square and $E$ a point on side $CD$. The circle inscribed in triangle $ADE$ touches $DE$ at $F$, and the circle inside quadrilateral $ABCE$, tangent to sides $AB$, $BC$, $EA$, touches $AB$ at $G$. Prove that lines $AE$, $BD$, and $FG$ meet in a point. | [
"Solution:\n\nExtend lines $BC$ and $AE$ to intersect at $H$. Then the circle inside quadrilateral $ABCE$, tangent to $AB$, $BC$, and $EA$, is really the inscribed circle of $\\triangle HBA$. (Actually, this is only true if the circle lies inside $\\triangle HBA$ rather than outside it. However, the fact that $CE$ ... | United States | Berkeley Math Circle Monthly Contest #7 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0hta | Problem:
Prove that
$$
tan \left(\frac{3 \pi}{11}\right)+4 \sin \left(\frac{2 \pi}{11}\right)=\sqrt{11}
$$ | [
"Solution:\nWe work with complex roots of unity. Let $\\zeta=\\cos \\frac{\\pi}{11}+i \\sin \\frac{\\pi}{11}$; then we know that $\\zeta^{n}=\\cos \\frac{n \\pi}{11}+i \\sin \\frac{n \\pi}{11}$ and, in particular, $\\zeta^{11}=-1$. Now, the given expression is clearly positive (since each term is positive), so we n... | United States | Berkeley Math Circle Take-Home Contest #2 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
05f7 | Problem:
Trouver toutes les fonctions $f$ de $\mathbb{R}$ dans $\mathbb{R}$ telles que pour tous réels $x, y$,
$$
f(x f(x)+f(y))=f(x)^{2}+y
$$ | [
"Solution:\n\nEn évaluant l'égalité en $(0, y)$, $f(f(y))=f(0)^{2}+y$. En particulier $f \\circ f$ est bijective, donc $f$ est injective et surjective donc bijective. Soit $a$ tel que $f(a)=0$. En évaluant en $(a, a)$, $f(0)=f(a f(a)+f(a))=f(a)^{2}+a=a$ donc $f(0)=a$. En particulier on obtient que $f(f(0))=0$. Comm... | France | ENVOI 2 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x for all real x, and f(x) = -x for all real x | |
05jd | Problem:
2014 scientifiques participent à un congrès, chaque scientifique étant soit un mathématicien, soit un physicien. Bien sûr, les physiciens mentent toujours et les mathématiciens disent toujours la vérité, sauf quand ils se trompent. Lors du dîner final, tous sont assis en rond autour d'une table, et chacun pré... | [
"Solution:\n\nSi 2 physiciens sont assis côte à côte, comme ils ont menti, ils ne sont pas entre un physicien et un mathématicien, donc chacun est entre deux physiciens, et ainsi de suite, donc le congrès n'accueille que des physiciens, ce qui est absurde car il y a au moins un mathématicien (celui qui s'est trompé... | France | OFM 2013-2014 Envoi 2 | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 671 | |
091u | Problem:
Let $A$ and $B$ be disjoint nonempty sets with $A \cup B=\{1,2,3, \ldots, 10\}$. Show that there exist elements $a \in A$ and $b \in B$ such that the number $a^{3}+a b^{2}+b^{3}$ is divisible by 11. | [
"Solution:\nFor each $n=0,1,2, \\ldots$ the numbers $2^{n}, 2^{n+1}, 2^{n+2}, \\ldots, 2^{n+9}$ have different remainders when divided by 11.\nSuppose that for every $b \\in B$ there is no $a \\in A$ such that $a \\equiv 2 b\\ (\\bmod\\ 11)$.\nFrom the above statement there exists $n \\in\\{0,1, \\ldots, 9\\}$ such... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
09xr | A triangle $ABC$ has the property that $|AB| + |AC| = 3|BC|$. Let $T$ be the point on line segment $AC$ satisfying $|AC| = 4|AT|$. Let $K$ and $L$ be points on the interior of line segments $AB$ and $AC$, respectively, such that $KL \parallel BC$, and $KL$ is tangent to the incircle of $\triangle ABC$. Let $S$ be the i... | [
"Denote the radius of the incircle of $\\triangle ABC$ by $r$. Then the area of triangle $ABC$ is\n$$\n\\frac{1}{2}|AB| \\cdot r + \\frac{1}{2}|BC| \\cdot r + \\frac{1}{2}|AC| = \\frac{1}{2}r \\cdot (3|BC| + |BC|) = 2r|BC|.\n$$\nOn the other hand, the area of $ABC$ equals $\\frac{1}{2}h|BC|$, where $h$ is the altit... | Netherlands | BxMO Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing... | English | proof and answer | 2/3 | |
05ku | Problem:
Soient $a$, $b$, $c$ trois nombres réels strictement positifs tels que $a + b + c = 9$. Montrer que
$$
\frac{a^{3} + b^{3}}{a b + 9} + \frac{b^{3} + c^{3}}{b c + 9} + \frac{c^{3} + a^{3}}{c a + 9} \geqslant 9
$$ | [
"Solution:\n\nOn cherche une minoration de la forme $\\frac{a^{3} + b^{3}}{a b + 9} \\geqslant u(a + b) + v$, c'est-à-dire $f(a) \\geqslant 0$ où $f(a) = a^{3} + b^{3} - (u(a + b) + v)(a b + 9)$.\n\nEn sommant les trois inégalités, on en déduirait\n$$\n\\frac{a^{3} + b^{3}}{a b + 9} + \\frac{b^{3} + c^{3}}{b c + 9}... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
08hv | Problem:
In the triangle $ABC$, $R$ and $r$ are the radii of the circumcircle and the incircle, respectively; $a$ is the longest side and $h$ is the shortest altitude. Prove that $R / r > a / h$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0h7h | Determine all strictly increasing functions $f: \mathbb{R} \to \mathbb{R}$ that satisfies the conditions:
$$
\bullet \quad f(x^3 + y^2 + f(y)) = x^2 f(x) + y f(y) + y;
$$
$$ \bullet \quad \text{Equality } y^2 + f(y) = t \text{ has a solution for every } t \in \mathbb{R}. $$ | [
"**Answer.** Solution does not exist.\n\nLet us substitute $x = y = 0$ at the first condition $\\Rightarrow f(f(0)) = 0$.\nThen when $x=0$ and when $x=f(0)$ we obtain such equations:\n$$\nf(y^2 + f(y)) = y f(y) + y \\quad \\text{ta} \\quad f(f^3(0) + y^2 + f(y)) = y f(y) + y.\n$$\nSince $f$ is strictly increasing:\... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | no such function | |
0fcb | Problem:
Sea $\mathbb{R}^{+}$ el conjunto de los números reales positivos. Halla todas las funciones $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$, tales que $f(x+f(y))=y f(x y+1)$, para todo $x, y>0$. | [
"Solution:\n\nSupongamos que exista un número real $y$ tal que $y-1$ y $f(y)-1$ son ambos no nulos y del mismo signo. Entonces podemos tomar el número real positivo\n$$\nx=\\frac{f(y)-1}{y-1}\n$$\nde donde se obtiene $f(y)=x(y-1)+1$ y $x+f(y)=x y+1$. Por tanto, $f(x+f(y))=f(x y+1)$ y utilizando el enunciado, result... | Spain | LIV Olimpiada matemática Española (Concurso Final) | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/x for all x > 0 | |
0gzy | Find all real values of $x$ for which the value of the function $y = (\sqrt{x})^{2009} + (\sqrt{1-x})^{2010}$ is integer? | [
"Only $x = 0$ and $x = 1$.\n\nObviously $x \\in [0, 1]$ and $0 \\le (\\sqrt{x})^{2009} + (\\sqrt{1-x})^{2010} < 1$. On the other hand we have $x \\le 1$ and $1-x \\le 1$ which implies $(\\sqrt{x})^{2009} + (\\sqrt{1-x})^{2010} \\le x < (1-x) = 1$. The case $(\\sqrt{x})^{2009} + (\\sqrt{1-x})^{2010} = 0$ is impossib... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | x = 0 or x = 1 | |
0frb | Problem:
En un torneo de ajedrez participan 8 maestros durante 7 días. Cada día se disputan 4 partidas en las cuales participan todos los maestros, y al finalizar el torneo todos se han enfrentado contra todos exactamente una vez. Demostrar que al terminar el quinto día del torneo existe un conjunto de al menos 4 maes... | [
"Solution:\n\nSea $A_{1}$ un maestro cualquiera. Llamamos $B_{1}$ a su rival del día $6$, $A_{2}$ al rival de $B_{1}$ el día $7$, $B_{2}$ al rival de $A_{2}$ el día $6$, y así sucesivamente, hasta que se cierre un ciclo:\n$$\nA_{1} \\xrightarrow{6} B_{1} \\xrightarrow{7} A_{2} \\xrightarrow{6} B_{2} \\xrightarrow{7... | Spain | Spain | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
09u2 | Arthur has written down five distinct positive integers smaller than $10$. If you add any two of these five numbers, then the result will never be equal to $10$.
Which number did Arthur write down for sure?
A) $1$ B) $2$ C) $3$ D) $4$ E) $5$ | [
"E) $5$"
] | Netherlands | First Round, January 2019 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | MCQ | E | |
0238 | Problem:
Encontre a lei que forma a sequência e dê seus próximos 2 termos:
$425, 470, 535, 594, 716, 802, \ldots$ | [
"Solution:\nCada termo é a soma do termo precedente com os quadrados de cada um de seus algarismos:\n$$\n470 = 425 + 4^2 + 2^2 + 5^2, \\quad 535 = 470 + 4^2 + 7^2 + 0^2, \\ldots\n$$\nAssim, os próximos termos são: $870$ e $983$."
] | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 870, 983 | |
000p | Sea $ABC$ un triángulo con $AB = 30$, $BC = 50$, $CA = 40$. Las rectas $l_0, l_1, l_2$ son paralelas a $BC$, $CA$, $AB$, respectivamente, y cortan al triángulo. Las distancias entre $l_0$ y $BC$, $l_1$ y $CA$, $l_2$ y $AB$ son $1$, $2$, $3$, respectivamente. Hallar los lados del triángulo que determinan $l_0, l_1, l_2$... | [] | Argentina | XII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | español | proof and answer | The triangle formed by l0, l1, l2 has side lengths 245/6, 98/3, and 49/2, corresponding respectively to the lines parallel to BC, CA, and AB. | |
0l2k | Problem:
Compute the number of ways to pick a 3-element subset of
$$
\left\{10^{1}+1,10^{2}+1,10^{3}+1,10^{4}+1,10^{5}+1,10^{6}+1,10^{7}+1\right\}
$$
such that the product of the 3 numbers in the subset has no digits besides 0 and 1 when written in base 10. | [
"Solution:\nGiven a subset $\\left\\{10^{a}+1,10^{b}+1,10^{c}+1\\right\\}$, we can directly expand the product of its elements:\n$$\n\\left(10^{a}+1\\right)\\left(10^{b}+1\\right)\\left(10^{c}+1\\right)=10^{a+b+c}+10^{b+c}+10^{a+c}+10^{a+b}+10^{a}+10^{b}+10^{c}+1 .\n$$\nIn order for all digits to be 0 or 1, all 7 n... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | null | proof and answer | 26 | |
0fig | Problem:
Determinar los valores de $n$ para los que es posible construir un cuadrado $n \times n$ ensamblando piezas del tipo
| [
"Solution:\n\nSi $n$ es múltiplo de $4$, evidentemente existe solución, como indica la figura siguiente.\n\n\n\nRecíprocamente, supongamos que el cuadrado $n \\times n$ se puede recubrir. Como cada pieza tiene $4$ cuadrados, deberá ser $n^{2}$ múltiplo de $4$, y $n$ múltiplo de $2$.\n\nEl c... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | n is a multiple of 4 | |
055d | Let $ABC$ be a triangle with $AB = \frac{AC}{2} + BC$. Consider the two semicircles outside the triangle with diameters $AB$ and $BC$. Let $X$ be the orthogonal projection of $A$ onto the common tangent line of those semicircles. Find $\angle CAX$. | [
"Let $K$ and $L$ be the midpoints of the sides $AB$ and $BC$, respectively, and $M$ and $N$ the feet of perpendiculars from $K$ and $L$, respectively, to the common tangent of the semicircles (Fig. 28). Then $\\angle CAX = \\angle LKM$. As $KM$ and $LN$ are radii of the semicircles, $KM = \\frac{AB}{2}$ and $LN = \... | Estonia | IMO Team Selection Contest II | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 60° | |
00b3 | A chooses 13 different 3-digit numbers. Then B selects several of these 13 and tries to obtain, by using each selected number once and the operations $+, -, \times$, an expression with value strictly between 3 and 4. B wins if he succeeds in doing so; otherwise A wins. Who has a winning strategy? | [
"Player B has a winning strategy. One way to see this is to divide the 3-digit numbers into 8 groups with the following property: For every two numbers $a$, $b$ from the same group, $a > b$, one has $\\frac{a}{b} < \\frac{4}{3}$.\n\n$$\n\\begin{align*}\nG_1 &= \\{100, \\dots, 133\\}, \\\\\nG_2 &= \\{134, \\dots, 17... | Argentina | Argentina_2018 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Player B | |
0hkb | Problem:
You are given an $m \times n$ chocolate bar divided into $1 \times 1$ squares. You can break a piece of chocolate by splitting it into two pieces along a straight line that does not cut through any of the $1 \times 1$ squares. What is the minimum number of times you have to break the bar in order to separate ... | [
"Solution:\n\nWe note that the number of separate pieces of chocolate increases by $1$ at each cut. We begin with $1$ piece and end with $m n$ pieces, so we must make $m n - 1$ cuts. Thus $m n - 1$ is the minimum (and also the maximum) number of cuts necessary to separate all the $1 \\times 1$ squares."
] | United States | Berkeley Math Circle Monthly Contest 3 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | mn - 1 |
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