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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
05u9 | Problem:
Soit $P_{1} P_{2} \ldots P_{2021}$ un polygone convexe à 2021 sommets tel que, pour chaque sommet $P_{i}$, les 2018 diagonales issues de $P_{i}$ divisent l'angle $\widehat{P}_{i}$ en 2019 angles égaux.
Démontrer que $P_{1} P_{2} \ldots P_{2021}$ est un polygone régulier, c'est-à-dire un polygone dont tous le... | [
"Solution:\n\nDans la suite, on pose $n=2019$ et, pour tout sommet $P_{i}$, on note $a_{i}$ l'angle $\\widehat{P}_{i} / n$. Soit alors $P_{k}$, $P_{k+1}, P_{k+2}, P_{k+3}$ et $P_{k+4}$ quatre sommets consécutifs, les indices des sommets étant considérés modulo $n+2$.\nLa somme des angles d'un triangle vaut $180^{\\... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0f66 | Problem:
Let $A = \cos^2 a$, $B = \sin^2 a$. Show that for all real $a$ and positive $x$, $y$ we have $x^A y^B < x + y$. | [] | Soviet Union | 18th ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0hw0 | Problem:
Let $A$ be a set of size $2023$. Find the maximum number of pairs of elements $x, y \in A$ so that $x-y$ is a power of $e$. | [
"Solution:\nLet $a_{n}$ be the maximum possible number of such pairs for a set of size $n$. Let $s_{2}(n)$ be the number of ones in $n$'s binary representation. Let $S(n)=\\sum_{k=0}^{n-1} s_{2}(k)$. We show that $a_{n}=S(n)$.\n\nFor the construction, we can take the binary representations of all numbers from $0$ t... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 11043 | |
07qn | Suppose $a$, $b$, $c$ are the side lengths of a triangle $ABC$. Prove that the system of equations
$$
\begin{aligned}
by + cz + aw &= 1, \\
bx + az + cw &= 0, \\
cx + ay + bw &= 0, \\
ax + cy + bz &= 0,
\end{aligned}
$$
has a unique solution. | [
"$$\n\\begin{align*}\n-abx + bcy + (c^2 - a^2)z &= c, \\\\\n-acx + (b^2 - a^2)y + bcz &= b, \\\\\nax + cy + bz &= 0.\n\\end{align*}\n$$\nNext, eliminate $z$ from these by first subtracting $c^2-a^2$ times the second from $bc$ times the first, and then subtracting $c$ times the third from the second.\nThese operatio... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0hx7 | Problem:
Let $a_{1}, \ldots, a_{n}$ be distinct integers. Prove that the polynomial
$$
\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)-1
$$
cannot be written as the product of two nonconstant polynomials with integer coefficients (i.e. it is irreducible over the integers). | [
"Solution:\nAssume there exist polynomials $f$ and $g$ satisfying\n$$\nf(x) g(x)=\\left(x-a_{1}\\right) \\ldots\\left(x-a_{n}\\right)-1 .\n$$\nLet $h(x)=f(x)+g(x)$. Now, for every $a_{i}$ we have $f\\left(a_{i}\\right) g\\left(a_{i}\\right)=-1$, so\n$$\nh\\left(a_{i}\\right)=f\\left(a_{i}\\right)+g\\left(a_{i}\\rig... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0dpy | Distinct positive integers $A$ and $B$ are given. Prove that there exist infinitely many positive integers that can be represented both as $x_1^2 + Ay_1^2$ for some positive coprime integers $x_1$ and $y_1$, and as $x_2^2 + By_2^2$ for some positive coprime integers $x_2$ and $y_2$. (Golovanov A.S.) | [
"Without loss of generality $A > B$.\nChoose an arbitrary prime $p > 2$ and let's find $x_1$ and $x_2$ so that\n$$\nx_1^2 + A(2p)^2 = x_2^2 + B(2p)^2,\n$$\nHence, $x_2^2 - x_1^2 = 4Cp^2$ where $C = A - B$. Set $x_1 = Cp^2 - 1$ and $x_2 = Cp^2 + 1$. If $x_1$ and $x_2$ are both odd, then they are both coprime with $y... | Silk Road Mathematics Competition | XXI SILK ROAD MATHEMATICAL COMPETITION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof only | null | |
0ep6 | Calculate $2015 - (2015 - (2015 - (2015 - 1)))$. | [
"$2015 - (2015 - (2015 - (2015 - 1))) = 2015 - (2015 - (2015 - 2014)) = 2015 - (2015 - 1) = 2015 - 2014 = 1.$\n\nAlternatively,\n$2015 - (2015 - (2015 - (2015 - 1))) = 2015 - 2015 + 2015 - 2015 + 1 = 1.$"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 1 | |
05hx | Problem:
Soit $a, b, c, d$ des réels strictement positifs tels que $a+b+c+d=1$. Montrer que :
$$
\frac{a^{4}}{a^{3}+a^{2} b+a b^{2}+b^{3}}+\frac{b^{4}}{b^{3}+b^{2} c+b c^{2}+c^{3}}+\frac{c^{4}}{c^{3}+c^{2} d+c d^{2}+d^{3}}+\frac{d^{4}}{d^{3}+d^{2} a+d a^{2}+a^{3}} \geqslant \frac{1}{4}
$$
et déterminer les cas d'égali... | [
"Solution:\n\nDans cette solution, on va utiliser à plusieurs reprises l'inégalité $2\\left(x^{2}+y^{2}\\right) \\geqslant (x+y)^{2}$, valable pour tous réels $x, y$ avec égalité si et seulement si $x=y$.\n\nDéjà notons que $a^{3}+a^{2} b+a b^{2}+b^{3}=\\left(a^{2}+b^{2}\\right)(a+b)$. En particulier l'inégalité se... | France | ENVOI 2 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Lower bound is 1/4, with equality if and only if a = b = c = d = 1/4. | |
07fu | Consider lattice points of a $6 \times 7$ grid. We start with two points $A$, $B$. We say two points $X$, $Y$ are connected if one can reflect several times with respect to points $A$, $B$ and reach from $X$ to $Y$. What is the minimum number of connected components, over all choices of $A$, $B$? | [
"We claim the answer is $8$. Let us first find the points in a connected component. Let $l_P$ be the line passing through a point $P$ and parallel to $AB$. Let $l'_P$ be the reflection of $l_P$ with respect to $AB$. First note that the reflection of any point $P$ with respect to each of $A$ and $B$ lies on $l'_P$. ... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 8 | |
07tb | Find all prime numbers $p$ such that $2^p + 1$ is a perfect square. | [
"**Solution 1.** The only such prime is $p=3$, when $2^3 + 1 = 3^2$. We consider the remainder of $2^k + 1$ on division by $9$, for arbitrary integers $k$ and first note that $2^6 = 64 \\equiv 1 \\pmod{9}$ and so\n$$\n2^{k+6} + 1 \\equiv 2^6 2^k + 1 \\equiv 2^k + 1 \\pmod{9}.\n$$\nThe statement of the question clea... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 3 | |
034x | Problem:
In an internet chess tournament 2005 chess players took part and everyone played one game against any other. After the tournament it appeared that for every two players $A$ and $B$ who had drawn their game every other player had lost his game with $A$ or with $B$. Prove that if there were at least two draws i... | [
"Solution:\n\nNote that a chess player could not have more than one draw. Indeed, if $A$ had draws with $B$ and $C$, then the condition for $A$ and $B$ implies that $B$ defeated $C$ and the same condition for $A$ and $C$ implies that $C$ defeated $B$, a contradiction.\n\nLet $A_{1}, A_{2}, \\ldots, A_{k}$ be the lo... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0diz | Positive numbers $a, b$ satisfy the condition $a + b + \frac{1}{a} + \frac{1}{b} = 5$. Prove that $1 \le a + b \le 4$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
006t | En cada casilla de un tablero de $60 \times 60$ está escrito un número de valor absoluto menor o igual que $1$. La suma de todos los números del tablero es igual a $600$. Demostrar que el tablero contiene un cuadrado de $12 \times 12$ en el que la suma de los $144$ números de sus casillas tiene valor absoluto menor o i... | [
"Sea $a_{i,j}$ el número escrito en la casilla de la fila $i$ y columna $j$, con $1 \\leq i, j \\leq 60$. Se tiene que $|a_{i,j}| \\leq 1$ para todo $i, j$, y\n$$\n\\sum_{i=1}^{60} \\sum_{j=1}^{60} a_{i,j} = 600.\n$$\n\nConsideremos todos los subcuadrados de $12 \\times 12$ del tablero. Hay $(60 - 12 + 1)^2 = 49^2 ... | Argentina | Argentina 2009 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Expected values"
] | Spanish | proof only | null | |
08k1 | Problem:
Find all four digit numbers $A$ such that
$$
\frac{1}{3} A+2000=\frac{2}{3} \bar{A}
$$
where $\bar{A}$ is the number with the same digits as $A$, but written in opposite order. (For example, $\overline{1234}=4321$.) | [
"Solution:\nLet $A=1000 a+100 b+10 c+d$. Then we obtain the equality\n$$\n\\frac{1}{3}(1000 a+100 b+10 c+d)+2000=\\frac{2}{3}(1000 d+100 c+10 b+a)\n$$\nMultiply both sides by $3$ to clear denominators:\n$$\n1000 a+100 b+10 c+d + 6000 = 2000 d + 200 c + 20 b + 2 a\n$$\nBring all terms to one side:\n$$\n1000 a + 100 ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2004 | |
06bk | Let $f(x)$ be a continuous function defined on the set of real numbers such that for any real numbers $x$,
(i) $f(x + 1) = f(x) + 1$, and
(ii) $f(x^3) = (f(x))^3$.
Find all possible $f(x)$. | [
"The only solution is $f(x) = x$ for any $x \\in \\mathbb{R}$.\n\nUsing $f(x+1) = f(x) + 1$, it can be proved easily by induction that\n$$\nf(x+n) = f(x) + n \\quad (1)\n$$\nfor any integer $n$.\n\nConsider any rational number $\\frac{a}{b}$ where $a, b \\in \\mathbb{Z}$. For any positive integer $k$, we substitute... | Hong Kong | 1997-2023 IMO HK TST | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x for all real x | |
04my | Determine all pairs $(p, q)$ of prime numbers for which $p^{q-1} + q^{p-1}$ is a perfect square. | [
"Let $n$ be a positive integer such that\n$$\np^{q-1} + q^{p-1} = n^2.\n$$\nWe will divide our solution into cases, depending on the parity of numbers $p$ and $q$.\n\nIf both are even, the only possibility is that both are equal to $2$. This leads to a solution $(p, q) = (2, 2)$ for $n = 2$.\n\nIf both $p$ and $q$ ... | Croatia | Croatia_2018 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (2, 2) | |
06rb | Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the number $(f(m)+n)(m+f(n))$ is a square for all $m, n \in \mathbb{N}$.
(U.S.A.) | [
"First, it is clear that all functions of the form $f(n)=n+c$ with a constant nonnegative integer $c$ satisfy the problem conditions since $(f(m)+n)(f(n)+m)=(n+m+c)^{2}$ is a square.\n\nWe are left to prove that there are no other functions. We start with the following Lemma. Suppose that $p \\mid f(k)-f(\\ell)$ fo... | IMO | 51st IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All functions of the form f(n) = n + c, where c is a fixed nonnegative integer. | |
0g6e | $\triangle ABC$ 為銳角三角形。過 $BC$ 上一點 $L$ 作圓 $\omega$, 使其切 $AB$ 於 $B'$, 同時切 $AC$ 於 $C'$。此外, 假設 $\triangle ABC$ 的外接圓圓心 $O$ 位於 $\omega$ 上較短一側的 $B'C'$ 弧。試證: $\triangle ABC$ 的外接圓與 $\omega$ 交於兩點。 | [
"顯然 $B'$ 為 $L$ 對 $AB$ 之垂足, 故其位於 $AB$ 線段內。同理, $C'$ 位於線段 $AC$ 內。因此 $O$ 必位於 $\\triangle AB'C'$ 內, 從而 $\\angle COB < \\angle C'OB'$。\n\n現令 $\\alpha = \\angle CAB$. 易見 $\\angle COB = 2\\angle CAB = 2\\alpha$ 且 $2\\angle C'OB' = 360^\\circ - \\angle C'LB'$. 又, $\\angle C'LB' = 180^\\circ - \\angle C'AB' = 180^\\circ - \\... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 | |
0fod | Let $p$ be a prime number and let $m$ and $n$ be positive integers written in base $p$ as $n = a_0 + a_1p + \dots + a_kp^k$ and $m = b_0 + b_1p + \dots + b_kp^k$, respectively. Show that
$$
\binom{n}{m} \equiv \prod_{i=0}^{k} \binom{a_i}{b_i} \pmod{p}
$$ | [
"Since $\\binom{n}{m} = 0$ for $n < m$ then hereafter we assume that $n \\ge m$. Next we use the well-known and easily proved fact that $(x+1)^p \\equiv x^p + 1 \\pmod{p}$, meaning that each coefficient of the polynomial $(x+1)^p - (x^p + 1)$ is divisible by $p$. Thus,\n$$\n\\begin{align*}\n(x+1)^n &= (x+1)^{a_0+a_... | Spain | International Mathematical Arhimede Contest | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Spanish | proof only | null | |
0jl7 | Problem:
Let $\omega$ be a root of unity and $f$ be a polynomial with integer coefficients. Show that if $|f(\omega)|=1$, then $f(\omega)$ is also a root of unity. | [
"Solution:\n\nSuppose $\\omega$ is a primitive $n$th root, so that $\\Phi_{n}(x)$ is the minimal polynomial of $\\omega$ (over $\\mathbb{Q}$). Thus $f(\\omega) f\\left(\\omega^{-1}\\right)=1$ implies $\\Phi_{n}(x) \\mid f(x) f\\left(x^{-1}\\right)-1$ (divisibility as Laurent polynomials). Hence $\\left|f\\left(\\om... | United States | HMMT 2014 HMIC | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Algebraic Number Theory > Algebraic numbers",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
0jrc | Problem:
Let $\pi$ be a permutation of $\{1,2, \ldots, 2015\}$. With proof, determine the maximum possible number of ordered pairs $(i, j) \in \{1,2, \ldots, 2015\}^{2}$ with $i<j$ such that $\pi(i) \cdot \pi(j)>i \cdot j$. | [
"Solution:\nAnswer: $\\dbinom{2014}{2}$\n\nLet $n=2015$. The only information we will need about $n$ is that $n>5$.\n\nFor the construction, take $\\pi$ to be the $n$-cycle defined by\n$$\n\\pi(k)= \\begin{cases}k+1 & \\text{ if } 1 \\leq k \\leq n-1 \\\\ 1 & \\text{ if } k=n\\end{cases}\n$$\nThen $\\pi(i)>i$ for $... | United States | HMMT February 2015 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | C(2014, 2) | |
0hzm | Problem:
What fraction of the Earth's volume lies above the $45$ degrees north parallel? You may assume the Earth is a perfect sphere. The volume in question is the smaller piece that we would get if the sphere were sliced into two pieces by a plane. | [
"Solution:\nWithout loss of generality, look at a sphere of radius $1$ centered at the origin. If you like cartesian coordinates, then you can slice the sphere into discs with the same $z$ coordinate, which have radius $\\sqrt{1-z^{2}}$, so the region we are considering has volume\n$$\n\\int_{\\sqrt{2} / 2}^{1} \\p... | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | (8 - 5√2) / 16 | |
0iss | Problem:
Let $f(x) = x^{3} + x + 1$. Suppose $g$ is a cubic polynomial such that $g(0) = -1$, and the roots of $g$ are the squares of the roots of $f$. Find $g(9)$. | [
"Solution:\n\nLet $a, b, c$ be the zeros of $f$. Then $f(x) = (x - a)(x - b)(x - c)$. Then, the roots of $g$ are $a^{2}, b^{2}, c^{2}$, so $g(x) = k(x - a^{2})(x - b^{2})(x - c^{2})$ for some constant $k$. Since $a b c = -f(0) = -1$, we have $k = k a^{2} b^{2} c^{2} = -g(0) = 1$. Thus,\n$$\ng(x^{2}) = (x^{2} - a^{2... | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 899 | |
06tb | Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences
$$
\begin{aligned}
& P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text{ and } \\
& Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-2014)... | [
"For convenience, we set $k=2015=2 \\ell+1$.\n\n**a.**\nConsider the following polynomials of degree $n+1$ :\n$$\nP(x)=\\prod_{i=0}^{n}(x-i k) \\quad \\text{ and } \\quad Q(x)=\\prod_{i=0}^{n}(x-i k-1) .\n$$\nSince $Q(x)=P(x-1)$ and $P(0)=P(k)=P(2 k)=\\cdots=P(n k)$, these polynomials are block-similar (and distinc... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
02xy | Problem:
Considere o quadrado $ABCD$ com lados de comprimento $1$, como no desenho a seguir. Um segmento horizontal $EW$ e um segmento vertical $NS$, ambos de comprimento $1/2$, estão inteiramente dentro do quadrado e se intersectam no ponto $X$ formando um ângulo de $90^\circ$.
a) Qual a área do quadrilátero $NESW$?
... | [
"Solution:\n\na) A área do quadrilátero $NESW$ é dada por\n$$\n\\begin{aligned}\nA_{NESW} & = A_{ENW} + A_{ESW} \\\\\n& = \\frac{EW \\cdot NX}{2} + \\frac{EW \\cdot XS}{2} \\\\\n& = \\frac{EW \\cdot (NX + XS)}{2} \\\\\n& = \\frac{1/2 \\cdot 1/2}{2} \\\\\n& = \\frac{1}{8}\n\\end{aligned}\n$$\n\nb) A soma das áreas d... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | a) 1/8; b) 3/8 | |
069m | If $a$, $b$, $c$ are positive real numbers, prove that:
$$
\frac{1}{ab(b+1)(c+1)} + \frac{1}{bc(c+1)(a+1)} + \frac{1}{ca(a+1)(b+1)} - \frac{3}{(1+abc)^2}
$$ | [
"The inequality is equivalent to\n$$\n\\frac{c(a+1)+a(b+1)+b(c+1)}{abc(a+1)(b+1)(c+1)} \\ge \\frac{3}{(1+abc)^2},\n$$\nor after simplifications to\n$$\n(1+abc)^2(ab+bc+ca+a+b+c) \\ge 3abc(ab+bc+ca+a+b+c+abc+1).\n$$\nWe put $m = a+b+c$, $n = ab+bc+ca$ and $x^3 = abc$, to obtain:\n$$ (m+n)(1+x^3)^2 \\ge 3x^3(x^3 + m+... | Greece | SELECTION EXAMINATION | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0kot | Problem:
Candice starts driving home from work at 5:00 PM. Starting at exactly 5:01 PM, and every minute after that, Candice encounters a new speed limit sign and slows down by 1 mph. Candice's speed, in miles per hour, is always a positive integer. Candice drives for $2/3$ of a mile in total. She drives for a whole n... | [
"Solution:\n\nSuppose that Candice starts driving at $n$ miles per hour. Then she slows down and drives $(n-1)$ mph, $(n-2)$ mph, and so on, with her last speed being $(m+1)$ mph.\n\nThen the total distance traveled is\n$$\n\\begin{aligned}\n\\frac{n}{60} + \\frac{n-1}{60} + \\cdots + \\frac{m+1}{60} &= \\frac{1}{6... | United States | HMMT November 2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 5:05 pm | |
0iiv | Problem:
Find the smallest positive integer $k$ such that $z^{10}+z^{9}+z^{6}+z^{5}+z^{4}+z+1$ divides $z^{k}-1$. | [
"Solution:\nLet $Q(z)$ denote the polynomial divisor. We need that the roots of $Q$ are $k$-th roots of unity. With this in mind, we might observe that solutions to $z^{7}=1$ and $z \\neq 1$ are roots of $Q$, which leads to its factorization. Alternatively, we note that\n$$\n(z-1) Q(z)=z^{11}-z^{9}+z^{7}-z^{4}+z^{2... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 84 | |
09pq | Problem:
Bewijs dat er oneindig veel paren positieve gehele getallen $(x, y)$ zijn met
$$
\frac{x+1}{y} + \frac{y+1}{x} = 4.
$$ | [] | Netherlands | TOETS TRAININGSKAMP | [
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
07h7 | Is it possible to write a permutation of positive integers on the cells of an infinite table (infinite from all sides) such that the sequence of numbers in each column from bottom to top and also in each row from left to right be increasing? | [
"The answer is yes. First, we choose a cell as origin and fill it with zero, then we fill the table in a way that the numbers in the cells of $(2k+1) \\times (2k+1)$ table centered at the origin be a permutation of numbers\n$$\n(-2k^2 - 2k, -2k^2 - 2k + 1, \\dots, 0, \\dots, 2k^2 + 2k - 1, 2k^2 + 2k),\n$$\nWe proce... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | Yes | |
06vg | Alice has a map of Wonderland, a country consisting of $n \geqslant 2$ towns. For every pair of towns, there is a narrow road going from one town to the other. One day, all the roads are declared to be "one way" only. Alice has no information on the direction of the roads, but the King of Hearts has offered to help her... | [
"We will show Alice needs to ask at most $4 n-7$ questions. Her strategy has the following phases. In what follows, $S$ is the set of towns that Alice, so far, does not know to have more than one outgoing road (so initially $|S|=n$ ).\n\nPhase 1. Alice chooses any two towns, say $A$ and $B$. Without loss of general... | IMO | IMO 2019 Shortlisted Problems | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
02xr | Problem:
Luísa faz experiências com uma rolha e um copo d'água. Por conta de sua densidade, uma rolha fica com apenas $60\%$ de seu volume imerso na água. A rolha da experiência de Luísa tem formato de um prisma hexagonal regular, ou seja, sua base é um hexágono regular, com $3~\mathrm{cm}$ de altura e $2~\mathrm{cm}$... | [
"Solution:\n\na) Como a altura da rolha é de $3~\\mathrm{cm}$, e a parte fora da água corresponde a $40\\%$, então esta altura corresponde a $0,4 \\cdot 3 = 1,2~\\mathrm{cm}$.\n\nb) Vamos analisar a figura abaixo, que representa uma seção transversal da rolha com sua parte inferior imersa. Vamos lembrar que um hexá... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | a) 1.2 cm; b) (√255 − 5√3) / 10 cm | |
0bxk | Let $ABCD$ be a convex quadrangle such that $\angle ABC = \angle ADC < 90^\circ$. The internal bisectrices of the angles $ABC$ and $ADC$ cross the diagonal $AC$ at $E$ and $F$, respectively, and meet at $P$. Let $M$ be the midpoint of the diagonal $AC$. The segments $BM$ and $DM$ cross the circle $BDP$ again at $X$ and... | [
"We first show that $Y$ lies on the circle $ABC$. To this end, let $Y'$ be the point on the ray $MD$, emanating from $M$, such that $MY' \\cdot MD = MA^2$. The triangles $MAY'$ and $MDA$ are therefore similar and have opposite orientations. Since $MY' \\cdot MD = MA^2 = MC^2$, so are the triangles $MCY'$ and $MDC$.... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geomet... | English | proof only | null | |
08au | Problem:
Giovanni disegna a matita un 9-agono regolare e collega ciascuno dei suoi vertici al centro, tracciando un totale di 18 segmenti e ottenendo in questo modo nove triangoli. Ripassa quindi a penna alcuni dei segmenti tracciati, facendo in modo che alla fine ognuno dei nove triangoli abbia esattamente un lato ri... | [
"Solution:\n\nChiameremo raggi i 9 segmenti che ammettono il centro del 9-agono come vertice.\nSi noti anzitutto che scegliere l'insieme dei segmenti da ripassare a penna equivale a scegliere un sottoinsieme dei 9 raggi che non contenga due raggi consecutivi: naturalmente Giovanni non può ripassare due raggi consec... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | D | |
0k9s | Problem:
Let $P(x)$ be a polynomial with real coefficients so that $P(x) \geq 0$ for all real $x$. Prove that there exist polynomials $Q_{1}(x)$ and $Q_{2}(x)$ with real coefficients such that $P(x) = Q_{1}^{2}(x) + Q_{2}^{2}(x)$ for all $x$. | [
"Solution:\n\nSince $P(x) \\geq 0$ for all $x$, it can have no real roots except double roots, so we can write it as a product\n$$\nP(x) = c \\prod_{k=1}^{n} \\left(x^{2} + p_{k} x + q_{k}\\right)\n$$\nof quadratics with nonpositive discriminant, i.e. $p_{k}^{2} - 4q_{k} \\leq 0$. But then completing the square in ... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05tf | Problem:
Soit $a_{0}, a_{1}, a_{2}, \ldots$ une suite d'entiers naturels non nuls, et soit $b_{0}, b_{1}, b_{2}, \ldots$ la suite telle que $b_{n}=\operatorname{PGCD}\left(a_{n}, a_{n+1}\right)$ pour tout entier $n \geqslant 0$. Est-il possible que tout entier naturel non nul soit égal à exactement un des termes $b_{0}... | [
"Solution:\nDans la suite, on dira qu'une suite $\\left(b_{k}\\right)_{k \\geqslant 0}$ est agréable si, pour tout $k \\geqslant 0$, les entiers $b_{k}$ et $b_{k+2}$ sont premiers entre eux. Étant donnée une suite $\\left(b_{k}\\right)_{k \\geqslant 0}$ agréable, posons $a_{0}=b_{0}$ puis $a_{k}=b_{k-1} b_{k}$ pour... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Yes | |
0ik5 | Problem:
In the plane, what is the length of the shortest path from $(-2,0)$ to $(2,0)$ that avoids the interior of the unit circle (i.e., circle of radius $1$) centered at the origin? | [
"Solution:\n\nThe path goes in a line segment tangent to the circle, then an arc of the circle, then another line segment tangent to the circle. Since one of these tangent lines and a radius of the circle give two legs of a right triangle with hypotenuse the line from $(0,0)$ to $(-2,0)$ or $(2,0)$, the length of e... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2*sqrt(3) + pi/3 | |
03m8 | Amy has divided a square into finitely many white and red rectangles, each with sides parallel to the sides of the square. Within each white rectangle, she writes down its width divided by its height. Within each red rectangle, she writes down its height divided by its width. Finally, she calculates $x$, the sum of the... | [
"Let $a_i$ and $b_i$ denote the width and height of each white rectangle, and let $c_i$ and $d_i$ denote the width and height of each red rectangle. Also, let $L$ denote the side length of the original square.\n\n**Lemma:** Either $\\sum a_i \\ge L$ or $\\sum d_i \\ge L$.\n\n**Proof of lemma:** Suppose there exists... | Canada | Kanada 2011 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 2.5 | |
0buu | Problem:
Fie $x, y, z \in \mathbb{R} - \{1\}$ numere reale astfel încât $x + y + z = 4$, $x^{2} + y^{2} + z^{2} = 6$ şi $xyz = a$, $a \in \mathbb{R} - \{2\}$.
Calculaţi valoarea expresiei
$$
E = \frac{1}{xy + z - 3} + \frac{1}{yz + x - 3} + \frac{1}{zx + y - 3}.
$$ | [] | Romania | OLIMPIADA DE MATEMATICĂ ETAPA LOCALĂ | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 1/(a - 2) | |
0kcx | Problem:
Let $a_{1}, a_{2}, a_{3}, \ldots$ be a sequence of positive integers where $a_{1}=\sum_{i=0}^{100} i!$ and $a_{i}+a_{i+1}$ is an odd perfect square for all $i \geq 1$. Compute the smallest possible value of $a_{1000}$. | [
"Solution:\nNote that $a_{1} \\equiv 1+1+2+6 \\equiv 2 \\pmod{8}$. Since $a_{1}+a_{2}$ must be an odd perfect square, we must have $a_{1}+a_{2} \\equiv 1 \\pmod{8} \\Longrightarrow a_{2} \\equiv 7 \\pmod{8}$. Similarly, since $a_{2}+a_{3}$ is an odd perfect square, we must have $a_{3} \\equiv 2 \\pmod{8}$. We can c... | United States | HMMO 2020 | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 7 | |
0dtv | Jack and Jill play the following game: Jack throws 3 dice and Jill can select some of them, possibly none, and turn each of them to the opposite side. Jill wins if the sum of the values on the dice is a multiple of 4. Can Jill always win? (Note the game is played with standard dice where the sum of the numbers on oppos... | [
"Jill can always turn the dice so that the numbers are $2a$, $2b$, $2c$, i.e., all even. On the other hand she can also achieve $7 - 2a$, $7 - 2b$, $2c$.\nLet $S = 2a + 2b + 2c$ and $T = (7 - 2a) + (7 - 2b) + 2c$. Then both $S$, $T$ are even and $S + T = 14 + 4c$. Since $S + T$ is even and not a multiple of 4, one ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Yes, Jill can always win. | |
0kow | Problem:
Consider the paths from $(0,0)$ to $(6,3)$ that only take steps of unit length up and right. Compute the sum of the areas bounded by the path, the $x$-axis, and the line $x=6$ over all such paths.
(In particular, the path from $(0,0)$ to $(6,0)$ to $(6,3)$ corresponds to an area of 0.) | [
"Solution:\nWe see that the sum of the areas under the path is equal to the sum of the areas above the path. Thus, the sum of the areas under the path is half the area of the rectangle times the number of paths, which is $\\frac{18\\binom{9}{3}}{2}=756$."
] | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 756 | |
0aqq | Problem:
Let $a$, $b$, and $c$ be the roots of the equation $2x^{3} - x^{2} + x + 3 = 0$. Find the value of
$$
\frac{a^{3} - b^{3}}{a - b} + \frac{b^{3} - c^{3}}{b - c} + \frac{c^{3} - a^{3}}{c - a}
$$ | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | -1 | |
0f4f | Problem:
1000 people live in a village. Every evening each person tells his friends all the news he heard during the day. All news eventually becomes known (by this process) to everyone. Show that one can choose 90 people, so that if you give them some news on the same day, then everyone will know in 10 days. | [] | Soviet Union | 15th ASU | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
06k2 | Let $\{a_n\}$ be a sequence of integers satisfying the following condition for all positive integral values of $n$: $a_n + a_{n+1} = 2a_{n+2}a_{n+3} + 2016$. Find all possible values of $a_1$ and $a_2$. | [
"$(a_1, a_2)$ can be $(1, -2015), (15, -69), (70, -14), (2016, 0)$ or their permutations.\n\nFirstly, taking the difference of the two relations $a_n + a_{n+1} = 2a_{n+2}a_{n+3} + 2016$ and $a_{n+1} + a_{n+2} = 2a_{n+3}a_{n+4} + 2016$, we obtain\n$$\na_{n+2} - a_n = 2a_{n+3}(a_{n+4} - a_{n+2}).\n$$\nInductively, th... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [(1, -2015), (-2015, 1), (15, -69), (-69, 15), (70, -14), (-14, 70), (2016, 0), (0, 2016)] | |
0ewe | Problem:
Given four positive real numbers $a$, $b$, $c$, $d$ such that $abcd = 1$, prove that
$$
a^2 + b^2 + c^2 + d^2 + ab + ac + ad + bc + bd + cd \geq 10.
$$ | [
"Solution:\nApplying the arithmetic/geometric mean result to the 10 numbers gives the result immediately."
] | Soviet Union | 2nd ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0c41 | Problem:
Considerăm mulţimea $M=\left\{\left.\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) \in \mathcal{M}_2(\mathbb{C}) \right\vert\, a b=c d\right\}$.
a) Daţi exemplu de matrice $A \in M$ astfel încât $A^{2017} \in M$ şi $A^{2019} \in M$, dar $A^{2018} \notin M$.
b) Arătaţi că, dacă $A \in M$ şi există numă... | [
"Solution:\na) Luăm $A \\in M$ astfel încât $A^{2} \\notin M$ şi $A^{2}+A+I_{2}=0_{2}$, deci $A^{3}=I_{2}=A^{2019} \\in M$, $A^{2017}=A \\in M$ şi $A^{2018}=A^{2} \\notin M$. Un exemplu este $A=\\left(\\begin{array}{cc}1 & \\sqrt{6} \\\\ -\\sqrt{3 / 2} & -2\\end{array}\\right)$\n\nb) Din teorema Hamilton-Cayley rei... | Romania | Olimpiada Nationala de Matematica | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | Part a: A = [[1, sqrt(6)], [−sqrt(3/2), −2]] satisfies A in M, A^2017 in M, A^2019 in M, but A^2018 not in M. Part b: If A in M and there exists k ≥ 1 with A^k, A^{k+1}, A^{k+2} in M, then A^n in M for all integers n ≥ 1. | |
0fqt | Problem:
Fijamos un número natural $k \geq 1$. Encuentra todos los polinomios $P(x)$ que cumplan
$$
P\left(x^{k}\right)-P(k x)=x^{k} P(x)
$$ | [
"Solution:\nFijémonos que una solución trivial es $P(x)=0$ para cualquier valor de $k \\geq 1$.\nPara encontrar otras soluciones, fijamos primero $k=1$. En ese caso, la ecuación queda\n$$\nP(x)-P(x)=x P(x)\n$$\npor lo que $P(x)=0$, que es la solución anterior.\nSi $k \\geq 2$ y $P(x)$ es una constante $c$, tendremo... | Spain | OME fase local | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | If the parameter equals two, all polynomials are of the form a times (x squared minus four) for any real a; otherwise the only solution is the zero polynomial. | |
09mp | Let $a$, $b$, $c$ be real numbers with sum $a + b + c = 6$. Prove that
$$
\frac{a^4 + 1}{b^2 + 1} + \frac{b^4 + 1}{c^2 + 1} + \frac{c^4 + 1}{a^2 + 1} \ge \frac{51}{5}.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
04yz | Find all pairs of integers $(m, n)$ such that for all positive real numbers $x$ and $y$ the inequality $x^m + y^n \ge x^n y^m$ holds. | [
"If $m = 0$, then the inequality is $1 + y^n \\ge x^n$. This holds for all positive real numbers $x$ and $y$ iff $n = 0$. Hence $(0, 0)$ is a solution.\n\nLet now both $m$ and $n$ be different from zero. If the pair $(m, n)$ satisfies the condition, then substituting $x$ and $y$ by $\\frac{1}{x}$ and $\\frac{1}{y}$... | Estonia | Estonija 2010 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (0, 0) | |
0ie5 | Problem:
A true-false test has ten questions. If you answer five questions "true" and five "false," your score is guaranteed to be at least four. How many answer keys are there for which this is true? | [
"Solution:\nSuppose that either nine or ten of the questions have the same answer. Then no matter which five questions we pick to have this answer, we will be right at least four times. Conversely, suppose that there are at least two questions with each answer; we will show that we can get a score less than four. B... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 22 | |
0jm6 | Problem:
Consider the set of 5-tuples of positive integers at most $5$. We say the tuple $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ is perfect if for any distinct indices $i, j, k$, the three numbers $a_{i}, a_{j}, a_{k}$ do not form an arithmetic progression (in any order). Find the number of perfect 5-tuples. | [
"Solution:\n\nAnswer: $780$\n\nThere are two situations.\n\n- 1. The multiset is $a\\, a\\, b\\, b\\, c$; the only condition here is $c \\neq \\frac{1}{2}(a+b)$, for $\\left(\\binom{5}{3}-|S|\\right) \\cdot \\binom{3}{1}=18$ such triples, where $S$ is the set of unordered triples $(a, b, c)$ which do not satisfy th... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 780 | |
01ms | Points $M$ and $K$ are marked on the side $AB$ of the triangle $ABC$ so that $AM = MK$, $CM = CB$, $\angle AKC = \frac{1}{2} \angle CAK + 90^\circ$.
Find $AC$ if $MB = 8$. | [] | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry >... | null | proof and answer | 8 | |
087m | Problem:
Ogni numero naturale, zero incluso, è colorato di bianco o di rosso, in modo che:
- vi siano almeno un numero bianco ed almeno un numero rosso;
- la somma tra un numero bianco ed un numero rosso sia bianca;
- il prodotto tra un numero bianco ed un numero rosso sia rosso.
Dimostrare che il prodotto di due num... | [
"Solution:\n\nLo zero è un numero rosso: infatti, se $0$ fosse bianco, dato che esiste un numero rosso $x$, avremmo che $0 + x = x$ è bianco per la seconda proprietà, contraddizione.\n\nUno è un numero bianco: infatti, se uno fosse rosso, dato che esiste un numero bianco $y$, avremmo che $y \\cdot 1 = y$ è rosso pe... | Italy | Cesenatico | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0gsc | There are $2018$ distinguishable boxes on the table. Starting Writer, Writer and Braker take turn writing a box pair to the table (each pair can be written at most once). They stop when there are $4032$ written pairs on the table. After that Braker numerates box pairs by numbers $1, 2, \dots, 4032$ and for each $k = 1,... | [
"Yes, Braker can guarantee that any two boxes will contain different number of balls. Suppose that Writer at the first move writes a pair $(A_1, A_2)$. At each move Braker chooses pairs containing box $A_1$ (if possible). By doing that he can guarantee that all pairs $(A_1, A_i)$, $i = 2,3,\\ldots,2018$ are on the ... | Turkey | 26th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Yes | |
05t0 | Problem:
Soit $ABC$ un triangle et soit $\Gamma$ son cercle circonscrit. Soit $P$ le point d'intersection de la droite $(BC)$ et de la tangente à $\Gamma$ en $A$. Soit $D$ et $E$ les symétriques respectifs des points $B$ et $A$ par rapport à $P$.
Soit alors $\omega_{1}$ le cercle circonscrit au triangle $DAC$ et soit ... | [
"Solution:\n\nCommençons par tracer une figure.\n\n\nUne première remarque que l'on peut formuler est que, puisque $P$ est le milieu des segments $[AE]$ et $[BD]$, le quadrilatère $ABED$ est un parallélogramme. Les droites $(AB)$ et $(DE)$ sont donc parallèles, de même que les droites $(AD)... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0k7l | Problem:
How many positive integers $2 \leq a \leq 101$ have the property that there exists a positive integer $N$ for which the last two digits in the decimal representation of $a^{2^{n}}$ is the same for all $n \geq N$ ? | [
"Solution:\n\nIt suffices to consider the remainder mod 100. We start with the four numbers that have the same last two digits when squared: $0,1,25,76$.\nWe can now go backwards, repeatedly solving equations of the form $x^{2} \\equiv n (\\bmod\\ 100)$ where $n$ is a number that already satisfies the condition.\n0... | United States | HMMT February 2019 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n"
] | null | proof and answer | 36 | |
0ihk | Let $p$ be a prime number and let $s$ be an integer with $0 < s < p$. Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and
$$
\{\frac{sm}{p}\} < \{\frac{sn}{p}\} < \frac{s}{p}
$$
if and only if $s$ is not a divisor of $p-1$.
(For $x$ a real number, let $\lfloor x \rfloor$ denote the greatest integer les... | [
"First suppose that $s$ is a divisor of $p-1$; write $d = (p-1)/s$. As $x$ varies among $1, 2, \\dots, p-1$, $\\{sx/p\\}$ takes the values $1/p, 2/p, \\dots, (p-1)/p$ once each in some order. The possible values with $\\{sx/p\\} < s/p$ are precisely $1/p, \\dots, (s-1)/p$. From the fact that $\\{sd/p\\} = (p-1)/p$,... | United States | USAMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
07uw | Let $a$, $b$, $c$ be positive integers. Show that
$$
\left( 1 + \frac{a-c}{b} \sum_{i=1}^{b} \frac{1}{c+i} \right)^{-b} \le \frac{a!\,(b+c)!}{(a+b)!\,c!} \le \left( 1 + \frac{c-a}{b} \sum_{i=1}^{b} \frac{1}{a+i} \right)^{b}.
$$ | [
"The fraction in the middle can also be written this way\n$$\nx = \\frac{a!\\,(b+c)!}{(a+b)!\\,c!} = \\frac{(c+1)(c+2)\\cdots(c+b)}{(a+1)(a+2)\\cdots(a+b)} = \\prod_{i=1}^{b} \\frac{c+i}{a+i}.\n$$\nBecause $1 + \\frac{c-a}{a+i} = \\frac{c+i}{a+i}$, we have\n$$\n1 + \\frac{c-a}{b} \\sum_{i=1}^{b} \\frac{1}{a+i} = \\... | Ireland | IRL_ABooklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0kwk | Problem:
Let $n > 1$ be a positive integer. Claire writes $n$ distinct positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$ in a row on a blackboard. In a move, William can erase a number $y$ and replace it with either $\frac{1}{y}$ or $y+1$ at the same location. His goal is to make a sequence of moves such that after ... | [
"Solution:\n\nWe use divide and conquer. The base case $n=1$ is clear. Let $f(n)$ denote the number of moves required for $n$ numbers. Let $x=\\lceil n / 2\\rceil$ and $y=\\lfloor n / 2\\rfloor$. Then, William can reach his goal by the following process:\n- Use $f(x)$ moves to make the first $x$ numbers a strictly ... | United States | HMIC 2023 | [
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0ct4 | In NBA, there are 30 teams; in a regular season every team is to participate in 82 games. Is it possible to split the teams into two conferences — Western and Eastern, and compose the schedule of the regular season so that the number of games between the teams from different conferences will be exactly the half of the ... | [
"No.\nThe total number of games between the teams from different conferences should be $30 \\cdot 82/4$, so it should be odd. This cannot happen, since each team from the Western conference participates in an even number of games, and each internal game is accounted twice.\n\nSolution:\nНет, не сможет.\nПусть $x$ и... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof and answer | No | |
0a5b | Problem:
Let $\{x_{1}, x_{2}, x_{3}, \ldots, x_{n}\}$ be a set of $n$ distinct positive integers, such that the sum of any 3 of them is a prime number. What is the maximum value of $n$? | [
"Solution:\nFirst we show that $n = 4$ is possible with an example. The example $\\{x_{1}, x_{2}, x_{3}, x_{4}\\} = \\{1, 3, 7, 9\\}$ satisfies the problem because:\n\n$\\cdot\\ 1 + 3 + 7 = 11$ is prime,\n\n$\\cdot\\ 1 + 3 + 9 = 13$ is prime,\n\n$\\cdot\\ 1 + 7 + 9 = 17$ is prime, and\n\n$\\cdot\\ 3 + 7 + 9 = 19$ i... | New Zealand | NZMO Round Two | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 4 | |
05k4 | Problem:
Partant d'un triplet d'entiers relatifs $(x, y, z)$, une opération consiste à ajouter à l'un de ces trois entiers un multiple de l'un des deux autres (ce multiple peut être positif ou négatif). Prouver que si $a, b, c$ sont des entiers premiers entre eux dans leur ensemble, on peut passer du triplet $(a, b, c... | [
"Solution:\n\nSi $b$ et $c$ sont nuls alors, puisque $\\operatorname{pgcd}(a, b, c)=1$, on doit avoir $a=1$ ou $a=-1$. Dans le premier cas, il n'y a rien à faire. Dans le second, par trois opérations successives, on peut passer de $(-1,0,0)$ à $(-1,1,0)$, puis à $(1,1,0)$ et enfin à $(1,0,0)$.\n\nOn peut donc suppo... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0do9 | Problem:
На планети $X$ облика лопте се налази $2n$ бензинских пумпи. Притом је свака пумпа упарена с по једном другом пумпом и сваке две упарене пумпе се налазе на дијаметрално супротним тачкама планете. На свакој пумпи се налази одређена количина бензина. Познато је следеће: уколико аутомобил с претходно празним (до... | [
"Solution:\n\nОдговор је $n \\leqslant 3$.\nПумпу дијаметрално супротну пумпи $X$ означаваћемо са $X'$\n\nЗа $n \\leqslant 1$ тврђење је тривијално. Нека је $n=2$ и нека је $AB=A'B'$ најмање међу свим растојањима између две пумпе. Од пумпе $A$ до $A'$ се може стићи, рецимо путем $AB'A'$ (случај пута $ABA'$ је слича... | Serbia | 13. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Non-Euclidean Geometry > Spherical Geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | n ≤ 3 | |
0bme | Let $ABC$ and $ABD$ be coplanar triangles with equal perimeters. The lines of support of the internal bisectors of the angles $CAD$ and $CBD$ meet at $P$. Show that the angles $APC$ and $BPD$ are congruent. | [
"Extend the segment $AC$ beyond $C$ by a segment $CE$ congruent to the segment $CB$, so $C$ lies on the perpendicular bisector of the segment $BE$. Similarly, extend the segment $BD$ beyond $D$ by a segment $DF$ congruent to the segment $AD$, so $D$ lies on the perpendicular bisector of the segment $AF$. Since the ... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hw9 | Problem:
Let $ABC$ be a triangle with $\angle A = 120^\circ$. The bisector of $\angle A$ meets side $BC$ at $D$. Prove that
$$
\frac{1}{AD} = \frac{1}{AB} + \frac{1}{AC}
$$ | [
"Solution:\nThe area of $\\triangle ABC$ is the sum of the areas of triangles $ABD$ and $ADC$, so\n$$\n\\begin{aligned}\n\\frac{1}{2} AB \\cdot AC \\cdot \\sin 120^\\circ & = \\frac{1}{2} AB \\cdot AD \\cdot \\sin 60^\\circ + \\frac{1}{2} AD \\cdot AC \\cdot \\sin 60^\\circ \\\\\n\\frac{1}{2} AB \\cdot AC \\cdot \\... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0iga | Problem:
Write down an integer $N$ between $0$ and $20$ inclusive. If more than $N$ teams write down $N$, your score is $N$; otherwise it is $0$. | [] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | 20 | |
04dg | Find real number $a$ such that $x = \frac{1}{2}$ is a solution of the equation
$$
\left( \frac{x+1}{1-x} - \frac{1-x}{x+1} - \frac{4x^2}{x^2-1} \right) \left( \frac{1}{a^3+a^2} - \frac{1-a}{a^2} - 1 \right) = \frac{1}{2}
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | -1/9 | |
0bj4 | a) Find all pairs of positive integers $(m, n)$, with $m \le n$, for which
$$
p(2m + 1) \cdot p(2n + 1) = 400.
$$ | [
"a) Since $400 = 1 \\cdot 400 = 4 \\cdot 100 = 16 \\cdot 25$, we analyze three cases.\nIf $p(2m+1) = 1$, $p(2n+1) = 400$, we obtain $1 \\le 2m+1 < 4$ and $400 \\le 2n+1 < 441$, hence $m \\in \\{1, 2\\}$ and $n \\in \\{200, 201, \\dots, 219\\}$, giving 40 pairs $(m, n)$. Similarly, in the second case we obtain 20 pa... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All pairs (m, n) with m ≤ n are exactly those in the following three families:
- p(2m+1) = 1 and p(2n+1) = 400, which gives m = 1 and n ∈ {200, 201, ..., 219};
- p(2m+1) = 4 and p(2n+1) = 100, which gives m ∈ {2, 3} and n ∈ {50, 51, ..., 59};
- p(2m+1) = 16 and p(2n+1) = 25, which gives m ∈ {8, 9, 10, 11} and n ∈ {12, ... | |
0a60 | Problem:
Let $p$ be a prime and let $f(x) = a x^{2} + b x + c$ be a quadratic polynomial with integer coefficients such that $0 < a, b, c \leq p$. Suppose $f(x)$ is divisible by $p$ whenever $x$ is a positive integer. Find all possible values of $a + b + c$. | [
"Solution:\nFirst substitute $x = p$, to get $p \\mid a p^{2} + b p + c$ so $p \\mid c$. Therefore $c = p$.\n\nNext substitute $x = 1$, to get $p \\mid a + b + c$. Since $c = p$, this gives us $p \\mid a + b$.\n\nFinally substitute $x = p - 1$, to get $(p - 1)^{2} a - (p - 1) b + c \\equiv 0$ (mod $p$). Thus $a - b... | New Zealand | NZMO Round One | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | For odd primes: 3p. For p = 2: 4 or 6. | |
0gd9 | 令 $a$ 和 $b$ 為相異正整數。有一個無限大的棋盤, 每格可以被填上一個非負整數。棋盤一開始是空的。我們持續進行以下操作:
(i) 如果棋盤上能找到一對相同的數字, 則我們挑選相同數字的兩格, 把其中一格的數字增加 $a$, 另一格增加 $b$。
(ii) 如果棋盤上找不到一對相同的數字, 則我們在兩個空的格子裡各填上一個 0。
試證: 不論如何, 自某次操作後, 都只能再操作 (i)。 | [
"不失一般性假設 $\\gcd(a, b) = 1$, 因為我們將 $a$ 和 $b$ 同除 $\\gcd(a, b)$ 即可。同時, 不失一般性假設 $b > a$。\n\n1. 首先, 假設在經過若干次 (i) 和 $N$ 次 (ii) 後, 我們被迫要再進行一次 (ii)。對於所有整數 $k$, 令 $f_N(k)$ 為到目前為止 $k$ 在棋盤上總計出現的次數。顯然 $f_N(0) = 2N$, 且對於所有 $k < 0$ 有 $f_N(k) = 0$。\n\n注意到對於所有 $k$, 由於我們本回合被迫進行 (ii), 這表示棋盤上目前至多只有一個 $k - a$; 而如果之前回合中有出現其他 $k - a$, 表... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequ... | null | proof only | null | |
0dl3 | Let $n$ be a positive integer and $P(x)$ be a polynomial of degree $n$ with real coefficients. Prove that all the roots of the polynomial $x^3P(x) + 1$ can not be real at the same time. | [] | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
023l | Problem:
Numa Maratona de Matemática, o número de questões é muito grande. O valor de cada questão é igual à sua posição na prova: 1 ponto para a questão 1, 2 pontos para a questão 2, 3 pontos para a questão 3, 4 pontos para a questão 4, \ldots, 10 pontos para a questão 10, \ldots\ e assim por diante. Joana totalizou ... | [
"Solution:\n\n25 e 63, respectivamente."
] | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | She missed question 25; there were 63 questions. | |
0cyx | Prove that for each $n \geq 4$ a parallelogram can be dissected in $n$ cyclic quadrilaterals. | [
"Let $ABCD$ be a parallelogram. If $ABCD$ is a rectangle, then it is clear that we can dissect it into $n$ rectangles by parallel lines to its sides.\n\nAssume that $ABCD$ is not a rectangle. Without loss of generality we can assume that $AB \\geq AD$. Let $E$ and $F$ be the midpoints of $BC$ and $AD$, respectively... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0bk9 | Let $k$ be a positive integer and let $m$ be a positive odd integer. Show that there exists a positive integer $n$ such that $m^n + n^m$ has at least $k$ distinct prime factors. | [
"Design a set of $k$ primes $p_1 < p_2 < \\cdots < p_k$ as follows. Begin by choosing $p_1 > 2m$. Having selected $p_j$, use Dirichlet's theorem to choose a prime\n$$\np_{j+1} \\equiv -1 \\pmod{p_1(p_1-1)p_2(p_2-1)\\cdots p_j(p_j-1)}.\n$$\nIf $i < j$, then $p_i < p_j$, so $p_j$ does not divide $p_i - 1$; further, $... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | null | proof only | null | |
00zb | Problem:
For which positive real numbers $a, b$ does the inequality
$$
x_{1} \cdot x_{2}+x_{2} \cdot x_{3}+\cdots+x_{n-1} \cdot x_{n}+x_{n} \cdot x_{1} \geq x_{1}^{a} \cdot x_{2}^{b} \cdot x_{3}^{a}+x_{2}^{a} \cdot x_{3}^{b} \cdot x_{4}^{a}+\cdots+x_{n}^{a} \cdot x_{1}^{b} \cdot x_{2}^{a}
$$
hold for all integers $n>2$... | [
"Solution:\nSubstituting $x_{i}=x$ easily yields that $2 a+b=2$. Now take $n=4$, $x_{1}=x_{3}=x$ and $x_{2}=x_{4}=1$. This gives $2 x \\geq x^{2 a}+x^{b}$. But the inequality between the arithmetic and geometric mean yields $x^{2 a}+x^{b} \\geq 2 \\sqrt{x^{2 a} x^{b}}=2 x$. Here equality must hold, and this implies... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | a = 1/2, b = 1 | |
0l08 | A model used to estimate the time it will take to hike to the top of a mountain on a trail is of the form $T = aL + bG$, where $a$ and $b$ are constants, $T$ is the time in minutes, $L$ is the length of the trail in miles, and $G$ is the altitude gain in feet. The model estimates that it will take 69 minutes to hike to... | [
"The given data from the first two hikes yield the system of equations\n$$\n\\begin{aligned}\n1.5a + 800b &= 69 \\\\\n1.2a + 1100b &= 69.\n\\end{aligned}\n$$\nTo solve this system, first subtract the second equation from the first equation to get $0.3a - 300b = 0$, which implies $a = 1000b$. Then the first equation... | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0ikw | Problem:
How many ways are there to label the faces of a regular octahedron with the integers $1$–$8$, using each exactly once, so that any two faces that share an edge have numbers that are relatively prime? Physically realizable rotations are considered indistinguishable, but physically unrealizable reflections are ... | [
"Solution:\n\nWell, instead of labeling the faces of a regular octahedron, we may label the vertices of a cube. Then, as no two even numbers may be adjacent, the even numbers better form a regular tetrahedron, which can be done in $2$ ways (because rotations are indistinguishable but reflections are different). The... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 12 | |
0ipc | Problem:
On the circumcircle of $A B C$, let $A'$ be the midpoint of arc $B C$ (not containing $A$).
a. Show that $A, I, A'$ are collinear.
b. Show that $A'$ is the circumcenter of $BIC$. | [
"Solution:\n\n\n\na. Since $A'$ bisects the arc $B C$, the two arcs $A' B$ and $A' C$ are equal, and so $\\angle B A A' = \\angle C A A'$. Thus, $A'$ lies on the angle bisector of $B A C$. Since $I$ also lies on the angle bisector of $B A C$, we see that $A, I, A'$ are collinear.\n\nb. We h... | United States | Harvard-MIT Mathematics Tournament | [
"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 | |
0fmh | Dado un número entero $n$ escrito en el sistema de numeración decimal, formamos el número entero $k$ restando del número formado por las tres últimas cifras de $n$ el número formado por las cifras anteriores restantes. Demostrar que $n$ es divisible por 7, 11 o 13 si y sólo si $k$ también lo es. | [
"Sea $A$ el número formado por las tres últimas cifras de $n$ y $B$ el número formado por las cifras anteriores. Entonces $n = 1000B + A$ y $k = A - B$. Tenemos $n - k = 1001B = 7 \\cdot 11 \\cdot 13B$ y $n$ y $k$ son congruentes módulo 7, 11 y 13."
] | Spain | Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof only | null | |
07km | For positive real numbers $a$, $b$, $c$ and $d$ such that $a^2 + b^2 + c^2 + d^2 = 1$ prove that
$$
a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2 \le \frac{3}{32},
$$
and determine the cases of equality. | [
"We have\n$$\n(a)\\ (6, 0, 0, 0) > (2, 2, 1, 1)\n$$\n$$\n(b)\\ (2, 2, 2, 0) > (2, 2, 1, 1)\n$$\n$$\n(c)\\ (4, 2, 0, 0) > (2, 2, 1, 1)\n$$\nBy Muirhead and the above majorizations we have the following\n$$\n1.\\ 6(a^6 + b^6 + c^6 + d^6) \\geq 4(a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a ... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | Maximum value is 3/32, achieved when a = b = c = d = 1/2. | |
0i0e | Problem:
Find all the roots of $\left(x^{2}+3 x+2\right)\left(x^{2}-7 x+12\right)\left(x^{2}-2 x-1\right)+24=0$. | [
"Solution:\nWe re-factor as $(x+1)(x-3)(x+2)(x-4)\\left(x^{2}-2 x-1\\right)+24$, or $\\left(x^{2}-2 x-3\\right)\\left(x^{2}-2 x-8\\right)\\left(x^{2}-2 x-1\\right)+24$, and this becomes $(y-4)(y-9)(y-2)+24$ where $y=(x-1)^{2}$.\n\nNow, $(y-4)(y-9)(y-2)+24=(y-8)(y-6)(y-1)$, so $y$ is $1, 6$, or $8$.\n\nThus the root... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 0, 2, 1 - sqrt(6), 1 + sqrt(6), 1 - 2*sqrt(2), 1 + 2*sqrt(2) | |
0ioc | Problem:
A student at Harvard named Kevin
Was counting his stones by $11$
He messed up $n$ times
And instead counted $9$s
And wound up at $2007$.
How many values of $n$ could make this limerick true? | [
"Solution:\nAnswer: $21$. The mathematical content is that $9n + 11k = 2007$, for some nonnegative integers $n$ and $k$. As $2007 = 9 \\cdot 223$, $k$ must be divisible by $9$. Using modulo $11$, we see that $n$ is $3$ more than a multiple of $11$. Thus, the possibilities are $n = 223, 212, 201, \\ldots, 3$, which ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | final answer only | 21 | |
02si | Problem:
A figura abaixo contém um quadrado e dois triângulos retângulos congruentes.
Com esses polígonos formamos um retângulo e um trapézio como mostra a figura seguinte:
Sabendo que o perímetro do retângulo é 58, e que o perímetro do trapézio é 60, calcule o lado d... | [
"Solution:\nFica claro pela figura no enunciado desse exercício que o comprimento do lado do quadrado deve coincidir com o comprimento de algum dos catetos dos triângulos. Chamemos de $a$ tal comprimento. Além disso, chamemos de $b$ o comprimento da hipotenusa e de $x$ o comprimento do outro cateto. Pela figura aba... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 12 | |
07wl | A game of coins is played as follows: You start with $1$ head and $1$ tail on a table. At each turn, you can perform any one of the following moves:
a. You can turn over all the coins on the table.
b. You can triple the numbers of heads and tails on the table.
c. If there are at least $4$ tails on the table, you can... | [
"Let $x$ be the number of heads and $y$ the number of tails at each point of the game. The pair $(x, y)$ completely describes the state of the game. Let $D(x, y) = x - y$. This quantity behaves as follows under each possible move:\n\na. $(x, y) \\to (y, x)$ changes $D(x, y)$ to $-D(x, y)$.\n\nb. $(x, y) \\to (3x, 3... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | All nonnegative multiples of 8 | |
0kvb | Problem:
A subset $S$ of the set $\{1,2, \ldots, 10\}$ is chosen randomly, with all possible subsets being equally likely. Compute the expected number of positive integers which divide the product of the elements of $S$. (By convention, the product of the elements of the empty set is $1$.) | [
"Solution:\n\nFor primes $p=2,3,5,7$, let the random variable $X_{p}$ denote the number of factors of $p$ in the product of the elements of $S$, plus $1$. Then we wish to find $\\mathbb{E}\\left(X_{2} X_{3} X_{5} X_{7}\\right)$.\n\nIf there were only prime powers between $1$ and $10$, then all $X_{p}$ would be inde... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 375/8 | |
03ly | Problem:
Given an $m \times n$ grid with squares coloured either black or white, we say that a black square in the grid is stranded if there is some square to its left in the same row that is white and there is some square above it in the same column that is white (see Figure).
A $4 \times 5$... | [
"Solution:\n\nThere is no condition for squares in the first row. A square in the second row can be black only if the square above it is black or all squares to the left of it are black. Suppose the first $k$ squares in the second row are black and the $(k+1)$-st square is white or $k=n$. When $k<n$ then for each o... | Canada | CANADIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 2*3^n - 2^n | |
0f40 | Problem:
Eighteen teams are playing in a tournament. So far, each team has played exactly eight games, each with a different opponent. Show that there are three teams none of which has yet played the other. | [] | Soviet Union | 15th ASU | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0dt9 | Determine all real numbers $x$ between $0$ and $180$ such that it is possible to partition an equilateral triangle into finitely many triangles, each of which has an angle of $x^\circ$. | [
"Consider the sum of the angles of the triangles at each type of the vertices. At type 0, the sum is $180^\\circ$. At type 1, the sum is $180t_1$. At type 2, the sum is $360t_2$. Therefore\n$$\n180 + 180t_1 + 360t_2 = 180n \\Rightarrow 1 + t_1 + 2t_2 = n.\n$$\nNow each Type 1 vertex has at most 1 angle of $x^\\circ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 0 < x ≤ 120 | |
0hb6 | Given two distinct natural numbers $a$ and $b$ greater than $1$,
a) Prove that there are infinitely many natural numbers $n$ such that $s_n = a^n + b^{n+1}$ is composite.
b) Prove that there are infinitely many prime numbers $p$ such that $s_n$ is divisible by $p$ for some $n$. | [
"a) If $s_n = p$ is prime for some $n$ larger than $a$ and $b$, then for some $k$ and $l$ the numbers $a^k - 1$ and $b^l - 1$ are divisible by $p$. If we set $m = kl$, then both $a^{mt} - 1$ and $b^{mt} - 1$ are divisible by $p$ for all natural numbers $t$. This implies that the number\n$$\ns_{n+mt} = a^{n+mt} + b^... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0ges | 魔術師準備表演一個魔術。她會先指定一個正整數 $n$，以及 $2n$ 個實數 $x_1 < \cdots < x_{2n}$。然後隨機地請一位觀眾秘密地寫下一個 $n$ 次實係數多項式 $P(x)$ 並計算 $P(x_1), \dots, P(x_{2n})$ 的值，再以“非遞減”的順序在黑板上寫下這 $2n$ 個值。最後，她必須從這 $2n$ 個值公布正確的 $P(x)$。
確定所有的 $n$ 使得魔術師可以找到策略來執行這樣的魔術。
A magician intends to perform the following trick. She announces a positive integer $n$, along wit... | [
"There doesn't exist such a $n$.\nLet $x_1 < x_2 < \\cdots < x_{2n}$ be real numbers chosen by the magician. We will construct two distinct polynomials $P(x)$ and $Q(x)$, each of degree $n$, such that the member of audience will write down the same sequence for both polynomials. This will mean that the magician can... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Linear Algebra > Matrices"
] | null | proof and answer | no positive integer n | |
0hvn | Problem:
Circles $j$ and $k$, centered at $O$ and $P$ respectively, do not intersect. The two tangent rays from $O$ to $k$ meet $j$ at $A$ and $B$, respectively, and the two tangent rays from $P$ to $j$ meet $k$ at $C$ and $D$, respectively. Prove that $A, B, C$, and $D$ are the vertices of a rectangle.
![](attached_... | [
"Solution:\n\nWithout loss of generality, we may assume that $A, B, C$, and $D$ have the relative positions shown. We label the points of tangency $W, X, Y$, and $Z$. We also note that the entire construction is symmetric about the line of centers $OP$, which therefore perpendicularly bisects segments $AB$, $CD$, $... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09lh | Consider the points $A(-1, 0)$ and $B(1, 0)$ on the coordinate plane. Find the maximum possible value of the angle $\angle ACB$ where $C$ moves along the hyperbola with equation $2xy = 1$. Note that angles are measured in degrees and lie in the interval $[0^\circ, 180^\circ]$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 90° | |
0auv | Problem:
A line intersects the $y$-axis, the line $y=2x+2$, and the $x$-axis at the points $A$, $B$, and $C$, respectively. If segment $AC$ has a length of $4\sqrt{2}$ units and $B$ lies in the first quadrant and is the midpoint of segment $AC$, find the equation of the line in slope-intercept form. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | y = -7x + 28/5 | |
0j48 | Problem:
Tom, Dick, and Harry play a game in which they each pick an integer between $1$ and $2011$. Tom picks a number first and informs Dick and Harry of his choice. Then Dick picks a different number and informs Harry of his choice. Finally, Harry picks a number different from both Tom's and Dick's. After all the p... | [
"Solution:\n\nAnswer: $503$, $1509$\n\nLet $x$ denote the number Tom chooses. By the symmetry of the problem, picking $x$ and picking $2012-x$ yield the same expected profit. If Tom picks $1006$, Dick sees that if he picks $1007$, Harry's best play is to pick $1005$, and Dick will win with probability $\\frac{1005}... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 503 and 1509 | |
087h | Problem:
Trovare tutte le terne ordinate di numeri interi positivi $(p, q, n)$ tali che $p, q$ siano primi e $p^{2}+q^{2}=p q n+1$. | [
"Solution:\nSupponiamo $p=q$. Sostituendo otteniamo $p^{2}(2-n)=1$ che è impossibile perché 1 non è diviso da nessun primo. Quindi necessariamente $p$ e $q$ sono diversi; poiché l'equazione è simmetrica in $p$ e $q$ possiamo supporre che $q>p$, cioè $q \\geq p+1$. Scriviamo ora la nostra equazione come:\n$$\np^{2}-... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2,3,2) and (3,2,2) | |
0cjm | Let $A = \{1, 2, 3, \dots, 2025\}$. A subset $B$ of the set $A$ will be called *nice* if it has 3 elements, one of them being the arithmetic mean of the other two, and there exists $b \in B$ such that $5 \cdot b \in B$.
a) Find how many nice sets have the element $225$.
b) Find how many nice subsets has $A$. | [
"Let $B = \\{b, 5b, a\\}$. The possible cases are:\n\nI. $a = \\frac{b+5b}{2} \\Rightarrow B = \\{b, 3b, 5b\\}$;\n\nII. $5b = \\frac{a+b}{2} \\Rightarrow B = \\{b, 5b, 9b\\}$;\n\nIII. $b = \\frac{a+5b}{2} \\Rightarrow a+3b=0$ - impossible.\n\n\na) Since $225$ is divisible by $3$, $5$ and $9$, it can be any element ... | Romania | 75th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | a) 6; b) 630 | |
000n | 1. En un torneo de fútbol entre cuatro equipos, $A$, $B$, $C$ y $D$, cada equipo juega con cada uno de los otros una sola vez.
a) Decidir si es posible que, al finalizar el torneo, las cantidades de goles anotados y recibidos por los equipos sean:
| | A | B | C | D |
|-------------------|---|---|---... | [] | Argentina | XIV Olimpiada Matemática de Países del Cono Sur | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | español | proof and answer | a) Yes. One possible set of match results is: AB 0–0, AC 0–2, AD 1–2, BC 0–3, BD 3–1, CD 1–4. b) No. Impossible because the team D is claimed to have conceded eleven goals, but the other three teams together scored only ten in total. | |
0en2 | Prove that the following equation holds for all non-negative integers $n$:
$$
\lfloor \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} \rfloor = \lfloor \sqrt{9n} + 8 \rfloor.
$$ | [
"For each non-negative integer $n$,\n$$\n2\\sqrt{n+1} > \\sqrt{n} + \\sqrt{n+2}.\n$$\nIndeed,\n$$\n\\begin{aligned}\n\\sqrt{n+1} - \\sqrt{n} &= \\frac{1}{\\sqrt{n+1} + \\sqrt{n}} \\\\\n&> \\frac{1}{\\sqrt{n+2} + \\sqrt{n+1}} \\\\\n&= \\sqrt{n+2} - \\sqrt{n+1},\n\\end{aligned}\n$$\nfrom which the result follows.\n\n... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0fvd | Problem:
Seien $a$, $b$, $c$ drei ganze Zahlen, sodass $a+b+c$ durch $13$ teilbar ist. Zeige, dass auch
$$
a^{2007} + b^{2007} + c^{2007} + 2 \cdot 2007 a b c
$$
durch $13$ teilbar ist. | [
"Solution:\n\nIst $x$ nicht durch $13$ teilbar, dann gilt $x^{12} \\equiv 1 \\pmod{13}$ nach dem kleinen Satz von Fermat. Daraus folgt unmittelbar, dass für alle ganzen $x$ gilt $x^{13} \\equiv x \\pmod{13}$. Wiederholte Anwendung dieser Gleichung liefert $x^{2007} \\equiv x^{3} \\pmod{13}$ wegen $2007 = 167 \\cdot... | Switzerland | SMO Finalrunde | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null |
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