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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0c81 | Let $n$ be an integer, $n \ge 2$, and let $\alpha_1, \alpha_2, \dots, \alpha_n$ be non-zero complex numbers such that $|\alpha_i| < 1$ for $i = 1, \dots, n-1$, and the coefficients of the polynomial $\prod_{i=1}^n (X - \alpha_i)$ are all integral. Show that, if $\alpha_i, \alpha_j, \alpha_k$ form a geometric progressio... | [
"Suppose now, if possible, that $\\alpha_{i_0}, \\alpha_{i_1}, \\alpha_{i_2}$ form a geometric progression for some indices $i_0, i_1, i_2$ of which at least two are distinct; say $\\alpha_{i_1}^2 = \\alpha_{i_0}\\alpha_{i_2}$. The condition on absolute values forces all three indices to be different from $n$.\n\nS... | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Algebraic Number Theory > Algebraic numbers",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
02us | Problem:
Considere o seguinte tabuleiro quadriculado onde todos os números naturais foram escritos em diagonal.
| $\ddots$ | | | | | |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 10 | $\ddots$ | | | | |
| 6 | 9 | $\ddots$ | | | |
| 3 | 5 | 8 | 12 | $\ddots$ | |
| 1 | 2 | 4 | 7 | 11 | $\ddots$ |
... | [
"Solution:\n\na. Podemos preencher mais casas do tabuleiro exibido para encontrar a casa $(4,4)$ :\n\n| 21 | 27 | | | | | |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 15 | 20 | 26 | | | | |\n| 10 | 14 | 19 | 25 | | | |\n| 6 | 9 | 13 | 18 | 24 | | |\n| 3 | 5 | 8 | 12 | 17 | 23 | |\n... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a) 25; b) 2033136; c) 8116423 | |
00c0 | Let $ABCD$ be a parallelogram. Construct a square $BDXY$ with no interior points in common with the triangle $ABD$ and a square $ACZW$ with no interior points in common with the triangle $ADC$.
Let $P$ and $Q$ be the centers of the squares $BDXY$ and $ACZW$ respectively. Prove that $AP = DQ$. | [
"Let $N$ be the intersection point of the diagonals of $ABCD$. As $ABCD$ is a parallelogram, we have that $N$ is the midpoint of $AC$ and the midpoint of $BD$.\n\n\n\nSince $P$ is the center of the square $BDXY$ and $N$ is the midpoint of its side $BD$, then $DN = NP$ (both equal to a half ... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0159 | Problem:
Suppose that the positive integers $a$ and $b$ satisfy the equation
$$
a^{b}-b^{a}=1008 .
$$
Prove that $a$ and $b$ are congruent modulo 1008. | [
"Solution:\nObserve that $1008=2^{4} \\cdot 3^{2} \\cdot 7$. First we show that $a$ and $b$ cannot both be even. For suppose the largest of them were equal to $2x$ and the smallest of them equal to $2y$, where $x \\geq y \\geq 1$. Then\n$$\n\\pm 1008=(2x)^{2y}-(2y)^{2x}\n$$\nso that $2^{2y}$ divides $1008$. It foll... | Baltic Way | Baltic Way 2008 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, ineq... | null | proof only | null | |
0ad5 | Two lines intersect in one point and form four angles, two acute and two obtuse. The sum of the two acute angles is half of the one obtuse angle. Calculate these angles? | [
"Let the size of the two acute angles be $x$ (they are equal). Then each of the obtuse angles equals $180^{\\circ} - x$. From the condition in the problem we have that $2x = \\frac{1}{2}(180^{\\circ} - x)$, i.e. $4x = 180^{\\circ} - x$, from where we obtain $x = 36^{\\circ}$. Hence each of the obtuse angles equals ... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Acute angles: 36 degrees each; Obtuse angles: 144 degrees each | |
04qk | Let $m$ be a positive integer, and $p$ be a prime number such that $p > m$. Prove that the number of positive integers $n$, for which
$$
m^2 + n^2 + p^2 - 2mn - 2mp - 2np
$$
is a square of some positive integer, does not depend on $p$. (Bulgaria 2013) | [
"Let $m^2 + n^2 + p^2 - 2mn - 2mp - 2np = k^2$ for some positive integer $k$. Since\n$$\nk^2 + 4mp = m^2 + n^2 + p^2 - 2mn + 2mp - 2np = (m - n + p)^2\n$$\nis a perfect square, there exists a positive integer $l$ such that $(k+l)^2 = k^2 + 4mp$, i.e. $2kl + l^2 = 4mp$, and hence $l$ is even. Thus $l = 2a$ and $a^2 ... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
04zj | Every unit square of a $n \times n$ board is colored either red or blue so that among all $2 \times 2$ squares on this board all possible colorings of $2 \times 2$ squares with these two colors are represented (colorings obtained from each other by rotation and reflection are considered different).
a) Find the least p... | [
"a) Since there are $2^4 = 16 = 4^2$ possibilities to color a $2 \\times 2$ square in two colors and a $n \\times n$ square contains $(n-1)^2$ such subsquares, we must have $n-1 \\ge 4$, or $n \\ge 5$. For $n = 5$ a suitable coloring is given in Fig. 20.\n\n\nFig. 20\n\nb) Fig. 20 presents ... | Estonia | Estonija 2010 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n = 5; minimum number of red unit squares = 10 | |
04xy | Triangular lattice cuts an equilateral triangle with a side of length $n$ into $n^2$ triangular cells (Fig. 1). Some of the cells are infected. A cell, which is not infected yet, could be infected if it is neighbouring (by side) with at least two already infected cells. Determine the minimal amount of initially infecte... | [
"Notice, that with a contamination of one cell, the perimeter of infected area decreases at least by 1. Let $k$ cells be infected at the beginning. Then the perimeter is at most $3k$. It takes $n^2 - k$ contaminations to get the whole triangle infected. The perimeter of the (infected) area is then $3n$. Thus $3n \\... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2013 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 45 | |
0chs | Let $n \in \mathbb{N}^*$. Determine all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy:
$$
f(x + y^{2n}) = f(f(x)) + y^{2n-1}f(y),
$$
for all $x, y \in \mathbb{R}$, and for which the equation $f(x) = 0$ has a unique solution. | [
"For $y = 0$, the given relation reduces to $f(x) = f(f(x))$, which means that the given relation becomes:\n$$\nf(x + y^{2n}) = f(x) + y^{2n-1}f(y), \\quad \\forall x, y \\in \\mathbb{R}. \\qquad (1)\n$$\nIf we consider $x = 0$ and $y = 1$ in (1), then $f(0) = 0$. Since the equation $f(x) = 0$ has a unique solution... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x | |
0ksn | Let $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$, $i$ be distinct integers from $1$ to $9$. The minimum possible positive value of
$$
\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}
$$
can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | [
"Solution:\nFirst consider the case when $abc = def + 1$. Let $X = abc$. Then\n$$\n\\frac{abc - def}{ghi} = \\frac{1}{ghi} = \\frac{abcdef}{9!} = \\frac{X \\cdot (X-1)}{9!}\n$$\nBecause $X > 1$, this is an increasing function of $X$. Note that $X(X - 1) = abcdef \\ge 6!$, and therefore $X \\ge 28$. Either $X$ or $X... | United States | 2022 AIME I | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 289 | |
0cog | Determine if there exist non-zero real numbers $a_1, a_2, \dots, a_{10}$ such that
$$ \left(a_1 + \frac{1}{a_1}\right) \cdots \left(a_{10} + \frac{1}{a_{10}}\right) = \left(a_1 - \frac{1}{a_1}\right) \cdots \left(a_{10} - \frac{1}{a_{10}}\right). $$
Существуют ли такие ненулевые действительные числа $a_1, a_2, \dots, ... | [
"Рассмотрим произвольные ненулевые числа $a_1, \\ldots, a_{10}$. Заметим, что числа $a_k$ и $\\frac{1}{a_k}$ имеют одинаковый знак. Значит,\n$$\n\\left| a_k + \\frac{1}{a_k} \\right| = \\left| a_k \\right| + \\frac{1}{\\left| a_k \\right|} > \\max \\left( \\left| a_k \\right|, \\frac{1}{\\left| a_k \\right|} \\righ... | Russia | Final round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Other"
] | English; Russian | proof and answer | No | |
07a5 | Let $f: \mathbb{R}^{\ge 0} \to \mathbb{R}^{\ge 0}$ be a function such that for all $a, b \in \mathbb{R}^{\ge 0}$:
i) $f(a) = 0 \Leftrightarrow a = 0$.
ii) $f(ab) = f(a)f(b)$.
iii) $f(a + b) \le 2 \max\{f(a), f(b)\}$.
Prove that for every $a,b \in \mathbb{R}^{\ge 0}$, $f(a+b) \le f(a) + f(b)$. | [
"We claim that for every $k \\in \\mathbb{N}$ and real numbers $a_1, a_2, \\dots, a_{2^k}$:\n$$\nf(a_1 + a_2 + \\cdots + a_{2^k}) \\le 2^k \\max\\{f(a_1), f(a_2), \\dots, f(a_{2^k})\\}\n$$\nProof is done by induction on $k$. Basis is obviously the condition (iii). Suppose the claim is true for $k$. For $k+1$ we hav... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
01b3 | For fixed positive integers $a$ and $b$, find all strictly increasing functions $f$ from positive integers to positive integers such that, for any positive integer $n > a$,
$$ f(f(n - a) + n) = n + b. $$ | [
"**Answer:** There are no such functions.\n\nObserve that $f(n) \\ge n$ for each $n$, since $f$ is strictly increasing. Consequently, $n + b = f(f(n - a) + n) \\ge f(n - a) + n$. Hence, $b \\ge f(n - a) \\ge n - a$ holds for each positive integer $n > a$, which is impossible. ▼"
] | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | No such functions exist. | |
09zl | In a tournament with the four teams $A$, $B$, $C$ and $D$, every team played against every other team in three rounds of two simultaneous games. No team won or lost all their games and no game ended in a draw. It is known that team $A$ won in the first and third round. Also, team $C$ won in the first round and team $D$... | [] | Netherlands | First Round | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | MCQ | B | |
04mo | Marin labels the vertices of a cube with numbers $1, 2, \ldots, 8$, and then labels each edge with the sum of numbers in the vertices joined by that edge. Can he arrange the numbers in the vertices so that all the edges are labelled with different numbers? | [] | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Other"
] | English | proof only | null | |
084l | Problem:
Cinque amici fanno, rispettivamente, le seguenti affermazioni.
"Comunque si scelga uno di noi, gli altri 4 mentono".
"Comunque si scelga uno di noi, gli altri 4 dicono il vero".
"Comunque si scelga uno di noi, ce n'è un altro che dice il vero".
"C'è uno di noi tale che ogni altro dice il vero".
"C'è uno di no... | [
"Solution:\n\nLa risposta è (A). La prima affermazione non può essere vera, perché scegliendo uno degli altri si autocontraddice.\nQuindi anche la seconda non può essere vera, perché scegliendo chi la dice, vi è il primo che mente. Quindi anche la quarta non può essere vera, perché ve ne sono già due false.\nInfine... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | A | |
02np | Problem:
Operação - Dados dois números reais $a$ e $b$, considere $a b = a^{2} - a b + b^{2}$. Quanto vale $1$?
(a) 1
(b) 0
(c) 2
(d) -2
(e) -1 | [
"Solution:\n\nFazendo $a = 1$ e $b = 0$ em $a b = a^{2} - a b + b^{2}$, obtemos $1 = 1^{2} - 1 \\times 0 + 0^{2} = 1$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | (a) | |
0dtk | Prove that for every positive integer $n$, there is an unique $n$-digit integer $A(n)$ which is a multiple of $5^n$ and whose digits are all odd. | [
"We shall prove by induction.\nThe assertion is true for $n=1$ as $A(1) = 5$.\nAssume that the assertion is true when $n = m$.\nLet $A(m) = \\overline{a_1a_2\\cdots a_m}$ be the unique integer with all digits odd such that $5^m \\mid A(m)$.\nWhen $n = m+1$, let $A(1, m) = \\overline{1a_1a_2\\cdots a_m}$, $A(3, m) =... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
046c | (1) Given coprime positive integers $a, b$. Prove: There exist real numbers $\lambda, \beta$ such that for any positive integer $m$, it holds that
$$
\lambda m - \beta \le \sum_{k=1}^{m-1} \left\{ \frac{ak}{m} \right\} \cdot \left\{ \frac{bk}{m} \right\} \le \lambda m + \beta.
$$
(2) Prove: there exists a positive int... | [
"(1) When $m$ is large, we expect that the summation in (i) is very close to the following integral:\n$$\n\\int_{0}^{m} \\left\\{ \\frac{ax}{m} \\right\\} \\left\\{ \\frac{bx}{m} \\right\\} dx = m \\cdot \\int_{0}^{1} \\{ax\\}\\{bx\\}dx.\n$$\nHence, we will prove the inequality (i) for\n$$\n\\lambda = \\int_{0}^{1}... | China | 2023 Chinese IMO National Team Selection Test | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and ... | English | proof only | null | |
0kqo | Problem:
Let $A$ and $B$ be diagonally opposite vertices of a cube. An ant is crawling on a cube starting from $A$, and each second it moves at random to one of the three vertices adjacent to its current one. Find the expected number of steps for the ant to get to vertex $B$. | [
"Solution:\n\nLet $C_{1}, C_{2}, C_{3}$ be the vertices of the cube adjacent to $A$, and $D_{1}, D_{2}, D_{3}$ the vertices adjacent to $B$. Let $x$ be expected time to get to $B$ starting from $A$. From $A$, you have to go to $C_{1}, C_{2}$, or $C_{3}$, so the expected time to get to $B$ from one of these vertices... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 10 | |
0dpt | A sequence $s$ consisting of zeroes and ones is given. For each positive integer $k$ we define $v_k$ as the maximum number of ways to find consecutive digits forming the sequence $s$ in a sequence of $k$ digits. (For example, if $s = 0110$, then $v_7 = v_8 = 2$, since in the sequences 0110110 and 01101100 consecutive d... | [
"Let the sequence $s = c_1c_2 \\dots c_m$ have length $m$, and some sequence $X$ of length $n$ be $a_1a_2 \\dots a_n$. If two occurrences of $s$ in $X$ begin with $a_p$ and $a_{p+k}$, then $a_{p+i} = a_{p+i+k} = c_{i+1}$ for $0 \\le i < m$. When $k \\le m$ this means that $c_i = c_{i+k}$ for $1 \\le i \\le m-k$, th... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICS COMPETITION XX | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other",
"Discrete Mathematics > Other"
] | English | proof only | null | |
0i9e | Problem:
Farmer John is inside of an ellipse with reflective sides, given by the equation $x^{2} / a^{2} + y^{2} / b^{2} = 1$, with $a > b > 0$. He is standing at the point $(3, 0)$, and he shines a laser pointer in the $y$-direction. The light reflects off the ellipse and proceeds directly toward Farmer Brown, traveli... | [
"Solution:\nThe points where the farmers are standing must be the foci of the ellipse, so they are $(3, 0)$ and $(-3, 0)$. If the total distance traveled is $10$, then $a$ must be half of that, or $5$, since the distance traveled by a ray reflecting off the wall from when it leaves one focus to when it reaches the ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (5, 4) | |
0l8t | Let be given a sequence of positive integers $\{a_n\}$, $n = 1, 2, 3, \ldots$, satisfying the condition
$$
0 < a_{n+1} - a_n \le 2001
$$
for every $n = 1, 2, 3, \ldots$. Prove that there exists an infinite number of couples of positive integers $(p, q)$ such that $p < q$ and $a_p$ is a divisor of $a_q$. | [] | Vietnam | VIETNAMESE MATHEMATICAL COMPETITION FOR TEAM SELECTION | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0bbg | Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that, on each non degenerated interval $I$, the function reaches its maximum or its minimum in an interior point of $I$. Prove that $f$ is a constant. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Precalculus > Functions"
] | null | proof only | null | |
064q | A triangle $AB\Gamma$ is given with $\hat{A} = 105°$ and $\hat{\Gamma} = \frac{\hat{B}}{4}$.
α. Determine the measures of the angles $\hat{B}$ and $\hat{\Gamma}$.
β. If $O$ is the center of the circumcircle of the triangle $AB\Gamma$ and $\Delta$ is the antipodal of $B$, prove that the distance of $\Gamma$ from $B\De... | [
"α. Since $\\hat{A} + \\hat{B} + \\hat{\\Gamma} = 180°$ and $\\hat{A} = 105°$, $\\hat{\\Gamma} = \\frac{\\hat{B}}{4}$, we have\n$$\n105° + \\hat{B} + \\frac{\\hat{B}}{4} = 180° \\Leftrightarrow \\hat{B} = 60°, \\text{ and hence } \\hat{\\Gamma} = 15°.\n$$\n\nβ. Since $OB = O\\Delta = O\\Gamma$, it follows that $\\w... | Greece | Selection Examination for Juniors | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | B = 60°, Γ = 15°; and the perpendicular distance from Γ to the line BΔ equals BΔ/4. | |
0hzq | Problem:
A sphere of radius $1$ is covered in ink and rolling around between concentric spheres of radii $3$ and $5$. If this process traces a region of area $1$ on the larger sphere, what is the area of the region traced on the smaller sphere? | [
"Solution:\nThe figure drawn on the smaller sphere is just a scaled down version of what was drawn on the larger sphere, so the ratio of the areas is the ratio of the surface area of the spheres. This is the same as the ratio of the squares of the radii, which is $\\frac{9}{25}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | final answer only | 9/25 | |
0da6 | Find all positive integers $n$ such that $\varphi(n)$ is a divisor of $n^{2}+3$. | [
"First assume $n$ is a prime number. Thus $\\varphi(n)=n-1 \\mid n^{2}+3=(n-1)(n+1)+4$, implying $n-1 \\mid 4$. We deduce that $n=2,3$ or $5$.\n\nFrom now on, assume $n$ is composite and set $n=\\prod_{i=1}^{k} p_{i}^{\\alpha_{i}}$, where $k$ is the number of distinct prime divisors of $n$. Since $n \\geq 3$, $\\va... | Saudi Arabia | Team selection tests for BMO 2018 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | {1, 2, 3, 5, 9, 21} | |
05ec | Problem:
Let $ABC$ be an acute triangle. Points $B$, $D$, $E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Let $H$ be the orthocentre of triangle $ADE$. Let $P$ and $Q$ be points on lines $BM$ and $CN$, respectively, such that $D$, $H... | [
"Solution:\n\nDenote by $B'$ and $C'$ the reflections of $B$ and $C$ in $M$ and $N$, respectively. Points $C'$, $A$, $B'$ are clearly collinear and $DEB'A$ is a parallelogram. Since $EH \\perp AD$, we have $EH \\perp EB'$. Also $HA \\perp AB'$, so points $H, E, B', A$ are concyclic. This gives\n\n$$\n\\angle C'QH =... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, n... | null | proof only | null | |
0jnp | Problem:
Find the smallest integer $n \geq 5$ for which there exists a set of $n$ distinct pairs $(x_{1}, y_{1}), \ldots, (x_{n}, y_{n})$ of positive integers with $1 \leq x_{i}, y_{i} \leq 4$ for $i=1,2, \ldots, n$, such that for any indices $r, s \in \{1,2, \ldots, n\}$ (not necessarily distinct), there exists an in... | [
"Solution:\n\nAnswer: $8$\n\nIn other words, we have a set $S$ of $n$ pairs in $(\\mathbb{Z} / 4 \\mathbb{Z})^{2}$ closed under addition. Since $1+1+1+1 \\equiv 0 \\pmod{4}$ and $1+1+1 \\equiv -1 \\pmod{4}$, $(0,0) \\in S$ and $S$ is closed under (additive) inverses. Thus $S$ forms a group under addition (a subgrou... | United States | HMMT February | [
"Algebra > Abstract Algebra > Group Theory",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 8 | |
0gy9 | A board $2009 \times 2009$ is divided into unit squares. Two players play the following game. They take turns to paint in yellow colour an unpainted unit segment which is a side of a unit square. If after a turn a unit square with all four sides painted in yellow is obtained, then the player who made this turn wins. Wh... | [
"The second player can follow the following rules for his steps.\n\n1) If he can, he paints the side which is the fourth yellow side in some square. He does it and he wins in this game.\n\n2) If the first player paints some side, the second player paints the side which is symmetric to the one just painted about the... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | The second player has a winning strategy. | |
03pv | Let $x \in \left[-\frac{5\pi}{12}, -\frac{\pi}{3}\right]$. Then the maximum value of
$$
y = \tan\left(x + \frac{2\pi}{3}\right) - \tan\left(x + \frac{\pi}{6}\right) + \cos\left(x + \frac{\pi}{6}\right)
$$
is ( ).
(A) $\frac{12}{5}\sqrt{2}$
(B) $\frac{11}{6}\sqrt{2}$
(C) $\frac{11}{6}\sqrt{3}$
(D) $\frac{12}{5}\sqrt{3}... | [
"Let $z = -x - \\frac{\\pi}{6}$. Then $z \\in [\\frac{\\pi}{6}, \\frac{\\pi}{4}]$, and $2z \\in [\\frac{\\pi}{3}, \\frac{\\pi}{2}]$. We have\n$$\n\\tan\\left(x + \\frac{2\\pi}{3}\\right) = -\\cot\\left(x + \\frac{6}{\\pi}\\right) = \\cot z.\n$$\nThen\n$$\ny = \\cot z + \\tan z + \\cos z = \\frac{2}{\\sin 2z} + \\co... | China | China Mathematical Competition (Shaanxi) | [
"Precalculus > Trigonometric functions"
] | English | MCQ | C | |
0ano | Problem:
How many perfect squares divide the number $2!5!6!$?
(a) 18
(b) 15
(c) 20
(d) 25 | [] | Philippines | Qualifying Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | c | |
0fnz | Sean $B$ y $C$ dos puntos fijos de una circunferencia de centro $O$, que no sean diametralmente opuestos. Sea $A$ un punto variable sobre la circunferencia, distinto de $B$ y $C$, y que no pertenece a la mediatriz de $BC$. Sean $H$, el ortocentro del triángulo $ABC$; y $M$ y $N$ los puntos medios de los segmentos $BC$ ... | [] | Spain | L Olimpiada Matemática Española | [
"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"
] | Spanish | proof only | The locus of P is the ellipse with foci at O and M consisting of all points X such that the sum of the distances from X to O and to M equals the circumradius of the circle. | |
01cu | Find all positive integers $n$ for which
$$
3x^n + n(x + 2) - 3 \ge nx^2
$$
holds for all real numbers $x$. | [
"**Answer:** The inequality holds if and only if $n$ is even.\n\nFirst suppose that $n$ is odd. Setting $x = -1$ in the inequality, the left-hand side becomes $3 \\cdot (-1)^n + n - 3 = n - 6$, while the right-hand side becomes $n \\cdot (-1)^2 = n$, and this is a contradiction.\n\nThen let $n$ be even. Since $|x| ... | Baltic Way | Baltic Way 2016 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | all even positive integers | |
01f4 | Does there exist a positive integer $n$ such that the first and the second digit of $2^n$ (in decimal notation) are 3 and 9, respectively? | [
"Yes. Let us observe powers of two of the form $2^{10k} = (2^{10})^k = 1024^k = (1,024)^k \\cdot 10^{3k}$. Let $m$ be the smallest integer such that $(1,024)^m \\ge 3,9$ ($m$ must exist since $(1,024)^k$ is an exponential function and is unbounded from above). From the definition of $m$ we know that $(1,024)^{m-1} ... | Baltic Way | Baltic Way 2019 | [
"Number Theory > Other",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Yes | |
0je0 | Problem:
The walls of a room are in the shape of a triangle $ABC$ with $\angle ABC = 90^{\circ}$, $\angle BAC = 60^{\circ}$, and $AB = 6$. Chong stands at the midpoint of $BC$ and rolls a ball toward $AB$. Suppose that the ball bounces off $AB$, then $AC$, then returns exactly to Chong. Find the length of the path of ... | [
"Solution:\n\nLet $C'$ be the reflection of $C$ across $AB$ and $B'$ be the reflection of $B$ across $AC'$. Note that $B'$, $A$, $C$ are collinear by angle chasing. The image of the path under these reflections is just the line segment $MM'$, where $M$ is the midpoint of $BC$ and $M'$ is the midpoint of $B'C'$. So ... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3*sqrt(21) | |
04dx | Prove that the equation
$$
3x^4 + 2013 = 25y^2 - 24x^2
$$
has no integer solutions. | [
"Rewrite the equation as:\n$$\n3x^4 + 2013 + 24x^2 = 25y^2\n$$\nwhich gives\n$$\n25y^2 = 3x^4 + 24x^2 + 2013\n$$\nSo $25y^2$ is congruent to $3x^4 + 24x^2 + 2013$.\n\nLet us consider the equation modulo $25$:\n$$\n3x^4 + 24x^2 + 2013 \\equiv 0 \\pmod{25}\n$$\nNote that $2013 \\equiv 13 \\pmod{25}$, so:\n$$\n3x^4 + ... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0c04 | Let $ABCD$ be a cyclic quadrangle and let $P$ be a point on the side $AB$. The diagonal $AC$ crosses the segment $DP$ at $Q$. The parallel through $P$ to $CD$ crosses the extension of the side $BC$ beyond $B$ at $K$, and the parallel through $Q$ to $BD$ crosses the extension of the side $BC$ beyond $B$ at $L$. Prove th... | [
"We show that the circles $BKP$ and $CLQ$ are tangent at the point $T$ where the line $DP$ crosses the circle $ABCD$ again.\nSince $BCDT$ is cyclic, we have $\\angle KBT = \\angle CDT$. Since $KP \\parallel CD$, we get $\\angle CDT = \\angle KPT$. Thus, $\\angle KBT = \\angle CDT = \\angle KPT$, which shows that $T... | Romania | Eleventh Romanian Master of Mathematics | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
004j | En un año que tiene 53 sábados, ¿qué día de la semana es el 12 de mayo?
Dar todas las posibilidades. | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Discrete Mathematics > Other"
] | Español | proof and answer | Thursday or Friday | |
097h | Problem:
Fie numerele complexe $z_{1}, z_{2}, z_{3}$, astfel încât $|z_{1}|=|z_{2}|=|z_{3}|=1$ şi $z_{1} z_{2} z_{3} \neq -1$. Arătați că
$$
w=\frac{z_{1}+z_{2}+z_{3}+z_{1} z_{2}+z_{1} z_{3}+z_{2} z_{3}}{1+z_{1} z_{2} z_{3}}
$$
este un număr real. | [
"Solution:\nObservăm că\n$$\n|z_{i}|=1 \\Rightarrow z_{i} \\bar{z}_{i}=1, \\ i=1,2,3\n$$\nAtunci\n$$\n\\begin{aligned}\n& \\bar{w}=\\overline{\\left(\\frac{z_{1}+z_{2}+z_{3}+z_{1} z_{2}+z_{1} z_{3}+z_{2} z_{3}}{1+z_{1} z_{2} z_{3}}\\right)}=\\frac{\\overline{z_{1}}+\\overline{z_{2}}+\\overline{z_{3}}+\\overline{z_{... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0jrv | Problem:
Rachel measures the angles of a certain pentagon $ABCD E$ in degrees. She finds that $\angle A < \angle B < \angle C < \angle D < \angle E$, and also that the angle measures form an arithmetic progression, meaning that $\angle B - \angle A = \angle C - \angle B = \angle D - \angle C = \angle E - \angle D$.
Wh... | [
"Solution:\n\nThe answer is $108$ degrees. Indeed, in a pentagon the sum of the angles is $180 \\cdot 3 = 540$ degrees. Also, in an arithmetic progression the middle term is the average, which is $\\frac{1}{5} \\cdot 540 = 108$."
] | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 108 | |
00um | Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$ lying in the interior. Let $E$ and $F$ be the midpoints of the segments $BC$ and $AD$, respectively. Let $X$ be the point lying on the same side of the line $EF$ as the vertex $C$ such that $\triangle EXF$ and $\triangle BOA$ are similar. Prove that $XC = XD$. | [
"Let $M$ be the midpoint of $CD$. Let $EM$ intersect $AD$ at $Z$ and $FM$ intersect $BC$ at $Y$. As $F$, $M$, and $E$ are midpoints of $AD$, $DC$, and $BC$ respectively, we have that $FM \\parallel AC$ and $EM \\parallel BD$. Thus $\\angle EZF = \\angle BDA = \\angle ACB = \\angle FYE$, hence $FEYZ$ is cyclic.\n\nN... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08rc | Find every $f : \mathbb{R} \to \mathbb{R}$ such that for any $x$ and $y$,
$$
f(x)^2 + 2y f(x) + f(y) = f(y + f(x)).
$$ | [
"The given equation clearly holds if $f(x) = 0$ for all $x$. Assume that there exists $a$ such that $f(a) \\neq 0$.\nSubstituting $y = -f(x)$ into the given equation and letting $c = f(0)$, we obtain\n$$\nf(-f(x)) = c + f(x)^2. \\quad (1)\n$$\nSubstituting $y = -f(y)$ into the equation and using (1),\n$$\n\\begin{a... | Japan | The 16th Japanese Mathematical Olympiad - The Final Round | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | Either the identically zero function, or the family f(x) = x^2 + c for any real constant c. | |
0c2y | Fix a circle $\Gamma$, a line $\ell$ tangent to $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ cross $\ell$ at $Y$ and $Z$. Prove that, as $X$ traces $\Omega$, the circle $XYZ$ is ... | [
"Assume $\\Gamma$ of unit radius and invert with respect to $\\Gamma$. No reference will be made to the original configuration, so images will be denoted by the same letters. Letting $\\Gamma$ be centered at $G$, notice that inversion in $\\Gamma$ maps tangents to $\\Gamma$ to circles of unit diameter through $G$ (... | Romania | Eleventh Romanian Master of Mathematics | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Ve... | null | proof only | null | |
051r | Find the smallest natural number $n$ for which there exist integers $a_1, \dots, a_n$ (that do not have to be different) such that $a_1^4 + \dots + a_n^4 = 2013$. | [
"**Answer:** 14.\n\nNote that the fourth powers of even numbers are divisible by 16 and the fourth powers of odd numbers are congruent to 1 modulo 16. As $2013 \\equiv 13 \\pmod{16}$, the desired representation must contain at least 13 odd summands.\n\nSuppose that no more summands are needed. As $7^4 = 2401 > 2013... | Estonia | Final Round of National Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 14 | |
08ex | Problem:
I partecipanti a un convegno di furfanti (che mentono sempre) e cavalieri (che dicono sempre la verità) sono numerati da $1$ a $2021$. Ciascuno di essi dichiara: "si possono formare almeno $i$ terne di partecipanti di cui io faccia parte e che contengano esattamente due cavalieri", dove $i$ è il numero assegn... | [
"Solution:\n\nLa risposta è (B). Dividiamo in casi a seconda del numero di cavalieri:\n\na. La configurazione con $0$ cavalieri, ovvero tutti furfanti, è sicuramente valida.\n\nb. Una configurazione con esattamente $1$ cavaliere è impossibile, perché questi affermerebbe di poter trovare almeno un altro cavaliere.\n... | Italy | Italian Mathematical Olympiad - February Round | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | MCQ | B | |
0bg8 | Problem:
a) Demonstraţi că, oricare ar fi numărul real $x$, are loc inegalitatea
$$
x^{4}-x^{3}-x+1 \geq 0
$$
b) Rezolvaţi în mulţimea numerelor reale sistemul
$$
\left\{
\begin{array}{l}
x_{1}+x_{2}+x_{3}=3 \\
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=x_{1}^{4}+x_{2}^{4}+x_{3}^{4}
\end{array}
\right.
$$ | [] | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities"
] | null | proof and answer | x1 = x2 = x3 = 1 | |
0ghy | 是否存在不等邊三角形 $ABC$, 使得三角形 $ABC$ 與三角形 $IHO$ 相似, 其中 $I, H, O$ 分別為三角形 $ABC$ 的內心、垂心及外心?
Is there a scalene triangle $ABC$ similar to triangle $IHO$, where $I$, $H$, $O$ are the incenter, orthocenter, and circumcenter, respectively, of triangle $ABC$? | [
"由於 $ABC$ 為不等邊三角形, 不妨假設 $\\angle A > \\angle B > \\angle C$。事實上, 我們可以證明 $\\angle OIH > \\angle A$。\n\n**Claim.** $A, H$ 位於 $OI$ 同側。\n\n*Proof.* 令 $M$ 為 $AI$ 與外接圓 $\\odot(ABC)$ 的第二個交點, $J$ 為 $\\triangle ABC$ 的 $A$-旁心, $S$ 為 $OI$ 與 $AH$ 的交點。雞爪定理告訴我們 $M$ 為 $IJ$ 中點。考慮 $I, M, I_a$ 關於 $AB$ 的投影點 $F_I, F_M, F_J$, 我們有\n$$\n... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"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 > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometr... | Chinese (Traditional) | proof and answer | No | |
04sr | In the field of real numbers solve a system of equations
$$
\begin{align*}
a(b^2 + c) &= c(c + ab), \\
b(c^2 + a) &= a(a + bc), \\
c(a^2 + b) &= b(b + ca).
\end{align*}
$$ | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | All real triples with a=b=c (any real value). | |
0jlu | Problem:
Let $n$ be a positive integer. A sequence $(a_{0}, \ldots, a_{n})$ of integers is acceptable if it satisfies the following conditions:
a. $0 = |a_{0}| < |a_{1}| < \cdots < |a_{n-1}| < |a_{n}|$.
b. The sets $\{ |a_{1} - a_{0}|, |a_{2} - a_{1}|, \ldots, |a_{n} - a_{n-1}| \}$ and $\{ 1, 3, 9, \ldots, 3^{n-1} \... | [
"Solution:\n\nWe actually prove a more general result via strong induction on $n$.\n\nFirst, we state the more general result we wish to prove.\n\nFor $n > 0$, define a great sequence to be a sequence of integers $(a_{0}, \\ldots, a_{n})$ such that\n\n1. $0 = |a_{0}| < |a_{1}| < \\cdots < |a_{n-1}| < |a_{n}|$\n\n2.... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (n+1)! | |
01lk | Non-zero real numbers $a, b, c$ satisfy the equality
$$
\frac{ab}{b-c} + \frac{bc}{c-a} + \frac{ca}{a-b} = \frac{ab}{b+c} + \frac{bc}{c+a} + \frac{ca}{a+b} + 6abc.
$$
Find all possible values of the expression
$$
\frac{1}{(a^2 - b^2)^2} + \frac{1}{(b^2 - c^2)^2} + \frac{1}{(c^2 - a^2)^2}
$$ | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 9 | |
0kgk | Problem:
Amelia wrote down a sequence of consecutive positive integers, erased one integer, and scrambled the rest, leaving the sequence below. What integer did she erase?
$$
6,12,1,3,11,10,8,15,13,9,7,4,14,5,2
$$ | [
"Solution:\n\nThe sequence of positive integers exactly contains every integer between $1$ and $15$, inclusive. $16$ is the only positive integer that could be added to this sequence such that the resulting sequence could be reordered to make a sequence of consecutive positive integers. Therefore, Amelia must have ... | United States | HMMT Spring 2021 Guts Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 16 | |
0bzx | Determine the least real number $c$ satisfying the condition $\sum_{k=1}^{n} x_{k}^{2} \le cn$, for all positive integers $n$ and all real numbers $x_1, \dots, x_n$ greater than or equal to $-1$ such that $\sum_{k=1}^{n} x_{k}^{3} = 0$. | [
"The required number is $c = 4/3$. We first show that if $n$ is a positive integer and $x_1, \\dots, x_n$ are real numbers greater than or equal to $-1$ such that $\\sum_{k=1}^{n} x_k^3 = 0$, then $\\sum_{k=1}^{n} x_k^2 \\le 4n/3$. Indeed, since $x_k^3 - 3x_k^2 + 4 = (x_k + 1)(x_k - 2)^2 \\ge 0$, $k = 1, \\dots, n$... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | c = 4/3 | |
0351 | Problem:
Let $ABC$ be an isosceles triangle such that $AC = BC = 1$ and $AB = 2x$, $x > 0$.
a) Express the inradius $r$ of $\triangle ABC$ as a function of $x$.
b) Find the maximum possible value of $r$. | [
"Solution:\n\na) It follows from the Pythagorean theorem that the altitude of $\\triangle ABC$ through $C$ is equal to $\\sqrt{1 - x^2}$. Then\n$$\nr = \\frac{S}{p} = \\frac{x \\sqrt{1 - x^2}}{1 + x} = x \\sqrt{\\frac{1 - x}{1 + x}}\n$$\n\nb) We have to find the maximum of the function\n$$\nf(x) = \\frac{x^2 (1 - x... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | r(x) = x sqrt((1 - x)/(1 + x)); the maximum inradius is sqrt((5 sqrt(5) - 11)/2), attained at x = (sqrt(5) - 1)/2 | |
0cd3 | Let $a$, $b$, $c$, $d$ be real numbers such that $ab(c+d) = cd(a+b)$. Prove that
$$
\frac{a+1}{a^2+3} + \frac{b+1}{b^2+3} \ge \frac{c-1}{c^2+3} + \frac{d-1}{d^2+3}.
$$ | [
"Replace $(c, d)$ with $(-c, -d)$ to rephrase the problem as: subject to the constraint $ab(c+d) + cd(a+b) = 0$, prove that $\\sum \\frac{a+1}{a^2+3} \\ge 0$ (1).\n\nSuppose one of the numbers equals $0$ and notice that (at least) another is $0$ as well – let them be $c$ and $d$. The claim (1) rewrites as\n$$\n\\fr... | Romania | THE Sixteenth STARS OF MATHEMATICS Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0buz | Problem:
a) Să se demonstreze că $\frac{1}{2}(a+b)^{2}+\frac{1}{4}(a+b) \geq a \sqrt{b}+b \sqrt{a}$, $a, b \geq 0$
b) Să se arate că pentru $\forall x, y, z \in\left[0, \frac{\pi}{2}\right)$, are loc inegalitatea:
$$
\begin{aligned}
& (\operatorname{tg} x+\operatorname{tg} y)^{2}+(\operatorname{tg} y+\operatorname{tg... | [] | Romania | Olimpiada de Matematică Etapa Locală | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0g5g | 令實數 $a, b, c, d$ 滿足 $a + b + c + d = 6$ 與 $a^2 + b^2 + c^2 + d^2 = 12$. 試證
$$
36 \le 4(a^3 + b^3 + c^3 + d^3) - (a^4 + b^4 + c^4 + d^4) \le 48.
$$ | [
"觀察\n$$\n\\begin{aligned}\n& 4(a^{3} + b^{3} + c^{3} + d^{3}) - (a^{4} + b^{4} + c^{4} + d^{4}) \\\\\n&= -((a - 1)^{4} + (b - 1)^{4} + (c - 1)^{4} + (d - 1)^{4}) \\\\\n& \\quad +6(a^{2} + b^{2} + c^{2} + d^{2}) - 4(a + b + c + d) + 4 \\\\\n&= -((a - 1)^{4} + (b - 1)^{4} + (c - 1)^{4} + (d - 1)^{4}) + 52.\n\\end{ali... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
04ur | Is it possible to fill the $8 \times 8$ table with numbers $6$ and $7$ such that the sum of the numbers in each column is a multiple of $5$ and the sum of the numbers in each row is a multiple of $7$? (Josef Tkadlec) | [] | Czech Republic | Second Round | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | No | |
0dtl | Suppose for some positive integer $n$, the numbers $2^n$ and $5^n$ have equal first digit. What are the possible values of this first digit? | [
"From $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 16$, $2^5 = 32$, $5^1 = 5$, $5^2 = 25$, $5^3 = 125$, $5^4 = 625$, $5^5 = 3125$, $3$ is a possible value. We shall show that there are no others.\n\nSuppose the first digits of $2^n$ and $5^n$ are both $a$ and that they have $s$ and $t$ digits, respectively.\n\nWhen $n >... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3 | |
04wv | Find out if there is a convex pentagon $A_1A_2A_3A_4A_5$, such that for $i = 1, 2, 3, 4, 5$ the lines $A_iA_{i+3}$, $A_{i+1}A_{i+2}$ are not parallel and intersect in a point $B_i$ and also the points $B_1, B_2, B_3, B_4, B_5$ are collinear. (We assume $A_6 = A_1, A_7 = A_2, A_8 = A_3$.) | [
"We will find such a pentagon. The obstruction is, symmetric pentagons (for which it could be easier to show) have always at least one pair of lines $A_iA_{i+3}$, $A_{i+1}A_{i+2}$ parallel. First we will solve more elementary problem - we will find pentagon $A_1A_2A_3A_4A_5$ with only four points $B_i$ collinear. H... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2006 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
017h | For any positive integer $n$, define
$$
X_n = \frac{n!}{2010n^2 + 2010n + 1}
$$
Show that $X_n$ is an integer for infinitely many $n$. | [
"The idea is to find infinitely many $n$ such that $2010n^2 + 2010n + 1$ has a factor that is close to, but not greater than $n$, and then close the deal by finding some additional finite factors, resulting in a factorization of $2010n^2 + 2010n + 1$.\nIf $n$ is big, we can, for any (small) $k$ reduce the number $2... | Baltic Way | BALTIC WAY | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
01nx | Ten points are marked in the plane so that no three of them lie on the same straight line. All points are connected with segments. Each of these segments is painted one of the $k$ colors.
For what positive integer $k$ ($1 \le k \le 5$) is it possible to paint the segments so that for any $k$ of the given 10 points ther... | [
"Answer: $k=5$.\n\nFirst we show that for $1 \\le k \\le 4$ the required colouring does not exist.\n\n1. There is nothing to prove for $k=1$ and $k=2$.\n\n2. Let $k=3$. Consider one (say $A$) of these 10 points. We have 3 colours and 9 segments connecting $A$ with other points. So there are two segments (say $AB$ a... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 5 | |
0l98 | For each integer $n > 1$, denote by $s_n$ the number of permutations $(a_1, a_2, ..., a_n)$ of $n$ first positive integers such that each permutation satisfies the condition:
$$
1 \le |a_k - k| \le 2 \quad \text{for every } k = 1, 2, ..., n.
$$
Prove that: $1.75 \cdot s_{n-1} < s_n < 2 \cdot s_{n-1}$ for all integers $... | [
"• Firstly, we construct an inductive relation for $s_n$.\nConsider an integer $n > 4$. Denote by $P_n$ the set of permutations $(a_1, a_2, \\dots, a_n)$ satisfying the condition of the problem.\nConsider the partition:\n$$\nP_n = \\bigcup_{k=1}^{n} S_k \\qquad (1)\n$$\nwhere $S_k$ is the set of permutations $(a_1,... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
02qi | Problem:
Uma caixa contém 105 bolas pretas, 89 bolas cinzentas e 5 bolas brancas. Fora da caixa há bolas brancas em quantidade suficiente para efetuar repetidamente o seguinte procedimento, até que sobrem duas bolas na caixa:
- retiram-se, sem olhar, duas bolas da caixa;
- se as bolas retiradas forem de cores diferent... | [
"Solution:\n\nQuando se retiram duas bolas pretas da caixa, elas não retornam; mas quando as bolas retiradas são uma preta e outra de cor distinta, a preta retorna. Isso mostra que o número de bolas pretas na caixa diminui de dois em dois. Como o número inicial de bolas pretas é ímpar, sempre haverá um número ímpar... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | MCQ | D | |
0iev | Problem:
Compute
$$
\sqrt[{2 \sqrt{2 \sqrt[3]{2 \sqrt[4]{2 \sqrt[5]{2 \cdots}}}}}]{ }
$$ | [
"Solution:\nTaking the base $2$ logarithm of the expression gives\n$$\n1+\\frac{1}{2}\\left(1+\\frac{1}{3}\\left(1+\\frac{1}{4}(1+\\cdots)\\right)\\right)=1+\\frac{1}{2!}+\\frac{1}{3!}+\\frac{1}{4!}+\\cdots=e-1.\n$$\nTherefore the expression is just $2^{e-1}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2^(e-1) | |
09wm | Given a positive integer $n$, we denote by $n!$ (‘n factorial’) the number we get if we multiply all integers from $1$ to $n$. For example: $5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$.
a. Determine all integers $n$ with $1 \le n \le 100$ for which $n! \cdot (n+1)!$ is a perfect square. Also, prove that you have fou... | [
"a. We observe that $(n+1)! = (n+1) \\cdot n!$, and therefore that $n! \\cdot (n+1)! = (n!)^2 \\cdot (n+1)$. That product is a perfect square if and only if $n+1$ is a perfect square, since $(n!)^2$ is a perfect square. For $1 \\le n \\le 100$ this is the case for $n = 3, 8, 15, 24, 35, 48, 63, 80, 99$ (perfect squ... | Netherlands | Second Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a) n equals 3, 8, 15, 24, 35, 48, 63, 80, 99. b) No such positive integer exists. | |
0aq5 | Problem:
What is the remainder when the sum
$$
1^{5}+2^{5}+3^{5}+\cdots+2007^{5}
$$
is divided by $5$? | [
"Solution:\nBy Fermat's Little Theorem, we have $a^{5} \\equiv a \\pmod{5}$ for any integer $a$. Modulo $5$, we have\n$$\n1^{5}+2^{5}+3^{5}+\\cdots+2007^{5} \\equiv 1+2+3+\\cdots+2007 = 2007 \\cdot 1004.\n$$\nNow, $2007 \\equiv 2 \\pmod{5}$ and $1004 \\equiv 4 \\pmod{5}$, so\n$$\n2007 \\cdot 1004 \\equiv 2 \\cdot 4... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 3 | |
0fd7 | Problem:
Denotamos por $\mathbb{N}=\{1,2,3, \ldots\}$ el conjunto de números naturales excluido el cero y por $\mathbb{N}^{*}=\{0,1,2,3, \ldots\}$ el conjunto de números naturales incluido el cero. Encontrar todas las funciones $f: \mathbb{N} \rightarrow \mathbb{N}^{*}$ que sean crecientes, es decir $f(n) \geq f(m)$ s... | [
"Solution:\n\n- La función nula: $f(n)=0$, para todo $n \\in \\mathbb{N}$ verifica evidentemente lo anterior.\n\n- Sea $f$ una función no nula verificando las condiciones del enunciado. Entonces\n\n1. $f$ no es constante, ni está acotada. En efecto, si $f(a) \\neq 0$ entonces $f\\left(a^{n}\\right)=n f(a)>f(a)$ par... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(n) = 0 for all n in N | |
08va | Suppose for a quadrilateral $ABCD$, $\angle DAB = 90^\circ$, $\angle ABC = \angle BCD = 60^\circ$. If $AB = 5$ and $CD = 4$, what is the value of $BC$? Here for a line segment $XY$, its length is also denoted by $XY$.
 | [
"Let $E$ be the point of intersection of the lines $AB$ and $CD$. Then, since $\\angle EBC = \\angle ECB = 60^\\circ$, the triangle $EBC$ is an equilateral triangle. If we let $x = EA$, then the triangle $ADE$ is a right triangle with $\\angle EAD = 90^\\circ$ and since $\\angle AED = 60^\\circ$, we have $DE = 2x$.... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 6 | |
03be | Let $a$ be a positive integer. Denote by $\tau(a)$ and $\varphi(a)$ respectively the number of all positive integers that divide $a$ and the number of all positive integers not greater than $a$ and relatively prime to $a$. Find all positive integers $n$ having only two prime divisors and such that $\varphi(\tau(n)) = \... | [
"Let $n = p^k q^l$, where $p < q$ are prime numbers, $k, l \\in \\mathbb{N}$ and let $u = \\varphi(\\tau(n)), v = \\tau(\\varphi(n))$. We have that $u = \\varphi((k+1)(l+1))$ and $v = \\tau(p^{k-1} q^{l-1} (p-1)(q-1))$. Obviously $u < kl + k + l$ and\n$$\nv \\geq \\tau(p^{k-1} q^{l-1} (p-1)) + 1 = kl \\tau(p-1) + 1... | Bulgaria | Bulgaria | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | All such n are n = 2^{r^t - 1} · 3^{r - 1}, where r is a prime and t is a positive integer. | |
092v | Problem:
We consider the equation $a^{2}+b^{2}+c^{2}+n=a b c$, where $a, b, c$ are positive integers.
Prove:
(a) There are no solutions $(a, b, c)$ for $n=2017$.
(b) For $n=2016$, $a$ must be divisible by $3$ for every solution $(a, b, c)$.
(c) The equation has infinitely many solutions $(a, b, c)$ for $n=2016$. | [
"Solution:\n\n(a) We distinguish cases depending on the parity of $a, b, c$ :\n- If all three are odd, we have $a^{2}+b^{2}+c^{2}+2017 \\equiv 0\\pmod{2}$ and $a b c \\equiv 1\\pmod{2}$.\n- If exactly one of them is even, we have $a^{2}+b^{2}+c^{2}+2017 \\equiv 1\\pmod{2}$ and $a b c \\equiv 0$ $\\pmod{2}$.\n- If e... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
... | null | proof only | null | |
0aqx | Problem:
The line from the origin to the point $\left(1, \tan 75^\circ\right)$ intersects the unit circle at $P$. Find the slope of the tangent line to the circle at $P$. | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | √3 − 2 | |
0dra | Let $n = \overline{30x070y03}$ be a 9-digit integer. Find all possible values of the pair $(x, y)$, so that $n$ is a multiple of 37. | [
"We have\n$$\nn = 300070003 + 10^6x + 10^2y = 37(8110000 + 27027x + 3y) + (3 + x - 11y).\n$$\nSince $0 \\le x, y \\le 9$, we have $-96 \\le 3 + x - 11y \\le 12$. Also $37 \\mid 3 + x - 11y$. Thus $3 + x - 11y = 0, -37$ or $-74$ and we get $(x, y) = (8, 1), (4, 4), (0, 7)$."
] | Singapore | Singapur 2015 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | (x, y) = (8, 1), (4, 4), (0, 7) | |
0f7d | Problem:
Let $d(n)$ be the number of (positive integral) divisors of $n$. For example, $d(12) = 6$. Find all $n$ such that $n = d(n)^2$. | [] | Soviet Union | 20th ASU | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 1 and n = 9 | |
00xb | Problem:
Denote by $d(n)$ the number of all positive divisors of a positive integer $n$ (including $1$ and $n$). Prove that there are infinitely many $n$ such that $\frac{n}{d(n)}$ is an integer. | [
"Solution:\n\nConsider numbers of the form $p^{p^{n}-1}$ where $p$ is an arbitrary prime number and $n=1,2, \\ldots$"
] | Baltic Way | Baltic Way 1992 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0717 | Problem:
The numbers $1, 2, \ldots, 2002$ are written in order on a blackboard. Then the 1st, 4th, 7th, $\ldots$, $3k+1$th, $\ldots$ numbers in the list are erased. Then the 1st, 4th, 7th, $\ldots$, $3k+1$th numbers in the remaining list are erased (leaving $3, 5, 8, 9, 12, \ldots$). This process is carried out repeat... | [
"Solution:\n\nLet $a_n$ be the first number remaining after $n$ iterations, so $a_0 = 1$, $a_1 = 2$, $a_2 = 3$, $a_3 = 5$, etc. We claim that:\n\n$\\displaystyle a_{n+1} = \\frac{3}{2} a_n$ if $a_n$ is even, and\n$\\displaystyle a_{n+1} = \\frac{3}{2}(a_n + 1) - 1$ if $a_n$ is odd.\n\nWe use induction on $n$.\n\nSu... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1598 | |
0ifs | Problem:
Let $a_{1} = 3$, and for $n \geq 1$, let $a_{n+1} = (n+1) a_{n} - n$. Find the smallest $m \geq 2005$ such that $a_{m+1} - 1 \mid a_{m}^{2} - 1$. | [
"Solution:\nWe will show that $a_{n} = 2 \\cdot n! + 1$ by induction. Indeed, the claim is obvious for $n = 1$, and $(n+1)(2 \\cdot n! + 1) - n = 2 \\cdot (n+1)! + 1$.\n\nThen we wish to find $m \\geq 2005$ such that $2(m+1)! \\mid 4(m!)^{2} + 4 m!$, or dividing by $2 \\cdot m!$, we want $m+1 \\mid 2(m! + 1)$.\n\nS... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2010 | |
0f7w | Problem:
An $L$ is an arrangement of $3$ adjacent unit squares formed by deleting one unit square from a $2 \times 2$ square. How many $L$s can be placed on an $8 \times 8$ board (with no interior points overlapping)? Show that if any one square is deleted from a $1987 \times 1987$ board, then the remaining squares ca... | [] | Soviet Union | 21st ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 21; and after deleting any one square from the one-thousand-nine-hundred-eighty-seven by one-thousand-nine-hundred-eighty-seven board, the remaining region can be tiled entirely by L-shaped trominoes. | |
04x0 | Find for which
$$
n \in \{3\,900, 3\,901, 3\,902, 3\,903, 3\,904, 3\,905, 3\,906, 3\,907, 3\,908, 3\,909\}
$$
the set $\{1, 2, 3, \dots, n\}$ can be partitioned into (disjoint) triples in such a way that one of the three numbers in any triple is the sum of the other two. | [
"From the possibility of partitioning the set into disjoint triples it follows that $3 \\mid n$. In each triple $\\{a, b, a+b\\}$ the sum of its elements is $2(a+b)$, hence an even number; thus also the sum of all numbers from $1$ to $n$ must be even, i.e. the product $n(n+1)$ must be divisible by four. Altogether ... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 3900 and 3903 | |
09bn | $ABC$ зөв гурвалжны $BC$ тал дээр $M$ ба $N$ цэгүүдийг ($M$ нь $B$ ба $N$-ийн хооронд оршино) $\angle MAN = 30^\circ$ байхаар авав. $AMC$ ба $ANB$ гурвалжнуудыг багтаасан тойрог $K$ цэгт огтлолцоно. $AK$ шулуун, $AMN$ гурвалжинг багтаасан тойргийн төвийг дайрахыг батал. | [
"$O_1, O_2$ нь $\\triangle ABN$ ба $\\triangle AMC$ багтаасан тойргийн төвүүд гэе. $\\angle ABN = \\frac{AN}{2} = 60^\\circ$, $\\angle AO_1N = 120^\\circ$ болно. $\\angle AO_1N = 120^\\circ$ ба $AO_1 = O_1N$ гэдгээс $O_1AN = \\angle O_1NA = 30^\\circ$ болно. $\\angle MAN = 30^\\circ$ ба $\\angle O_1AN = 30^\\circ \... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Mongolian | proof only | null | |
0duq | Problem:
Poišči najmanjše naravno število, ki ga lahko zapišemo kot vsoto 9, 10 in 11 zaporednih naravnih števil. | [
"Solution:\n\nKer lahko število zapišemo kot vsoto 9 zaporednih naravnih števil, je enako devetkratniku srednjega števila v tem zaporedju 9 števil. Podobno sklepamo, da je enako enajstkratniku srednjega števila v zaporedju 11 zaporednih števil. Ker se da število zapisati tudi kot vsoto 10 zaporednih naravnih števil... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 495 | |
03nh | Problem:
David and Jacob are playing a game of connecting $n \geq 3$ points drawn in a plane. No three of the points are collinear. On each player's turn, he chooses two points to connect by a new line segment. The first player to complete a cycle consisting of an odd number of line segments loses the game. (Both endp... | [
"Solution:\n\nAnswer: David has a winning strategy if and only if $n \\equiv 2 (\\bmod 4)$.\n\nCall a move illegal if it would cause an odd cycle to be formed for the first time. First we show that if $n$ is odd, then any strategy where Jacob picks a legal move if one is available to him causes him to win. Assume f... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n ≡ 2 (mod 4) | |
0eny | Find all functions $f : \mathbb{N} \to \mathbb{R}$ ($\mathbb{N}$ denotes the set of all positive integers, $\mathbb{R}$ the set of all real numbers) such that
$$
f(km) + f(kn) - f(k)f(nm) \geq 1
$$
for all $k, m, n \in \mathbb{N}$. | [
"Plugging in $k = n = m = 1$ yields\n$$\nf(1)^2 - 2f(1) + 1 = (f(1) - 1)^2 \\leq 0,\n$$\nwhich implies $f(1) = 1$. Plugging in $k = 1, n = m$ and $k = n, m = 1$, respectively, we obtain the two inequalities\n$$\n2f(n) - f(n^2) \\geq 1, \\qquad (1)\n$$\n$$\nf(n^2) + f(n) - f(n)^2 \\geq 1. \\qquad (2)\n$$\n\nWe add t... | South Africa | South African Mathematical Olympiad Third Round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(n) = 1 for all n | |
0fzf | Problem:
Bestimme alle natürlichen Zahlen $n$ mit folgender Eigenschaft:
Für alle Primzahlen $p < n$ ist $n - \left\lfloor \frac{n}{p} \right\rfloor p$ nicht durch das Quadrat einer natürlichen Zahl grösser als $1$ teilbar. | [
"Solution:\n\nOn remarque tout d'abord, que la condition avec la partie entière revient à considérer le reste de la division de $n$ par $p$.\nSi $n$ n'est pas premier, alors il existe un premier $q < n$ tel que $q \\mid n$ et donc $n - \\left\\lfloor \\frac{n}{q} \\right\\rfloor q = 0$ et en particulier est divisib... | Switzerland | IMO-Selektionsprüfung | [
"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 | 3, 5, 7, 13 | |
06eq | A convex quadrilateral $ABCD$ with $AC \neq BD$ is inscribed in a circle with centre $O$. Let $E$ be the intersection of diagonals $AC$ and $BD$. If $P$ is a point inside $ABCD$ such that $\angle PAB + \angle PCB = \angle PBC + \angle PDC = 90^\circ$, prove that $O, P$ and $E$ are collinear. | [
"We only work on the configuration as shown since the other cases are similar.\nWe have\n$$\n\\angle CPA = \\angle BAP + \\angle CBA + \\angle PCB = 90^\\circ + \\angle CBA.\n$$\nLet $O_1$ be the centre of $(APC)$. Then we find that\n$$\n\\angle AO_1C = 360^\\circ - 2\\angle CPA = 180^\\circ - 2\\angle CBA = 180^\\... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0870 | Problem:
a) Qual è il minimo intero positivo $c$ tale che esista almeno una coppia $(a, b)$ di interi positivi distinti tali che $2 c^{2} = a^{2} + b^{2}$?
b) Dimostrare che esistono infinite terne $(a, b, c)$ di interi positivi distinti tali che $2 c^{2} = a^{2} + b^{2}$. | [
"Solution:\n\na.\nPoiché $a$ e $b$ hanno la stessa parità, posso porre $a = x + y$ e $b = x - y$ e l'equazione diventa $2 c^{2} = 2 x^{2} + 2 y^{2}$ e quindi il più piccolo $c$ è dato dalla più piccola terna pitagorica $(x, y, c) = (3, 4, 5)$, che dà $(a, b, c) = (7, 1, 5)$.\n\nb.\nBasta osservare che anche $(7k, k... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 5 | |
0dd5 | Do there exist two polynomials $P$ and $Q$ with integer coefficient such that
i) both $P$ and $Q$ have a coefficient with absolute value bigger than $2021$,
ii) all coefficients of $P \cdot Q$ by absolute value are at most $1$. | [
"Note that the polynomial\n$$\n(1 - x^2)(1 - x^4)(1 - x^8) \\dots (1 - x^{2n})\n$$\nhas all coefficients equal $0$, $+1$ or $-1$. Also note, that\n$$\n\\begin{aligned}\n& (1 - x^2)(1 - x^4)(1 - x^8) \\dots (1 - x^{2n}) \\\\\n&= \\prod_{i=0}^{n-1} (1 - x^{2i}) \\cdot \\prod_{i=0}^{n-1} (1 + x^{2i}) \\\\\n&= (1 - x)^... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Yes | |
04w5 | For a natural number $n$, consider the sequence $(a_k)_{k=1}^{\infty}$ given by $a_1 = \frac{1}{n}$ and the recurrent relation
$$
a_{k+1} = 3a_k - \lfloor 2a_k \rfloor - \lfloor a_k \rfloor,
$$
for all $k \ge 1$. Determine all the values of $n$ for which the sequence is eventually constant. | [
"Let $f$ be the function $f(x) = 3x - \\lfloor 2x \\rfloor - \\lfloor x \\rfloor$, then the defining relation can be conveniently written as $a_{k+1} = f(a_k)$.\n\nNote that since $f(0) = f(1) = 0$ and $f(\\frac{1}{2}) = \\frac{1}{2}$, the sequence will be eventually constant whenever $\\frac{1}{2}$ or $1$ occurs i... | Czech Republic | Final Round of the 73rd Czech and Slovak Mathematical Olympiad (March 17–20, 2024) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | n = 3^\alpha or n = 2 \cdot 3^\alpha for some non-negative integer \alpha | |
075g | Let $ABCD$ be a trapezium with $AB \parallel CD$. Let $P$ be a point on $AC$ such that $C$ is between $A$ and $P$; and let $X$, $Y$ be the mid-points of $AB$, $CD$ respectively. Let $PX$ intersect $BC$ in $N$ and $PY$ intersect $AD$ in $M$. Prove that $MN \parallel AB$. | [
"Observe that\n$$\n\\frac{BN}{NC} = \\frac{[PNB]}{[PNC]}\n$$\nHowever $[PNB] + [XNC] = [PXB] = [PXA] = [PCN] + [ACN] + [AXN]$. Since $[XNB] = [AXN]$, we obtain $[PNB] = [PNC] + [ACN]$. Thus\n$$\n\\frac{BN}{NC} = \\frac{[PNC] + [ACN]}{[PNC]} = 1 + \\frac{[ACN]}{[PNC]} = 1 + \\frac{AC}{PC}.\n$$\nSimilarly, we can pro... | India | Indija TS 2012 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0j9l | Problem:
Triangle $ABC$ has $AB = 4$, $BC = 5$, and $CA = 6$. Points $A'$, $B'$, $C'$ are such that $B'C'$ is tangent to the circumcircle of $\triangle ABC$ at $A$, $C'A'$ is tangent to the circumcircle at $B$, and $A'B'$ is tangent to the circumcircle at $C$. Find the length $B'C'$. | [
"Solution:\n\nAnswer: $\\frac{80}{3}$\n\nNote that by equal tangents, $B'A = B'C$, $C'A = C'B$, and $A'B = A'C$. Moreover, since the line segments $A'B'$, $B'C'$, and $C'A'$ are tangent to the circumcircle of $ABC$ at $C$, $A$, and $B$ respectively, we have that $\\angle A'BC = \\angle A'CB = \\angle A$, $\\angle B... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 80/3 | |
0fl4 | Problem:
Sean $N_{0}$ y $Z$ el conjunto de todos los enteros no negativos y el conjunto de todos los enteros, respectivamente. Sea $f: N_{0} \rightarrow Z$ la función que a cada elemento $n$ de $N_{0}$ le asocia como imagen el entero $f(n)$ definido por
$$
f(n) = -f\left(\left\lfloor\frac{n}{3}\right\rfloor\right) - 3\... | [
"Solution:\nSe prueba fácilmente por inducción que, si $n = (\\overline{a_{k} a_{k-1} \\ldots a_{0}})_{3}$, entonces\n$$\nf(n) = \\sum_{\\substack{j=0 \\\\ j \\text{ impar}}}^{k} a_{j} - \\sum_{\\substack{j=0 \\\\ j \\text{ par}}}^{k} a_{j}\n$$\nEn efecto, $f(0) = 0$, $f(1) = -1$, $f(2) = -2$.\nSupongamos que, para... | Spain | XLVI Olimpiada Matemática Española Fase nacional | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 3 * (3^{2010} - 1) / 4 | |
0ipv | Problem:
Determine the last two digits of $17^{17}$, written in base 10. | [
"Solution:\nWe are asked to find the remainder when $17^{17}$ is divided by $100$. Write the power as $(7+10)^{17}$ and expand with the binomial theorem:\n$$\n(7+10)^{17} = 7^{17} + 17 \\cdot 7^{16} \\cdot 10 + \\ldots\n$$\nWe can ignore terms with more than one factor of $10$ because these terms are divisible by $... | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 77 | |
0c8z | Find all positive integers $a, b, c$ such that $2^a + 2^b + 2^c + 3$ is a square. | [
"We can assume $a \\le b \\le c$. If $a \\ge 2$, then $2^a + 2^b + 2^c + 3 \\equiv 3 \\pmod{4}$, hence it cannot be a square. We deduce that $a = 1$, and we must find $b \\le c$ such that $2^b + 2^c + 5$ is a square.\n\nIf $b \\ge 3$, then $2^b + 2^c + 5 \\equiv 5 \\pmod{8}$, again not a square. It follows that $b$... | Romania | RMC 2020 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a = b = c = 1, or {a, b, c} = {1, 2, 4} | |
0cx7 | Find all integers $n$ for which $9 n+16$ and $16 n+9$ are both perfect squares. | [
"It is clear that $n=0$ and $n=1$ are solutions. Let $9 n+16 = x^2$, $16 n+9 = y^2$. Then\n$$\n16 x^2 - 9 y^2 = 16^2 - 9^2\n$$\nthat is\n$$\n(4 x - 3 y)(4 x + 3 y) = 7 \\cdot 5^2.\n$$\nWe can assume that $x, y > 0$, hence we have $4 x - 3 y > 0$. Considering the following possibilities\n$$\n\\left\\{\n\\begin{array... | Saudi Arabia | SAMC | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 0, 1, 52 | |
0amk | Problem:
Find the area of the region bounded by the graph of $|x| + |y| = \frac{1}{4} |x + 15|$. | [] | Philippines | 18th PMO Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 30 | |
0806 | Problem:
Sia $M$ il minimo comune multiplo di tutti gli interi compresi fra $1$ e $100$. Quale dei seguenti numeri è un divisore di $M$?
(A) $1990$
(B) $2000$
(C) $2002$
(D) $2004$
(E) $2020$. | [] | Italy | Italy Febbraio Contest | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | C | |
075v | In a triangle $ABC$, with $AB \neq BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \neq B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. ... | [
"We need the following well-known result.\n**Lemma.** In a triangle $KLM$ with $KL \\neq KM$, $R$ is a point on the $LM$ such that $KR$ is perpendicular to $LM$. Let $U$ be a point on the line $KR$. Let the lines $LU$ and $MU$ intersect $KM$ and $KL$, respectively, at $S$ and $T$ respectively. Then $L, T, S, M$ are... | India | Indija TS 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
08fx | Problem:
Sia $ABCD$ un trapezio isoscele di base maggiore $AB$ tale che la bisettrice dell'angolo in $D$ passi per $B$. Supponiamo che la bisettrice dell'angolo in $A$ intersechi il lato $BC$ nel punto $P$. Dimostrare che $AB = AP$ se e solo se la bisettrice di $\widehat{PAD}$ passa per $C$. | [
"Solution:\n\nSia $\\alpha = \\widehat{DAB} = \\widehat{ABC}$ (poiché il trapezio è isoscele, i due angoli sono uguali); la condizione $AB = AP$ equivale al fatto che si abbia $\\widehat{BPA} = \\widehat{PBA} = \\alpha$. A sua volta, poiché $\\widehat{PAB} = \\alpha / 2$ ($AP$ è bisettrice dell'angolo in $A$), ques... | Italy | Olimpiadi di Matematica - Febbraio | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0e0a | a. Prove: if $a - \frac{1}{b} + b(b + \frac{3}{a})$ is an integer for some positive integers $a$ and $b$, then it is a perfect square.
b. Find two integers $a$ and $b$ such that $a - \frac{1}{b} + b(b + \frac{3}{a})$ is a positive integer but not a perfect square. | [
"Let $I = a - \\frac{1}{b} + b(b + \\frac{3}{a})$. If we want $I = a + b^2 + \\frac{3b}{a} - \\frac{1}{b}$ to be an integer, then $N = \\frac{3b}{a} - \\frac{1}{b}$ has to be an integer as well. Since $N = \\frac{3b^2 - a}{ab}$, we conclude that $ab$ must divide $3b^2 - a$. This implies that $b$ divides $3b^2 - a$,... | Slovenia | Selection Examinations for the IMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a) It is always a perfect square for positive integers satisfying the integrality condition. (b) Examples: a = 4, b = −2 gives 7, and a = −4, b = −2 gives 2. | |
00jt | Solve the following system of equations in the set of rational numbers:
$$
\begin{aligned}
(x^2 + 1)^3 &= y + 1 \\
(y^2 + 1)^3 &= z + 1 \\
(z^2 + 1)^3 &= x + 1.
\end{aligned}
$$ | [
"We first note that $(0, 0, 0)$ is obviously a solution of the system of equations. We will now show that there are no others.\nLet $x = \\frac{p}{q}$ with relatively prime integer values of $p$ and $q$ and $q > 0$. We then have\n$$\ny = \\left( \\left( \\frac{p}{q} \\right)^2 + 1 \\right)^3 - 1 = \\frac{(p^2 + q^2... | Austria | AustriaMO2013 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (0, 0, 0) |
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