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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0hd6 | Determine which of the following numbers is greater: $2$ or $\operatorname{tg} 1$? | [
"Consider the following inequalities: $2 > \\sqrt{3} = \\operatorname{tg} \\frac{\\pi}{3} > \\operatorname{tg} 1$."
] | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | 2 > \tan(1) | |
0ddy | Consider the sequence $a_1, a_2, a_3, \dots$ defined by $a_1 = 9$ and
$$
a_{n+1} = \frac{(n+5)a_n + 22}{n+3}
$$
for $n \ge 1$. Find all positive integers $n$ for which $a_n$ is a perfect square. | [
"First of all, we will prove by induction on $n$ that $a_n = n^2 + 7n + 1$. The base case is trivial. Assume that $a_k = k^2 + 7k + 1$, so\n$$\na_{k+1} = \\frac{(k + 5)a_k + 22}{k + 3} = \\frac{(k + 5)(k^2 + 7k + 1) + 22}{k + 3} = (k + 1)^2 + 7(k + 1) + 1\n$$\nWe distinguish two cases regarding the value of $n$\n* ... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n = 1, 8 | |
09v7 | In a tall hat there are one hundred notes, numbered from $1$ to $100$. You want to have three notes with the property that each of the three numbers is smaller than the sum of the other two. For example, the three notes numbered $10$, $15$, and $20$ would be suitable (as $10 < 15 + 20$, $15 < 10 + 20$, and $20 < 10 + 1... | [
"11"
] | Netherlands | Second Round, March 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 11 | |
0iua | Problem:
Let $a$, $b$, and $c$ be real numbers. Consider the system of simultaneous equations in variables $x$ and $y$:
$$
\begin{aligned}
a x + b y &= c - 1 \\
(a + 5) x + (b + 3) y &= c + 1
\end{aligned}
$$
Determine the value(s) of $c$ in terms of $a$ such that the system always has a solution for any $a$ and $b$. | [
"Solution:\nWe have to only consider when the determinant of $\\begin{pmatrix} a & b \\\\ a+5 & b+3 \\end{pmatrix}$ is zero. That is, when $b = \\dfrac{3a}{5}$. Plugging in $b = \\dfrac{3a}{5}$, we find that $(a+5)(c-1) = a(c+1)$ or that $c = \\dfrac{2a}{5} + 1$."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | c = 1 + (2a)/5 | |
05hp | Problem:
Soit $ABC$ un triangle. On désigne par $D$ le pied de la bissectrice de $\widehat{BAC}$, et par $E$ le pied de la hauteur issue de $A$. La médiatrice de $[AD]$ rencontre les demi-cercles de diamètres respectifs $[AB]$ et $[AC]$ construits extérieurement à $ABC$, en $X$ et $Y$.
Prouver que les points $X, Y, D... | [
"Solution:\n\n\nOn désigne par $M, N$ et $P$ les milieux respectifs de $[AB],[AC]$ et $[AD]$. D'après le théorème des milieux, on a $N, P, M$ alignés et, compte-tenu de ce que $(AP)=(AD)$ est la bissectrice de $\\widehat{BAC}$, on a\n$$\n\\frac{MP}{NP}=\\frac{BD}{CD}=\\frac{AM}{AN}=\\frac{A... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > ... | null | proof only | null | |
04zb | Does there exist a prime number $p$ such that both $p^3 + 2008$ and $p^3 + 2010$ are primes as well? | [
"Let $p$ be any prime number. If $p$ is not divisible by $7$, then $p^3$ is congruent to either $1$ or $-1$ modulo $7$. Since $2008 \\equiv -1 \\pmod{7}$ and $2010 \\equiv 1 \\pmod{7}$, either of the numbers $p^3 + 2008$ and $p^3 + 2010$ is divisible by $7$ and hence composite. If $p$ is divisible by $7$, then $p =... | Estonia | Estonija 2010 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No | |
0l33 | Problem:
Let $a$, $b$, and $c$ be real numbers such that
$$
\begin{aligned}
a + b + c &= 100 \\
ab + bc + ca &= 20, \text{ and } \\
(a + b)(a + c) &= 24
\end{aligned}
$$
Compute all possible values of $bc$. | [
"Solution:\nWe first expand the left-hand side of the third equation to get $(a + b)(a + c) = a^2 + ac + ab + bc = 24$. From this, we subtract the second equation to obtain $a^2 = 4$, so $a = \\pm 2$.\n\nIf $a = 2$, plugging into the first equation gives us $b + c = 98$ and plugging into the second equation gives u... | United States | HMMT February 2024 Guts Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 224, -176 | |
0cim | Consider the acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$. Outside the triangle, construct the segments $CF \perp BC$ with $CF = \frac{AC}{2}$ and $CG \perp AC$ with $CG = \frac{BC}{2}$. Let $X$ be the point of intersection $OC \cap FG$. Prove that:
$$
\overrightarrow{OX} = \frac{\overrightarrow{OH} -... | [] | Romania | 75th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0itv | Problem:
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$. | [
"Solution:\nWe can factor the equation as $(a-b^{2})(a+b^{2})=41 \\cdot 49$, from which it is evident that $a=45$ and $b=2$ is a possible solution. By examining the factors of $2009$, one can see that there are no other solutions."
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 47 | |
0259 | Problem:
O número $119$ é muito interessante porque deixa resto $1$ ao ser dividido por $2$, deixa resto $2$ ao ser dividido por $3$, deixa resto $3$ ao ser dividido por $4$, deixa resto $4$ ao ser dividido por $5$ e, finalmente, deixa resto $5$ ao ser dividido por $6$. Existem outros números de três algarismos com es... | [
"Solution:\n\nSuponhamos que $N$ seja um dos números procurados. Como $N$ e $119$ deixam os mesmos restos quando divididos por $2$, $3$, $4$, $5$ e $6$, temos que a diferença $N-119$ entre eles deixa resto zero quando dividido por esses números. Portanto, $N-119$ é um múltiplo de $2,3,4,5$ e $6$. Como $60$ é o míni... | Brazil | Nível 2 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 179, 239, 299, 359, 419, 479, 539, 599, 659, 719, 779, 839, 899, 959 | |
00qb | Let $M = \{1, 2, \dots, 2013\}$ and let $\Gamma$ be a circle. For every nonempty subset $\mathcal{A}$ of the set $M$, denote by $S(\mathcal{A})$ the sum of elements of the set $\mathcal{A}$, and define $S(\emptyset) = 0$ ($\emptyset$ is the empty set). Is it possible to join every subset $\mathcal{A}$ of $M$ with some ... | [
"We will prove that this is possible. Total number of subsets of the set $M$ is $2^{2013}$. On circle $\\Gamma$ we arbitrarily choose $2^{2013}$ points which are vertices of a regular $2^{2013}$-gon. We join subsets of the set $M$ and chosen points in the following manner: if we join subset $\\mathcal{A}$ with some... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Geometry > Plane Geometry > Transformations > Rotation",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0d54 | Let be given the sequence $\left(x_{n}\right)$ defined by $x_{1}=1$ and $x_{n+1}=3 x_{n}+\left\lfloor x_{n} \sqrt{5}\right\rfloor$ for all $n=1,2,3, \ldots$, where $\lfloor x\rfloor$ denotes the greatest integer that does not exceed $x$. Prove that for any positive integer $n$ we have
$$
x_{n} x_{n+2}-x_{n+1}^{2}=4^{n-... | [
"It is easy to prove by induction that $x_{n}$ is an integer for all $n \\geq 1$. Therefore, we have $x_{n+1}=\\left\\lfloor(3+\\sqrt{5}) x_{n}\\right\\rfloor$ for all $n \\geq 1$, and we deduce that\n$$\n\\begin{aligned}\nx_{n+2} & =\\left\\lfloor(3+\\sqrt{5}) x_{n+1}\\right\\rfloor=6 x_{n+1}+\\left\\lfloor-(3-\\s... | Saudi Arabia | SAMC 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English, Arabic | proof only | null | |
0j5c | Problem:
On Facebook, there is a group of people that satisfies the following two properties:
(i) there exists a positive integer $k$ such that any subset of $2k-1$ people in the group contains a subset of $k$ people in the group who are all friends with each other, and
(ii) every member of the group has 2011 friends ... | [
"Solution:\n\na.\nAnswer: 4024\n\nIf $k=2$, then among any three people at least two of them are friends. Clearly if we have 4024 people divided into two sets of 2012 such that everyone is friends with everyone in their set but no one in the other set, then any triple of three people will contain two people from th... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4024 for both (a) and (b) | |
03a7 | Consider the functions $f(x) = |x-1| - |x-2|$ and $g(x) = |x-3|$.
a) Draw the graph of the function $f(x)$.
b) Find the area of the figure determined by the graphs of $f(x)$ and $g(x)$. | [] | Bulgaria | Fall Mathematical Competition | [
"Precalculus > Functions",
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | English | proof and answer | 1 | |
05mw | Problem:
Trouver toutes les fonctions $f: \mathbb{R} \rightarrow \mathbb{R}$ telles que $f(f(x)+3y)=12x+f(f(y)-x)$ pour tous $x, y \in \mathbb{R}$. | [
"Solution:\n\nSoit $f$ une fonction satisfaisant à la condition de l'énoncé. L'idée est de montrer que $f$ est injective en utilisant le fait qu'elle est surjective.\n\nMontrons d'abord que $f$ est surjective. En prenant $y=-f(x)/3$ dans l'équation fonctionnelle, on obtient $f(f(y)-x)=f(0)-12x$. En faisant varier $... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES - ENVOI No. 3 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | All functions of the form f(x) = a + 3x for any real constant a. | |
011q | Problem:
In a triangle $A B C$, the bisector of $\angle B A C$ meets the side $B C$ at the point $D$. Knowing that $|B D| \cdot |C D| = |A D|^{2}$ and $\angle A D B = 45^{\circ}$, determine the angles of triangle $A B C$. | [
"Solution:\n\n$\\angle B A C = 60^{\\circ}$, $\\angle A B C = 105^{\\circ}$ and $\\angle A C B = 15^{\\circ}$.\n\nSuppose the line $A D$ meets the circumcircle of triangle $A B C$ at $A$ and $E$ (see Figure 5). Let $M$ be the midpoint of $B C$ and $O$ the circumcentre of triangle $A B C$. Since the arcs $B E$ and $... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Dist... | null | proof and answer | ∠A = 60°, ∠B = 105°, ∠C = 15° | |
0fl8 | Problem:
Sean $a, b, c$ tres números reales positivos. Demuestra que
$$
\frac{a+b+3c}{3a+3b+2c} + \frac{a+3b+c}{3a+2b+3c} + \frac{3a+b+c}{2a+3b+3c} \geq \frac{15}{8}
$$ | [
"Solution:\n\nHaciendo $a = x_{1}$, $b = x_{2}$, $c = x_{3}$ y llamando $s = x_{1} + x_{2} + x_{3}$, resulta que el lado izquierdo de la desigualdad se escribe como\n$$\nS = \\frac{s + 2x_{1}}{3s - x_{1}} + \\frac{s + 2x_{2}}{3s - x_{2}} + \\frac{s + 2x_{3}}{3s - x_{3}}\n$$\nPor otro lado,\n$$\nS + 6 = \\sum_{k=1}^... | Spain | XLVI Olimpiada Matemática Española Fase nacional | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
02u2 | Problem:
Um jogo comum de dominó é composto por 28 peças. Cada peça é formada por dois números inteiros que variam de $0$ a $6$, inclusive. Todas as possibilidades de combinações possíveis $(a, b)$, com $a \leq b$, são listadas exatamente uma vez. Note que a peça $(4,2)$ é listada como a peça $(2,4)$, pois $2 \leq 4$.... | [
"Solution:\n\na) Basta começar contando pelos maiores denominadores e não repetir quando aparecerem os menores.\n\ni) Para $b=6$, temos\n$$\n\\left(\\frac{0}{6}, \\frac{1}{6}, \\frac{2}{6}, \\frac{3}{6}, \\frac{4}{6}, \\frac{5}{6}, \\frac{6}{6}\\right)=\\left(0, \\frac{1}{6}, \\frac{1}{3}, \\frac{1}{2}, \\frac{2}{3... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | a) 13; b) 13/2 | |
08i4 | Problem:
In the square $ABCD$ the point $N$ is the middle point of the side $[AB]$ and the point $M$ lies on the diagonal $(AC)$ so that $AC = 4CM$. Prove that the angle $DMN$ is right. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0cyu | Pentagon $A B C D E$ is inscribed in a circle. Distances from point $E$ to lines $A B$, $B C$ and $C D$ are equal to $a$, $b$ and $c$, respectively. Find the distance from point $E$ to line $A D$. | [
"Let $K$, $L$, $M$ and $N$ be the feet of perpendiculars dropped from point $E$ to lines $A B$, $B C$, $C D$ and $D A$, respectively.\n\n\n\nPoints $K$ and $N$ lie on the circle with diameter $A E$, hence $\\widehat{E K N} \\equiv \\widehat{E A N}$. Similarly, $\\widehat{E L M} \\equiv \\widehat{E C M} \\equiv \\wi... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | ac/b | |
04c5 | For a given prime number $p$ determine all integers $n$ such that $\sqrt{n^2 + pn}$ is an integer. | [
"Let $\\sqrt{n^2 + pn} = m$, $m \\in \\mathbb{N}_0$. Then $n^2 + pn - m^2 = 0$.\nThe solutions of this quadratic equation are\n$$\nn = \\frac{1}{2} \\left( -p \\pm \\sqrt{p^2 + 4m^2} \\right).\n$$\nThe solutions are integers, so $p^2 + 4m^2 = w^2$, for some $w \\in \\mathbb{N}_0$.\nSince $w + 2m \\ge 0$ and $w + 2m... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | For any prime p, the solutions are n = 0 and n = -p. If p is odd (p > 2), there are two additional solutions: n = ((p−1)/2)^2 and n = −((p+1)/2)^2. | |
05tn | Problem:
Un mauvais sorcier a enfermé $n$ mathématiciens. Il dispose de $n$ couleurs. Le sorcier place sur la tête de chaque mathématicien un chapeau d'une des $n$ couleurs; deux chapeaux peuvent avoir la même couleur. Chaque mathématicien peut voir la couleur du chapeau de chacun de ses collègues mais pas la sienne. ... | [
"Solution:\n\nTout d'abord, nous allons coder les différentes informations. Par exemple, on numérote les couleurs de $1$ à $n$ et on numérote les mathématiciens de $1$ à $n$ également. On note $i_{1}, \\ldots, i_{n}$ les couleurs des chapeaux des mathématiciens.\n\nPuisqu'un mathématicien ne peut pas voir son propr... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0dbm | Let $d$ be a positive divisor of the number $A = 1024^{1024} + 5$ and suppose that $d$ can be expressed as $d = 2x^{2} + 2xy + 3y^{2}$ for some integers $x, y$. Which remainder can we have when dividing $d$ by $20$? | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 3 or 7 | |
0j1d | Problem:
In a group of people, there are $13$ who like apples, $9$ who like blueberries, $15$ who like cantaloupe, and $6$ who like dates. (A person can like more than $1$ kind of fruit.) Each person who likes blueberries also likes exactly one of apples and cantaloupe. Each person who likes cantaloupe also likes exac... | [
"Solution:\n\nAnswer: $22$\n\nEveryone who likes cantaloupe likes exactly one of blueberries and dates. However, there are $15$ people who like cantaloupe, $9$ who like blueberries, and $6$ who like dates. Thus, everyone who likes blueberries or dates must also like cantaloupes (because if any of them didn't, we wo... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 22 | |
0fk8 | Problem:
Sea $P$ un punto interior a un triángulo $ABC$. Por $P$ se trazan paralelas $KP$, $MP$ y $NP$ a los lados $AB$, $AC$ y $BC$ que dividen el triángulo inicial en tres triángulos y tres paralelogramos. Sean $S_{1}$, $S_{2}$, $S_{3}$ las áreas de los nuevos triángulos y $S$ el área del triángulo $ABC$. Probar que... | [
"Solution:\n\nSea $L$ el punto de intersección de $KP$ con el lado $BC$ y sean $h_{i}$, $i=1,2,3$ las alturas de los nuevos triángulos y $h$ la altura de $\\triangle ABC$. Dado que cada uno de los triángulos son semejantes con $ABC$, se tiene\n$$\n\\begin{gathered}\n\\frac{S_{1}}{S}=\\frac{KP \\cdot h_{1}}{AB \\cdo... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
045z | Find all positive integers $n$ with the following property: there exist nonempty finite sets of integers $A, B$, such that for every integer $m$, exactly one of the following three statements is true,
(i) there exists $a \in A$, such that $m \equiv a \pmod{n}$;
(ii) there exists $b \in B$, such that $m \equiv b \pmod{n... | [
"*Proof.* Let $A+B = \\{a+b \\mid a \\in A, b \\in B\\}$. The problem can be understood as the remainders of $A$ modulo $n$, the remainders of $B$ modulo $n$, and the remainders of $(A+B)$ modulo $n$ form a partition of all remainders modulo $n$.\n\n(1) If $n > 1$ is an odd number, let $n = 2k+1$, $k \\in \\mathbb{... | China | 2022 CGMO | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All positive integers except 1, 2, and 4 | |
0auz | Problem:
Find the last three digits of $2016^{3} + 2017^{3} + 2018^{3} + \ldots + 3014^{3}$. | [
"Solution:\nNote that $\\{2016, 2017, 2018, \\ldots, 3014, 3015\\}$ comprise 1000 consecutive integers, and thus produce all possible residues modulo 1000. Thus,\n$$\n\\begin{aligned}\n2016^{3} + 2017^{3} + 2018^{3} + \\ldots + 3014^{3} + 3015^{3} &\\equiv 0^{3} + 1^{3} + 2^{3} + \\ldots + 999^{3} \\pmod{1000} \\\\... | Philippines | 18th PMO National Stage Oral Phase | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 625 | |
0j4d | Problem:
Determine the number of integers $D$ such that whenever $a$ and $b$ are both real numbers with $-1/4 < a, b < 1/4$, then $|a^{2} - D b^{2}| < 1$. | [
"Solution:\n\nWe have\n$$\n-1 < a^{2} - D b^{2} < 1 \\Rightarrow \\frac{a^{2} - 1}{b^{2}} < D < \\frac{a^{2} + 1}{b^{2}}\n$$\nWe have $\\frac{a^{2} - 1}{b^{2}}$ is maximal at $-15 = \\frac{0.25^{2} - 1}{0.25^{2}}$ and $\\frac{a^{2} + 1}{b^{2}}$ is minimal at $\\frac{0^{2} + 1}{0.25^{2}} = 16$. However, since we can... | United States | Harvard-MIT November Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 32 | |
0bi7 | a) $\left(\frac{1}{2}\right)^3 + \left(\frac{2}{3}\right)^3 + \left(\frac{5}{6}\right)^3 = 1$;
b) $3^{33} + 4^{33} + 5^{33} < 6^{33}$. | [
"a) A straightforward computation proves the claim.\n\nb) It suffices to show that\n$$\n\\frac{3^{33}}{6^{33}} + \\frac{4^{33}}{6^{33}} + \\frac{5^{33}}{6^{33}} < 1,\n$$\nthat is,\n$$\n\\left(\\frac{1}{2}\\right)^{33} + \\left(\\frac{2}{3}\\right)^{33} + \\left(\\frac{5}{6}\\right)^{33} < 1.\n$$\nBut\n$$\n\\left(\\... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0bfk | For each complex $z$ define $A_z = \{1+z+z^2+\dots+z^n \mid n \in \mathbb{N}\}$.
a) Find all $z$ for which $A_z$ is finite.
b) How many complex numbers $z$ have the property that $A_z$ has 2013 elements? | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | a) Exactly those z with z = 0 or z a root of unity different from 1 (the set has size equal to the order of z). b) 1200. | |
07rz | Mary and Pat play the following number game. Mary picks an initial integer greater than $2017$. She then multiplies this number by $2017$ and adds $2$ to the result. Pat will add $2019$ to this new number and it will again be Mary's turn. Both players will continue to take alternating turns. Mary will always multiply t... | [
"Let $m$ be the initial integer Mary has picked. Because\n$$\n2017x + 2 \\equiv 2 - x \\pmod{2018} \\text{ and}\n$$\n$$\nx + 2019 \\equiv x + 1 \\pmod{2018},\n$$\nthe numbers produced (mod $2018$) are\n$$\nm \\xrightarrow{M} 2 - m \\xrightarrow{P} 3 - m \\xrightarrow{M} 2 - (3 - m) = m - 1 \\xrightarrow{P} m\n$$\nw... | Ireland | Irish | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2022 | |
0j7u | Problem:
Rachelle picks a positive integer $a$ and writes it next to itself to obtain a new positive integer $b$. For instance, if $a=17$, then $b=1717$. To her surprise, she finds that $b$ is a multiple of $a^{2}$. Find the product of all the possible values of $\frac{b}{a^{2}}$. | [
"Solution:\n\nSuppose $a$ has $k$ digits. Then $b = a(10^{k} + 1)$. Thus $a$ divides $10^{k} + 1$. Since $a \\geq 10^{k-1}$, we have $\\frac{10^{k} + 1}{a} \\leq 11$. But since none of $2, 3$, or $5$ divide $10^{k} + 1$, the only possibilities are $7$ and $11$. These values are obtained when $a = 143$ and $a = 1$, ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 77 | |
02ds | Show that $1 \le n^{1/n} \le 2$ for all positive integers $n$. Find the smallest $k$ such that $1 \le n^{1/n} \le k$ for all positive integers $n$. | [
"If $k < 1$, then $k^n < 1$, so $n^{1/n} \\ge 1$. We have $2^1 > 1$, and $2^{n+1} - 2^n = 2^n > 1 = (n+1) - n$, so a trivial induction shows that $2^n > n$. Hence $2 > n^{1/n}$.\n\nNote that $3^2 > 2^3$, so taking 6th roots, $3^{1/3} > \\sqrt{2}$. Note also that $4^{1/4} = \\sqrt{2}$.\n\nWe show that $n^{1/n} < \\s... | Brazil | V OBM | [
"Algebra > Intermediate Algebra > Exponential functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 3^{1/3} | |
0047 | Úrsula dibujó un polígono convexo de $11$ lados y llamó $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8, A_9, A_{10}$ y $A_{11}$ a sus vértices. Luego, trazó algunas diagonales del polígono (tales que dos cualesquiera no se cortaban en el interior del polígono) hasta que el polígono quedó dividido en triángulos.
Para cada vér... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | Español | proof and answer | Missing counts: A4 = 1, A7 = 1, A10 = 1. The diagonals are A3A5, A6A8, A9A11, A3A6, A1A9, A2A6, A1A8, A1A6. | |
065m | Find all triples of real numbers $(x, y, z)$ which are greater than $3$ and satisfy the equality:
$$
\frac{(x+2)^2}{y+z-2} + \frac{(y+4)^2}{z+x-4} + \frac{(z+6)^2}{x+y-6} = 36.
$$ | [
"Since $x, y, z$ are greater than $3$, it follows that $y+z-2, z+x-4, x+y-6$ are positive. Thus, from Cauchy-Schwarz inequality we get:\n$$\n\\left( \\frac{(x+2)^2}{y+z-2} + \\frac{(y+4)^2}{x+z-4} + \\frac{(z+6)^2}{x+y-6} \\right) \\left( (y+z-2) + (x+z-4) + (x+y-6) \\right) \\geq (x+y+z+12)^2 \\\\\n\\Leftrightarro... | Greece | SELECTION EXAMINATION | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (10, 8, 6) | |
06x2 | Alice fills the fields of an $n \times n$ board with numbers from $1$ to $n^{2}$, each number being used exactly once. She then counts the total number of good paths on the board. A good path is a sequence of fields of arbitrary length (including $1$) such that:
(i) The first field in the sequence is one that is only a... | [
"We will call any field that is only adjacent to fields with larger numbers a well. Other fields will be called non-wells. Let us make a second $n \\times n$ board $B$ where in each field we will write the number of good sequences which end on the corresponding field in the original board $A$. We will thus look for... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2n^2 - 2n + 1 | |
0dxk | Problem:
Za celo število $x$ velja
$$
|\ldots||| x-1|-10|-10^{2}\left|\ldots-10^{2006}\right|=10^{2007}
$$
Poišči stoto števko števila $|x|$. | [
"Solution:\n$$\nx_{n}=|\\ldots||| x-1|-10|-10^{2}\\left|\\ldots-10^{n-1}\\right|-10^{n}\n$$\n\nEnačba iz naloge nam torej pove, da velja $\\left|x_{2006}\\right|=10^{2007}$, torej je $x_{2006}= \\pm 10^{2007}$. Ker pa lahko ocenimo\n$$\nx_{2006}=\\left|x_{2005}\\right|-10^{2006} \\geq-10^{2006}\n$$\nod tod sledi, d... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1 | |
0ivw | For each positive integer $n$, let $c(n)$ be the largest real number such that
$$
c(n) \leq \left| \frac{f(a) - f(b)}{a - b} \right|
$$
for all triples $(f, a, b)$ such that
* $f$ is a polynomial of degree $n$ taking integers to integers, and
* $a, b$ are integers with $f(a) \neq f(b)$.
Find $c(n)$. | [
"Let $L(n) = \\operatorname{lcm}(1, 2, \\dots, n)$. We claim that $c(n) = \\frac{1}{L(n)}$.\n\nFirst we show that this $c(n)$ is a lower bound. For any choice of $f(x)$ and $(a, b)$, we can translate $f(x)$ vertically so that $f(b) = 0$, and then translate $f(x)$ horizontally so that $b = 0$. We only deal with this... | United States | Team Selection Test 2009 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization >... | null | proof and answer | 1 / lcm(1, 2, ..., n) | |
0itb | Problem:
Compute $\sum_{n=1}^{\infty} \sum_{k=1}^{n-1} \frac{k}{2^{n+k}}$. | [
"Solution:\n\nAnswer: $\\frac{4}{9}$ We change the order of summation:\n$$\n\\sum_{n=1}^{\\infty} \\sum_{k=1}^{n-1} \\frac{k}{2^{n+k}} = \\sum_{k=1}^{\\infty} \\frac{k}{2^{k}} \\sum_{n=k+1}^{\\infty} \\frac{1}{2^{n}} = \\sum_{k=1}^{\\infty} \\frac{k}{4^{k}} = \\frac{4}{9}.\n$$"
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 4/9 | |
0ijp | Problem:
A dot is marked at each vertex of a triangle $A B C$. Then, $2$, $3$, and $7$ more dots are marked on the sides $A B$, $B C$, and $C A$, respectively. How many triangles have their vertices at these dots? | [
"Solution:\nAltogether there are $3+2+3+7=15$ dots, and thus $\\binom{15}{3}=455$ combinations of $3$ dots. Of these combinations, $\\binom{2+2}{3}+\\binom{2+3}{3}+\\binom{2+7}{3}=4+10+84=98$ do not give triangles because they are collinear (the rest do give triangles). Thus $455-98=357$ different triangles can be ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 357 | |
0fk6 | Problem:
Demostrar que, en un triángulo, la distancia de un vértice cualquiera al ortocentro es el doble de la distancia del circuncentro al lado opuesto a ese vértice. | [
"Solution:\nSean el triángulo $\\triangle ABC$, su ortocentro $H$ y su circuncentro $O$. Sean $H'$ y $O'$ sus simétricos respecto del lado $BC$.\n\n\n\ni) Puesto que los triángulos $\\triangle BPA$ y $\\triangle BCR$ son rectángulos y comparten el ángulo $\\subset BA$, son semejantes y, por... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00ri | Let positive integers $K$ and $d$ be given. Prove that there exists a positive integer $n$ and a sequence of $K$ positive integers $b_1, b_2, \dots, b_K$ such that the number $n$ is a $d$-digit palindrome in all number bases $b_1, b_2, \dots, b_K$. | [
"$$\n\\left\\langle i^{d-1} \\binom{d-1}{d-1}, i^{d-1} \\binom{d-1}{d-2}, i^{d-1} \\binom{d-1}{d-3}, \\dots, i^{d-1} \\binom{d-1}{1}, i^{d-1} \\binom{d-1}{0} \\right\\rangle_{\\frac{n!}{i}-1}\n$$\nWe first show that, for each large enough $n$, all these digits are smaller than the considered base, that is, they are... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
01kp | Some elements of an eight-element set are colored white while the others are colored black. Let $N$ be a number of all subsets containing the elements of both colors.
Find the greatest possible value of $N$. | [
"Suppose $k$ elements are colored white and $8 - k$ elements are colored black. A subset contains elements of both colors if it contains at least one white and at least one black element.\n\nThe total number of subsets of the eight-element set is $2^8 = 256$.\n\nThe subsets that do not contain elements of both colo... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 225 | |
02sh | Problem:
Um comerciante recebeu quatro sacos de batatas e deseja medir o peso de cada um deles. Ele sabe que os pesos desses sacos em quilogramas são quantidades inteiras e distintas. Suponha que os pesos dos sacos (em quilogramas) sejam $a, b, c$ e $d$, com $a<b<c<d$.
a) Mostre que, ao pesar os sacos de dois em dois... | [
"Solution:\n\na) Somando $a$ em cada lado da desigualdade $b<c$, obtemos a desigualdade $a+b < a+c$. Com argumentos similares, podemos obter as desigualdades:\n$$\na+c < a+d, \\quad a+c < b+c, \\quad a+d < b+d, \\quad b+c < b+d \\quad \\text{e} \\quad b+d < c+d\n$$\nUsando essas desigualdades, podemos concluir que:... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a=41, b=45, c=56, d=71 | |
0g4j | Problem:
Let $ABC$ and $AMN$ be two similar, non-overlapping triangles with the same orientation, such that $AB = AC$ and $AM = AN$. Let $O$ be the circumcentre of the triangle $MAB$. Prove that the points $O$, $C$, $N$ and $A$ lie on a circle if and only if the triangle $ABC$ is equilateral. | [
"Solution:\n\n\n\nLet $O'$ be the circumcentre of triangle $NAC$. Consider the rotation of center $A$ and angle $\\angle BAC$: the conditions of the problem imply that it maps $B$ to $C$ and $M$ to $N$. Moreover, because it fixes $A$ we know that it maps $O$ to $O'$, and triangle $AOO'$ is ... | Switzerland | Switzerland Selection Solution | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
06w4 | Let $n$ be a positive integer. Find the number of permutations $a_{1}, a_{2}, \ldots, a_{n}$ of the sequence $1,2, \ldots, n$ satisfying
$$
a_{1} \leqslant 2 a_{2} \leqslant 3 a_{3} \leqslant \ldots \leqslant n a_{n}
$$ | [
"Denote by $P_{n}$ the number of permutations that satisfy (*). It is easy to see that $P_{1}=1$ and $P_{2}=2$.\n\nLemma 1. Let $n \\geqslant 3$. If a permutation $a_{1}, \\ldots, a_{n}$ satisfies (*) then either $a_{n}=n$, or $a_{n-1}=n$ and $a_{n}=n-1$.\n\nProof. Let $k$ be the index for which $a_{k}=n$. If $k=n$... | IMO | IMO 2020 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | F_{n+1} | |
0izs | Problem:
Distinct points $A$, $B$, $C$, $D$ are given such that triangles $ABC$ and $ABD$ are equilateral and both are of side length $10$. Point $E$ lies inside triangle $ABC$ such that $EA=8$ and $EB=3$, and point $F$ lies inside triangle $ABD$ such that $FD=8$ and $FB=3$. What is the area of quadrilateral $AEFD$? | [
"Solution:\n\nSince $AEB \\cong DFB$, we have $\\angle EBA = \\angle FBD$. Thus, $\\angle EBF = \\angle EBA + \\angle ABF = \\angle FBD + \\angle ABF = \\angle ABD = 60^\\circ$. Since $EB = BF = 3$, this means that $EBF$ is an equilateral triangle of side length $3$.\n\nNow we have\n$$\n[AEFD] = [AEBD] - [EBF] - [F... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 91*sqrt(3)/4 | |
04ld | Let $R$ be a positive integer less than $2017$. Exactly $R$ vertices of a regular $2017$-gon are red, while the remaining ones are blue. Prove that the number of isosceles triangles having all three vertices of the same colour does not depend on the arrangement of the blue and the red vertices. (Russia 1994) | [
"First, notice that none of the triangles formed by taking three vertices of the given polygon is equilateral. If there were one, we would have the same number of vertices of the original polygon between any two of the triangle's vertices. This would imply that $3$ divides $2014$, which is false.\n\nLet us now prov... | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
02rm | The real numbers $a$ and $b$ satisfy $(a+b)(a+1)(b+1) = 2$ and $a^3 + b^3 = 1$. Find $a+b$. | [
"Let $S = a+b$ and $P = ab$. Then $(a+1)(b+1) = ab + a + b + 1 = P + S + 1$ and $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = S(S^2 - 3P)$. So\n$$\n\\left| \n\\begin{array}{l} \nS(P + S + 1) = 2 \\\\ \nS(S^2 - 3P) = 1 \n\\end{array} \n\\right. \n\\iff \n\\begin{array}{l} \n3SP + 3S^2 + 3S + S^3 - 3PS = 3 \\cdot 2 + 1 \\\\ \... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1 | |
0kuo | Problem:
Elisenda has a piece of paper in the shape of a triangle with vertices $A$, $B$, and $C$ such that $AB = 42$. She chooses a point $D$ on segment $AC$, and she folds the paper along line $BD$ so that $A$ lands at a point $E$ on segment $BC$. Then, she folds the paper along line $DE$. When she does this, $B$ la... | [
"Solution:\n\nLet $F$ be the midpoint of segment $DC$.\n\nEvidently $\\angle ADB = 60^{\\circ} = \\angle BDE = \\angle EDC$. Moreover, we have $BD = DF = FC$, $AD = DE$, and $AB = BE$. Hence angle bisector on $BDC$ gives us that $BE = 42$, $EC = 84$, and hence angle bisector on $ABC$ gives us that if $AD = x$ then ... | United States | HMMT February 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 168 + 48√7 | |
0es7 | For which integers $n \ge 2$ is it possible to draw $n$ distinct straight lines in the plane in such a way that there are at least $n-2$ points where exactly three of the lines intersect? | [
"For $n=2$, any two lines satisfy the condition, and for $n=3$, we can take any three lines passing through a common point.\n\nFor $n=4$, there is no feasible choice of four lines: suppose there are two points where exactly three lines meet. At most one of the lines can pass through both, so we need at least $1+2 \... | South Africa | South African Mathematics Olympiad Third Round | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | all integers n ≥ 2 with n ≠ 4, 5 | |
090k | Determine how many integers between $1$ and $1000$ (inclusive) have the property that the remainders upon division by $2$, $3$, $4$, $5$, and $6$ are all distinct. | [
"Let $n$ be an integer with $1 \\le n \\le 1000$, and for each integer $i$ with $2 \\le i \\le 6$, let $r_i$ denote the remainder when $n$ is divided by $i$. We require that $r_2, r_3, r_4, r_5, r_6$ be pairwise distinct. Since $0 \\le r_i \\le i - 1$, in particular $r_2$ can only be $0$ or $1$.\n\n**Case 1:** Assu... | Japan | The 35th Japanese Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | 49 | |
0ikz | Problem:
Given a regular $n$-gon with sides of length $1$, what is the smallest radius $r$ such that there is a non-empty intersection of $n$ circles of radius $r$ centered at the vertices of the $n$-gon? Give $r$ as a formula in terms of $n$. Be sure to prove your answer. | [
"Solution:\n\nIt is easy to see that, with this $r$, all the circles pass through the center of the $n$-gon. The following proves that this $r$ is necessary even if the word \"circle\" is replaced by the word \"disk.\"\n\nFor $n$ even, it is easy to see using symmetry that containing the center point is necessary a... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | r = 1 / (2 sin(pi/n)) | |
06qt | A positive integer $N$ is called balanced, if $N=1$ or if $N$ can be written as a product of an even number of not necessarily distinct primes. Given positive integers $a$ and $b$, consider the polynomial $P$ defined by $P(x)=(x+a)(x+b)$.
a. Prove that there exist distinct positive integers $a$ and $b$ such that all t... | [
"Define a function $f$ on the set of positive integers by $f(n)=0$ if $n$ is balanced and $f(n)=1$ otherwise. Clearly, $f(n m) \\equiv f(n)+f(m) \\bmod 2$ for all positive integers $n, m$.\n\na. Now for each positive integer $n$ consider the binary sequence $(f(n+1), f(n+2), \\ldots, f(n+50))$. As there are only $2... | IMO | IMO Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | English | proof only | null | |
0ei1 | Problem:
Kateri izraz je enakovreden izrazu $\log _{a b}\left(a^{-1} \cdot \sqrt{a^{3} b^{-1}}: \sqrt[3]{b^{-1} a^{2}}\right)$ ?
(A) $a b$
(B) $\frac{a}{b}$
(C) 1
(D) $-\frac{1}{6}$
(E) $-\frac{1}{2}$ | [
"Solution:\nIzraz v oklepaju preoblikujemo na skupni korenski eksponent $a^{-1} \\cdot \\sqrt{a^{3} b^{-1}}: \\sqrt[3]{b^{-1} a^{2}}=$ $\\sqrt[6]{a^{-6}} \\cdot \\sqrt[6]{a^{9} b^{-3}}: \\sqrt[6]{b^{-2} a^{4}}$. Izraz poenostavimo in damo pod skupni koren: $\\sqrt[6]{a^{-1} b^{-1}}=\\sqrt[6]{\\frac{1}{a b}}=$ $\\le... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | D | |
0g0m | Problem:
Cunégonde a $n$ blocs de hauteur 1 à $n$ et souhaiterait les agencer, les uns après les autres, de telle manière que son chat puisse se déplacer en sautant d'un bloc à l'autre, de la gauche vers la droite. Son chat peut sauter d'un bloc au suivant si celui-ci est soit moins haut, soit plus haut de 1 que le bl... | [
"Solution:\n\nSupposons que Cunégonde place le bloc de hauteur $n$ à la position $k$. Alors le bloc à la position $k-1$ doit avoir la hauteur $n-1$. En continuant comme ça, on voit que les $k$ premiers blocs doivent être $n-k+1, n-k+2, \\ldots, n$. Il nous reste ensuite les $n-k$ premiers blocs à placer sur $n-k$ p... | Switzerland | SMO - Vorrunde | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2^{n-1} | |
09w2 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$
f(x^2y) + 2f(y^2) = (x^2 + f(y)) \cdot f(y)
$$
for all $x, y \in \mathbb{R}$. | [
"$$\n2f(y^2) = f(y)^2. \\qquad (4)\n$$\nUsing this, we can cancel the $2f(y^2)$ on the left hand side of the original functional equation against the $f(y)^2$ on the right hand side:\n$$\nf(x^2y) = x^2f(y).\n$$\nSubstituting $y = 1$ in this equation yields $f(x^2) = x^2f(1)$, and substituting $y = -1$ yields $f(-x^... | Netherlands | BxMO Team Selection Test, March 2020 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = 2x for all real x; f(x) = 2|x| for all real x | |
0ei4 | Problem:
Mama je na vprašanje, koliko je star sin, odgovorila z uganko: Če od sedanje dvakratne sinove starosti odštejemo trikrat toliko let, kot jih je imel pred šestimi leti, dobimo, koliko je star danes. Koliko let je sin star danes?
(A) 3 leta
(B) 18 let
(C) 9 let
(D) 6 let
(E) 27 let | [
"Solution:\nPredpostavimo, da je danes star $x$ let in zapišemo enačbo $2x - 3(x - 6) = x$. Odpravimo oklepaj, enačbo uredimo in dobimo rešitev $x = 9$."
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0gbu | 設 $x, y$ 為相異正數, 且 $n$ 為大於 1 的正整數。若
$$
x^n - y^n = x^{n+1} - y^{n+1}.
$$
證明:
$$
1 < x + y < \frac{2n}{n + 1}.
$$ | [
"首先證明：$x + y > 1$.\n注意到，\n$$\n(x + y)(x^n - y^n) = x^{n+1} - y^{n+1} + xy(x^{n-1} - y^{n-1}).\n$$\n$$\n\\text{則 } x + y = \\frac{x^{n+1} - y^{n+1}}{x^n - y^n} + xy \\times \\frac{x^{n-1} - y^{n-1}}{x^n - y^n} = 1 + xy \\times \\frac{x^{n-1} - y^{n-1}}{x^n - y^n}.\n$$\n由 $n > 1$, 知 $x^n - y^n$ 與 $x^{n-1} - y^{n-1}$ ... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
00pv | Let $ABCD$ be a quadrilateral, let $O$ be the intersection point of diagonals $AC$ and $BD$, and let $P$ be the intersection point of sides $AB$ and $CD$. Consider the parallelograms $AODE$ and $BOCF$. Prove that $E$, $F$ and $P$ are collinear. | [
"\nFigure 1\nClearly, $EZFY$ is a parallelogram. Let $ZY \\cap AD = \\{X_2\\}$ and $ZY \\cap BC = \\{X_1\\}$.\n\nFrom Menelaos Theorem in triangle $ZYF$ and the line $(X_1, B, C)$ it follows\n$$\n\\frac{ZB}{BF} \\cdot \\frac{FC}{CY} \\cdot \\frac{YX_1}{X_1Z} = 1,\n$$\nhence\n$$\n\\frac{X_1Y... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0k8z | Problem:
Contessa is taking a random lattice walk in the plane, starting at $(1,1)$. (In a random lattice walk, one moves up, down, left, or right 1 unit with equal probability at each step.) If she lands on a point of the form $(6m, 6n)$ for $m, n \in \mathbb{Z}$, she ascends to heaven, but if she lands on a point of... | [
"Solution:\n\nLet $P(m, n)$ be the probability that she ascends to heaven from point $(m, n)$. Then $P(6m, 6n) = 1$ and $P(6m+3, 6n+3) = 0$ for all integers $m, n \\in \\mathbb{Z}$. At all other points,\n$$\n4P(m, n) = P(m-1, n) + P(m+1, n) + P(m, n-1) + P(m, n+1)\n$$\nThis gives an infinite system of equations. Ho... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 13/22 | |
0h7u | Let $x$, $y$, $z$ be real numbers from segment $[0; 1]$. Prove that
$$
(x^4 + y^4 + z^4) + (x^5 + y^5 + z^5) + (x - y)^6 + (y - z)^6 + (z - x)^6 \le 6.
$$ | [
"Firstly, we will prove that if $\\alpha$, $\\beta \\in [0, 1]$, then $\\alpha^4 + \\beta^5 + (\\alpha - \\beta)^6 \\le 2$. Indeed, $t^n \\le t$ for all real $t \\in [0, 1]$ and any positive integer $n$. Suppose that $\\alpha \\ge \\beta$ (case $\\alpha \\le \\beta$ is considered analogously). Then $0 \\le \\alpha ... | Ukraine | UkraineMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
04jz | In a quadrilateral $ABCD$ we have $\angle BAD = 110^\circ$, $\angle CBA = 50^\circ$, $\angle DCB = 70^\circ$. Let $M$ and $N$ be the midpoints of segments $\overline{AB}$ and $\overline{CD}$ respectively. Let $P$ be a point on the segment $\overline{MN}$ such that $|AM| : |CN| = |MP| : |NP|$ and $|AP| = |CP|$. Determin... | [
"Let $K$ be the intersection of lines $AD$ and $BC$. Since $ABCD$ is a cyclic quadrilateral, triangles $AKB$ and $CKD$ are similar. As $M$ and $N$ are midpoints of respective segments, we have $\\angle AKM = \\angle CKN$. It follows that triangles $AKM$ and $CKN$ are similar, too.\n\n---\n\nHence $|KM| : |KN| = |AM... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 160° | |
0jmq | Problem:
Compute the side length of the largest cube contained in the region
$$
\left\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 25 \text{ and } x \geq 0\right\}
$$
of three-dimensional space. | [
"Solution:\nAnswer: $\\sqrt{\\frac{5 \\sqrt{6}}{3}}$\n\nThe given region is a hemisphere, so the largest cube that can fit inside it has one face centered at the origin and the four vertices of the opposite face on the spherical surface. Let the side length of this cube be $s$. Then, the radius of the circle is the... | United States | HMMT 2014 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 5√6/3 | |
0k6p | Problem:
Let $ABC$ be a triangle with $AB = 5$, $BC = 8$, $CA = 11$. The incircle $\omega$ and $A$-excircle $\Gamma$ are centered at $I_1$ and $I_2$, respectively, and are tangent to $BC$ at $D_1$ and $D_2$, respectively. Find the ratio of the area of $\triangle A I_1 D_1$ to the area of $\triangle A I_2 D_2$. | [
"Solution:\nLet $D_1'$ and $D_2'$ be the points diametrically opposite $D_1$ and $D_2$ on the incircle and $A$-excircle, respectively. As $I_x$ is the midpoint of $D_x$ and $D_x'$, we have\n$$\n\\frac{[A I_1 D_1]}{[A I_2 D_2]} = \\frac{[A D_1 D_1']}{[A D_2 D_2']}\n$$\nNow, $\\triangle A D_1 D_1'$ and $\\triangle A ... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 1/9 | |
0fzy | Problem:
Sei $ABC$ ein Dreieck mit $AB > AC$ und sei $M$ der Mittelpunkt der Seite $AC$. Weiter sei $D$ ein Punkt auf der Seite $AB$, sodass $DB = DC$ gilt. Die Parallele zu $BC$ durch $D$ und die Gerade $BM$ schneiden sich im Punkt $K$. Zeige, dass $\angle KCD = \angle DAC$ gilt. | [
"Solution:\n\nSei $\\beta = \\angle CBA$. Da $DBC$ ein gleichschenkliges Dreieck ist, gilt $\\angle CBD = \\angle DCB = \\beta$. Der Außenwinkel $\\angle CDA$ beträgt somit $2\\beta$. Da $KD$ parallel zu $BC$ ist, gilt $\\angle CDK = \\angle DCB = \\beta$, woraus folgt, dass $\\angle KDA = \\angle CDA - \\angle CDK... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0i8b | Problem:
You have a list of real numbers, whose sum is $40$. If you replace every number $x$ on the list by $1-x$, the sum of the new numbers will be $20$. If instead you had replaced every number $x$ by $1+x$, what would the sum then be? | [
"Solution:\n\nLet $n$ be the number of numbers on the list. If each initial number is replaced by its negative, the sum will then be $-40$, and adding $1$ to every number on this list increases the sum by $n$, so $n - 40 = 20 \\Rightarrow n = 60$. Then, if we had simply added $1$ to each of the initial numbers (wit... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 100 | |
0kwx | Problem:
Let $S$ be the set of all points in the plane whose coordinates are positive integers less than or equal to $100$ (so $S$ has $100^{2}$ elements), and let $\mathcal{L}$ be the set of all lines $\ell$ such that $\ell$ passes through at least two points in $S$. Find, with proof, the largest integer $N \geq 2$ f... | [
"Solution:\n\nLet the lines all have slope $\\frac{p}{q}$ where $p$ and $q$ are relatively prime. Without loss of generality, let this slope be positive. Consider the set of points that consists of the point of $S$ with the smallest coordinates on each individual line in the set $L$. Consider a point $(x, y)$ in th... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 4950 | |
00ng | Mr. Precise wants to take his tea cup out of the microwave precisely at the front. The microwave of Mr. Precise is not precisely cooperative.
More precisely, the two of them play the following game:
Let $n$ be a positive integer. The rotating plate of the microwave takes $n$ seconds for a full turn. Each time the micro... | [
"**Answer.** Mr. Precise can ensure his victory when $n$ is a power of 2.\n\nWe label the positions consecutively $0$, $1$, ..., $n-1$ where $0$ is the front position.\nIf $n$ is a power of $2$, say $n = 2^k$, Mr. Precise can simply always put in the current position as number of seconds. If the microwave turns the... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | Exactly when the number of positions is a power of two | |
00wk | Problem:
Two equal triangles are inscribed into an ellipse. Are they necessarily symmetrical with respect either to the axes or to the centre of the ellipse? | [
"Solution:\n\nNo, not necessarily (see Figure 3 where the two ellipses are equal).\n\n\nFigure 3"
] | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | No | |
040p | Find the minimum positive integer $n$ such that
$$
\sqrt{\frac{n-2011}{2012}} - \sqrt{\frac{n-2012}{2011}} < \sqrt[3]{\frac{n-2013}{2011}} - \sqrt[3]{\frac{n-2011}{2013}}
$$ | [
"We see that if $2012 \\le n \\le 4023$, then $\\sqrt{\\frac{n-2011}{2012}} - \\sqrt{\\frac{n-2012}{2011}} \\ge 0$ and $\\sqrt[3]{\\frac{n-2013}{2011}} - \\sqrt[3]{\\frac{n-2011}{2013}} < 0$.\n\nOtherwise,\n$$\n\\begin{aligned}\n\\sqrt{\\frac{n-2011}{2012}} &\\le \\sqrt{\\frac{n-2012}{2011}} &&\\Leftrightarrow n > ... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 4024 | |
094g | Problem:
Let $\mathbb{Z}$ denote the set of all integers and $\mathbb{Z}_{>0}$ denote the set of all positive integers.
a. A function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ is called $\mathbb{Z}$-good if it satisfies $f\left(a^{2}+b\right)=f\left(b^{2}+a\right)$ for all $a, b \in \mathbb{Z}$. Determine the largest pos... | [
"Solution:\nThe answer is (a) 2 and (b) 1077.\n\na.\nNote that\n$$\nf\\left(a^{2}+b\\right)=f\\left(b^{2}+a\\right)=f\\left((-b)^{2}+a\\right)=f\\left(a^{2}-b\\right)\n$$\nIn particular, by setting $a \\in \\{0,1\\}$ we get $f(b)=f(-b)$ and $f(1+b)=f(1-b)$. This then yields\n$$\nf(2+b)=f(1+(1+b))=f(1-(1+b))=f(-b)=f... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (a) 2; (b) 1077 | |
0ff5 | Problem:
Dada la ecuación $x^{2}+a x+1=0$, determinar
a) El intervalo en que debe mantenerse el número real $a$ para que las raíces de esa ecuación sean imaginarias.
b) El lugar geométrico de los puntos representativos de esas raíces en la representación gráfica habitual de los números complejos, cuando $a$ recorre ... | [
"Solution:\n\nEl discriminante de la ecuación es $a^{2}-4$, que tiene que ser estrictamente menor que cero; por lo tanto el intervalo pedido para $a$ es\n$$\n-2<a<2\n$$\nEn esas condiciones, las raíces de la ecuación son\n$$\n-\\frac{a}{2} \\pm \\frac{i \\sqrt{4-a^{2}}}{2}\n$$\nasí que las ecuaciones paramétricas d... | Spain | Olimpiadas Matemáticas Españolas | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | a ∈ (−2, 2); locus: x^2 + y^2 = 1 with the points (1,0) and (−1,0) excluded. | |
08sp | Let $ABCD$ be a square of side length $1$. Let $O$ be the circle having the side $AD$ of the square as its diameter, and pick a point $E$ on the side $AB$ of the square in such a way that the line $CE$ becomes a tangent line to the circle $O$. Determine the area of the triangle $CBE$. | [
"Let $M$ be the mid-point of the side $AD$ of the square, and let $N$ be the point of tangency of the line $CE$ to the circle $O$. Since $MD = MN$, and $\\angle MDC = \\angle MNC = 90^\\circ$, we have $\\triangle MNC \\equiv \\triangle MDC$. Similarly, we have $\\triangle MNE \\equiv \\triangle MAE$. From $\\triang... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 3/8 | |
0k8l | Problem:
Let $\overline{S T}$ be a chord of a circle $\omega$ which is not a diameter, and let $A$ be a fixed point on $\overline{S T}$. For which point $X$ on minor arc $\widehat{S T}$ is the length $A X$ minimized? | [
"Solution:\nExtend the circular segment to make a whole circle, and let $O$ be its center. Draw $O A$ and let it meet the circle at $X$. Then the circle with center $A$ and radius $A X$ is tangent to the larger circle at $A$, and thus lies entirely inside it. Therefore, the distance from $A$ to any other point on t... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0h3q | Each of $1005$ citizens of the Emerald city wears glasses (with two lenses). The lenses in the glasses are colored rose, blue and green evenly: in total there are $670$ lenses of each color. All citizens composed a circle so that no two neighbors in this circle have lenses of a common color. What is the maximum possibl... | [
"Будемо позначати кольори лінз літерами Р, Б, З. Зрозуміло, що в колі не можуть стояти поруч мешканці з різнокольоровими окулярами, тому таких мешканців не більше за $502$. Якщо їх $502$, то рівно двоє людей з однокольоровими окулярами стоять поруч (нехай це мешканці з лінзами РР та ББ), а далі по колу «різнокольор... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 501 | |
02pz | Problem:
Considere um poliedro convexo com $100$ arestas. Todos os vértices foram aparados próximos a eles mesmos, usando uma faca plana afiada (isto foi feito de modo que os planos resultantes não se intersectassem no interior ou na fronteira do poliedro). Calcule para o poliedro resultante:
(a) o número de vértices.... | [
"Solution:\n\n(a) Quando realizamos os cortes, cada aresta antiga estará ligada a dois vértices novos, enquanto os vértices antigos desaparecem. Assim o novo poliedro tem $200$ vértices.\n\n(b) Quando realizamos um corte, de cada novo vértice surgem duas arestas novas (correspondentes a duas arestas consecutivas na... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | Vertices: 200; Edges: 300; Faces: 102 | |
0k81 | Problem:
For any angle $0 < \theta < \pi / 2$, show that
$$
0 < \sin \theta + \cos \theta + \tan \theta + \cot \theta - \sec \theta - \csc \theta < 1
$$ | [
"Solution:\nWe use the following geometric construction, which follows from the geometric definition of the trigonometric functions: Let $Z$ be a point on the unit circle in the coordinate plane with origin $O$. Let $X_{1}, Y_{1}$ be the projections of $Z$ onto the $x$- and $y$-axis respectively, and let $X_{2}, Y_... | United States | HMMT February 2019 Team Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0cqx | Стозначное натуральное число $n$ назовём необычным, если десятичная запись числа $n^3$ заканчивается на $n$, а десятичная запись числа $n^2$ не заканчивается на $n$. Докажите, что существует не менее двух значных необычных чисел. | [
"Например, такими числами являются $n_1 = 10^{100} - 1 = 99...9$ и $n_2 = \\frac{10^{100}}{2} - 1 = 49...9$. Действительно, числа\n$$n_1^3 - n_1 = (n_1 + 1)n_1(n_1 - 1) = 10^{100} \\cdot n_1(n_1 - 1)$$\nи\n$$n_2^3 - n_2 = (n_2 + 1)n_2(n_2 - 1) = 10^{100} \\cdot n_2 \\cdot \\frac{n_2 - 1}{2}$$\nделятся на $10^{100}$... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0j5z | Problem:
A classroom has 30 students and 30 desks arranged in 5 rows of 6. If the class has 15 boys and 15 girls, in how many ways can the students be placed in the chairs such that no boy is sitting in front of, behind, or next to another boy, and no girl is sitting in front of, behind, or next to another girl? | [
"Solution:\n\nAnswer: $2 \\cdot 15!^{2}$\n\nIf we color the desks of the class in a checkerboard pattern, we notice that all of one gender must go in the squares colored black, and the other gender must go in the squares colored white. There are 2 ways to pick which gender goes in which color, $15!$ ways to put the... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 2 * (15!)^2 | |
0cte | $n$ boys and $n$ girls stand along a circle. We call a pair of a boy and a girl good if one of two arcs joining them contains equal numbers of girls and boys (in particular, neighboring boy and girl form a good pair). It appears that there exists a girl belonging to exactly 10 good pairs. Prove that there also exists a... | [
"Let $D$ be a girl participating in 10 good pairs. Let $K_1, K_2, \\dots, K_{2n}$ be all the children in a clockwise order, with $K_1 = D$; set $K_{i \\pm 2n} = K_i$. Denote by $d_i$ the difference between the numbers of boys and girls among $K_1, K_2, \\dots, K_i$; in particular, $d_1 = 1 - 0 = 1$ and $d_{2n} = 0$... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English; Russian | proof only | null | |
0042 | Demuestre que no existen dos primos $p$ y $q$, con $q-p=2$, y un entero $k$, tales que $pq+10^k$ sea un número primo. | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Español | proof only | null | |
06p7 | Consider those functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy the condition
$$
f(m+n) \geq f(m)+f(f(n))-1
$$
for all $m, n \in \mathbb{N}$. Find all possible values of $f(2007)$.
($\mathbb{N}$ denotes the set of all positive integers.) | [
"Suppose that a function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfies (1). For arbitrary positive integers $m>n$, by (1) we have\n$$\nf(m)=f(n+(m-n)) \\geq f(n)+f(f(m-n))-1 \\geq f(n),\n$$\nso $f$ is nondecreasing.\n\nFunction $f \\equiv 1$ is an obvious solution. To find other solutions, assume that $f \\no... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | all integers from 1 to 2008 inclusive | |
09h1 | Define the sequences $\{a_n\}_{n \ge 0}$, $\{b_n\}_{n \ge 0}$ as follows: $a_0, b_0$ are natural numbers and
$$a_{n+1} = a_n + \lfloor \sqrt{b_n} \rfloor$$
$$b_{n+1} = b_n + \lfloor \sqrt{a_n} \rfloor$$
for $n = 0, 1, 2, \dots$. Prove that there exists a nonnegative $n$ such that $a_n = b_n$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0gim | 令 $\mathbb{N}$ 為所有正整數所成集合。找出所有的函數 $f: \mathbb{N} \to \mathbb{N}$ 使得對於所有正整數 $m, n$，都有 $mf(m) + (f(f(m)) + n)^2$ 整除 $4m^4 + n^2 f(f(n))^2$。
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that $mf(m) + (f(f(m)) + n)^2$ divides $4m^4 + n^2 f(f(n))^2$ for all p... | [
"**Answer:** The only possible $f$ is $f(n) = n$.\n\nLet $A(m, n) = mf(m) + (f(f(m)) + n)^2$, $B(m, n) = 4m^4 + n^2 f(f(n))^2$, and $C(m, n) = m^2 + (m - n^2)$. We can check $A(m, n)C(m, n) = B(m, n)$ if $f(n) = n$ and $f(m) = m$. So $f(k) = k$ is a solution.\n\nTo show that there are no other solutions. First, we ... | Taiwan | IMO 2J, Mock Exam 1 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Divisibility / Factorizati... | Chinese; English | proof and answer | f(n) = n | |
04o5 | Let $n$ be a positive integer. A good word is a sequence of $3n$ letters, in which each of the letters $A$, $B$ and $C$ appears exactly $n$ times. Prove that for every good word $X$ there exists a good word $Y$ such that $Y$ cannot be obtained from $X$ by swapping neighbouring letters fewer than $\frac{3}{2}n^2$ times.... | [
"Let us define the distance of good words $X$ and $Y$, denoted by $d(X, Y)$, as the smallest number of swaps of neighbouring letters necessary to obtain $Y$ from $X$ (or vice versa). Note that $d(X, Y) = d(Y, X)$, and that for any three good words $X$, $Y$ and $Z$ we have\n$$\nd(X, Y) + d(Y, Z) \\ge d(X, Z).\n$$\nF... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
06ft | Show that the equation $y^{37} \equiv x^3 + 11 \pmod{p}$ is solvable for every prime $p$, where $p \le 100$. | [
"Since $p \\le 100$, we have $(p - 1, 37) = 1$ (by checking $1, 38, 75$ are not primes). Thus, there exists $k \\in \\mathbb{Z}^+$ such that $37k \\equiv 1 \\pmod{p-1}$. Thus, by the Fermat little theorem, we have\n$$\n(11^k)^{37} = 11^{37k} \\equiv 11 \\pmod{p}.\n$$\nTherefore, $(x, y) = (0, 11^k)$ is a solution t... | Hong Kong | null | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0cmh | A quadratic trinomial $f(x)$ is chosen so that the equation
$$(f(x))^3 - f(x) = 0$$
has exactly $3$ real roots. Find the $y$-coordinate of the vertex of the graph of $f(x)$. | [
"Ответ. $0$.\n\nТак как $g(x) = (f(x))^3 - f(x) = f(x)(f(x) - 1)(f(x) + 1)$,\nто корнями многочлена $g(x)$ являются корни трехчленов $f(x)$,\n$f(x) - 1$ и $f(x) + 1$. Ясно, что любое число может быть корнем\nтолько одного из них.\n\nПусть $y_0$ — искомая ордината вершины. Предположим, что\n$y_0 \\neq 0$. Будем счит... | Russia | Russian mathematical olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English; Russian | proof and answer | 0 | |
02gg | In a convex quadrilateral, the *altitude* relative to a side is defined to be the line perpendicular to this side through the midpoint of the opposite side. Prove that the four altitudes have a common point if and only if the quadrilateral is cyclic, that is, if and only if, there exists a circle which contains its fou... | [
"Consider the following\n**Lemma.** Let $ABCD$ be a convex quadrilateral. The lines that connect the midpoints of opposite sides meet in $E$ and the perpendicular bisectors of opposite sides $AB$ and $CD$ meet in $O$. Then the altitudes relative to $AB$ and $CD$ meet in a point $O'$ symmetric to $O$ with respect to... | Brazil | XXIII OBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0f0w | Problem:
Arrange the numbers $1, 2, \ldots, 32$ in a sequence such that the arithmetic mean of two numbers does not lie between them. (For example, $\ldots, 3, 4, 5, 2, 1, \ldots$ is invalid, because $2$ lies between $1$ and $3$.) Can you arrange the numbers $1, 2, \ldots, 100$ in the same way? | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0jz5 | Problem:
Each of the integers $1,2, \ldots, 729$ is written in its base-3 representation without leading zeroes. The numbers are then joined together in that order to form a continuous string of digits: $12101112202122 \ldots$. How many times in this string does the substring 012 appear? | [
"Solution:\n\nIgnore $729=3^{6}=1000000_{3}$ since it will not contribute to a 012 substring. Break into cases on how 012 appears: (i) when an individual integer contains the string 012; (ii) when 01 are the last two digits of an integer and 2 is the first digit of the next integer; and (iii) when 0 is the last dig... | United States | HMMT November 2017 | [
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 148 | |
0e94 | Let $O$ be the origin of the coordinate system. The point $A(\frac{5}{2}, -\frac{5\sqrt{3}}{2})$ is rotated around $O$ by $2013\pi$ into the point $B$. The point $B$ is reflected across the bisector of the odd quadrants into the point $C$. Find the size of the angle $\angle AOC$. | [
"The coordinates of the point $B$ are $(-\\frac{5}{2}, \\frac{5\\sqrt{3}}{2})$ and the coordinates of $C$ are $(\\frac{5\\sqrt{3}}{2}, -\\frac{5}{2})$. All three points lie on a circle with the centre at $O$ and radius $5$. From the values of the trigonometric functions $\\sin(-30^\\circ) = -\\frac{1}{2}$ and $\\co... | Slovenia | National Math Olympiad 2013 - First Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 30° | |
07k5 | Let $p$ be a prime number. Find all positive integers $x, y > 1$ such that
$$
\frac{x^2 - 1}{y^2 - 1} = (p + 1)^2.
$$ | [
"Reformulate the original expression as $x^2 - 1 = (p+1)^2(y^2 - 1)$. $x^2 - 1 = ((p+1)y)^2 - (p+1)^2 \\Rightarrow (p+1)^2 - 1 = ((p+1)y)^2 - x^2$. $p^2 + 2p + 1 - 1 = ((p+1)y - x)((p+1)y + x)$. So, $p(p+2) = ((p+1)y - x)((p+1)y + x)$. Let $A = (p+1)y - x$ and $B = (p+1)y + x$. Then $AB = p(p+2)$. Since $x, y > 1$,... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | For odd primes p: x = (p^2 + 2p − 1) / 2 and y = (p + 1) / 2. There is no solution for p = 2. | |
03a4 | Determine the locus of the points $M$ which are interior for the convex quadrilateral $ABCD$ and the areas of the quadrilaterals $ABCM$ and $AMCD$ are equal. | [
"The locus is the segment $PQ$ such that $PQ \\parallel AC$ and $d(PQ, AC) = \\frac{|S_{ABC} - S_{ACD}|}{AC}$, where $S_{ABC}$ and $S_{ACD}$ are the areas of triangles $ABC$ and $ACD$, respectively."
] | Bulgaria | Fall Mathematical Competition | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | The locus is the segment PQ inside ABCD where PQ is parallel to AC and at perpendicular distance |S_ABC − S_ACD| / AC from AC, with S_ABC and S_ACD the areas of triangles ABC and ACD. | |
02fa | Given any convex polygon, show that there are three consecutive vertices such that the polygon lies inside the circle through them. | [
"Let $V_1$ and $V_2$ be consecutive vertices of the polygon. Let $V_k$ be a vertex such that $\\angle V_1 V_k V_2$ is minimal. Since all angles $\\angle V_1 V_j V_2 \\ge \\angle V_1 V_k V_2$ for all $j \\ne k$ of the polygon, then all vertices are contained in the circumcircle of the triangle $V_1 V_2 V_k$.\n\nIf $... | Brazil | XVI OBM | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
087g | Problem:
Il monumento a Mathenkamen è a forma di piramide che poggia sulla sua base quadrata di lato $18~\mathrm{m}$. La sua altezza misura $15~\mathrm{m}$, e il piede dell'altezza cade nel centro del quadrato. La piramide è orientata in modo che, quando i raggi del sole arrivano da sud inclinati di $45^\circ$ rispett... | [
"Solution:\n\nLa risposta è $27$. Chiamiamo $ABCD$ la base della piramide e $V$ il suo vertice lontano da terra. Si noti che l'ombra $V'$ del punto $V$ cade su una circonferenza di raggio $15~\\mathrm{m}$ centrata nel centro di $ABCD$, e a seconda dell'orientazione del sud rispetto alla piramide (che consideriamo f... | Italy | Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 27 | |
044n | Prove that there exists a constant $\lambda > 0$, such that: for any positive integer $m$, in the rectangular coordinates system, if all vertices of $\triangle ABC$ are integral points, and there is a unique interior (not on the side) integral point whose $x$ and $y$ coordinates are multiples of $m$, then the area of $... | [
"We call $(x, y)$ an “$m$-integral point” if $x, y$ are integers and $m|x$, $m|y$. By a proper translation, we may assume that the unique interior $m$-integral point is at $(0, 0)$. Extend $AO$, $BO$, and $CO$ to meet the opposite sides at points $D$, $E$, and $F$, respectively. Denote\n$$\np = \\frac{OD}{AD} = \\f... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Combinatorial Geometry > Minkowski's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0ktk | Problem:
Alice and Bob play the following "point guessing game." First, Alice marks an equilateral triangle $A B C$ and a point $D$ on segment $B C$ satisfying $B D=3$ and $C D=5$. Then, Alice chooses a point $P$ on line $A D$ and challenges Bob to mark a point $Q \neq P$ on line $A D$ such that $\frac{B Q}{Q C}=\frac... | [
"Solution:\n\nFirst, if $P=A$ then clearly Bob cannot choose a $Q$. So we can have $B P: P C=1$.\n\nOtherwise, we need $A P$ to be tangent to the Apollonius Circle. The key claim is that $A B=A C=A P$. To see why, simply note that since $B$ and $C$ are inverses with respect to the Apollonius Circle, we get that $\\... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | {1/sqrt(3), 1, 3*sqrt(3)/5} | |
00ye | Problem:
Let $a_{1}, a_{2}, \ldots, a_{9}$ be any non-negative numbers such that $a_{1}=a_{9}=0$ and at least one of the numbers is non-zero. Prove that for some $i$, $2 \leq i \leq 8$, the inequality $a_{i-1}+a_{i+1}<2 a_{i}$ holds. Will the statement remain true if we change the number $2$ in the last inequality to ... | [
"Solution:\n\nSuppose we have the opposite inequality $a_{i-1}+a_{i+1} \\geq 2 a_{i}$ for all $i=2, \\ldots, 8$. Let $a_{k}=\\max_{1 \\leq i \\leq 9} a_{i}$. Then we have $a_{k-1}=a_{k+1}=a_{k}$, $a_{k-2}=a_{k-1}=a_{k}$, etc. Finally we get $a_{1}=a_{k}$, a contradiction.\n\nSuppose now $a_{i-1}+a_{i+1} \\geq 1.9 a... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes, it remains true for one point nine. | |
00qh | In triangle $ABC$ with $AB = AC$, $M$ is the midpoint of $BC$, $H$ is the projection of $M$ onto $AB$ and $D$ is arbitrary point on the side $AC$. Let $E$ be the intersection point of the parallel line through $B$ to $HD$ with the parallel line through $C$ to $AB$. Prove that $DM$ is the bisector of $\angle ADE$. | [
"Let $\\omega$ be the circle of center $M$ and radius $MH$ and let the tangent to $\\omega$ through $D$ different from $DC$ meets the line $AB$ at $F$. It suffices to show that $E$ lies on $DF$.\nLet $E'$ be the intersection of $DF$ and the line through $C$ parallel to $AB$ and let $P$ be the projection of $D$ onto... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"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",
"Geometry > Plane Geometry > Miscellaneous > Constructions an... | English | proof only | null |
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