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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
00m0 | Es sei eine reelle Zahl $\alpha$ gegeben.
Man bestimme in Abhängigkeit von $\alpha$ alle Funktionen $f: \mathbb{R} \to \mathbb{R}$ mit
$$
f(f(x+y)f(x-y)) = x^2 + \alpha y f(y)
$$
für alle $x, y \in \mathbb{R}$. | [
"Wir zeigen: Für $\\alpha = -1$ ist $f(x) = x$ die einzige Lösung; sonst gibt es keine Lösung.\nBeweis. Mit $x = y = 0$ erhalten wir $f(f(0)^2) = 0$. Mit $x = 0$ und $y = f(0)^2$ erhalten wir $f(0) = 0$. Mit $y = x$ erhalten wir $f(0) = x^2 + \\alpha x f(x)$. Für $\\alpha = 0$ ergibt das einen Widerspruch, wir nehm... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Algebra > Algebraic Expressions > Functional Equations"
] | German | proof and answer | alpha = -1 with f(x) = x for all real x; for all other alpha there is no solution | |
00ho | Let $ABCD$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $AC$ and $BD$, let $L$ be the center of the circle tangent to sides $AB$, $BC$, and $CD$, and let $M$ be the midpoint of the arc $BC$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter... | [
"Let $L$ be the intersection of the bisectors of $\\angle ABC$ and $\\angle BCD$. Let $N$ be the $E$-excenter of $\\triangle BCE$. Let $\\angle BAC=\\angle BDC=\\alpha$, $\\angle DBC=\\beta$ and $\\angle ACB=\\gamma$.\nWe have the following:\n$$\n\\begin{array}{r}\n\\angle CBL=\\frac{1}{2} \\angle ABC=90^\\circ-\\f... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2021 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter... | English | proof only | null | |
060b | Problem:
Aurélien découpe une feuille de papier en 7 morceaux. Une étape consiste ensuite à choisir un morceau et à le découper en 4, 7 ou 10 morceaux. Aurélien peut-il obtenir ainsi 2021 morceaux ? | [
"Solution:\n\nL'exercice décrit une suite d'opérations et demande s'il est possible de passer d'un état initial à un état final, on peut donc légitimement chercher un invariant du système.\n\nIci, nous allons montrer que le nombre de morceaux d'Aurélien est toujours de la forme $3k+1$, avec $k$ un entier naturel.\n... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | No | |
06qo | Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P \subset R$ we have
$$
\frac{|R|}{|P|} \leq \sqrt{2}
$$
where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. | [
"We will construct two parallelograms $R_{1}$ and $R_{3}$, each of them containing $P$, and prove that at least one of the inequalities $|R_{1}| \\leq \\sqrt{2}|P|$ and $|R_{3}| \\leq \\sqrt{2}|P|$ holds (see Figure 1).\nFirst we will construct a parallelogram $R_{1} \\supseteq P$ with the property that the midpoin... | IMO | IMO Problem Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ijo | Problem:
Compute $\sum_{k=1}^{\infty} \frac{k^{4}}{k!}$. | [
"Solution:\n\nDefine, for non-negative integers $n$,\n$$\nS_{n}:=\\sum_{k=0}^{\\infty} \\frac{k^{n}}{k!}\n$$\nwhere $0^{0}=1$ when it occurs. Then $S_{0}=e$, and, for $n \\geq 1$,\n$$\nS_{n}=\\sum_{k=0}^{\\infty} \\frac{k^{n}}{k!}=\\sum_{k=1}^{\\infty} \\frac{k^{n}}{k!}=\\sum_{k=0}^{\\infty} \\frac{(k+1)^{n}}{(k+1)... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 15e | |
05uh | Problem:
Pour tout entier $k \geqslant 0$, on note $F_{k}$ le $k^{\text{ème}}$ nombre de Fibonacci, défini par $F_{0}=0$, $F_{1}=1$, et $F_{k}=F_{k-2}+F_{k-1}$ lorsque $k \geqslant 2$. Soit $n \geqslant 2$ un entier, et soit $S$ un ensemble d'entiers ayant la propriété suivante :
Pour tout entier $k$ tel que $2 \leqsl... | [
"Solution:\n\nTout d'abord, soit $m=\\lceil n / 2\\rceil$, et soit $S$ l'ensemble $\\{F_{2 \\ell}: 0 \\leqslant \\ell \\leqslant m\\}$. Pour tout entier $k$ tel que $2 \\leqslant k \\leqslant n$, on choisit $x=F_{k}$ et $y=F_{0}=0$ si $k$ est pair, ou bien $x=F_{k+1}$ et $y=F_{k-1}$ si $k$ est impair. Dans les deux... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | ceil(n/2)+1 | |
04ew | How many complex numbers $z$ are there that satisfy the following two conditions:
$$
|z| = 1, \quad \text{Re}(z^{100}) = \text{Im}(z^{200}) \quad ? \qquad (\text{Ilko Brnetić})
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 400 | |
0j5d | Problem:
A small fish is holding 17 cards, labeled 1 through 17, which he shuffles into a random order. Then, he notices that although the cards are not currently sorted in ascending order, he can sort them into ascending order by removing one card and putting it back in a different position (at the beginning, between ... | [
"Solution:\nAnswer: 256\nInstead of looking at moves which put the cards in order, we start with the cards in order and consider possible starting positions by backtracking one move: each of 17 cards can be moved to 16 new places. But moving card $k$ between card $k+1$ and card $k+2$ is equivalent to moving card $k... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 256 | |
06bu | Prove that there are infinitely many positive integers $n$ such that $2^n + 1$ is divisible by $n$. Find all such $n$'s that are prime numbers. | [
"All integers $n = 3^k$ with $k \\in \\mathbb{Z}^+$ satisfy $n \\mid 2^n + 1$. Indeed, since $3 \\mid 2 + 1$, by the lifting the exponent lemma, we have\n$$\nv_3(2^{3^k} + 1^{3^k}) = v_3(2 + 1) + v_3(3^k) = 1 + k.\n$$\nThis implies $3^{k+1} \\mid 2^n + 1$, and hence $n \\mid 2^n + 1$.\n\nThe only prime number $n$ s... | Hong Kong | IMO HK TST | [
"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"
] | null | proof and answer | Infinitely many examples: n = 3^k for any positive integer k. Among primes, the only solution is n = 3. | |
026q | Problem:
No quadrilátero $ABCD$, tem-se: $AB=5$, $BC=17$, $CD=5$, $DA=9$, e a medida do segmento $DB$ é um inteiro. Determine $DB$.
 | [
"Solution:\n\nLembre que, num triângulo, qualquer lado é maior que a diferença e menor que a soma dos outros dois. Do triângulo $ADB$ temos $AD-AB < BD < AD+AB$, e do triângulo $CBD$ segue que $BC-CD < BD < BC+CD$. Substituindo os valores conhecidos obtemos:\n$$\n9-5 < BD < 5+9 \\text{ e } 17-5 < BD < 17+5\n$$\nou ... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 13 | |
0ker | Problem:
The elevator buttons in Harvard's Science Center form a $3 \times 2$ grid of identical buttons, and each button lights up when pressed. One day, a student is in the elevator when all the other lights in the elevator malfunction, so that only the buttons which are lit can be seen, but one cannot see which floo... | [
"Solution:\n\nWe first note that there are $2^{6}-1=63$ possibilities for lights in total. We now count the number of duplicates we need to subtract by casework on the number of buttons lit. To do this, we do casework on the size of the minimal \"bounding box\" of the lights:\n\n- If the bounding box is $1 \\times ... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 44 | |
004b | Dado un entero positivo $N$, las operaciones permitidas para obtener otro son:
a. Dividir a $N$ por 2 ó por 3, si el resultado es un número entero.
b. Agregar un 0 ó un 8 a la derecha de la cifra de las unidades de $N$.
Decimos que un número es "rioplatense" si se puede obtener a partir del número 8 mediante una sucesi... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Español | proof and answer | i) 2006 es rioplatense (por ejemplo: 8 → 4 → 48 → 24 → 12 → 128 → 64 → 32 → 320 → 3208 → 1604 → 16048 → 8024 → 4012 → 2006 usando únicamente las operaciones permitidas). ii) Sí, todo entero positivo es rioplatense: trabajando hacia atrás, para cualquier número se puede multiplicar por potencias de dos hasta que su últi... | |
01s7 | Given positive real numbers $a, b, c, d$, with
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 1,
$$
prove that
$$
\frac{a+b}{a^2 - ab + b^2} + \frac{b+c}{b^2 - bc + c^2} + \frac{c+d}{c^2 - cd + d^2} + \frac{d+a}{d^2 - da + a^2} \le 2.
$$ | [
"Let $x = 1/a$, $y = 1/b$, $z = 1/c$, $u = 1/d$. Then $x + y + z + u = 1$, $xyzu \\ne 0$, and the required inequality has the form\n$$\n\\frac{xy(x+y)}{x^2-xy+y^2} + \\frac{yz(y+z)}{y^2-yz+z^2} + \\frac{zu(z+u)}{z^2-zu+u^2} + \\frac{ux(u+x)}{u^2-ux+x^2} \\le 2, \\quad (1)\n$$\nsince\n$$\n\\frac{a+b}{a^2-ab+b^2} = \... | Belarus | FINAL ROUND | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
00qp | Let $a$, $b$ and $c$ be the sides of a triangle and $r$, $R$ and $s$ be the inradius, the circumradius and the semiperimeter of the triangle respectively. Prove that
$$
\frac{1}{a+b} + \frac{1}{a+c} + \frac{1}{b+c} \le \frac{r}{16Rs} + \frac{s}{16Rr} + \frac{11}{8s}
$$ | [
"The following are well known identities relating $r$, $R$, $s$:\n$$\na+b+c=2s, \\quad ab+bc+ca=s^2+r^2+4Rr, \\quad abc=4Rrs\n$$\nThen\n$$\n\\frac{1}{a+b} + \\frac{1}{a+c} + \\frac{1}{b+c} \\le \\frac{r}{16Rs} + \\frac{s}{16Rr} + \\frac{11}{8s}\n$$\n$$\n\\Leftrightarrow 4 \\left( \\frac{1}{a+b} + \\frac{1}{a+c} + \... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, ... | null | proof only | null | |
06xd | Prove that $5^{n}-3^{n}$ is not divisible by $2^{n}+65$ for any positive integer $n$. | [
"Notice that if $n$ is even, then $3 \\mid m$, but $3 \\nmid 5^{n}-3^{n}$, contradiction. So, from now on we assume that $n$ is odd, $n=2k+1$. Obviously $n=1$ is not possible, so $n \\geqslant 3$. Notice that $m$ is coprime to $2$, $3$ and $5$.\nLet $m_1$ be the smallest positive multiple of $m$ that can be written... | IMO | International Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof only | null | |
0e90 | Find all quadruples of non-zero digits $a$, $b$, $c$ and $d$ such that $\overline{ab20} - \overline{13cd} = \overline{cdab}$. | [
"Rewrite the equation as $cdab + 13cd = ab20$. Since $b$ and $d$ are non-zero digits the equation for the ones is $b + d = 10$. Subtracting this on both sides of the equation we get $cda0 + 13c0 = ab10$. Dividing the equation by $10$ we obtain $cda + 13c = ab1$. Since $a$ and $c$ are non-zero digits, their sum cann... | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | (a, b, c, d) = (6, 7, 5, 3) | |
0itf | Problem:
How many integers between $2$ and $100$ inclusive cannot be written as $m \cdot n$, where $m$ and $n$ have no common factors and neither $m$ nor $n$ is equal to $1$? Note that there are $25$ primes less than $100$. | [
"Solution:\n\nA number cannot be written in the given form if and only if it is a power of a prime. We can see this by considering the prime factorization. Suppose that $k = p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{n}^{e_{n}}$, with $p_{1}, \\ldots, p_{n}$ primes. Then we can write $m = p_{1}^{e_{1}}$ and $n = p_{2}^... | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 35 | |
0dlb | A large rectangle is subdivided into smaller rectangles, each of which has at least one pair of sides of integer length. Prove that the large rectangle also has at least one pair of sides of integer length. | [
"Label the large rectangle by $R$ and place it in the plane so that its lower-left corner is at $(0, 0)$ with sides parallel to $Ox, Oy$.\nLet $S$ be the set of vertices of the small rectangles whose both coordinates are integers, and let $T$ be the set of all the small rectangles. We form a bipartite graph on the ... | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0f4u | Problem:
Do there exist polynomials $p(x)$, $q(x)$, $r(x)$ such that
$$
p(x - y + z)^3 + q(y - z - 1)^3 + r(z - 2x + 1)^3 = 1
$$
for all $x$, $y$, $z$?
Do there exist polynomials $p(x)$, $q(x)$, $r(x)$ such that
$$
p(x - y + z)^3 + q(y - z - 1)^3 + r(z - x + 1)^3 = 1
$$
for all $x$, $y$, $z$? | [] | Soviet Union | 16th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | Yes for both equations; for example, take p to be identically one and q and r to be identically zero. | |
0i3v | Problem:
Find the real solutions of $(2x+1)(3x+1)(5x+1)(30x+1) = 10$. | [
"Solution:\n$(2x+1)(3x+1)(5x+1)(30x+1) = [(2x+1)(30x+1)][(3x+1)(5x+1)] = (60x^2 + 32x + 1)(15x^2 + 8x + 1) = (4y+1)(y+1) = 10$, where $y = 15x^2 + 8x$.\n\nThe quadratic equation in $y$ yields $y = 1$ and $y = -\\frac{9}{4}$.\n\nFor $y = 1$, we have $15x^2 + 8x - 1 = 0$, so $x = \\frac{-4 \\pm \\sqrt{31}}{15}$.\n\nF... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-4 + sqrt(31))/15 and (-4 - sqrt(31))/15 | |
051i | Let $ABC$ be a triangle where $|AB| = |AC|$. Points $P$ and $Q$ are different from the vertices of the triangle and lie on the sides $AB$ and $AC$, respectively. Prove that the circumcircle of the triangle $APQ$ passes through the circumcenter of $ABC$ if and only if $|AP| = |CQ|$. | [
"Without loss of generality, we can assume that $|AP| \\le |AQ|$. Let $O$ be the circumcenter of $ABC$. Let $R$ be the intersection point of the bisector of $\\angle BAC$ with the circumcircle of the triangle $PAQ$ — we then have $|RB| = |RC|$ (Fig. 21).\n\nAlso, $\\angle APR = 180^\\circ - \\angle AQR = \\angle CQ... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
015y | In a quadrilateral $ABCD$ we have $AB \parallel CD$ and $AB = 2CD$. A line $\ell$ is perpendicular to $CD$ and contains the point $C$. The circle with the centre $D$ and the radius $DA$ intersects the line $\ell$ at points $P$ and $Q$. Prove that $AP \perp BQ$. | [
"Let $S$ be the intersection of diagonals of the given quadrilateral.\n\nThe segment $AC$ is a median of the triangle $APQ$ and by the Tales theorem $AS = 2SC$, so $S$ is the centroid of the triangle $APQ$. The point $D$ is the circumcenter of the triangle $APQ$, thus the line $e = DS$ is t... | Baltic Way | Baltic Way SHL | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0eyp | Problem:
Prove that:
$$
\frac{2}{x^{2} - 1} + \frac{4}{x^{2} - 4} + \frac{6}{x^{2} - 9} + \ldots + \frac{20}{x^{2} - 100} =
$$
$$
\frac{11}{(x - 1)(x + 10)} + \frac{11}{(x - 2)(x + 9)} + \ldots + \frac{11}{(x - 10)(x + 1)}.
$$ | [
"Solution:\n$$\n\\text{lhs} = \\frac{1}{x - 1} - \\frac{1}{x + 1} + \\frac{1}{x - 2} - \\frac{1}{x + 2} + \\ldots + \\frac{1}{x + 10} - \\frac{1}{x - 10} =\n$$\n$$\n\\frac{1}{x - 1} - \\frac{1}{x + 10} + \\frac{1}{x - 2} - \\frac{1}{x + 9} + \\ldots + \\frac{1}{x - 10} - \\frac{1}{x + 1} = \\text{rhs}.\n$$"
] | Soviet Union | 2nd ASU | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0bkb | A 10 digit positive integer is called a *cute number* if its digits are from the set $\{1, 2, 3\}$ and every two consecutive digits differ by 1.
a) Prove that exactly 5 digits of a cute number are equal to 2.
b) Find the total number of cute numbers.
c) Prove that the sum of all cute numbers is divisible by 1408. | [
"a) In the decimal representation of a cute number, the parity of the digits clearly alternates, hence exactly 5 digits are even, that is, equal to 2.\n\nb) There are $2^5 = 32$ cute numbers of the form $\\overline{2a2b2c2d2e}$ and another 32 of the form $\\overline{a2b2c2d2e2}$, therefore the requested number is 6... | Romania | 65th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 5; b) 64; c) sum divisible by 1408 | |
0f9y | Problem:
Does there exist a triangle in which two sides are integer multiples of the median to that side? Does there exist a triangle in which every side is an integer multiple of the median to that side? | [
"Solution:\n\nAnswer yes, no\n\nThe obvious approach is to make the triangle isosceles. So suppose the sides are $a$, $b$, $b$. Then the length $m$ of a median to one of the sides length $b$ satisfies: $a^2 + b^2 = 2m^2 + b^2/2$. The simplest possibility is to take $m = b$, so $a^2 = 3b^2/2$. Thus if $b = 2$, $a = ... | Soviet Union | 25th ASU | [
"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 | yes, no | |
0kfb | Problem:
A bar of chocolate is made of 10 distinguishable triangles as shown below:
How many ways are there to divide the bar, along the edges of the triangles, into two or more contiguous pieces? | [
"Solution:\nEvery way to divide the bar can be described as a nonempty set of edges to break, with the condition that every endpoint of a broken edge is either on the boundary of the bar or connects to another broken edge.\nLet the center edge have endpoints $X$ and $Y$. We do casework on whether the center edge is... | United States | HMMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1689 | |
045k | Given a positive integer $n$, find all $n$-tuples of real numbers $(x_1, x_2, \dots, x_n)$ such that
$$
f(x_1, x_2, \dots, x_n) = \sum_{k_1=0}^{2} \sum_{k_2=0}^{2} \cdots \sum_{k_n=0}^{2} \left| k_1 x_1 + k_2 x_2 + \cdots + k_n x_n - 1 \right|
$$
attains its minimum. | [
"We divide the set $A = \\{0, 1, 2\\}^n$ into: $A = A_0 \\cup A_1 \\cup \\cdots \\cup A_{2n}$, where\n$$\nA_k = \\{\\beta = (i_1, i_2, \\dots, i_n) \\in A : i_1 + i_2 + \\dots + i_n = k\\}, \\quad k = 0, 1, \\dots, 2n.\n$$\nLet $a_k$ denote the number of elements in $A_k$. Then\n$$\n(1 + t + t^2)^n = a_0 + a_1 t + ... | China | 2022 China Team Selection Test for IMO | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | x1 = x2 = ... = xn = 1/(n+1) | |
0kp7 | Problem:
How many ways are there to color every integer either red or blue such that $n$ and $n+7$ are the same color for all integers $n$, and there does not exist an integer $k$ such that $k$, $k+1$, and $2k$ are all the same color? | [
"Solution:\n\nIt suffices to color the integers from $0$ through $6$ and do all arithmetic mod $7$. WLOG, say that $0$ is red (we'll multiply by $2$ in the end). Then $1$ must be blue because $(0,0,1)$ can't be monochromatic. $2$ must be red because $(1,2,2)$ can't be monochromatic. Then we have two cases for what ... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 6 | |
0ghh | For every positive integer $M \ge 2$, find the smallest real number $C_M$ such that for any integers $a_1, a_2, \dots, a_{2023}$, there always exists some integer $1 \le k < M$ such that
$$
\left\{ \frac{ka_1}{M} \right\} + \left\{ \frac{ka_2}{M} \right\} + \dots + \left\{ \frac{ka_{2023}}{M} \right\} \le C_M.
$$
Here... | [
"The answer is $1011 + \\frac{1}{M}$.\n\nWe can see that $C_M \\ge 1011 + \\frac{1}{M}$ by picking $a_{2i+1} = 1$ and $a_{2i} = -1$. This way, we get for all $k = 1, \\dots, M-1$,\n$$\n\\left\\{ \\frac{ka_1}{M} \\right\\} + \\left\\{ \\frac{ka_2}{M} \\right\\} + \\dots + \\left\\{ \\frac{ka_{2023}}{M} \\right\\} = ... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Chinese (Traditional) | proof and answer | 1011 + 1/M | |
07ne | Find all pairs of integers $(a, b)$ such that $a^3 + b^3 + 3ab = 53$. | [
"**First Solution:** The key to this solution is the identity\n$$\na^3 + b^3 - 1 + 3ab = (a+b-1)(a^2 + b^2 - ab + a + b + 1).\n$$\nEven though this can be verified readily, we explain how to find it. We start with a polynomial of degree 3 which has $a$, $b$ and an auxiliary number $c$ as its roots $f(x) = (x-a)(x-b... | Ireland | Ireland | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta'... | English | proof and answer | (-3, 5), (5, -3), (2, 3), (3, 2) | |
09e0 | Let $H$ be orthocenter of the triangle $ABC$. A certain circle with diameter $AC$ intersects circumcircle of the triangle $ABH$ at point $K$ which is different from $A$. Prove that intersection point of $CK$ and $BH$ divides the line segment $BH$ into equal parts.
(proposed by B. Battsengel) | [
"Let $AD$, $CF$ be altitudes dropped from vertices $A$, $C$ respectively. Consequently points $D$, $F$ lie on the circle with diameter $AC$. Let $X$ be intersection point of line $CK$ with line segment $BH$.\n\nSince points $A$, $H$, $K$, $B$ lie on a circle, $\\angle KAH = \\angle KBH$.\n\nOn the other hand points... | Mongolia | Mongolian 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 > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0ioy | Problem:
Let $f(x) = 1 + x + x^{2} + \cdots + x^{100}$. Find $f'(1)$. | [
"Solution:\nNote that $f'(x) = 1 + 2x + 3x^{2} + \\cdots + 100x^{99}$, so $f'(1) = 1 + 2 + \\cdots + 100 = \\frac{100 \\cdot 101}{2} = 5050$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 5050 | |
0gl6 | Let $n$ be a positive integer. We want to make up a collection of cards with the following properties:
* each card has a number of the form $m!$ written on it, where $m$ is a positive integer;
* for any positive integer $t \le n!$, we can select some card(s) from this collection such that the sum of the number(s) on th... | [
"We need at least $\\frac{n(n-1)}{2} + 1$ cards.\n\nFor example, we can have $i$ cards with number $i!$ for each $i = 1, 2, \\dots, n-1$ and another card with number $n!$. For $t = n!$, we can simply choose the card with number $n!$. Suppose that $1 \\le t < n!$. Let $r_0 = t$, and for $i = 1, 2, \\dots, n-1$, defi... | Thailand | Tajland 2014 | [
"Number Theory > Other",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n(n-1)/2 + 1 | |
06bj | Let $n$ be a positive integer with the following property: $2^n - 1$ divides a number of the form $m^2 + 81$, where $m$ is a positive integer. Find all possible $n$. | [
"$n$ can be any nonnegative integral power of $2$.\n\nIf $n$ has an odd divisor $d \\ge 3$, then $2^d - 1 \\mid 2^n - 1 \\mid m^2 + 81$. Since $2^d - 1 \\equiv 3 \\pmod{4}$ and $3 \\nmid 2^d - 1$, there exists an odd prime $p > 3$ such that $p \\equiv 3 \\pmod{4}$ and $p \\mid 2^d - 1$. This implies $p \\mid m^2 + ... | Hong Kong | 1997-2023 IMO HK TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modul... | null | proof and answer | n = 2^k for any nonnegative integer k | |
0g4k | Problem:
Determine the smallest possible value of the expression
$$
\frac{a b+1}{a+b}+\frac{b c+1}{b+c}+\frac{c a+1}{c+a}
$$
where $a, b, c \in \mathbb{R}$ satisfy $a+b+c=-1$ and $a b c \leq -3$. | [
"Solution:\n\nThe minimum is $3$, which is obtained for $(a, b, c) = (1, 1, -3)$ and permutations of this triple.\n\nAs $a b c$ is negative, the triple $(a, b, c)$ has either exactly one negative number or three negative numbers. Also, since $|a b c| \\geq 3$, at least one of the three numbers has absolute value gr... | Switzerland | Swiss Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3 | |
056r | Let $p_1, p_2, p_3, \dots$ be all prime numbers in increasing order. Prove that $p_2 + p_4 + \dots + p_{2n} > 3n^2 - 2n + 1$ for every positive integer $n$. | [
"For any positive integer $k$, six consecutive integers $6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5$ can contain at most two prime numbers. Hence $p_i \\ge p_{i-2}+6$ for all $i \\ge 5$, implying $p_{2i} \\ge p_4+6(i-2) = 6i-5$ for all $i \\ge 2$. Thus $p_2 + p_4 + \\dots + p_{2n} > 2 + (7+13+\\dots+(6n-5)) = 3n^2 - 2n + 1$.... | Estonia | Final Round of National Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0gbf | 令 $n$ 為某個大於 1 的奇數, 而 $f(x)$ 為 $x$ 的 $n$ 次多項式。已知 $f(k) = 2^k$ 對於 $k = 0, 1, \dots, n$ 均成立。試證: 使 $f(x)$ 的值為 2 的幂次的整數 $x$ 僅為有限多個。 | [
"由於 $n+1$ 個值已可唯一決定一個 $n$ 次多項式，且\n$$\nf(k) = 2^k = (1+1)^k = C(k, 0) + C(k, 1) + \\dots + C(k, n)\n$$\n對於 $k = 0, 1, \\dots, n$ 都成立，而右式為一 $n$ 次多項式，故知\n$$\nf(x) = C(x, 0) + C(x, 1) + \\dots + C(x, n).\n$$\n又因為 $n$ 是奇數，將上式兩兩合併，可得\n$$\n\\begin{aligned}\nf(x) &= C(x + 1, 1) + C(x + 1, 3) + \\dots + C(x + 1, n) \\\\\n&= ... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techni... | null | proof only | null | |
0dzi | Find all functions $f: \mathbb{R} \to \mathbb{R}$, such that
$$
x + f(xf(y)) = f(y) + yf(x)
$$
for all real $x$ and $y$. | [
"If $x = 0$ we get $f(0) = f(y) + y f(0)$. So, $f$ is a linear function of the form $f(x) = a - a x$ for some real $a$. Inserting this into the functional equation we see that for all $x, y \\in \\mathbb{R}$ we have $x + a - a x(a - a y) = a - a y + y(a - a x)$, so\n$$\nx - a^2 x + a^2 x y = -a x y.\n$$\nNow, let $... | Slovenia | Slovenija 2008 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x - 1 | |
07fi | Given a function $g : [0, 1] \to \mathbb{R}$ such that for every non-empty partition of the interval $[0, 1]$ into subsets $A, B$ either $\exists x \in A : g(x) \in B$ or $\exists x \in B : g(x) \in A$. Furthermore, $g(x) > x$ for all $x \in [0, 1]$. Prove that $g(x) = 1$, for infinitely many $x$ in its domain. | [
"Let $S_1 = \\{1\\}$ and for every $i \\ge 1$,\n$$\nS_{i+1} = \\{x \\mid g(x) \\in S_i\\} - \\{1\\}.\n$$\nAlso,\n$$\n\\bigcup_{i=1}^{\\infty} S_i = A = \\{x \\mid \\exists n \\ge 0 : g^n(x) = 1\\}.\n$$\nClearly $A$ is non-empty. If $[0, 1] - A = B$ is also non-empty, partitions $A, B$ of $[0, 1]$ will lead to a con... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
0ioj | Problem:
A set of six edges of a regular octahedron is called Hamiltonian cycle if the edges in some order constitute a single continuous loop that visits each vertex exactly once. How many ways are there to partition the twelve edges into two Hamiltonian cycles?
 | [
"Solution:\nAnswer: 6. Call the octahedron $A B C D E F$, where $A, B$, and $C$ are opposite $D, E$, and $F$, respectively. Note that each Hamiltonian cycle can be described in terms of the order it visits vertices in exactly 12 different ways. Conversely, listing the six vertices in some order determines a Hamilto... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 6 | |
02za | Problem:
Seja $ABC$ um triângulo retângulo com $\angle BAC = 90^{\circ}$ e $I$ o ponto de encontro de suas bissetrizes. Uma reta por $I$ corta os lados $AB$ e $AC$ em $P$ e $Q$, respectivamente. A distância de $I$ para o lado $BC$ é $1~\mathrm{cm}$.
a) Encontre o valor de $PM \cdot NQ$.
b) Determine o valor mínimo pos... | [
"Solution:\n\na. Se $\\angle APQ = \\alpha$, segue que $\\angle MIP = 90^{\\circ} - \\alpha$ e\n$$\n\\angle NIQ = 180^{\\circ} - \\angle MIN - \\angle MIP = \\alpha\n$$\nPortanto, os triângulos $IMP$ e $NIQ$ são semelhantes e daí\n$$\n\\frac{IM}{NQ} = \\frac{PM}{IN} \\Rightarrow PM \\cdot NQ = IM \\cdot IN = 1\n$$\... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequ... | null | proof and answer | PM·NQ = 1 and the minimum area of triangle APQ is 2 (attained when AP = AQ = 2). | |
09eg | Let $a_n$ be an arithmetic progression with integer terms. Find all polynomials with integer coefficients such that $\frac{a_n^n + 1}{P(a_n)}$ is a whole number for any natural $n$. | [
"$$\n|P(a_n)| \\neq 1 \\text{ if } 6a(n, P(a_0)) = 1. \\quad (*)\n$$\nLet $p = |P(a_n)|$. Then for any natural number $s$ the congruence $P(a_{n+ps}) = P(a_n + psd) \\equiv 0 \\pmod p$ holds. Therefore from $a_{n+ps}^{n+ps} + 1 \\equiv 0 \\pmod p$ follows $a_n^{n+ps} + 1 \\equiv 0 \\pmod p$. Let's choose $s$ such t... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Exactly the constant polynomials P(x) = ±1 for any arithmetic progression; and if every term of the progression is odd, also P(x) = ±2. | |
0a0o | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ waarvoor geldt dat
$$
f(a-b) f(c-d)+f(a-d) f(b-c) \leq (a-c) f(b-d)
$$
voor alle reële getallen $a, b, c$ en $d$. | [
"Solution:\nDe oplossingen zijn $f(x)=0$ voor alle $x$ en $f(x)=x$ voor alle $x$. Voor $f(x)=0$ vinden we eenvoudig dat altijd gelijkheid geldt. Voor $f(x)=x$ controleren we dat\n$$\n\\begin{aligned}\n(a-b)(c-d)+(a-d)(b-c) & = a c - a d - b c + b d + a b - a c - b d + c d \\\\\n& = -a d - b c + a b + c d \\\\\n& = ... | Netherlands | Selectietoets | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
0jte | Problem:
Chris and Paul each rent a different room of a hotel from rooms $1$-$60$. However, the hotel manager mistakes them for one person and gives "Chris Paul" a room with Chris's and Paul's room concatenated. For example, if Chris had $15$ and Paul had $9$, "Chris Paul" has $159$. If there are $360$ rooms in the ho... | [
"Solution:\n\nThere are $60 \\cdot 59 = 3540$ total possible outcomes, and we need to count the number of these which concatenate into a number at most $360$. Of these, $9 \\cdot 8$ result from both Chris and Paul getting one-digit room numbers. If Chris gets a two-digit number, then he must get a number at most $3... | United States | HMMT November | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 153/1180 | |
00g5 | Suppose $ABCD$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\ell_{1}$ and $\ell_{2}$, which are also $a$ units apart. The square $ABCD$ is placed on the plane so that sides $AB$ and $AD$ intersect $\ell_{1}$ at $E$ and $F$ respectively. Also, sides $CB$ and $CD$ intersect $\el... | [
"Let $EH$ intersect $FG$ at $O$. The distance from $G$ to line $FD$ and line $EF$ are both $a$. So $FG$ bisects $\\angle EFD$. Similarly, $EH$ bisects $\\angle BEF$. So $O$ is an excentre of $\\triangle AEF$. Similarly, $O$ is an excentre of $\\triangle CGH$.\n\nConstruct these excircles with centre $O$. Let $M, N,... | Asia Pacific Mathematics Olympiad (APMO) | XV APMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > A... | null | proof only | null | |
0kf8 | Problem:
Roger the ant is traveling on a coordinate plane, starting at $(0,0)$. Every second, he moves from one lattice point to a different lattice point at distance $1$, chosen with equal probability. He will continue to move until he reaches some point $P$ for which he could have reached $P$ more quickly had he take... | [
"Solution:\nRoger is guaranteed to be able to take at least one step. Suppose he takes that step in a direction $u$. Let $e_{1}$ be the expectation of the number of additional steps Roger will be able to take after that first move. Notice that Roger is again guaranteed to be able to make a move, and that three type... | United States | HMMO | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 1103 | |
04gk | Let $ABCD A_1 B_1 C_1 D_1$ be a rectangular cuboid with edge-lengths $|AB| = |AD| = a$ and $|AA_1| = 2a$. Let $A', B', C'$ and $D'$ be the midpoints of $AA_1, BB_1, CC_1$ and $DD_1$ respectively.
Find the volume of the intersection of the cube $ABCD A'B'C'D'$ and the pyramid $A_1A'B'B_1D$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | a^3/8 | |
07zv | Problem:
Diciamo che un numero naturale è equilibrato se si scrive con tante cifre quanti sono i suoi divisori primi distinti (per esempio, $15$ è equilibrato, mentre $49$ non lo è).
Dimostrare che c'è solo un numero finito di numeri equilibrati. | [
"Solution:\n\nNon esistono numeri equilibrati di $c$ cifre se $c > 100$. Infatti un tale numero sarebbe prodotto di $c$ numeri primi, almeno metà dei quali maggiori di $100$. Quindi il numero sarebbe maggiore di $100^{c / 2} = 10^{c}$, il che è assurdo. In effetti il più grande numero equilibrato ha esattamente die... | Italy | XV GARA NAZIONALE DI MATEMATICA | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0jzc | Problem:
Let $S$ be the set $\{-1,1\}^n$, that is, $n$-tuples such that each coordinate is either $-1$ or $1$. For
$$
s = (s_1, s_2, \ldots, s_n),\ t = (t_1, t_2, \ldots, t_n) \in \{-1,1\}^n
$$
define $s \odot t = (s_1 t_1, s_2 t_2, \ldots, s_n t_n)$.
Let $c$ be a positive constant, and let $f: S \rightarrow \{-1,1\}$... | [
"Solution:\n\nWe use finite Fourier analysis, which we explain the setup for below. For a subset $T \\subseteq S$, define the function $\\chi_T: S \\rightarrow \\{-1,1\\}$ to satisfy $\\chi_T(s) = \\prod_{t \\in T} s_t$. Note that $\\chi_T(s \\odot t) = \\chi_T(s) \\chi_T(t)$ for all $s, t \\in S$. Therefore, we sh... | United States | HMIC | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
08ru | There is a village with a population of $2007$. This village has no name. You are God of this village and you want villagers to decide the name of this village. Every villager has one idea of the village's name.
Each villager can send a letter to each villager (including himself). And every villager can send any numbe... | [
"If $0 \\le T \\le 668$, we will prove that there exists an instruction which fulfills the conditions. Give the following instruction to every villager.\n\nDefine today as 0th day. All the villagers must prepare a notebook and a memo pad.\n\nToday, each villager $p$ should write the idea of the village's name $m$ i... | Japan | Japanese Mathematical Olympiad | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 668 | |
091k | Problem:
Initially, only the integer $44$ is written on a board. An integer $a$ on the board can be replaced with four pairwise different integers $a_{1}, a_{2}, a_{3}, a_{4}$ such that the arithmetic mean $\frac{1}{4}\left(a_{1}+a_{2}+a_{3}+a_{4}\right)$ of the four new integers is equal to the number $a$. In a step ... | [
"Solution:\n\nLet us first prove an auxiliary statement.\n\nLemma. If $a_{1}, a_{2}, a_{3}, a_{4}$ are four different integers such that their average $a=\\left(a_{1}+a_{2}+a_{3}+a_{4}\\right) / 4$ is also an integer, then\n$$\n\\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}}{4}-a^{2} \\geqslant \\frac{5}{2}\n$$\nPr... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0dss | Each of the squares in a $2 \times 2018$ grid of squares is to be coloured black or white such that in any $2 \times 2$ block, at least one of the $4$ squares is white. Let $P$ be the number of ways colouring the grid. Find the largest $k$ so that $3^k$ divides $P$. | [
"Let $P_n$ be the number of colourings of a $2 \\times n$ grid, $A_n$ be the number of colourings in which the first two squares are coloured black and $B_n$ be the number of colourings in which at least one of the first two squares is coloured white. Then $P_1 = 4$, $P_2 = 15$ and for $n \\ge 3$,\n$$\n\\begin{alig... | Singapore | Singapore Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1009 | |
0enu | Dylan has a list of all 25-digit numbers consisting of the digits 1, 2, 3 and 4 such that there are an equal number of 1s and 2s. Robert has a list of all 50-digit numbers consisting of 25 digits 1 and 25 digits 2. Show that the number of numbers on each list is the same. | [
"Let $D$ be the set of 25-digit numbers with the same number of 1s and 2s, and let $R$ be the set of 50-digit numbers with 25 digits 1 and 25 digits 2. We define a bijection between the two sets.\n\nLet $d = d_1d_2\\dots d_{25} \\in D$. We define a function on the digits of $d$ as follows:\n$$\nf(d_i) = \\begin{cas... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0jeh | Problem:
Let $ABCD$ be an isosceles trapezoid such that $AD = BC$, $AB = 3$, and $CD = 8$. Let $E$ be a point in the plane such that $BC = EC$ and $AE \perp EC$. Compute $AE$. | [
"Solution:\n\nAnswer: $2 \\sqrt{6}$\n\nLet $r = BC = EC = AD$. $\\triangle ACE$ has right angle at $E$, so by the Pythagorean Theorem,\n\n$$\nAE^{2} = AC^{2} - CE^{2} = AC^{2} - r^{2}\n$$\n\nLet the height from $A$ of $\\triangle ACD$ intersect $DC$ at $F$. Once again, by the Pythagorean Theorem,\n\n$$\nAC^{2} = FC... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2 sqrt(6) | |
0bjl | Let $p$ and $n$ be positive integers, with $p \ge 2$, and let $a$ be a real number such that $1 \le a < a + n \le p$. Prove that the set
$$
\{ \lfloor \log_2 x \rfloor + \lfloor \log_3 x \rfloor + \dots + \lfloor \log_p x \rfloor \mid x \in \mathbb{R},\ a \le x \le a + n \}
$$
has exactly $n + 1$ elements. | [
"Let $f(x) = \\sum_{k=2}^{p} \\lfloor \\log_k x \\rfloor$ and let $M = \\{f(x) \\mid x \\in [a, a+n]\\}$. It is easy to show that if $k \\ge 2$ is a positive integer, then $\\lfloor \\log_k \\lfloor x \\rfloor \\rfloor = \\lfloor \\log_k x \\rfloor$. This implies that $f(x) = f(\\lfloor x \\rfloor)$, for all $x \\i... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
0esr | Prove that for all natural numbers $n$, $\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$ is also a natural number. | [
"$$\n\\begin{aligned}\n\\frac{n}{3} + \\frac{n^2}{2} + \\frac{n^3}{6} &= \\frac{2n + 3n^2 + n^3}{6} \\\\\n&= \\frac{n(2 + 3n + n^2)}{6} \\\\\n&= \\frac{n(n + 1)(n + 2)}{6}\n\\end{aligned}\n$$\nSince $n$, $n + 1$, $n + 2$ are three consecutive integers, at least one is divisible by $2$ and one is divisible by $3$. H... | South Africa | South African Mathematics Olympiad Third Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0h04 | There are three runners in different vertices of an equilateral triangle with side $1$: First, Second and Third. They start moving simultaneously in the same direction (Second in First's direction, Third in Second's direction, First in Third's direction). Is it necessary that they all meet in one point at the same time... | [
"Answer: a) necessary; b) not necessary.\n\na) We first write the condition, which implies that they eventually meet at one point: $2008t = 2010t + 1 - 3m = 2009t + 2 - 3n$, where $m, n \\in \\mathbb{Z}, t \\in \\mathbb{R}$. We have: $t = 3n - 2$ or $2t = 6n - 4$. Moreover, $2t = 3m - 1 \\Rightarrow 3m - 1 = 6n - 4... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | a) necessary; b) not necessary. | |
0jz3 | Problem:
Given that $a, b, c$ are integers with $a b c = 60$, and that complex number $\omega \neq 1$ satisfies $\omega^{3} = 1$, find the minimum possible value of $\left| a + b \omega + c \omega^{2} \right|$. | [
"Solution:\n\nSince $\\omega^{3} = 1$, and $\\omega \\neq 1$, $\\omega$ is a third root of unity. For any complex number $z, |z|^{2} = z \\cdot \\bar{z}$. Letting $z = a + b \\omega + c \\omega^{2}$, we find that $\\bar{z} = a + c \\omega + b \\omega^{2}$, and\n$$\n\\begin{aligned}\n|z|^{2} & = a^{2} + a b \\omega ... | United States | HMMT November 2017 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | sqrt(3) | |
0gfc | 設四邊形 $ABCD$ 內接於一圓 $\Omega$。過點 $D$ 與 $\Omega$ 相切的直線分別交射線 $BA$, $BC$ 於點 $E$, $F$。於三角形 $ABC$ 內部選取一點 $T$ 使得 $TE \parallel CD$ 且 $TF \parallel AD$。設點 $K$ 異於 $D$ 且落在線段 $DF$ 上，滿足 $TD = TK$。
試證：直線 $AC$, $DT$, $BK$ 三線共點。 | [
"Let the segments $TE$ and $TF$ cross $AC$ at $P$ and $Q$, respectively. Since $PE \\parallel CD$ and $ED$ is tangent to the circumcircle of $ABCD$, we have\n$$\n\\angle EPA = \\angle DCA = \\angle EDA,\n$$\nand so the points $A$, $P$, $D$, and $E$ lie on some circle $\\alpha$. Similarly, the points $C$, $Q$, $D$, ... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 國際競賽實作(一) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese; English | proof only | null | |
0ap7 | Problem:
It is given that $\triangle C A B \cong \triangle E F D$. If $A C = x + y + z$, $A B = z + 6$, $B C = x + 8z$, $E F = 3$, $D F = 2y - z$, and $D E = y + 2$, find $x^{2} + y^{2} + z^{2}$. | [
"Solution:\nSince $\\triangle C A B \\cong \\triangle E F D$, it follows that $A C = E F$, $A B = F D$, and $B C = E D$. Thus, we need to solve the following system of linear equations:\n$$\n\\left\\{\\begin{aligned}\nx + y + z & = 3 \\\\\nz + 6 & = 2y - z \\\\\nx + 8z & = y + 2\n\\end{aligned}\\right.\n$$\nSolving... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 21 | |
0fnk | Let $a$, $k$ be positive integers and let $n$ be a nonnegative integer. Show that $(ka^2 + 1)^{2n+1}$ can be expressed as a sum of $k + 1$ squares and $(ka^2 + 1)^{2n+2}$ can be expressed as a sum of $(k + 1)^2$ squares. | [
"First, we have\n$$\n(ka^2+1)^{2n+1} = (ka^2+1)(ka^2+1)^{2n} = \\underbrace{(a^2+a^2+\\dots+a^2+1^2)}_{k} (ka^2+1)^{2n} \\\\\n= \\underbrace{(a(ka^2+1)^n)^2 + (a(ka^2+1)^n)^2 + \\dots + (a(ka^2+1)^n)^2}_{k} + \\underbrace{((ka^2+1)^n)^2}_{k}\n$$\nwhich is the sum of $k + 1$ squares.\n\nLikewise,\n$$\n(ka^2 + 1)^{2n... | Spain | BarcelonaTech Mathcontest | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof only | null | |
0abt | Does there exist a sequence $a_1, a_2, \ldots, a_n, \ldots$ of positive real numbers satisfying both of the following conditions:
$\sum_{i=1}^{n} a_i \le n^2$, for every positive integer $n$;
$\sum_{i=1}^{n} \frac{1}{a_i} \le 2008$, for every positive integer $n$? | [
"The answer is no. It is enough to show that if $\\sum_{i=1}^{n} a_i \\le n^2$ for any $n$, then $\\sum_{i=2}^{2n} \\frac{1}{a_i} > \\frac{n}{4}$. (or any other precise estimate)\n\nFor this, we use that $\\sum_{i=2^k+1}^{2^{k+1}} a_i \\le \\sum_{i=2^k+1}^{2^{k+1}} \\frac{1}{a_i} \\ge 2^{2k}$ for any $k \\ge 0$ by ... | North Macedonia | Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | No | |
03gi | Problem:
Show that if $a_{1} / b_{1}=a_{2} / b_{2}=a_{3} / b_{3}$ and $p_{1}, p_{2}, p_{3}$ are not all zero, then
$$
\left(\frac{a_{1}}{b_{1}}\right)^{n}=\frac{p_{1} a_{1}^{n}+p_{2} a_{2}^{n}+p_{3} a_{3}^{n}}{p_{1} b_{1}^{n}+p_{2} b_{2}^{n}+p_{3} b_{3}^{n}}
$$
for every positive integer $n$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0cyi | Let $a$ and $b$ be integers such that $a-b=a^{2} c-b^{2} d$ for some consecutive integers $c$ and $d$. Prove that $|a-b|$ is a perfect square. | [
"Let $d = c + 1$. The equality $a-b = a^{2} c - b^{2} (c + a)$ implies\n$$\n(a-b)[c(a+b)-1] = b^{2}.\n$$\nBut $c(a+b)-1$ and $a-b$ are relatively prime. Indeed, if $p$ is a prime dividing $a-b$, then the above equality shows that $p$ also divides $b$, so $p$ will divide $a+b = (a-b) + 2b$. Hence $p$ cannot divide $... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
00ws | Problem:
Let's expand a little bit three circles, touching each other externally, so that three pairs of intersection points appear. Denote by $A_{1}, B_{1}, C_{1}$ the three so obtained "external" points and by $A_{2}, B_{2}, C_{2}$ the corresponding "internal" points. Prove the equality
$$
|A_{1} B_{2}| \cdot |B_{1}... | [
"Solution:\n\nFirst, note that the three straight lines $A_{1}A_{2}$, $B_{1}B_{2}$ and $C_{1}C_{2}$ intersect in a single point $O$. Indeed, each of the lines is the locus of points from which the tangents to two of the circles are of equal length (it is easy to check that this locus has the form of a straight line... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
02vl | Problem:
As peças a seguir são chamadas de $L$-triminós.
Essas peças são usadas para cobrir completamente um tabuleiro $6 \times 6$. Nessa cobertura, cada $L$-triminó cobre exatamente 3 quadradinhos do tabuleiro $6 \times 6$ e nenhum quadradinho é coberto por mais de um $L$-triminó.
a) Quant... | [
"Solution:\n\na) Seja $x$ o número de $L$-triminós usados para cobrir um tabuleiro $6 \\times 6$. Como esse tabuleiro possui exatamente $6 \\cdot 6=36$ quadradinhos e cada um deve ser coberto por exatamente um dos $L$-triminós, então $3x=36$, ou seja, $x=12$.\n\n\nb) Seja $y$ a quantidade de $L$-triminós que cada f... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 12; b) 4; c) impossible | |
06ri | Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_{0}$ be the midpoint of $AC$ and let $C_{0}$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$, and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_{0}$ and $C_{0}$ that is tangent to the circle $\Om... | [
"If $AB = AC$, then the statement is trivial. So without loss of generality we may assume $AB < AC$. Denote the tangents to $\\Omega$ at points $A$ and $X$ by $a$ and $x$, respectively.\nLet $\\Omega_{1}$ be the circumcircle of triangle $AB_{0}C_{0}$. The circles $\\Omega$ and $\\Omega_{1}$ are homothetic with cent... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, ci... | null | proof only | null | |
05gr | Problem:
Soit $\Gamma$ un cercle, $P$ un point à l'extérieur du cercle. Les tangentes au cercle $\Gamma$ passant par le point $P$ sont tangentes au cercle $\Gamma$ en $A$ et $B$. Soit $M$ est le milieu du segment $[BP]$ et $C$ le point d'intersection de la droite $ (AM) $ et du cercle $\Gamma$. Soit $D$ la deuxième in... | [
"Solution:\n\nEn utilisant la puissance du point $M$ par rapport au cercle $\\Gamma$, $MB^2 = MC \\cdot MA$. Puisque $M$ est le milieu du segment $[BP]$, $MP^2 = MB^2$ donc $MP^2 = MC \\cdot MA$. On déduit de la réciproque de la puissance d'un point par rapport à un cercle que la droite $(PM)$ est tangente au cercl... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g9b | 三角形 $ABC$ 中, $A'$, $B'$, $C'$ 分別是 $BC$, $AC$, $AB$ 邊的中點。$B^*$, $C^*$ 分別在 $AC$, $AB$ 上, 使得 $BB^*$, $CC^*$ 是三角形 $ABC$ 的高。再令 $B^\#$, $C^\#$ 分別為 $BB^*$, $CC^*$ 的中點。設 $B'B^\#$ 與 $C'C^\#$ 交於 $K$ 點, $AK$ 交 $BC$ 於 $L$ 點。證明: $\angle BAL = \angle CAA'$. | [
"同樣定義 $A^*$, $A^\\#$. 因 $A'B' \\parallel AB$, $B'C' \\parallel BC$, $C'A' \\parallel CA$, 且三高共點, 得\n$$\n\\frac{C'A^\\sharp}{A^\\sharp B'} \\cdot \\frac{B'C^\\sharp}{C^\\sharp A'} \\cdot \\frac{A'B^\\sharp}{B^\\sharp C'} = 1 = \\frac{BA^*}{A^*C} \\cdot \\frac{AC^*}{C^*B} \\cdot \\frac{CB^*}{B^*A} = 1,\n$$\n故 $K$ 亦在 ... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Pla... | null | proof only | null | |
05np | Problem:
Deux cercles $\omega_{1}$ et $\omega_{2}$ sont tangents en $S$, avec $\omega_{1}$ à l'intérieur de $\omega_{2}$. On note $O$ le centre de $\omega_{1}$. Une corde $[AB]$ de $\omega_{2}$ est tangente à $\omega_{1}$ en $T$. Montrer que $(AO)$, la perpendiculaire à $(AB)$ passant par $B$ et la perpendiculaire à $... | [
"Solution:\n\nOn note $(Bx)$ la perpendiculaire à $(AB)$ passant par $B$, et $(Sy)$ la perpendiculaire à $(ST)$ passant par $S$. Pour obtenir le résultat grâce au théorème de Ceva trigonométrique dans le triangle $ABS$, on doit montrer :\n\n$$\n\\frac{\\sin \\widehat{BAO}}{\\sin \\widehat{SAO}} \\cdot \\frac{\\sin ... | France | French Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometr... | null | proof only | null | |
0jjk | Problem:
There are 10 people who want to choose a committee of 5 people among them. They do this by first electing a set of $1, 2, 3$, or $4$ committee leaders, who then choose among the remaining people to complete the 5-person committee. In how many ways can the committee be formed, assuming that people are distingu... | [
"Solution:\n\nThere are $\\binom{10}{5}$ ways to choose the 5-person committee. After choosing the committee, there are $2^{5} - 2 = 30$ ways to choose the leaders (since any nonempty proper subset of the 5 can be the set of leaders, i.e., any subset except the empty set and the full set). So the answer is $30 \\cd... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 7560 | |
09a3 | Let $A$ be a nonempty subset of the positive integers. If $x \in A$, then $[\sqrt[3]{x}] \in A$ and $[9x] \in A$ holds for any $x$. Prove that $A$ is the set of all positive integers. ($[x]$ denotes the integer part of $x$) | [
"Since $A$ is a nonempty subset of the positive integers, $A$ has a minimum element $m$. If $m > 1$ then $m > \\sqrt[3]{m} \\ge [\\sqrt[3]{m}]$ and $[\\sqrt[3]{m}] \\in A$. It is contrary to that $m$ is the minimum element. So $m = 1$.\n\nSince $1 \\in A$, $9^k \\in A$. From this $[\\sqrt[3]{81}] = 4 \\in A$ and $4... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
0ck9 | Let $f : [0, 1] \to \mathbb{R}$ be a differentiable function, with integrable derivative on $[0, 1]$, such that $f(1) = 0$. Prove that
$$
\int_{0}^{1} (x f'(x))^2 dx \geq 12 \cdot \left( \int_{0}^{1} x f(x) dx \right)^2 .
$$ | [
"Let $g : [0, 1] \\to \\mathbb{R}$ defined by $g(x) = x f(x)$, for $x \\in [0, 1]$. We have $g(0) = 0 = g(1)$, $g'(x) = f(x) + x f'(x)$, so that:\n$$\n\\begin{align*}\n\\int_0^1 x^2 (f'(x))^2 dx &= \\int_0^1 (g'(x) - f(x))^2 dx \\\\\n&= \\int_0^1 (g'(x))^2 dx - 2 \\int_0^1 g'(x) f(x) dx + \\int_0^1 (f(x))^2 dx \\\\... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
09nx | A triangular fortress has guard towers at each vertex and at the midpoint of each side. A guard stationed at a vertex can defend both adjacent sides, while a guard stationed at a midpoint can defend only that side. How many ways can guards be assigned to the six towers so that each side is defended by exactly one guard... | [] | Mongolia | MMO2025 Round 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 4 | |
0959 | Problem:
Fie $P$ un punct, situat în interiorul unui triunghi $ABC$, astfel încât $\angle CAP \equiv \angle CBP$. Fie $D$ mijlocul laturii $AB$, iar $M$ şi $N$ proiecţiile punctului $P$ pe laturile $BC$ şi $AC$, respectiv. Să se demonstreze că $DM = DN$. | [
"Solution:\n\nFie $E$ şi $F$ mijloacele segmentelor $AP$ şi $BP$, respectiv. Întrucât $DE$ şi $DF$ sunt linii mijlocii ale triunghiului $ABP$, atunci $DEP F$ este paralelogram, iar $FM$ şi $EN$ sunt mediane corespunzătoare ipotenuzelor triunghiurilor dreptunghice $BPM$ şi $APN$, respectiv. Au loc egalităţile:\n1. $... | Moldova | Olimpiada de Matematică a Republicii Moldova | [
"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 | |
09fh | Let $H$ denote the intersection of the altitudes $AD$ and $CE$ of an acute triangle $ABC$. Let $M$ and $N$ denote the midpoints of the sides $AB$ and $BC$ respectively. The rays $MH$ and $NH$ intersect the circumcircle $\omega$ of $ABC$ at points $K$ and $L$ respectively. If the circumcircles of the triangles $EHK$ and... | [
"Let $O$ denote the center of the circumcircle $\\omega$ and denote $\\alpha := \\angle BAC$, $\\beta := \\angle ABC$ and $\\gamma := \\angle BCA$. Let $A' := (AO) \\cap \\omega$. First we show that $N \\in A'H$.\n\n\n\nIndeed, let $N' := AH \\cap BC$. Since $O$ is the midpoint of $AA'$, th... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 | |
05ws | Problem:
Félix souhaite colorier les entiers de $1$ à $2023$ tels que si $a, b$ sont deux entiers distincts entre $1$ et $2023$ et $a$ divise $b$, alors $a$ et $b$ sont de couleur différentes. Quel est le nombre minimal de couleurs dont Félix a besoin? | [
"Solution:\n\nOn peut essayer de colorier de manière gloutonne les nombres : $1$ peut être colorié d'une couleur qu'on note $a$, $2$ et $3$ de la même couleur $b$ (mais pas de la couleur $a$), puis $4, 6$ de la couleur $c$ (on peut aussi colorier $5$ et $7$ de la couleur $c$), etc. Il semble donc qu'une coloration ... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 11 | |
03em | Given the equation
$$
2 + x\sqrt{9 + 6\sqrt{2}} = x\sqrt{5 - 2\sqrt{6}} + \sqrt{6} - 2\sqrt{3} + \sqrt{2}.
$$
a) Write the root of the equation in the form $m - \sqrt{n}$, where $m$ and $n$ are natural numbers.
b) Factor the expression $a^3 - 3a^2 - 5a + 7$ into two non-constant factors with integer coefficients and ca... | [
"$$\n\\sqrt{9 + 6\\sqrt{2}} = \\sqrt{3}\\sqrt{2 + 2\\sqrt{2} + 1} = \\sqrt{3}(\\sqrt{2} + 1) = \\sqrt{6} + \\sqrt{3}\n$$\n$$\n\\sqrt{5 - 2\\sqrt{6}} = \\sqrt{3 - 2\\sqrt{6} + 2} = |\\sqrt{3} - \\sqrt{2}| = \\sqrt{3} - \\sqrt{2}.\n$$\nThe equation takes the form\n$$\nx(\\sqrt{6} + \\sqrt{3} - \\sqrt{3} + \\sqrt{2}) ... | Bulgaria | Bulgarian Winter Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | a) x = 1 − √2; b) a^3 − 3a^2 − 5a + 7 = (a − 1)(a^2 − 2a − 7), and for a = 1 − √2 the value is 6√2. | |
0a74 | Problem:
Find a polynomial $P$ of lowest possible degree such that
(a) $P$ has integer coefficients,
(b) all roots of $P$ are integers,
(c) $P(0) = -1$,
(d) $P(3) = 128$. | [
"Solution:\nLet $P$ be of degree $n$, and let $b_{1}, b_{2}, \\ldots, b_{m}$ be its zeroes. Then\n$$\nP(x) = a(x - b_{1})^{r_{1}} (x - b_{2})^{r_{2}} \\cdots (x - b_{m})^{r_{m}}\n$$\nwhere $r_{1}, r_{2}, \\ldots, r_{m} \\geq 1$, and $a$ is an integer. Because $P(0) = -1$, we have $a b_{1}^{r_{1}} b_{2}^{r_{2}} \\cd... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 3 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | P(x) = (x - 1)(x + 1)^3 | |
0aei | Дали постои природен број кај кој: првите 2009 цифри се тројки, наредните 2009 цифри се двојки, на наредните 2009 се единици, а останатите нули и е точен куб на природен број? (Одговорот да се образложи) | [
"Значи дадениот број е од облик $n = \\overline{33...3} \\overline{22...2} \\overline{11...1} 000...$. Нека $n = k^3$, каде $k \\in N$. Збирот на цифри на дадениот број е $6 \\cdot 2009 = 12054$. Според тоа $3|n$, од каде следува дека $3|k$, и $3^3|k^3 = n$. Тоа не е можно бидејќи збирот на цифри на $n$ не се дели ... | North Macedonia | Републички натпревар по математика за основно образование | [
"Number Theory > Divisibility / Factorization"
] | Macedonian, English | proof and answer | No, such a number does not exist. | |
0dsg | Consider a regular cube with side length $2$. Let $A$ and $B$ be two vertices that are furthest apart. Construct a sequence of points on the surface of the cube $A_1, A_2, \dots, A_k$ so that $A_1 = A$, $A_k = B$ and for any $i = 1, \dots, k-1$, the distance from $A_i$ to $A_{i+1}$ is $3$. Find the minimum value of $k$... | [
"The sphere with centre $A$ and radius $3$ intersects the three edges at $B$ in three points $K, L, N$. By straightforward calculation using Pythagoras' Theorem, it's easy to show that they are the midpoints of the edges. Let $M$ be an interior point of any of the arcs $KL, LN, KN$ arising from the intersection wit... | Singapore | Singapore Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 7 | |
0gxv | Find the number of pairs of positive integer numbers $(n, m)$, satisfying $(2^k)! = 2^n m$. | [
"Note that if $(2^k)! : 2^l$, then the pair $(l, (2^k)!)$ satisfies the equation and vice versa: if a pair $(n, m)$ is the solution of equation, then $(2^k)! : 2^n$. That is, the number of solutions equals to the number of divisors of the kind $2^l$ of $(2^k)!$. By the Legendre theorem the latter equals: $[2^k/2] +... | Ukraine | The Problems of Ukrainian Authors | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2^k - 1 | |
0l6y | Problem:
Trapezoid $ABCD$, with $AB \parallel CD$, has side lengths $AB = 11$, $BC = 8$, $CD = 19$, and $DA = 4$. Compute the area of the convex quadrilateral whose vertices are the circumcenters of $\triangle ABC$, $\triangle BCD$, $\triangle CDA$, and $\triangle DAB$. | [
"Solution:\n\n\nLet $O_{A}$, $O_{B}$, $O_{C}$, and $O_{D}$ be the circumcenters of $\\triangle BCD$, $\\triangle CDA$, $\\triangle DAB$, and $\\triangle ABC$, respectively. Note that $O_{B}O_{C}$ is the perpendicular bisector of $\\overline{AD}$. Similarly, $O_{B}O_{D} \\perp AC$ and $O_{C}... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 9√15 | |
0ay7 | Problem:
Find the largest value of $x$ such that $\sqrt[3]{x} + \sqrt[3]{10-x} = 1$. | [
"Solution:\n\nCubing both sides of the given equation yields\n$$\nx + 3 \\sqrt[3]{x(10-x)} (\\sqrt[3]{x} + \\sqrt[3]{10-x}) + 10 - x = 1,\n$$\nwhich then becomes\n$$\n10 + 3 \\sqrt[3]{x(10-x)} = 1\n$$\nor\n$$\n\\sqrt[3]{x(10-x)} = -3. \\tag{1}\n$$\nCubing both sides of equation (1) gives\n$$\nx(10-x) = -27\n$$\nor\... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 5 + 2 sqrt(13) | |
0kd8 | Problem:
Each unit square of a $4 \times 4$ square grid is colored either red, green, or blue. Over all possible colorings of the grid, what is the maximum possible number of L-trominos that contain exactly one square of each color? (L-trominos are made up of three unit squares sharing a corner, as shown below.)
$$
\... | [
"Solution:\n\nNotice that in each $2 \\times 2$ square contained in the grid, we can form 4 L-trominoes. By the pigeonhole principle, some color appears twice among the four squares, and there are two trominoes which contain both. Therefore each $2 \\times 2$ square contains at most 2 L-trominoes with distinct colo... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 18 | |
0hi7 | Anna has placed real numbers with sum $S$ in the cells of a row. It turned out that she cannot cut the row into two parts so that the sum of the numbers in one part is positive and in the other part is negative. Prove that the modulus of $S$ is not less than any of Anna's numbers.
(Oleksii Masalitin) | [
"Let's assume that $S = 0$. Let's choose an arbitrary division of the string into two parts. It is clear that the sum of the numbers in the two parts is $0$, so one of them is not less than $0$, and the other is not greater than $0$. If they are not $0$, we get a contradiction, so the sum of the numbers of any smal... | Ukraine | Problems from Ukrainian Authors | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0btz | Problem:
Fie $ABCD$ un pătrat şi $E$ un punct situat pe diagonala $BD$, diferit de mijlocul acesteia. Se notează cu $H$ şi $K$ ortocentrele triunghiurilor $ABE$, respectiv $ADE$. Arătaţi că $\overline{BH} + \overline{DK} = 0$. | [
"Solution:\n\n\n\nSe observă că punctele $H$ şi $K$ se află pe diagonala $AC$, deoarece $AC$ este perpendiculară pe $BE$ şi $DE$.\n\nDe asemenea, $H$ şi $K$ se află pe înălţimile duse din $E$ în cele două triunghiuri, care sunt perpendiculare pe laturile pătratului iniţial.\n\nDeducem că tr... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Me... | null | proof only | null | |
0j7w | Problem:
Charlie folds an $\frac{17}{2}$-inch by $11$-inch piece of paper in half twice, each time along a straight line parallel to one of the paper's edges. What is the smallest possible perimeter of the piece after two such folds? | [
"Solution:\n\n$\\boxed{\\frac{39}{2}}$\n\nNote that when a piece of paper is folded in half, one pair of opposite sides is preserved and the other pair is halved. Hence, the net effect on the perimeter is to decrease it by one of the side lengths. The original perimeter is $2\\left(\\frac{17}{2}\\right) + 2 \\cdot ... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | 39/2 | |
0izz | Problem:
Jeff has a 50 point quiz at 11 am. He wakes up at a random time between 10 am and noon, then arrives at class 15 minutes later. If he arrives on time, he will get a perfect score, but if he arrives more than 30 minutes after the quiz starts, he will get a 0, but otherwise, he loses a point for each minute he'... | [
"Solution:\n\nIf Jeff wakes up between 10:00 and 10:45, he gets 50. If he wakes up between 10:45 and 11:15, and he wakes up $k$ minutes after 10:45, then he gets $50 - k$ points. Finally, if he wakes up between 11:15 and 12:00 he gets 0 points. So he has a $\\frac{3}{8}$ probability of 50, a $\\frac{3}{8}$ probabil... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 55/2 | |
0lac | Denote by $M$ the set of the first $2008$ positive integers. The numbers in $M$ are colored blue, yellow and red such that each number is of one color and each color is used at least once. Consider the following sets:
$$
S_1 = \{(x, y, z) \in M^3 \mid x, y, z \text{ are of the same color and } (x + y + z) \equiv 0 \pmo... | [] | Vietnam | Vijetnam 2008 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0c2z | A nonempty finite set $S$ of complex numbers has the property: $xy \in S$, for every $x, y \in S$.
a) Prove that, for every $x \in S$, $\frac{1}{x} \in S$.
b) Let $a, b$ be two fixed elements of $S$ and $m, n \in \mathbb{N}$, $m, n \ge 2$. Find the number of $m \times n$ matrices, with elements from $S$, having the p... | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | Number of matrices = 0 if a^m ≠ b^n; otherwise it equals |S|^{(m−1)(n−1)}. | |
0bb7 | Consider a regular prism $ABCA'B'C'$. A plane $\alpha$ containing point $A$ meets the rays $BB'$ and $CC'$ at points $E$ and $F$ such that
$$
\text{area } [ABE] + \text{area } [ACF] = \text{area } [AEF].
$$
Find the angle determined by the planes $AEF$ and $BCC'$.
| [
"Let $M$ be the midpoint of $BC$ and let $u$ be the angle determined by the planes $AEF$ and $BCC'$. Since triangle $MEF$ is the projection of the triangle $AEF$ onto $BCC'$, we have\n$$\n\\begin{aligned}\n\\cos u &= \\frac{[MEF]}{[AEF]} = \\frac{[BCFE]}{2[AEF]} = \\frac{[BCFE]}{2([ABE] + [ACF])} = \\\\\n&= \\frac{... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 60° | |
008q | Let $ABC$ be an acute triangle and $O$ be its circumcenter. The line $AO$ meets the side $BC$ at the point $D$. It is known that $OD = BD = 1$ and $CD = 1 + \sqrt{2}$. Calculate the lengths of the sides of the triangle. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | BC = 2 + sqrt(2), AB = sqrt((2 + sqrt(2)) (2 + sqrt(2 + sqrt(2)))), AC = sqrt((2 + sqrt(2)) (2 − sqrt(2 − sqrt(2)))) | |
08cb | Problem:
Il polinomio $P(x)$, di grado 42, assume il valore 0 nei primi 21 numeri primi dispari e nei loro reciproci (si ricorda che il reciproco di un intero positivo $n$ è il numero razionale $1 / n$). Quanto vale il rapporto $P(2) / P(1 / 2)$?
(A) 0
(B) 1
(C) $2^{21}$
(D) $3^{21}$
(E) $4^{21}$ | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. Osserviamo che l'espressione $Q(x) = P(x) - x^{42} P(1 / x)$ è un polinomio, dal momento che il monomio $x^{42}$ semplifica il denominatore di $P(1 / x)$. Inoltre, esso ha grado al più 42, e se $r$ è uno dei primi 21 numeri primi dispari, $Q(x)$ si annulla in $r$ e in $1... | Italy | Gara di Febbraio | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | E | |
05ed | Problem:
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB$, $AB \parallel QR$, $AC \parallel SP$, a... | [
"Solution:\n\nWe will prove that $\\triangle BIR \\sim \\triangle CIS$, since the statement then follows from $\\angle TRI = \\angle BRI = \\angle CSI = \\angle TSI$.\n\n\n\nStep 1. Let us prove $\\angle RBI = \\angle SCI$. We will use directed angles:\n\n$$(BR,BI) = (BR,AB) + (AB,BI) = (AQ... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transf... | null | proof only | null | |
0bzn | Find all $x$ that satisfy
$$
\log_2(x^2 + 4) - \log_2 x + x^2 - 4x + 2 = 0.
$$ | [
"First, note that $x > 0$ since $\\log_2 x$ is defined only for $x > 0$.\n\nRewrite the logarithmic terms:\n$$\n\\log_2(x^2 + 4) - \\log_2 x = \\log_2\\left(\\frac{x^2 + 4}{x}\\right)\n$$\nSo the equation becomes:\n$$\n\\log_2\\left(\\frac{x^2 + 4}{x}\\right) + x^2 - 4x + 2 = 0\n$$\nLet $y = x^2 - 4x + 2$. Then:\n$... | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2 | |
0hh8 | On the plane 2022 points $A_1, A_2, \dots, A_{2022}$ are given, no three of which lie on the same line. Consider all the angles $A_i A_j A_k$ for the triples of distinct points $A_i, A_j, A_k$. What largest number of these angles can be right? | [
"Consider any point $A_i$ and count the number of pairs of points $(A_j, A_k)$ such that $\\angle A_i A_j A_k = 90^\\circ$. For each point $A_j$ there exists at most one point $A_k$ (because on the line through $A_j$ perpendicular to $A_iA_j$ there can be at most one point other than $A_j$). Also note that if $X$ i... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof and answer | 2042220 | |
01gd | Let $a$, $b$, $c$ be positive real numbers with $abc = 1$. Prove that
$$
\frac{a}{b(c+1)} + \frac{b}{c(a+1)} + \frac{c}{a(b+1)} \ge \frac{3}{2}.
$$ | [
"By Cauchy-Schwarz we have\n$$\n\\left( \\frac{a}{b(c+1)} + \\frac{b}{c(a+1)} + \\frac{c}{a(b+1)} \\right) \\left( (c+1) + (a+1) + (b+1) \\right) \\ge \\left( \\sqrt{\\frac{a}{b}} + \\sqrt{\\frac{b}{c}} + \\sqrt{\\frac{c}{a}} \\right)^2.\n$$\nTherefore it suffices to prove that\n$$\n\\begin{aligned}\n& \\frac{a}{b}... | Baltic Way | Baltic Way 2020 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
01vb | For a fixed integer $n \ge 2$ consider the sequence
$$
a_k = \text{lcm}(k, k+1, \dots, k+(n-1)).
$$
Find all integers $n \ge 2$ for which the sequence $a_k$ increases starting from some number. | [
"Answer: $n = 2$.\n\nNote that if $n = 2$, the sequence has the form $a_k = k(k+1)$ since consecutive numbers are always coprime. It is clear that $k(k+1) < (k+1)(k+2)$, so the sequence is increasing.\n\n*First solution.* Let us show that if $n \\geq 3$ the sequence is not increasing from any number. Choose $k = np... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | n = 2 | |
06jm | Let $\Gamma$ be a circle and $AB$ be a diameter. Let $\ell$ be a line outside the circle, and is perpendicular to $AB$. Let $X, Y$ be two points on $\ell$. If $X'$ and $Y'$ are two points on $\ell$ such that $AX$ and $BX'$ intersect on $\Gamma$ and such that $AY$ and $BY'$ intersect on $\Gamma$, prove that the circumci... | [
"Let $AX$ meet $\\Gamma$ again at $P$, and let $AY$ meet $\\Gamma$ again at $Q$. If $PQ \\parallel \\ell$, the figure is symmetric with respect to $AB$, and so $(AXY)$ and $(AX'Y')$ are tangent at $A$. In the following, we only consider the configuration as shown.\n\nFirstly, since\n$$\n\\angle AQP = \\angle ABP = ... | Hong Kong | Year 2016 | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null |
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