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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07or | Let $0 < a < 1$ be a real number. What point $z$ on the upper half of the unit circle in the complex plane maximises the sum $|z + a| + |z - a|$? | [
"Noting that if $x, y > 0$, then $x + y \\le \\sqrt{2} \\sqrt{x^2 + y^2}$, with equality iff $x = y$, then for all complex numbers $z$ satisfying $|z| = 1$\n$$\n\\begin{align*}\n|z - a| + |z + a| &\\le \\sqrt{2} \\sqrt{|z + a|^2 + |z - a|^2} \\\\\n&= \\sqrt{2} \\sqrt{(z + a)(\\bar{z} + a) + (z - a)(\\bar{z} - a)} \... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | i | |
0bcp | Problem:
Un număr infinit de persoane sunt înregistrate ca useri pe reţeaua socială Mutrăbook. Unele perechi de useri (diferiţi) sunt înregistrate ca prieteni, dar fiecare persoană are cel mult un număr finit de prieteni. Fiecare user are cel puţin un prieten. (Relaţia de prietenie este simetrică; dacă $A$ este priete... | [] | Romania | Olimpiada europeana de matematica a fetelor | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Other"
] | null | proof only | null | |
05ic | Problem:
Soit $k \geqslant 2$ un entier.
1) Soit $n > k$ un entier.
Existe-t-il $n$ entiers strictement positifs tels que $k$ quelconques ne sont jamais premiers entre eux dans leur ensemble, mais $k+1$ quelconques sont toujours premiers entre eux dans leur ensemble?
2) Existe-t-il une suite infinie d'entiers strict... | [
"Solution:\n\n1) Soit $A$ l'ensemble des parties de $\\{1, \\ldots, n\\}$ de cardinal $k$. Comme l'ensemble des nombres premiers est infini, il existe une famille $\\left(p_{I}\\right)_{I \\in A}$ de nombres premiers deux à deux distincts indexée par $A$. Pour tout $i$, notons $A(i)$ l'ensemble des parties $I$ de $... | France | Test EGMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0ax6 | Problem:
How many real number solutions does the equation $\frac{1}{3} x^{4} + 5|x| = 7$ have? | [
"Solution:\nThe solutions correspond to intersection points of $y = 7 - \\frac{1}{3} x^{4}$ and $y = 5|x|$. Clearly, there are only two intersection points."
] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2 | |
04gr | Find all prime numbers $p$ for which there exists a positive integer $n$ such that both $n^2 + 3$ and $(n + 1)^2 + 3$ are divisible by $p$. (Stipe Vidak) | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 13 | |
0lai | Given real numbers $a$, $b$, $c$ satisfying the condition: for each positive integer $n$, $a^n + b^n + c^n$ is an integer. Show that there exist integers $p$, $q$, $r$ such that $a$, $b$, $c$ are 3 solutions to the equation $x^3 + px^2 + qx + r = 0$. | [
"For each positive integer $n$, let $T_n = a^n + b^n + c^n$. By assumption, $T_n \\in \\mathbb{Z}$ for all $n \\ge 1$.\nWe will show that the numbers $p = -(a + b + c)$, $q = ab + bc + ca$ and $r = -abc$ satisfy the requirement for the problem.\nIndeed, according to the theorem of Viet, $a$, $b$, $c$ are 3 solution... | Vietnam | Vijetnam 2009 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof only | null | |
01d6 | In a computer game a $4 \times 4 \times 4$ cube is built using $4^3$ unit cubes. At the beginning of the game each unit cube contains an integer. In each turn of the game, you choose a unit cube and increase by 1 all the integers in the cubes having a face in common with the chosen cube. You win the game if you reach a... | [
"Answer: No.\n\nTwo unit cubes with a common face are called neighbours. Colour the cubes either black or white in such a way that two neighbours always have different colours. Notice that the integers in the white cubes only change when a black cube is chosen. Now recolour the white cubes that have exactly 4 neigh... | Baltic Way | Baltic Way 2016 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
09vx | Ward and Gabrielle are playing a game on a large sheet of paper. At the start of the game, there are 999 ones on the sheet of paper. Ward and Gabrielle each take turns alternatingly, and Ward has the first turn. During their turn, a player must pick two numbers $a$ and $b$ on the sheet such that $\gcd(a, b) = 1$, erase... | [
"Gabrielle can always win using the following strategy: during each of her turns, she picks the largest two numbers on the sheet as $a$ and $b$. Using induction on $k$, we will prove that she is always allowed to do so, and that after her $k$-th turn, the sheet contains the number $2k + 1$ and $998 - 2k$ ones.\n\nI... | Netherlands | IMO Team Selection Test 2, June 2020 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Gabrielle | |
09gg | Тойргоор тоглосон тэмцээнд 52 сурагч оролцжээ. Хэдэн тоглолтын дараа аль ч сурагчийг сонгон авахад түүнтэй тоглосон сурагчдын тоглолтын тоо ялгаатай байсан бол хамгийн олондоо хэдэн тоглолт явагдсан байсан бэ? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 650 | |
0l7d | Problem:
A checkerboard is a rectangular grid of cells colored black and white such that the top-left corner is black and no two cells of the same color share an edge. Two checkerboards are distinct if and only if they have a different number of rows or columns. For example, a $20 \times 25$ checkerboard and a $25 \tim... | [
"Solution:\nSince there is a black corner on the checkerboard, the number of white squares is at most the number of black squares. So, the board either has 40 or 41 white squares. Therefore, we want to compute the number of ordered pairs $(r,c)$ with a product of 81 or 82. Since $81 = 3^{4}$ has 5 divisors and $82 ... | United States | HMMT February | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 9 | |
0j4y | Problem:
Sarah and Hagar play a game of darts. Let $O_{0}$ be a circle of radius $1$. On the $n$th turn, the player whose turn it is throws a dart and hits a point $p_{n}$ randomly selected from the points of $O_{n-1}$. The player then draws the largest circle that is centered at $p_{n}$ and contained in $O_{n-1}$, an... | [
"Solution:\n\nAnswer: $\\boxed{\\frac{6\\pi}{7}}$\n\nLet $f(r)$ be the average area colored red on a dartboard of radius $r$ if Sarah plays first. Then $f(r)$ is proportional to $r^{2}$. Let $f(r) = (\\pi x) r^{2}$ for some constant $x$. We want to find $f(1) = \\pi x$.\n\nIn the first throw, if Sarah's dart hits a... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 6π/7 | |
0esb | The lengths of all sides of a triangle are prime numbers. If two sides are $5$ and $7$ what is the longest possible length of the third side? | [
"The length of the third side of a triangle cannot exceed the sum of the lengths of the first two sides, which is $5 + 7 = 12$. The largest prime number less than $12$ is $11$."
] | South Africa | South African Mathematics Olympiad Second Round | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | final answer only | 11 | |
0ki9 | Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the 24-sided polygon shown in the figure below. The area of this polygon c... | [
"Let $O$ be the center of the polygon, and label 11 points as shown in the figure. Let $a = AB = BC$.\n\n\n\nTriangle $BCK$ is a $30$-$60$-$90^\\circ$ triangle, so $BK = 2a$ and $CK = KG = a\\sqrt{3}$. Then $AG = 3a + a\\sqrt{3} = 6$, so $a = 3 - \\sqrt{3}$. The area of the 24-sided polygon... | United States | Fall 2021 AMC 10 B | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles"
] | English | final answer only | 147 | |
04sn | Determine all pairs $(p, q)$ of integers such that $p$ is an integer multiple of $q$ and quadratic equation $x^2 + px + q = 0$ has at least one integer root. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All pairs (p, q) are: (i) p = 0 and q = −t^2 for some integer t (including t = 0), and (ii) (p, q) ∈ {(2, 1), (−2, 1), (4, 4), (−4, 4)}. | |
0ahx | Find all numbers $p$, $q$ and $r$, such that $p$ and $r$ are prime, $q$ is a positive integer so that they satisfy the equation:
$$
(p+q+r)^2 = 2p^2 + 2q^2 + r^2.
$$ | [
"After simplifying the equation we get $2r(p+q) = (p-q)^2$. Since $r$ is prime, it follows that $r$ is a divisor of $p-q$, so that $r^2$ is a divisor of the right-hand side of the last equality. This implies that $r$ is a divisor of $2(p+q)$. If $r>2$, then $r$ is a divisor of $(p+q)$, so that $r$ has to be a divis... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No solutions; there are no primes p and r and positive integer q that satisfy the equation. | |
0h7c | Find all positive real numbers $a, b, c$, which satisfy the equality:
$$
ab\left(1 - \frac{c^2}{(a+b)^2}\right) = bc\left(1 - \frac{a^2}{(b+c)^2}\right) = ca\left(1 - \frac{b^2}{(c+a)^2}\right).
$$ | [
"Rewrite the given equality in the following way:\n$$\n\\frac{a((a+b)^2-c^2)}{(a+b)^2} = \\frac{c((b+c)^2-a^2)}{(b+c)^2} \\quad \\text{or} \\quad \\frac{a(a+b-c)(a+b+c)}{(a+b)^2} = \\frac{c(b+c-a)(b+c+a)}{(b+c)^2}.\n$$\nAfter dividing by $a+b+c$ we obtain\n$$\n\\frac{a(a+b-c)}{(a+b)^2} = \\frac{c(b+c-a)}{(b+c)^2}.\... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | (t, t, t) for any t > 0 | |
0gae | Let $\mathbb{Z}$ be the set of all integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that
$$
f(x - f(y)) = f(f(x)) - f(y) - 1
$$
holds for all $x, y \in \mathbb{Z}$.
令所有整數所成的集合為 $\mathbb{Z}$。求所有函數 $f : \mathbb{Z} \to \mathbb{Z}$, 滿足：
$$
f(x - f(y)) = f(f(x)) - f(y) - 1
$$
對所有整數 $x, y$ 都成立。 | [
"There are two solutions: $f(x) = -1$ and $f(x) = x + 1$.\nThe functional equation is\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 1 \\quad (1)\n$$\n1. Take $x = 0$ and $y = f(0)$ into (1), we learn that $z = -f(f(0))$ satisfies $f(z) = -1$.\n\n2. Set $y = z$ into (1), we have\n$$\nf(x + 1) = f(f(x)). \\quad (2)\n$$\n\n3. T... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(x) = -1 and f(x) = x + 1 | |
04rt | In the real numbers, solve the following system of equations:
$$
\begin{aligned}
\sin^2 x + \cos^2 y &= \tan^2 z, \\
\sin^2 y + \cos^2 z &= \tan^2 x, \\
\sin^2 z + \cos^2 x &= \tan^2 y.
\end{aligned}
$$ | [
"Substituting $\\cos^2 x = a$, $\\cos^2 y = b$, $\\cos^2 z = c$ leads to the system\n$$\n\\begin{aligned}\n1 - a + b &= \\frac{1}{c} - 1, \\\\\n1 - b + c &= \\frac{1}{a} - 1, \\\\\n1 - c + a &= \\frac{1}{b} - 1,\n\\end{aligned}\n\\qquad (1)\n$$\nwhere $a, b, c \\in (0, 1)$.\nAdding these equations together yields\n... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | (x, y, z) = (\pi/4 + k\pi/2, \pi/4 + l\pi/2, \pi/4 + m\pi/2) for any integers k, l, m | |
0kf2 | Problem:
Find the sum of all positive integers $n$ for which
$$
\frac{15 \cdot n!^{2}+1}{2 n-3}
$$
is an integer. | [
"Solution:\nIt is clear that $n=1$ and $n=2$ work so assume that $n>2$. If $2 n-3$ is composite then its smallest prime factor is at most $\\frac{2 n-3}{2}<n$ so will be coprime to $15 \\cdot n!^{2}+1$. Therefore assume that $2 n-3=p$ is prime. We can rewrite the numerator as\n$$\n(-1)^{n} \\cdot 15 \\cdot\\left(1 ... | United States | HMMT February 2020 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 90 | |
09vc | Problem:
Op een middelbare school zit in elke klas een oneven aantal leerlingen. Verder heeft elke leerling een beste vriend (mogelijk uit een andere klas). Iedereen is de beste vriend van zijn of haar beste vriend. Bij de stedenreis gaat elke leerling naar Rome of Parijs. Bewijs dat de leerlingen over de twee bestemm... | [
"Solution:\n\nNoem een losse leerling een leerling van wie de beste vriend in een andere klas zit. We gaan nu eerst alle losse leerlingen een bestemming geven. Begin met een willekeurige losse leerling $A_{1}$ en stuur deze naar Parijs samen met zijn beste vriend $A_{2}$. Kies in de klas van $A_{2}$ een andere loss... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07fj | Consider a triangle $\triangle ABC$ with incenter $I$. Let $D$ be the intersection point of $BI$ and $AC$, and let $CI$ intersect the circumcircle of $\triangle ABC$ at $M$. Point $K$ lies on the line $MD$ such that $\angle KIA = 90^\circ$. Let $F$ be the reflection of $B$ with respect to $C$. Prove that $BIKF$ is cycl... | [
"Let $E$ be the intersection point of line $MK$ and the circumcircle of $\\triangle ABC$, then, since $\\angle BEM = \\frac{\\angle C}{2}$ and $\\angle DIK = 90 - \\angle AID$, so $\\angle BEM = \\angle DIK$, it follows that the points $B$, $I$, $K$, $E$ lie on a circle. Suppose $BC$ intersects this circle at point... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscella... | English | proof only | null | |
0hco | A figure is called *polyomino* if it is formed by joining one or more $1 \times 1$ squares edge to edge. It is known that a rectangle which is not a square can be split into 8 pairwise distinct polyominos. Polyominos are considered equal if one can be transformed into another by translations and rotations. What is the ... | [
"We will start with counting polyominos of smallest area. There is only 1 polyomino of area 1, 1 polyomino of area 2, and 2 polyominos of area 3. Thus, the smallest area of a rectangle that consists of 8 polyominos is $1 \\cdot 1 + 1 \\cdot 2 + 2 \\cdot 3 + 4 \\cdot 4 = 25$. But then it has to be of a size $25 \\ti... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 26 | |
062z | Problem:
Die Ecken und Kanten eines regulären $n$-Ecks seien im Uhrzeigersinn jeweils so von 1 bis $n$ nummeriert, dass die Kante Nr. $i$ auf die Ecke Nr. $i$ folgt ($1 \leq i \leq n$).
Nun werden die Ecken mit nichtnegativen ganzen Zahlen $e_{i}$ und die Kanten mit nichtnegativen ganzen Zahlen $k_{i}$ so belegt, dass... | [
"Solution:\n\na) Ein mögliches Beispiel für $n \\geq 3$ ist $e_{1} = e_{2} = 1$, $e_{i} = 0$ für $3 \\leq i \\leq n$. Dies bedingt $k_{2} = k_{n} = 1$ und $k_{i} = 0$ sonst. Offensichtlich sind alle Bedingungen erfüllt.\n\nb) Wir beweisen, dass stets $n = m + 2$ gilt. Der Beweis erfolgt in zwei Schritten.\n\ni) Es ... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | m + 2 | |
0cjy | Determine the set of rational numbers $r$ for which there exist non-negative integers $a$ and $b$, such that $\frac{a+b}{2} - \sqrt{ab} = r$. | [
"For $a$, $b$, $r$ as in the statement, $r = \\frac{(\\sqrt{a} - \\sqrt{b})^2}{2}$, therefore $r \\ge 0$. Moreover, $ab$ must be a square of a natural number $n$, so $r = \\frac{a+b-2n}{2} = \\frac{m}{2}$, where $m$ is a non-negative integer. Therefore, $r = \\frac{m}{2}$, with $m$ a non-negative integer.\n\nConver... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | { m/2 : m is a non-negative integer } | |
0i39 | Problem:
There are two red, two black, two white, and a positive but unknown number of blue socks in a drawer. It is empirically determined that if two socks are taken from the drawer without replacement, the probability they are of the same color is $\frac{1}{5}$. How many blue socks are there in the drawer? | [
"Solution:\n\nLet the number of blue socks be $x > 0$. Then the probability of drawing a red sock from the drawer is $\\frac{2}{6 + x}$ and the probability of drawing a second red sock from the drawer is $\\frac{1}{6 + x - 1} = \\frac{1}{5 + x}$, so the probability of drawing two red socks from the drawer without r... | United States | Harvard-MIT Math Tournament | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 4 | |
09cy | Цифрүүдийн нийлбэр нь $9$-тэй тэнцүү ба $4$ орон нь $1$, $0$, $0$, $4$ цифрүүдээс тогтсон $6$ оронтой тоо хичнээн байх вэ? | [
"Уг $6$ оронтой тооны үлдэх $2$ цифр нь $1$, $2$ эсвэл $0$, $3$ байна. Эхний тохиолдолд уг $6$ оронтой тоо $1$, $2$, $4$-ийн аль нэгээр эхлэх ба $1$-ээр эхэлсэн бол бусад цифрүүд нь $1$, $2$, $4$, $0$, $0$ болж эдгээрийг сэлгэх боломжийн тоо нь $\\frac{5!}{2!} = 60$.\n\n$2$-оор эхэлсэн бол бусад цифрүүд нь $1$, $1$... | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | Mongolian | proof and answer | 180 | |
08mv | Problem:
Can we divide an equilateral triangle $\triangle ABC$ into 2011 small triangles using 122 straight lines? (There should be 2011 triangles that are not themselves divided into smaller parts and there should be no polygons which are not triangles.) | [
"Solution:\nFirstly, for each side of the triangle, we draw 37 equidistant, parallel lines to it. In this way we get $38^{2} = 1444$ triangles.\n\nThen we erase 11 lines which are closest to the vertex $A$ and parallel to the side $BC$ and we draw 21 lines perpendicular to $BC$, the first starting from the vertex $... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | Yes | |
0iyp | Problem:
Let $\mathcal{R}$ be the region in the plane bounded by the graphs of $y = x$ and $y = x^{2}$. Compute the volume of the region formed by revolving $\mathcal{R}$ around the line $y = x$. | [
"Solution:\nWe integrate from $0$ to $1$ using the method of washers. Fix $d$ between $0$ and $1$. Let the line $x = d$ intersect the graph of $y = x^{2}$ at $Q$, and let the line $x = d$ intersect the graph of $y = x$ at $P$. Then $P = (d, d)$, and $Q = (d, d^{2})$.\n\nNow drop a perpendicular from $Q$ to the line... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | π/(30√2) | |
078s | Let $ABC$ be an acute-angled triangle with $AB < AC$, and let $O, H$ be its circumcentre and orthocentre respectively. Points $Z, Y$ lie on segments $AB, AC$ respectively, such that
$$
\angle ZOB = \angle YOC = 90^\circ.
$$
The perpendicular line from $H$ to line $YZ$ meets lines $BO$ and $CO$ at $Q, R$ respectively. L... | [
"Define $K$ to be the point on $YZ$ such that $HK \\perp YZ$. Let $A'$ be a point on the circumcircle of $\\triangle ABC$ such that $AA' \\parallel BC$.\n\n**Lemma 1.** $YZ$ is the perpendicular bisector of $HA'$.\n*Proof.* Let $H_B$ denote the reflection of $H$ in $AC$. Then note that $A'H_B \\parallel CO$. This i... | India | IMO TST | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geomet... | null | proof only | null | |
0d8b | For each pair of positive integers $x$, $y$ a nonnegative integer $x \Delta y$ is defined. It's known that for all positive integers $a$ and $b$ the following equalities hold:
i. $(a+b) \Delta b = a \Delta b + 1$.
ii. $(a \Delta b) \cdot (b \Delta a) = 0$.
Find values of the expressions $2016 \Delta 121$ and $2016 \... | [
"Let us analyze the properties:\n\ni. $(a+b) \\Delta b = a \\Delta b + 1$\n\nii. $(a \\Delta b) \\cdot (b \\Delta a) = 0$\n\nFrom i., for fixed $b$, the function $f(a) = a \\Delta b$ satisfies $f(a+1) = f(a) + 1$ for $a \\geq 1$. This means $f(a)$ is an affine function in $a$:\n\nLet $f(a) = a \\Delta b$. Then $f(a... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2016 Δ 121 = 16; 2016 Δ 144 = 13 | |
03zk | Are there any positive integers $m, n$ such that $m^{20} + 11^n$ is a square number? Prove your conclusion. (posed by Yuan Hanhui) | [
"Assuming there are positive integers $m, n$ such that $m^{20} + 11^n = k^2$ with $k \\in \\mathbb{Z}$, we then have\n$$\n11^n = k^2 - m^{20} = (k - m^{10})(k + m^{10}),\n$$\nwhich means that there are integers $\\alpha, \\beta \\ge 0$ such that\n$$\n\\begin{cases} k - m^{10} = 11^{\\alpha}, \\\\ k + m^{10} = 11^{\... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No; there are no positive integers m and n such that m^20 + 11^n is a perfect square. | |
0ev1 | Let $ABC$ be a triangle satisfying $\angle B > \angle C$ and let $D$ be the point on the side $AC$ such that $\angle ADC = \angle C$. Let $I$ be the incenter of $ABC$ and $E$ be the intersection of the circumcircle of $CDI$ and the line $AI$ which is not $I$. Let $P$ be the intersection of the line $BD$ and the line wh... | [
"First, we will show that the line $PJ$ passes the midpoint of a side $AB$.\nLet $M$ be the intersection of two lines $PJ$ and $AB$, and $S$ be the intersection of two lines $BP$ and $AI$. By using Menelaus' theorem to the triangle $ABS$ and the line $PJ$, we have\n$$\n\\frac{AM}{BM} \\cdot \\frac{BP}{PS} \\cdot \\... | South Korea | The 26th Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geom... | null | proof only | null | |
05e8 | Problem:
Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the following assertions are true for every pair of positive integers $(x, y)$:
- $x$ and $f(x)$ have the same number of positive divisors.
- If $x$ does not divide $y$ and $y$ does not div... | [
"Solution:\nLet $f$ be a function satisfying these properties. From the first one, it follows that $f(1)=1$ and for each prime number $p$, $f(p)=q_p$ for some prime number $q_p$. Assume $p \\neq 2$, the pair $(2, p)$ in property 2 gives\n$$\n\\operatorname{gcd}\\left(q_2, q_p\\right) = \\operatorname{gcd}(f(2), f(p... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2024 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All functions of the form f(n) = q^{d(n)-1} for an arbitrary fixed prime q. | |
0a7y | Problem:
The real-valued function $f$ is defined for positive integers, and the positive integer $a$ satisfies
$$
\begin{gathered}
f(a)=f(1995), \quad f(a+1)=f(1996), \quad f(a+2)=f(1997) \\
f(n+a)=\frac{f(n)-1}{f(n)+1} \quad \text{ for all positive integers } n
\end{gathered}
$$
(i) Show that $f(n+4a)=f(n)$ for all po... | [
"Solution:\nTo prove (i), we use the formula $f(n+a)=\\frac{f(n)-1}{f(n)+1}$ repeatedly:\n$$\n\\begin{gathered}\nf(n+2a)=f((n+a)+a)=\\frac{\\frac{f(n)-1}{f(n)+1}-1}{\\frac{f(n)-1}{f(n)+1}+1}=-\\frac{1}{f(n)} \\\\\nf(n+4a)=f((n+2a)+2a)=-\\frac{1}{-\\frac{1}{f(n)}}=f(n)\n\\end{gathered}\n$$\n\n(ii) If $a=1$, then $f(... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 10 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 3 | |
021k | Problem:
Seja $P$ um ponto no interior do triângulo equilátero $ABC$ tal que:
$$
PA = 5, \quad PB = 7, \quad \text{e} \quad PC = 8
$$
Determine a medida do lado do triângulo $ABC$. | [] | Brazil | Desafios | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | sqrt(129) | |
0758 | Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$. Let $E$, $F$, $G$, $H$ be the midpoints of the arcs $AB$, $BC$, $CD$, $DA$ of the circle $\Gamma$. Suppose $AC \cdot BD = EG \cdot FH$. Prove that $AC$, $BD$, $EG$, $FH$ are concurrent. | [
"\n\nLet $R$ be the radius of the circle $\\Gamma$. Observe that $\\angle EDF = \\frac{1}{2}\\angle D$. Hence $EF = 2R \\sin \\frac{D}{2}$. Similarly, $HG = 2R \\sin \\frac{B}{2}$. But $\\angle B = 180^\\circ - \\angle D$.\nThus $HG = 2R \\cos \\frac{D}{2}$. We hence get\n$$\nEF \\cdot GH =... | India | Indija mo 2011 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
09lc | Let $\omega$ be the circumcircle of a triangle $ABC$. Let $P$ be a point on the arc $AB$ of $\omega$, not containing $C$. The circle inscribed in triangle $ABC$ has center $I$, and the line $AI$ intersects $\omega$ at a point $M$, distinct from $A$. The circumcircle of triangle $API$ intersects segment $AB$ at a point ... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Inversion"
] | English | proof only | null | |
062l | Problem:
Das Dreieck $A B C$ sei bei $A$ rechtwinklig. Es bezeichne $M$ den Mittelpunkt der Strecke $B C$. Der Punkt $D$ liege auf der Seite $A C$ und erfülle $\overline{A D}=\overline{A M}$. Der von $C$ verschiedene Schnittpunkt der Umkreise der Dreiecke $A M C$ und $B D C$ heiße $P$. Man beweise, dass $C P$ den bei ... | [
"Solution:\n\nDer Winkel $\\varangle P D C$ ist einerseits Nebenwinkel von $\\varangle A D P$, andererseits liegt er im Sehnenviereck $B P D C$ dem Winkel $\\varangle C B P$ gegenüber. Folglich $\\varangle A D P = \\varangle C B P$. Analog ist der Winkel $\\varangle C M P$ Nebenwinkel von $\\varangle P M B$ und lie... | Germany | 1. IMO-Auswahlklausur | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
0hjz | Problem:
Find a positive integer satisfying the following three clues:
- When multiplied by $2$, it yields a perfect square.
- When multiplied by $3$, it yields a perfect cube.
- When multiplied by $5$, it yields a perfect fifth power. | [
"Solution:\n\nIt is reasonable to try a number of the form $2^{a} 3^{b} 5^{c}$, where $a$, $b$, and $c$ are nonnegative integers. We want\n$$\n\\begin{array}{ll}\n2^{a+1} 3^{b} 5^{c} \\text{ is a square, i.e. } & 2|a+1,\\ 2| b, \\quad \\text{ and } 2 \\mid c \\\\\n2^{a} 3^{b+1} 5^{c} \\text{ is a cube, i.e. } & 3|a... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 2^15 * 3^20 * 5^24 | |
0fne | Sea $n \ge 3$ un número entero. Se considera una circunferencia en la que se han marcado $n+1$ puntos igualmente espaciados. Cada punto se etiqueta con uno de los números $0, 1, \dots, n$ de manera que cada número se usa exactamente una vez. Dos distribuciones de etiquetas se consideran la misma si una se puede obtener... | [
"Antes de empezar, nótese que si $x + y = n$, y $\\text{mcd}(x, y) = 1$, entonces $\\text{mcd}(x, n) = \\text{mcd}(y, n) = 1$, ya que si por ejemplo $\\text{mcd}(x, n) = d > 1$, entonces $y = n - x$ es divisible por $d$, luego $\\text{mcd}(x, y) \\ge d$, contradicción. Recíprocamente, si una de entre $\\text{mcd}(x... | Spain | Olimpiada Internacional de Matemáticas | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | Spanish | proof only | null | |
077t | Alice has an integer $N > 1$ written on a blackboard. Each minute, she deletes the current number $x$ on the blackboard and writes $2x + 1$ if $x$ is not the cube of an integer, or the cube root of $x$ otherwise. Prove that at some point in time, she writes a number larger than $10^{100}$. | [] | India | EGMO TST Day 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0aki | A collection of $n$ squares on the plane is called tri-connected if the following criteria are satisfied:
(i) All the squares are congruent
(ii) If two squares have a point $P$ in common, then $P$ is a vertex of each of the squares.
(iii) Each square touches exactly three other squares.
How many positive integers $n$ a... | [] | North Macedonia | Asian-Pacific Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof and answer | 501 | |
01z9 | The polynomial $p(x)$ with integer coefficients satisfies the equality
$$p(\sqrt{2} + \sqrt{3}) = \sqrt{2} - \sqrt{3}.$$
a) Find all possible values of $p(\sqrt{2} - \sqrt{3})$.
b) Give an example of at least one polynomial $p(x)$ that satisfies the condition. | [
"a) Consider any monomial $a_n x^n$ of $p(x)$ and substitute $x = \\sqrt{2} + \\sqrt{3}$ and $x = \\sqrt{2} - \\sqrt{3}$ into it. When opening the brackets in the expression $a_n(\\sqrt{2} + \\sqrt{3})^n$, all terms will look like $a_n C_n^m \\sqrt{2}^k \\sqrt{3}^m$, where $k + m = n$. If $m$ is even then if $\\sqr... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | English | proof and answer | a) √2 + √3. b) One example is p(x) = x^3 − 10x. | |
0gsm | Find all pairs $(p, q)$ of prime numbers satisfying
$$
2^p = 2^{q-2} + q!
$$ | [
"Answer: $(p, q) = (3, 3), (7, 5)$.\n\nIf $q = 2$ then no pair $(p, 2)$ satisfies the equation. If $q = 3$ and $q = 5$ then the only pairs satisfying the equation are $(3, 3)$ and $(5, 7)$, respectively.\n\nLet us show that there is no solution for $q \\ge 7$. Consider the binary representation of $q$: $q = 2^{a_1}... | Turkey | Team Selection Test | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | [(3, 3), (7, 5)] | |
0gka | Consider the sequence $\{a_n\}_{n \ge 1}$ of positive integers with $a_n a_{n+3} = a_{n+2} a_{n+5}$ for all positive integer $n$. Determine the largest integer that always divides $\sum_{k=1}^{2550} a_{2k} a_{2k-1}$. | [
"From $a_n a_{n+3} = a_{n+2} a_{n+5}$ for every positive integer $n$, we have\n$a_{n+1} a_{n+4} = a_{n+3} a_{n+6}$ and $a_{n+2} a_{n+5} = a_{n+4} a_{n+7}$. Then\n$$\na_n a_{n+3} \\cdot a_{n+1} a_{n+4} \\cdot a_{n+2} a_{n+5} = a_{n+2} a_{n+5} \\cdot a_{n+3} a_{n+6} \\cdot a_{n+4} a_{n+7}.\n$$\nTherefore, $a_n a_{n+1... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 850 | |
0kob | Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$. | [
"Let the spheres with radii $11$, $13$, and $19$ have centers $P$, $Q$, and $R$, respectively, and let the three circles have common radius $r$. Segments $\\overline{AP}$, $\\overline{BQ}$, and $\\overline{CR}$ are perpendicular to the plane of $\\triangle ABC$, so $\\overline{AB}$ is the projection of $\\overline{... | United States | 2022 AIME I | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 756 | |
04bb | Let $ABCDE$ be a regular pentagon. The lines $BC$ and $DE$ intersect at point $F$. Determine the angles of triangle $BEF$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof and answer | ∠EBF = 108°, ∠BEF = 54°, ∠BFE = 18° | |
08jw | Problem:
A polygon having $n$ sides is arbitrarily decomposed in triangles having all the vertices among the vertices of the polygon. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is side of the polygon and white those triangles that hav... | [
"Solution:\n\nDenote by $b, r, w$ the number of black, red, and white triangles respectively.\nIt is easy to prove that the polygon is divided into $n-2$ triangles, hence\n$$\nb + r + w = n - 2\n$$\nEach side of the polygon is a side of exactly one triangle of the decomposition, and thus\n$$\n2b + r = n\n$$\nSubtra... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0con | Let $\alpha, \beta, \gamma$ be the angles in triangle $\Delta$. Suppose that $\sin \alpha > \cos \beta$, $\sin \beta > \cos \gamma$, and $\sin \gamma > \cos \alpha$. Prove that $\Delta$ is acute-angled.
Углы треугольника $\alpha, \beta, \gamma$ удовлетворяют неравенствам $\sin \alpha > \cos \beta, \sin \beta > \cos \g... | [
"Предположим противное; пусть для определённости $\\gamma \\ge 90^\\circ$. Тогда $\\alpha + \\beta \\le 90^\\circ$, и углы $\\alpha$ и $\\beta$ острые. Поэтому $0 < \\beta \\le 90^\\circ - \\alpha < 90^\\circ$, откуда $\\cos \\beta \\ge \\cos(90^\\circ - \\alpha) = \\sin \\alpha$, что противоречит условию."
] | Russia | Regional round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English; Russian | proof only | null | |
0duw | Problem:
Skupina geometrov želi določiti višino hriba. Iz kraja $A$ v ravnini vznožja vidijo vrh hriba pod kotom $\alpha=19^\circ 40^\prime$, ko pa se po ravnem približajo hribu za $250 \mathrm{~m}$, vidijo vrh pod kotom $\beta=24^\circ$. Kako visok je hrib? Vmesne rezultate zaokroži na štiri decimalke, rezultat pa na... | [
"Solution:\n\nZapisana zveza $\\operatorname{tg} \\alpha=\\frac{h}{250+x}$\n\nZapisana zveza $\\operatorname{tg} \\beta=\\frac{h}{x}$\n\nZapisana ugotovitev $x \\cdot \\operatorname{tg} \\beta=(x+250) \\cdot \\operatorname{tg} \\alpha$\n\nIzražena spremenljivka $x: x=\\frac{250 \\cdot \\operatorname{tg} \\alpha}{\\... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 453 m | |
0jve | Problem:
Seven lattice points form a convex heptagon with all sides having distinct lengths. Find the minimum possible value of the sum of the squares of the sides of the heptagon. | [
"Solution:\nAnswer: $42$\nConsider the vectors corresponding to the sides of the heptagon, and call them $[x_{i}, y_{i}]$ for $i$ between $1$ and $7$. Then since $\\sum x_{i} = \\sum y_{i} = 0$, and $a^{2}$ has the same parity as $a$, we have that $\\sum x_{i}^{2} + y_{i}^{2}$ must be an even number. A side length ... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 42 | |
09yp | Find all quadruples $(a, b, c, d)$ of non-negative integers such that $ab = 2(1 + cd)$ and there exists a non-degenerate triangle with sides of length $a - c$, $b - d$, and $c + d$. | [
"Note that $a > c$ and $b > d$, as $a - c$ and $b - d$ are sides of a non-degenerate triangle. So $a \\ge c + 1$ and $b \\ge d + 1$, as they are integers. Consider two cases: $a > 2c$ and $a \\le 2c$.\n\nSuppose that $a > 2c$. Then $ab > 2bc \\ge 2c \\cdot (d+1) = 2cd + 2c$. We also have $ab = 2 + 2cd$, so $2c < 2$... | Netherlands | IMO Team Selection Test 3 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof and answer | [(1,2,0,1),(2,1,1,0)] | |
0a1x | Problem:
Vind alle functies $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}$ zodanig dat
$$
f(x)+f(y)=\left(f(x+y)+\frac{1}{x+y}\right)(1-x y+f(x y))
$$
voor alle $x, y \in \mathbb{Q}^{+}$. | [
"Solution:\n\nAntwoord: de enige functie die voldoet is $f\\left(\\frac{m}{n}\\right)=\\frac{m}{n}-\\frac{n}{m}$. Deze functie voldoet inderdaad want\n$$\n\\begin{aligned}\n\\left(f(x+y)+\\frac{1}{x+y}\\right)(1-x y+f(x y)) & =\\left((x+y)-\\frac{1}{x+y}+\\frac{1}{x+y}\\right)\\left(1-x y+x y-\\frac{1}{x y}\\right)... | Netherlands | IMO-selectietoets III | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(x) = x - 1/x for all positive rational x | |
04kn | Let $S$ and $T$ be points inside the triangle $ABC$. The distance from $S$ to lines $AB$, $BC$ and $CA$ is $10$, $7$ and $4$, respectively. The distance from $T$ to these lines is $4$, $10$ and $16$, respectively.
Determine the radius of the incircle of triangle $ABC$. | [
"Denote $a = |BC|$, $b = |CA|$, $c = |AB|$. Let $P$ be the area, $s = \\frac{1}{2}(a + b + c)$ the semiperimeter, and $r$ the radius of the incircle of triangle $ABC$.\nWe can divide triangle $ABC$ into three smaller triangles by connecting point $S$ to vertices $A$, $B$ and $C$.\n\n\nBy ad... | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 8 | |
0joo | Problem:
Three ants begin on three different vertices of a tetrahedron. Every second, they choose one of the three edges connecting to the vertex they are on with equal probability and travel to the other vertex on that edge. They all stop when any two ants reach the same vertex at the same time. What is the probabili... | [
"Solution:\n\nAt every second, each ant can travel to any of the three vertices they are not currently on. Given that, at one second, the three ants are on different vertices, the probability of them all going to the same vertex is $\\frac{1}{27}$ and the probability of them all going to different vertices is $\\fr... | United States | HMMT November 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/16 | |
0ecn | Using eight isosceles right-angled triangles we assemble a regular octagon with an octagonal hole, as shown in the figure. The side length of the regular octagon equals $1$ cm. How many centimetres is the side length of the octagonal hole?
(A) $\frac{\sqrt{2}}{2}$
(B) $\frac{1}{2}$
(C) $2 - ... | [
"The legs of each right-angled triangle are of length $1$ cm and its hypotenuse is of length $\\sqrt{2}$ cm. Thus, the side of the octagonal hole is of length $\\sqrt{2} - 1$ cm."
] | Slovenia | National Math Olympiad 2015 – First Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
03oq | Consider a square on the complex plane. The complex numbers corresponding to its four vertices are the four roots of some equation of the fourth degree with one unknown and integer coefficients $x^4+px^3+qx^2+rx+s=0$. Find the minimum value of the area of such square. (posed by Xiong Bin) | [
"Suppose the complex number corresponding to the center of the square is $a$. Then after translating the origin of the complex plane to $a$, the vertices of the square distribute evenly on the circumference. That is, they are the solutions of equation $(x-a)^4 = b$, where $b$ is a complex number. Hence,\n\n$$\n\\be... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof and answer | 2 | |
0l9l | Find the least value and the greatest value of the expression
$$
P = x + y
$$
where $x, y$ are real numbers satisfying the condition
$$
x - 3\sqrt{x + 1} = 3\sqrt{y + 2} - y.
$$ | [
"Write the given condition in the form\n$$\nx + y = 3(\\sqrt{x+1} + \\sqrt{y+2}).\n$$\nDenote by $G$ the set of values of $P$.\nIt is easily seen that:\n$a \\in G \\Leftrightarrow a$ is a real number so that the following system of equations (with unknowns $x, y$) has solutions\n$$\n\\begin{cases} 3(\\sqrt{x+1} + \... | Vietnam | Vietnam Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | min P = (9 + 3√21) / 2, max P = 9 + 3√15 | |
0jrx | Problem:
Dodecagon $Q W A R T Z S P H I N X$ has all side lengths equal to $2$, is not self-intersecting (in particular, the twelve vertices are all distinct), and moreover each interior angle is either $90^{\circ}$ or $270^{\circ}$. What are all possible values of the area of $\triangle S I X$? | [
"Solution:\n\nThe dodecagon has to be a \"plus shape\" of area $20$, then just try the three non-congruent possibilities."
] | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | null | |
03wq | Prove that for any given positive integer $k$, there exist infinitely many positive integers $n$, such that the numbers
$$
2^n + 3^n - 1, 2^n + 3^n - 2, \dots, 2^n + 3^n - k
$$
are all composite. | [
"For any given positive integer $k$, choose positive integer $m$ sufficiently large such that $2^m + 3^m - k > 1$. Consider the following $k$ integers:\n$$\n2^m + 3^m - 1, 2^m + 3^m - 2, \\dots, 2^m + 3^m - k,\n$$\nall of which are larger than $1$. From each of these integers, pick a prime factor: $p_1, p_2, \\dots... | China | China Western Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0i71 | Problem:
A circle is inscribed in a square dartboard. If a dart is thrown at the dartboard and hits the dartboard in a random location, with all locations having the same probability of being hit, what is the probability that it lands within the circle? | [
"Solution:\n\nThe answer is the area of the circle over the area of the square, which is $\\frac{\\pi}{4}$."
] | United States | Harvard-MIT Math Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | π/4 | |
0k5c | As usual, let $\mathbb{Z}[x]$ denote the set of single-variable polynomials in $x$ with integer coefficients.
Find all functions $\theta : \mathbb{Z}[x] \to \mathbb{Z}$ such that for any polynomials $p, q \in \mathbb{Z}[x]$,
* $\theta(p + 1) = \theta(p) + 1$, and
* if $\theta(p) \neq 0$ then $\theta(p)$ divides $\theta... | [
"The answer is $\\theta : p \\mapsto p(c)$, for each choice of $c \\in \\mathbb{Z}$. Obviously these work, so we prove these are the only ones. In what follows, $x \\in \\mathbb{Z}[x]$ is the identity polynomial, and $c = \\theta(x)$.\n\n**First solution (Merlijn Staps)** Consider an integer $n \\neq c$. Because $x... | United States | USA TSTST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All such functions are evaluations at a fixed integer: there exists c in the integers such that θ(p) = p(c) for all polynomials p. | |
0f92 | Problem:
A convex polygon is such that any segment dividing the polygon into two parts of equal area which has at least one end at a vertex has length $< 1$. Show that the area of the polygon is $< \pi /4$. | [] | Soviet Union | 23rd ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
09vl | Problem:
Aan een wiskundewedstrijd doen 300 deelnemers mee. Na de wedstrijd spelen sommige deelnemers wat potjes schaak. Elke twee deelnemers spelen hooguit één keer tegen elkaar. Er zijn geen drie deelnemers bij deze wedstrijd die onderling allemaal tegen elkaar schaken. Bepaal de maximale $n$ waarvoor het mogelijk i... | [
"Solution:\n\nWe laten zien dat $n$ maximaal 200 is. We geven eerst een voorbeeld waarin $n=200$ geldt. Bekijk spelers $A_{1}, A_{2}, \\ldots, A_{200}$ en $B_{1}, B_{2}, \\ldots, B_{100}$. Dit zijn 300 spelers in totaal. Laat $B_{i}$ schaak spelen tegen de spelers $A_{j}$ met $1 \\leq j \\leq i+100$, terwijl verder... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 200 | |
04il | The sum of squares of all solutions of the equation $x^4 + a x^2 + b = 0$ is $32$, and the product of all solutions of that equation is $4$. Determine $a$ and $b$.
(Tamara Srnec) | [
"**2.4.** By using the condition $a+b+c=1$ and the inequality between arithmetic and geometric means, we have that\n$$\n\\frac{a}{a+b^2} = \\frac{a}{a(a+b+c)+b^2} = \\frac{a}{a^2+b^2+ab+ac} \\le \\frac{a}{2ab+ab+ac} = \\frac{1}{3b+c}.\n$$\nBy applying the inequality between harmonic and arithmetic means, it follows... | Croatia | First round – City competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | a = -16, b = 4 | |
03xx | Let $G = G(V; E)$ be a simple graph with vertex set $V$ and edge set $E$, and assume that $|V| = n$. A map $f: V \rightarrow \mathbb{Z}$ is said to be good if $f$ satisfies:
$$
(1) \quad \sum_{v \in V} f(v) = |E|;
$$
(2) If one colors arbitrarily some vertices into red, there exists a red vertex $v$, such that $f(v)$ i... | [
"Given an ordering $\\tau = (v_1, v_2, ..., v_n)$ on the vertices in $V$, we associate a map $f_\\tau: V \\to \\mathbb{Z}$ as follows: $f_\\tau(v)$ is equal to the number of vertices in $V$ that are ordered preceding $v$. We claim that $f_\\tau$ is good.\n\nEach edge is counted exactly once in $\\sum_{v \\in V} f_\... | China | China National Team Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0dmm | Problem:
Нека је $n \in \mathbb{N}$ и $A_{n}$ скуп свих пермутација $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ скупа $\{1,2, \ldots, n\}$ таквих да важи
$$
k \mid 2\left(a_{1}+a_{2}+\cdots+a_{k}\right) \quad \text{ за свако } 1 \leqslant k \leqslant n
$$
Одредити број елемената скупа $A_{n}$. | [
"Solution:\n\nОзначимо са $F_{n}$ број елемената скупа $A_{n}$. Имамо $F_{1}=1, F_{2}=2$ и $F_{3}=6$. За $n>3$, посматрајмо било коју пермутацију $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ у $A_{n}$. Како $n-1$ дели $2\\left(a_{1}+\\cdots+a_{n-1}\\right)=n(n+1)-2 a_{n} \\equiv 2-2 a_{n}(\\bmod n-1)$, следи да је... | Serbia | Српска математичка олимпијада | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | F1 = 1, F2 = 2, and for n ≥ 3: Fn = 3 · 2^(n−2) | |
04cq | Let $n$ and $d$ be positive integers such that $2n^2$ is divisible by $d$. Prove that $n^2 + d$ is not a perfect square. | [
"Let $k$ be an integer such that $2n^2 = dk$. Then we have\n$$\n\\begin{aligned}\nk^2(n^2 + d) &= k^2n^2 + k^2d = k^2n^2 + 2kn^2 \\\\\n&= n^2(k^2 + 2k) = n^2((k+1)^2 - 1).\n\\end{aligned}\n$$\nExpression $(k+1)^2 - 1$ is not a square of an integer for $k \\in \\mathbb{N}$. Then, $k^2(n^2 + d)$ is not a square of an... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0k7v | Problem:
Alice and Bob play a game. There are 9 cards numbered 1 through 9 on a table, and the players alternate taking the cards, with Alice going first. A player wins if at any point they hold three cards with sum $15$; if all nine cards are taken before this occurs, the game is a tie. Does either player have a winn... | [
"Solution:\n\nNeither player has a winning strategy. Arrange the cards in the following $3 \\times 3$ array:\n$$\n\\left[\\begin{array}{lll}\n8 & 3 & 4 \\\\\n1 & 5 & 9 \\\\\n6 & 7 & 2\n\\end{array}\\right]\n$$\nIn this setup, a sum of $15$ corresponds to three-in-a-row. Thus we see that Alice and Bob are actually p... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Neither player has a winning strategy; with optimal play the game is a tie. | |
0hyf | Problem:
Prove that among any 12 consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (The proper divisors of a positive integer $n$ are all positive integers other than 1 and $n$ which divide $n$. For example, the proper divisors of 14 are 2 and 7.) | [
"Solution:\n\nOne of the twelve numbers must be a multiple of twelve; call it $a = 12 n$. Among the proper divisors of $a$ are the integers $2 n$, $3 n$, $4 n$, $6 n$. These sum to $15 n > a$."
] | United States | 1st Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | null | proof only | null | |
04xf | Let $p$ be a prime number. Show that one can choose $p^3$ fields of a $p^2 \times p^2$ chessboard such that centres of no four chosen fields are vertices of a rectangle with sides parallel to the sides of the chessboard. | [
"Let us label the $p^2$ rows and the $p^2$ columns of the chessboard by all the pairs $(a, b)$ and $(c, d)$, respectively, where $a, b, c, d \\in \\{0, 1, \\dots, p-1\\}$. A field lying in the row $(a, b)$ and in the column $(c, d)$ will be called good if and only if\n$$\nac \\equiv b+d \\pmod{p}. \\qquad (1)\n$$\n... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
06ra | Let $a, b$ be integers, and let $P(x) = a x^{3} + b x$. For any positive integer $n$ we say that the pair $(a, b)$ is $n$-good if $n \mid P(m) - P(k)$ implies $n \mid m - k$ for all integers $m, k$. We say that $(a, b)$ is very good if $(a, b)$ is $n$-good for infinitely many positive integers $n$.
a. Find a pair $(a,... | [
"a. We show that the pair $(1, -51^{2})$ is $51$-good but not very good. Let $P(x) = x^{3} - 51^{2} x$. Since $P(51) = P(0)$, the pair $(1, -51^{2})$ is not $n$-good for any positive integer that does not divide $51$. Therefore, $(1, -51^{2})$ is not very good.\n\nOn the other hand, if $P(m) \\equiv P(k) \\pmod{51}... | IMO | 51st IMO Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynom... | English | proof and answer | a: (1, -51^2). b: Every 2010-good pair is very good (indeed it is 67^i-good for all positive integers i). | |
0ghv | II-C. 兩隻魷魚被迫參加遊戲。在開始前，牠們會被告知所有的規則，並可以充分討論策略。之後，牠們會被關進各自的房間，並被分別告知一個不超過 $2023$ 的相異正整數做為其編號。魷魚接著輪流進行回合；在某一方的回合，牠從以下行動中擇一進行：
1. 喊出一個正整數；這會被另外一隻魷魚聽到。
2. 宣稱哪一支魷魚的編號較大。若正確，牠們共同獲勝並獲釋；否則，牠們共同失敗並變成炒魷魚。
試求最小的正整數 $N$，使得不論牠們被告知的編號為何，魷魚皆能保證於有限回合內獲勝，且在遊戲過程中喊出的數字總和不大於 $N$。
II-C. Two squids are forced to participate in a game. Before... | [
"答案是 $N = 10$，其中「」為上高斯符號。一般性地，如果被告知的編號必不超過 $M$，則最小的 $N = \\lceil \\log_2 M \\rceil - 1$。\n\n構造：將兩隻魷魚編號 $1$ 與 $2$，被告知的編號分別為 $D_1$ 與 $D_2$。當 $M \\le 2^{N+1}$ 時，編號可表示成\n$$\nD_i = p_{i,1} 2^N + p_{i,2} 2^{N-1} + \\dots + p_{i,N} 2^1 + 1 = 1 + \\sum_{j=1}^{N} p_{i,j} 2^{N+1-j},\n$$\n其中 $p_{i,j} \\in \\{0,1\\}$。方便起見，用 $p... | Taiwan | 2023 數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Chinese (Traditional) | proof and answer | 10 | |
0bok | Given an integer $n \ge 3$ and a permutation $a_1, a_2, \dots, a_n$ of the first $n$ positive integers, show that at least $\sqrt{n}$ distinct residue classes modulo $n$ occur in the list $a_1, a_1 + a_2, \dots, a_1 + a_2 + \dots + a_n$. | [
"Let $r_k$ be the remainder of the partial sum $a_1 + a_2 + \\dots + a_k$ upon division by $n$, and let $m$ be the number of distinct residues in the list $r_1, r_2, \\dots, r_n$. Since there are $m$ distinct integers in the list, the number of ordered pairs with different entries that can be formed from these $m$ ... | Romania | 2015 Ninth STARS OF MATHEMATICS Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | English | proof only | null | |
01su | There are $n \geq 1$ cities on a horizontal line. Each city is guarded by a pair of stationary elephants, one just to the left and one just to the right of the city, and facing away from it. The $2n$ elephants are of different sizes. If an elephant walks forward, it will knock aside any elephant that it approaches from... | [] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0gub | Let $p$ be a given prime number and $n, k \ge 2$ be integers. For each integer $i = 1, 2, \dots, n$, the set $S_i$ is constructed by choosing exactly one square from each column of a given $p \times k$ chessboard. Given that $|S_i \cap S_j| = 1$ for all $1 < i < j \le n$ and for every pair of squares belonging to diffe... | [
"Only $(p^2, p+1)$.\n\nObviously $|S_i| = k$ for every $i$. Also note that given conditions imply that for every pair of squares belonging to different columns there exists a unique $S_i$ containing these two squares. Therefore, by a double counting argument we get $\\binom{k}{2}p^2 = n\\binom{k}{2}$, which yields ... | Turkey | Team Selection Test for EGMO 2024 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | (p^2, p+1) | |
0ajx | Let be given $\triangle ABC$ and let $AA_1$, $BB_1$ and $CC_1$ are the medians in the triangle which intersect in the point $T$ and $\overline{AD_1}=\overline{AT}$. On the continuation of $CC_1$ we choose a point $C_2$ such that $\overline{C_1C_2}=\frac{CC_1}{3}$, and on the continuation of $BB_1$ we choose a point $B_... | [
"Since $AA_1$ is a median in the $\\triangle ABC$ and $\\overline{BA_1}=\\overline{AT}$, we get that $\\overline{AT}=\\frac{BC}{2}$. $A_1$ is the circumcenter of the circumcircle of $\\triangle BC_1T$. So according to the Thales theorem $\\angle BT_1C=90^\\circ$. We have $\\angle B_2T_1C_2=90^\\circ$ (as vertically... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08yh | Find the smallest integer $n$ with the following property:
Remainders obtained when $n$ is divided by $97$, $100$, $103$ are $32$, $33$, $34$, respectively. | [
"$333033$\nLet $n$ be a positive integer satisfying the conditions of the problem. Then, $n-32$, $n-33$, $n-34$ are multiples of $97$, $100$, $103$, respectively. Then, we have\n$$\n3n+1 = 3(n-32) + 97 = 3(n-33) + 100 = 3(n-34) + 103,\n$$\nfrom which it follows that $3n + 1$ is a multiple of $97$, $100$, $103$. Any... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 333033 | |
04r0 | Find all pairs of integers $a, b$ such that
$$
\frac{a^2+1}{2b^2-3} = \frac{a-1}{2b-1}
$$ | [
"Obviously $a \\neq 1$, thus we can rewrite the equation as\n$$\n\\frac{a^2+1}{a-1} = \\frac{2b^2-3}{2b-1} \\quad (1)\n$$\nThe numerator of the fraction on the left is positive, the numerator on the right is negative just for $b \\in \\{-1, 0, 1\\}$.\nFor $b = -1$ we get $3a^2 - a + 4 = 0$, which has no real soluti... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(0, 1), (-1, 1), (0, -2), (-1, -2), (7, 8)] | |
09ln | For $n \ge 1$, let $P_n(x)$ be the polynomial defined by $P_1(x) = 1$, $P_2(x) = 2x$, and
$$
P_n(x) = 2x P_{n-1}(x) - (x^2 + 1) P_{n-2}(x)
$$
for $n \ge 3$. Prove that we have $|P_n(x)| \le (1 + x^2)^{n/2}$ for any positive integer $n$ and real number $x$, and determine the conditions under which equality holds. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | For all positive integers n and real x, |P_n(x)| ≤ (1 + x^2)^{n/2}. Equality holds if and only if (x + i)^n is purely imaginary, equivalently there exists an integer k such that x = cot((2k + 1)π/(2n)). | |
04kq | Let $a$ and $b$ be positive integers such that $8a^2 + 1 = b^2$. Prove that $ab$ is divisible by $3$. | [
"We have $8a^2 + 1 = b^2$, so $b^2 - 8a^2 = 1$.\n\nConsider the equation modulo $3$:\n\n$8a^2 + 1 \\equiv b^2 \\pmod{3}$\n\nSince $8 \\equiv 2 \\pmod{3}$, we have:\n\n$2a^2 + 1 \\equiv b^2 \\pmod{3}$\n\nNow, $a^2$ modulo $3$ can be $0$ or $1$:\n- If $a \\equiv 0 \\pmod{3}$, then $a^2 \\equiv 0$.\n- If $a \\equiv 1 ... | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0anf | Problem:
Evaluate $\cos \frac{\pi}{10} + \cos \frac{2\pi}{10} + \cos \frac{3\pi}{10} + \ldots + \cos \frac{19\pi}{10}$.
(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) $-1$
(d) $1$ | [] | Philippines | Qualifying Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | MCQ | c | |
0a75 | Problem:
Three sides of a tetrahedron are right-angled triangles having the right angle at their common vertex. The areas of these sides are $A$, $B$, and $C$. Find the total surface area of the tetrahedron. | [
"Solution:\n\nLet $PQRS$ be the tetrahedron of the problem and let $S$ be the vertex common to the three sides which are right-angled triangles. Let the areas of $PQS$, $QRS$, and $RPS$ be $A$, $B$, and $C$, respectively. Denote the area of $QRS$ by $X$. If $SS'$ is the altitude from $S$ (onto $PQR$) and $\\angle R... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 3 | [
"Geometry > Solid Geometry > Surface Area",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | A + B + C + sqrt(A^2 + B^2 + C^2) | |
0ji6 | Problem:
It is known that exactly one of the three (distinguishable) musketeers stole the truffles. Each musketeer makes one statement, in which he either claims that one of the three is guilty, or claims that one of the three is innocent. It is possible for two or more of the musketeers to make the same statement. Af... | [
"Solution:\n\nAnswer: 99\n\nWe divide into cases, based on the number of distinct people that statements are made about.\n\n- The statements are made about 3 distinct people. Then, since exactly one person is guilty, and because exactly one of the three lied, there are either zero statements of guilt or two stateme... | United States | HMMT | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Logic"
] | null | proof and answer | 99 | |
0l5g | Find the sum of all positive integers $n$ such that $n + 2$ divides the product $3(n + 3)(n^2 + 9)$. | [
"Solution:\nNote that $n \\equiv -2 \\pmod{(n+2)}$ and thus $3(n+3)(n^2+9) \\equiv 3 \\cdot 1 \\cdot 13 = 39 \\pmod{(n+2)}$. Therefore $n$ is a positive solution if and only if $n+2$ is a divisor of $39$ that is at least $3$. Those divisors are $3$, $13$, and $39$, and the corresponding values of $n$ are $1$, $11$,... | United States | AIME II | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 49 | |
03cg | Find all primes $p$ and all positive integers $a$ and $m$ such that $a \le 5p^2$ and $(p-1)! + a = p^m$. | [
"Answer: $(p, a, m) = (2, 1, 1), (2, 3, 2), (2, 7, 3), (3, 1, 1), (3, 7, 2), (3, 25, 3), (5, 1, 2), (2, 15, 4), (5, 101, 3)$.\nDirect checks for $p = 2, 3, 5$ give the above solutions.\nFor $p \\ge 7$ the Wilson theorem implies $a \\equiv 1 \\pmod p$. Moreover, obviously $p-1|a-1$. Therefore $a = kp(p-1)+1$ for som... | Bulgaria | BULGARIAN NATIONAL MATHEMATICAL OLYMPIAD | [
"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... | English | proof and answer | (p, a, m) = (2, 1, 1), (2, 3, 2), (2, 7, 3), (3, 1, 1), (3, 7, 2), (3, 25, 3), (5, 1, 2), (2, 15, 4), (5, 101, 3) | |
0jwt | Problem:
In a certain college containing $1000$ students, students may choose to major in exactly one of math, computer science, finance, or English. The diversity ratio $d(s)$ of a student $s$ is defined as the number of students in a different major from $s$ divided by the number of students in the same major as $s$... | [
"Solution:\n\nIt is easy to check that if $n$ majors are present, the diversity is $1000(n-1)$. Therefore, taking $n = 1, 2, 3, 4$ gives us all possible answers."
] | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 0, 1000, 2000, 3000 | |
017n | In a group of $30$ people, every member initially had a hat. One day each member sent his hat to a different member (a member could have received more than one hat). Prove that there exists a subgroup of $10$ people such that no member of the subgroup has received a hat from another member of the subgroup. | [
"Let $S$ be the given group of $30$ people. Consider all subsets $A \\subset S$ such that no member of $A$ received a hat from a member of $A$. Among such subsets, let $T$ be a subset of maximal cardinality. The assertion of the problem is that $|T| \\ge 10$.\n\nLet $U \\subset S$ consist of all people that have re... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0avb | Problem:
Find the number of pairs of positive integers $(n, k)$ that satisfy the equation $(n+1)^{k} - 1 = n$. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | infinitely many | |
0ff4 | Problem:
Decimos que tres números naturales distintos forman una terna aditiva si la suma de los dos primeros de ellos es igual al tercero. Hallar, razonadamente, el máximo número de ternas aditivas que puede haber en un conjunto dado de 20 números naturales. | [
"Solution:\n\nSean veinte números naturales mutuamente distintos, arbitrariamente elegidos:\n$$\nx_{1} < x_{2} < \\cdots < x_{20}\n$$\nVamos a obtener una cota superior para el número de ternas aditivas que se pueden encontrar entre ellos, es decir, ternas $\\{x_{i}, x_{j}, x_{k}\\}$ que verifiquen $1 \\leq i < j <... | Spain | TANDA III | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 90 | |
0gxq | Find all functions $f$, mapping the set of integer numbers into itself, such that for all integers $m$ and $n$ the following equality holds: $f(n|m|) + f(n(|m|+2)) = 2f(n(|m|+1))$. | [
"Put $n=1$: $f(|m|) + f(|m|+2) = 2f(|m|+1)$. If we denote $l=|m| \\ge 0$, then we can easily get: $f(l+2) - f(l+1) = f(l+1) - f(l)$, so our sequence is an arithmetic progression, hence $f(l) = al + b$, where $a,b \\in \\mathbb{Z}$, $l \\in \\mathbb{Z}^+$. \n\nIn the same way, substituting $n=-1$ we obtain: $f(-|m|)... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | All solutions are piecewise linear with a common intercept: f(n) = a n + b for n > 0, f(0) = b, and f(n) = c n + b for n < 0, where a, b, c are integers. | |
0aoc | Problem:
What is the sum of the first 800 terms of $3,4,4,5,5,5,6,6,6,6, \ldots$ ? | [] | Philippines | Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 22940 | |
01wz | Given the equation
$$
a^b \cdot b^c = c^a
$$
in positive integers $a$, $b$ and $c$.
(i) Prove that any prime divisor of $a$ divides $b$ as well.
(ii) Solve the equation under the assumption $b \ge a$.
(iii) Prove that the equation has infinitely many solutions. | [
"(ii) $a = b = c = 1$.\n\nIt is easy to verify that $a = n^n$, $b = n^{n-1}$, $c = n^n$ is a solution of the given equation for all $n \\in \\mathbb{N}$, thus (iii) is solved.\n\nNote that if $a$, $b$ or $c$ equals $1$, then $a = b = c = 1$, which is a solution. Now we prove that if $(a, b, c)$ is a solution of\n$$... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (i) Every prime factor of a also divides b. (ii) Under the assumption b at least a, the only solution is a = b = c = 1. (iii) There are infinitely many solutions, for example (a, b, c) = (n^n, n^{n-1}, n^n) for all positive integers n. | |
097s | Problem:
În triunghiul $A B C$, punctul $M$ este mijlocul laturii $B C$. Pe bisectoarea unghiului $B A C$ se ia punctul $N$ astfel încât $B N \perp A N$. Determinați lungimea laturii $A C$, dacă $A B=10~\mathrm{cm}$ și $M N=2~\mathrm{cm}$. | [
"Solution:\n\nFie $B N \\cap A C=\\{L\\}$.\nDistingem următoarele cazuri potențiale:\n\n1. Dacă $N \\in(B C)$, atunci $A N$ este bisectoare și înălțime în $\\triangle A B C$, adică și mediană. Prin urmare $N \\equiv M$, ceea ce contrazice ipotezei $M N=2~\\mathrm{cm}$. Cazul este imposibil.\n\n2. Presupunem că punc... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 14 cm or 6 cm | |
0a9h | Problem:
A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n = a_{1} \pm a_{2} \pm a_{3} \pm \cdots \pm a_{k}$, where $a_{1}, \ldots, a_{k}$ are simple. | [
"Solution:\nWe can always write $n = a_{1} + a_{2} + \\cdots + a_{9}$ where $a_{j}$ has 1's in the places where $n$ has digits greater or equal to $j$ and 0's in the other places. So $k \\leq 9$.\n\nTo show that $k \\geq 9$, consider $n = 10203040506070809$. Suppose $n = a_{1} + a_{2} + \\cdots + a_{j} - a_{j+1} - ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 24 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 9 | |
03y9 | Two persons roll two dice in turn. Whoever gets the sum number greater than $6$ first will win the game. The probability for the person rolling first to win is ________. | [
"The probability for rolling two dice to get the sum number greater than $6$ is $\\frac{21}{36} = \\frac{7}{12}$. Therefore, the required probability is\n$$\n\\frac{7}{12} + \\left(\\frac{5}{12}\\right)^2 \\frac{7}{12} + \\left(\\frac{5}{12}\\right)^4 \\frac{7}{12} + \\cdots = \\frac{7}{12} \\times \\frac{1}{1 - \\... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 12/17 | |
0k7r | Problem:
Let $P$ be a polynomial with positive real coefficients. Prove that if
$$
P\left(\frac{1}{x}\right) \geq \frac{1}{P(x)}
$$
holds for $x=1$, then it holds for every $x>0$. | [
"Solution:\nLet $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{1} x+a_{0}$, so $a_{k}>0$ for every $k$. Since the statement holds for $x=1$, $P(1) \\geq 1$. Then by Cauchy-Schwarz,\n$$\nP(x) P\\left(\\frac{1}{x}\\right)=\\left(\\sum_{k=0}^{n}\\left(\\sqrt{a_{k} x^{k}}\\right)^{2}\\right)\\left(\\sum_{k=0}^{n}\\left(\... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
05mg | Problem:
Soient $n$ et $k$ des entiers strictement positifs. Il y a $n k$ objets (de même taille) et $k$ boîtes qui peuvent contenir chacune $n$ objets. Chaque objet est colorié en une couleur parmi $k$ couleurs possibles. Montrer qu'il est possible de ranger les objets dans les boîtes de sorte que chaque boîte contie... | [
"Solution:\n\nOn le montre par récurrence sur $k$. Si $k=1$ c'est évident. Supposons la propriété vraie pour tous les entiers $<k$.\n\nSi l'une des couleurs $c$ est prise exactement $n$ fois, on range tous les objets dont la couleur est $c$ dans la dernière boîte. L'hypothèse de récurrence permet de ranger les $n(k... | France | Olympiades FRANçaises DE Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0iwt | Problem:
How many subsets $A$ of $\{1,2,3,4,5,6,7,8,9,10\}$ have the property that no two elements of $A$ sum to $11$? | [
"Solution:\nFor each element listed, there is exactly one other element such that the two elements sum to $11$. Thus, we can list all the $10$ numbers above as $5$ pairs of numbers, such that each pair sums to $11$.\n\nThe problem then can be solved as follows: in any given subset with no two elements summing to $1... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 243 | |
05fp | Problem:
Prouver que tout ensemble de 90 nombres choisis dans $\{1,2, \cdots, 100\}$ en contient 10 qui forment une progression arithmétique. | [
"Solution:\nOn dispose les entiers de 1 à 100 dans un tableau carré $10 \\times 10$ comme suit :\n\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|\n| 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |\n| 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 ... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null |
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